Chapter 1 Problems and Solution jan30

Chapter 1 Problems and Solution

Energy and Civilization 1.1 Perform the following: write a 3000 word report summarizing the Kyoto Protocol (http://unfccc.int/kyoto_protocol/items/2830.php) i) Compute the simple operating margin CO2 footprint factor for a power grid load of 6000 MW, if it is supplied by 50 coal fired units with a capacity of 100 MW, 10 oil-fired generators with a capacity of 50 MW, and 10 gas-fired generators with a capacity of 50 MW. ii) If the grid loads of 6000 MW is supplied equally from gas-fired, wind and solar powered generators each.

Solution:

i) The simple operating margin CO2 footprint factor is calculated as the amount of CO2 emissions of all units supplying power to a grid during a year. The simple operating margin footprint factor has a unit of ton of CO2 per MWh (tCO2/MWh). According to Kyoto Protocol,

∑

∑

=

== n

1ii

n

1iii

x

y.xZ

where Z = Simple operating margin CO2 emission factor (lb-CO2/kWh) xi= Net electricity power generated by unit i (kWh) yi = CO2 emission factor of unit i (lb-CO2/kWh) n = number of units in operation The simple operating margin can be calculated from two sources: table 1.2 and table 1.4 using two different unit systems.

Conversion in Metric system to English system is needed to solve this problem. The conversion from BTU to Watt is given below. P = energy / time Watt = Joule/ Sec. Watt-Sec= Joule A BTU (British thermal unit) measures the amount of heat needed to raise the temperature of 1 lb of water by one degree Fahrenheit (1°F). It is equivalent to the measures used for energy or work, i.e., a joule or W·sec. Energy needed to raise the temperature in BTU = (specific heat of the material). (mass in lb) . (change in temperature in F) Or in metric system, the specific heat of water is given as

Specific heat of water = C.g

J186.4°

Therefore, heat needed to raise 1 g of water by 1 °C = 4.186 J

Therefore, heat needed for 453.5 g (= 1 lb) by 1 °C = J18985.453186.4 =× Converting one degree F to metric system in degree C, we have One degree F = 0.556 C Heat needed for 1 lb of water to raise by 1 °F (= 0.556 °C) is expressed as One BTU = J1055556.01898 =× From definition of BTU, one BTU is equal to amount of heat needed to raise 1 lb of water by one degree Fahrenheit (F). Therefore, from above, we have 1 BTU = 1055 J One BTU= 1055 joule 1055 watt . sec = 1 BTU = BTU1

3600hour1W1055 =×

= BTU10553600Wh =

= BTU41.3Wh = = BTU1041.3kWh 3×=

ii) Based on Table 1.2 we have the following data on CO2 carbon footprints. Table 1.2. Carbon Footprint of Various Fossil Fuels for Production of 1 kWh of Electric Energy27 (Low Impact Life Onboard. Carbon footprints. Available at http://www.liloontheweb.org.uk/handbook/carbonfootprint. Accessed 2010 Nov 9) Fuel Type CO2 Footprint

(lb CO2/kWh) Wood 3.306 Coal-fired plant 2.117 Gas-fired plant 1.915 Oil-fired plant 1.314 Combined-cycle gas

0.992

Table 1.4. Fossil Fuel Emission Levels in Pounds of pollutant per Billion BTU of Energy Input.27

(Low Impact Life Onboard. Carbon footprints. Available at http://www.liloontheweb.org.uk/handbook/carbonfootprint. Accessed 2010 Nov 9)

Pollutant

Natural Gas(lb /billion BTU)

Oil (lb/billion BTU)

Coal (lb/billion BTU)

Carbon dioxide (CO2 )

117,000 164,000 208,000

Carbon Monoxide, CO

40 33 208

Nitrogen oxides 92 448 457

Sulfur dioxide 1 1,122 2,591

Particulates 7 84 2,744 Mercury 0.000 0.007 0.016 Number of hours in one year = h876036524 =× kWh generated by coal in one year = kWh108.4310876010050 93 ×=××× kWh generated by gas in one year = kWh1038.41087605010 93 ×=××× kWh generated by oil in one year = kWh1038.41087605010 93 ×=××× From table 1.2, the simple operating margin =

999

999

j 1038.41038.4108.43 1.3141038.4 1.9151038.4 2.117108.43Z

×+×+×××+××+××

=

kWh/COoflb2.0332 2= Total kWh in one year = kWh105.25608760106000 103 ×=×× Total CO2 in one year = 2

910 COoflb10106.860332.2105.2560 ×=×× Total BTU = BTU1017.9230103.41105.2560 13310 ×=××× 1 kWh = 3.41 thousand BTU We can rewrite the simple operating margin in term BTU by converting lb of Co2/kWh. to lb of co2/BTU

BTU/COoflb1059625.0100041.3

0332.2Z 23-

j ×=×

=

BTUbillion/COoflb250,596101059625.0Z 29-3

j =××= From table 1.4, BTU generated by coal in one year = BTUbillion1035.1491041.3108.43 339 ×=××× BTU generated by gas in one year = BTUbillion10935.141041.31038.4 339 ×=××× BTU generated by oil in one year = BTUbillion10935.141041.31038.4 339 ×=×××

Simple operating margin,

333

333

j 10935.1410935.1410935.14000,11710935.14000,16410935.14000,2081035.149Z

×+×+×××+××+××

=

BTUbillion/COoflb196,750 2= Total CO2 in one year = 2

9913 COoflb1035.26410/1017.9230750,196 ×=×× The results obtained form table 1.2 and 1.4 are different. That is because the two sources have used different qualities of coal, natural gas and oil to analyze the CO2 content.

iii) The simple operating margin can be calculated from the data in table 1.3 Table 1.3. Carbon Footprint of Green and Renewable Sources for Production of 1 kWh of Electric Energy.27 Fuel Type CO2 Footprint

(lb Co2/kWh) Hydroelectric 0.0088 PV 0.2204 Wind 0.03306

The power generated by each of the units = MW20003

6000=

kWh generated by each unit in one year = kWh1052.171087602000 93 ×=×× From table 1.2 and 1.3,

Simple Operating margin 999

999

j 1052.171052.171052.170.22041052.17 03306.01052.17 1.9151052.17Z

×+×+×××+××+××

=

kWh/COoflb0.7228 2= Total CO2 in one year = 2

910 COoflb1099.377228.0105.2560 ×=×× BTU41.3Wh1 =

1 kWh = 3.41 thousand BTU Therefore,

BTU/COoflb10212.0100041.3

7228.0Z 23-

j ×=×

=

BTUbillion/COoflb000,21210100.212Z 29-3

j =××=

Total kWh in one year = kWh105.25608760106000 103 ×=×× Total BTU in one year = BTU1017.9230103.41105.2560 13310 ×=××× Total CO2 in one year = 2

9913 COoflb1037.99710/1017.9230000,212 ×=××

1.2. For one year of operation, using the data in table 1.4 perform the following:

i) The carbon footprint of 500 W if coal is used to produce the electric power.

ii) The carbon footprint of a 500 W bulb if natural gas is used to produce the electric power.

iii) The carbon footprint of a 500 W bulb if wind is used to produce the electric power.

iv) The carbon footprint of a 500 W bulb if PV energy is used to produce the electric power.

Solution Since hWhWBTU .293.0.

41.311 == , therefore, one thousand BTU= 293 Wh and one

million BTU= 293 kWh. From the table 1.4, pounds of CO2 emission per billion BTU of energy input for coal is 208,000 lb, and that of natural gas is: 117,000 lb.

Hours in one year = h760,836524 =×

Energy consumed in one year = kWh380,4876010500 3 =×× −

BTU108.935,14380,41041.3 33 ×=××= i) From table 1.4, Carbon foot print with coal as fuel =

29

3

COoflb106,310

208000108.935,14=

××

1 kg = 2.2 lb. or 1 lb. = 0.45 kg Therefore, the carbon footprint 2COofkg395,1lb45.03106 =×= From table 1.2, Carbon foot print with coal fuel =

2COoflb272,9117.2380,4 =× The difference in the value may be due to the quality of fuels might differ.

ii) From Table 1.4, Carbon foot print with gas as fuel =

29

3

COoflb747,110

117000108.935,14=

××

Therefore, the carbon footprint 2COofkg786lb45.0747,1 =×= From table 1.2, Carbon foot print with gas as fuel =

2COoflb387,8915.1380,4 =× iii) From table 1.3, Carbon foot print with wind farm generating power

2COoflb8.14403306.0380,4 =× Therefore, the carbon footprint 2COofkg16.6545.08.144 =×=

iv) Carbon foot print with PV farm generating power

2COoflb3.9652204.0380,4 =× Therefore, the carbon footprint 2COofkg38.43445.03.965 =×=

1.3 Compute the money saved in one month by using a CFL (compact fluorescent light) bulb (18

W) instead of using an incandescent light bulb (60 W) if the cost of electricity is $ 0.12 per kWh.

Assume the lights are used for 10 hours a day.

Solution Power saved when CFL light bulb is used instead of incandescent light bulb = W421860 =− Number of hours the bulb is used in a year = Hours per day ×number of days in a month =

h3003010 =× Energy saved in one month = power×hours = kWh6.121030042 3 =×× − Money saved in one month = cost per unit×no. of units = 51.1$6.1212.0 =×

1.4 Compute the carbon footprint of the light bulb of Problem 1.3 if natural gas is used as fuel to

generate electricity. How much more will the carbon footprint are increased if the fuel used is

coal?

Solutions

From the table 1.4, pounds of CO2 emission per billion BTU of energy input for coal is: 208,000 lb of CO2/ billion BTU And that of natural gas is: 117,000 lb of CO2/ billion BTU 1 kWh = 3.41 thousand BTU Hours used in one year = h365036510 =×

Energy consumed by incandescent light bulb = kWh21910603650 3 =×× −

BTU1079.7462191041.3 33 ×=××= Energy consumed by CFL light bulb = kWh7.6510183650 3 =×× − BTU1004.2247.651041.3 33 ×=××= Carbon foot print for incandescent light bulb with natural gas as fuel =

29

3

COoflb374.8710

1170001079.746=

××

Carbon foot print for CFL light bulb with natural gas as fuel =

29

3

COoflb300.2610

1170001004.224=

××

Carbon foot print for incandescent light bulb with coal as fuel =

29

3

COoflb332.15510

2080001079.746=

××

Carbon foot print for CFL light bulb with coal as fuel =

29

3

COoflb756.4610

2080001004.224=

××

Amount of carbon footprint more with coal for incandescent light bulb = 374.87332.155 − 2COoflb957.67= Amount of carbon footprint more with coal for CFL light bulb = 212.26600.46 − 2COoflb387.20= From table 1.2, the difference in carbon foot print in coal and natural gas 915.1117.2 −=

kWh/COoflb202.0 2=

Therefore, amount of carbon footprint more with incandescent light bulb =

2COoflb23.44219202.0 =×=

Therefore, amount of carbon footprint more with CFL lamp = 2COoflb27.137.65202.0 =×

1.5 Will an electric oven rated at 240 V and 1200 W provide the same heat if connected to a

voltage of 120 V? If not, how much power will it consume now?

Solution No, the oven will not give the same power output at 120 V Power consumed by a device ( )2voltage∝

Therefore, the power consumed by the oven at 120 V = W3002401201200

2

=⎟⎠⎞

⎜⎝⎛×

1.6 If the emission factor of producing electric power by PV cells is 100 g of CO2 per kWh, by

wind power is 15 g of CO2 per kWh, and by coal is 1000 g of CO2 per kWh, then find the ratio of

CO2 emission when (a) 15% of power comes from wind farms, (b) 5% from a PV source, and (c)

the rest from coal as opposed to when all power is supplied by coal-run power stations.

Solution

Emission when wind power, PV and coal is used to produce 1 kWh of energy =

1000)05.015.01(10005.01515.0 ×−−+×+× 2COofg25.807=

The emission with all-coal power plant 2COofg1000=

Therefore, the emission with renewable energy is reduced to 80.7% as opposed to coal.

1.7 Compute the operating margin of the emission factor of a power plant with three units with

the following specifications over one year:

Unit Generation

(MW)

Emission Factor

(lb. of CO2/MWh)

1 160 1000

2 200 950

3 210 920

Solution

Hours in one year = h876036524 =×

Energy consumed in one year by 160 MW unit = MWh1040.18760160 6×=×

Energy consumed in one year by 200 MW unit = MWh1075.18760160 6×=×

Energy consumed in one year by 210 MW unit = kWh1084.18760210 6×=×

∑

∑

=

== n

1ii

n

1iii

x

y.xZ

where Z = Simple operating margin CO2 emission factor (lb-CO2/MWh) xi= Net electricity power generated by unit i (MWh) yi = CO2 emission factor of unit i (lb-CO2/MWh) n = number of units in operation

∑

∑

=

== n

1ii

n

1iii

x

y.xZ

MWh/COoflb98.9521084.11075.11040.1

9201084.19501075.110001040.12666

666

=×+×+×

××+××+××=

1.8 If the initial installation cost of a thermal power plant of 100 MW is 2 million dollars and

that of a PV farm of the same capacity is 300 million dollars, and the running cost of the thermal

power plant is $90 per MWh and that of PV farm is $12 per MWh then find the time in years

needed for the PV farm to become economical if 90% of the plant capacity is utilized in each

case.

Solution

The total cost of installation and running a thermal power plant for t years = t90365241009.0102 6 ×××××+×

The total cost of installation and running a PV farm for t years=

t12365241009.010300 6 ×××××+× For the cost to be equal,

tt ×××××+×=×××××+× 12365241009.01030090365241009.0102 66

Or 66 10210300t12365241009.0t90365241009.0 ×−×=×××××−×××××

Or 610298t200,495,61 ×=×

Therefore, the time required for PV to be more economical = years85.4200,495,61

10298t6

=×

=

1.9 Consider a feeder that is rated 120 V and serving five light bulbs. Loads are rated 120 V and

120 W. All light loads are connected in parallel. If the feeder voltage is dropped by 20%,

compute the following:

i) The power consumption by the loads on the feeder in watts

ii) The percentage of reduction in illumination by the feeders

iii) The amount of carbon footprint if coal is used to produce the energy

Solution

i) The power carried by feeder at 120 V = W6001205 =×

The power carried at 20% reduced voltage = W384120

1208.06002

=⎟⎠⎞

⎜⎝⎛ ×

×

ii) Percent reduction in power = %36600

384600=

−

Since illumination is directly proportional to the power, Percent reduction in illumination %36=

iii) The energy consumed in one year = kWh84.33633652410384 3 =××× − BTU4.694,470,1184.33631041.3 3 =××

Hence, carbon footprint for coal from table 1.4 =

29 COoflb9.385,210

2080004.694,470,11=

×=

Hence, carbon footprint for coal from table 1.2 = 2COoflb7121117.284.3363 =×

1.10 The same as Problem 1.9, except a refrigerator rated 120 V and 120 W is also connected to

the feeder and voltage is dropped by 30%.

i) Compute the power consumption by the loads on the feeder in watts

ii) Compute the percentage of reduction in illumination by the feeders

iii) Do you expect any of the loads on the feeder to be damaged?

iv) Compute the amount of carbon footprint if coal is used to produce the energy

(Hint: a 40 W incandescent light bulb produces approximately 500 lumens of light)

Solution

i) The power carried by feeder at 120 V = W7201201205 =+×

The power carried at 20% reduced voltage = W8.352120

1207.07202

=⎟⎠⎞

⎜⎝⎛ ×

×

ii) Percent reduction in power = ( ) %517.01100 22 =− Since illumination is directly proportional to the power, Percent reduction in illumination %51= iii) The motors connected to the feeder will also be subjected under-voltage and will cause reduction in torque to support its load and motor will be running at much lower speed and motor will heat up and over time will burn the motor coils. iv) The energy consumed in one year = kWh53.309036524108.352 3 =××× −

BTU3.707,538,1053.30901041.3 3 =×× Hence, carbon footprint for coal from tab. 1.4 =

29 COoflb05.192,210

2080003.707,538,10=

×=

Hence, carbon footprint for coal from table 1.2 = 2COoflb542,6117.253.3090 =×

1.11 The same as Problem 1.9, except a refrigerator rated 120 W is also connected to the feeder

and voltage is raised by 30%.

i) Compute the power consumption by the loads on the feeder in watts

ii) Compute the percentage of reduction in illumination by the feeders

iii) Do you expect any of the loads on the feeder to be damaged?

iv) Compute the amount of carbon footprint if coal is used to produce the energy

Solution

i) The power carried by the feeder at 120 V = W7201201205 =+×

The power carried at 30% increased voltage = W8.216,1120

1203.17202

=⎟⎠⎞

⎜⎝⎛ ×

×

ii) Percent increase in power = ( ) %6913.1100 22 =−

Since illumination is directly proportional to the power, Percent increase in illumination %69=

iii) The devices connected to the feeder may be affected in two ways: a) there

may be voltage breakdown of insulators, b) the current through the devices

will increase and contribute towards power loss and increase the

temperature of the device which may cause burnout of light bulb and

motor over time..

iv) The energy consumed in one year =

kWh17.659,1036524108.1216 3 =××× −

BTU7.769,347,3617.659,101041.3 3 =×× Hence, carbon footprint for coal from table 1.4=

29 COoflb33.560,710

2080007.769,347,36=

×

Hence, carbon footprint for coal from table 1.2 = 2COoflb565,22117.217.659,10 =×

1.12 Compute the CO2 emission factor in pounds of CO2 per BTU for a unit in a plant that is

supplied by coal, oil, and natural gas if 0.3 million tons of coal, 0.1 million barrels of oil, and 0.8

million cubic feet of gas have been consumed over one year. The average power produced over

the period was 210 MW. Use the following data and the data of Table 1.4 for computation: a ton

of coal has 25 million BTU; a barrel (i.e., 42 gallons) of oil has 5.6 million BTU; a cubic foot of

natural gas has 1030 BTU.

Solution

Total number of hours in one year = 876036524 =×

The energy produced in one year = kWh106.1839876010210 63 ×=××

BTU1004.273,61041.3106.1839 936 ×=×××=

x

y.Q.fZ

n

1iiii∑

==

where,

Z = CO2 emission factor of generating unit (lb-CO2/BTU)

fi = fuel of type i consumed by a generating unit per year (mass or volume unit)

Qi = heat energy of fuel type i in BTU per ton of coal, in BTU per gallon of oil or in BTU per

cubic feet of natural gas

yi = CO2 emission factor of fossil fuel type i (lb-CO2/BTU)

x = Net power delivered to power grid in BTU

n = total number of units.

Here,

BTU1004.273,6x 9×=

26

coal COoftons103.0f ×=

coalofton/BTU1025Q 6coal ×= ,

From table 1.4, BTUbillionperlb000,208ycoal =

oilofbarrels101.0f 6oil ×=

oilofbarrel/BTU106.5Q 6oil ×= ,

From table 1.4, BTUbillionperCOoflb000,164y 2oil =

gasoffeetcubic108.0f 6gas ×=

gasoffeetcubic/BTU1030Qgas = ,

From table 1.4, BTUbillionperCOoflb000,117y 2gas =

Therefore,

9

96

966966

1004.273,6101170001030108.0

10164000106.5101.0102080001025103.0

Z×

⎟⎟⎠

⎞⎜⎜⎝

⎛

××××+

×××××+×××××

=−

−−

BTUperCOoflb10263.0 23−×=

Total kWh in one year = kWh10839.1876010210 93 ×=××

Total BTU in one year = BTU106.271103.4110839.1 1239 ×=×××

Total CO2 in one year = 29123 COoflb10.6491106.27110263.0 ×=××× −

Power Grids 2.1 Assume a =120 + j 100, b = 100 + j 150 and c = 50 + j 80. Compute the following complex numbers: i) ( ) cba /×

ii) ( ) cba ×/

iii) ( ) ( )acba −− / Solutions

i) 184234 j+ ii) 5461 j+ iii) 74.0075.0 j+−

2.2 Assume ( )°+= 5377cos120 tV .If a voltmeter is used to measure this voltage, what value should the voltmeter read? Express this voltage in polar form. Solution The voltmeter will read V85.842/120 =

°∠= 5120V 2.3 The operation of AC machines (in particular, transformers and induction machines) can be studied with the aid of the T-circuit shown in Fig. 2.65 below.

Chapter 2

Figure 2.65 Equivalent Circuit for Problem 2.3.

Assume the frequency is 60 Hz. The circuit elements are given in table.

Case

R1 L1 Rf Lm 'R 2 'L2 RL LL V1 V2

I1

I2

’

If

1 1 0.01 1000 8 1 0.01 Open 480 ? ? ? ?

2 1 0.01 1000 8 1 0.01 200 0 480 ? ? ? ?

3 0.02

0.00265

Open 0 0 Open 1 ? ? ? ?

4 0.02

0.00265

Open 0 0 1 0 1 ? ? ? ?

5 0.02

0.00265

Open 0 0 0.707 1.875

310−×

? 1 ? ? ?

6 0 0 100 0.1 0.01 106610−×

1 0 1 ? ? ? ?

7 0 0 100 0.01 0.01 106610−×

1.414 3.75310−×

1 ? ? ? ?

8 0.3 1.33 Ope 3.45 0.15 0.56 7.35 0 127 ? ? ? ?

310−× n 210−× 310−×

9 10 5.2210−×

Open

0 0 200 0.4 0 500 ? ? ? ?

10 0.15

2.54310−×

Open

1.57 6.24310−×

98.5 0.178 0 240 ? ? ? ?

11 0.3 0.003 1 4.25210−×

0.2 0.003 10 0 440 ? ? ? ?

12 0.3 0.003 0 4.25210−×

0.2 0.003 1 0 380 ? ? ? ?

Use the polar form for all complex numbers and solve set 1 through set 12. Show your calculations separately.

Solutions

Case 1:

openXjRZ LLL =+=

Ω+=×π×+=+= 3.77j160201.0j1XjRZ 111

Ω=+×

==ϕ 298.73j+900.95XjRXjR

Xj||RZmf

mfmf

A18.5407-0.504ZZ

VII

1

11 °∠=

+==

ϕϕ

( ) =°∠+=ϕ= ϕ 18.5407-0.50473.298j95.900IZV2 478.91∠-0.19o V

0I 2 =

Case 2:

0ZV

'ZVV

ZVV

2

2

1

1 =+−

+−

ϕ

ϕϕϕ

0'Z

V'ZVV

L

2

2

2 =+− ϕ

( ) ( ) 0YV'YVVYVV 2211 =+−+− ϕϕϕϕ

( ) 0'YV'YVV L222 =+− ϕ

Where ϕ

ϕ ===Z1Y,

'Z1'Y,

Z1Y

22

11

Rearranging,

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−

−++ ϕϕ

0YV

VV

'Y'Y'Y'YY'YY 11

2L22

221

⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡+−

−++=⎥

⎦

⎤⎢⎣

⎡−

ϕϕ

0YV

'Y'Y'Y'YY'YY

VV 11

1

L22

221

2

The MATLAB program for solving the above (Case 2) is given below.

X1 = L1*w;

Z1 = R1+X1*1j;

Y1 = 1/Z1;

Rf = 1000;

Lm = 8;

Xm = Lm*w;

Zf = Rf*Xm*1j/(Rf+Xm*1j);

Yf = 1/Zf;

R2 = 1;

L2 = 0.01;

X2 = L2*w;

Z2 = R2+X2*1j;

Y2 = 1/Z2;

Rload = 200;

Lload = 0;

Xload = Lload*w;

Zload = Rload + Xload*1j;

Yload = 1/Zload;

Ybus = [Y1+Y2+Yf -Y2; -Y2 Y2+Yload];

VfV2 = inv(Ybus)*[V1*Y1; 0];

Vf = VfV2(1);

V2 = VfV2(2);

abs_V2 = abs(V2)

angle_V2 = angle(V2)*180/pi

Iphi = Vf*Yf;

abs_Iphi = abs(Iphi)

angle_Iphi = angle(Iphi)*180/pi

I_f = Iphi*Xm*1j/(Rf+Xm*1j);

abs_I_f = abs(I_f)

angle_I_f = angle(I_f)*180/pi

I2 = V2*Yload;

abs_I2 = abs(I2)

angle_I2 = angle(I2)*180/pi

I1 = (V1-Vf)*Y1;

abs_I1 = abs(I1)

angle_I1 = angle(I1)*180/pi

The above program can be used to solve other cases by using the data from the problem by substituting the values of impedances and voltages. When a part of circuit is specified as open, its value can be substituted with a large number and when as a short with zero impedance. The results are tabulated in table 2 blow:

SOLUTION TABLE 2.1 The results for all cases.

case V1 V2 I1 I2’ If

1 480 478.91

-0.19o

0.505

-18.5o

0 0.48

-0.2o

2 480 473.82

-2.33o

2.85

-5.32o

2.37

-2.33o

0.48

-1.26o

3 1 1

0o

0 0 0

4 1 0.7

‐44.4o

0.7

‐44.4o

0.7

‐44.4o

0

5 1.85

‐21.9o

1 0o 1

‐45o

1

‐45o

0

6 1 0.989

‐2.27o

1

‐3.76o

0.989

‐2.27o

0.01

0o

7 1 0.983

‐0.6o

0.71

-60o

0.491

‐45.6o

0.01

0o

8 127 115.1

‐3.87o

18.29

‐33.4o

15.65

‐3.87o

0

9 500 0 22.71

‐62.97

o

0 0

10 240 0.0011

‐89.86o

0.411

‐89.98

o

0.0065

‐89.86

o

0

11 440 229.90

-47.12o

259

-44.61o

22.99

-47.13o

235.93

-40.80o

12 380 0 324.76

-75.14o

0 324.76

-75.14o

2.4. Assume a balanced three-phase load. Each phase load is rated 120 Ω with a phase angle of 10° lagging and the load is connected as a Y bank. A three-phase balance source rated 240 V is applied to this three-phase load. Perform the following: i) Give the three-phase circuit of this load. ii) Compute the active and reactive power consumed by this load. Solutions

Figure 2.4.1 The Three Phase Wye connected load of problem 2.4.

( ) W7.47210cos120

3/2403cosZV.3

P22

ph =×

=θ=

( ) VAr3.8310sin120

3/2403sinZV.3

Q22

ph =×

=θ=

The load consumes472.7 W of active power and 83.3 VAr . 2.5 Assume a single-phase load has inductance of 10 millihenry (mH) and resistance 3.77 Ω and it is connected to a 60 Hz source at 120 V. Compute the active and reactive power consumed or generated by the load. Solution

Ω=×π××=ω= − 77.36021010LX 3

Ω=+= 33.5XRZ 22

°==θ − 4577.377.3tan 1

W910,145cos33.5

120cosZ

VP

22

==θ=

VAr910,145sin33.5

120sinZ

VQ

22

==θ=

The load consumes 1,910 W of active power and 1,910 VAr of reactive power.

2.6 Assume a single-phase load has a capacitance of 10 millifarad (mF) and resistance of 2.0 Ω and it is connected to a 60 Hz source at 120 V. Compute the active and reactive power consumed or generated by the loads.

Solution

Ω=×π××

=ω

= − 265.06021010

1C1X 3

Ω=+= 02.2XRZ 22

°−==θ − 5.72265.0tan 1

W1006.75.7cos02.2

120cosZ

VP 322

×==θ=

VAr4.9305.7sin02.2

120sinZ

VQ22

−=−=θ−=

The load consumes 7060 W of active power and generates 930.4 VAr of reactive power.

2.7 A single-phase load consumes 10 kW at 0.9 power factor, p.f. lagging at 220 V. What is the magnitude and phase angle of current drawn from the source? If another single-phase load rated at 10 kW and 0.8 leading power factor is connected in parallel to the same sources, what is the magnitude of current drawn? What is the current drawn from the source? Give your answers in polar form. Solution

A84.255.509.0cos9.0220

000,10coscos.VPI 11

lag °−∠=−∠×

=θ−∠θ

= −−

A8.368.568.0cos8.0220

000,10coscos.VPI 11

lead °∠=∠×

=θ∠θ

= −−

A28.703.91III laglagsource °−∠=+=

2.8. A single-phase transformer is rated 200 kVA, 120/220 V, 5% reactance. Compute the reactance of the transformer from the high-voltage side. Give the single-phase equivalent circuit. Solution

Ω=== 242.0000,200

220SV

Z2

b

2b

b

Ω=×=×= 0121.0j05.0242.0jZZZ u.pb

Solution Figure 2.8.1 The single-phase equivalent circuit of problem 2.8

Solution Figure 2.8.2 The per-unit equivalent circuit of problem 2.8. 2.9 Compute the per unit model of Problem 2.7 based on 40 kVA and a base voltage of 440 V. Solution

Lagging load in p.u = 9.0cosSSS 1

blag,u.p

−∠=

kVAu.p84.25277.09.0cos40

9.0/10 1 °∠=∠= −

Lagging load in p.u = 8.0cosSSS 1

blead,u.p

−−∠=

kVAu.p86.363125.09.0cos40

8.0/10 1 °−∠=−∠= −

Per unit voltage Vu.p5.0440220

VVV

b,u.p ====

kVAu.p86.363125.0 °−∠=kVAu.p84.25277.0 °∠=

Solution Figure 2.9.1 The per-unit -model diagram for problem 2.9 2.10 Three single-phase transformers are each rated 460 V/ 13.2 kV, 400 kVA and 5% short circuit reactance. The three single-phase transformers are connected as a three-phase Y–Y transformer. Perform the following: i) Compute the line-to-line voltages of three-phase Y–Y transformer from the high-voltage-side and the low-voltage-side voltage and kVA ratings. ii) Compute the per unit model of the three-phase transformer. Solutions i) Low voltage side rating VVph 7.79634603 ===

High voltage side rating kVVph 86.2232.133 ===

kVA rating kVAS 120040033 1 =×=×= φ

ii) Base values: kV86.22V,V7.796V,kVA1200S HV,bLV,bb ====

Base impedance at low voltage side: Ω=

×=== 5289.0

1012007.796

SV

Z 3

2

b

2b

b

The per unit transformer impedance = 5%

Solution Figure 2.10.1 The per-unit equivalent model of the three-phase transformer of problem 2.10. 2.11A bank of three-phase Y-connected resistive load of 30 Ω is connected to the low-voltage side of the transformer of Problem 2.10 using a feeder with resistance of 1 + j10 Ω. Perform the following: i) Give the three-phase equivalent model. ii) Give a one-line diagram. iii) Compute the per unit model based on the transformer rating. Solutions

i) Base impedance on high voltage side: Ω=×

=== 5.435101200

22860SV

Z 3

2

b

2b

b

Transformer impedance on high voltage side: Ω=×=×= 7.2105.05.435XjZb

Solution Figure 2.11.1 The three-phase equivalent circuit of problem 2.11

ii)

Solution Figure 2.11.2 The one-line diagram of problem 2.11

iii) Base values: kV86.22V,V7.796V,kVA1200S HV,bLV,bb ====

Base impedance low voltage side: Ω=

×== 5289.0

1012007.796

SV

Z 3

2

b

2b

b


Per unit transformer impedance Ω==== 72.565289.030

ZZZ

bu.p,load

Per unit load impedance Ω==== u.p72.565289.030

ZZZ

bu.p,load

Per unit line impedance: Ω+=+

=== u.p9.18j89.15289.0

10j1ZZZ

bu.p,line

Solution Figure 2.11.3 The per-unit equivalent circuit of problem 2.11

2.12 Three single transformers are each rated 460 V/ 13.2 kV 400 kVA, 5% short circuit reactance. The three single-phase transformers are connected as a three-phase Y–Δ transformer. Perform the following: i) Compute the line-to-line voltage of the three-phase Y–Δ transformer from high-voltage-side and low-voltage-side voltage and kVA ratings. ii) Compute the per unit model of the three-phase transformer. Solutions Assuming that the low voltage side is Y connected and high voltage side is Δ connected, we will have: i) Low-voltage-side rating V7.79634603Vph ===

High-voltage-side rating kV2.132.13Vph ===

kVA rating kVA12004003S3 1 =×=×= φ

ii) Base values: kV2.13V,V7.796V,kVA1200S HV,bLV,bb ====

Base impedance on low voltage side: Ω=

×=== 5289.0

1012007.796

SV

Z 3

2

b

2b

b


Solution Figure 2.12.1 The per-unit equivalent circuit of problem 2.12 2.13 A bank of three-phase Y-connected resistive load of 30 Ω is connected to the low-voltage side of the transformer of Problem 2.12 using a feeder with resistance of 1 + j10 Ω. Perform the following: i) Give the three-phase equivalent model. ii) Give the one phase of a Y-connected equivalent circuit. iii) Give a one-line diagram. iv) Compute the per unit model based on the transformer rating. Solutions

i) Assuming low voltage side is Y connected,

Solution Figure 2.13.1 The three-phase equivalent circuit of problem 2.13 ii) Low-voltage-side line to neutral rating V460Vph ==

High-voltage-side line to neutral rating kV7.62

32.13

3V LL === −

Base impedance on high voltage side: Ω=×

=== 2.145101200

13200SV

Z 3

2

b

2b

b

Transformer impedance on high voltage side: Ω=×=×= 26.705.02.145XjZb

Solution Figure 2.13.2 The single-phase equivalent circuit of problem 2.13

iii)

Solution Figure 2.13.3 The per-unit equivalent circuit of problem 2.13 iv) Basevalues: kV2.13V,V7.796V,kVA1200S HV,bLV,bb ====


×== 5289.0

1012007.796

SV

Z 3

2

b

2b

b

Theper unit transformer impedance = 5%


ZZZ

bu.p,load


=== u.p9.18j89.15289.0

10j1ZZZ

bu.p,line

Solution Figure 2.13.4 The per-unit equivalent circuit of problem 2.13 2.14 Three single-phase transformers are each rated 460 V/ 13.2 kV 400 kVA, 5% short circuit reactance. The three single-phase transformers are connected as a three-phase Δ–Δ transformer. Perform the following: i) Compute the line-to-line voltage of three-phase Δ–Δ transformer from the high-voltage-side and the low-voltage-side voltage and kVA ratings. ii) Give the three-phase equivalent model. iii) Give the one-phase of a Y-connected equivalent circuit. iv) Compute the per unit model based on the transformer rating.

Solutions i) Low-voltage-side rating V460460Vph ===

High-voltage-side rating kV2.132.13Vph ===


ii)


iii) Low-voltage-side rating of Y equivalent V265.53

460Vph ===

High-voltage-side rating of Y equivalent kV7.6232.13Vph ===



=== 2.145101200

13200SV

Z 3

2

b

2b

b

Transformer impedance on high voltage side: Ω=×=×= 26.705.02.145jXZb

Solution Figure 2.14.2 The single-phase equivalent circuit of problem 2.14 iv) Base values: V7.796V,kVA1200S bb ===

Base impedance on low voltage side: Ω=×

=== 1763.0101200

460SV

Z 3

2

b

2b

b

Low-voltage-side base voltage V460V LV,b ==

High-voltage-side base voltage kV13.2V HV,b ===

The per unit transformer impedance = 5% ohm

Solution Figure 2.14.3 The per-unit equivalent circuit of problem 2.14 2.15 A bank of three-phase Δ–connected resistive load of 30 Ω is connected to the low-voltage side of the transformer of Problem 2.14 using a feeder with resistance of 1 + j10 Ω.Perform the following: i) Give the three-phase equivalent model. ii) Give the one phase of a Y-connected equivalent circuit. iii) Give a one-line diagram. iv) Compute the per unit model based on the transformer rating.

Solutions i)


ii) Low-voltage-side line to neutral voltage V265.53

460V low,NL === −

High-voltage-side line to neutral voltage kV7.6232.13V high,NL === −


=== 2.145101200

13200SV

Z 3

2

b

2b

b

Transformer impedance on high voltage side: Ω=×=×= 26.705.02.145jXZb

The Y-equivalent load Ω=×=== Δ 90303Z.3ZY

Solution Figure 2.15.2 The single-phase equivalent circuit of problem 2.15 iii)

Solution Figure 2.15.3 The one-line equivalent circuit of problem 2.15 iv) Base values: kV2.13V,V460V,kVA1200S HV,bLV,bb ====


×=== 1763.0

101200460

SV

Z 3

2

b

2b

b



ZZZ

bu.p,load


=== u.p7.56j67.51763.0

10j1ZZZ

bu.p,line

Solution Figure 2.15.4 The per-unit equivalent circuit of problem 2.15 2.16 A balanced three-phase, three-wire feeder has three balanced loads as shown in Fig. 2.66.

Figure 2.66 The three-phase diagram of problem 2.16. Each lamp is rated at 100 W and 120 V. The line-to-line voltage on the feeder is 240 V and remains constant under the loads. Find the source current in the feeder lines and the power delivered by the source. i) Compute the line current in phase a. ii) Compute the active and reactive power supplied by the sources. Solutions

Ω== 144100

120R2

lamp

A96.03144

240Ilamp ==

A38.13100

240IR ==

A901220j

240I 20j −∠==

A901220j

240I 20j ∠=−

=−

034.2IIIII 20j20jRlampS ∠=+++= −

kW973.0134.22403cosV.I.3P sSs =×××=θ=

kVar0Qs=

2.17. Consider a three-phase distribution feeder as shown in Fig. 2.67.

Figure 2.67 The single–line diagram of Problem 2.17. For the three-phase system, all voltages are given line-to-line and the complex power is given as three-phase power. Compute the following: i) The source voltage Vs, if VR is to be maintained at 4.4 kV (VR = 4.4 kV V line value) ii) The source current and the power factor at the source iii) The total complex power supplied by the source iv) How much reactive power should be connected to the source bus to obtain a unity power factor at the source bus? Solutions

kVA80j60S1 −= kVA33.133j100S2 +=

kVAjS 33.13303 −= kVA100jS4 =

100100432 jSSSS R +=++=

AjjI R 14.1314.133/4.43

100100*

−=⎟⎟⎠

⎞⎜⎜⎝

⎛×+

=

i) ( ) kV35.1612.4100j14.13j14.133

4400ZIVV RRS °∠=×−+=+=

ii) A48.6909.835.1612.43

80j60I*

1L ∠=⎟⎠⎞

⎜⎝⎛

∠×−

=

°−∠=+= 56.559.15III R1LS , power factor = 94.05.18cos = , lagging

iii) kVA90.71j77.178I.V.3S *SSS +==

iv) Reactive power to be connected to the source for unity power factor = -j 71.90kVar

2.18 Assume a three-phase transformer is Y-grounded on the low-voltage side and Δ-connected winding on the high-voltage side as shown in Fig. 2.68. Each winding of tap-changing winding has 2000 turns on the low-voltage side, and a variable number of turns on the high-voltage winding side. Assume the high-voltage winding side has a maximum number of turns equal to 7300 turns and a minimum number of turns equal to 5300. Compute the minimum and maximum line-to-line voltage that can be maintained on the high-voltage side. Assume the winding voltage across the low-voltage side is set equal to 36.4 kV.

Figure 2.68 Figure for Problem 2.18.

Solution

T/V5.102000

3/104.362000V 3

p =×

=

kV65.76V,7300V

5.10N

V2000V

secsec

maxsec

secp ====

kV65.55V,5300V

5.10N

V2000V

secsec

minsec

secp ====

Thus, secondary voltage can be changed to between 55.65 and 76.65 kV.

2.19. Write a term paper on the capital cost of a one megawatt solar system and its payback over 25 years. Assume the cost of a coal-fired power plant is $.10 per kWh and PV, to generate electricity the cost is about $0.25–0.40 per kWh.

2.20. A three-phase generator rated 480 V and 400 kVA connected to a three-phase transformer rated 3.2 kV/480 V, 200 kVA, with reactance of 10% is used to serve a load over a line of 1 + j20 Ω. At the load site, the voltage is stepped down to 220 V from 3.2 kV using a three-phase transformer rated 150 kVA with reactance of 7%. The voltage at the load is to be maintained at 220 V. Perform the following:

i) Give a one-line diagram.

ii) Compute a per unit equivalent circuit based at 200 kVA and 480 V.

iii) Compute the required generator voltage for unity power factor load. Is this design viable?

Solutions

i)

Solution Figure 2.20.1 The one-line diagram of Problem 2.20.

ii)Base impedance on the high voltage side = Ω=== 2.51000,200

200,3 22

b

bb S

VZ

The per unit impedance of the feeder = Ω+=+

upjj

.39.0019.02.51201

The per unit impedance of the transformer on the generator side= Ωupj .1.0

The per unit impedance of the distribution transformer = Ω=× u.p093.015020007.0j

Solution Figure 2.20.1 The per unit equivalent circuit of problem 2.20

iii) The generator voltage in per-unit =

( ) Vu.p8.2917.1139.0j019.0093.0j1.0j1 °∠=×++++=

Because the generator voltage is not within ±5% of the rated value, the design is unacceptable.

The magnitude of the generator voltage = V6.56148017.1 =×

CHAPTER 3 MODELING CONVERTERS IN MICROGRID

POWER SYSTEMS

3.1 Consider the microgrid of Figure 3.66. A three-phase transformer, T1, is rated 500 kVA, 220 Y grounded / 440 VΔ, transformer with the reactance of 3.5%.The microgrid is supplied from an AC bus of a PV generating station with its DC bus rated at 540 V. The distribution line is 10 miles long and has a series impedance of 0.1 + j 1.0 Ω per mile and local load of 100 kVA at 440 V. The microgrid is connected to the local power grid using a three-phase transformer T2, rated at 440V Y grounded/13.2 kV Δ, 500 kVA with the reactance of 8%. Compute the per-unit impedance diagram of the microgrid system. Assume the voltage base of 13.2 kV on the local power grid side and a kVA base of 500.

Figure 3.66 A One-Line Diagram for Problem 3.1. Solution

kVA500Sb = , kV2.13Vb =

kVAu.p2.0500100

SS

Sb

loadload,pu ===

Assuming that power factor is unity, the load is equal to 0.2 p.u kW

Ω=×

== 38.010500

440VAV

Z 3

2

base

2base

trans,b

Ω+=+

== u.p3.26j63.238.010j1

ZZ

Ztrans,b

transtrans,pu

Ω= u.p08.0jX 2T,pu , Ω= u.p035.0jX 1T,pu

The AC bus-base voltage on the LV side of T2 = V4402.132.13

440VVV

HV,bHV

LV =×=

The AC bus-base voltage on the LV side of T1 = V220440440220V

VV

HV,bHV

LV =×=

Vb AC side of inverter = base voltage for low voltage side of T1 =VLL,b = 220 V

The current rating of the inverter on DC side = A926540

10500V

PI3

DC

=×

==

For three phase power, I.V.3VA = where VA is the volt-ampere V is the line to line voltage and I is the line current.

