Photo Voltaic Power Systems and the National Electrical Code
15698118 Design of Photo Voltaic Systems
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Transcript of 15698118 Design of Photo Voltaic Systems
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A Typical PV SystemIntroduction
DC LOADPV ARRAY
DIODE
BATTERY
FANS,
LAMPS
etc.
AC LOAD
I N V
E
R
T
E
R
GRID
Power
Conditioner
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Construction of PV Array1. PV Array consists of severalModules2. Single, polycrystalline oramorphous
silicon3. Packing Factor
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Losses1. The transmission of the radiation is
reduced because of reflection of the
protective glass sheet on top andabsorption in it, and2. The packing factor. That is, the entire
area of the module is not covered by
the solar cells but there are large gapsbetween the adjacent solar cells.
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Space Wasted by Round Solar
CellsSuppose that the radius of the cells is r.
Then the total area required to place fourcells is:At = (2r + 2r) X (2r+2r) = 4r X 4r = 16r2..(1)However, the area covered by the fourcells, the cell area, is:
4X ( r2) = 4 r2. Therefore, the ratio of the cell area to thetotal area At, is:4 r2 /16r2 = /4 = 0.7854.
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Connection of Array
A1
A2
A3
B1
B2
B3
C1
C2
C3
D1
D2
D3
S1
R2 R4R3R1
S3
S2
Figure 4. A typical array of solar modules with bypass diodes.
T1
T2
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Bypass DiodesIf a module in a string fails due to somereason, or comes under the shadow of someobject then the current in that module will
reduce drastically and will limit the currentfrom the other two which pass through it. Inshort the current through a string willreduce. In such a case the bypass diodeassociated with that module will allow the
current to pass through itself. For example if module D2 fails then S2 will bypass thecurrent generated by D1 and D3.
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Hot Spot Formation1. Hot spots in a module.2. A module consists of a large number
of solar cells connected in series. If one cell is shaded and the module iseither short circuited or connected to aheavy load then the current from theother cells will cause i2R heat to begenerated in it. The cell under shadowwill present a high resistance.
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Hot Spots (continued)1. The other nine cells will approach open
circuit voltage Voc
2. This Voc will then be applied across theshaded cell and force a current in thereverse direction
3. This will not only reverse bias the junction,
which may cause a breakdown, but alsoforce the current through its combinedseries and shunt resistors, R = (Rs + Rsh)
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Hot Spot in Shaded Cell
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Equivalent Circuit of a
Solar Cell
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The Maximum Power Point
Tracker (MPPT)
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Converters and Algorithms1. Buck Converter based MPPT2. Boost Converter based MPPT
Common Algorithms for Converters1. Perturb and Observe (P&O), and2. Incremental Conductance
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Block Diagram of MPPTMicrocontrollers and DSPs
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A Typical P & O Algorithm
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A 2.2kW MPPT Response
Curve The response time of a buck basedMPPT with P&O algorithm
V OC
V MPP
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Battery charge/discharge
controller
R 1 6
U 3 CL M 3 2 4
1 0
9
4
1 1
8
+
-
V +
V -
O U T
R 1
F 1
R 2 9
R 2 6
1
3
2
D 9
R 1 3
1 2
R 2 7
R 1 4
R 1 1
1
3
2
R 62
1
3
D 1 4
L E DR 2 8
R 1 5
+ C 1
R 1 7
R 4
Q 1
B C 1 0 7 A 32
1
R 2 3
R 1 0
D 5
R 2 1
U 1 A
L M 3 2 4
3
2
4
1 1
1
+
-
V +
V -
O U T
R 2 5
U 4 DL M 3 2 4
1 2
1 3
4
1 1
1 4
+
-
V +
V -
O U T
U 2 BL M 3 2 4
5
6
4
1 1
7
+
-
V +
V -
O U T
R 1 2
D 8
C 3
C 2
C 7
R 2 0
R 1 8
D 1 1