The current rating of the inverter on AC side = A312,13220

10500I3

=×

=

Using equation aDC

LL M2

V23V = (3.81), for three-phase inverter, the modulation index,

can be computedas:

The modulation index of the inverter = 665.0540220

322

VV

322M

DC

LLa =×==

Vb DC side assuming the Ma remains fixed at 0.665 = a

b,LLb,DC M

V322V =

DCV540665.0

220322 =×=

The inverter and the PV are designed to be rated at 500 kW.

The current rating of the PV bus = A926540

10500V

PI3

DC

=×

==

Solution Figure 3.1.1 A One line diagram of problem 3.1

Solution Figure 3.1.2 The Per Unit Diagram of Problem 3.1.

3.2 Consider the microgrid of Figure 3.67. Assume the inverter AC bus voltage of 240 V and transformer T1 is rated 5% impedance, 240 V Δ/120 V Y-grounded and 150 kVA. The transformer T2 is rated at 10% impedance, 240 V Δ/3.2 kV Y-grounded and 500 kVA. The local loads are rated at 100 kVA and a power factor of 0.9 lagging. The inverter modulation index is 0.9.Compute the following:

i) The DC bus voltage and inverter rating ii) The boost converter PV bus input voltage and input current ratings for the required DC bus voltage of the inverter iii) The size of the microgrid PV generating station iv) The per unit model of microgrid

Inverter

PV Generating Station DC/DC DC/AC

DC Bus

Load-1

Boost Converter

Local PowerGrid

T2

DC BusPV

AC Bus

Load-2

T1

Figure 3.67.A One-Line Diagram of Problem 3.2. Solutions

i) Using equation aDC

LL M2

V23V = (3.81), for three-phase inverter,

V4359.0

240322

MV

322V

a

LLDC =×==

Rearranging DC-DC converter equation (3.47), Vin = Vo (1 – D),as presented in the book, we will have

Assuming the duty ratio of the DC-DC converter is fixed at 0.7,

( ) ( ) V5.1307.01435D1VV DCPV =−=−=

The rating of the PV bus is 130.5 V.

The power rating of the PV, boost converter, and inverter should be more than the sum of the transformer ratings = kW650500150 =+

The converter rating for a lossless converter on high and low voltage sides can be expressed as

(Vdc. Idc) high voltage side= ( Vdc.Idc) low voltage side

The current rating of the inverter on the DC side = kA49.1435650

VPIDC

===


The current rating of the inverter on the AC side = kA56.13240

650I ==

ii) The PV bus voltage changes with a change in environmental conditions. The rating given above is treated as the maximum PV bus voltage. As the voltage of the PV bus drops, the duty ratio of the boost converter should be adjusted to maintain the DC bus voltage of the inverter.

Table 3.2.1 Boost Converter Ratings

Output voltage = 435 V

Power = 650 kW

Output current = 1.49 kA

Input Voltage, V

D

Duty Ratio

Input Current, kA

100 0.77 6.50

75 0.82 8.67

50 0.88 13.0

iii) The size of the microgrid is 650 kVA

iv) Sb = 650 kVA, Vb = 3.2 kV on the utility side The voltage base of all sides of the transformer is the same as the rating because the base voltage has been chosen at the rated value.

The AC bus-base voltage on the LV side of T2 = V2402.132.13

240VVV

HV,bHV

LV =×=

The AC bus-base voltage on the LV side of T1 = V120240240120V

VV

HV,bHV

LV =×=

The base voltage of the AC side of the inverter

= base voltage of the LV side of T2 = VLL,b = 240 V The base voltage of the DC side of the inverter and the PV side are also the same at the rated value assuming the modulation index and the duty ratio are fixed at 0.9

The base value of DC side of the inverter is therefore = a

b,LLb,DC M

V322V =

V4359.0

240322 =×=

With the duty ratio of the boost converter fixed at 0.7, the base voltage at the PV bus is ( ) ( ) V5.1307.01435D1VV b,DCb,PV =−=−=

kVAu.p153.0650100

SS

Sb

loadload,pu ===

Assuming that the load power factor is unity, the load is 0.153 p.u KW.

Ω=×== u.p22.0j15065005.0j

SS

ZZold

newold1T,pu1T,pu

Ω=×== u.p13.0j5006501.0j

SS

ZZold

newold2T,pu2T,pu

The AC bus base voltage of the inverter = 240 V The DC side base voltage of the inverter = 435 V The PV side base voltage of the converter = 130.5 V

Solution Figure 3.2.1 A One line diagram of problem 3.2

Solution Figure 3.2.2 The Per Unit Model of Problem 3.2

3.3 For Problem 3.2 depicted in Figure 3.67, assume the DC bus voltage of the inverter is equal to 800 V DC. The transformer T1 is rated 9% at impedance, 400 VΔ/220 V Y groundedand 250 kVA. The transformer T2 is rated at 10% impedance, 400 V Y grounded /13.2 kV Δ and 500 kVA. The Load 1 is rated at 150 kVA with a power factor of 0.85 lagging. The Load 2 is rated at 270 kVA with a power factor of .95 leading. Compute the following:

i) The modulation index and inverter rating ii) The boost converter PV bus voltage and input current for the required DC bus voltage of the inverter iii) The minimum size of the microgrid PV station

Solutions


LL M2

V23V = (3.81), for three-phase inverter, the

modulation index, can be computed as:

81.0800400

322

322 =×==

DC

LLa V

VM

Sizing the inverter as a sum of the rating of the transformers = 500 + 250 = 750 kW

ii) The high voltage DC bus of boost converter is equal to 800 V. Assuming the duty ratio of the DC-DC boost converter is 50%, the PV bus voltage on low side of boost converter rearranging DC-DC converter equation (3.47), Vin = Vo (1 – D), as presented in the book, we will have

( ) ( ) V4005.01800D1VV DCPV =−=−=

The output voltage rating = 800 V

The input voltage rating = 400 V

The rating of the converter and the inverter is so chosen that they are able to supply full load to the two transformers at the same time.The power rating of the inverter and the boost converter must be equal to the ratings of transformerT1 and transformer T2.

Boost Converter rating= 750 kW

Inverter rating= 750 kW

iii) Since the converters are rated 750 kW, the size of the PV-generating station cannot be greater than 750 kW with an output voltage of 400 V.


VPIDC

===



750I ==

The PV bus voltage changes with achange in environmental conditions. The rating given above is treated as the maximum PV bus voltage. As the voltage of the PV bus drops, the duty ratio of the boost converter should be adjusted to maintain the DC bus voltage of the inverter.

Table 3.3.1 Boost Converter Ratings


Power = 750 kW


Input Voltage, V

D

Duty Ratio

Input Current, kA

300 0.625 2.5

200 0.75 3.75

100 0.875 7.5

Size of the PV station is the sum of the transformer ratings = 500 + 250 = 750 kW The voltage rating of the PV = input bus voltage of the boost converter = 400 V

3.4 For Problem 3.3 depicted in Figure 3.67, assume transformer T1 is rated at 5% impedance, 440 V Y grounded /120 V Δ and 150 kVA. The transformer T2 is rated at 10% impedance, 440 V Y grounded/3.2 kV Δand 500 kVA. The inverter modulation index is 0.9. Select the power base of 500 kVA and the voltage base of 600V. The local power grid internal reactance is 0.2 per-unit based on 10 MVA and 3.2 kV.Assume the microgrid is not loaded.Assume PV has 50% short circuit resistance based on its own rating. Compute the following:

i) The per-unit equivalent impedance model ii) The fault current if a balanced three-phase fault occurs on the inverter AC bus

Solutions

kVASb 500= , V600Vb = on the AC side of the inverter i) The transformer high and low voltages are as follows:

Voltage base of the low-voltage side of T1 =

V6.163600440120V

VV

HV,bHV

LV =×=

Voltage base of high-voltage side of T2 kV3.4600440

3200VVV

LV,bLV

HV =×==

Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p089.0j

600440

15050005.0j

VV

SS

ZZ22

new

old

old

newold1T,pu1T,pu

Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p053.0j

600440

5005001.0j

VV

SS

ZZ22

new

old

old

newold2T,pu2T,pu

Ω×=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= − u.p105j

600440

000,10500002.0

VV

SS

ZZ 522

new

old

old

newold,Utility,puUtility,pu

Using equation aDC

LL M2

V23V = (3.81), for three-phase inverter,

The base voltage of the AC side of the inverter = base voltage of LV side of T2 VLL,b = 600 V

V088,19.0

600322

MV

322V

a

b,LLb,DC =×== = base voltage of the DC side of the inverter

assuming that the modulation index remains fixed.

Assuming a duty ratio of 0.6 for the DC-DC converter,

Rearranging DC-DC converter equation (3.47), Vin = Vo (1 – D), as presented in the book, we will have

( ) ( ) V2.4356.011088D1VV b,DCb,PV =−=−= = base voltage on the PV side assuming

duty ratio of the converter remains fixed.

The inverter, converter, and the PV voltages should be rated at 500 + 150 = 650 kV. The PV cells and DC-AC converters have losses and the PV-generating station as seen from AC side has a net short circuit resistance. Assume the PV-generating station has a short circuit resistance of 50% based on its own rating.

Ω= u.p5.0Z PV,pu

Ω=×== u.p38.06505005.0

SS

ZZold

newold,PV,punew,PV,pu

Assuming the AC bus of inverter is at its rate voltage before the fault, the per-unit voltage of AC bus can be computed as

The per unit voltage of inverter AC bus load voltage = Vu.p73.0600440Vpu ==

The inverter AC bus voltageis rated at 440 V AC

The DC bus voltage is rated at DCV794108873.0 =×

The boost converter high DC bus voltage is rated at 794 V DC and

Low DC bus voltage is rated at ( ) ( ) =×−=− 7946.01VD1 o 317 V DC.

Therefore, the PV bus is rated at 317 V DC.


VPIDC

===



650I ==

The PV bus voltage changes with achange in environmental conditions. The rating given above is treated as the maximum PV bus voltage. As the voltage of the PV bus drops, the duty ratio of the boost converter should be adjusted to maintain the DC bus voltage of the inverter.

Table 3.4.1 Boost Converter Rating


Power = 650 kW


Input Voltage, V

D

Duty Ratio

Input Current, kA

100 0.87 6.50

75 0.91 8.67

50 0.94 13.0

Solution Figure. 3.4.1 one line diagram of problem 3.4

Solution Figure 3.4.2 The Per Unit Model of Problem 3.4 ii) The figure below shows the per unit model for short-circuit studies.

Solution Figure 3.4.3 The Per Unit Model for Fault Studies Neglecting the Loads.

To calculate the fault current if a balanced three-phase fault occurs on the inverter AC bus, we can use superposition concept one source acting at a time. We have two sources of fault currents when AC side is subjected to three-phase faults. The calculation of the short-circuit current is to estimate the expected short circuit current to rate the circuit breakers and to adjust the settings for protection. The maximum inverter short-circuit current is if the inverter acts as short. Then we have two sources of short-circuit currents. One source is a PV DC bus; its short-circuit current is limited by its internal resistance of the PV source. The other source is a short-circuit current contribution from the grid side. The total fault current is summation of the current contribution from the PV (DC) and the utility (AC).

Solution Figure 3.4.4 Equivalent circuit before fault

Solution Figure 3.4.5 Fault current contribution from PV

Solution Figure 3.4.6 Fault current contribution from utility

Solution Figure 3.4.7 Fault current as superposition of both PV and utility

Au.p92.138.073.0

RV

IPV,pu

puPV,F ===

Au.p7.13j105j053.0j

73.0XZ

VI 5

utility,pu2T,pu

puutility,F −=

×+=

+= −

IF,PV is DC and IF,utility is AC and therefore the fault current is the two currents superposed.

tcos7.1392.1III utility,FPV,FF ω+=+=

The rms value of the fault current = 2utility,F

2PV,Frms,F III +=

A.u.p83.137.1392.1 22 =+=

The base current = kA44.06503

500V3

SI

b

bbase =

×==

Fault current = kA08.644.083.13 =× 3.5 Consider the microgrid of Figure 3.68. A three-phase 500 kVA, 440 V Y grounded/ 3.2 kV Δ transformer with the per-unit reactance of 3.5% feeds from an AC source of a PV generating station. The distribution line is 10 miles long and has a series impedance of 0.01 + j 0.09 Ω per mile. The local load is 250 kVA. Balance of power can be injected into the local utility using a 500 kVA, 3.2kV Y grounded/ 13.2 kV Δ transformer with a per-unit reactance of 8%. Assume the voltage base of 13.8 kV on the local power grid side, kVA base of 500, and the DC bus voltage of 800 V. Compute the following:

i) The inverter and the PV generating station ratings ii) The per unit impedance diagram of the microgrid

Figure 3.68 A One-Line Diagram of Problem 3.5.

Solution


LL M2

V23V = (3.81), for three-phase inverter, the

modulation index, can be computed as:

90.0800440

322

322 =×==

DC

LLa V

VM

The PV bus is rated at 500 kVA, 800 V. The inverter is rated at 500 kVA, 800 V DC, and 440 V AC.

ii) kVA500Sb = , kV8.13Vb = on the utility side

Voltage base of the low-voltage side of T2 kV3.38.132.132.3V

VV

b,HVHV

LV =×==

Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p073.0j

8.132.13

50050008.0j

VV

SS

ZZ22

new

old

old

newold2T,pu2T,pu

Voltage base of the low-voltage side of T1 V45333003200440V

VV

b,HVHV

LV =×==

Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p03.0j

3.32.3

500500035.0j

VV

SS

ZZ22

new

old

old

newold1T,pu1T,pu

The base voltage of the AC side of the inverter = base voltage of the LV side of T1 = VLL,b = 453 V

Using equation aDC

LL M2

V23V = (3.81), for three-phase inverter, the modulation index,

can be computed as:

90.0800440

322

322 =×==

DC

LLa V

VM

The PV bus is rated at 500 kVA, 800 V. The inverter is rated at 500 kVA, 800 V DC, and 440 V AC.

The per unit voltage = Vu.p96.08.132.13Vpu ==

The base voltage of the PV side assuming modulation index is fixed at 0.9 =

a

b,LLb,DC M

V322V = DCV822

9.0453

322 =×=

The per unit impedance of the transmission line ( ) 2

3

2b

b

33001050009.0j01.0

VS

Z ××+=×=

Ω+= u.p004.0j00046.0

Assuming unity power factor, the per-unit load is given as

The per unit load = kVAu.p5.0500250

SS

b

load ==


VPIDC

===

For three phase power, I.V.3VA = where VA is the volt-ampere V is the line to line voltage and Iis the line current.


500I ==

Solution Figure 3.5.1. One-line diagram of problem 3.5

Solution Figure 3.5.2 The Per Unit Model of problem 3.5

3.6 Consider the microgrid of Figure 3.68 and the data of Problem 3.5. The local power grid net input reactance can be approximated as 9% based on the voltage base of 13.2 kV and 500 kVA base. Assume DCbus voltage of 880 V. Compute the following:

i) The inverter modulation index and inverter rating ii) The short-circuit current of the AC bus if a three-phase short circuit occurs on

the AC bus of the inverter;ignore the PV station fault current contribution. iii) The same as part (ii), except assume that the internal resistance of the PV

generating station can be approximated as 1 per-unit based on the inverter rating.

Solutions

i) 81.0880440

322

VV

322M

DC

LLa =×==

The inverter is rated at 440 V AC, 880 V DC and 500 kVA

The current rating of the inverter on the DC side = kA56.0880500I ==


500I ==

ii) With Vb = 13.8 kV and Sb = 500 kVA,

Voltage base of the low-voltage side of T2 kV3.38.132.132.3V

VV

b,HVHV

LV =×==

Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p073.0j

8.132.13

50050008.0j

VV

SS

ZZ22

new

old

old

newold2T,pu2T,pu

Voltage base of the low-voltage side of T1 V45333003200440V

VV

b,HVHV

LV =×==

Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p03.0j

3.32.3

500500035.0j

VV

SS

ZZ22

new

old

old

newold1T,pu1T,pu

Base voltage of the AC side of the inverter = base voltage of LV side of T1 = VLL,b = 453 V

The p.u voltage = Vu.p96.08.132.13V u.p ==

The base voltage of the PV side assuming modulation index is fixed at 0.81 =

a

b,LLb,DC M

V322V = DCV917

81.0453

322 =×=

Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p082.0j

8.132.13

50050009.0j

VV

SS

ZZ22

new

old

old

newold1T,puutility,pu

Ω= u.p09.0jZ Utility,pu

Base impedance for the transmission line =( )

Ω=××

== u.p78.2110500

103.3SV

Z 3

23

b

2b

b

The p.u impedance of the transmission line = ( )

Ω+=+

== u.p041.0j005.078.21

09.0j01.010Z

ZZ

b

trantran,u.p

Solution Figure 3.6.1 One-line diagram of problem 3.6


( )Au.p73.8824.4

082.0j073.0j041.0j005.003.0j96.0

ZZZZV

Iutility,pu2T,puTran,pu1T,pu

puutility,F

°−∠=

++++=

+++=


500V3

SI

b

bbase =

×==

Fault current = kA67.263.024.4 =× iii) To calculate the fault current if a balanced three-phase fault occurs on the

inverter AC bus, we can use superposition concept one source acting at a time. We have two sources of fault currents when AC side is subjected to three-phase faults.

Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p92.0j

917880

5005001

VV

SS

ZZ22

new

old

old

newold1T,puPV,pu

Au.p04.192.096.0

ZV

IPV,pu

puPV,F ===

We assume that the inverter acts as a short circuit after the switches get destroyed by the high current and the current contributed by the PV source is limited by its internal resistance. The total fault current is summation of the current contribution from the PV (DC) and the utility (AC).

Solution Figure 3.6.3 Equivalent circuit before fault

Solution Figure 3.6.4 Fault current contribution from PV


Solution Figure 3.6.6 Fault current as superposition of both PV and utility

IF,PV is DC and IF,utility is AC and therefore the fault current is the two currents superposed.

tcos24.404.1III utility,FPV,FF ω+=+= The rms value of the fault current =

A.u.p36.424.404.1III 222utility,F

2PV,Frms,F =+=+=


500V3

SI

b

bbase =

×==

Fault current = kA79.264.036.4 =× 3.7 Consider a three-phase DC/AC inverter with a DC bus voltage rate at 600 V. Assume the DC bus voltage can drop during a discharge cycle to 350 V DC. Determine the AC bus voltage range for a modulation index of 0.9.

Solution

aDC

LL M2

V23V =

V3319.02

60023M

2V

23V a

DCLL =××==

V1939.02

35023M

2V

23V a

DCLL =××==

V331VV193 LL ≤≤

3.8 Consider a rectifier mode of operation for the following system data of a three-phase rectifier. Assume the data given in Table 3.8.

TABLE 3.8 Data for Problem 3.8.

VAN f1 P X p.f. Ma

120 V RMS 60 Hz 1500 W

15 Ω 1 0.85

Perform the following:

i) Find the angle δ and the DC link voltage. Ignore the resulting harmonics.

ii) Write a MATLAB program to simulate the rectifier operation to plot the current Iaand the voltage VAN considering VPWMa and δ are known and have the same values as the results of part i). Assume the PWM voltage is calculated from the incoming AC power. Ignore the resulting harmonics.

Solutions

i) The three-phase power is given by

θ= cos.I.V.3P aAN

Therefore, the phase current,

A017.411203

1500cos.V.3

PIAN

a ∠=××

=θ

=

δaI

X.I a

ANV

The PWM voltage of the rectifier input is given by

jXIVV aANPWMa −=

V°−∠=∠×∠−= 5.273.1359015017.4120

Hence, the angle °= 5.27δ

The line to neutral voltage of the input of the rectifier is given by

2V

2M

V dcarms,NL =−

Hence, the rectified voltage is

V22.45085.0

3.13522MV.22

Va

PWMaDC =

×==

ii) The MATLAB program is given below:

clc; clear all; Van=120; % the values assigned

Vpwma=135.3; f=60; delta=27.53; X=15i; t=0:0.0001:0.2; Vpwma=Vpwma*(cosd(-delta)+sind(-delta)*i); % expresses in cartesian coordinate Ia=(Van-Vpwma)/X; % Ia determined Ia_r=real(Ia); Ia_im=imag(Ia); [theta, Ia]=cart2pol(Ia_r,Ia_im); % Ia expressed in polar form Van=Van*sqrt(2)*sin(2*pi*f*t); % Van expressed as afunction of time Ia=Ia*sqrt(2)*sin(2*pi*f*t-theta); % Ia expressed as afunction of time plot(t,Van,t,10*Ia) grid on; ylabel('Input voltage (V) and Current (A)'); xlabel('time(sec)'); title('Input Current (magnified 10 times) and Voltage');

Solution Figure 3.8.1 Voltage and Current

3.9 A three-phase rectifier has a supply-phase voltage of 120 V.The reactance between the AC source and the rectifier is 11 Ω. Write a MATLAB program to plot the modulation index necessary to keep the VDC constant at 1200 V when the input power factor angle is varied from -60o to 60o. Assume that the active power supplied by the system remains fixed at 5.5 kW. Solutions Steps to write the program:

1. The values of AC supply voltage (VAN), reactance (X), active power (P), and DC link voltage, (VDC) are specified.

2. The range of the power factor angle is set. 3. The value of Ia is computed for the specific power factor angle and active power:

θcos..3 ANa V

PI =

4. VPWMa is computed: jXIVV aANPWMa −= 5. From the magnitude of VPWMa, the modulation index is computed:

DC

PWMaa V

Vm

.2.2=

6. The plot of the modulation index versus the power factor angle is determined. The MATLAB code: % MODULATION INDEX vs POWER FACTOR ANGLE FOR FIXED Vdc and P clc; clear all; P=5.5e3; Van=120; X=11i; Vdc=1500; theta=-60:1:60; % power factor angle is varied from -60 to 60 degree Ia=P/3/Van./cosd(theta); Vpwma=Van-X*Ia.*(cosd(theta)+i*sind(theta)); % Vpwma is computed for the power factor Vpwma=abs(Vpwma); % The magnitude of Vpwma is found out ma=2*sqrt(2)*Vpwma/Vdc; % modulation index is determined to keep Vdc fixed plot(theta,ma) % modulation index vs power factor angle axis([-60 60 0 1]); grid on; ylabel('m_a');

xlabel('power factor angle (degree)'); title('m_a versus power factor angle');

Result:

Solution Figure 3.9.1 The Plot of Ma versus the Power Factor Angle to Keep VDC and the Active P Fixed.

3.10 Develop a MATLAB testbed and compute the triangular waveVT using the Fourier series. Assume the following specifications: peak 48 V, frequency 1 kHz, and order of harmonics ≤ 30. Plot the waveform.

Hint:VT(t) can be constructed using the Fourier series of a triangular wave. An advantage to this method is that you can create the waveform over a long period. The equation and solution of a Fourier series can be found on many sites online. It should be quickly noticed that the triangular waveform has an odd symmetry, thus: a0 = 0 and ak = 0. The Fourier equation becomes

).sin(.).cos(.)(1

0 tkbtkaatF kk

k ωω ++= ∑∞

=

).sin(.)(1

tkbtF kk

ω∑∞

=

=

bkis all that needs to be solved for. The triangular function can then be split into sections, somewhat like the first method, to findbk. The equation to form a triangular wave has been discussed in section 3.5.1.

⎟⎠⎞

⎜⎝⎛−−

=2

.sin.

)1.(822

ππ

kk

bk

k

Now, we will substitute bkand we will obtain the equations for the triangular wave as shown by equation:

-60 -40 -20 0 20 40 600

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

ma

power factor angle (degree)

ma versus power factor angle

( )∑∞

=⎟⎠⎞

⎜⎝⎛−−

=5,3,1

22 .sin.2

.sin.

)1.(8.)(k

k

tkkk

AtF ωππ

whereA is amplitude. This equation can then be put into a MATLABsimulation:We can use this equation to describe the operation of a converter in a MATLAB simulation. For example, let us define the frequency, sampling time, and amplitude as fs = 5000; Ts = 1 / fs; VTMax = 48 To see the full waveform, the step size should be at least half of Ts and a power of 10 smaller to create enough points. t = 0 :Ts / 20 : 10 The above MATLAB codes will create a waveform 10 seconds long. We can use a loop

to iterate the equation constantly adding it to itself as needed by a series ∑∞

= 5,3,1k

.

Solution Following the hint,Matlab program: clc; clear all; %defining the parameters Vt_max = 48; fs = 1000; dt = 1/fs/100; t = 0:dt:2/fs; fori = 1:length(t) Vt(i) = 0; for j = 1 : 2 : 30 %VT is defined using fourier series Vt(i) = Vt(i)+(-8*(-1)^j)/(j^2*pi^2)*sin (pi*j/2)*sin (j*2*pi*fs*t(i)); end end Vt = Vt*Vt_max; %Plotting the result plot(t,Vt); grid on; ylabel('V_T'); xlabel('time (mili-sec)')

Solution Figure 3.10.1 The Waveforms of Problem 3.10 and 3.11

3.11. Develop a MATLAB testbed and compute the triangular wave for PWM using the identity mapping method. Assume a peak value of 48 and a frequency of 1 kHz. Plot the waveform. Solution It is the same program as was created forProblem 3.10. The only difference isthat the triangular waveisdefined using identity mapping:

Vt(i) = -2/pi*(asin(sin(2*pi*fs*t(i)))); The waveform is the same as that for Problem 3.10.

3.12Develop a MATLAB testbed for the DC/AC inverter given below.

Figure 3.69 The Linear and Overmodulated Operation of a Three-Phase Converter.

Assume Vidc is equal to 560 V (DC), VC (max) is equal to 220 V, fe is equal to 60 Hz a modulation index of 0.5 ≤ Ma ≤ 1.0.Change Mf from 2 to 20. Perform the following:

i) The ratio of the fundamental–frequency of the line–line (RMS) to the input DC voltage, Vidc, ( V(lin –line (RMS)) / Vidc.)

ii) Plot the V(lin –line (RMS)) / Vidcas a function of Ma for Mf= 2 and Mf= 20. Solutions

i) The ratio of the fundamental output voltage and the input DC voltage is

aidc

o MVV

223

=

ii) From the relationship in part i), the following plot is derived.

Solution Figure 3.12 Plot of the Ratio of the Output Voltage to the Input Voltage

versustheAmplitude Modulation Index With the changing frequency modulation index, Mf, the harmonic content in the voltage

of the inverter will change.

The Solution Table 3.12.1 shows the harmonics for different modulation indices.

SOLUTION TABLE 3.12.1 The Harmonics for the Different Modulation Indices.

Ma = 0.8, Mf = 5 Ma = 0.8, Mf = 15

Harmonic % of fundamental Harmonic % of fundamental

3rd 27.08 3rd 0

5th 6.51 5th 0.01

7th 28.8 7th 0.01

It is observed that as the frequency modulation index is increased, the harmonics in the

voltage gets reduced to insignificant value.

3.13 Consider a PV source of 60 V. A single-phase inverter with four switches is used to convert DC to 50 Hz AC using a unipolar scheme. Select the following modulation indices:

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Ma

Vo /

Vid

c

a. Ma = 0.5 and Mf = 7 b. Ma = 0.5 and Mf = 10 c. Ma = 0.9 and Mf = 4

Write a MATLAB program to generate the waveforms of the inverter showing Van, Vbn, Vcn voltages. Make tables and discuss your results.

Solution Matlab program for part a. is given below, similar programs can be written for part b. and c.

%% Sine PWM Half Bridge Inverter clc; clear all; %defining the parameters Vidc = 60; fe = 50; Ma = 0.5; Mf = 7; fs = fe*Mf; dt = 1/fs/100; t = 0:dt:2/fe; %Deciding the output volatge fori = 1:length(t) Vc(i) = Ma*sin(2*pi*fe*t(i)); Vt(i) = 0; for j = 1 : 2 : 30 %VT is defined using fourier series Vt(i) = Vt(i)+(-8*(-1)^j)/(j^2*pi^2)*sin (pi*j/2)*sin (j*2*pi*fs*t(i)); end ifVc(i) >Vt(i) Vo(i) = Vidc; else Vo(i) = -Vidc; end end %Plotting the result subplot(2,1,1) plot(t,Vc); holdon; plot(t,Vt); gridon; ylabel('V_c and V_T'); subplot(2,1,2) plot(t,Vo) gridon; xlabel('time (sec)') ylabel('output voltage (V)') axis([0 2/60 -(Vidc+10) (Vidc+10)]);

a.

Solution Figure 3.13.1 Part a. Ma = 0.5 and Mf = 7.

b.

Solution Figure 3.13.2 Part b. Ma = 0.5 and Mf = 10.

c.

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-1

-0.5

0

0.5

1

Vc a

nd V

T

0 0.005 0.01 0.015 0.02 0.025 0.03

-60

-40

-20

0

20

40

60

time (sec)

outp

ut v

olta

ge (V

)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-1

-0.5

0

0.5

1

Vc a

nd V

T

0 0.005 0.01 0.015 0.02 0.025 0.03

-60

-40

-20

0

20

40

60

time (sec)

outp

ut v

olta

ge (V

)

solution Fig. 3.13.3 Part c. Ma = 0.9 and Mf = 4

Solution table 3.13.1 The table below shows the harmonics for different modulation

indices.

Ma = 0.5, Mf = 7 Ma = 0.5, Mf = 10 Ma = 0.9, Mf = 4

Harmonic % of

fundamental

Harmonic % of

fundamental

Harmonic % of

fundamental

3rd 0.23 3rd 0.01 3rd 2.38

5th 18.74 5th 0.03 5th 0.99

7th 10.63 7th 0.04 7th 28.62

Therefore we see that as the frequency modulation index is increased, the harmonic voltages are reduced. However, the modulation index should be limited by the properties of the switches.

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-1

-0.5

0

0.5

1

Vc a

nd V

T

0 0.005 0.01 0.015 0.02 0.025 0.03

-60

-40

-20

0

20

40

60

time (sec)

outp

ut v

olta

ge (V

)

1

Chapter 4 Smart Power Grid Systems

4.1 A three-phase generator rated 440 V, 20 kVA is connected through a cable with impedance of 4 + j15 Ω to two loads: a. A three-phase, Y-connected motor load rated 440 V, 8 kVA, p.f. of 0.9 lagging. b. A three-phase, Δ-connected motor load rated 440 V, 6 kVA, p.f. of 0.85 lagging If the motor load voltage is to be 440 V, find the required generator voltage. Solution

E

ZVph

I

Load 1

E

Load 2

I1 I2

Solution Figure 4.1.1 The Single Phase Equivalent Circuit

Cable impedance = °∠=+= 07.7552.1515j4Z

( ) ( ) A84.255.109.0cos4403

108.f.pcosV3

SI o13

1

LL1 −∠=−∠

××

=−∠×

= −−

−

V003.2543

4403

VV LL

ph ∠=== − , taking the infinite bus as the reference

( ) ( ) A78.3187.785.0cos4403

106.f.pcosV3

SI o13

1

LL2 ∠=∠

××

=∠×

= −−

−

A53.114.1678.3187.784.255.10III ooo21 −∠=∠+−∠=+=

V4.3640407.7552.1553.114.16003.254IZVE ooph ∠=°∠×−∠+∠=+=

The line voltage of the generator = V7003404 = 4.2 A generator is rated at 100 MVA, 20 kV, 60 Hz, 0.8 p.f. lagging and reactance of 10%. Compute the following: i) The generator per unit model if it is loaded at 50% ii) The generator per unit model if it is loaded at 100%

2

iii) The number of poles in the generators if the shaft power is supplied at 1200 rpm Solutions

i) Let the base voltage Vb = 20 kV and base power Sb = 100 MVA Per unit reactance = 0.1 p.u Ω For 50% load, the per unit load = MVAu.p86.365.08.0cos5.0 1 °−∠=−∠ −

ii) For 100% load, the per unit load = MVAu.p86.3618.0cos1 1 °−∠=−∠ −

Generator

j0.1ΩVt

IL

P + jQ (load)

E

Solution Figure 4.2.1 Per unit model for part i) and ii)

iii) Number of poles = 61200

60120Nf120P =

×==

4.3 Develop a table showing the speed of magnetic field rotation in AC machines with two, four, and six poles operating at frequencies 50, 60, and 400 Hz. Solution

rpmPf120N = . The table shows the speed of the magnetic field.

SOLUTION TABLE 4.3.1 The Speed of the Magnetic Field Rotation with Two, Four, and Six Poles Operating at Frequencies 50, 60, and 400 Hz. 2 Poles 4 Poles 6 Poles 50 Hz 3000 rpm 1500 rpm 1000 rpm 60 Hz 3600 rpm 1800 rpm 1200 rpm 400 Hz 24000 rpm 12000 rpm 8000 rpm 4.4 A 20 MVA machine-rated 20 kV, three-phase, 60 Hz generator is supplying power to the local power grid at rated machine voltage. The machine is delivering the rate power to the local power grid. The machine’s synchronous reactance is equal to 8 Ω with negligible resistance. Compute the following: i) The machine excitation voltage when the machine is operating at 0.85 lagging power factor ii) The machine excitation voltage when the machine is operating at 0.85 leading power factor iii) The machine excitation voltage when the machine is operating at unity power factor

3

iv) The maximum power the machine can deliver for i, ii, and iii. Solutions

EA

jXsV∞

IL

Pmech Pɸ+jQɸ (load)

EA

Solution figure 4.4.1 The One-Phase Equivalent Circuit Diagram

i) Assuming a full load, the current = A7.3157785.0cos203

1020I 13

L °−∠=−∠××

= −

Excitation voltage = kV69.1551.148j7.3157731020jX.IVE

3

SLA °∠=×°−∠+×

=+= ∞

δθ

LI

LS I.XAE

The Figure Phasor Diagram for a Lagging Load

ii) Assuming a full load, the current = A7.3157785.0cos203

1020I 13

L °∠=∠××

= −

Excitation voltage = kV29.2393.98j7.3157731020jX.IVE

3

SLA °∠=×°∠+×

=+= ∞

δθ

LI LS I.XAE

V∞ The Figure Phasor Diagram for a Leading Load

iii) Assuming a full load, the current = A0577203

1020I3

L °∠=××

=

4

Excitation voltage = kV7.2143.128j057731020jX.IVE

3

SLA °∠=×°∠+×

=+= ∞

δLI

LS I.XAE

The Figure Phasor Diagram for a Unity Power Factor Load

iv) Active power for part i) = kW1785.020P =×=

Reactive power for part i) = kVar5.107.31sin20Q =×= The torque angle for part i) is °=δ 69.15 Active power for part ii) = kW1785.020P =×= Reactive power for part ii) = kVar5.107.31sin20Q −=×−= The torque angle for part ii) is °=δ 29.23 Active power for part iii) = kW20120P =×= Reactive power for part iii) = kVar00sin20Q =×= The torque angle for part iii) is °=δ 7.21

4.5 For Problem 4.4, assume the load is equal to 5000 W at unity p.f. Compute the p.u equivalent circuit. Assume Sb = 100 MV A, Vb = 345 kV. Solution

Per unit load = Wu.p00005.010100

5000S 6load =×

=

Base impedance = Ω=×

= u.p0012.010100

345Z 6

2

base

Per unit synchronous reactance = Ω== u.p667,60012.08Z u.p

5

Solution figure 4.5.1 Per Unit Model for Problem 5.5

4.6 A two-pole, Y-connected generator rated at 13.8 kV, 20 MVA, 0.8 p.f. leading is running at 1800 rpm. The generator has a synchronous reactance of 8 Ω per phase (at 60 Hz) and a negligible armature resistance per phase. The generator is operated in parallel with an interconnected power network. Compute the following: i) What is the torque angle of the generator at rated conditions? ii) What is the maximum power possible out of this generator? Solutions

i) Assuming full-load, the current = A8.368378.0cos8.133

1020I 13

L °∠=∠××

= −

Excitation voltage = kV57.5367.68j8.368373108.13jX.IVE

3

SLA °∠=×°∠+×

=+= ∞

Therefore, the torque angle = °=δ 57.53

ii) Active power = kW168.020P =×= Reactive power = kVar98.118.36sin20Q −=×−=

4.7 A three-phase, eight pole, 220 MVA, 13.2 kV, 0.9 leading power factor, Y-connected synchronous generator is running at 1200 rpm. Its synchronous reactance is 0.8 Ω per phase at 60 Hz. The generator is fully loaded and supplies power to a network at rated voltage. Compute the generator voltage regulation. Solution

Assuming a full load, the current = A84.2596239.0cos2.133

10220I 13

°∠=∠××

= −

6

Excitation voltage = kV36.5814.88.0j84.2596233102.13jX.IVE

3

f °∠=×°∠+×

=+=

Voltage regulation = %81.61002.13

2.13314.8=×

−

4.8 The one line diagram of power grid is depicted by Fig. 4.47. The data for the system are as follows: Transmission line between bus 2 and bus 3 is j10 Ω and between bus 3 and bus 5 is j6 Ω G1 wind generating system 25 MVA, 13.2 kV, and reactance of 0.20 per unit G2 gas turbine generating system 50 MVA, 20 kV, and reactance of 0.20 per unit Transformer T1 100 MVA, 220 Y/13.8 Δ kV, and reactance of 10% Transformer T2 200 MVA, 220/20 kV, and reactance of 10% Transformer T3 100 MVA, 220 Y/22Y kV, and reactance of 10%

Figure 4.47 The System for Problem 4.8.

Develop the following:

i) Per unit impedance model for short-circuit studies

ii) Per unit impedance model for power flow studies Solution

MVA100Sbase = , kV220Vbase = on the line between bus 2 and 3

Base impedence of transmission line between bus 2 and bus 3 = Ω=×

= 48410100

000,220SV

6

2

base

2base

The p.u reactance in the line between bus 2 and 3 = Ω===− .u.p021.048410

ZXXbase

32,u.p

7

The p.u reactance in the line between bus 3 and 5 = Ω===− u.p012.04846

ZXXbase

53,u.p

Per unit reactance remains same because the base values are same as rated for transformer T1 = Xp.u,T1 = 0.1

New per unit reactance of T2 = 05.02001001.0

SS

XXold

newold2T,u.p ===

Per unit reactance is same as the base values are same as rated T3 = Xp.u,T3 = j0.1

Base voltage of bus 1 = kV8.13220220

8.13VVV

b,HVHV

LV =×=

Base voltage of bus 4 = kV2222022022V

VV

b,HVHV

LV =×=

Base voltage of bus 6 = kV2022022020V

VV

b,HVHV

LV =×=

Per unit generation of G1 = kVAu.p25.010025

SS

b

==

New per unit reactance of G1 = Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p73.0

8.132.13

251002.0

VV

SS

XX22

new

old

old

newold1G,u.p

Per unit generation of G2 = kVAu.p5.0100

5SS

b

==

New per unit reactance of G2 = Ω=⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛= u.p4.0j

2020

501002.0j

VV

SS

XX22

new

old

old

newold2G,u.p

The load at bus 4 is assumed to be 1 0° per unit kVA

The load impedance = ( )

Ω∠=∠

== u.p0101

1S

VZ *

2

*u.p

2u.p

load,u.p

i) The short circuit model:

8

Solution figure 4.8.1 The Model for short circuit studies.

ii) The model for power flow studies In this model, the buses 1, 2, 3, 4, 5, and 6 are considered. The buses 1′ and 2′ are

neglected.

Solution figure 4.8.2 Fig. Model for power flow studies.

4.9 The one-line diagram of a power grid is depicted by Fig. 4.48. The data for the grid are given in the figure.

9

CB CB

CB

CB

CBCBCB

CB

CB

CBCB

CBCBCB

1.2 MW0.62 MVar

11

7

To Feeder

1

HV BusInfiniteBus

LocalUtility

Utility EMS

NetMetering

8

9

10

12

0.05

+j0

.5

0.04 +j0.51

0.04 +j0.4

0.01

+j0

.12

0.04 +j0.5

A Zone with DGs(Micro-Grid System)

1.18 MW0.62 Mvar

DGEMS

0.03 +j0.32CBCB

CB

CB

CB

CB

0.04 +j0.45CBCB

1716LocalLoads

18

0.9 MW0.48 Mvar

15

1.06 MW0.56 Mvar

Gas turbineSync. Gen.

1.8 MVA

14

13

2

3

CB

CB CB

CB CB

CB

CB

LocalLoads

LocalLoads

Variable SpeedWind Turbine with

DFIG (2 MW)

CBCB

T1

13.2 / 3.3 kV20 MVA

Xt = 12%4.5 MW

2.2 MVar

0.99 MW 0.54 MVar

0.96 MW 0.51 MVar

0.9 MW 0.48 MVar

CBCB

G

G

DC/AC

PVStation

DC/AC

PVStation

DC/AC

PVStation

1 MW1 MW1 MW

54 6

T2

T3

T4 T5 T6

T7

10


Assume an MVA base of 100 and base voltage of 13.2 kV; also assume the input reactance of the local power grid is 10% based on the transformer T1. The input reactance of all the sources is 7% with a base same as their rating.

The system data are

Transformer T1: 20 MVA, 33 /13.2 kV, 10% reactance

Transformer T2: 20 MVA, 13.2 / 3.3 kV, 12% reactance

Transformer T3: 5 MVA, 3.3 / 460 V, 6.5% reactance

Transformer T4, T5, and T6: 2 MVA, 3.3 / 460 V, 6.5% reactance

Transformer T7: 5 MVA, 3.3 / 460 V, 6% reactance

Transmission line impedance is given in Table 4.7

TABLE 4.7 Transmission Line Data for Problem 4.9.

Line Resistance (Ω) Series Reactance (Ω)

8–9 .05 0.5

9–10 0.04 0.4

9–11 0.04 0.51

11–12 0.04 0.5

11–13 0.01 0.12

13–14 0.03 0.32

14–15 0.04 0.45

The local loads are 1 MVA at 0.85 power factor lagging.

Develop the following:

i) Per unit model for power flow studies

ii) Per unit model for short-circuit studies

11

Solutions

The new p.u impedances of the transformers:

For T1, Ω=×= u.p5.020

1001.0X u.p

For T2, Ω=×= u.p6.020

10012.0X u.p

For T3, Ω=×= u.p3.15

100065.0X u.p

For T4, T5, and T6, Ω=×= u.p25.32

100065.0X u.p , Ω=×= u.p2.15

10006.0X u.p

For T7, Ω=×= u.p2.15

10006.0X u.p

Base impedance for 3.3 kV side = Ω=== u.p109.0100

3.3SV

Z2

b

2b

3.3,b

The p.u. impedance = bZ

ZΩ

SOLUTION TABLE 4.9.1 the p.u line impedance.

Line Resistance (Ω) Series Reactance (Ω)

Per Unit Resistance (p.u Ω)

Per Unit Series Reactance (p.u Ω)

8–9 .05 0.5 0.45 4.5

9–10 0.04 0.4 0.37 3.7

9–11 0.04 0.51 0.37 4.67

11–12 0.04 0.5 0.37 4.5

11–13 0.01 0.12 0.09 1.1

13–14 0.03 0.32 0.28 2.9

14–15 0.04 0.45 0.37 4.1

12

i)

11

71

HV BusInfiniteBus

Utility EMS

8

9

10

12

0.45

+j0

.45

p.uΩ

0.37 + j4.67 p.u Ω

0.37 + j3.7 p.u Ω

0.09

+ j1

.1 p

.uΩ

0.37 + j4.5 p.u Ω

A Zone with DGs(Microgrid System)

DGEMS

0.28 + j2.9 p.u Ω 0.37 + j4.1 p.u Ω

1716LocalLoads

18

15

14

13

2

3

LocalLoads

LocalLoads

T1

-0.0099 p.u MW

-0.0054 p.u MVar

54 6

T2

T3

T4 T5 T6

T7

j0.5 p.u Ω j0.6 p.u Ω

j1.3 p.u Ω

j3.25 p.u Ω j3.25 p.u Ω j3.25 p.u Ω

P = 0.01 p.u MW P = 0.01 p.u MW P = 0.01 p.u MW

P = 0.02 p.u MW

S = 0.018 p.u MVA

j1.2

-0.0096 p.u MW

-0.0051 p.u MVar

-0.009 p.u MW

-0.0048 p.u MVar

-0.012 p.u MW

-0.0062 p.u MVar

-0.045 p.u MW

-0.022 p.u MVar

-0.0118 p.u MW

-0.0062 p.u MVar

-0.009 p.u MW

-0.0048 p.u MVar -0.0106 p.u MW

-0.0056 p.u MVar

P8 + jQ8P1 + jQ1

Vb = 33 kV

Vb = 13.2 kV

Vb = 3.3 kV

Vb = 460 V

Vb = 460 V Vb = 460 V Vb = 460 V

Vb = 460 V

Vb = 3.3 kV

Vb = 3.3 kV

Vb = 3.3 kV

Vb = 3.3 kV

Vb = 3.3 kV

Vb = 3.3 kV Vb = 3.3 kV

Vb = 3.3 kV Vb = 3.3 kV

Solution figure 4.9.1 Per Unit Model for Power Flow Studies.

13

ii) The load impedance can be calculated by ( )jQPV

S

VZ

2u.p

*u.p

2u.p

load,u.p −==

For bus 7 = ( ) Ω+=−

= u.p76.8j93.17022.0j045.0

1Z2

7,u.p

For bus 9 = ( ) Ω+=−

= u.p2.43j2.810051.0j0096.0

1Z2

9,u.p

For bus 11 = ( ) Ω+=−

= u.p1.46j5.860048.0j009.0

1Z2

11,u.p

For bus 12 = ( ) Ω+=−

= u.p4.42j8.770054.0j0099.0

1Z2

12,u.p

For bus 13 = ( ) Ω+=−

= u.p9.34j4.660062.0j0118.0

1Z2

13,u.p

For bus 14 = ( ) Ω+=−

= u.p1.46j5.860062.0j0118.0

1Z2

14,u.p

For bus 15 = ( ) Ω+=−

= u.p96.38j75.730056.0j0106.0

1Z2

15,u.p

14

11

71

HV BusInfiniteBus

Utility EMS

8

9

10

12

0.45

+j0

.45

p.uΩ

0.37 + j4.67 p.u Ω

0.37 + j3.7 p.u Ω

0.09

+ j1

.1 p

.uΩ

0.37 + j4.5 p.u Ω

A Zone with DGs(Microgrid System)

DGEMS

0.28 + j2.9 p.u Ω 0.37 + j4.1 p.u Ω

1716 18

15

14

13

2

3

T1

54 6

T2

T3

T4 T5 T6

T7

j0.5 p.u Ω j0.6 p.u Ω

j1.3 p.u Ω

j3.25 p.u Ω j3.25 p.u Ω j3.25 p.u Ω

7 p.u Ω

j1.2 p.u Ω

G6'

7 p.u Ω

G5'

7 p.u Ω

G4'

j3.5 p.u Ω

G2'

Gj3.85 p.u Ω

Gj0.5 p.u Ω

3'

1'

Ω+ u.p4.42j8.77

Ω+ u.p76.8j93.17

Ω+ u.p2.43j2.81

Ω+ u.p1.46j5.86

Ω+ u.p9.34j4.66

Ω+ u.p1.46j5.86

Ω+ u.p96.38j75.73

Vb = 33 kV

Vb = 13.2 kV Vb = 3.3 kV

Vb = 460 V Vb = 460 V Vb = 460 V

Vb = 460 V

Vb = 460 V

Vb = 3.3 kV

Vb = 3.3 kV

Vb = 3.3 kV

Vb = 3.3 kV

Vb = 3.3 kV Vb = 3.3 kV

Vb = 3.3 kV Vb = 3.3 kV

Solution figure 4.9.2 Per Unit Model for Short-Circuit Studies.

15

4.10 Consider the microgrid given in Fig. 4.49.

Figure 4.49 Photovoltaic (PV) Microgrid of Problem 4.10.

Assume the following data:

a. Transformers connected to the PV-generating station are rated at 460 V Y-grounded/13.2 kV Δ, have 10% reactance, and 10 MVA capacity. The transformer connected to the power grid is 13.2 / 63 kV, 10 MVA, 10% reactance.

b. Assume the load on bus 4 is 1.5 MW at a power factor of 0.9 lagging, on bus 5 is 2.5 MW at a power factor 0.9 lagging, on bus 6 is 1.0 MW at a power factor of 0.95 lagging, on bus 7 is 2 MW at a power factor of 0.95 leading, and on bus 8 is 1.0 MW at a power factor of 0.9 lagging.

c. Transmission line has a resistance of 0.0685 Ω/mile, reactance of 0.40 Ω/mile and half of line charging admittance (Y′/2) of 61011 −× Ω-1/mile. The line 4–7 is 5 miles, 5–6 is 3 miles, 5–7 is 2 miles, 6–7 is 2 miles, and 6–8 is 4 miles long. The transmission line model is given in Fig. 4.50

16

Figure 7.50 Transmission Line Pie Model.

d. Assume PV-generating station 1 is rated at 0.75 MW and PV generating station 2 is rated at 3 MW. Assume PV generating stations are operating at unity power factors.

e. Assume Sbase of 10 MVA and a voltage base of 460 V in PV generators. Assume the local power grid bus voltage has 5% tolerance. Develop the following:

i) Per unit model for power flow studies

ii) Per unit model for short-circuit studies

Solutions

i)

17

460 V

13.2 kV

S=‐0.15‐j0.07

13.2 kV

13.2 kV

13.2 kV

460 V

LocalUtility

1

47

2

5

6

3

8

63 kV

j0.1

j0.1

j0.1

0.17+j0.96

0.13+j0.75

0.06+j0.37

0.06+j0.37

0.1+j0.57

P = 0.075

P = 3

S=‐0.25‐j0.01

S=‐0.1‐j0.08

S=‐0.2+j0.065

S=‐0.1‐j0.096


ii)

Assume internal resistance of PV station are 50% based on 10 MVA and the local power grid has internal short circuit reactance of 10% based on 10 MVA. For short circuit studies. The short circuit model is given as

The load impedance can be calculated by ( )jQPV

S

VZ

2u.p

*u.p

2u.p

load,u.p −==

For bus 4 = ( ) Ω+=−

= u.p55.2j47.507.0j15.0

1Z2

4,u.p

For bus 5 = ( ) Ω+=−

= u.p16.0j401.0j25.0

1Z2

5,u.p

For bus 6 = ( ) Ω+=−

= u.p87.4j09.608.0j1.0

1Z2

6,u.p

18

For bus 7 = ( ) Ω+=−

= u.p47.1j52.4065.0j2.0

1Z2

7,u.p

For bus 8 = ( ) Ω+=−

= u.p5j2.5096.0j1.0

1Z2

8,u.p

460 V

13.2 kV 13.2 kV

13.2 kV

13.2 kV

460 VAll values are

in per unit

1

47

2

5

6

3

8

63 kV

j0.1

j0.1

j0.1

0.17+j0.96

0.13+j0.75

0.06+j0.37

0.06+j0.37

0.1+j0.57

1'2'

2

0.5

0.5

j0.1G

G

G

55.2j47.5 +

16.0j4 +

87.4j09.6 +

47.1j52.4 +

5j2.5 +

Solution figure 4.10.2 Per Unit Model for Short-Circuit Studies.

4.11 Consider the PV power grid given by Fig. 4.51 below. The PV1 has an internal resistance of 0.8 p.u and inject 1 per unit power into the grid; PV2 has an internal resistance of 0.4 p.u and inject 0.5 p.u power into the grid. The load is 1 p.u active and 1 p.u reactive power.

19

Perform the following: i) Per unit model for short-circuit studies ii) Per unit model for power flow studies

j0.2

Figure 4.51 The One-Line Diagram of Problem 4.11.

Solutions

Assume the infinite bus of utility has a short-circuit reactance of 0.1 p.u.

i)

Solution figure 4.11.1 Per Unit Model for Short Circuit Studies.

ii)

20

j0.2


4.12 Compute the load factor of a feeder assuming that the maximum load is 8 MW and the average power is 6 MW.

Solution

The load factor %7510086

=×==powerpeak

poweraverage

4.13 Compute the load factor of a feeder for daily operation for one month assuming the same daily profile. Assume the average power is 170 MW and the peak is 240 MW.

Solution

The load factor %83.70100240170

powerpeakpoweraverage

=×==

4.14 If the feeder of Example 4.13 is supplied from a wind source rated 80 MW and a central power-generating station rated 500 MW, assume the capital cost of wind power is $500 per KW and the central station $100 per KW. Compute the EUC if the maintenance cost for the wind source is free, except maintenance is 1 cent per kWh and the central power-generating station fuel and maintenance cost is 3.2 cents per kWh. Give a figure for EUC from zero load factors to unity over 5 year’s utilization. Solution

21

Steps for the program:

1. The capacities and the various costs of the wind farm and the central power station are assigned.

2. The energy consumed at full load for 5 years, E5 is calculated:

( ) yearsinhoursofnoplantcentralofratingfarmwindofratingE 5.5 ×+=

3. The amortized fixed cost is determined by 5

cos

E

tcapitalcapacityplantFC

central

windiii

amor

∑−

×=

4. The variable cost, VC, is determined from the load supplied. If the load is less than or equal to the capacity of the wind farm, the variable cost is the same as that of the wind farm. Otherwise, it is equal to that of the central plant.

5. Define energy unit cost (EUC) as

6. LF

CostfixedamortizedVCEUC += is calculated and plotted.

Assuming load factor of 10%,

The energy in one year at full capacity = ( ) kWh10800,080,53652450080 3×=××+

Amortized fixed cost = yearoneinenergy

capital.capacitycapital.capacity centralcentralwindwind +

kWh/$017.010800,080,5

1050010010805003

33

=×

××+××

LFCostfixedamortizedVCEUC +=

kWh/$18.01.0

017.001.0 =+=

The energy in two year at full capacity = ( ) kWh10600,161,1036524500802 3×=××+×


capital.capacitycapital.capacity centralcentralwindwind +

22

kWh/$009.010600,161,10

1050010010805003

33

=×

××+××


kWh/$095.01.0

009.001.0 =+=

Solution Table 4.14.1 the EUC for different years

Year EUC ($/kWh)

1st 0.18

2nd 0.095

3rd 0.067

4th 0.052

5th 0.044

The MATLAB program:

clc; clear all;

W_cap=80e3; % capacity of wind farm in kW

C_cap=500e3; % capacity of central plant in kW

W_capital=500; % capital cost of wind farm in $/kW

C_capital=100; % capital cost of central plant in $/kW

W_VC=0.01; % variable cost for wind energy in $/kWh

C_VC=0.032; % variable cost for central plant in $/kWh

Energy5=(W_cap+C_cap)*24*365*5; % energy used in 5 years in kWh

23

FC_amor=(W_cap*W_capital+C_cap*C_capital)/Energy5;

a=W_cap/(W_cap+C_cap); % the percentage capacity of wind farm

for LF=0.01:0.001:1

if LF<=a

VC=W_VC;

else

VC=C_VC;

end

EUC=VC+FC_amor./LF; % EUC in $/kWh

EUC=EUC*100; % EUC in cents/kWh

plot(LF,EUC,'.') % plotting EUC versus load factor

hold on;

end

grid on; % labeling the axes

xlabel('Load Factor');

ylabel('EUC (in cents/kWh)');

title('EUC vs Load Factor');

The results are plotted in Solution Figure 4.14.1.

24

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

20

25

30

35

40

Load Factor

EU

C (i

n ce

nts/

kWh)

EUC vs Load Factor

Solution figure 4.14.1 The Plot of Energy unit cost, EUC (in cents/kWh) versus Load Factor for Problem 4.14.

Since the variable cost of fuel cell is higher than that of micro-turbine, during low loads, only micro-turbines are used. However, as the demand rises, the fuel cells are also connected. At the point where the load exceeds the capacity of micro-turbine, fuel cell is turned on which gives rise to discontinuity in the curve.

4.15 If the feeder of Problem 4.13 is supplied from 10 fuel cell sources rated for a total of 2 MW and 20 microturbines rated for 6 MW, assume the capital cost of a fuel cell is $1000 per kW and a microturbine is $200 per kW. Compute the fuel source EUC if the variable cost for the fuel cell is 15 cents per kWh and microturbine is 2.2 cents per kWh, assume 5 years of operation. Show EUC as a function of time from zero load factors to unity in a figure.

Solution

Assuming load factor of 50%,

The energy in one year at full capacity = ( ) kWh10400,26,1236524620210 3×=×××+×


capital.capacitycapital.capacity micromicrofuelfuel +

kWh/S035.010400,226,1

200106201000102103

33

=×

×××+×××

25


kWh/$092.05.0

035.0022.0 =+=

The energy in two year at full capacity = ( ) kWh10800,52,24236524620210 3×=××××+×


capital.capacitycapital.capacity micromicrofuelfuel +

kWh/$018.010800,452,2

200106201000102103

33

=×

×××+×××


kWh/$058.05.0

018.0022.0 =+=

Solution Table 4.15.1 the EUC for different years

Year EUC ($/kWh)

1st 0.092

2nd 0.058

3rd 0.045

4th 0.039

5th 0.036

The steps to solve the problem are the same as those of Problem 4.14.

The MATLAB program:

clc; clear all;

F_cap=2e3; % capacity of fuel cell in kW

26

M_cap=6e3; % capacity of micro-turbine in kW

F_capital=1000; % capital cost of fuel cell in $/kW

M_capital=200; % capital cost of micro-turbine in $/kW

F_VC=0.15; % variable cost for fuel cell in $/kWh

M_VC=0.022; % variable cost for micro-turbine in $/kWh

Energy5=(F_cap+M_cap)*24*365*5; % energy used in 5 years in kWh

FC_amor=(F_cap*F_capital+M_cap*M_capital)/Energy5;

a=M_cap/(F_cap+M_cap); % the percentage capacity of micro-turbine

for LF=0.01:0.001:1

if LF<=a

VC=M_VC;

else

VC=F_VC;

end

EUC=VC+FC_amor./LF; % EUC in $/kWh

EUC=EUC*100; % EUC in cents/kWh

plot(LF,EUC,'.') % plotting EUC versus load factor

hold on;

end

grid on; % labeling the axes

xlabel('Load Factor');

ylabel('EUC (in cents/kWh)');

27

title('EUC vs Load Factor');

The results are plotted in the graph.

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Load Factor

EU

C (i

n ce

nts/

kWh)

EUC vs Load Factor

Solution figure 4.15.1 The Plot of energy unit cost EUC (in cents/kWh) versus Load Factor for Problem 4.15.

Since the variable cost of fuel cell is higher than that of micro-turbine, during low loads, only micro-turbines are used. However, as the demand rises, the fuel cells are also connected. At the point where the load exceeds the capacity of micro-turbine, fuel cell is turned on which gives rise to discontinuity in the curve.

CHAPTER 5 MICROGRID SOLAR ENERGY SYSTEMS

5.1 Search the Internet and specify four PV modules. Give a table as shown below (table 5.45)and compare the rated voltage, cost, width, length, and weight. Solutions We find BP 3230T solar panel the data sheet of which is available at http://www.solarelectricsupply.com/images/content/pdf/panels/spec-sheets/BP-3230T.pdf, accessed Oct 8, 2011 And the cost of which found from: http://www.google.com/products/catalog?q=cost+of+BP+3230T&um=1&ie=UTF-8&tbm=shop&cid=17740868210462196047&sa=X&ei=r_uQTr6KLoijsQKO9JGLAQ&ved=0CDkQ8wIwAw, accessed Oct 8, 2011 Voltage temperature coefficient = kv = -0.36 %/°C Current temperature coefficient = ki = 0.065 %/°C Open circuit voltage at 10°C=

( ) ( ) V38.6825107.36100

36.07.36TTV.kV STCocvOC =−××−=−+=

Short circuit current at 10°C=

( ) ( ) A8.3225104.8100065.04.8TTI.kI STCocvOC =−××+=−+=

Open circuit voltage at 20°C ( ) V37.3625207.36

10036.07.36 =−××−=

Short circuit current at 20°C ( ) A8.3725204.8100065.04.8 =−××+=


10036.07.36 =−××−=



36.07.36 =−××−=



10036.07.36 =−××−=


TABLE 5.45.The Voltage and Current Characteristic of a typical PV module Power (max) 230 W

Voltage at maximum power point (MPP) 29.1 V

Current at maximum power point (MPP) 7.90 A

VOC (open circuit voltage) 36.70 V

ISC (short circuit current) 8.40 A

Efficiency 13.8%

Cost $410.71

Width 39.4 inch

Length 65.6 inch

Height 2.0 inch

Weight 42.8 lb

5.2 Search the Internet to find the voltage-current characteristic of four PV modules. Make a table of input impedances as current varies for each operating temperature. Develop a plot of input impedance as a function of PV load current for each operating temperature. Solutions From the PV parameters of PV-MF165EB3 found in http://www.wholesalesolar.com/pdf.folder/module%20pdf%20folder/mf165eb3.pdf, accessed Oct 8, 2011 we the internal resistance can be derived. The table below lists the input impedances:

Output current, A

Thevenin’s resistance, Ω

0 to 0.9624 1.2469

0.9624 to 1.7514

1.5210

1.7514 to 1.9371

2.3709

2.3709 to 2.8361

2.5797

2.8361 to 3.1709

3.5836

3.1709 to 3.4033

5.1635

3.4033 to 3.5600

7.6583

3.5600 to 3.6634

11.6002

3.6634 to 3.8700

157.3446

The internal PV resistancevs current 5.3 Design a microgrid of PV rated at 100 kW of power at 230 V AC using a PV module with the voltage and current characteristics.

TABLE 5.46.Photovoltaic Module Data for Problem 5.3.

Power (max) 320W

Voltage at Maximum Power Point (MPP) 52.6 V

Current at MPP 6.1 A

0 0.5 1 1.5 2 2.5 3 3.5 40

20

40

60

80

100

120

140

160

I (A)

R (o

hm)

Voc (Open-circuit voltage) 63.2 V

Isc (Short-circuit current) 7.0 A

Determine the following: i) Number of modules in a string for each PV type ii) Number of strings in an array for each PV type iii) Number of arrays iv) Inverter specifications v) One-line diagram of this system

Solutions It is assumed that the system is single phase and the data is given in table 5.46. The following table lists the variables used: Table 5.4. The PV design nomenclature Terms Abbreviations Descriptions String Voltage SV String voltage for series connected

modules Power of a module PM Power produced by a module String Power SP Power that can be generated in one

string Number of strings NS Number of strings per array Number of Arrays NA Number of arrays in a design Surface area of a module SM Surface area of a module Total Surface area TS Total surface area Array Power AP Array power is generated by

connecting a number of strings in parallel

Number of modules NM Number of modules per string Total number of modules TNM Total number of modules in all

arrays put together Array voltage for maximum power point tracking

VAMPP The operating voltage for maximum power point tracking of an array

Array current for maximum power point tracking

IAMPP The operating current for maximum power point tracking of an array

Array maximum power point

PAMPP The maximum operating power of an array

Number of converters NC Total number of DC-DC converters

Number of rectifiers NR Total number of AC-DC rectifiers Number of inverters NI Total number of inverters i) The load voltage is specified as 230 V single-phase AC. To acquire maximum power from the PV array, we select a modulation index of 9.0=aM .The inverter input voltage is given by

a

acidc M

V2V =

idcV = V4.3619.02302

=×

The inverter is designed to operate at the maximum power point (MPP) tracking of the PV array. Therefore, the number of modules to be connected in series in a string is given by

NM= MPP

idc

VV

WhereVMPP is the voltage at the MPP of the PV array of the module.

76.524.361NM ≈=

ii) The string voltage (SV) is given as

SV = NM MPPV×

Using this module, string voltage for this design is SV V2.3686.527 =×= The power generated by one string (SP) is given by SP = NM MPPP× WherePMPP is the nominal power generated at the MPP. The power generated by a string for this design is given as kW per string W246,21.66.527 =××= iii) If we design each array to generate a power of 20 kW, then the number of strings, NS, in an array is given by

NS = stringoneofpowerarrayoneofpower

NS 10246.220

≈=

The number of array(NA) for total power generation is

NA= arrayoneofpower

generationPV

Therefore, NA 520

100==

SOLUTION: The PV Generating Station Specifications. Modules per

String Strings per

Array Number of

Arrays String Voltage

(V) 7 10 5 368.2

iv) In the final design, the inverter should be rated such that it is able to process generation of 20 kW for each array and supply the load at 230 V AC from its array at its MPP tracking. Based on the PV module, the string voltage is specified as

V2.368Vidc = and the modulation index is given as follows:

idc

aca V

VM

2=

Ma 88.02.368

2302=

×=

Let us select a switching frequency of 6 kHz. Therefore, the frequency modulation index is given by

10060

6000===

e

Sf f

fM

NI = inverteroneofpower

generationPV

Therefore, NI 520

100==

SOLUTION The Inverter Specifications.

Input Voltage Vidc(V)

Power Rating (kW)

Output Voltage,

VAC

Amplitude Modulation Index, Ma

Frequency Modulation Index, Mf

Number of Inverters

(V) 368.2 20 230 0.88 100 5

v)

AC BusDC Bus

230 V368.2 V

1

2

5 PV arrays:10 string

7 modules per string

5

Amplitude modulation index = 0.88

100 kW

20 kW each

DC/AC3 phase InverterDC/AC3 phase Inverter

DC/AC3 phase Inverter

1

2

5

The One-Line Diagram of Problem 5.3.

5.4Design a microgrid of a PV system rated at 600 kW of power at 460 V AC (three-phase) using a PV module with the data given in Table 5.46

TABLE 5.46. Photovoltaic Module Data for Problem 5.4.

Power (max) 320W

Voltage at maximum power point (MPP) 52.6 V

Current at MPP 6.1 A

Voc (open-circuit voltage) 63.2 V

Isc (short-circuit current) 7.0 A

Determine the following:

i) Number of modules in a string for each PV type

ii) Number of strings in an array for each PV type iii) Number of arrays iv) Inverter specifications v) One-line diagram of this system

Solutions i) The load is 600 kW rated at 460 V AC. Based on the voltage of the load and an amplitude modulation index of 0.9,we will have the following input DC voltage for a three-phase inverter:

a

LLidc M

VV

322

=

V8359.03

46022=

××

=

We will limit the maximum string voltage, SV, to 600 V DC. Therefore, we can use a boost converter to boost the SV to 835 V. If we select an approximateSV of 550 V, we will have the number of modules in a string is given by

116.52

550VV

MPP

string ≈=

whereVMPP is the voltage of a module at MPP tracking. ii) The string power, SP, can be computed as

SP = NM MPPP× SP W529,31.66.5211 =××=

The string voltage, SV, is given as

SV = NM MPPV×

Therefore, the SV for this design is SV V6.5786.5211 =×=

iii) If we design each array to generate a power of 20 kW, then the number of strings, NS, in an array is given by


NS 6578.320

≈=

The number of arrays, NA, for total power generation is

NA= arrayoneofpower

generationPV

Therefore,

NA 3020

600==

kWkW

iv) SOLUTION

The PV Specifications. Modules per

String Strings per

Array Number of


(V) 11 6 30 578.6

The inverters should be rated to withstand the output voltage of the boost converter and should be able to supply the required power. The inverter is rated at 100 kW with input voltage of 835 V DC and anamplitude modulation index of 0.9. The output voltage of the inverter is 460 V AC. The number of inverters, NI, needed to process a generation of 600 kW is given by

NI=inverteroneofpower

generationPV

Therefore,

NI 6100600

==

Hence, we need to connect six inverters in parallel to supply the load of 600 kW. If a switching frequency is set at 5.04 kHz, the frequency modulation index, Mf is given by

8460

5040===

e

Sf f

fM

SOLUTION: The Inverter Specifications.

Number of

Inverters

Input Voltage Vidc (V)

Power Rating (kW)

Output Voltage,

VAC (V)



6 835 100 460 0.90 84

The number of boost converters needed is the same as the number of arrays, which is 30, and the power rating of each boost converter is 20 kW. The boost converter input voltage, Vi, is equal to the SV:

6.578Vi = V The output voltage, Vo, of the boost converter, Vidc, is equal to the inverter input voltage:

VVV oidc 835== The duty ratio, D, of the boost converter is given by

o

i

VV

D −= 1

D 31.0835

6.5781 =−=

SOLUTION The Boost Converter Specifications for a 500 kW System. Number of

Boost Converters

Input Voltage,

Vi(V)

Power Rating (kW)

Output Voltage, Vo

(V)

Duty Ratio, D

30 578.6 20 835 0.31

v) DC Bus

460 V

DC Bus AC Bus

DC/AC

3 phase Inverter

DC/AC

DC/DC

DC/DC

Boost

DC/DC

Boost

Boost

835 V578.6 V

1

2



30

1

6

Duty ratio = 0.31 Amplitude

modulation index = 0.9

600 kW

3 phase Inverter

20 kW each 100 kW

each


5.5 Search the Internet for four single-phase inverters and summarize the operating conditions in a table and discuss the results.

Solutions The table below gives the specification of single phase inverters http://www.solaredge.com/files/pdfs/products/inverters/se-single-phase-us-inverter-datasheet.pdf, Accessed Oct 8, 2011 Inverter SE3300US SE3800US SE6000US SE7000US Maximum AC power output

3300 W 3800 W 6000 W 7000 W

AC output voltage

183-294 183-294 183-294 183-294

Maximum continuous output current

16 A 18.5 A 25 A 25 A

Maximum input DC voltage

500 V 500 V 500 V 500 V

Maximum input DC current

10.5 A 12 A 17.5 A 18 A

Efficiency 97.2% 97.3% 97% 97% Weight 52 lb 52 lb 52 lb 52 lb Length 27.5 inch 27.5 inch 27.5 inch 27.5 inch Width 12.5 inch 12.5 inch 12.5 inch 12.5 inch Height 7.5 inch 7.5 inch 7.5 inch 7.5 inch

5.6 Search the Internet for DC/DC boost converters and DC/AC inverters and create a table and summarizing the operating conditions of four DC/DC boost converters and DC/AC inverters in a table and discuss the results and operations.

Solutions DC/DC boost converter, PC12480.8, specification is obtained from http://www.l-com.com/item.aspx?id=20899, accessed Oct 8, 2011 SOLUTION TABLE 5.6.1 DC/DC boost converter ratings Input voltage range 12 V to 26 V DC Output current 0.8 Amps Max. Output voltage 48 VDC Weight 0.9 lbs

Dimensions 4.1×2.6×1.4 inches The inverter specifications are given in the solution of problem 5.5

5.7 Design a microgrid of 50 kW, rated at 230 V AC. Use the PV module of Problem 5.3 and the converters of Problem 5.5. The design should use the least number of converters and inverters. Determine the following:

i) Number of modules in a string for each PV type ii) Number of strings in an array for each PV type iii) Number of arrays iv) Converter and inverter specifications v) One-line diagram of this system

Solutions i) The load voltage is specified as 230 V single-phase AC. To acquire maximum power from the PV array, we select a modulation index of 9.0=aM .The inverter input voltage is given by

a

acidc M

VV 2=

idcV = V4.3619.02302

=×

The inverter is designed to operate at the MPP tracking of PV array. Therefore, the number of modules, NM, to be connected in series in a string is given by

NM = MPP

idc

VV

whereVMPPis the voltage at MPP of PV of the module.

76.524.361NM ≈=

The string voltage is given as

SV = NM MPPV×

Using this module, the SV for this design is SV V2.3686.527 =×= The power generated by one string is given by SP = NM MPPP× WherePMPP is the nominal power generated at the MPP.

The power generated by a string for this design is given as kW per string W246,21.66.527 =××= ii) If we design each array to generate a power of 25 kW, then the number of strings, NS, in an array is given by:


NS 12246.225

==

iii) The number of arrays, NA, for total power generation is

NA = arrayoneofpower

generationPV

Therefore,

NA 22550

==

iv) SOLUTION: The PV Specifications. Modules per

String Strings per

Array Number of


(V) 7 12 2 368.2

In the final design, the inverter should be rated such that it is able to process the generation of 25 kW for each array and supply the load at 230 V AC from its array at its MPP tracking. Based on the PV module, the string voltage is specified as

V2.368Vidc = and the amplitude modulation index, Ma, is given as follows:

idc

aca V

VM

2=

Ma 88.02.368

2302=

×=

Let us select a switching frequency of 6 kHz. Therefore, the frequency modulation index,Mf, is given by

10060

6000===

e

Sf f

fM


generationPV

Therefore,

NI 22550

==


Input Voltage Vidc

(V)

Power Rating (kW)

Output Voltage, VAC

(V)



Number of Inverters

368.2 25 230 0.88 100 2

v)

The One-Line Diagram of the System.

5.8 Design a microgrid of 600 kW of power rated at 230 V AC. Use the PV module

of Problem 5.3. The design should use the least number of converters and inverters. Determine the following:

i) Number of modules in a string for each PV type ii) Number of strings in an array for each PV type iii) Number of arrays iv) Converter and inverter specifications

v) One-line diagram of this system Solutions i) The load is 600 kW rated at 230 V AC. Based on the voltage of the load and an amplitude modulation index of 0.9,we will have the following input DC voltage for a three-phase inverter:

a

LLidc M

VV

322

=

V4179.03

23022=

××

=

Let us select the string voltage at approximately 420 V,we will have the number of modules in a string given by

86.52

420VV

MPP

string ≈=

whereVMPP is the voltage of a module at MPP. ii) The string power, SP, can be computed as

SP = NM MPPP×

SP W566,21.66.528 =××= The string voltage, SV, is given as SV = NM MPPV×

Therefore, the SV for this design is SV V8.4206.528 =×=

iii) If we design each array to generate a power of 20 kW, then the number of strings, NS, in an array is given by


NS 8566.220

≈=

The number of arrays, NA, for total power generation is

NA= arrayoneofpower

generationPV

Therefore,

NA 3020

600==

kWkW

iv) SOLUTION: The PV Specifications.

Modules per String

Strings per Array

Number of Arrays

String Voltage (V)

8 8 30 420.8

The inverters should be rated to withstand the output voltage of the PV and should be able to supply the required power. The inverter is rated at 100 kW with input voltage of 420 V DC and anamplitude modulation index of 0.9. The output voltage of the inverter is 230 V AC. The number of inverters, NI, needed to process a generation of 600 kW is given by

NI=inverteroneofpower

generationPV

Therefore,

NI 6100600

==

Hence, we need to connect six inverters in parallel to supply the load of 600 kW. If the switching frequency is set at 5.04 kHz, the frequency modulation index, Mf is given by

8460

5040===

e

Sf f

fM

89.08.4203

23022V3V22

Midc

LLa =

××

==


Number of

Inverters

Input Voltage Vidc (V)

Power Rating (kW)

Output Voltage,

VAC (V)



6 420 100 230 0.89 84

v)

DC Bus

230 V

DC Bus AC Bus

DC/AC

3 phase Inverter

DC/AC

420.8 V

1

2



30

1

6

Amplitude modulation index = 0.89

600 kW

3 phase Inverter

100 kW each


5.9 Design a microgrid of a PV system rated at 2 MW and connected through a smart net metering to the local utility at 13.2 kV. The local loads consist of 100 kW of lighting loads rated at 120 V and 500 kW of AC load rated at 220 V. The system has a 700 kWh storage system. Local transformer specifications are 13.2 kV/460 V, 2 MVA, and 10% reactance; 460 V/230 V, 250 kVA, and 7% reactance; and 460V/120 V, 150 kVA, and 5% reactance. The data for this problem are given in Table 5.47.

TABLE 5.47 Photovoltaic Module Data.

Panel Type 1 Type 2 Type 3 Type 4 Power (Max), W 190 200 170 87 Voltage at max. power point (MPP), V 54.8 26.3 28.7 17.4 Current at MPP, A 3.47 7.6 5.93 5.02 VOC (open-circuit voltage), V 67.5 32.9 35.8 21.7 ISC (short-circuit current, A 3.75 8.1 6.62 5.34 Efficiency 16.40% 13.10% 16.80% >16% Cost $870.00 $695.00 $550.00 $397.00 Width 34.6" 38.6" 38.3" 25.7" Length 51.9" 58.5" 63.8" 39.6" Thickness 1.8" 1.4" 1.56" 2.3" Weight 33.07 lb 39 lb 40.7 lb 18.3 lb

i) Search the Internet for four DC/DC boost converters, rectifier, and inverters and create a table. Summarize the operating conditions in a table and discuss the results and operations as relates to this design problem. Develop a MATLAB testbed to perform the following: ii) Select boost converter, bidirectional rectifier, and inverters for the design of a microgrid from commercially available converters. If commercial converters are not available, specify the data for a new design of a boostconverter, bidirectional rectifier, and inverters. iii) Give the one-line diagram of your design. Make tables and give the number of modules in a string for each PV type, number of strings in an array for each PV type, number of arrays, converters, weight, and surface area required for each PV module type. iv) Design a 700 kWh storage system. Search online and select a deep-cycle battery storage system. Give the steps in your design and includethe dimension and weight of the storage system. v) Develop a per unit model of the PV microgrid system

Solutions

i) The following MATLAB program is used for the design:

clc; P = 5000; Vmpp = [54.8 26.3 28.7 17.4]; Impp = [3.47 7.6 5.93 5.02]; Wt = [33.07 39 40.7 18.3]; Pmpp = Vmpp.*Impp; Wt_p = Wt./Pmpp; [lightestPV_type] = min(Wt_p) P_inv = [100 250 500 1000]; No_inv = ceil(P./P_inv); [min_invinv_type] = min(No_inv)

SOLUTIONT; Data for PV Modules.

Module Type 1 Type 2 Type 3 Type 4 Power (Max), W 190 200 170 87 Voltage at Max. Power 54.8 26.3 28.7 17.4

Point, V Current at MPP, A 3.47 7.6 5.93 5.02 VOC (open-circuit voltage), V 67.5 32.9 35.8 21.7 ISC (short-circuit current, A 3.75 8.1 6.62 5.34 Efficiency 16.40% 13.10% 16.80% >16% Cost $870.00 $695.00 $550.00 $397.00 Width 34.6" 38.6" 38.3" 25.7" Length 51.9" 58.5" 63.8" 39.6" Thickness 1.8" 1.4" 1.56" 2.3" Weight 33.07 lb 39 lb 40.7 lb 18.3lb

SOLUTION: Data Single-Phase Inverter.

Inverter Type 1 Type 2 Type 3 Type 3 Power 500 W 5 kW 15 kW 4.7 kW

Input Voltage DC 500 V 500 V max 500 V 500 V Output Voltage AC

230 VAC/60 Hz @ 2.17 A

230 VAC/ 60 Hz @ 27 A

220 VAC/ 60 Hz@ 68 A

230 VAC/ 60 Hz @ 17.4 A

Efficiency Min. 78% @ full

load 97.60% > 94% 96%

Length 15.5" 315 mm 625 mm 550 mm Width 5" 540 mm 340 mm 300 mm Height 5.3" 191 mm 720 mm 130 mm Weight 9 lb 23 lb 170 kg 21 kg

SOLUTION: Data forBoost Converters. Input Voltage (V) Output Voltage (V) Power (kW) 24–46 26–48 9.2 24–61 26–63 12.2 24–78 26–80 11.23 24–78 26–80 13.1 24–98 26–100 12.5 80–158 82–160 15.2 80–198 82–200 14.2 80–298 82–300 9.5

200–600 700–1000 20.0 ii)

The load is 2 MW rated at 460 V AC. Based on the voltage of the load and an amplitude modulation index of 0.9, we will have the following input DC voltage for the inverter:

VMV

Va

LLidc 835

9.0346022

.322

=××

==

Selecting an inverter rated 250 kW, the total number of inverters, NI, for processing of 2 MW is given.

NI = invertersofrating

generationPV

NI 82502000

==

For this design, eight inverters should be connected in parallel. If we select a switching frequency of 5.04 kHz, the frequency modulation index is

8460

5040===

e

Sf f

fM

SOLUTION The Three-Phase Inverter Specifications. Number of Inverters

Input Voltage Vidc

(V)

Power Rating (kW)

Output Voltage, VAC

(V)



8 835 250 460 0.90 84 Other designs are also possible. The input DC voltage of PV specifies the output AC voltage of inverters. The PV specifications are given below. iii) SOLUTION: The design data for each type PV.

PV

Type

Surface Area

of One Module

(ft2)

Power

Rating

(W)

Area per Unit

Power (ft2per W)

1 12.47 190 0.066

2 15.68 200 0.078

3 16.97 170 0.100

4 7.07 87 0.081

It can be seen from the above table, the PV module of Type 1 requires the minimum surface area. Selecting PV Type 1 and string open circuit voltage of 550 V DC, the number of modules, NM, is

NM = MPPVvoltagestring

whereVMPP is the voltage at maximum power point of PV module from the PV data

NM 108.54

550≈= for type 1 PV

The string voltage, SV, under load is given as: SV = NM MPPV×

SV V5488.5410 =×=

The string power, SP, is given as SP = NM MPPP×

wherePMPP is the power generated by a PV module at MPP. SP kW9.119010 =×= for Type 1

If we design each array to generate a power of 20 kW, then the number of strings, NS, is given by:


NS 119.1

20==

The number of arrays, NA, is given by


generationPV

NA 10020

2000==

The total number of PV modules, TNM, in an array is given by

NANSNMTNM ××=

WhereNM is number of modules in a string and NS is the number of strings; NA is

number of arrays in a PV station.

NA 110001001110 =××= for PV module of Type 1

The total surface area needed, TS, for Type 1 PV module is as

TS acreftsq 14.3172,137144

9.516.3411000==

××=

The total weight, TW, needed for a Type 1 PV module is the product of the number of modules and the weight of each module.

TW lb770,36307.3311000 =×= The total cost for a PV module is the product of the number of modules and the cost of each module:

Total cost million57.9$87011000 =×= for PV module Type 1

SOLUTION:

The PV Specifications.

PV

Type

Number

of

Modules

per String

Number

of

Strings

per

Array

Number

of

Arrays

String

Voltage

(V)

Total

Area of

the PV

(ft2)

Total

Weight

of the

PV

(lb)

Total

Cost of

the PV

(million

$)

1 10 11 100 548 137,172 363,770 9.57

The output voltage of the boost converter, Vo, is the same as the input voltage, of the inverter, Vidc..

VVV idco 835== The boost input voltage, Vi , is same as the string voltage, SV= 548=iV V The duty ratio of the boost converter is given by

o

i

VV

D −= 1

For this design, it is,

D 34.08355481 =−=

We need one boost converter for each array. Therefore, the number of boost converters is 100 and each is rated 20 kW.

SOLUTION: TheBoostConverter Specifications. Number of

Boost Converters

Input Voltage Vi(V)

Power Rating (kW)

Output Voltage, Vo

(V)

Duty Ratio, D

100 548 20 835 0.34

iv)

SOLUTION:

The Battery Storage Specifications. Class 1 34–40 Ah 12 V

Class 2 70–85 Ah 12 V

Class 3 85–105 Ah 12 V

Class 4 95–125 Ah 12 V

Class 5 180–215 Ah 12 V

Class 6 225–255 Ah 12 V

Class 7 180–225 Ah 6 V

Class 8 340–415 Ah 6 V

The battery specification given above presents a number of batteries for storing 700 kWh of energy. In storage design, we need to limit the number of batteries in a string and limit the number of arrays to three. These limitations are imposed on lead-acid-type batteries to extend the life of storage system. We select the Class 6 batteries that are rated at 255 Ah at 12 V. In this design, three batteries per string and three strings in each array are used. The string voltage, SV, of a storage system is

SV= V36123 =× The string energy stored, SES, in each battery is given by the product of the Ah and the battery voltage. SES = kWh06.312255 =× Each array has nine batteries. Therefore, the array energy stored, AES, is given as:

AES = kWh54.2706.39 =× The number of arrays, NA, needed to store 700 kWh is given by

NA =arrayeachinenergy

energytotal

NA 2654.27

700≈=

SOLUTION: TheBattery Storage Array Specification. Battery Class

Number of Batteries per String

Number of Strings per Array

Number of

Arrays

String Voltage

(V)

Energy Stored per

Array (kWh) 6 3 3 26 36 27.54

Becausewe have 26 storage arrays, we needone buck-boost converter for each array storage system. Therefore, we need a total of 26 buck-boost converters. The buck-boost converters are used to charge–discharge the battery storage system. In this design, the buck-boost converter input is 835 V of the DC bus and its output must be 36 V DC to charge the battery storage system. If the storage systems are to be used for 8 hours, they can be discharged up to 50% of their capacity; hence, they can supply 350 kWh. The power, P, supplied by the storage system is given by

P = hourkWh

P kW75.438

350==

The array power, AP, rating is given by

AP=arraysofnumber

power

AP kW68.126

75.43==

Let us select a buck-boost converter rated at 1.68 kW. The duty ratio is given by

oi

o

VVV

D+

=

D 04.036835

36=

+=

SOLUTION:

The Buck-Boost Converter Specifications. Number of

Buck-Boost

Converters

Input Voltage,Vi

(V)

Power Rating (kW)

Output voltage, Vo

(V)

Duty ratio, D

26 835 1.68 36 0.04

The One-line diagramof the PV system.

v) To compute the per unit system, we let the base volt-ampere be MVASb 2= . The base values for the system of Problem5.9 are:

The base value of Wh is therefore, MWhEb 2= The base voltage on the utility side is13.2 kV. The base voltage on the low -voltage side of the transformer is 460 V. The new p.u. reactance of the transformer on the new 1MVA base is given by

2

)(

)(

)(

)(),(,),(, ⎟

⎟⎠

⎞⎜⎜⎝

⎛××=

newb

oldb

oldb

newboldtranspunewtranspu V

VSS

XX

120/460),(,, newtranspuX Ω=⎟⎠⎞

⎜⎝⎛××= up.0038.0

460460

200015005.0

2

,

220/460),(,, newtranspuX Ω=⎟⎠⎞

⎜⎝⎛××= up.0088.0

460460

200025005.0

2

,

The per unit power, Pp.u, rating of the inverters is given by

Pp.u= bSratingpower

Pp.u.inverter Wup.125.0102

102506

3

=××

=

And, the per unit power rating of the boost converters

WupP uboostp .010.01021020

6

3

. =××

=

The base voltage of the DC side of the inverter is 835 V. Therefore, the p.u voltage of the DC side, Vp.u, of the inverter is

Vp.u= Vup.1835835

=

Becausethe base voltage of the low-voltage side of the boost converter is the same as rated voltage, the per unit value is 1per unit. The same is true for the storage system. Similarly, the per unit power for buck-boost and energy storage system can be computed.

Pp.u buck-boost Wu.p0008.0102

1068.16

3

=××

=

Energy stored Wup.0135.0102

1054.276

3

=××

=

Per unit load WupS .05.010210100

6

3

1 =××

=

Per unit load WupS .25.0102

105006

3

2 =××

=

The Per Unit Model of the System.

5.10 Design a PV microgrid system operating at a voltage of 400 V DC serving a load

of 50 kW and at 220 V AC. Use the datasets given in Tables 5.48 to 5.51. Perform the following:

i) Select a deep-cycle battery to store 100 kWh ii) Select a boost converter, bidirectional rectifier, and inverters for the design of a microgrid from commercially available converters (see data in Tables 5.48–5.51). If commercial converters are not available, specify the data for the new design of a boost converter, bidirectional rectifier, and inverters. iii) Give the one-line diagram of your design. Make tables and give the number of modules in a string for each PV type; number of strings in an array for each PV type; number of arrays, converters, weight, and surface area required for each PV module type.

TABLE 5.48 Typical Deep-Cycle Battery Data.

Part Number

Volts

Overall Dimensions Unit Wtlb(kg)

Capacity Ampere-Hours

Length (mm)

Weight (mm)

Height (mm)

1-H Rate

2-H Rate

4-H Rate

8-H Rate

24-H Rate

48-H Rate

72-H Rate

120-H Rate

PVX-340T

12 7.71 (196)

5.18 (132)

6.89 (175)

25 (11.4) 21 27 28 30 34 36 37 38

PVX-420T

12 7.71 (196)

5.18 (132)

8.05 (204)

30 (13.6) 26 33 34 36 42 43 43 45

PVX-490T

12 8.99 (228)

5.45 (138)

8.82 (224)

36 (16.4) 31 39 40 43 49 52 53 55

PVX-560T

12 8.99 (228)

5.45 (138)

8.82 (224)

40 (18.2) 36 45 46 49 56 60 62 63

PVX-690T

12 10.22 (260)

6.60 (168)

8.93 (277)

51 (23.2) 42 53 55 60 69 73 76 79

PVX-840T

12 10.22 (260)

6.60 (168)

8.93 (277)

57 (25.9) 52 66 68 74 84 90 94 97

PVX-1080T

12 12.90 (328)

6.75 (172)

8.96 (228)

70 (31.8) 68 86 88 97 108 118 122 126

PVX-1040T

12 12.03 (306)

6.77 (172)

8.93 (227)

66 (30.0) 65 82 85 93 104 112 116 120

PVX-890T

12 12.90 (328)

6.75 (172)

8.96 (228)

62 (28.2) 55 70 72 79 89 95 98 102

TABLE 5.49 The Specification for Boost Converters. Input Voltage (V) Output Voltage (V) Power (kW) 24–46 26–48 9.2 24–61 26–63 12.2 24–78 26–80 11.23 24–78 26–80 11.23 24–78 26–80 13.1 24–98 26–100 12.5 80–158 82–160 15.2 80–198 82–200 14.2 80–298 82–400 12.5 TABLE 5.50 The Specification for Single-Phase Inverter Data.

Inverter Type 1 Type 2 Type 3 Type 3 Power 500 W 5 kW 15 kW 4.7 kW Input Voltage DC 500 V 500 V max 500 V 500 V Output Voltage AC

230 VAC/60 Hz @ 2.17 A

230 VAC/ 60 Hz @ 27 A

220 VAC/ 60 Hz@ 68 A

230 VAC/ 60 Hz @ 17.4 A

Efficiency Min. 78% @ full

load 97.60% > 94% 96%

Length 15.5" 315 mm 625 mm 550 mm Width 5" 540 mm 340 mm 300 mm Height 5.3" 191 mm 720 mm 130 mm Weight 9 lb 23 lb 170 kg 20 lb

TABLE 5.51 The Specification for Three-Phase Inverters.

Inverter Type 1 Type 2 Type 3 Type 3

Power 100 kW 250 kW 500 kW 1 MW

Input voltage

DC

900 V 900 V max 900 V 900 V

Output voltage

AC

660 VAC/60 Hz 660 VAC/ 60 Hz 480 VAC/ 60 Hz 480 VAC/ 60 Hz

Efficiency Peak efficiency

96.7%

Peak efficiency

97.0%

Peak efficiency

97.6%

Peak efficiency

96.0%

Depth 30.84" 38.2" 43.1" 71.3"

Width 57" 115.1" 138.8" 138.6"

Height 80" 89.2" 92.6" 92.5"

Weight 2350 lb 2350 lb 5900 lb 12000lb

Solutions

The load is 50 kW rated at 220 V AC. Based on the voltage of the load and an amplitude modulation index of 0.9, the input DC voltage for the inverter is

VMV

Va

LLidc 400

9.0322022

.322

=××

==

Let us select a boost converter with an output voltage of 400 V and an input voltage of 250 V, and a power rating of 12.5 kW.

If we select string voltage, SV of 250 V, the number of modules is


Where VMPP is the voltage at MPP of the PV module.

NM 58.54

250≈= Type 1

103.26

250≈= for Type 2

97.28

250≈= for Type 3

154.17

250≈= for Type 4

The string voltage, SV, is given as: SV = NM MPPV× Therefore, the string voltage, SV, for this design is: SV V2748.545 =×= for Type 1

V2633.2610 =×= for Type 2

V3.2587.289 =×= for Type 3

V2614.1715 =×= for Type 4

Let us select the 12.5 kW boost converter, the number of boost converter, NC, is

NC = ratingpowerconverterboost

generationPV

NC 45.12

50==

Therefore, the design should have fourarrays: each with its boost converter. The array

power, AP, is

AP = arraysofnumber

generationPV

AP kW5.124

50==

String power, SP, is given as SP = NM MPPP× where PMPP is the power generated by PV module at MPP.

SP kW95.01905 =×= for Type 1

kW0.220010 =×= for Type 2

W53.11709 =×= for Type 3

kW305.18715 =×= for Type 4

The number of strings, NS, is given by

NS = stringperpowerarrayperpower

NS 1495.05.12== for Type 1

72

5.12== for Type 2

953.15.12== for Type 3

10305.1

5.12== for Type 4

The total number of module, TNM, is given by:

NANSNMTNM ××=

TNM 2804145 =××= for Type 1

2804710 =××= for Type 2

324499 =××= for Type 3

60041015 =××= for Type 4

The surface area, TS, needed by each PV type is given by the product of the total number

of modules, and the length and the width of one PV module:

TS ftsq3492144

9.516.34280=

××= for Type 1

ftsq4391144

5.586.38280=

××= for Type 2

ftsq5498144

8.633.38324=

××= for Type 3

ftsq4241144

6.397.25600=

××= for Type 4

The total weight needed for each type of PV is the product of the number of modules and

the weight of one module:

The total weight lb926007.33280 =×= for Type 1

lb1092000.39280 =×= for Type 2

lb1318770.40324 =×= for Type 3

lb1104040.18600 =×= for Type 4

The total cost for each type of PV is the product of the number of modules and the cost of

one module:

The total cost 600,243$870280 =×= for Type 1

600,194$695280 =×= for Type 2

200,178$550324 =×= for Type 3

200,238$397600 =×= for Type 4

SOLUTION:

The Specifications of PV Designs.

PV

Type

Number

of

Modules

per String

Number

of

Strings

per

Array

Number

of

Arrays

String

Voltage

(V)

Total

Area of

the PV

(ft2)

Total

Weight

of the

PV

(lb)

Total

Cost of

the PV

($)

1 5 14 4 274 3,492 9,260 243,600

2 10 7 4 263 4,391 10,920 194,600

3 9 9 4 258.3 5,498 13,187 178,200

4 15 10 4 261 4,241 11,040 238,200

Number of boost converters required 45.12

50==

Selecting the boost converter output voltage of VVV oidc 400== and input voltage equal to string voltage:

274=iV VforType 1, 263=iV V for Type 2, 3.258=iV V for Type 3 and 261=iV V for

Type 4

The duty ratio of the boost converter is given by

o

i

VV

D −= 1

D 315.04002741 =−= for Type 1 PV

342.04002631 =−= for Type 2 PV

355.0400

3.2581 =−= for Type 3 PV

348.04402611 =−= for Type 4 PV

SOLUTION: The Boost Converter Specifications. PV Type Number of

Boost Converters

Input Voltage,Vi

(V)

Power Rating (kW)

Output Voltage, Vo

(V)

Duty ratio, D

1 4 274 12.5 400 0.315 2 4 263 12.5 400 0.342 3 4 258.3 12.5 400 0.355 4 4 261 12.5 400 0.348

The inverters should be rated to withstand the output voltage of the boost converter and should be able to supply the required power. Let us design with each inverter having a rating of 10 kW. The input voltage of the inverter is VVidc 400= with an amplitude modulation index of 0.90. and the output voltage of the inverter is at 220 V AC.

The number of inverters, NI, to process a generation of 50 kW is given by


generationPV

NI 51050

==

Hence, we need to connect five inverters in parallel to supply the load of 50 kW. Selecting a switching frequency of 5.1 kHz, the frequency modulation index will be given as

8560

5100===

e

Sf f

fM


Number of

Inverters

Input Voltage, Vidc (V)

Power Rating (kW)

Output Voltage,

VAC (V)



5 440 10 220 0.90 85

i) SOLUTION TABLE 5.10.4 presents the number of batteries for storing 100 kWh of energy. In storage design, we need to limit the number of batteries in a string and limit the number of arrays to three. These limitations are imposed on lead-acid-type batteries to extend the life of storage system. We select the Class 6 batteries that are rated at 255 Ah at 12 V. In this design, three batteries per string and three strings in each array are used. The string voltage, SV, of the storage system is

SV= V36123 =× The string energy stored, SES, in each battery is given by the product of the ampere-hour and the battery voltage. SES = kWh06.312255 =× Each array has nine batteries. Therefore, the array energy stored, AES, is given as: AES = kWh54.2706.39 =× The number of arrays, NA, needed to store 100 kWh is given by

NA = arrayeachinenergy

energytotal

NA 454.27

100≈=

SOLUTION TABLE 5.10.4: The Battery Storage Array Specification. Battery Number of Number Number String Energy

Class Batteries per String

of Strings per Array

of Arrays

Voltage (V)

Stored per Array (kWh)

6 3 3 4 36 27.54 ii) Because we have four storage arrays, we use one buck-boost converter for each array storage system. We need a total of four buck-boost converters. The buck-boost converters are used to charge–discharge the battery storage system. In this design, the buck-boost converter input is 440 V of the DC bus and its output must be 36 V DC to charge the battery storage system. If the storage systems are to be used for 8 hours, they can be discharged up to 50% of their capacity. Therefore, they can be used to supply 50 kWh. The power, P, supplied by the storage system is given by

P = hourkWh

P kW25.6850

==


AP = arraysofnumber

power

AP kW56.1425.6

==


oi

o

VVV

D+

=

D 075.036440

36=

+=

SOLUTION: The Buck-Boost Converter Specifications. Number of

Buck-Boost

Converters

Input Voltage,Vi

(V)

Power Rating (kW)

Output Voltage, Vo

(V)

Duty Ratio, D

4 440 1.71 36 0.075

iii)

The One-Line Diagram

5.11.Write a MATLAB testbed for design PV system with minimum weight and

minimum number of inverters using the data of Tables 5.47–5.51.

Perform the following: i) PV system for 5000 kW at 3.2 kV AC: Specify the inverter operating condition ii) PV system for 500 kW at 460 V AC: Specify the inverter operating condition iii) PV system for 50 kW at 120 V AC: Specify the inverter operating condition

Solutions For selecting the type of PV and the type of inverter the following MATLAB program is used: clc; P = 5000; Vmpp = [54.8 26.3 28.7 17.4]; Impp = [3.47 7.6 5.93 5.02]; Wt = [33.07 39 40.7 18.3]; Pmpp = Vmpp.*Impp; Wt_p = Wt./Pmpp; [lightestPV_type] = min(Wt_p) P_inv = [100 250 500 1000];

No_inv = ceil(P./P_inv); [min_invinv_type] = min(No_inv)

i) From the program above, we find that the PV with minimum weight is of type 1 and minimum number of inverter can be used with type 4. With the above type of PV, the design of the microgrid is presented below:

The load is 5000 kW rated at 3.2kV AC. Let a 480 V/ 3.2 kV transformer be used to step up the voltage of the AC bus of the inverter. Based on this voltage and an amplitude modulation index of 0.9, the input DC voltage for the inverter is:

V8719.03

48022M.3V22

Va

LLidc =

××

==

Let us select a boost converter with an output voltage of 871 V and an input voltage of 400 V, and a power rating of 100 kW.


NM =MPPVvoltagestring

whereVMPP is the voltage at MPP of the PV module

NM 78.54

400≈= Type 1

The string voltage, SV, is given as: SV = NM MPPV×

Therefore, the string voltage, SV, for this design is

SV V6.3838.547 =×= for Type 1

Let us select the 100 kW, the number of boost converters, NC, is


generationPV

NC 501005000

==

Therefore, the design should have fifty arrays: each with its boost converter. The array

power, AP, is

AP = arraysofnumber

generationPV

AP kW10050

5000==

String power, SP, is given as SP = NM MPPP× WherePMPP is the power generated by the PV module at MPP.

SP kW13301907 =×= for Type 1



NS 75330.1

100== for Type 1

SOLUTION:

The specifications of minimum weight PV design.

PV

Type

Number

of

Modules

per String

Number

of

Strings

per

Array

Number

of

Arrays

String

Voltage

(V)

1 7 75 50 383.6

Number of boost converters required 501005000

==

Selecting the boost converter output voltage of V871VV oidc == and input voltage equal to string voltage:

V6.383Vi = for type 1


o

i

VV

D −= 1

D 56.0871

6.3831 =−= for Type 1 PV

SOLUTION: The boost converter specifications. Number of

Boost Converters

Input Voltage,Vi(V)

Power Rating (kW)

Output Voltage, Vo

(V)

Duty Ratio, D

50 383.6 100 871 0.56

The inverters should be rated to withstand the output voltage of the boost converter and should be able to supply the required power. Let us design with each inverter having a rating of 1 MW. The input voltage of the inverter is V871Vidc = with an amplitude modulation index of 0.90. The output voltage of the inverter is at 480 V AC. The number of inverters, NI, to process a generation of 5000 kW is given by


generationPV

NI 510005000

==

Hence, we need to connect five inverters in parallel to supply the load of 50 kW. Selecting a switching frequency of 5.1 kHz, the frequency modulation index is given as

8560

5100===

e

Sf f

fM

SOLUTION: The inverter specifications.

Number of

Inverters


Power Rating (kW)

Output Voltage,

VAC (V)



5 871 1000 480 0.9 85

The one-line diagram. ii)The load is 500 kW rated at 460 V AC. Based on this voltage and an amplitude modulation index of 0.9, the input DC voltage for the inverter is:

V8349.03

46022M.3V22

Va

LLidc =

××

==

Let us select a boost converter with an output voltage of 834 V and an input voltage of 400 V, and a power rating of 100 kW. If we select string voltage, SV of 400 V, the number of modules is



NM 78.54

400≈= Type 1



SV V6.3838.547 =×= for Type 1

Let us select the 100 kW, the number of boost converters, NC, is


generationPV

NC 5100500

==

Therefore, the design should have five arrays: each with its boost converter. The array

power, AP, is

AP = arraysofnumber

generationPV

AP kW1005

500==

String power, SP, is given as SP = NM MPPP× wherePMPP is the power generated by the PV module at MPP.

SP kW13301907 =×= for Type 1



NS 75330.1

100== for Type 1

SOLUTION:

The PV generating station specifications.

PV

Type

Number

of

Modules

per String

Number

of

Strings

per

Array

Number

of

Arrays

String

Voltage

(V)

1 7 75 5 383.6

Number of boost converters required 5100500

==


V6.383Vi = for type 1


o

i

VV

D −= 1

D 54.0834

6.3831 =−= for Type 1 PV

SOLUTION: The boost converter specifications. Number of

Boost Converters

Input Voltage,Vi(V)

Power Rating (kW)

Output Voltage, Vo

(V)

Duty Ratio, D

5 383.6 100 834 0.54

The inverters should be rated to withstand the output voltage of the boost converter and should be able to supply the required power. Let us design with each inverter having a rating of 100 kW. The input voltage of the inverter is V834Vidc = with an amplitude modulation index of 0.90. The output voltage of the inverter is at 460 V AC. The number of inverters, NI, to process a generation of 500 kW is given by


generationPV

NI 5100500

==


8560

5100===

e

Sf f

fM

SOLUTION: The inverter specifications.

Number Input Power Output Amplitude Frequency

of Inverters

Voltage, Vidc (V)

Rating (kW)

Voltage, VAC (V)

Modulation Index, Ma

Modulation Index, Mf

5 834 100 460 0.9 85

The one-line diagram. iii) The load is 50 kW rated at 120 V AC. Based on this voltage and an amplitude modulation index of 0.9, the input DC voltage for the inverter is:

V2189.03

12022M.3V22

Va

LLidc =

××

==




NM 48.54

220≈= Type 1



SV V2.2198.544 =×= for Type 1

Let us have 2 arrays, therefore, the rating of each array is:

AP = arraysofnumber

generationPV

AP kW252

50==

String power, SP, is given as SP = NM MPPP× wherePMPP is the power generated by the PV module at MPP.

SP W7601904 =×= for Type 1



NS 3376.0

25== for Type 1

SOLUTION:

The Specifications of PV Design.

PV

Type

Number

of

Modules

per String

Number

of

Strings

per

Array

Number

of

Arrays

String

Voltage

(V)

1 4 33 2 219.2

The inverters should be rated to withstand the output voltage of the PV and should be able to supply the required power. Let us design with each inverter having a rating of 5 kW.

The input voltage of the inverter is V2.219Vidc = with an amplitude modulation index of 0.90. The output voltage of the inverter is at 120 V AC. The number of inverters, NI, to process a generation of 50 kW is given by


generationPV

NI 105

50==


8560

5100===

e

Sf f

fM


Number of

Inverters


Power Rating (kW)

Output Voltage,

VAC (V)



10 219.2 5 120 0.9 85

The One-Line Diagram.

5.12. Consider the residential home of Fig.5.42. Perform the following: i) Estimate the load consumption of the house. ii) Plot the daily load cycle operation of the house’s loads over 24 hours and calculate the total energy consumption.

Tree

Tree

Tree

Tree

Tree

Lighting

Dryer

Washer

Air Conditioning

TV

DC

NetSmartMeter

Transformer

DC Bus

AC Bus

AC

DCAC

Storage

Inverter

Bi-DirectionalConverter

PV ArrayBox

EV

Figure 5.42.Figure for Problem 5.12.

iii) Search the Internet and select a PV module and design the PV array for the house. Compute cost, weight of PV array, and roof areas needed for the PV system. Search the Internet and select an inverter, battery storage, and bidirectional converter.

Solutions i) Let us assume that the house has 1 TV set of 200 W, air conditioner of 1200

W, lighting of 100 W, washer of 1000 W and an 800 W dryer. If it is a summer evening and all the loads are on, then the peak load that the house consumes: Peak load = W330080010001001200200 =++++

ii) Let us assume that the air conditioner is on for the entire day, the TV is on from 10 AM to 10 PM, the light bulb is on from 7 PM to 11 PM, washer is on

from noon to 1 PM and drier from 1 PM to 3 PM, then we have the following load cycle: Load from midnight to 10 AM = air conditioner W1200=

Load from 10 AM to 12 PM = air conditioner + TV = 1200 + 200 = 1400 W Load from 12 PM to 1 PM = air conditioner + TV + washer = 1200 + 200 + 1000 = 2400 W Load from 1 PM to 3 PM = air conditioner + TV + drier = 1200 + 200 + 800 = 2200 W Load from 3 PM to 7 PM = air conditioner + TV = 1200 + 200 = 1400 W Load form 7 PM to 10 PM = air conditioner + TV + light bulb = 1200 + 200 + 100= 1500 W Load from 10 PM to 11 PM = air conditioner + lamp = 1200 + 100 = 1300 W Load from 11 PM to 12 PM = air conditioner = 1200 W

Pow

er (i

n W

)

A typical load cycle for one summer day

iii) Let us design a 2400 W 120 V AC PV system with 600 kWh of storage

The load is 2.4kW rated at 120 V AC. Based on the voltage of the load and an amplitude modulation index of 0.9, we will have the following input DC voltage for the inverter:

V2179.03

12022M.3V22

Va

LLidc =

××

==

Selecting an inverter rated 2.4 kW, the total number of inverters, NI, for processing of 2.4kW is given.

NI = invertersofrating

generationPV

NI 15.25.2==

For this design, eight inverters should be connected in parallel. If we select a switching frequency of 5.04 kHz, the frequency modulation index is

8460

5040===

e

Sf f

fM

SOLUTION: The Inverter Specifications. Number of Inverters

Input Voltage Vidc

(V)

Power Rating (kW)

Output Voltage, VAC

(V)



1 217 2.4 120 0.90 84 Selecting PV Type 1 from the above tableand string voltage at MPP at 220 V, the number of modules in a string, NM, is given by:


where VMPP is the voltage at maximum power point of PV module from the PV data

NM 48.54

220≈= for type 1 PV

The string voltage, SV, under load is given as: SV = NM MPPV×

SV V2.2198.544 =×=

The string power, SP, is given as SP = NM MPPP×

wherePMPP is the power generated by a PV module at MPP. SP kW76.01904 =×= for Type 1

If we design each array to generate a power of 2.4 kW, then the number of strings, NS, is given by:


NS 376.04.2==

The number of arrays, NA, is given by


generationPV

NA 14.24.2==

The total number of PV modules, TNM, in an array is given by

NANSNMTNM ××=

Where NM is number of modules in a string and NS is the number of strings; NA is

number of arrays in a PV station.

NA 12134 =××= for PV module of Type 1

The total surface area needed, TS, for Type 1 PV module is as

TS 2ft150144

9.516.3412=

××=

The total weight, TW, needed for a Type 1 PV module is the product of the number of modules and the weight of each module.

TW lb39707.3312 =×= The total cost for a PV module is the product of the number of modules and the cost of each module:

Total cost 440,10$87012 =×=

SOLUTION:

The PV Generating Station Specifications.

PV

Type

Number

of

Modules

per String

Number

of

Strings

per

Array

Number

of

Arrays

String

Voltage

(V)

Total

Area of

the PV

(ft2)

Total

Weight

of the

PV

(lb)

Total

Cost of

the PV

($)

1 4 3 1 217 150 397 10,440

SOLUTION TABLE 5.12.3 presents a number of batteries for storing 600 kWh of energy. In storage design, we need to limit the number of batteries in a string and limit the number of arrays to three. These limitations are imposed on lead-acid-type batteries to extend the life of storage system. We select the Class 6 batteries that are rated at 255 Ah at 12 V. In this design, three batteries per string and three strings in each array are used. The string voltage, SV, of a storage system is

SV= V36123 =× The string energy stored, SES, in each battery is given by the product of the Ah and the battery voltage. SES = kWh06.312255 =× Each array has nine batteries. Therefore, the array energy stored, AES, is given as: AES = kWh54.2706.39 =× The number of arrays, NA, needed to store 700 kWh is given by


energytotal

NA 2254.27

600≈=

SOLUTION TABLE 5.12.3 The Battery Storage Array Specifications. Battery Class



Number of

Arrays

String Voltage

(V)

Energy Stored per

Array (kWh) 6 3 3 22 36 27.54

Because we have 22 storage arrays, we need one buck-boost converter for each array storage system. Therefore, we need a total of 22 buck-boost converters. The buck-boost converters are used to charge–discharge the battery storage system. In this design, the buck-boost converter input is 217 V of the DC bus and its output must be 36 V DC to charge the battery storage system. If the storage systems are to be used for 8 hours, they can be discharged up to 50% of their capacity; hence, they can supply 300 kWh. The power, P, supplied by the storage system is given by

P = hourkWh

P kW5.378

300==


AP=arraysofnumber

power

AP kW70.122

5.37==


oi

o

VVV

D+

=

D 14.036217

36=

+=


Buck-Boost

Converters

Input Voltage, Vi (V)

Power Rating (kW)

Output voltage, Vo

(V)

Duty ratio, D

22 217 1.70 36 0.14

Let a 120 V / 240 V, 2.4 kVA transformer be used to connect to the utility.

TheOne-Line Diagramof the PV System.

5.13. For Problem 5.11, if only 25% of the load is operated during the night, use the data of Problem 5.10 and specify a battery storage system to store the required energy for operating 25% of the load during the night.

Solutions With 25% load, the load = kWh1250500025.0 =× SOLUTION TABLE 5.13.1 for storage system presents a number of batteries for storing 1250 kWh of energy. In storage design, we need to limit the number of batteries in a string and limit the number of arrays to three. These limitations are imposed on lead-acid-type batteries to extend the life of storage system. We select the Class 6 batteries that are

rated at 255 Ah at 12 V. In this design, three batteries per string and three strings in each array are used. The string voltage, SV, of a storage system is

SV = V36123 =× The string energy stored, SES, in each battery is given by the product of the Ah and the battery voltage. SES = kWh06.312255 =× Each array has nine batteries. Therefore, the array energy stored, AES, is given as: AES = kWh54.2706.39 =× The number of arrays, NA, needed to store 700 kWh is given by


energytotal

NA 4554.27

1250≈=

SOLUTION TABLE 5.13.1: The Battery Storage Array Specification. Battery Class



Number of

Arrays

String Voltage

(V)

Energy Stored per

Array (kWh) 6 3 3 45 36 27.54

Because we have 45 storage arrays, we need one buck-boost converter for each array storage system. Therefore, we need a total of 45 buck-boost converters. The buck-boost converters are used to charge–discharge the battery storage system. In this design, the buck-boost converter input is 871 V of the DC bus and its output must be 36 V DC to charge the battery storage system. If the storage systems are to be used for 1 hour, they can be discharged up to 50% of their capacity; hence, they can supply 625 kWh. The power, P, supplied by the storage system is given by

P = hourkWh

P kW6251

625==


AP =arraysofnumber

power

AP kW88.1345

625==


oi

o

VVV

D+

=

D 04.036871

36=

+=


Buck-Boost

Converters


Power Rating (kW)

Output voltage, Vo

(V)

Duty ratio, D

45 871 13.88 36 0.04

Fig. One line diagram

5.14. If the price of kWh from a utility company is $0.3 for buying or selling energy,

estimate the net operating cost or revenue for the house of Problem 5.12.

Solutions Using the data from problem 5.12, Total energy consumed by the loads in one day = timepower × ( )

kWh2.341011200113003150041400222001240021400101200 3

=××+×+×+×+×+×+×+×= −

Assuming that the PV is able to generate 2.4 kW of power for 8 hours a day Total energy generated in one year kWh008,736584.2timepower =××=×=

Total energy consumed in one year by the load kWh483,123652.34 =×= Energy drawn from the utility kWh475,5008,7483,12 =−=

Total price for electricity 50.1642$3.0475,5 =×= 5.15 Design a PV system rated 50 kW using a boost converter and a DC/AC inverter. The

system operates as a standalone and supports a water pumping system with a rated load voltage of 120 V AC. Use the data given in Problem 5.10. Solution

The load is 50 kW rated at 220 V AC. Based on the voltage of the load and an amplitude modulation index of 0.9, the input DC voltage for the inverter is:

V2189.03

12022M.3V22

Va

LLidc =

××

==

Let us select a boost converter with an output voltage of 218 V and an input voltage of 150 V, and a power rating of 12.5 kW.


NM=MPPVvoltagestring

Where VMPP is the voltage at MPP of the PV module

NM 38.54

150≈= Type 1



SV V4.1648.543 =×= for Type 1

Let us select the 12.5 kW boost converters, the number of boost converters, NC, is


generationPV

NC 45.12

50==

Therefore, the design should have four arrays: each with its boost converter. The array

power, AP, is

AP = arraysofnumber

generationPV

AP kW5.124

50==

String power, SP, is given as SP = NM MPPP× Where PMPP is the power generated by the PV module at MPP.

SP kW57.01903 =×= for Type 1



NS 2257.05.12== for Type 1

The total number of modules, TNM, is given by

NANSNMTNM ××=

TNM 2644223 =××= for Type 1

The surface area, TS needed by each PV type is given by the product of the total number

of modules, and the length and the width of 1PV module:

TS ftsq3292144

9.516.34264=

××= for Type 1

The total weight needed for each type of PV is the product of the number of modules and

the weight of one module:

The total weight lb873007.33264 =×= for Type 1

The total cost for each type of PV is the product of the number of modules and the cost of

one module:

The total cost 680,229$870264 =×= for Type 1

SOLUTION:

The Specifications of PV Types.

PV

Type

Number

of

Modules

per String

Number

of

Strings

per

Array

Number

of

Arrays

String

Voltage

(V)

Total

Area of

the PV

(ft2)

Total

Weight

of the

PV

(lb)

Total

Cost of

the PV

($)

1 3 22 4 164.4 3,292 8,730 229,680

Number of boost converters required 45.12

50==


4.164Vi = V for type 1


o

i

VV

D −= 1

D 25.0220

4.1641 =−= for Type 1 PV

SOLUTION: The Boost Converter Specifications. PV Type Number of

Boost Converters

Input Voltage,

Vi(V)

Power Rating (kW)

Output Voltage, Vo

(V)

Duty Ratio, D

1 4 164.4 12.5 220 0.25

The inverters should be rated to withstand the output voltage of the boost converter and should be able to supply the required power. Let us design with each inverter having a rating of 10 kW. The input voltage of the inverter is V220Vidc = with an amplitude modulation index of 0.80. The output voltage of the inverter is at 120 V AC. The number of inverters, NI, to process a generation of 50 kW is given by


generationPV

NI 51050

==


8560

5100===

e

Sf f

fM

89.0220312022

V.3V22M

idc

LLa =

××

==


Number of

Inverters


Power Rating (kW)

Output Voltage,

VAC (V)



5 220 10 120 0.89 85

The One-Line Diagram. 5.16. Design a residential PV system. The load cycle is 10 kW from 11 P.M.until8 A.M.

and 14 kW for the remaining 15 hours. Determine the following:

i) Total kWh energy consumption for 24 hours ii) What is the roof space needed to generate adequate kWh for 24 hours operation iii) Assume the maximum kWh to be used during the night is 40% of the total daily load. Search the Internet to select a battery storage system and compute the required energy for nightly operation. Give your design data.

Solutions i) The load is 10 kW for 9 hours (11:00 P.M. to 8:00 A.M.) and it is 14 kW for 15

hours (8:00 A.M. to 11:00 P.M.). The load cycle is given below:

Plot of the daily load cycle

The total kWh energy consumption for 24 hours is the area under the curve of daily load cycle and is given by kWh = kW × hours Therefore, the energy consumption kWh3001514910 =×+×=

ii) Assuming 0.5 sun and sun is available for 8 hours a day. Type 1 PV is selected because it needs the minimum area per unit power produced.

The amount of power produced by type 1 PV is equal to 190 W per module for 1 sun. Therefore, the energy produced for 0.5 sun for 8 hours is given by:

kWh76.081905.0 =×× The number of modules, NM, needed is given by:

NM = paneloneofenergy

demandenergytotal

39476.0

300≈=NM

The surface area, SM, of one module is given by lengthwidth × SM ftsq.47.12144/9.516.34 =×=

The total area, TS, for 300 kWh is given by the product of the number of modules and the area of one module: TS = ftsq .18.491347.12394 =×

iii) The energy used during the night is 40% of the total energy. Therefore, the energy demand for one night = kWh1203004.0 =× .

A Class 6battery storage system is chosen to store the kWh needed for the night. The batteries should not be discharged more than 50% of their capacity. The energy stored per battery is given by voltageAh× . Therefore, the energy

stored in one battery kWh06.312255 =×= . The number of batteries, NB, needed is given by

NB = batteryperstoredenergy

demandenergy×2

NB= 7906.31202

=× .

We can use three batteries in a string and the maximum number of strings in an array is equal to three. Therefore, the maximum number of batteries in the array is equal to 933 =× . The number of arrays of battery is given by

NA=arrayperbatteriesofnumber

batteriesofnumbertotal

NA 9979

≈=

5.17 Design amicrogridfora PV system rated 1MW of power at 220 V, 60 Hz, with all the PV strings connected to the same DC bus. The transformer data are 220/460 V, 250 kVA, and 5% reactance; and 460 V/13.2 kV of 1 MVA, and 10% reactance. Use the data given in Tables 5.47–5.51.

Determine the following:

i) Number of modules in a string for each PV type, number of strings in an array for each PV type, number of arrays and surface area, weight and cost for each PV type. ii) Boost converter and inverter specifications and the one-line diagram of this system

Solutions Let the inverter output voltage be 460 V AC at a modulation index of 0.85

VMV

Va

LLidc 884

85.0346022

322

=××

==

We will limit the maximum voltage that a string is allowed to have to 600 V. Therefore, we use a boost converter to boost the string voltage to 884 V.

i) If we select an approximatestring voltage of 550 V, the number of modules in a string, NM, is given as

NM= 108.54

550≈=

MPP

string

VV

WhereVMPP is the voltage at MPP of the PV module.

SP = NM MPPP×

WherePMPP is the power generated by a PV module at MPP.

SP W190019010 =×= And the string voltage, SV, is

SV = NM MPPV×

Therefore, the string voltage for this design is SV V5488.5410 =×=

If each array is to have a rating of 20 kW, the number of strings, NS, in an array is

NS = SPAP

NS 119.1

20==

The number of arrays for this design is


generationPV

NA 5020

1000==

The total number of PV modules, TNM, is given by the product of the number of modules per string, the number of strings per array, and the number of arrays:

NSNSNMTNM ××= TNM = 500,5501110 =×× The surface area of one module, SM, is given by the product of its length and width.

SM= ..5.12144

9.516.34 ftsq=×

The total surface area, TS, is therefore given by the total number of modules and the surface area of each module.

TS = acreftsq 57.1560,43750,68..750,685.125500 ===×

The total cost of PV modules is given by the product of the number of PV modules and the cost of one module. The total cost = million78.4$8705500 =× The total weight of PV modules is given by the product of the number of PV modules and the weight of one module.

The total weight = lb885,18107.335500 =× SOLUTION: The PV Specifications for 1000 kW Generating Station.

Modules per String

Strings per

Array

Number of Arrays

String Voltage

(V)

Total Area (ft2)

Total weight

(lb)

Total cost

(million $)

10 11 50 548 68,750 181,885 4.78

ii) The inverters should be rated to withstand the output voltage of the boost converter and should be able to supply the required power. Selecting an inverter rated at 250 kW, we have the number of inverters, NI, needed to process thegeneration of 1000 kW given by


generationPV

NI 4250

1000==

Hence, we need to connect four inverters in parallel to supply the load of 1000

kW. Selecting a switching frequency of 5.40 kHz, the frequency modulation index is

given by

9060

5400===

e

Sf f

fM

SOLUTION: TheInverterSpecifications.

Number of

Inverters

Input Voltage,

Vidc (V)

Power Rating (kW)

Output Voltage,

VAC (V)



4 884 250 460 0.85 90 The number of boost converters needed is the same as the number of arrays, which is 50. Select a boost converter rating of 20 kW and let the boost converter input voltage be equal to the string voltage:

548=iV V The output voltage of the boost converter is equal to the inverter input voltage:

VVV oidc 884== The duty ratio of the boost converter is given by

o

i

VV

D −= 1

D 38.08845481 =−=

SOLUTION: The Boost Converter Specifications. Number of

Boost Converters


Power Rating (kW)

Output Voltage, Vo

(V)

Duty Ratio, D

50 548 20 884 0.38

DC Bus

460 V

DC Bus

AC Bus

DC/AC

3 phase Inverter

DC/AC

DC/DC

DC/DC

Boost

DC/DC

Boost

Boost

884 V548 V

1

2

50 PV arrays:11 string array,


50

1

4

Duty ratio = 0.38 Amplitude

modulation index = 0.85

3 phase Inverter

1

2

50

20 kW each 250 kW

each

250 kW

460 V / 120 V250 kVAX = 5%

250 kW

460 V / 120 V250 kVAX = 5%

250 kW

460 V / 120 V250 kVAX = 5%

250 kW

460 V / 120 V250 kVAX = 5%

The One-Line Diagram

5.18 Assume a sample value for the global daily irradiation,G = [1900, 2690, 4070, 5050, 6240, 7040, 6840, 6040, 5270, 3730, 2410, 1800],for 12 months of the year. Assume a reflectivity of 0.25. Perform the following:

i) Write a MATLAB M-file program to (a) compute the irradiation on different inclination angles, (b) tabulate the irradiance for each month at different inclination angles, (c) tabulate the overall irradiance per year for different inclination angles, and (d) find the optimum inclination angle for each month and a year. ii) If the sun irradiance is 0.4 sun for 8 hours daily for this location, what is the roof space needed to capture 20 kW at an optimal angle?

iii) If the sun irradiance is 0.3 sun on the average over a year for 5 hours daily for this location, what total kW can be captured over 1500 square feet at the optimum inclination angle?

Solutions

1. The solar declination angle, in degrees, can be found out from an empirical

formula, ( )

⎥⎦

⎤⎢⎣

⎡ +=δ

365284360

sin45.23 nd, as mentioned in [P. I. Cooper, The

Absorption of Solar Radiation in Solar Stills, Sol. Energy 12 (3), 333 – 346 (1969)]

2. The sunset hour angle, in radian, is found from [M. Iqbal, An Introduction to Solar Radiation, New York: Academic Press, 1983, pp. 1 – 84.]:

[ ]1tan.tan

1tan.tantan.tancos 1

>φδπ=

≤φδφδ−=ω − fors

3. The extraterrestrial daily irradiance, Bo, in Wh/m2, is calculated from [M. Iqbal, An Introduction to Solar Radiation, New York: Academic Press, 1983, pp. 1 – 84.]

( )δφω+ωδφ⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛ π

+π

= sin.sin.sin.cos.cos.365

2cos033.01..24

ssn

od

SB

4. The clearness index, KT is calculated oT BGK /=

5. The diffusion factor, GD / , is calculated empirically from TKGD 13.11/ −= [ J. K. Page, “The estimation of monthly mean values of daily total short-wave radiation on vertical and inclined surfaces from sunshine records for latitudes 40°N – 40°S”, in Proc. United Nations on New Sources of Energy, vol. 4, 1961, pp. 378 – 390]

6. The beam irradiation is calculated from DGB −= 7. The sunset angle for the tilted module is determined using:

( )[ ]1tan.tan

1tan.tantan.tancos 1,

>φδπ=

≤φδβ−φδ−=ω − fortilts

8. The effective sunset angle is determined from ( )stiltss ωω=ω ,min' ,

9. The value of the beam irradiation is adjusted for the tilt angle from [ T. Markvart, Solar Electricity, West Sussex, England: Wiley, 1994, pp. 11 – 16.]:

( ) ( ) ( )δφω+ωδφ

δβ−φω+ωδβ−φ=β

sin.sin.sin.cos.cossin.sin'.'sin.cos.cos

ss

ssBB

10. Assuming that the diffusion is isotropic, the diffusion irradiation on tilted surface is given by [ T. Markvart, Solar Electricity, West Sussex, England: Wiley, 1994,

pp. 11 – 16]: ( ) ( )DD .cos121

β+=β

11. The isotropically modeled albedo is given by [ T. Markvart, Solar Electricity,

West Sussex, England: Wiley, 1994, pp. 11 – 16]: ( ) ( ) DR ..cos121

ρβ−=β

12. The global irradiation on the tilted surface is the sum of beam irradiation, diffusion irradiation and albedo [T. Markvart, Solar Electricity, West Sussex, England: Wiley, 1994, pp. 11 – 16.]: ( ) ( ) ( ) ( )β+β+β=β RDBG

i) The Steps 1–12 given above are used to calculate irradiance for each month.

The tilt angle β is varied from zero to 90 degrees in steps of 12 degrees. The results are tabulated below.

The MATLAB program: clc; clearall; % solar irradiation data on horizontal surface in Wh/m^2/day GI = [1.90 2.69 4.07 5.05 6.24 7.04 6.84 6.04 5.27 3.73 2.41 1.80]*1e3; % latitude of Los Angeles, California, in degree L=33.93; S=1367; %W/m2 global irradiation deg = 5; % tilt angle adjustment degree g = 0.25; % for semi bare ground reflectivity month = zeros(1,12); % initialising betha = zeros(1,floor(90/deg)+1); % initialising fori=1:12 % finding the day number for the middle of each month month(i)=15+(i-1)*30; end fori=1:floor(90/deg)+1 % Varying the tilt angle betha(i)=(i-1)*deg; end Ir(floor(90/deg)+1,13)=zeros; % initialising irradiation matrix fori=1:floor(90/deg)+1 GG=0; for j=1:12 delta=23.45*sind(360*(284+month(j))/365);% Calculating declin. angle % Calculate the sunset angle x = -tand(L)*tand(delta); if abs(x) > 1 if x > 0 x = 1; else x = -1; end end

ws = acos(x); % Finding sunset angle y = -tand(L-betha(i))*tand(delta); if abs(y) > 1 if y > 0 y = 1; else y = -1; end end ws1=acos(y); % Finding sunset angle for tilted surface wo=min(ws,ws1); % Calculationg extraterrestrial irradiation Bo=(24/pi)*S*(1+.0333*cos(2*pi*month(j)/365))*(cosd(L)*cosd(delta)*sin(ws)+ws*sind(L)*sind(delta)); if Bo == 0 % no sunrise DD = GI(j); % radiation on surface is all diffusion B = 0; else KT=GI(j)/Bo; % clearness index DD=GI(j)*(1-1.13*KT); % diffuse irradiation BB=GI(j)-DD; % beam(direct irradiation) B=BB*(cosd(L-betha(i))*cosd(delta)*sin(wo)+wo*sind(L-betha(i))*sind(delta))/(cosd(L)*cosd(delta)*sin(ws)+ws*sind(L)*sind(delta)); end D=DD*0.5*(1+cosd(betha(i))); % diffused irradiation on tilted surface R=DD*0.5*g*(1-cosd(betha(i))); % reflection irradiation on tilted surface G=B+D+R; % total global irradiation GG=GG+G; Ir(i,j)=G; % irradiation for each month at different inclination angles end Ir(i,13)=GG/12; % irradiation for entire year end a=Ir(:,13); mm=max(a); % the maximum yearly average irradiation for optimum angle fori=1:floor(90/deg)+1 % Finding optimum angle if a(i)== mm nn=i; end end % Display the irradiation of each month and the year Ir optimal_angle = betha(nn) % optimum angle for the entire year

SOLUTION: The Winter Irradiation at Different Tilt Angles Tilt Angle

January W/m2

February W/m2

March W/m2

0 1900 2690 4070

12 2194 3016 4414

24 2425 3253 4616

36 2582 3390 4667

48 2660 3421 4567

60 2653 3344 4316

72 2563 3164 3929

84 2393 2887 3421

90 2281 2716 3129

SOLUTION: The SpringIrradiation at Different Tilt Angles. Tilt Angle

April W/m2

May W/m2

June W/m2

0 5050 6240 7040

12 5237 6231 6851

24 5252 6016 6433

36 5093 5595 5787

48 4768 4986 4940

60 4288 4214 3928

72 3678 3320 2807

84 2965 2356 1653

90 2582 1870 1099

SOLUTION: The SummerIrradiation on Different Tilt Angles.

Tilt July August Sept.

Angle W/m2 W/m2

W/m2

0 6840 6040 5270

12 6721 6180 5699

24 6368 6095 5915

36 5784 5784 5910

48 4991 5259 5682

60 4024 4542 5241

72 2935 3671 4608

84 1795 2689 3810

90 1237 2175 3361

SOLUTION: The Autumn and Yearly Average Irradiation on Different Tilt Angles.

Tilt Angle

Oct. W/m2

Nov. W/m2

Dec. W/m2

Yearly Average W/m2

0 3730 2410 1800 4233

12 4221 2803 2117 4640

24 4570 3110 2371 4702

36 4760 3316 2553 4602

48 4785 3415 2653 4344

60 4641 3398 2667 3938

72 4337 3270 2595 3406

84 3885 3035 2440 2778

90 3609 2880 2333 2439

The above tables list the irradiance in W/m2 for tilt angles varying from zero to 90 degreeswith an interval of 12 degrees. The maximum irradiation occurs at a tilt angle of 24 degrees if the PV modules are fixed and not free to change the tilt angle. At this angle, irradiance energy is 4702 Wh/m2. This amount is almost 469 Wh/m2 greater than the irradiation on a horizontal surface. If the angle of tilt can be changed every month of the year, the yield will be almost 4923.5 Wh/m2. Using the optimum angle of irradiance, we

can increase substantially the average of the maximum value of each month’s irradiation in the PV system.

ii) With the optimum angle, G = 4702 W / m2,the power received is 2

s m/W188147024.0P =×=

Selecting a module of Type 1, which has an efficiency of 16.4%, the electrical power developed from the solar power =

2/3081881164.0 mWPe =×=

The area needed 23

65308

1020 mA =×

=

iii) The power received 2/6.141047023.0 mWPs =×=

With an efficiency of 16.4%, the electrical power developed from the solar power =

2/2316.1410164.0 mWPe =×=

Therefore, with 1500 m2, the electrical power, Pe kW5003462311500 .=×=

With 5 hours of sunshine, the energy produced kWh5.173255.346 =×=

5.19 Assume the global daily irradiation (G) forthe city of Columbus solar irradiation data, G, on the horizontal surface is as follows:

G = [1800, 2500, 3500, 4600, 5500, 6000, 5900, 5300, 4300, 3100, 1900, 1500] for 12 months of the year. The latitudinal location of Columbus is 40 degrees. Assume a reflectivity of 0.25. Perform the following:

i) Write a MATLAB M-file to (a) compute the irradiation on different inclination angles, (b) tabulate the irradiance for each month at different inclination angles, (c) tabulate the overall irradiance per year for different inclination angles, and (d) find the optimum inclination angle for each month and a year. ii) If the sun irradiance is 0.4 sun for 8 hours daily for this location what is the roof space needed to capture 50 kW at an optimum inclination angle?

iii) If the sun irradiance is 0.3 sun on average over a year for 5 hours daily for this location what is the total kW that can be captured over 1500 ft2 at the optimum inclination angle?

Solutions i) The algorithm and the MATLAB program is given in problem 5.18

SOLUTION: The Winter Irradiation Data for Columbus, Ohio. Tilt Angle

January W/m2

February W/m2

March W/m2

0 1800 2500 3500

12 2112 2822 3782

24 2363 3061 3948

36 2541 3206 3992

48 2638 3250 3911

60 2650 3193 3708

72 2577 3035 3394

84 2422 2784 2981

90 2315 2627 2743

SOLUTION: The Spring Irradiation Data for Columbus, Ohio. Tilt Angle

April W/m2

May W/m2

June W/m2

0 4600 5500 6000

12 4776 5512 5882

24 4802 5353 5585

36 4675 5022 5109

48 4399 4534 4470

60 3987 3908 3698

72 3457 3176 2832

84 2835 2378 1927

90 2499 1971 1484

SOLUTION: The Summer IrradiationData for Columbus,Ohio

Tilt Angle

July W/m2

Aug W/m2

Sept W/m2

0 5900 5300 4300

12 5832 5431 4614

24 5579 5378 4772

36 5140 5138 4764

48 4531 4720 4593

60 3779 4143 4265

72 2922 3434 3794

84 2013 2630 3202

90 1561 2206 2869

SOLUTION: The Autumn and Yearly Average Irradiation Data for Columbus, Ohio.

Tilt Angle

Oct W/m2

Nov W/m2

Dec W/m2

Yearly Average W/m2

0 3100 1900 1500 3825

12 3479 2182 1762 4016

24 3748 2400 1973 4080

36 3895 2546 2125 4013

48 3912 2613 2211 3815

60 3800 2597 2227 3496

72 3563 2500 2172 3071

84 3211 2326 2049 2563

90 2997 2212 1964 2287

With the tilt angle set at the optimal value of 24°,

2/163240804.0 mWPs =×=

Selecting a module of Type 1, which hasan efficiency of 16.4%, the electrical power developed from the solar power =

2/2681632164.0 mWPe =×=

The area needed is 23

187268

1050 mA =×

=

iii) The power received is 2/122440803.0 mWPs =×=

With an efficiency of 16.4%, the electrical power developed from the solar power =

2/2001224164.0 mWPe =×=

Therefore, with 1500 m2, the electrical power,

PekW3002001500 =×=

With 5 hours of sunshine, the energy produced kWh15005300 =×=

5.20For your city, search the Internet for solar irradiation data, G, on the horizontal surface and its latitudinal location. Perform the following:

i) Write a MATLAB M-file to (a) compute the irradiation on different inclination angle, (b) tabulate the irradiance for each month at different inclination angles, (c) tabulate the overall irradiance per year for different inclination angles, and (d) find the optimum inclination angle for each month and a year. ii) If the sun irradiance is 0.3 sun on the average over a year for 5 hours daily for

this location what is the total kWh that can be captured over 1500 ft2 at the optimum inclination angle? Solutions Los Angeles, California

The optimum tilt angle for each month and the entire year is found for Los Angles, California. The city is located at 33.93° N. The global irradiation of the city is given in table below: Table Mean monthly Global Radiant Exposure (kWh/m2/day) over Los Angeles, California http://rredc.nrel.gov/solar/pubs/redbook/PDFs/CA.PDF, accessed Oct 8, 2011 Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec Annual 2.8 3.6 4.8 6.1 6.4 6.6 7.1 6.5 5.3 4.2 3.2 2.6 4.9 Following the above algorithm, the solar irradiation on Los Angeles for different tilt angles for each month and annual average is given in tables below. The optimum irradiations are highlighted for each month in the tables.

Table Irradiation data for Los Angeles, California: Winter Tilt Angle January Wh/m2/day February Wh/m2/day

March Wh/m2/day

0 2800 3600 4800 5 3068 3848 4995 10 3318 4074 5160 15 3548 4276 5294 20 3756 4451 5396 25 3940 4599 5465 30 4100 4719 5500 35 4234 4809 5502 40 4341 4868 5469 45 4420 4898 540350 4471 4896 5304 55 4493 4864 5172 60 4486 4801 5009 65 4450 4708 4816 70 4385 4585 4594 75 4292 4434 4345 80 4171 4256 4071 85 4025 4052 3774 90 3852 3823 3457

Table Irradiation data for Los Angeles, California: Spring Tilt Angle April Wh/m2/day May Wh/m2/day

June Wh/m2/day

0 6100 6400 6600 5 6202 6394 6545 10 6268 6356 6459 15 6296 6284 6342 20 6287 6178 6193 25 6240 6040 6013 30 6155 5869 5804 35 6033 5668 5567 40 5876 5438 5303 45 5683 5180 501550 5458 4896 4705 55 5201 4590 4376 60 4914 4263 4030 65 4601 3919 3672 70 4262 3561 3305 75 3903 3193 2934 80 3524 2819 2563 85 3131 2444 2201 90 2727 2076 1857

Table Irradiation data for Los Angeles, California: Summer Tilt Angle July Wh/m2/Day

August Wh/m2/day

September Wh/m2/day

0 7100 6500 5300 5 7057 6547 5448 10 6977 6557 5564 15 6862 6529 5647 20 6710 6464 5695 25 6522 6361 5709 30 6300 6221 5687 35 6045 6046 5632 40 5758 5836 5542 45 5443 5593 5419 50 5102 5319 5263 55 4737 5016 5076 60 4352 4687 4859 65 3951 4335 4614 70 3539 3963 4343 75 3119 3574 4047 80 2698 3172 3729 85 2282 2763 3392 90 1883 2350 3039

Table Irradiation data for Los Angeles, California: Autumn Tilt Angle October Wh/m2/day

November Wh/m2/day

December Wh/m2/day

Annual monthly average Wh/m2/day

0 4200 3200 2600 4933 5 4436 3477 2869 5074 10 4646 3734 3121 5186 15 4828 3967 3355 5269 20 4980 4176 3567 5321 25 5101 4359 3758 5342 30 5190 4514 3925 5332 35 5247 4641 4066 5291 40 5270 4737 4182 5218 45 5261 4804 4271 5116 50 5219 4839 4332 4984 55 5144 4843 4365 4823 60 5036 4815 4370 4635 65 4898 4757 4346 4422 70 4729 4668 4294 4186 75 4531 4549 4215 3928 80 4306 4401 4108 3652 85 4055 4225 3974 3360 90 3780 4022 3816 3057

ii) With the tilt angle set at the optimal value of 25°, day/m/Wh603,153423.0P 2

s =×=

The total area = 22 m13910.761500ft1500 ==

kWh received = kWh103.81101603139365nirradiatioareadays 33 ×=×××=×× −

5.21. For a PV module given below (Table 5.52), write a MATLAB simulation testbed

using the Gauss–Seidel iterative approximation toestimate the module parameters .

TABLE 5.52 Data for Problem 5.21. (Isc) 3.87 A (Voc) 42.1 V (VMPP) 33.7 V (IMPP) 3.56 A (nc) 72

Solutions Let the known and unknown quantities be represented with the following variables:

Datasheet values

Isc a1 Short‐circuit current

Voc a2 Open circuit voltage

Vmpp a3 Voltage at MPP

Impp a4 Current at MPP

ns a5 Number of cells in series in a module

Unknown parameters

Iph x1 Photo‐generated current

Io x2 Dark saturation current

Vt x3 Junction voltage

Rs x4 Series resistance

Rsh x5 Parallel resistance

Rsho x6 Effective resistance at short circuit

Output quantities of the PV source

I y1 Output current

V y2 Output voltage

P y3 Output power

The Gauss-Seidel method is an iterative method, by which transcendental equations can be

solved by expressing the equations in the following form,

( )kk xfx =+ 1 (Sol.5.1)

where x is the unknown variable whose value is to be determined and k is the kth iteration. The iterations are repeated until the solution reaches below an acceptable tolerance.

The relationship between the voltage and current of PV is given by (5.26) in the text.

The data sheet provides three remarkable points: open circuit voltage, short circuit current and the

maximum power point (MPP). By substituting these points in (5.26), we get the equations (5.29)

through (5.33) given in the text.

Expressing these equations in the form of (Sol.5.1), the following algorithm is followed to

estimate the parameters of the PV:

Fig. Algorithm for estimating the parameters of PV from Gauss Seidel method. [Abir Chatterjee, Ali Keyhani, Dhruv Kapoor, "Identification of Photovoltaic Source Models", IEEE Transactions on Energy Conversion, Vol. 26, No. 3, Sep. 2011. pp. 883 – 889]

The f1(x),…………f5(x) can be expressed in residual from as

–

–

The initial values of x1,….x5 can initialized to zero. The Gauss-Seidel is an iterative approximation that increment the variable x by Xnew = Xold + Δx until residual is approximately equal to zero and all five equations for a value of x1 through x5 are satisfied. The derivation of f1(x) through f5(x) is given in reference below.

[Abir Chatterjee, Ali Keyhani, Dhruv Kapoor, "Identification of Photovoltaic Source Models", IEEE Transactions on Energy Conversion, Vol. 26, No. 3, Sep. 2011. pp. 883 – 889]

The Matlab program for determining the parameters of single diode model of PV is given below:

a1 = 3.87; a2 = 42.1; a3 = 33.7; a4 = 3.56; a5 = 72; k = 1.3806503e-23; Tstc = 298; q = 1.60217646e-19; convergentSolution = 0; i = 1; tolerance = 1e-6; x4 = 0; x5 = 0; while((i< 1000000) && (convergentSolution == 0))

x3 = (a3+a4*x4-a2)/log(-(a4*x5-a1*x5+a3+a4*x4-a1*x4)/(a1*x5- a2+a1*x4))/a5;

x4 = (-a3+a2+log(a5*x3*(a4*x5+a4*x4-a3)/(-a4*a1*x5*x4+a4*a2*x4-a4*a1*x4^2+a3*x5*a1-a2*a3+a3*a1*x4))*a5*x3)/a4;

x5 = (a5*x3*x5+exp((a1*x4-a2)/a5/x3)*a1*x5*x4-exp((a1*x4-a2)/a5/x3)*a2*x4+exp((a1*x4-a2)/a5/x3)*a1*x4^2+x4*a5*x3)/(exp((a1*x4-a2)/a5/x3)*a1*x5-exp((a1*x4-a2)/a5/x3)*a2+exp((a1*x4-a2)/a5/x3)*a1*x4+a5*x3);

ifi ~= 1

if (norm(x5 - x5Previous) < tolerance) && (norm(x4 - x4Previous) < tolerance) && (norm(x3 - x3Previous) < tolerance)

convergentSolution = 1; end end x3Previous = x3; x4Previous = x4; x5Previous = x5; i = i + 1; end

ifconvergentSolution == 0 disp('Convergent solution not found') else x2 = (a1 - (a2 - a1*x4)/x5) * exp(-1*a2/(a5*x3)); x1 = x2 * exp(a2/(a5*x3)) + a2/x5; A = x3 * q / (k * Tstc); x = [x1; x2; A; x4; x5] end

Table below gives lists the parameters estimated from the above program. Parameters Values Iph 3.87 Io 0.322 mA A 1.398 Rs 0.473 Ω Rsh 1.367 kΩ

CHAPTER 6

MICROGRID WIND ENERGY SYSTEMS

Notes to Instructors: Additional Problems

Please note that Problems D.1 and D.2 are based on Appendix D, Wind Power.

Additional Problems

D.1 The Rayleigh distribution functions for different mean wind speeds are shown in Solution Figure D.1.1. The curve moves to the right for greater mean wind speeds (also greater values of the shape parameter (a), which means more days have high winds—hence, potentially more wind energy revenue.

Solution Figure D.1.1 Rayleigh Distribution Functions for Three Different Mean Wind Speeds.

Solution Figure D.1.2 Rayleigh Wind Speed Frequency Distribution Functions for Two Sites Each Having the Same Average Wind Speeds.

The wind speed frequency distributions shown in Solution Figure D.1.2 are obtained from wind speed data for a year of 10-minute means for two sites with similar average wind speed.

i) Compute the mechanical power generated for an hour of operation assuming s/m12v = and the air density is 1.2 kg/m3.

Solution

i) A MATLAB code has been written to solve the problem. The steps followed are

1. The values of v and ρ assigned. 2. ( )vf is defined.

3. The integration ( )∫∞

0

3 dv.v.vf is done using a trapezoidal rule where the differential

areas are considered to be small trapezoids. The area of each trapezoid is added to get the integration over the entire range.

4. ( )3

0

3RMS dv.v.vfV ∫

∞

= is solved.

5. 3max, ..

41

RMST VP ρ= is solved.

6. 36524max, ××= TPE is solved.

MATLAB Code:

clc;

vbar=12;

rho=1.2; % density of air

a=vbar/0.9;

v=0:1:50;

fv=2*v/14.77.*exp(-(v/14.77).^2); % f(v) is defined;

plot(v,fv);

grid on;

xlabel('v in m/s');

ylabel('f(v)');

% Integration using trapezoidal rule

I=0;

for v=0:50

fv_vt3=2*v/14.77.*exp(-(v/14.77).^2).*v.^3; % f(v)*v^3

fvp1_vp1t3=2*(v+1)/14.77.*exp(-((v+1)/14.77).^2).*(v+1).^3; % f(v+1)*(v+1)^3

I=I+(fv_vt3+fvp1_vp1t3)/2;

end

Vrms=I^(1/3)

PtMax=0.25*rho*Vrms^3

E=PtMax*365*24

Solution Figure D.1.3 depicts the wind speed data for Problem D.1.

0 5 10 15 20 25 30 35 40 45 500

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

v in m/s

f(v)

f(v) versus v

Solution Figure D.1.3 Wind Speed Data for Problem D.1.

( ) s/m84.39dv.v.vfV 3

0

3RMS == ∫

∞

23max, /975.18..

41 mkWVP RMST == ρ

yrmMWhPE T //22.16636524 2max, =××=

D.2. Assuming the annual wind speed data for a site are given in Solution Figure D.2.1 with an annual mean speed of , estimate the average annual power and energy density for that site.

Solution Figure D.2.1 Wind Speed Data for Problem D.2.

Solution

To find the average annual power and energy density, the RMC speed needs to be calculated .The annual average wind speed is given as

Assuming a Rayleigh distribution with a shape parameter a equal to

the Rayleigh distribution function which fits the wind data is

where f is expressed in relative frequency. The RMC wind speed is then found by evaluating the following integral using trapezoidal numerical integration

.

Hence,

The wind power density which can be captured by the wind turbine assuming maximum theoretical rotor power coefficient cp = 0.5 and air density ρ = 1.225 kg/m3 is

and the site estimated annual energy yield per square meter of the turbine rotor swept area is

6.1 Consider a wind microgrid given by Figure 6.45. The system has a local load rated 100 kVA at a power factor rated 0.8 lagging.

Figure 6.45 The Schematic of Problem 6.1

The three-phase transformer is rated 11 kV/0.44 kV; 300 kVA; X = 0.06 p.u. The induction generator is rated as 440 V, 60 Hz, three-phase, eight-pole; stator resistance of 0.08 ohm/phase; rotor referred resistance in the stator side of 0.07 Ω/phase, stator reactance of 0.2 Ω/phase; rotor referred reactance of X2 0.1 Ω/phase. Compute the following:

i) Compute the p.u equivalent model based on kVA base of 300 kVA and 440 V

ii) The shaft mechanical power if the shaft speed is at 1200 rpm

iii) The amount of power injected into the local utility

iv) The flow of reactive power between the grid and the local microgrid

v) How much reactive power must be placed at the local grid to have unity power factor at the local power grid?

Solutions

Because the base chosen is that of the transformer, its p.u values remain unchanged.

i) The base impedance of the induction machine: Ω=×

== 645.010300

440SV

Z 3b

2b

b

The p.u value of stator resistance = Ω=== u.p124.0645.008.0

ZR

Rb

spu,s

The p.u value of rotor resistance referred to primary = Ω=== u.p109.0645.007.0

ZR

Rb

'r'

pu,r

The p.u value of stator reactance = Ω=== u.p310.0645.0

2.0ZX

Xb

spu,s

The p.u value of rotor resistance referred to primary = Ω=== u.p155.0645.0

1.0ZXX

b

'r'

pu,r

The p.u load = kVAu.p87.36333.08.0cos333.0300100

SS

S o1

b

loadpu ∠=∠=== −

The p.u transformer impedance = Ω= u.p06.0jZ pu,trans

Solution Figure 6.1.1 Per Unit Model for power flow studies of Problem 6.1

1R

V

1Xj 2Xj

grid

+

-

motorIQP

genI

2R

LTemT

backE+

-

Xj tran

load

Fig. Per unit model for power flow studies with the back emf represented by a voltage source

ii) Synchronous speed = rpm9008

60120P

f120Ns =×

==

Slip = 33.0900

1200900N

NNs

s

s −=−

=−

=

The supply voltage = 440 V = 1 p.u

The base current = A65.393440310300

V3VA

I3

b

bb =

××

=×

=

p.u impedance = ( ) ( )'pu,rpu,spu,tran'

pu,rpu,su.p XXXjs/RRZ ++++=

( ) ( ) Ω°∠=+++−= u.p45.11156.0155.031.006.0j33.0/109.0124.0

Let the voltage at the utility bus be selected to be the reference Vu.p01 °∠

The p.u current in motor convention = ou.p

u.p,gridu.p,motor 45.11156.0

01Z

VI

∠°∠

==

Au.p45.11179.1 o−∠=

With generator convention, the direction of current is reversed:

( ) Au.p55.6879.145.11118079.1II ou.p,motoru.p,gen °∠=−∠=−=

Back emf in p.u = ( )s

s1R.IE u.p,ru.p,motoru.p,back−

=

( ) Vu.p55.6878.0333.0333.01109.045.11179.1 °∠=

−+

××°−∠=

It is found that the back emf, Eback is in phase with the current Imotor. That is because this

voltage is over the resistance ( )s

s1R r− . Since slip is negative, the power from this

resistance is negative. In other words, power is generated. Now, since the voltage and current are in phase, the power is completely active without any reactive component. The reactive power required by the transformer and the stator and rotor reactances comes from the grid.

The complex power flowing from the back emf towards the grid = *u.p,genu.p,back I.E

( ) kVAu.p0j40.155.6879.155.6878.0 * +=°∠×∠=

The power from the back emf of the generator is only active power and does not have any reactive component.

The complex power flow from the grid towards the generator = *u.p,motoru.p,grid I.V

( ) kVAu.p67.1j65.045.11179.101 * +−=−∠×∠=

Therefore, the generator consumes 1.67 p.u kVar from the grid while supplying 0.65 p.u kW of power.

Fig. Phasor Diagram for induction generator

The current in generator convention = °∠×=×= 55.6879.165.393III .u.pbgen

A55.68705 °∠=

The complex power injected by the generator *u.p,genu.p,genu.p,gen I.VS ==

( ) ( ) kVAu.p67.1j65.055.6879.10155.6879.101 * −=−∠×∠=∠×∠=

The power in absolute values ( ) kVA500j19667.1j65.0300SS u.p,genb −=−×=×=

Algebraically, we find that the reactive power is negative. That indicates that the flow of reactive power is from the grid towards the induction generator.

Therefore active power injected from the generator = 196 kW

Therefore, the reactive power drawn by the generator = 500 kVar

The copper loss = ( ) ( ) kW7507.008.0705RRI 2'rs

2 =+×=+

The shaft mechanical power input (neglecting the mechanical and iron losses) =

Power injected + the copper losses = kW27175196 =+

iii) The speed of the machine is above the synchronous speed. Therefore, the machine generates active power.

The power generated by the machine = 196 kW

The power consumed by the local load = kW808.0000,100 =×

Power factor angle of the load = °=− 87.368.0cos 1

Because there are no loss components in the transformer, the power supplied to the utility is the balance of the power produced by the generator and the power consumed by the load = .kW11680196 =−

iv) The reactive power supplied by the utility is the sum of the reactive powers of the load, the transformer, and the generator = ( ) 50087.36sin100 +

kVar560=

v) Since the microgrid consumes reactive power, a reactive power source of 560 kVar should be placed at the terminals of the microgrid to obtain unity power factor for microgrid of wind generator.

6.2 The microgrid of Figure 6.46 is supplied by an induction generator. The system has a local load rated 100 kVA at a power factor rated 0.8 lagging. The three-phase transformer is rated 11 kV/0.44 kV; 300 kVA, and reactance of 6%. The induction machine rated at 440 V, 60 Hz, three-phase, eight-pole; 500 kVA, 440V, 60 Hz; stator resistance of 0.1 Ω/phase; rotor referred resistance in the stator side of 0.1 Ω/phase, stator reactance of 0.8 Ω/phase; rotor referred reactance of 0.4 Ω/phase. Compute the following:

i) Compute the per unit power flow model and short circuit model based on a base of 500 kVA and 440 V

ii) If the speed of the induction generator is 1000 rpm, what is the rotor frequency?

iii) The flow of active and reactive power between the microgrid and the local power grid

Figure 6.46 The System of Problem 6.2.

i) Sb = 500 kVA, Vb = 440 V on the LV side of the transformer

The base impedance of the induction machine: Ω=×

== 387.010500

440VAV

Z 3

2

b

2b

b

The p.u value of stator resistance = Ω=== u.p258.0387.0

1.0ZR

Rb

spu,s

The p.u value of rotor resistance referred to primary =

Ω=== u.p258.0387.0

1.0ZR

Rb

'r'

pu,r

The p.u value of stator reactance = Ω=== u.p07.2387.0

8.0ZX

Xb

spu,s

The p.u value of rotor resistance referred to primary= Ω=== u.p03.1387.0

4.0ZXX

b

'r'

pu,r

The new p.u value of the transformer reactance = 2

new,b

old,b

old,b

new,bold,punew,pu V

V.

SS

.XX ⎟⎟⎠

⎞⎜⎜⎝

⎛=

Ω=⎟⎠⎞

⎜⎝⎛×

××

×= u.p1.0440440

103001050006.0

2

3

3

The p.u load = kVAu.p87.362.08.0cos2.0500100

SS

S o1

b

loadpu ∠=∠=== −

It is assumed that the local utility has a p.u reactance of Xutility

The p.u model for short circuit studies will include the Thevenin’s equivalent impedance of the utility however, the model for power flow studies, it will not be accounted for.

Solution Figure 6.2.1 Per Unit Model for the short circuit studies of Problem 6.2.

Solution Figure Per Unit Model for the power flow studies of Problem 6.2.

ii) Synchronous speed = rpm9008

60120P

f120Ns =×

==

Slip = 111.0900

1000900−=

−=

−=

s

s

NNN

s

The rotor voltage frequency = Hz67.660111.0f.s =×=

iii) The supply voltage = 440 V = 1 p.u

The base current = A656440310500

V3VA

I3

b

bb =

××

=×

=

p.u impedance = ( ) ( )'pu,rpu,spu,trans'

pu,rpu,su.p XXXjs/RRZ ++++=

( ) ( ) Ω°∠=+++−= u.p12381.303.107.21.0j111.0/258.0258.0

The p.u current in motor convention = °−∠=°∠

== 12326.012381.3

1ZV

Iu.p

u.ppu,motor

With generator convention, the direction of current is reversed:

( ) Au.p5726.012318026.0II ou.p,motoru.p,gen °∠=−∠=−=

Back emf in p.u current = ( )s

s1R.IE u.p,rmotoru.p,back−

=

( )°∠==

−+

××°−∠= 5767.0111.0111.01258.012326.0

It is found that the back emf, Eback is in phase with the current Imotor. That is because this

voltage is over the resistance ( )s

s1R r− . Since slip is negative, the power from this

resistance is negative. In other words, power is generated. Now, since the voltage and current are in phase, the power is completely active without any reactive component. The reactive power required by the transformer and the stator and rotor reactances comes from the grid.

The complex power flowing from the back emf towards the grid = *u.p,genu.p,back I.E

( ) kVAu.p0j17.05767.05726.0 * +=°∠×∠=

The power from the back emf of the generator is only active power and does not have any reactive component.

The complex power flow from the grid towards the generator = *u.p,motoru.p,grid I.V

( ) kVAu.p218.0j142.012326.001 * +−=−∠×∠=

Therefore, the generator consumes 0.218 p.u kVar while supplying 0.142 p.u kW of power.

Fig. Phasor Diagram for induction generator

The absolute value of current in motor convention = u.p,motorbmotor III ×=

A1236.17012326.0656 °−∠=°−∠×=

With generator convention, the current is reversed ( )1231806.170Igen −∠=

A576.170 °∠=

The complex power injected by the generator *u.p,genu.p,gridu.p,gen I.VS ==

( ) kVAu.p218.0j142.05726.001 * −=∠×∠=

The power in absolute values ( ) kVA109j71218.0j142.0500SS u.p,genb −=−×=×=

Algebraically, we find that the reactive power is negative. That indicates that the flow of reactive power is from the grid towards the induction generator.

Therefore active power injected by the generator = 71 kW

Therefore, the reactive power drawn by the generator = 109 kVar

The power consumed by the local load = kW808.0000,100 =×

The power factor angle of the load = °=− 87.368.0cos 1

Therefore, the power supplied from the utility = kW97180 =−

The reactive power supplied by the utility = kVar16987.36sin100109 =×+

6.3 A six-pole wound rotor induction machine rated at 60 Hz, 380 V, 160 kVA. The induction machine has a stator and referred rotor resistance of 0.8 Ω/phase and stator and rotor reactance of 0.6 Ω/phase. The generator shaft speed is 1500 rpm. Determine how much resistance the rotor circuit must have to operate the generator at 1800 rpm.

Solution

Synchronous speed = rpm12006

60120P

f120Ns =×

==

Slip at 1500 rpm = 25.01200

15001200N

NNs

s

s1500 −=

−=

−=

Slip at 1800 rpm = 5.01200

18001200N

NNs

s

s1800 −=

−=

−=

For the current to stay fixed at the same value, the effective impedance should remain the same. Let '

extR be the external resistance used to control speed.

( ) ( ) ( ) ( )2'rs

2

1800'

ext'

rs

2'rs

2

1500'

rs XXs/RRRXXs/RR ++++=+++

( ) ( ) Ω=−−×−=−×= 8.08.025.0/8.05.0Rs/RsR 'r1500

'r1800

'ext

6.4 A 400 V, 3-phase Y-connected induction generator has the following data.

The generator is connected to a local power grid. Perform the following:

i) The maximum active power that the generator can supply.

( ) phasejZ /2.16.01 Ω+=

( ) phasejZ /3.15.0'2 Ω+=

ii) The reactive power flow between the induction generator and the local power grid.

Solutions

From (6.60), the stator current =

( )2'21

2'2

1

1

21

1'1

XXs

RR

VZZ

VI

++⎟⎟⎠

⎞⎜⎜⎝

⎛+

=+

=

The phase voltage = V2313

400V1 ==

( )22

'1

3.12.1s5.06.0

231I

++⎟⎠⎞

⎜⎝⎛ +

=

From (6.55), the power supplied by the grid =

( ) ( )sfunctionIRs

s1Ps1RIPP 2'2

'2AG

'2

2'2AGem =

−=−=−=

To find the slip for maximum power, the derivative of Pgrid is found and equated to zero:

( ) ( )[ ]( )

( ) ( )[ ]( )

3036.0or2015.03.12.16.0

6.03.12.16.05.06.05.06.0s

XXRRXXR'RR'RR

s

0ds

dP

22

2222

211

21

22

211

21

22121

grid

−=+−

×+−−×±×−=

+−

+−−±−=⇒

=

For generator operation, s is negative.

Therefore, s = -0.2015

Writing Kirchhoff’s voltage law (KVL) for the circuit, the current in motor convention,

( )'XXjs/'RRV

I2121

gridm +++=

( ) °−∠=++−

= 127743.12.1j2015.0/5.06.0

3/400

With current following the generator convention and flowing from an induction generator to the grid, the current ( ) .537412718074IG °∠=°−∠=

The power flow at the grid terminal = *Ggridgrid I.V3S = ( )*53653/4003 °∠××=

°−∠××= 53653/4003 VAk36j27 −=

i) Therefore, the active power injected to the grid = 27 kW

ii) Therefore, the reactive power injected by the grid = 36 kVar

6.5 Design a 15 kW wind power generator that is supplied from a variable wind speed. The designed system must provide 220 VAC, single-phase AC power. Compute the DC bus voltage.

Solution

The system will consist of an inverter and a rectifier.

For the inverter from (3.4),

2/V.MV dcainv,ac =

V7.3459.0/2202M/V.2V ainv,acdc =×== , assuming Ma = 0.9

For the rectifier from (3.77),

π= /V.23V acdc

V25623/7.34523/VV dcac =π=π=

Hence, the rectifier should be rated at DC voltage of 345 V and AC voltage of 256 V

The inverter should be rated at 345.7 V DC and 220 V AC, 15 kVA assuming unity power factor.

Both the inverter and the rectifier should be rated at 15 kVA.

6.6 The same as Problem 6.5, except the wind generating system must provide three- phase AC nominal voltage of 210 VAC . Compute the DC bus voltage.

Solution

For the three-phase inverter (from 3.83),

2V

23.MV dc

ainv,LL,ac =−

Assuming 0.8 modulation index,

V66.42821032

8.02V

32.

M2V inv,LL,ac

adc =××==⇒ −


π= /V.23V acdc

V31723/66.42823/VV dcac =π=π=

Hence, the rectifier should be rated at DC voltage of 429 V and AC voltage of 317 V

The inverter should be rated at 429 V DC and 210 V AC, 15 kVA assuming unity power factor.

Both the inverter and the rectifier should be rated at 15 kVA.

6.7 Design a 2 MW wind systems using a variable speed system. The DC bus voltage is at a nominal value of 600 VDC. The generators are located 5 miles from the local utility. The utility voltage is three-phase AC rated at 34.5 kV.

TABLE 6.2 13.2–132 kV Class One Phase–Neutral Return Line Model.

Conductor DC Resistance

(Ω/km)

Inductance (Ω/km)

XL

Susceptance (S/km)

YC

Current Ratings

Magpie 1.646 j 0.755 j 1.45e-7 100 Amp

Squirrel 1.3677 j 0.78 j 6.9e-7 130 Amp

Gopher 1.0933 j 0.711 j 7.7e-7 150 Amp

The data for the transmission line are given in Table 6.2. The data for the transformers are 460 V/13.2 kV 250 kVA 10% impedance and 13.2 kV–34.5 kV, 1 MVA 8.5% impedance. Perform the following:

i) Give the one-line diagram ii) Per unit model based on a rated wind generator

Solution

i) The DC voltage is 600 V. Therefore, the DC side of rectifier must be rated at 600 V.

Fig. The one line diagram

The utility voltage = 34.5 kV

The inverter voltage on the AC-side is chosen to be 460 V to match with the available transformer rating.

For the three-phase inverter (from 3.83),

2V

23.MV dc

ainv,LL,ac =−

If the DC side of the inverter is rated at 600 V, the modulation index,

25.146032

6002V

32.

V2M inv,LL,acdc

a =××== −

Since the modulation index works out to be greater than 1, a boost converter must be used to boost the voltage of the rectifier to a sufficient level for the inverter.

Let the modulation index of the inverter be chosen to be 0.9. Therefore, the DC side

voltage of the inverter = V83546032

9.02V

32.

M2V inv,LL,ac

adc =××== −

Let the inverters be rated at 1 MVA each.

The number of inverters in parallel 212

invertereachofratingMVAMVAtotal

===

Therefore, each inverter should be rated at 835 V DC, 460 V AC, 1 MVA.

Let the boost converter be rated at 835 V DC on the output side and 600 V on the input side.

The duty ratio of the boost converter = 28.08356001

VV1D

o

in =−=−=

Let each boost converter be rated at 1 MW

The number of boost converters in parallel 212

convertereachofratingMWMWtotal

===

Therefore, each boost converter should be rated at 600 V on the input, 835 V on the output, 1 MW of power.


π= /V.23V acdc

V44423/60023/VV dcac =π=π=

Therefore the rectifier should be rated 444 V AC and 600 V DC

Let each rectifier be rated at 1 MVA,

The number of rectifiers in parallel 212

invertereachofratingMVAMVAtotal

===

The current rating = A48.87102.133

102V.3

SI3

6

Lrated =

×××

==

Hence, the conductor chosen is “Magpie.”

The value of impedance = ( ) Ω+=×+×=+= 775.3j23.8755.0j646.15jXRZ

The value of admittance = S10725.01045.15Y 67 −− ×=××=

The number of 250 kVA transformers connected in parallel =

825.02

.traneachofratingMVAMVAtotal

===

The number of 1 MVA transformers connected in parallel =

212

.transeachofratingMVAMVAtotal

===

ii) Let Sb = 2 MVA and Vb = 34.5 kV on the local power grid side.

The base voltage on LV side of 13.2 / 34.5 kV transformer

= kV2.135.345.342.13V

VV

b,HVHV

LV =×=

The per unit impedance of the transformer = ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

old,b

new,b

2

new,b

old,bold,b S

SVV

Z

Ω=⎟⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛×= u.p170.0

12

2.132.13085.0j

2

The base voltage on LV side of 0.46 / 13.2 kV transformer

= V4602.132.13

46.0VVV

b,HVHV

LV =×=

The per unit impedance of the transformer = ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

old,b

new,b

2

new,b

old,bold,b S

SVV

Z

Ω=⎟⎠⎞

⎜⎝⎛×⎟

⎠⎞

⎜⎝⎛×= u.p8.0

25.02

2.132.131.0j

2

The base impedance on the transmission line = Ω=== 12.8722.13

SV

Z2

b

2b

b

The p.u impedance on the transmission line =

Ω+=+

= u.p043.0j094.012.87

775.3j23.8Z u.p,tran

The base admittance on the transmission line = S011.0Z1Y

bb ==

The p.u admittance on the transmission line =

Su.p109.65011.0

10725.0Y 66

u.p,tran−

−

×=×

=

The base voltage on DC side of inverter = V835460460835V

VV

b,ACAC

DC =×=

The base voltage on DC side of rectifier = V600835835600V

VV

b,inv,ACinv,DC

rec,DC =×=

The base voltage on generator bus = V444600600444V

VV

b,DCDC

gen =×=

Solution Figure 6.7.1 Per Unit model

6.8 A wound rotor six-pole 60 Hz induction generator has stator resistance of 1.1 Ω/phase and rotor resistance of 0.8 Ω and runs at 1350 rpm. The prime mover torque remains constant at all speeds. How much resistance must be inserted in the rotor circuit to change the speed to 1800 rpm? Neglect the motor leakage reactance, X1 and X2.

Solution

The air-gap power from (6.55 and 6.60) is given by:

( )2'21

2'2

1

21

'2

'22'

2g

XXs

RR

Vs

R3

sR

I3P

++⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

Torque =

( )2'21

2'2

1

21

s

'2

s

g

XXs

RR

Vs

R3P

++⎟⎟⎠

⎞⎜⎜⎝

⎛+

ω=

ω

For torque to remain same, s'R 2 must remain constant.

Synchronous speed = rpm12006

60120P

f120Ns =×

==

Slip at 1500 rpm = 5.01200

180012001800 −=

−=

−=

s

s

NNN

s

Slip at 1500 rpm = 125.01200

135012001350 −=

−=

−=

s

s

NNN

s

Let 'insertR be the external inserted resistance

5.0RR

125.0R '

insert'2

'2

−+

=−

Ω=−=−= 4.28.0)8.0(4RR125.0

5.0R '2

'2

'insert

6.9 Consider a three-phase Y-wound rotor connected induction generator rated 220 V, 60 Hz, 16 hp, eight-pole with the following parameters:

R1 = 1 Ω/phase and X1 = 1.6 Ω/phase

R2’ = 0.36 Ω/phase and X2=1.8 Ω/phase

The generator is connected to the local power grid.

Write a MATLAB simulation testbed to plot slip and speed as a function of machine torque and various external inserted resistance in the rotor circuit. Make the plot for a value of external resistance of 0.0, 0.4, 0.8, and 1.2 Ω.

Solution

Program Steps:

1. The values of supply voltage, the stator and rotor resistances and reactances are assigned

2. The supply frequency is defined and the synchronous speed is calculated. 3. The value of external resistance is varied from zero to 1 in the steps of the 0.5

calculation 4. The speed of the machine is varied from zero to the synchronous speed 5. The plot is made for the different values of Rext.

MATLAB Program:

%TORQUE vs SPEED

clc; clear all;

v1=220/sqrt(3);

f=60;

P=8;

r1=1;

x1=1.6;

r2d=.36;

x2d=1.8; % The electrical quantities are defined

ws=120*f/P;

Tmax=-(3/2/ws)*v1^2/(r1+sqrt(r1^2+(x1+x2d)^2))

Tmax_gen=-(3/2/ws)*v1^2/(r1-sqrt(r1^2+(x1+x2d)^2))

w=0:1:2*ws;

for r_ext=0:0.4:1.2 % the value of external resistance is varied

Tstart=-(3/ws)*((r2d+r_ext)/1)*v1^2/((r1+(r2d+r_ext)/1)^2+(x1+x2d)^2)

smax=(r2d+r_ext)/sqrt(r1^2+(x1+x2d)^2)

for j = 1:length(w)

s(j)=(ws-w(j))/ws;

Tem(j)=-(3/ws)*((r2d+r_ext)/s(j))*v1^2/((r1+(r2d+r_ext)/s(j))^2+(x1+x2d)^2);

end

plot(w,Tem,'k','linewidth',2)

hold on;

end

grid on;

xlabel('Speed')

ylabel('Electromagnetic Torque')

axis([0 2*ws 1.1*Tmax 1.1*Tmax_gen])

gtext('R_e_x_t=0')

gtext('R_e_x_t^,=0.4')

gtext('R_e_x_t^,^,=0.8')

gtext('R_e_x_t^,^,^,=1.2')

Solution Figure 6.9.1 depicts the plot of the torque versus the speed of the induction motor.

0 200 400 600 800 1000 1200 1400 1600 1800

-6

-4

-2

0

2

4

6

8

10

Speed

Ele

ctro

mag

netic

Tor

que

Rext=0 Rext, =0.4 Rext

,, =0.8 Rext,,, =1.2

Solution Figure 6.9.1 The plot of the Torque Versus Speed of the Induction Motor,

The results are tabulated in Solution Table 6.9.1.

Solution Table 6.9.1 Variation of starting torque and slip at maximum torque with external resistance

External Resistance

(Ω)

Starting Torque

(in N-m)

Slip at Maximum Torque

Maximum Torque or Breaking Torque (in N-m)

0 -1.44 0.10

-5.92 (Motor)

10.56 (Generator)

0.4 -2.79 0.21

0.8 -3.84 0.33

1.2 -4.63 0.44

CHAPTER 7

LOAD FLOW ANALYSIS OF POWER GRIDS AND MICROGRIDS

7.1 A three-phase generator rated 440 V, 20 kVA is connected through one cable with impedance of 1+j0.012 Ω to a motor load rated 440 V, 15 kVA, 0.9 p.f. lagging. Assume the load voltage is set at 5% above its rated value. Perform the following:

i) Give three-phase circuit if the load is Y connected

ii) Give the three-phase circuit if the load in Δ connected

iii) Give a one-line diagram

iv) Compute the generator voltage

Solutions

i)

ii)

iii)

Solution Figure 7.1.1The One-Line Diagram of Problem 7.1.

iv) For both the Y and Δ connections:

√ √ ..

.√

. . . .

. . °

7.2 A three-phase generator rated 440 V, 20 kVA is connected through one cable with impedance of 1+j0.012 Ω to a Δ-connected motor load rated 440 V, 10 kVA, 0.9 p.f. lagging. Assume the generator voltage is controlled at its rated voltage and its phase angle is used as the reference angle. Perform the following:

i) What is the number of unknown variables?

ii) How many equations are needed to solve for bus voltage? Give the expressions.

iii) Compute the load bus voltage

Solutions

Let Sb = 20 kVA, Vb = 440 V

Ω== 6892

.S

VZ

b

bb

Ω+=+

== u.p.j..

.jZ

ZZ

b

cablecable,u.p 0012010330

68901201

Load power factor angle = °==θ − 8425901 ..cos

Load active power = u.p..P load,u.p 45020

9010=

×=

Load reactive power = u.p..sinQ load,u.p 218020

842510=

×=

Fig. One-line diagram for problem 7.2

Solution Figure 7.2.2. Per unit injection model for power flow studies for problem 7.2

i) The number of unknowns is two: the bus voltage and its angle at load bus 2 assuming the generator bus 1 is a swing bus.

( ) 011221 =−− IyVV

where y12 = 1/ z1-2

*

1

11 V

SI ⎟⎟

⎠

⎞⎜⎜⎝

⎛= ,

*

2

22 V

SI ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

1212112 Ivyvy =−

The above can be written as,

⎥⎦

⎤⎢⎣

⎡−

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

2

1

2

1

2221

1211

II

VV

.YYYY

where 1211 yY = , 1212 yY −= , 1221 yY −= , 1222 yY =

The matrix equation represents the bus admittance matrix, which is also known as the YBus model for Problem7.2.

ii) The YBus matrix is described as

BusBusBus V.YI =

If the system has n buses, IBus is a vector of 1n× current injection, VBus is a voltage vector of 1n× and YBus is a matrix of nn× .

For Problem 7.2, we have two buses. Therefore, the YBus is complex matrix with dimension of 22×

Generally, for a power grid with n buses, for each bus k, we have

*kkk IVS = 2,1k =

where Ik is the current injection into the power grid at bus k. Therefore, from the row k of the YBus matrix, we have

∑=

=2

1jjkjk VYI

Substituting, we have

*2

1jjkjkk VYVS ⎟⎟⎠

⎞⎜⎜⎝

⎛= ∑

=

2,1k =

Where Sk is the net injection.

For each bus, k, we have a complex equation. Therefore, we have two nonlinear complex equations.

⎭⎬⎫

⎩⎨⎧

= ∑=

2

1j

*j

*kjkk VYVReP

⎭⎬⎫

⎩⎨⎧

= ∑=

2

1j

*j

*kjkk VYVImQ

where ( )jjjj sinjcosVV θ+θ=

( )kkkk sinjcosVV θ+θ=

For Problem 7.2, N=2, we have two complex nonlinear equations or four nonlinear equations in real domain. However, because the generator bus voltage magnitude is given and it is used as a reference with a phase angle of zero, we have two nonlinear equations in real domain.

iii) In Problem 7.2 (See Solution Figure 7.2.2), we have the feeder impedance, z1-2, and load, S2. To find the bus load voltage, we need to solve the two nonlinear equations for V2, and θ2. After calculating the bus voltages, we can calculate the complex power (S1 = P1 + jQ1) injected by the local utility feeder.

The Gauss-Seidel method formulation in real domain is given in problem, 7.13. In the complex domain the steps are as follow:

The calculated real and imaginary power injections at each bus are as follows

( )⎭⎬⎫

⎩⎨⎧

= ∑=

2

1j

*j

*kjkcalculatedk VYVReP

( )⎭⎬⎫

⎩⎨⎧

= ∑=

2

1j

*j

*kjkcalculatedk VYVImQ

After calculating the real and imaginary powers, the mismatches are calculated as

follows:

( ) ( )calculated,kscheduledk PPP −=Δ

( ) ( )calculated,kscheduledk QQQ −=Δ

After, the power flow problem converges and bus voltages are known, we can calculate the power losses using the power flow balance equation as given below. We use the load bus voltage and swing bus voltage to calculate the line flow.

21 SSSSS loadgenloss +=−= ∑∑

( )[ ] ( )[ ]*,sh*

,sh* V.yVV.Y.VV.yVV.y.VI.VS 121211211212121121121 −−−−− +−−=+−==

( ) *,sh

*** y.VVV.Y.V 212

121121 −+−−=

The Matlab program for Gauss Seidel YBus method to calculate load bus voltage and power losses and line flow is given below:

clc; clear all; tolerance=1e-5; PV=0; %no. of PV buses Vb=440; %base voltage VAb=20e+3; %base volt-amp Zb=Vb^2/VAb; %base impedance Yb=1/Zb; %base admittance % Scheduled active and reactive powers P_sch = [0 -10e3/VAb*sin(acos(0.9))]'; Q_sch = [0 P_sch(2)*tan(acos(0.9))]'; %% Admittance matrix formation % line data line_data = [ 1 2 1 0.012 0]; N = max(max(line_data(:,1:2))); Y_BUS = zeros(N,N); % assemble admittance matrix for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); Z = (line_data(line,3) + 1j*line_data(line,4))/Zb; Y_shunt = 1j*line_data(line,5)/Yb; Y_BUS(i,i) = Y_BUS(i,i) + 1/Z + Y_shunt; Y_BUS(j,j) = Y_BUS(j,j) + 1/Z + Y_shunt; Y_BUS(i,j) = Y_BUS(i,j) - 1/Z;

Y_BUS(j,i) = Y_BUS(j,i) - 1/Z; end %% Power flow V=ones(N,1); theta=zeros(N,1); V_cplx = V .* exp(1j*theta); iteration=0; while (iteration <= 1000) iteration=iteration+1; for m=2:N VY=0; for n=1:N if m~=n VY=VY+Y_BUS(m,n)*V_cplx(n); end end Vm = ((P_sch(m)-1j*Q_sch(m))/conj(V_cplx(m))-VY)/Y_BUS(m,m); theta(m)=angle(Vm); % for P-V and P-Q buses if ( m >= 2+PV ) % for P-Q buses only V(m)=abs(Vm); end end V_cplx = V .* exp(1j*theta); S_calc = V_cplx .* conj(Y_BUS*V_cplx); P_calc = real(S_calc); Q_calc = imag(S_calc); mismatch=[P_sch(2:N)-P_calc(2:N); Q_sch(2+PV:N)-Q_calc(2+PV:N)]; if (norm(mismatch,'inf') < tolerance) break; end end %% output solution data %Bus quantities S_loss = 0; for i = 1:N fprintf(1, 'Bus %d:\n', i); fprintf(1, ' Voltage = %f p.u., %.1f deg.\n', ... abs(V_cplx(i)), angle(V_cplx(i))*180/pi); fprintf(1, ' Injected P = %f p.u.\n', P_calc(i)) if i > 1 fprintf(1, ' Scheduled P = %f p.u.\n', P_sch(i)); end fprintf(1, ' Injected Q = %f p.u.\n', Q_calc(i)) if i > PV+1 fprintf(1, ' Scheduled Q = %f p.u.\n', Q_sch(i));

end %Calculating the power loss S_loss = S_loss + S_calc(i); end P_loss = real(S_loss); Q_loss = imag(S_loss); %Line quantities fprintf(1, '\nPower flow from:\n'); for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); I_line = (V_cplx(i)-V_cplx(j))*(-Y_BUS(i,j)); Y_shunt = 1j*line_data(line,5)/Yb; Shunt_power_line = -abs(V_cplx(i))^2*Y_shunt; S_line = V_cplx(i)*conj(I_line) + Shunt_power_line; P_line = real(S_line); Q_line = imag(S_line); fprintf(1, 'bus %d to bus %d\n', i, j); fprintf(1, ' P = %.4f, Q = %.4f\n', P_line, Q_line); end %Total loss fprintf(1, '\nThe total active power loss = %f\nThe total reactive power loss = %f\n'... , P_loss, Q_loss); SOLUTION TABLE 7.2.1 Bus voltages using Gauss-Seidel YBus method for problem 7.2

kVA20Sb = , V440Vb =

Number of iterations is 3

Bus #

Volts

(p.u)

Angle

(Degrees)

Generation Load PΔ QΔ

MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 0.224 0.106 0 0 0 0

2 0.977 0.6 0 0 0.218 0.106 510416.0 −× 510005.0 −×

Calculation of losses

V.YI Bus=

[ ]BusYReG= , [ ]BusYImB=

YBus = G+j B Matrix

Using the YBus algorithm, we can calculate YBus model.

jijiji jxrz −−− +=

Ω+=+= −−− u.p.j.jxrz 0012010330212121

222222

11

jiji

ji

jiji

ji

jiji

jiji

jijijiji xr

xj

xr

r

xr

jxrjxrz

y−−

−

−−

−

−−

−−

−−−−

+

−+

+=

+

−=

+==

Now we formulate

[ ] [ ]BjGYYYY

YBus +=⎥⎦

⎤⎢⎣

⎡=

2221

1211

212211 −−− == yYY

211221 −−− −== yYY

For the YBus we have,

⎥⎦

⎤⎢⎣

⎡−

−=

689689689689

....

G , ⎥⎦

⎤⎢⎣

⎡−

−=

120120120120..

..B

⎥⎦

⎤⎢⎣

⎡°∠°−∠

=⎥⎦

⎤⎢⎣

⎡°∠

°∠

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−

−+⎥

⎦

⎤⎢⎣

⎡−

−=⎥

⎦

⎤⎢⎣

⎡8154240225240

60977001

120120120120

689689689689

2

1

....

......

j....

II

*kkk I.VS =

( ) kVAu.p10564.0j2243.02.2524.001I.VS **111 +=−∠×∠==

( ) kVAu.p10555.0j2179.08.15424.06.0977.0I.VS **222 −−=∠×∠==

( ) ( ) 0000800064010556021790105640224302

1.j..j..j.SS

nkloss +=−−++==∑

=

Line flow:

The line follow can be calculated as follows

( ) *,sh

*** y.VVV.Y.VS 212

12112121 −− +−−=

( ) ( ) 10602240016097700112068901 2 .j....j. +=×+−∠−∠×+×∠=

SOLUTION TABLE 7.2.2 The Power Flow through the feeder for problem 7.2.

Ploss (p.u) = 0.0064, Qloss (p.u) = 0.00008

From Bus #

To Bus #

MW Flow (p.u)

MVar Flow (p.u)

1 2 0.224 0.106

The power supplied by bus 1, the swing bus is equal to the total load minus the total generation by all other busses plus the losses.

2240006002180221 ...PPPP loss,G,Lbusswing =+−=+−=

106000008001060221 ...QQQQ loss,G,Lbusswing =+−=+−=

The bus voltage of the swing bus is 01∠ and the p.u power is 224.0P1 = p.u and Q1 is 0.106 p.u supplied from the generator.

7.3 The radial feeder of Fig. 7.24 is connected to a local utility rated at 11.3 kV distribution. Assume power base of 10 kVA and voltage base of 11.3 kV. Perform the following:

i) Compute the per unit model

ii) Write the number of equations that are needed to solve for the bus load voltages

iii) Use the Gauss-Seidel method and compute the bus voltages

iv) Compute the power at bus 1. Assume the power mismatch of 0.00001 per unit.

v) Compute the total active and reactive power losses

Figure 7.24The Radial Feeder for Problem 7.3.

Solutions

i) , Vb = 11.3 kV

,

, . , . . . .

, ..

. . . . .

, ..

. . . . .

Power flow injection model for problem 7.3

Per Unit model

ii) We have two load buses, therefore, we need to solve two complex nonlinear equations or four nonlinear equations in real domain as discussed in problem 7.2

iii) We use the Gauss-Seidel program given in solution 7.2. The YBus model is computed using YBus algorithm as given in problem 7.2. However, in this problem we have two load buses

jBGYBus +=

jijiji jxrz −−− +=

For example,

Ω+=+= −−− u.p.j.jxrz 0012000330212121

222222

11

jiji

ji

jiji

ji

jiji

jiji

jijijiji xr

xj

xr

r

xr

jxrjxrz

y−−

−

−−

−

−−

−−

−−−−

+

−+

+=

+

−=

+==

Now we formulate

[ ] [ ]BjGYYYYYYYYY

YBus +=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

333231

232221

131211

2111 −− = yY , 322122 −−− += yyY , 3233 −− = yY

211221 −−− −== yYY , 322332 −−− −== yYY

01331 == −− YY

Where G and B are given below

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−=

21221202124232120212212

G

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

795795079515907950795795

B

Using the program in solution 7.2, we solve the two complex nonlinear equations using

the Gauss-Seidel YBus method.

SOLUTION TABLE 7.3.1 Bus Voltages using Gauss-Seidel YBus method for problem 7.3

kVASb 10= , kV.Vb 311=

Number of iterations is 33

Bus #

Volts

(p.u)

Angle

(Degrees)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 1.231 0.668 0 0 0 0

2 0.999 -0.1 0 0 0.720 0.349 51079.0 −× 51049.0 −×

3 0.999 -0.1 0 0 0.510 0.316 51056.0 −× 51027.0 −×

Complex power from bus i to bus j:

( )[ ] ( )[ ]*jji,shjiiji*

iji,shjijii*

jiiji VyVV.Y.VVyVV.y.VI.VS −−−−− +−−=+−==

( ) *ji,shi

*j

*i

*iji y.VVV.Y.V −+−−= 2

SOLUTION TABLE 7.3.2 ThePower Flow through the feeders for problem 7.3

Ploss (p.u) = 0.001, Qloss (p.u) = 0.003

From Bus #

To Bus #

MW Flow (p.u)

MVarFlow (p.u)

1 2 1.231 0.668

2 3 0.510 0.317

iv) The bus voltage of the swing bus is 01∠ and the p.u power supplied from the grid is

( ) ( ) 23110007000510072003

2

3

21 ....PPPP loss

ii,G

ii,Lbusswing =++−+=+−= ∑∑

==

( ) ( ) 6680003000316034903

2

3

21 ....QQQQ loss

ii,G


==

v)

321 SSSSSS loadgenloss ++=−= ∑∑

The active and reactive power losses are:

000503

1

3

1.PPP

ii,L

ii,Gloss =−= ∑∑

==

00180

3

1

3

1.QQQ

ii,L


==

7.4 The radial feeder of Fig. 7.24 is connected to a local power grid rated at 11.3 kV distribution. Assume the base voltage of 15 kVA and a voltage base of 11.3 kV. Perform the following:

i) Compute the per unit model

ii) Write the number of equations that are needed to solve for the bus load voltages

iii) Compute the Y bus matrix

iv) Compute the matrix B′and B′′

v) Compute the bus voltages. Assume the power mismatch of 0.00001 per unit.

vi) Compute the power at bus 1

vii) Compute the total active and reactive power losses

Solutions

i) Sb = 15 kVA, Vb = 11.3 kV

, ,

, . , . . . .

, ..

. . . . .

, ..

. . . . .

Power flow Per unit injection model for problem 7.4

ii) We have to solve two nonlinear complex equations, one complex equation for each load bus. The number of nonlinear equations in real domain is equal to four.

iii) Using the YBus algorithm of problem 7.2, we can compute the YBus model asjBGYBus +=

Where G and B are given below:

2111 −− = yY , 322122 −−− += yyY , 3233 −− = yY

211221 −−− −== yYY , 322332 −−− −== yYY

01331 == −− YY

jBGYBus +=

iv) The matrix is the imaginary part of YBus and matrix is formulated same as YBus except the resistances and line charging are ignored in the formulation. In this problem, we do not have line charging.

1 1 1

1 1

matrix is the imaginary part of YBus give above, Since bus 1 is swing bus and its voltage (magnitude and angle) is given, first row and first column are not needed in and martrix for bus voltage calculation.

v) To calculate the bus voltages, fast decoupled load flow (FDLF) is used

The steps for FDLF are given below:

Step 1. Newton-Raphson

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ΔΔ

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡ΔΔ

VV

LJ

NHQP θ

Where:

Δθ = θ0 - θnew

ΔV = V0 - Vnew

ΔP = PSch – PCal

ΔQ = QSch – QCal

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

LJ

NHJ is Jacobian Matrix

Sk =Pk + j Qk = VkIk*= Vk ( )∑

=

5

1m

*mkmVY

If Ykm = Gkm + j Qkm

)sinBcosG(VVP kmkmkmkm

5

1mmkk θ+θ= ∑

=

)cosBsinG(VVQ kmkmkmkm

5

1mmkk θ−θ= ∑

=

If we assume J = 0, N = 0, then we have will have decoupled Newton-Raphson method.

[ΔP] = [H] Δθ

[ΔQ] = [L] [ΔV/V]

km)cosBsinG(VVHP

kmkmkmkmmkkmm

k ≠θ−θ==δ∂∂

k2

kkkk QVBHm−−=

k2

kkkk QVBLm+−=

The angle θ12 is very small for well define system, then we will have

kmkkm BVVHm

−= m ≠ k

If 2kkk m

VB<<Θ

2m

VBH kkkk −=

''kkk

5

1mkmSiik

5

1mkmkkkkkmkmkmkm

5

1mmkkkkm

5

1mmkkk

k

kkk

BV)BB2(V

BVBV2)]cosBsinG(V[BV2YVBV2VQ

L

=+−=

+−≈δ−δ−−=−−=∂∂

=

∑

∑∑∑

=

===

''kmkkmkkmk

m

kkm BVBVYV

VQ

L =≈−=∂∂

=

[ΔP] = [V B’ V] [Δθ]

[ΔQ] = [V B’’ V] [ΔV/V]

Where B’ = [YBus] is obtained by formulating YBus when resistance of the transmission lines and line changing are neglected.

B’’ = Img [YBus] including resistance and reactance for YBus calculation.

The above formulation is called fast decoupled load flow.

Solution Algorithms:

At the end of calculating the voltages of each bus in an iteration, the active and reactive power injections at each bus are calculated using the formulas:

( )∑=

θ+θ=N

jkjkjkjkjjkk sinBcosGVVP

1

( )∑=

θ−θ=N

jkjkjkjkjjkk cosBsinGVVQ

1

where kjkjkj jBGY += , jkkj θ−θ=θ

Now, the mismatch of power is calculated for each load bus by using the following:



The fast decoupled load flow given above is coded in Matlab for computing bus voltages and line flows and power losses. The Matlab code is given below. The code also computes the line flows and the power loss after the solution converges. All the calculations are done in real domain including the calculations for line flows and power losses.

( )[ ] ( )[ ]*jji,shjiiji*

iji,shjijii*

jiiji VyVV.Y.VVyVV.y.VI.VS −−−−− +−−=+−==

Separating the real and imaginary parts of YBus

jBGYBus +=

ijijij,Bus jBGY +=

Separating the real and imaginary parts of the voltage,

( )iiii sinjcosVV θ−θ=

Substituting in the equation of line flow,

( ) ( ) [ ]jiijjiijjijiiij sinBcosGVG.VVP θ−θ+θ−θ+−=

( ) ( ) [ ]jiijjiijjij,shijiiij sinGcosBVyB.VVQ θ−θ−θ−θ−−=

%% Power Flow: FDLF clc; clear all; tolerance=1e-5; PV = 0; %no. of PV buses Vb=11.3e3; %base voltage VAb=15e+3; %base volt-amp Zb=Vb^2/VAb; %base impedance Yb=1/Zb; %base admittance % Scheduled active and reactive powers P_sch = [0 -8e+3*0.9 -6e+3*0.85]'/VAb; Q_sch = [0 P_sch(2)*tan(acos(0.9)) P_sch(3)*tan(acos(0.85))]'; %% Admittance matrix formation % line data line_data = [ 1 2 4 15 0 2 3 4 15 0];

N = max(max(line_data(:,1:2))); G = zeros(N,N); B = zeros(N,N); Bd = zeros(N,N); % assemble admittance matrix for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); r = line_data(line,3)/Zb; x = line_data(line,4)/Zb; y_shunt = line_data(line,5)/Yb; G(i,i) = G(i,i) + r/(r^2 + x^2); B(i,i) = B(i,i) - x/(r^2 + x^2) + y_shunt; G(j,j) = G(j,j) + r/(r^2 + x^2); B(j,j) = B(j,j) - x/(r^2 + x^2) + y_shunt; G(i,j) = G(i,j) - r/(r^2 + x^2); B(i,j) = B(i,j) + x/(r^2 + x^2); G(j,i) = G(j,i) - r/(r^2 + x^2); B(j,i) = B(j,i) + x/(r^2 + x^2); Bd(i,i) = Bd(i,i) - 1/x; Bd(j,j) = Bd(j,j) - 1/x; Bd(i,j) = Bd(i,j) + 1/x; Bd(j,i) = Bd(j,i) + 1/x; end %% Power flow % allocate storage the matrix B Bd = Bd(2:N,2:N); Bdd = B(2+PV:N,2+PV:N); % initial values V = ones(N,1); theta = zeros(N,1); P_calc = zeros(N,1); Q_calc = zeros(N,1); for k = 1:N for m = 1:N P_calc(k) = P_calc(k)... + V(k)*V(m)*(G(k,m)*cos(theta(k) - theta(m)) + B(k,m)*sin(theta(k) - theta(m))); Q_calc(k) = Q_calc(k)... + V(k)*V(m)*(G(k,m)*sin(theta(k) - theta(m)) - B(k,m)*cos(theta(k) - theta(m))); end end mismatch_P = [P_sch(2:N)-P_calc(2:N)]; mismatch_Q = [Q_sch(2+PV:N)-Q_calc(2+PV:N)]; iteration=0; % FDLF iteration

while (iteration < 1000) iteration=iteration+1; correction_theta = -Bd\mismatch_P; correction_V = -Bdd\mismatch_Q; theta(2:N) = theta(2:N) + correction_theta(1:(N-1)); V(PV+2:N) = V(PV+2:N) .* (1+correction_V(1:(N-1)-PV)); % calculate mismatch and stop iterating % if the solution has converged P_calc = zeros(N,1); Q_calc = zeros(N,1); for k = 1:N for m = 1:N P_calc(k) = P_calc(k)... + V(k)*V(m)*(G(k,m)*cos(theta(k) - theta(m)) + B(k,m)*sin(theta(k) - theta(m))); Q_calc(k) = Q_calc(k)... + V(k)*V(m)*(G(k,m)*sin(theta(k) - theta(m)) - B(k,m)*cos(theta(k) - theta(m))); end end mismatch_P = [P_sch(2:N)-P_calc(2:N)]; mismatch_Q = [Q_sch(2+PV:N)-Q_calc(2+PV:N)]; mismatch = [mismatch_P; mismatch_Q]; if (norm(mismatch,'inf') < 1e-5) break; end end %% output solution data %Bus quantities P_loss = 0; Q_loss = 0; for i = 1:N fprintf(1, 'Bus %d:\n', i); fprintf(1, ' Voltage = %f p.u., %.1f deg.\n', ... V(i), theta(i)*180/pi); fprintf(1, ' Injected P = %f p.u.\n', P_calc(i)) if i > 1 fprintf(1, ' Scheduled P = %f p.u.\n', P_sch(i)); end fprintf(1, ' Injected Q = %f p.u.\n', Q_calc(i)) if i > PV+1 fprintf(1, ' Scheduled Q = %f p.u.\n', Q_sch(i)); end %Calculating the power loss P_loss = P_loss + P_calc(i); Q_loss = Q_loss + Q_calc(i); end %Line quantities fprintf(1, '\nPower flow from:\n'); for line = 1:size(line_data,1)

i = line_data(line,1); j = line_data(line,2); Y_shunt = line_data(line,5)/Yb; P_line = V(i)*(V(i)*-G(i,j) - V(j)*(-G(i,j)*cos(theta(i)-theta(j)) - B(i,j)*sin(theta(i)-theta(j)))); Q_line = V(i)*(V(i)*(B(i,j) - Y_shunt) + V(j)*(-B(i,j)*cos(theta(i)-theta(j)) + G(i,j)*sin(theta(i)-theta(j)))); fprintf(1, 'bus %d to bus %d\n', i, j); fprintf(1, ' P = %.4f, Q = %.4f\n', P_line, Q_line); end %Total loss fprintf(1, '\nThe total active power loss = %f\nThe total reactive power loss = %f\n'... , P_loss, Q_loss); iteration mismatch

SOLUTION TABLE 7.4.1 Bus voltage using FDLF for problem 7.4

kVASb 15= , kV.Vb 311=

Number of iterations = 8

Bus #

Volts

(p.u)

Angle

(Degrees)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 0.821 0.445 0 0 0 0

2 0.999 -0.1 0 0 0.480 0.232 0.78 510−× 0.55 510−×

3 0.999 -0.1 0 0 0.340 0.211 0.55 510−× 0.36 510−×

SOLUTION TABLE 7.4.2Power flow through the feeders lines for problem 7.4

Ploss (p.u) = 0.0005, Qloss (p.u) = 0.0018

From Bus #

To Bus #

MW Flow (p.u)

MVarFlow (p.u)

1 2 0.821 0.445

2 3 0.340 0.211

vi) The active and reactive power from the swing bus:

( ) ( ) 8210000500034004800

3

2

3

21 ....PPPP loss

ii,G


==

( ) ( ) 44500018000211023203

2

3

21 ....QQQQ loss

ii,G


==

The bus voltage of the swing bus is 01∠ and the p.u power is

445082101 .j.S += . vii)


000503

1

3

1.PPP

ii,L


==

00180

3

1

3

1.QQQ

ii,L


==

Ploss (p.u) = 0.0005, Qloss (p.u) = 0.0018

7.5 For the power grid of Fig.7.25 perform the following:

i) Compute the bus admittance and bus impedance model for power flow studies

ii) Add a parallel line between bus 1 and bus 2 with the same impedance and compute the bus impedance model

iii) What is the driving point impedance of bus 1 before and after adding the line?

iv) Remove the shunt element to ground and compute the bus admittance and the bus impedance model

Figure 7.25 The Power Grid for Problem 7.5.

Solutions

Fig. Per unit injection model for power flow studies for problem 7.5

i) Using the procedure of problem 7.2, the YBus model can be formulated as follow:

YBus = G +jB

However, in this problem, we have assumed a DC case and reactances are zero and therefore, B = 0

. . . .

. . . .

. . . .

. . . .

Note that ZBus model can be computed from YBus model since the bus 1 has an impedance to the reference bus (ground). This ZBus model is used for power flow studies.

The above power flow problem is solved using Gauss-Seidel ZBus method. The flowchart is given below:

↓

Gauss

Since problem 7.5 has only real quantities, the complex power S is replaced by active power P and the conjugates are neglected. The Gauss-Seidel code in MATLAB is given below with only real quantities.

At the end of each iteration, the calculated active power injection at each bus are found out using:

( ) ∑=

=N

jjkjkcalculatedk IZIP

1

The mismatch is given by:


After the power flow has converged, the effect of the fictitious resistance is removed by subtracting the current and power flowing into it from the swing bus.

( )fic

N

jjjcalculated R

VIYIP

21

1111 −= ∑

=

Where Rfic is the fictitious resistance

In the program, the power flow in the lines and the total power loss is also calculated.

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −==

ij

jiiijiij r

VVVI.VP

For only resistive circuits, ij

ij rG 1

−= , therefore,

( ) ijijiij GVVVP −=

∑∑==

−=4

1

4

1 ii,L

ii,Gloss PPP

%% Gauss-Seidel Zbus method only real quantities clc; clear all; tolerance = 1e-5; P_sch=[1 -.5 -0.5 -1]'; %% Admittance matrix formation R_fic = 0.01; % line data line_data = [ 1 2 0.01; 1 4 0.01; 2 3 0.02; 3 4 0.03];

N = max(max(line_data(:,1:2))); Y_BUS = zeros(N,N); % assemble admittance matrix for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); r = line_data(line,3); Y_BUS(i,i) = Y_BUS(i,i) + 1/r; Y_BUS(j,j) = Y_BUS(j,j) + 1/r; Y_BUS(i,j) = Y_BUS(i,j) - 1/r; Y_BUS(j,i) = Y_BUS(j,i) - 1/r; end Y_BUS(1,1) = Y_BUS(1,1) + 1/R_fic; Z_BUS = inv(Y_BUS); I = zeros(N,1); V = ones(N,1); iteration=0; while (iteration < 999) iteration=iteration+1; VZ=0; for n=2:N VZ=VZ+Z_BUS(1,n)*I(n); end I(1)=(V(1)-VZ)/Z_BUS(1,1); for m=2:N V(m)=Z_BUS(m,:)*I; I(m)=P_sch(m)/V(m); end P_calc = Z_BUS*I .* I; mismatch=[P_sch(2:N)-P_calc(2:N)]; if (norm(mismatch,'inf') < tolerance) break; end end %Bus quantities P_loss = V'*I - V(1)^2/R_fic; P_calc(1)= P_calc(1)-V(1)^2/R_fic; for i = 1:N fprintf(1, 'Bus %d:\n', i); fprintf(1, ' Voltage = %f p.u.\n', ... V(i)); fprintf(1, ' Injected P = %f p.u.\n', P_calc(i)) if i > 1 fprintf(1, ' Scheduled P = %f p.u.\n', P_sch(i)); end end %Line quantities

fprintf(1, '\nPower flow from:\n'); for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); P_line = V(i)*(V(i)-V(j))*(-Y_BUS(i,j)); fprintf(1, 'bus %d to bus %d\n', i, j); fprintf(1, ' P = %.4f\n', P_line); end %Total loss fprintf(1, '\nThe total active power loss = %f\n\n', P_loss); iteration mismatch SOLUTION TABLE 7.5.1 Bus voltage using Gauss-Seidel ZBus method for problem 7.5 part (i)

Iterations = 3

Bus #

Volts

(p.u)

Angle

(Degrees)

Generation

MW (p.u)

Load

MW (p.u)

PΔ

1 1.000 0.0 2.024 0 0

2 0.991 0.0 0 0.5 0.02 510−×

3 0.984 0.0 0 0.5 0.05 510−×

4 0.988 0.0 0 1 0.18 510−×

( ) ( ) 024015050000002424

1

4

1....PPP

ii,L

ii,Gloss =−−−−+++=−= ∑∑

==

SOLUTION TABLE 7.5.2 Power flow through the lines for problem 7.5 part (i)

Ploss (p.u) = 0.024

From Bus #

To Bus #

MW Flow (p.u)

1 2 0.867

1 4 1.157

2 3 0.360

3 4 -0.143

ii) Adding a parallel line with same impedance will have the following changes

01412111

112

1

−−−

++=rr/r

Y ,Bus , 3221

221

21

−−

+=r/r

Y ,Bus , 21

212112 /r

YY ,Bus,Bus−

−==

jBGYBus +=

B = 0 because there is no reactive element

The rest of the elements of G will remain unchanged.

400 200 0 100200 250 50 500 50 83 33100 0 33 133

. . . .

. . . .

. . . .

. . . .

Solving the power flow problem by adding a parallel line between bus 1 and 2 using Gauss-Seidel ZBus method of programming given in part (i)

SOLUTION TABLE 7.5.3 Bus voltage using Gauss-Seidel ZBus method for problem 7.5 part (ii)

Iterations = 3

Bus #

Volts

(p.u)

Angle

(Degrees)

Generation

MW (p.u)

Load

MW (p.u)

PΔ

1 1.000 0.0 2.020 0 0

2 0.995 0.0 0 0.5 0.02 510−×

3 0.986 0.0 0 0.5 0.03 510−×

4 0.989 0.0 0 1 0.13 510−×

( ) ( ) 020015050000002024

1

4

1....PPP

ii,L

ii,Gloss =−−−−+++=−= ∑∑

==

SOLUTION TABLE 7.5.4 Power flow through the lines for problem 7.5 part (ii)

It is seen that the power loss has decreased when another parallel line is added. This is because the effective resistance for the path of the current has decreased. It is also seen that the voltages of the buses have gone up. By adding a parallel line between buses 1 and 2, the impedance between the two buses is halved and voltage drop is reduced.

iii) The driving point impedance before adding the line = Z(1,1) from part (i) = 0.01 p.u Ω.

The driving point impedance after adding the line = Z(1,1) from part (ii) = 0.01 p.u Ω.

This is due to the fact that the injection model of power grid only has one tie to ground at bus 1. However, the power flow solution will show the effect of adding a line to the network as can be seen from the solution using Z bus model. It is seen that only the first row and the first column remains unchanged but the other rows and columns change.

iv) 4121

1111

−−

+=rr

Y ,Bus

,

Ploss (p.u) = 0.020

From Bus #

To Bus #

MW Flow (p.u)

1 2 0.931

1 2 0.931

1 4 1.089

2 3 0.427

3 4 -0.077

jBGYBus +=

B = 0 because there is no reactive element

The other elements of G remain unchanged.

The impedance matrix does not exist without the shunt element.

7.6 Consider the power grid depicted in Fig. 7.26.


Assume the following data:

a. The transformers connected to the PV generating station are rated at 460 V Y grounded/13.2kVΔ, have 10% reactance and 8 MVA capacity. The transformer connected to the power grid is 13.2 kV/63 kV, 8 MVA, 10% reactance.

b. Assume the bus 5 load is 1.5 MW,0.85 p.f. lagging, the bus 6 load is 1.2 MW, 0.9p.f. lagging, the bus 7 load is 2.4 MW, 0.9 p.f. leading, the bus 4 load is 1.5 MW,0.85p.f. lagging, and the bus 8 load is 1.3 MW, 0.95 p.f. lagging.

c. Assume PV generating station 1 is rated at 0.95MW and PV generating station 2 is rated at 3.5 MW.

d. The transmission line has a resistance of 0.0685 Ω/mile, reactance of 0.40 Ω/mile and half of line charging admittance (Y′/2) of 61011 −× Ω-1/mile. The line 4–7 is 4 miles, 4–8 is 2 miles, 5–6 is 4 miles, 5–7 is 1 mile, 6–7 is 3 miles, and 6–8 is 5 miles long.


i) Find the per unit YBus matrix

ii) Write a MATLAB program and compute the load bus voltages using FDLF

iii) If line 6–7 is out of service, compute the power flow through each transformer

iv) If 500 kWis added to bus 5, compute the bus load voltages

Solutions

Solution Figure 7.6.1ThePer Unit Model of the YBusMatrix.

. on the transmission line side

.

The p.u. impedance of PV transformers is , , ,,

,

,

,.

.

..

The p.u. impedance of utility transformers is , , ,,

,

,

,

. ..

.

Line impedances in p.u:

Z4-7,p.u = . ..

. . . , , .

. .

Z4-8,p.u = . ..

. . . , , .

. .

Z5-6,p.u = . ..

. . . , , .

. .

Z5-7,p.u = . ..

. . . , , .

. .

Z6-7,p.u = . ..

. . . , , .

. .

Z6-8,p.u = . ..

. . . , , .

. .

i) jBGYBus +=

The first 3 buses have only transformers connected to them which has only reactance and no resistive elements. Therefore, the first 3 rows and columns of

the real part of YBus, G are all zeros. Using the algorithm for YBus matrix formation given in solution 7.2,

. . .. . .. . . .

. . . .

. . .

. . .. . .. . . .

. . . .

. . .

ii) The Matlab program to calculate the voltages using FDLF method is given in solution 7.4. However, YBUS matrix given as above and Psch = [0 0.095 0.35 -0.15 -0.15 -0.12 -0.24 -0.13]’; and Qsch = [0 0 0 -0.093 -0.093 -0.0581 0.1162 -0.0427]’;

The calculations are based on the generator buses as PV buses; the results are given in Solution Table 7.6.1.

SOLUTION TABLE 7.6.1Bus voltages using FDLF for problem 7.6 part (ii)

MVA10Sb = , kV2.13Vb = on the transmission line


Bus #

Volts

(p.u)

Angle

(Degrees)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 0.346 0.079 0 0 0 0

2 1.000 -2.3 0.095 0.059 0 0 0.0015510−×

0

3 1.000 -0.0 0.35 0.062 0 0 0.0031510−×

0

4 0.991 -2.5 0 0 0.15 0.093 0.0869510−×

0.2854510−×

5 0.993 -3.0 0 0 0.15 0.093 0.0089510−×

0.1491510−×

6 0.993 -2.6 0 0 0.12 0.058 0.1418510−×

0.3117510−×

7 0.994 -3.1 0 0 0.24 -0.116 0.0705510−×

0.5811510−×

8 0.990 -2.8 0 0 0.13 0.042 0.1545510−×

0.2265510−×

iii) If line 6-7 is out of service, the new YBus from the algorithm in solution 7.2 is: jBGYBus +=

. . .. . .. . .

. . .

. . .

. . .. . .. . .

. . .

. . .

SOLUTION TABLE 7.6.2 Bus voltages using FDLF method for problem 7.6 part (iii)



Bus #

Volts

(p.u)

Angle

(Degrees)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 0.346 0.079 0 0 0 0

2 1.000 -2.4 0.095 0.059 0 0 0.0021510−×

0

3 1.000 0.3 0.35 0.065 0 0 0.0036510−×

0

4 0.991 -2.5 0 0 0.15 0.093 0.1057510−×

0.2933510−×

5 0.993 -3.1 0 0 0.15 0.093 0.0510510−×

0.1452510−×

6 0.993 -2.2 0 0 0.12 0.058 0.0867510−×

0.2247510−×

7 0.994 -3.3 0 0 0.24 -0.116 0.0679510−×

0.5168510−×

8 0.989 -2.7 0 0 0.13 0.042 0.1585510−×

0.2313510−×

iv) Adding 500 kW to bus 5, the voltages are calculated again using the following scheduled active power:

P_sch = [0 0.095 0.35 -0.15 -0.15 -0.12 -0.24 -0.13]';

Solving the power flow problem by FDLF using the program given in solution 7.4

SOLUTION TABLE 7.6.3 Bus voltages using FDLF method for problem 7.6 part (iv)



Bus #

Volts

(p.u)

Angle

(Degrees)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 0.346 0.079 0 0 0 0

2 1.000 -2.3 0.095 0.056 0 0 0.0015510−×

0

3 1.000 0.0 0.35 0.062 0 0 0.0031510−×

0

4 0.991 -2.5 0 0 0.15 0.093 0.0869510−×

0.2854510−×

5 0.993 -3.0 0 0 0.15 0.093 0.0089510−×

0.1491510−×

6 0.993 -2.6 0 0 0.12 0.058 0.1418510−×

0.3117510−×

7 0.994 -3.1 0 0 0.24 -0.116 0.0705510−×

0.5811510−×

8 0.990 -2.8 0 0 0.13 0.043 0.1545 0.2265

510−× 510−×

7.7 Consider a power grid where the system is modeled as

IBus = YBus.VBus

For each bus k we ∑=

=n

1mmkmk VYI and where m is the number of buses and Ykm is the

element of the YBus matrix. For each bus k, we can also write:

*kkkk IVjQP =+

wherePk and Qk are real and reactive power is entering node k (* is a complex conjugate andj = 1− )

Let

*

1

*m

n

mkmkkk VYVjQP ∑

=

=+

Vk = Vkejθk= ek + j f k;θk = k

k1

ef

tan −

Ykm = Ykmejαkm= Gkm + j Bkm;αkm = km

km1

GB

tan −

Using the above in *m

n

1m

*kmkkk VYVjQP ∑

=

=+

obtain ∑=

θ−α−θ=+n

1m

jm

jkm

jkkk

mkmk eVeYeVjQP

Then using Taylorseries expansion, express the power flow problem as

∑∑==

Δ+θΔ=Δ

n

1m m

mkm

n

1mmkmk V

VNHP

∑∑==

Δ+θΔ=Δ

n

1m m

mkm

n

1mmkmk V

VLJQ

And in compact form as

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡ΔΔ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

∂∂

∂∂

∂

∂

∂

∂

=⎥⎦

⎤⎢⎣

⎡ΔΔ

VV

VVQQ

VVPP

QP θ

θ

θ .

Show that for m = k, we have

kk2

kkk

kkk BVQ

PH −−=

θ∂∂

=

kk2

kkk

kkk GVP

QJ −=

θ∂∂

=

Solution

k

kPθ∂∂ + j

k

kQθ∂

∂ = ⎟⎠

⎞⎜⎝

⎛+− ∑

=

θ−α−θθ−α−θn

1m

jm

jkm

jk

jk

jkk

jk

kkmkkkkk eVeYeVj)eVejY(eV

= )jQP(jeVeYejV kkj

kj

kkj

kkkmk ++− θ−α−θ

= )jQP(jeY)eV)(eV(j kkj

kkj

kj

kkmkk ++− α−θ−θ

= )jQP(j)jBG(jV kkkkkk2

k ++−−

= )GVP(j)BVQ( kk2

kkkk2

kk −+−−

kk2

kkk

kkk BVQ

PH −−=

θ∂∂

=

kk2

kkk

kkk GVP

QJ −=

θ∂∂

=

7.8 For the system given in Problem 7.7 show that for m=k, we have the following expressions

kkkkkk

kkk GVPV

VPN 2+=∂∂

=

kkkkkk

kkk BVQV

VQL 2−=∂∂

=

Solution

First, we need to take the partial derivative of the equation

kP + j kQ = ∑=

θ−α−θn

1m

jm

jkm

jk

kkmk eVeYeV

kk

k VVP

∂∂ + j k

k

k VVQ∂∂ = ∑

=

θ−αθθ−α−θ +n

1m

jm

jkm

jk

jjkk

j2k

kkmkkkkk eVeYeV)eeY(eV

= kkkkkk2

kkkj

kk2

k jQP)jBG(VjQPeYV kk ++−=++α−

= )BVQ(jGVP kk2

kkkk2

kk −++

For m = k

Separating the real and imaginary components,

kkkkkk

kkk GVPV

VPN 2+=∂∂

=

kkkkkk

kkk BVQV

VQL 2−=∂∂

=

Thus, we have calculated (general equations) the diagonal elements of the Jacobian matrix.

7.9 Consider the Problem 7.7, first let

Im = am + j bm = YkmVm

andYkmVm = (Gkm + j Bkm) (em + j f m)

YkmVm = (Gkm em– Bkmf m) + j (Bkm em + Gkmf m)

then using the following:

am = (Gkm em– Bkmf m)

bm = (Bkm em + Gkmf m)

derive the following expressions:

kmkmm

kkm ebfa

PH −=

θ∂∂

=

kmkmm

kkm fbea

QJ −−=

θ∂∂

=

Solution

We need to take the derivative of the following equation

∑=

θ−α−θ=+n

1m

jm

jkm

jkkk

mkmk eVeYeVjQP

m

kPθ∂∂ + j

m

kQθ∂∂ = mkmk j

mj

kmj

k eVeYejV θ−α−θ−

( )( ) *mm

*jm

jkm

jk I.jVeVeY.eVj mkmk −=−= θαθ

= – j (ek + jfk) (am– j bm)

= (amf k–bmek) – j (amek + bm f k)

Separating the real and imaginary components,

kmkmm

kkm ebfa

PH −=

∂∂

=θ

kmkmm

kkm fbea

QJ −−=

∂∂

=θ

7.10 For Problems 7.7 and 7.8 show that form≠ k, we have the following off-diagonal elements for theJacobian matrix:

kmkmmm

kkm fbeaV

VP

N +=∂∂

=

kmkmmm

kkm ebfaV

VQ

L −=∂∂

=

Solution

∑=

θ−α−θ=+n

1m

jm

jkm

jkkk

mkmk eVeYeVjQP

mm

k VVP

∂∂ + j m

m

k VVQ∂∂

mkmk jm

jkm

jk eVeYeV θ−α−θ=

( )( )mkmk jm

jkm

jk eVeY.eV θ−α−θ=

( )( )*jm

jkm

jk

mkmk eVeY.eV θαθ=

= (ek + j f k) (am– j bm) = (amek + bm f k) + j (amf k–bmek)

kmkmmm

kkm fbeaV

VP

N +=∂∂

=

kmkmmm

kkm ebfaV

VQ

L −=∂∂

=

7.11 For Problem 7.7 show that the off-diagonal elements of the Jacobian matrix for m≠ k are:

MatrixJacobianofElementsDiagonalOfffbeaJN

ebfaLH

kmkmkmkm

kmkmkmkm −⎭⎬⎫

+=−=−==

And diagonal elements of the Jacobian matrix for m = kare:

MatrixJacobianofElementsDiagonal

VGPJ

VGPN

VBQL

VBQH

kkkkkk

kkkkkk

kkkkkk

kkkkkk

⎪⎪

⎭

⎪⎪

⎬

⎫

−=

+=

−=

−−=

2

2

2

2

Solution

∑=

θ−α−θ=+n

1m

jm

jkm

jkkk

mkmk eVeYeVjQP

m

kPθ∂∂ + j

m

kQθ∂∂ = mj

mj

kmj

k eVeYejV kmk θ−α−θ−

( )( )*jm

jkm

jk

mkmk eVeY.eVj θαθ−=

= – j (ek+ j f k) (am– j bm)

= (amf k– bmek) – j (amek + bm f k)

kmkmm

kkm ebfa

PH −=

∂∂

=θ

kmkmm

kkm fbea

QJ −−=

∂∂

=θ

mm

k VVP

∂∂ + j m

m

k VVQ∂∂

mkmk jm

jkm

jk eVeYeV θ−α−θ=

( )( )mkmk jm

jkm

jk eVeY.eV θ−α−θ=

( )( )*jm

jkm

jk

mkmk eVeY.eV θαθ=

= (ek + j f k) (am– j bm) = (amek + bm f k) + j (amf k–bmek)

kmkmmm

kkm fbeaV

VP

N +=∂∂

=

kmkmmm

kkm ebfaV

VQ

L −=∂∂

=

Therefore,

⎭⎬⎫

+=−=−==

kmkmkmkm

kmkmkmkm

fbeaJNebfaLH

kP + j kQ = ∑=

θ−α−θn

1m

jm

jkm

jk

mkmk eVeYeV

kk

k VVP

∂∂ + j k

k

k VVQ∂∂ = ∑

=

θ−α−θθ−α−θ +n

1m

jm

jkm

jk

jjkk

j2k

mkmkkkkk eVeYeV)eeY(eV

= kkkkkk2

kkkj

kk2

k jQP)jBG(VjQPeYV kk ++−=++α−

= )BVQ(jGVP kk2

kkkk2

kk −++

andform = k

kkkkkk

kkk GVPV

VPN 2+=∂∂

=

kkkkkk

kkk BVQV

VQL 2−=∂∂

=

k

kPθ∂∂ + j

k

kQθ∂

∂ = ∑=

θ−α−θθ−α−θ +−n

1m

jm

jkm

jk

jk

jkk

jk

mkmkkkmk eVeYejV)eVejY(eV

= )jQP(jeVeYejV kkj

kj

kkj

kkkmk ++− θ−α−θ

= )jQP(jeY)eV)(eV(j kkj

kkj

kj

kkmkk ++− α−θ−θ

= )jBG(jV)jQP(j kkkk2

kkk −−+

= )GVP(j)BVQ( kk2

kkkk2

kk −+−−

kkkkk

kkk BVQ

PH 2−−=

∂∂

=θ

kkkkk

kkk GVP

QJ 2−=

∂∂

=θ

Therefore,

⎪⎪

⎭

⎪⎪

⎬

⎫

−=

+=

−=

−−=

2

2

2

2

kkkkkk

kkkkkk

kkkkkk

kkkkkk

VGPJ

VGPN

VBQL

VBQH

7.12 For Problem 7.7 compute the )Calculated(kP and )Calculated(kQ for each bus k as expressed

below.

∑=

++−=n

mkmmkmmkkmmkmmkk GfBefBfGeeP

1)]()([

∑=

+−−=n

mkmmkmmkkmmkmkkk GfBeeBfGefQ

1)]()([

wherevk = ek + j f k;Ykm* = Gkm– j Bkm; Vm

* = em– jf m

Solution

*m

n

1m

*kmk

n

1m

*mkkk VYVIVjQP ∑∑

==

==+

Let vk = ek + j f k;Ykm* = Gkm– j Bkm; Vm

* = em– j f m

then

∑=

−−+=+n

mmmkmkmkkkk jfejBGjfejQP

1))(()(

∑=

+−−+=+n

mkmmkmmkmmkmmkkkk GfBejBfGejfejQP

1)()()(

∑=

+−−+=n

mkmmkmmkmmkmmkk GfBejBfGejfe

1)]())[((

∑=

++−++−−=n

mkmmkmmkkmmkmkkkmmkmmkkmmkmmk GfBefBfGejfGfBejeBfGee

1)()()()(

Therefore, equating the real and imaginary parts of both sides,Pk and Qk are:

∑=

++−=n

mkmmkmmkkmmkmmkk GfBefBfGeeP

1)]()([

∑=

+−−=n

mkmmkmmkkmmkmkkk GfBeeBfGefQ

1)]()([

7.13. Assume that the power balance equation for a power system network can be written as

S = P + j Q = [ ] [ ]*T I.V T= [ ] [ ]*T V.Y.V

Where S is the complex power injection vector; P is the real power injection vector; Q is the reactive power injection vector; I is the current injection vector; V is the bus voltage vector, and Ykj = Gkj + j Bkj is the system admittance matrix.

Assume that in a polar coordinate system, the complex voltage can be written as

kV = Vk (cosθk + j sin θk)

Show that the calculated real and reactive powers can be expressed as

)sincos()( kjkjkjkj

n

1jjkCalculatedk BGVVP θ+θ= ∑

=

)cossin()( kjkjkjkj

n

1jjkCalculatedk BGVVQ θ−θ= ∑

=

Whereθkj≅θk– θj

And in the Cartesian coordinate system the calculated real and reactive powers can be expressed as

)()()( jkmjkj

n

1jjjkjjkj

n

1jkCalculatedk eBfGffBeGeP ++−= ∑∑

==

)()()( jkjjkj

n

1jkjkjjkj

n

1jkCalculatedk eBfGefBeGfQ +−−= ∑∑

==

wherethe bus voltage is given by

kV = ek + j fk

Solution

For each bus k,wehave ∑=

=n

1jjkjk VYI

wheren is the number of buses and Ykm is an element of the YBus matrix. For each bus k, we can also write:

*kkkk IVjQP =+

wherePk, and Qk are the real and reactive power entering node k (* is a complex conjugate

and j = 1− )

Let

**j

n

1jkjkkk VYVjQP ∑

=

=+

vk = Vkejθk= ek + j f k;θk = tan -1k

k

ef

Ykm = Ykmejαkj= Gkj + j Bkj;αkj = tan -1kj

kj

GB

Using the above in **j

n

1jkjkkk VYVjQP ∑

=

=+

( ) ( )

)cossin()sincos(

sincos.

kjkjkjkj

n

1jjkkjkjkjkj

n

1jjk

n

1jkjkjjkjkjk

n

1j

jj

jkjk

n

1j

jj

jkj

jkkk

BGVjVBGVV

jVjBGVeVeYV

eVeYeVjQP

kjkj

jkjk

θ−θ+θ+θ=

θ+θ−==

=+

∑∑

∑∑

∑

==

==

θα−

=

θ−α−θ

Therefore,

)sincos()( kjkjkjkj

n

1jjkCalculatesk BGVVP θ+θ= ∑

=

)cossin()( kjkjkjkj

n

1jjkCalculatesk BGVVQ θ−θ= ∑

=

By including kV = ek + jfk and Y kj = Gkj + jBkjin the equations below:

( ) ( ) ( ) ( ) ( )( )∑∑

∑

==

=

−−+=+++=

=+

n

1jjjkjkjkk

n

1jjjkjkjkk

j

n

1jkjkkk

jfejBGfejfejBGfe

VYVjQP

**

**

and by expanding the above and separating the real and imaginary parts, we have

( ) ( )∑∑==

++−=n

1jjkjjkjk

n

1jjkjjkjkCalculatesk eBfGffBeGeP )(

( ) ( )∑∑==

+−−=n

1jjkjjkjk

n

1jjkjjkjkCalculatesk eBfGefBeGfQ )(

For Gauss-Seidel iterations in Cartesian coordinates, the real and imaginary parts of the voltage at each bus are iteratively computed until the mismatch of the power at each bus is below an accepted level. The derivation is as given below:

( )( )∑=

++=n

1jjjkjkjk jfejBGI .

( )( )∑=

++=⎟⎟⎠

⎞⎜⎜⎝

⎛++ n

1jjjkjkj

kk

SchkSchk jfejBGjfejQP

.*

,,

( )( ) ( ) ( )∑∑==

++−=+

+− n

1jjkjjkj

n

1jjkjjkj2

k2

k

kkSchkSchk fGeBjfBeGfe

jfejQP....

.,,

( )( ) ( ) ( )⎥⎦

⎤⎢⎣

⎡++−

+

++=+ ∑∑

==

n

1jjkjjkj

n

1jjkjjkj2

Schk2

Schk

SchkSchk2

k2

kkk fGeBjfBeG

QPjQPfe

jfe .....

,,

,,

( ) ( )

( ) ( )⎭⎬⎫

⎩⎨⎧

++−++

+

⎭⎬⎫

⎩⎨⎧

+−−++

=+

∑∑

∑∑

==

==

n

1jjkjjkjSchk

n

1jjkjjkjSchk2

Schk2

Schk

2k

2k

n

1jjkjjkjSchk

n

1jjkjjkjSchk2

Schk2

Schk

2k

2k

kk

fGeBPfBeGQQPfe

j

fGeBQfBeGPQPfe

jfe

....

....

,,,,

,,,,

Separating out the real and imaginary parts,

( )( ) ( )

( ) ( )

( )( ) ( )

( ) ( )⎭⎬⎫

⎩⎨⎧

++−++

=

⎭⎬⎫

⎩⎨⎧

+−−++

=

∑∑

∑∑

==

==

n

1jjkjjkjSchk

n

1jjkjjkjSchk2

Schk2

Schk

2oldk

2oldknew

k

n

1jjkjjkjSchk

n

1jjkjjkjSchk2

Schk2

Schk

2oldk

2oldknew

k

fGeBPfBeGQQPfef

fGeBQfBeGPQPfe

e

....

....

,,,,

,,,,

Now, that the real and imaginary parts have been separated, the Gauss-Seidel program in solution 7.2 can be written in real domain without dealing with complex numbers. The algorithm in real domain is as follows:

( )( ) ( )

( ) ( )

( )( ) ( )

( ) ( )⎭⎬⎫

⎩⎨⎧

++−++

=

⎭⎬⎫

⎩⎨⎧

+−−++

=

∑∑

∑∑

==

==

n

1jjkjjkjSchk

n

1jjkjjkjSchk2

Schk2

Schk

2oldk

2oldknew

k

n

1jjkjjkjSchk

n

1jjkjjkjSchk2

Schk2

Schk

2oldk

2oldknew

k

fGeBPfBeGQQPfef

fGeBQfBeGPQPfee

....

....

,,,,

,,,,

After each iteration, the following formulas are used to calculated the active and reactive powers injected at each bus:

)()()( jkjjkj

n

1jkjkjjkj

n

1jkCalculatesk eBfGffBeGeP ++−= ∑∑

==

)()()( jkjjkj

n

1jkjkjjkj

n

1jkCalculatesk eBfGefBeGfQ +−−= ∑∑

==

Then for each load bus the power mismatch is calculated using the following formulas:



7.14 Consider the feeder given in Fig. 7.27

Figure 7.27 The Feeder of Problem 7.14.

System data: V1= 1∠0; Z25 = 5 + j 10 Ω; Z34 = 2 + j 8 Ω; Z23 = 5.41 + j 3.34 Ω;S2 = 3 MVA, p.f. = 0.75 lagging; S3 = 3 MVA, p.f. = 0.8 lagging; S4 = 4 MVA, p.f. = 0.9 lagging;S5 = 2 MVA, p.f. = 0.9 lagging; T1; 63/20 kV, 10% reactance, 20 MVA; T2 is the same as T1


i) Calculate the per unit equivalent circuit model. Use Sb equal to MVA rating of transformer T1.

ii) Let V1=V1∠0, VK = VRK + j VIK, K = 2, 3, 4, 5. Develop an equation in a Cartesian coordinate for VRK = f (VRK, VIK), ΔPK and ΔQK, K = 2, 3, 4, 5 for each bus. Use an iterative Gauss-Seidel approximation technique and compute bus voltages and system losses. Use five iterations and assume initial voltages are VK= 1∠0, K = 1, 2, 3, 4, 5. Put your results in table.

iii) Compute the YBus, B′ and B′′ matrices and the use the fast decoupled load flow technique to compute bus voltages. Make a table and compare your results with (ii) above. Use five iterations.

iv) Use the FDLF technique and compute bus voltages after correcting the power factor of each bus to unity. Make a table and compare your results with (iii) above.

Solutions

i) Calculate the per unit equivalent circuit model. Use Sb equal to an MVA rating of transformer T1 (= 20 MVA), assuming voltage base of 20 kV on the LV side of T1:

S2(p.u) = 3/20 = 0.15 p.u; p.f. = 0.75 lagging

S3(p.u) = 3/20 = 0.15 p.u; p.f. = 0.8 lagging

S4(p.u) = 4/20 = 0.2 p.u; p.f. = 0.9 lagging

S5(p.u) = 2/20 = 0.1 p.u; p.f. = 0.9 lagging

The per unit values of the impedance are

Zb = Vb 2/ Sb = (20 kV) 2/ (20 MVA) = 20 Ω

Z25 = 5 + j 10, Z25p.u = (5 + j 10)/20 = 0.25 + j 0.5 p.u

Z34 = 2 + j 8 ,Z34p.u = (2 + j 8)/20 = 0.1 + j 0.4 p.u

Z23 = 5.41 + j 3.34, Z23p.u = (5.41 + j 3.34)/20 = 0.27 + j 0.167 p.u

T1: j 10% = j 0.1 p.u

T

Fig.The per unit injection model for power flow studies for problem 7.14

ii) The results using the Gauss-Seidel YBus method(given is solution 7.2) for solving the load flow problem are obtained by simulations in MATLAB using Psch = [0 -0.15 -0.15 -0.2 -0.1]’; Qsch = [0 -0.132 -0.113 -0.097 -0.048]’;

SOLUTION TABLE 7.14.1For problem 7.14 part (ii) using Gauss-Seidel YBus Method

Sb = 20 MVA, Vb = 20 kV on the LV side of transformer


Iterations #

V1 V2 V3 V4 V5

1 1∠0 0.994∠-0.3 0.965∠-0.5 0.944∠-4.3 0.952∠-2.3

2 1∠0 0.989∠-0.8 0.922∠-1.9 0.901∠-5.0 0.941∠-2.6

3 1∠0 0.989∠-0.8 0.922∠-1.9 0.901∠-5.0 0.941∠-2.6

4 1∠0 0.986∠-1.4 0.873∠-2.4 0.827∠-7.3 0.934∠-3.6

5 1∠0 0.983∠-1.6 0.858∠-2.2 0.799∠-8.1 0.931∠-3.8

37 1∠0 0.974∠-2.0 0.801∠-1.5 0.712∠-8.6 0.920∠-4.5

SOLUTION TABLE 7.14.2 Power mismatch for problem 7.14 part (ii) with Gauss-Seidel YBus Method

Iteration #

mismatch

Bus 1 Bus 2 Bus 3 Bus 4 Bus 5

1 ΔP 0 0.21 0.22 0.05 0.02

ΔQ 0 0.08 0.09 0.09 0.01

2 ΔP 0 0.17 0.07 0.09 0.02

ΔQ 0 0.02 0.08 0.08 0.001

3 ΔP 0 0.07 0.10 0.02 0.01

ΔQ 0 0.04 0.07 0.06 0.001

4 ΔP 0 0.09 0.03 0.04 0.006

ΔQ 0 0.02 0.06 0.05 0.001

5 ΔP 0 0.04 0.04 0.01 0.006

ΔQ 0 0.03 0.05 0.03 0.002

37 ΔP 0 0.74 510−× 0.47 510−× 0.11 510−× 0.11 510−×

ΔQ 0 0.98 510−× 0.98 510−× 0.67 510−× 0.05 510−×

The problem converges after 37 iterations.

In the table for the voltages, the actual line voltages are the per unit in the table multiplied by the base voltage (20kV), and the losses multiplied by 20 MVA (base power).


093805

1

5

1.PPP

ii,L


==

1336605

1

5

1.QQQ

ii,L


==

SOLUTION TABLE 7.14.3Power flow through the transmission lines for problem 7.14 part (ii) after the solution converged using Gauss-Seidel YBus Method

Ploss (p.u) = 0.0938, Qloss (p.u) = 0.13366

From Bus #

To Bus #

MW Flow (p.u)

MVar Flow (p.u)

1 2 0.694 0.524

2 5 0.104 0.056

2 3 0.440 0.298

3 4 0.210 0.136

iii)

The YBus of the system using algorithm in solution 7.2 is jBGYBus +=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−−−

=

8.0008.0005882.05882.00005882.02671.36789.20

8.006789.24789.3000000

G

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−−

−−

−

=

6.1006.1003529.23529.20003529.20099.46569.106.106569.12569.2320

0002020

B

The matrix is the imaginary part of YBus and matrix is formulated same as YBus except the resistances and line charging are ignored in the formulation. In this problem, we do not have line charging. matrix is the imaginary part of YBus give above, Since bus 1 is swing bus and its voltage (magnitude and angle) is given, first row and first column are not needed in and martrix for bus voltage calculation.

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

=

00020000020500250020050024888988500020988598827

....

......

B'

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

=

61006103523520035246516106512523

....

.....

B ''

Using the FDLF algorithm and program given in solution 7.4, and Psch = [0 -0.15 -0.15 -0.2 -0.1]’; Qsch = [0 -0.132 -0.113 -0.097 -0.048]’; and YBus, B’ and B’’ as given above,

SOLUTION TABLE 7.14.4 Bus Voltages for problem 7.14 part (iii) using FDLF.



Iteration #

V1 V2 V3 V4 V5

1 1∠0 0.981∠-1.7 0.854∠-5.1 0.813∠-9.7 0.950∠-4.6

2 1∠0 0.976∠-1.9 0.720∠-5.2 0.650∠-11.2 0.918∠-4.9

3 1∠0 0.971∠-2.1 0.682∠-2.8 0.596∠-10.9 0.911∠-4.6

4 1∠0 0.971∠-2.3 0.732∠-0.5 0.632∠-10.0 0.916∠-4.6

5 1∠0 0.972∠-2.2 0.802∠-0.5 0.697∠-8.6 0.919∠-4.6

42 1∠0 0.974∠-2.0 0.801∠-1.5 0.713∠-8.6 0.920∠-4.5

SOLUTION TABLE 7.14.5 Power Mismatch for problem 7.14 part (iii) with FDLF

Iteration #

mismatch

Bus 1 Bus 2 Bus 3 Bus 4 Bus 5

1 ΔP 0 0.08 0.06 0.05 0.003

ΔQ 0 0.15 0.12 0.06 0.04

2 ΔP 0 0.25 0.20 0.05 0.01

ΔQ 0 0.01 0.03 0.04 0.01

3 ΔP 0 0.21 0.20 0.04 0.01

ΔQ 0 0.04 0.04 0.04 0.004

4 ΔP 0 0.12 0.15 0.02 9.15 710−×

ΔQ 0 0.06 0.07 0.03 0.003

5 ΔP 0 0.05 0.07 0.001 0.001

ΔQ 0 0.05 0.08 0.02 0.0005

42 ΔP 0 0.96 510−× 0.87 510−× 0.33 510−× 0.003 510−×

ΔQ 0 0.28 510−× 0.44 510−× 0.10 510−× 0.001 510−×


09386605

1

5

1

.PPPi

i,Li

i,Gloss =−= ∑∑==

1336705

1

5

1

.QQQi

i,Li

i,Gloss =−= ∑∑==

SOLUTION TABLE 7.14.6ThePower Flow through the Transmission Lines for problem 7.14 part (iii)

Ploss (p.u) = 0.093866, Qloss (p.u) = 0.13367

From Bus #

To Bus #

MW Flow (p.u)

MVarFlow (p.u)

1 2 0.694 0.524

2 5 0.104 0.056

2 3 0.440 0.298

3 4 0.210 0.136

It is seen that with Gauss-Seidel YBus method, the problem converges with 37 iterations while with FDLF, it takes 42 iterations.

iv) With unity power factor load,

P2 = 0.15 × 0.75 Q2 = 0

P2 = 0.15 × 0.8 Q3 = 0

P2 = 0.2 × 0.9 Q4 = 0

P5 = 0.1 × 0.9 Q5 = 0

SOLUTION TABLE7.14.7 Bus Voltage using FDLF method for problem 7.14 part (iv)



Bus #

Volts

(p.u)

Angle

(Degrees)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 0.652 0.076 0 0 0 0

2 0.997 -1.9 0 0 0.15 0 0.40 510−× 0.37 510−×

3 0.882 -5.4 0 0 0.15 0 0.50 510−× 0.48 510−×

4 0.853 -11.5 0 0 0.20 0 0.005510−×

0.07 510−×

5 0.970 -4.8 0 0 0.10 0 0.001510−×

0.001510−×


05230105

1

5

1.PPP

ii,L


==

07611605

1

5

1

.QQQi

i,Li

i,Gloss =−= ∑∑==

SOLUTION TABLE 7.14.8 The Power Flow through the Transmission Lines for problem 7.14 part (iv)

Ploss (p.u) = 0.052301, Qloss (p.u) = 0.076116

From Bus #

To Bus #

MW Flow (p.u)

MVar Flow (p.u)

1 2 0.652 0.076

2 5 0.103 0.005

2 3 0.400 0.049

3 4 0.206 0.022

It is seen with unity power factor, the bus voltages have a magnitude higher than that of lagging power factor operation. Reactive power causes the voltage to drop and when the load does not consume reactive power, voltage drop is very small.

7.15 Consider the five bus system below with the impedances in p.u given in the figure 7.28.

Figure 7.28 System for problem 7.15

Table 7.11 Five – Bus System: Bus Data (pu)

Bus

Name

Bus

Type

VN

(kV)

V

(p.u.)

θ

(rad)

PG

(MW)

PL

(MW)

QL

(Mvar)

Bus 1 PQ 138 1.0 - - 160 80

Bus 2 PQ 138 1.0 - - 200 100

Bus 3 PQ 138 1.0 - - 370 130

Bus 4 PV 1 1.05 - 500 - -

Bus 5 Swing 4 1.05 0.0 - - -


i) Calculate Y bus = Yi + jYr (internal Y bus) i.e. Compute Yi and Yr matrices

ii) Compute the Jacobian Matrix [H, N, J, L]

iii) Compute ΔP and ΔQ

iv) Solve the Newton-Raphson Power Flow for 5 bus system of Figure 7.28

Solutions

Fig. Per unit model of the system in problem 7.15

i) Assuming the voltage base of 138 kV on the transmission line and a power base of 1000 MVA and the one line diagram in p.u, using the YBus algorithm given in solution 7.2

jBGYBus +=

. . .. . .. . .

. . .. . .. . .

ii) The calculated injected active and reactive powers for each bus are given by:

( )∑=

θ+θ=N


1

( )∑=

θ−θ=N


1





The voltage of each bus is found out using Newton Raphson. After the solution has converged, the power flow in lines and the power losses are calculated using the same formula as given in solution 7.4.

The following MATLAB program is used to solve the power flow problem using Newton-Raphson method:

%% Power Flow: Newton-Raphson clc; clear all; Vb = 1; Sb = 1; Zb = Vb^2/Sb; Yb = 1/Zb; PV=1; %no. of PV buses tolerance=1e-5; %error tolerance % Scheduled active and reactive powers P_sch = [0 0.50 -.37 -.20 -.16]'; Q_sch = [0 0 -.13 -.10 -.08]'; %% Admittance matrix formation % line data line_data = [ 2 4 0.00 0.10 0; 4 3 0.01 0.10 0; 3 1 0.00 0.10 0; 4 5 0.02 0.21 0; 3 5 0.03 0.33 0]; N = max(max(line_data(:,1:2))); G = zeros(N,N); B = zeros(N,N); % assemble admittance matrix for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); r = line_data(line,3)/Zb; x = line_data(line,4)/Zb; y_shunt = line_data(line,5)/Yb;

G(i,i) = G(i,i) + r/(r^2 + x^2); B(i,i) = B(i,i) - x/(r^2 + x^2) + y_shunt; G(j,j) = G(j,j) + r/(r^2 + x^2); B(j,j) = B(j,j) - x/(r^2 + x^2) + y_shunt; G(i,j) = G(i,j) - r/(r^2 + x^2); B(i,j) = B(i,j) + x/(r^2 + x^2); G(j,i) = G(j,i) - r/(r^2 + x^2); B(j,i) = B(j,i) + x/(r^2 + x^2); end %% Power flow % allocate storage for Jacobian J11 = zeros(N-1,N-1); %dP/d th J12 = zeros(N-1,N-1-PV); %V*dp/dv J21 = zeros(N-1-PV,N-1); %dQ/d th J22 = zeros(N-1-PV,N-1-PV); %V*dQ/dv % initial values V = [1.05 1.05 1 1 1]'; theta = zeros(N,1); P_calc = zeros(N,1); Q_calc = zeros(N,1); for k = 1:N for m = 1:N P_calc(k) = P_calc(k)... + V(k)*V(m)*(G(k,m)*cos(theta(k) - theta(m)) + B(k,m)*sin(theta(k) - theta(m))); Q_calc(k) = Q_calc(k)... + V(k)*V(m)*(G(k,m)*sin(theta(k) - theta(m)) - B(k,m)*cos(theta(k) - theta(m))); end end mismatch = [P_sch(2:N)-P_calc(2:N); Q_sch(2+PV:N)-Q_calc(2+PV:N)]; iteration=0; % Newton-Raphson iteration while (iteration < 10) iteration=iteration+1; % calculate Jacobian for k = 2:N for m = 2:N if (k == m) J11(k-1,m-1) = -Q_calc(k) - V(k)^2*B(k,m); if (k >= 2+PV) J21(k-PV-1,m-1) = P_calc(k) - V(k)^2*G(k,m); end if (m >= 2+PV)

J12(k-1,m-PV-1) = P_calc(k) + V(k)^2*G(k,m); end if (k >= 2+PV && m >= 2+PV) J22(k-PV-1,m-PV-1) = Q_calc(k) - V(k)^2*B(k,m); end else em = V(m)*cos(theta(m)); fm = V(m)*sin(theta(m)); ek = V(k)*cos(theta(k)); fk = V(k)*sin(theta(k)); am = G(k,m)*em - B(k,m)*fm; bm = B(k,m)*em + G(k,m)*fm; J11(k-1,m-1) = am*fk - bm*ek; if (k >= 2+PV) J21(k-PV-1,m-1) = -am*ek - bm*fk; end if (m >= 2+PV) J12(k-1,m-PV-1) = am*ek + bm*fk; end if (k >= 2+PV && m >= 2+PV) J22(k-PV-1,m-PV-1) = am*fk - bm*ek; end end end end % calculate correction J = [J11, J12; J21, J22]; correction = J\mismatch; theta(2:N) = theta(2:N) + correction(1:(N-1)); V(PV+2:N) = V(PV+2:N) .* (1+correction(N:2*(N-1)-PV)); P_calc = zeros(N,1); Q_calc = zeros(N,1); for k = 1:N for m = 1:N P_calc(k) = P_calc(k)... + V(k)*V(m)*(G(k,m)*cos(theta(k) - theta(m)) + B(k,m)*sin(theta(k) - theta(m))); Q_calc(k) = Q_calc(k)... + V(k)*V(m)*(G(k,m)*sin(theta(k) - theta(m)) - B(k,m)*cos(theta(k) - theta(m))); end end mismatch = [P_sch(2:N)-P_calc(2:N); Q_sch(2+PV:N)-Q_calc(2+PV:N)]; if (norm(mismatch,'inf') < 1e-5) break; end end

%% output solution data %Bus quantities P_loss = 0; Q_loss = 0; for i = 1:N fprintf(1, 'Bus %d:\n', i); fprintf(1, ' Voltage = %f p.u., %.1f deg.\n', ... V(i), theta(i)*180/pi); fprintf(1, ' Injected P = %f p.u.\n', P_calc(i)) if i > 1 fprintf(1, ' Scheduled P = %f p.u.\n', P_sch(i)); end fprintf(1, ' Injected Q = %f p.u.\n', Q_calc(i)) if i > PV+1 fprintf(1, ' Scheduled Q = %f p.u.\n', Q_sch(i)); end %Calculating the power loss P_loss = P_loss + P_calc(i); Q_loss = Q_loss + Q_calc(i); end %Line quantities fprintf(1, '\nPower flow from:\n'); for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); Y_shunt = line_data(line,5)/Yb; P_line = V(i)*(V(i)*-G(i,j) - V(j)*(-G(i,j)*cos(theta(i)-theta(j)) - B(i,j)*sin(theta(i)-theta(j)))); Q_line = V(i)*(V(i)*(B(i,j) - Y_shunt) + V(j)*(-B(i,j)*cos(theta(i)-theta(j)) + G(i,j)*sin(theta(i)-theta(j)))); fprintf(1, 'bus %d to bus %d\n', i, j); fprintf(1, ' P = %.4f, Q = %.4f\n', P_line, Q_line); end %Total loss fprintf(1, '\nThe total active power loss = %f\nThe total reactive power loss = %f\n'... , P_loss, Q_loss); iteration

Using the MATLAB program above, and using the results of problem 7.11, the Jacobian matrix is computed for the first iteration:

⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−−−−−−−−−

−−−−−−−−−−−−−

−

=

72.772.401.372.044.027.0072.412.2490.945.044.199.0001.390.941.2227.099.02.10

72.045.027.072.772.43045.044.199.072.412.259.95.1027.099.026.101.39.94.230

00005.1005.10

J

iii) After 3 iterations the Newton-Raphson method converges with tolerance of 510 −

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

×=Δ −

0591.01002.00711.00289.0

10P 8

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡×=Δ −

0227.01008.00497.0

10Q 8

iv) Solving the power flow problem using above Newton Raphson method with MATLAB program,

SOLUTION TABLE 7.15.1Bus voltage using Newton-Raphson method for problem 7.15


Bus no. Volts

(pu)

Angle

(deg)

Generation Load ΔP ΔQ

MW (pu)

MVar (pu)

MW (pu)

MVar (pu)

1 1.022 -1.7 0 0 0.1600 0.0800 0.06810−×

0.02810−×

2 1.035 -0.3 0 0 0.2000 0.1000 0.10810−×

0.10810−×

3 1.033 -1.2 0 0 0.3700 0.1300 0.07810−×

0.05810−×

4 (P-V) 1.050 2.3 0.5000 0.1734 0 0 0.03 0

810−×

5

(swing)

1.050 0.0 0.2307 0.1768 0 0 0 0


00069205

1

5

1.PPP

ii,L


==

04022405

1

5

1.QQQ

ii,L


==

SOLUTION TABLE 7.15.2 Power flow through transmission lines for problem 7.15

Ploss (pu) = 0.000692, Qloss (pu) = 0.040224

From Bus no.

To Bus no.

MW flow (pu)

MVar flow (pu)

4 2 0.500 0.173

2 3 0.171 -0.003

3 5 -0.230 -0.169

2 1 0.129 0.051

3 1 0.031 0.033

7.16 Consider the power grid given in Fig. 7.29.

Figure 7.29 System for problem 7.16

i) Bus 1 is Swing bus V1= 1∠0

ii) Bus 2 is P – V bus ⎜V2 ⎜= 1.05 and P2(Scheduled) = 0.9 p.u

iii) Bus 2 Q2 is assumed to be – 1≤ Q2≤2.

Compute:

i) The bus voltages using the decoupled Newton-Raphson method

Solutions

Fig. Per unit injection model for power flow studies of problem 7.16

YBus matrix is formulated using the algorithm in solution 7.2

jBGYBus +=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−−

−−−−−

=

18.1017.10091.256.135.117.156.157.383.0

035.183.018.2

G

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−

=

27.5029.50052.1245.410.829.545.480.1211.3010.811.317.11

B

The calculated injected active and reactive powers for each bus are given by:

( )∑=

θ+θ=N


1

( )∑=

θ−θ=N


1





The voltage of each bus is found out using decoupled Newton Raphson. After the solution has converged, the power flow in lines and the power losses are calculated using the same formula as given in solution 7.4.

The following MATLAB code is used to solve the power flow problem using decoupled Newton-Raphson method:

%% Power Flow: decoupled Newton-Raphson clc; clear all; tolerance=1e-5; PV = 1; %no. of PV buses Vb=1; %base voltage VAb=1; %base volt-amp

Zb=Vb^2/VAb; %base impedance Yb=1/Zb; %base admittance % Scheduled active and reactive powers P_sch = [0 0.9 -0.6 -0.8]'; Q_sch = [0 0.89914 -0.2 -0.4]'; %% Admittance matrix formation % line data line_data = [ 1 2 0.08 0.30 0; 1 3 0.02 0.12 0; 2 3 0.07 0.20 0; 2 4 0.03 0.18 0]; Yshunt = [0.045 0.072 0.02 0.02]; N = max(max(line_data(:,1:2))); G = zeros(N,N); B = zeros(N,N); % assemble admittance matrix for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); r = line_data(line,3)/Zb; x = line_data(line,4)/Zb; y_shunt = line_data(line,5)/Yb; G(i,i) = G(i,i) + r/(r^2 + x^2); B(i,i) = B(i,i) - x/(r^2 + x^2) + y_shunt; G(j,j) = G(j,j) + r/(r^2 + x^2); B(j,j) = B(j,j) - x/(r^2 + x^2) + y_shunt; G(i,j) = G(i,j) - r/(r^2 + x^2); B(i,j) = B(i,j) + x/(r^2 + x^2); G(j,i) = G(j,i) - r/(r^2 + x^2); B(j,i) = B(j,i) + x/(r^2 + x^2); end for i = 1:N B(i,i) = B(i,i) + Yshunt(i); end %% Power flow % Initialization of Voltages V=ones(N,1); V(1) = 1; % Swing bus voltage V(2) = 1.05; % Initialization of P-V bus voltages theta=zeros(N,1); % Initialization of bus voltage angles P_calc = zeros(N,1);

Q_calc = zeros(N,1); for k = 1:N for m = 1:N P_calc(k) = P_calc(k)... + V(k)*V(m)*(G(k,m)*cos(theta(k) - theta(m)) + B(k,m)*sin(theta(k) - theta(m))); Q_calc(k) = Q_calc(k)... + V(k)*V(m)*(G(k,m)*sin(theta(k) - theta(m)) - B(k,m)*cos(theta(k) - theta(m))); end end % allocate storage for Jacobian H = zeros(N-1,N-1); %dP/d th L = zeros(N-1-PV,N-1-PV); %V*dQ/dv mismatch_P = [P_sch(2:N)-P_calc(2:N)]; mismatch_Q = [Q_sch(2+PV:N)-Q_calc(2+PV:N)]; % Decoupled Newton-Raphson iteration iteration=0; while (iteration < 10) iteration=iteration+1; % calculate Jacobian for k = 2:N for m = 2:N if (k == m) H(k-1,m-1) = -Q_calc(k) - V(k)^2*B(k,m); if (i >= 2+PV && m >= 2+PV) L(k-PV-1,m-PV-1) = Q_calc(k) - V(k)^2*B(k,m); end else em = V(m)*cos(theta(m)); fm = V(m)*sin(theta(m)); ek = V(k)*cos(theta(k)); fk = V(k)*sin(theta(k)); am = G(k,m)*em - B(k,m)*fm; bm = B(k,m)*em + G(k,m)*fm; H(k-1,m-1) = am*fk - bm*ek; if (k >= 2+PV && m >= 2+PV) L(k-PV-1,m-PV-1) = am*fk - bm*ek; end end end end % calculate correction correction_theta = H\mismatch_P; correction_V = L\mismatch_Q; theta(2:N) = theta(2:N) + correction_theta(1:(N-1)); V(PV+2:N) = V(PV+2:N) .* (1+correction_V(1:(N-1)-PV));

% calculate mismatch and stop iterating % if the solution has converged P_calc = zeros(N,1); Q_calc = zeros(N,1); for k = 1:N for m = 1:N P_calc(k) = P_calc(k)... + V(k)*V(m)*(G(k,m)*cos(theta(k) - theta(m)) + B(k,m)*sin(theta(k) - theta(m))); Q_calc(k) = Q_calc(k)... + V(k)*V(m)*(G(k,m)*sin(theta(k) - theta(m)) - B(k,m)*cos(theta(k) - theta(m))); end end mismatch_P = [P_sch(2:N)-P_calc(2:N)]; mismatch_Q = [Q_sch(2+PV:N)-Q_calc(2+PV:N)]; mismatch = [mismatch_P; mismatch_Q]; if (norm(mismatch,'inf') < 1e-5) break; end end %% output solution data %Bus quantities P_loss = 0; Q_loss = 0; for i = 1:N fprintf(1, 'Bus %d:\n', i); fprintf(1, ' Voltage = %f p.u., %.1f deg.\n', ... V(i), theta(i)*180/pi); fprintf(1, ' Injected P = %f p.u.\n', P_calc(i)) if i > 1 fprintf(1, ' Scheduled P = %f p.u.\n', P_sch(i)); end fprintf(1, ' Injected Q = %f p.u.\n', Q_calc(i)) if i > PV+1 fprintf(1, ' Scheduled Q = %f p.u.\n', Q_sch(i)); end %Calculating the power loss P_loss = P_loss + P_calc(i); Q_loss = Q_loss + Q_calc(i); end %Line quantities fprintf(1, '\nPower flow from:\n'); for line = 1:size(line_data,1) i = line_data(line,1); j = line_data(line,2); Y_shunt = line_data(line,5)/Yb; P_line = V(i)*(V(i)*-G(i,j) - V(j)*(-G(i,j)*cos(theta(i)-theta(j)) - B(i,j)*sin(theta(i)-theta(j)))); Q_line = V(i)*(V(i)*(B(i,j) - Y_shunt) + V(j)*(-B(i,j)*cos(theta(i)-theta(j)) + G(i,j)*sin(theta(i)-theta(j))));

fprintf(1, 'bus %d to bus %d\n', i, j); fprintf(1, ' P = %.4f, Q = %.4f\n', P_line, Q_line); end %Total loss fprintf(1, '\nThe total active power loss = %f\nThe total reactive power loss = %f\n'... , P_loss, Q_loss); iteration SOLUTION TABLE 7.16.1The voltages and power of each bus from decoupled Newton-Raphson method for problem 7.16


Bus no.

Volts

(pu)

Angle

(deg)


MW (pu)

MVar (pu)

MW (pu)

MVar (pu)

1 1.000 0.0 0.5392 -0.2496

0 0 0 0

2 1.050 -2.3 0.9000 0.8991 0 0 5102632.0 −× 0

3 0.995 -3.1 0 0 0.600 0.200 5103845.0 −× 6105151.0 −×

4 0.942 -10.0 0 0 0.800 0.400 16103307.3 −× 16104409.4 −×


039204

1

4

1.PPP

ii,L


==

049604

1

4

1

.QQQi

i,Li

i,Gloss =−= ∑∑==

SOLUTION TABLE 7.16.2 Power flow through transmission lines and transformers for problem 7.16

Ploss (p.u) = 0.0392, Qloss (p.u) = 0.0496

From Bus no.

To Bus no.

MW flow (p.u)

MVar flow (p.u)

1 2 0.090 -0.188

1 3 0.450 -0.017

2 3 0.160 0.236

2 4 0.827 0.542

7.17 Consider the power grid given in Fig. 30

1 3

2

0.6 +

j 0.18

0.08 + j 0.24

0.02 + j 0.06

All Values in per Unit

Figure 7.30 Power grid for problem 7.19

Consider bus 1 as swing bus with 1.05∠0, bus 2 as P – V with P2 = 0.20 and ⎜V2 ⎜= 0.96, P3 and Q3 are -0.3 and -0.4, Compute:

i) B’, B’’ and YBus matrices.

ii) Compute fast decoupled load flow solution and line follow from bus k to bus m and system losses.

Solutions

Fig.Per unit injection model for power flow studies for problem 7.17

i) The YBus matrix formulation algorithm from solution 7.2 is used. jBGYBus +=

. .. .

. . .

. .. .

. . .

The matrix is the imaginary part of YBus and matrix is formulated same as YBus except the resistances and line charging are ignored in the formulation. In this problem, we do not have line charging.

matrix is the imaginary part of YBus given above, Since bus 1 is swing bus and its voltage (magnitude and angle) is given, first row and first column are not needed in for the calculation of voltage angle. Since bus 1 is swing bus and bus 2 is a P-V bus, the voltage magnitude is known for the first two buses, in the first two rows and columns are eliminated and only the third bus is considered whose voltage magnitude is unknown.

′ . .. .

′′ .

ii) The result after first iteration using FDLF program in solution 7.4, YBus as above and Psch = [0 0.2 -0.3]’; Qsch = [0 0 -0.4]’;

SOLUTION TABLE 7.17.1Bus voltages using FDLF method for problem 7.17

After 1st iteration

Bus no.

Volts

(p.u)

Angle

(deg)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.00 0 0.222 2.162 0 0 0 0

2 0.9 2.3 0.167 -1.586 0 0 0.0332 0

3 0.867 -1.6 0 0 0.301 0.333 0.0015 0.0668


087603

1

3

1.PPP

ii,L


==

243103

1

3

1.QQQ

ii,L


==

SOLUTION TABLE 7.17.2 Power flow through transmission lines for problem 7.17

Ploss (pu) = 0.0876, Qloss (pu) = 0.2431

From Bus no.

To Bus no.

MW flow (p.u)

MVar flow (p.u)

1 2 -0.0365 1.6908

1 3 0.2587 0.4713

2 3 0.0732 -0.0667

7.18 Consider the power system given in Figure 7.31. Assume the bus 1 has a gas turbine generator. And it has a load of 1 p.u; the bus voltage is fixed at 1 p.u and it is the swing bus. The bus 2 has a number of PV generators with total injected power of1.5 p.u into the bus. The transmission data is given in per unit as specified. The bus three has a number of loads with total connected load of 2.0 per unit. Perform the following by writing a MATLAB simulation testbed. The maximum mismatch tolerance is 0.001. Perform the following:

i) Compute the system YBus

ii) Use Gauss–Seidel Y bus and compute the bus 2 and 3 voltages

iii) Use Gauss–Seidel ZBus and compute the bus 2 and 3 voltages

iv) Use the Newton–Raphson method and compute the bus 2 and 3 voltages

v) Make a table and compare the above methods

vi) Determine the power provided by swing generator

vii) Determine the total power losses.

o011 ∠=V

Figure 7.31 One line diagram of problem 7.18

Solutions


i) For Gauss-Seidel with YBus approach and Newton-Raphson methods,

The YBus formulation algorithm is given in solution 7.2.

jBGYBus +=

For only resistive elements, B = 0.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−

=8571.128571.2108571.28571.6410414

G

For Gauss-Seidel with ZBus approach, a fictitious shunt resistance of 0.01 is assumed.

The new G,⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−−−

=8571.128571.2108571.28571.64104114

G

Using Psch = [0 1.5 -2.0]’; Qsch = [0 0 0]’;

ii) Gauss-Seidel YBus method: Using programs in solution 7.2

The program and algorithm is given in solution 7.2. All reactive (imaginary) components are zero.

SOLUTION TABLE 7.18.1 Bus voltage using Gauss-Seidel YBus Method for problem 7.18 part (ii)


Bus no.

Volts

(p.u)

Angle

(deg)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 1.039 0 0 0 0 0

2 1.129 0.0 1.500 0 0 0 0.22310−×

0

3 0.844 0.0 0 0 1.999 0 0.42310−×

0

iii) Gauss-Seidel ZBus method algorithm and program is given in solution 7.5

SOLUTION TABLE 7.18.2 Bus voltage using Gauss-Seidel ZBus Method for problem 7.18 part (iii)


Bus no.

Volts

(p.u)

Angle

(deg)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 1.039 0 0 0 0 0

2 1.129 0.0 1.499 0 0 0 0.12310−×

0

3 0.844 0.0 0 0 1.999 0 0.26310−×

0

iv) Newton-Raphson method: using program in solution 7.15. All reactive (imaginary) components are zero.

SOLUTION TABLE 7.18.2 Bus voltage using Newton-Raphson Method for problem 7.18 part (iv)


Bus no.

Volts

(p.u)

Angle

(deg)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 1.039 0 0 0 0 0

2 1.129 0.0 1.500 0 0 0 0.01910−×

0

3 0.844 0.0 0 0 2.000 0 0.29910−×

0

v)

Results obtained are tabulated below for comparison:

SOLUTION TABLE 7.18.4Bus voltages for problem 7.18 obtained from different methods

Method V1 V2 V3 Iterations

Gauss – Seidel Y Bus

1.000 1.129 0.844 8

Gauss – Seidel Z Bus

1.000 1.129 0.845 6

Newton - Raphson

1.000 1.129 0.844 4

It is seen that the result obtained from the three methods is same up to the third place after decimal and the number of iterations is different.

vi) The power provided by the swing bus =

SOLUTION TABLE 7.18.5 Power from the swing bus for problem 7.18 using different solution methods

Method P1


1.039


1.038

Newton - Raphson

1.040

vii) Active total power loss are given by:

∑∑==

−=3

1

3

1 ii,L

ii,Gloss PPP

SOLUTION TABLE 7.18.6 Power loss for problem 7.18 using different solution methods

Method Line Losses


0.540


0.538

Newton - Raphson

0.540

7.19 Consider the power system given in Fig. 7.32.

Figure 7.32 Power grid of problem 7.19

Assume )BusSwing(.u.p01V o1 ∠= . Also, assume transmission line impedances are

given in per unit on 440 V, 100 MVA base ( MVA100Sb = for the entire system)

Assume generation and load schedules are as follows:

Bus 1:

Load # 1: 4 MVA, pf=0.85 (lagging)

Bus 2:

G2: 2 MW, pf=0.95 (lagging),

Load # 2: 4 MVA, pf=0.90 (leading)

Bus 3:

G3: 1 MW, pf=0.95 (leading),


Compute the bus voltages using decoupled Newton-Raphson and Gauss Seidel methods

Solutions

j0.12 p.u. j0.40

p.u.

Fig.Per unit injection model for problem 7.19

Sb = 100 MVA

Using the YBus formulation algorithm in solution 7.2,

jBGYBus +=

G = 0 since there is no resistive element in the network.

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

83.1050.233.850.207.657.333.857.39.11

B

Using the program for decoupled Newton-Raphson method in solution 7.16 and the above YBus and Psch = [0 -0.016 -0.0080]’; and Qsch = [0 0.0109 0.01120]’; given for this problem we solve problem 7.19.

SOLUTION TABLE 7.19.1 Bus voltage using decoupled Newton-Raphson Method for problem 7.19

Sb = 100 MVA, number of iterations = 3

Bus no.

Volts

(p.u)

Angle

(deg)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

1 1.000 0.0 0.064 -0.044 0.040 0.021 0 0

2 1.002 -0.2 0.020 -0.007 0.036 -0.174 0.0952610−×

0.1051610−×

3 1.002 -0.1 0.010 0.003 0.018 -0.087 0.0137610−×

0.0387610−×

03

1

3

1=−= ∑∑

== ii,L

ii,Gloss PPP

The active power loss is zero since there is no resistive element in the network.

0001103

1

3

1

.QQQi

i,Li

i,Gloss =−= ∑∑==

SOLUTION TABLE 7.19.2 Power flow through transmission lines using decoupled Newton-Raphson method for problem 7.19

Ploss (pu) = 0.00, Qloss (pu) = 0.00011

From Bus no.

To Bus no.

MW flow (pu)

MVar flow (pu)

1 2 0.012 -0.009

1 3 0.012 -0.014

2 3 -0.004 0.002

Using the MATLAB program in 7.2 with the above YBus and Psch = [0 -0.016 -0.0080]’; and Qsch = [0 0.0109 0.01120]’;

SOLUTION TABLE 7.19.3 Bus voltages using Gauss-Seidel YBus Method for problem 7.19

Sb = 100 MVA, number of iterations = 7

Bus no.

Volts

(pu)

Angle

(deg)


MW (pu)

MVar (pu)

MW (pu)

MVar (pu)

1 1.000 0.0 0.064 -0.044 0.040 0.021 0 0

2 1.002 -0.2 0.020 -0.007 0.036 -0.174 0.1607510−×

0.2118510−×

3 1.002 -0.8 0.010 0.003 0.018 -0.087 0.5617510−×

0.3682510−×

03

1

3

1=−= ∑∑

== ii,L

ii,Gloss PPP

The active power loss is zero since there is no resistive element in the network.

0001103

1

3

1

.QQQi

i,Li

i,Gloss =−= ∑∑==

SOLUTION TABLE 7.19.4 Power flow through transmission lines using Gauss-Seidel YBusMethod for problem 7.19

Ploss (pu) = 0.00, Qloss (pu) = 0.00011

From Bus no.

To Bus no.

MW flow (pu)

MVar flow (pu)

1 2 0.012 -0.008

1 3 0.012 -0.013

2 3 -0.004 0.002

7.20 Consider the system of figure 7.33.

0.08 + j 0.24

0.45 + j 0.45

All Values in P.U Unit

1 3

2

Figure 7.33 The system for problem 7.20

Assume that generators internal reactance in are 0.8 p.u. Assuming the load voltage is one P.U, compute the following:

i) YBus model for short circuit studies. ii) ZBus model for short circuit studies

Solutions

Fig. Per unit model for short circuit studies

( )Ω+=

+== u.p11.1j11.1

45.0j45.01

S

VZ *

2

*u.p

2u.p

load

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−+−+−+−−+−+−+−−

=4586.0j4558.00009.0j0030.00073.0j0024.00009.0j0030.02799.1j0126.00290.0j0097.00073.0j0024.00290.0j0097.02863.1j0121.0

YBus

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−++−++−++−+

=0970.1j0903.10036.0j0017.00083.0j0041.0

0035.0j0017.07816.0j0074.00178.0j0056.00083.0j0041.00178.0j0056.07778.0j0071.0

ZBus

7.21 For the radial power system given below, compute the following:

Figure 7.34 System of problem 7.21

Compute the following:

i) For the load of 500W, which bus should be considered as an infinite bus? ii) For the load of 5 MW, which bus should be considered as an infinite bus?

Solutions

Let the Sb = 100 MVA and Vb be 345 kV on the utility side

Ω=×== u.p1.0j1001001.0j

SS

XXold,b

new,bold,u.pu.p,1T

Ω=×== u.p2.0j50

1001.0jSS

XXold,b

new,bold,u.pu.p,2T

Ω=×== u.p5.0j1010005.0j

SS

XXold,b

new,bold,u.pu.p,2T

Ω×=×

== − u.p10510100

500S

SS 6

6b

loadu.p,load

Fig. Per unit model of the system for power flow studies

i) Let us assume that bus 2 is the infinite bus, then the problem reduces to a 2 bus system as given in the figure below:

°∠= 01V u.p

Fig. Per unit model with bus 2 as infinite bus

Using the YBus formulation algorithm in solution 7.2,

jBGYBus +=


⎥⎦

⎤⎢⎣

⎡−

−=

2222

B

Using Gauss Seidel YBus method given in solution 7.2, to solve the power flow problem with the program in solution 7.2 and YBus given as above and P_sch = [0 -5e-6]';Q_sch = [0 0]';

SOLUTION TABLE 7.21.1 Bus voltages using Gauss-Seidel YBus method form problem 7.21 part (i)

Sb = 100 MVA, Vb = 345 kV on the utility side


Bus no.

Volts

(p.u)

Angle

(deg)

Generation Load ΔP ΔQ

MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

2 1.000 0.0 6105 −× 0 0 0 0 0

1 1.000 0.0 0 0 6105 −× 0 221047.8 −× 101023.0 −×

It is seen that the voltage at bus 1 does not drop from voltage at bus 2 for a load of 500 W and therefore, bus 1 can be considered to be the infinite bus for a load of 500 W

ii) Let us assume that bus 3 is the infinite bus for a load of 5 MVA, then the problem reduces to a 3 bus system as given below:

°∠= 01V u.p

Fig. Per unit model assuming bus 3 as infinite bus

jBGYBus +=


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

220275055

B

Ω×=

××

== − u.p10510100

105S

SS 2

6

6

b

loadu.p,load

Using Gauss Seidel YBus method given in solution 7.2, to solve the power flow problem with the program in solution 7.2 and YBus given as above and P_sch = [0 0 -5e-2]';Q_sch = [0 0 0]';

SOLUTION TABLE 7.21.2 Bus voltages using Gauss-Seidel YBus method form problem 7.21 part (ii)

Sb = 100 MVA, Vb = 345 kV on the utility side


Bus no.

Volts Angle Generation Load ΔP ΔQ

MW MVar MW MVar

(p.u) (deg) (p.u) (p.u) (p.u) (p.u)

3 1.000 0.0 0.05 0.002 0 0 0 0

2 0.999 -0.6 0 0 0 0 101003.0 −× 101038.0 −×

1 0.999 -2.0 0 0 0.050 0 101046.0 −× 101001.0 −×

It is seen that the voltage at bus 2 and 1 drops from that of bus 3. Therefore, bus 3 is considered to be infinite bus for a load of 5 MVA

7.22 Consider the microgrid power system given in Fig. 7.35

Figure 7.35 The system of problem 7.22

The transmission line impedances are given in per unit on 100 MVA base (Sb 100MVA for the entire system.) The generation and load schedules are as follows:

Bus 1:

Load # 1: 4 MVA, pf=0.85 (lagging)

Bus 2:

G2: 2 MW, pf=0.95 (lagging), X=10%


Bus 3:

G3: 1 MW, pf=0.95 (leading), X=25%



i) Compute bus load voltages using FDLF power flow method. ii) Compute the load impedance of each load

Solutions

j0.12 p.u. j0.40

p.u.


The bus admittance matrix for power flow studies:

jBGYBus +=


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

83.1050.233.850.207.657.333.857.39.11

B

i) Using the FDLF program in solution 7.4 with the above YBus and Psch = [0 -0.016 -0.0080]’; and Qsch = [0 0.0109 0.01120]’;

SOLUTION TABLE 7.22.1 Bus Voltages using FDLF Method for problem 7.22

MVA100Sb =


Bus no.

Volts

(pu)

Angle

(deg)


MW (pu)

MVar (pu)

MW (pu)

MVar (pu)

1 1.000 0.0 0.064 -0.044 0.040 0.021 0 0

2 1.002 -0.2 0.020 -0.007 0.036 -0.174 0.3351610−×

0.5007610−×

3 1.001 -0.1 0.010 0.003 0.018 -0.087 0.0379610−×

0.2131610−×


03

1

3

1=−= ∑∑

== ii,L

ii,Gloss PPP

The active power loss is zero because there is no resistive element in the network.

0001103

1

3

1

.QQQi

i,Li

i,Gloss =−= ∑∑==

SOLUTION TABLE 7.22.2 Power flow through transmission lines using FDLF for problem 7.22

Ploss (pu) = 0.00, Qloss (pu) = 0.00011

From Bus no.

To Bus no.

MW flow (pu)

MVar flow (pu)

1 2 0.012 -0.009

1 3 0.012 -0.014

2 3 -0.004 0.002

ii) Assuming Sb = 100 MVA,

( )Ω+=

∠==

−u.p1696.13j2500.21

85.0cos04.0000000.1

S

VZ *1

2

*u.p

2u.p

1,load

( )Ω+=

∠==

−u.p9512.10j6113.22

9.0cos04.0002471.1

S

VZ *1

2

*u.p

2u.p

2,load

( )Ω+=

∠==

−u.p8676.21j1509.45

9.0cos02.0001675.1

S

VZ *1

2

*u.p

2u.p

3,load

1

CHAPTER 8

POWER GRID AND MICROGRID FAULT STUDIES

8.1 Consider a typical power system given in Fig. 8.53.All reactances are in p.u on a 100MVA base.

Xth = 0.01 when the maximum number of generators is in service

Xth = 0.015 when the minimum number of generators is in service

Figure 8.53 A Typical Power Grid System


i) The SCC of the 415 V bus when all transformers are in service, but generator G1 is not in service. Assume that the maximum number of generators are in service.

2

ii) The SCC of the 415 V bus when all the transformers and G1are in service. Assume one generator is in service.

Solutions

The MVA base is given as MVA100Sb = and 275 kV on the infinite bus bar

i) The equivalent circuit diagrams when generator G1 is out of service

Zf = 0 1

j 5.248Vb = 415 V

(c)Ground

3

Solution Figure 8.1.1(a)–(d) The Impedance Diagrams from problem 8.1 part (i)

thf Z

SCCI 1==

u.p..j

SCC 1905024851

==

ii) The equivalent circuit diagrams for fault analysis when one of the generators is in service:

4

Solution Figure 8.1.2(a)–(d) The Impedance Diagrams for problem 8.1 part (ii)

thf Z

I 1=

u.p..j.j

SCC 190802070330

1=

+=

8.2 Consider the power system given in Fig. 8.54.

Figure 8.54 The One-Line Diagram of Problem 8.2.

Generator A:

X″G(1) = 0.25, X″G(2) = 0.15, X″G(0) = 0.03 p.u

Generator B:

X″G(1) = 0.2, X″G(2) = 0.12, X″G(0) = 0.02 p.u

5

Transmission line C–D: Z1 = Z2 = j0.08, Z0 = j0.14 p.u

Transmission line D–E: Z1 = Z2 = j0.06, Z0 = j0.12 p.u

All values are given in per unit with an MVA base of 100.

Assume generator A is Y connected and ungrounded and generator B is Y connected and grounded, compute the single line to ground fault at bus D and the fault current and actual phase voltages (i.e., Va, Vb, Vc in p.u) of buses C, D, and E.

Solutions

Solution Figure 8.2.1TheThree-Phase Diagram and Single to Ground Fault at Bus D.

The equivalent circuit for a single line to ground fault at bus D is shown in Solution Fig. 8.2.2.

(a)The Positive Sequence Network.

6

(b)The Negative Sequence Network.

(c)The Zero Sequence Network.

(d) The Connection Diagram of Sequence Networks for a Single Line to Ground Fault at Bus D.

(e)The Simplified Circuit of Fig. 8.2.2(d).

Solution Figure 8.2.2 The Equivalent Circuit for a Single Line to Ground Fault at Bus D.

7

From the equivalent circuit,

oo

9056200 −∠=

∠=

++==== −+

++ .0.39j

01ZZZ

VIIIIDDD

p.u A

Therefore the fault current at bus D is given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ °−∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

00

90687

11

111 0

2

2

.

III

aaaa

III

cD

bD

aD

Fault current contributions from bus C to D is given by:

00 =aCDI p.u A

oo 90121330260

26090562 −∠=+

×−∠=+ ..j.j

.j.IaCD p.u A

oo 90121230180

18090562 −∠=+

×−∠=− ..j.j

.j.IaCD p.u A

The actual currents in the line C–D are as follows:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠

∠

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

12112124200

.j

.j.j

90-1.1290-1.12

a a 1

aa 1

1 1 1

I

I

I

T III

2

2

CD

CD

CD

s

cCD

bCD

aCD

o

o

Fault current in the line E–D are as follows:

o90-2.56II0aED ∠== p.u A

°−∠== ++ 90441I-II aCDaED . p.u A

°−∠== −− 90441I-II aCDaED . p.u A

The actual currents from bus E to D are given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠∠∠

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

j1.12-j1.12-j5.44-

90-1.4490-1.4490-2.56

a a 1

aa 1

1 1 1

I

I

I

T III

2

2

ED

ED

ED

s

cED

bED

aED

o

o

o0

8

The sequence voltages at bus C are given by:

360905621400 00 ...jIjXV AC −=−∠×−=×−= o p.u V

720901212501 ...jIjXEV aCDAC =−∠×−=×−= +++ o p.u V

170901211500 ...jIjXV aCDAC −=−∠×−=×−= −−− o p.u V

The actual voltages at bus C are given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠

∠=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

o

o

129.481129.48-1

.

0.17-0.720.36-

a a 1

aa 1

1 1 1

V

V

V

T VVV

2

2

C

C

C

s

cC

bC

aC 1900

The sequence voltages at bus D are given by:

360905621400 00 ...jIjXV 0DD −=−∠×−=×−= o p.u V

620905621501 ...jIjXEV DD =−∠×−=×−= ++ o p.u V

26090562100 ...jIjXV DD −=−∠×−=×−= −− o p.u V

The actual voltages at bus D are given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠

∠=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

o

o

125.40.93125.4-0.93

0.26-0.620.36-

a a 1

aa 1

1 1 1

V

V

V

T VVV

2

2

D

D

D

s

cD

bD

aD 00

The sequence voltages at bus E are given by:

050905620200 00 ...jIjXV 0EDBE −=−∠×−=×−= o p.u V

710904412001 ...jIjXEV aEDBE =−∠×−=×−= +++ o p.u V

170904411200 ...jIjXV aEDBE −=−∠×−=×−= −−− o p.u V

9

The actual voltages at bus E are given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠

∠=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

o

o

112.780.83112.78-0.83

.

0.17-0.710.05-

a a 1

aa 1

1 1 1

V

V

V

T VVV

2

2

E

E

E

s

cE

bE

aE 4900

8.3 Consider the power grid given in Fig. 8.55.


The sequence network data are

Generator: 20 kV, 100 MVA, 10% positive reactance, negative sequence reactance is equal to positive sequence reactance, zero sequence reactance is equal to 8% based on the generator’s rating.

Transmission line length: bus 1–2 = 50 km, reactance = 0.5 Ω/km

bus 2–3 = 100 km, reactance = 0.7 Ω/km

Transformer T1: 20 kV/138 kV, 8% reactance, 150 MVA

Transformer T2: 138kV / 13.8 kV, 10% reactance, 200 MVA

Transmission line sequence impedances: Z1 = Z2 = j0.06 p.u, Z0 = j0.12 in p.u on a 100 MVA base.

10

Transformers’ sequence impedances: positive = negative = zero

Load: SLoad = 50 MVA, p.f. = 0.95 lagging.

Generator ground impedance = j0.01 p.u based on its rating


i) Compute the per unit model for positive, negative and zero sequence networks based on 100 MVA.

Solution

MVASb 100= , kVVb 20= on the generator side

Base voltage on HV side of T1 = kVVVVV LV,b

LV

HVHV,b 13820

20138

=×==

Ω=== 19010013822

b

bb S

VZ

10.XX GG == −+ , Ω= u.p.X G 0800

Ω=×

== −+ u.p..XX 130190

50501212

Ω=×

== −+ u.p..XX 370190

701002323

Ω=×⎟⎠⎞

⎜⎝⎛×=×⎟⎟

⎠

⎞⎜⎜⎝

⎛=== −+ u.p..

SS

VV

XXXXold,b

new,b

new,b

old,boldTTT 0530

150100

138138080

22

0111

Ω=×⎟⎠⎞

⎜⎝⎛×=×⎟⎟

⎠

⎞⎜⎜⎝

⎛=== −+ u.p..

SS

VV

XXXXold,b

new,b

new,b

old,boldTTT 050

200100

13813810

22

0222

Load °∠=°∠

= 191850100

191850 ...SL

6245.09.119.18219.185.0

12

*

2

jSV

Zpu

puL +=°∠=

°−∠==

11

Solution Figure 8.3.1(a) Zero Sequence Network for problem 8.3

Where XGr0, XGr,load

0 is the reactance of the generator and load grounding reactances respectively, Rload

0, Xload0 are the zero sequence resistance and reactance of the load respectively.

Solution Figure 8.3.1(b) Positive Sequence Network for problem 8.3

Solution Figure 8.3.1(c) Negative Sequence Network for problem 8.3

8.4 For Problem 8.3, perform the following:

i) Compute the load voltage if the generator is set at 5% above its own rating.

ii) For a double line to ground fault at bus 3, find the fault currents flowing from bus 1 and bus 2 to bus 3 (faulted bus) when the load is ignored.

12

Solutions

i) Calculating load flow using Newton Raphson method program given in solution 7.15

Solution Table 8.4.1 Load flow solution for problem 8.4 part i

MVASb 100= , kVVb 20= on the generator side


Bus no.

Volts

(p.u)

Angle

(deg)


MW (p.u)

MVar (p.u)

MW (p.u)

MVar (p.u)

0 1.05 0.0 0.475 0.031 0 0 0 0

1 1.048 -0.3 0 0 0 0 0.61 810−× 0.68 810−×

2 1.042 -1.1 0 0 0 0 0.29 810−× 0.57 810−×

3 1.027 -3.5 0 0 0 0 0.004810−×

0.54 810−×

4 1.025 -3.8 0 0 0.475 0.156 0.97 810−× 0.02 810−×

Table 8.4.2 Power flow through transmission lines

Ploss (p.u) = 0.0000, Qloss (p.u) = 0.0359

From Bus no.

To Bus no.

MW flow (p.u)

MVar flow (p.u)

0 1 0.4750 0.1920

1 2 0.4750 0.1889

2 3 0.4750 0.1811

3 4 0.4750 0.1591

13

ii) The equivalent circuit for the fault ignoring the load:

Solution Figure 8.4.1(a) The Equivalent Circuit for the Fault for problem 8.4

Solution Figure 8.4.1(b) The Simplified Circuit of Solution Figure 8.4.1(a)

Solution Figure 8.4.1(c) The Simplified Circuit of Figure 8.4.1(b)

The sequence currents for the fault at bus 3 are given by:

14

431104606530

121 .j

.j.jII −=

+== +

−+

329102

023

0 .jjXjX

jXIIIT

=+

×−== −

−

−

10200221 .jIIII =+−== −

−−

The fault current at bus 3 is given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

o

o

56.32.40123.72.40

j0.102j1.431-

j1.329

a a 1

aa 1

1 1 1

I

I

I

T III

2

2s

c

b

a 00

The sequence currents from bus 1 and 2 to the fault at bus 3 are given by:

431104606530

121 .j

.j.jI −=

+=+

−

0021 =−I

10200221 .jIII =+−=−

−

The actual currents from bus 1 and 2 to the fault at bus 3 are given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠∠∠

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

+−

−

o

o

o

261.48153.41.48

90-.

j0.102j1.431-

0

a a 1

aa 1

1 1 1

I

I

I

T III

2

2s

c

b

a 321

21

21

021

8.5 For Problem 8.3, perform the following:

i) Compute the single line to ground fault current at bus 2, but take the load into consideration

ii) The same as part i, but assume the generator is not grounded

Solutions

15

Load °∠=°∠

= 191850100

191850 ...SL

6245.09.119.18219.185.0

12

*

2

jSV

Zpu

puL +=°∠=

°−∠==

i) The equivalent circuit for the fault:

Solution Figure 8.5.1(a) The Equivalent Circuit for single line to ground Fault for problem 8.5 part (i)

16

Solution Figure 8.5.1(b) The Simplified Circuit of Figure 8.5.1(a).

Solution Figure 8.5.1(c) The Simplified Circuit of Figure 8.5.1(b).

The sequence currents at bus 2 for single line to ground fault at bus 2 are given by:

°−∠===== −+ 179633122

02 ..

0.270j + 0.052IIII

The actual fault currents are bus 2 are given by:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ °−∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

°−∠°−∠°−∠

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

00

..

..

..

..

a a 1

aa 1

1 1 1

I

I

I

T III

2

2

2

2

2

s

2-c,1

2-b,1

2-a,1 1798910

1796331796331796330

The sequence currents from bus 1 and 3 to bus 2 are given by:

17

00 =2-1I

°−∠==− 179633023 ..II

°−∠=+++×+

=+ 335444404450912630

04451911 ...j..j

).j.(II 2-1

°∠=−= ++− 717890123 ..III 2-1

°−∠=++

+=− 884413

263004451910445191 ..

.j.j..j.II 2-1

°−∠=−= −−− 12441023 ..III 2-1

The actual fault current from bus 1 to bus 2:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠∠∠

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

°−∠°−∠

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

o

o

o

103.025.62136.902.23

67.52-.

....

a a 1

aa 1

1 1 1

I

I

I

T III

2

2

2-1

2-1

2-1

s

2-c,1

2-b,1

2-a,1 577

8844133354444

00

The actual fault current contribution from bus 3 to bus 2:

[ ]⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∠∠∠

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

°−∠°∠°−∠

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

o

o

o

021039036

587

12441071789011796330

.5.62

.12.2367.52-.

..

....

a a 1

aa 1

1 1 1

I

I

I

T III

2

2

2-3

2-3

2-3

s

2-c,3

2-b,3

2-a,3

ii) When the generator is not grounded, the zero sequence equivalent circuit will change and the single line to ground fault equivalent circuit will become:

18

Solution Figure 8.5.2(a) The Equivalent Circuit for single line to ground Fault for problem 8.5 part (ii) Since the generator is connected to bus 1 through a Y-Δ transformer, the zero sequence circuit of the generator side is isolated from bus 2. Therefore, the fault currents remain the same as in part (i) for a single line to ground fault.


Figure 8.56A One-Line Diagram for Problem 8.6.

The impedances of the transmission lines are given in the one-line diagram (Fig. 8.56).

19

The system data are:

PV generating station: 2 MW, 460 V AC; positive, negative, and zero sequence impedances=10%

Gas turbine generating station: 10 MVA, 3.2 kV, 10% reactance Sequence impedance: negative sequence = positive sequence, zero sequence = ½ positive sequence

Transformers’ sequence impedances: positive = negative = zero

Transformer T1: 10 MVA, 460 V/13.2 kV, 7% reactance

Transformer T2: 25 MVA, 13.2 kV/69 kV, 9% reactance

Transformer T3: 20 MVA, 13.2 kV/3.2 kV, 8% reactance

Loads: S4: 4 MW, p.f. = 0.9 lagging; S5: 8 MW, p.f. = 0.9 lagging; S6: 10 MVA, p.f. = 0.9 leading; S7: 5 MVA, p.f. = 0.85 lagging.

Local power grid: positive, negative, and zero sequence internal reactance = 10 Ω; negligible internal resistance


i) Per unit impedance model for positive, negative, and zero sequences

ii) Ignore the loads and compute a single line to ground fault current at bus 4.

iii) Compute the load voltages; use the load impedance models and compute the single line to ground fault current at bus 4.

Solutions

i) MVASb 10= , kV.Vb 213= on the transmission line side.

Ω=== 4241710

213 22

..S

VZ

b

bb

287000287042417

5505454 .j.

.j.ZZ u.p,,u.p,, +=

+== −+ ,

14350014402

2870002870054 .j..j.Z u.p,, +=

+=

20

573900574042417

1016464 .j.

.jZZ u.p,,u.p,, +=

+== −+ ,

28700028702

5739005740064 .j..j.Z u.p,, +=

+=

344400712042417

6306565 .j.

.j.ZZ u.p,,u.p,, +=

+== −+ ,

17220035602

3444007120065 .j..j.Z u.p,, +=

+=

147814241720

7676 .j.jZZ u.p,,u.p,, === −+ , 57390

2147810

76 .j.jZ u.p,, ==

5739042417100 .j.jZZZ u.p,utilu.p,utilu.p,util ==== −+

10.jZZ u.p,gasu.p,gas == −+

0502

100 .j.jZ u.p,gas ==

502

1010 ..ZZ u.p,PVu.p,PV =×== −+

2502500 ..Z u.p,PV ==

0701010

213213070

22

10111 .

.

..SS

VV

XXXXold,b

new,b

new,b

old,boldTnew,Tnew,Tnew,T =×⎟

⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛=== −+

0402510

213213090

22

10

222 ....

SS

VV

XXXXold,b

new,b

new,b


⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛=== −+

0402010

213213080

22

10

333 ....

SS

VV

XXXXold,b

new,b

new,b


⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛=== −+

21

Solution Figure 8.6.1(a) The Positive Sequence Network for problem 8.6

Solution Figure 8.6.1(b) The Negative Sequence Network for problem 8.4

22

Solution Figure 8.6.1(c) The Zero Sequence Network for problem 8.6

ii) It is assumed that the voltage at bus 4 before the fault occurred was 1 p.u since it should be within 5% of 1 p.u.

From the bus admittance matrix:

Z44+ =0.1340 + j0.2101

Z44- = 0.1340 + j0.2101

Z440 = 0.0192 + j0.7643

Fig. The equivalent circuit for single line to ground fault for problem 8.6

Since the PV sources have high internal resistance, the fault current contribution from the PVs is neglected.

23

Therefore, the sequence currents for a single line to ground fault =

( ) ( ) ( ) °−∠=+++++

=++

=== −+−+ 476820

76430019202101013400210101340011

0444444

0 ...j..j..j.ZZZ

III fff

The actual fault currents are given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ °−∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

00

476462

11

111 0

2

2

..

III

aaaa

III

c

b

a

iii) The loads in p.u and the equivalent impedance of the loads are given as follows:

1904009010

904 1

4 .j..

.cos S +=

×−∠

=−

, ( )

960042190400

12

4 .j..j.

Z * +=+

=

3808009010

908 1

5 .j..

.cos S +=

×∠

=−

, ( )

480021380800

12

5 .j..j.

Z * +=+

=

44090010

9010 1

6 .j..cos S −=

−∠=

−

, ( )

440900440900

12

6 .j..j.

Z * −=−

=

26043010

8505 1

7 .j..cos S +=

∠=

−

, ( )

031701260430

12

7 .j..j.

Z * +=+

=

With the load impedances included,


Z44+ =0.1363 + j0.1675

Z44- = 0.1363 + j0.1675

Z440 = 0.3025 + j0.3033

Since the PV sources have high internal resistance, the fault current contribution from the PVs is neglected.

Therefore, the sequence currents for a single line to ground fault =

24

( ) ( ) ( ) °−∠=+++++

=++

=== −+−+ 48161

30330302501675013630167501363011

0444444

0 ..j..j..j.ZZZ

III fff

The actual fault currents are given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ °−∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

00

48483

11

111 0

2

2

.

III

aaaa

III

c

b

a


Figure 8.57A Microgrid of Distributed Generation Connected to a Local Network.

The system data are:

Local power grid SCC = 1600 MVA

PV-generating stations: ungrounded, approximate internal impedance (resistive only) = 50% for positive, negative, and zero sequences, 100 MVA.

Transformers: 460 V Y-grounded/13.2 kV Δ, 10% reactance, 10 MVA

Local power grid transformer: 20 MVA, 63 kV / 13.2 kV, 7% reactance

25

Transmission line: resistance = 0.0685 Ω/mile, reactance = 0.40 Ω/mile, half of line charging admittance (Y′/2) = 11 mS/mile. The length of Line 4–7 : 10 miles, 4–8 : 7 miles, 5–6 : 12 miles, 5–7 : 7 miles, 6–7 : 6 miles, 6–8 : 8 miles.

Transmission line sequence impedance: positive = negative; zero sequence = 2× positive sequence impedance.


i) Per unit equivalent model for positive, negative, and zero sequence impedances based on 20 MVA base

ii) For three phase fault on bus 4, compute the bus 4 SCC

Solutions

i) MVASb 20= , Vb = 460 V on PV generator side

The voltage base on transmission line side kV..Vb 213460460

213=×=

The p.u. SCC of the utility is given by

8020

1600==

bSSCC

Therefore, the internal p.u impedance of the utility is

0125080110 .j

SCCZZZ

puutilutilutil ===== −+

The internal p.u impedance of the PV station at 20 MVA base is given by

( )

( )

( )

( )

2

⎟⎟⎠

⎞⎜⎜⎝

⎛××=

newb

oldb

oldb

newb)old(pu)new(pu V

VVAVA

ZZ

10460460

10100102050

2

6

60 ..ZZZ PVPVPV =⎟

⎠⎞

⎜⎝⎛×

××

×=== −+

26

200460460

1020100

20 ..ZZZ trantrantran =⎟

⎠⎞

⎜⎝⎛××=== −+

The per unit impedances of the lines are given as follows:

Base impedance on the transmission line side = Ω== 71820

213 22

..S

V

b

b

( ) 4592007860718

4006850107474 .j.

..j.ZZ u.p,u.p, +=

+== −

−+

− , 95871811110

74 j./

Y u.p, =×

=−

( ) 918401572045920078602074 .j..j.Z u.p, +=+=−

( ) 3215005510718

400685078484 .j.

..j.ZZ u.p,u.p, +=

+== −

−+

− , 6717181

11784 j

./Y u.p, =

×=−

( ) 643001102032150055102084 .j..j.Z u.p, +=+=−

( ) 5511009440718

4006850126565 .j.

..j.ZZ u.p,u.p, +=

+== −

−+

− , 115071811112

65 j./

Y u.p, =×

=−

( ) 102211888055110094402065 .j..j.Z u.p, +=+=−

( ) 3215005510718

400685077575 .j.

..j.ZZ u.p,u.p, +=

+== −

−+

− , 6717181

11775 j

./Y u.p, =

×=−

( ) 643001102032150055102075 .j..j.Z u.p, +=+=−

( ) 2755004720718

400685067676 .j.

..j.ZZ u.p,u.p, +=

+== −

−+

− , 5757181

11676 j

./Y u.p, =

×=−

( ) 551000944027550047202076 ...j.Z u.p, +=+=−

( ) 3674006290718

400685088686 .j.

..j.ZZ u.p,u.p, +=

+== −

−+

− , 7667181

11886 j

./Y u.p, =

×=−

( ) 734801258036740062902086 .j..j.Z u.p, +=+=−

27

Solution Figure 8.7.1 The Positive Sequence Network for problem 8.7

Solution Figure 8.7.2 The Negative Sequence Network for problem 8.7

28

Solution Figure 8.7.3 The Zero Sequence Network for problem 8.7

ii) From the ZBus matrix, the value of Z44+ = 0.0114+j0.1446

90611

44

.j0.1446+0.0114Z

SCC puAC === +

The contribution to the fault current from the DC sources should be neglected because the internal resistance of the PV is high.

8.8 For Problem 8.7, for a single line to ground fault on bus 1, compute the ground fault current.

The positive, negative and zero sequence network is given in Solution Fig. 8.7

From the bus admittance matrix,

Z11+ =0.00004+j0.01227

Z11- = 0.00004+j0.01227

Z110 = j0.0125

29

Fig. Equivalent circuit of single line to ground fault at bus 1 for problem 8.8

The sequence currents for fault at bus 1 are given by:

( ) ( ) °−∠=++++

=++

=== −+−+ 98927

01250012270000040012270000004011

0111111

0 ..j.j..j.ZZZ

III fff

The actual currents for fault at bus 1 are given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ °−∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

00

98981

11

111 0

2

2

.

III

aaaa

III

c

b

a

8.9 For Problem 8.7, to increase the security of the system two identical transformers are used at each distribution network and at the interconnection to the local power grid. Compute the SCC of bus 4.

Solution

From the impedance matrix, and equivalent circuits in solution 8.7,

Z44+ = 0.0054+ j0.0881

31111

44

.j0.0881 +0.0054 Z

SCC AC,pu === +

8.10 For Problem 8.7, assume that transformer T1 is grounded Y-Y. Compute the line to ground fault current at bus 4.

Solution

30

The positive and negative sequence networks remain unaltered as in Problem 8.7. However, the zero sequence network changes to as given below assuming Y-Y connection with grounding on both sides:

Solution Figure 8.10.1 The Zero Sequence Network for problem 8.10


Z44+ =0.0114 +j 0.1446

Z44- = 0.0114 +j 0.1446

Z440 =j0.2125

Fig. Equivalent circuit for single line to grind fault for problem 8.10

Therefore, the sequence current for a single line to ground fault neglecting the contribution from PV =

31

( ) ( ) °−∠=++++

=++

=== −+−+ 487991

212501446001140144600114011

0444444

0 ...j.j..j.ZZZ

III fff

The actual fault currents for single line to ground fault at bus 4 are given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ °−∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

00

487975

11

111 0

2

2

..

III

aaaa

III

c

b

a

8.11 For Problem 8.7, assume that transformer T1 is grounded Y-Y. Compute the double line to ground fault current at bus 4.

Solution

The sequence networks remain as they are given in solution 8.10. The fault current contribution from PV is ignored as the impedance of the PV is high.

Z4‐40

Z4‐4+

Z4‐4‐

I+

I‐ I0

Fig. Equivalent circuit for double line to ground fault at bus 4 for problem 8.11


Z44+ =0.0114 +j 0.1446

Z44- = 0.0114 +j 0.1446

Z440 =j0.2125

The equivalent impedance = 230800154044

044

440

4444 .j.

ZZZZZZeq +=

+×

+= −−−

−−−+

−

The sequence currents are given by:

°−∠=+

==+ 2863242308001540

11 ...j.Z

Ieq

32

°∠=+

−= −−−

−+− 659557244

044

044 ..ZZ

ZII

°∠=+

−= −−−

−−+ 1591761

440

44

440 ..ZZ

ZII

The actual fault currents for a double line to ground fault are given by:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

°∠°∠=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+

732767619160386

0

11

111 0

2

2

....

III

aaaa

III

c

b

a

Chapter 1 Problems and Solution jan30

Documents

Transcript of Chapter 1 Problems and Solution jan30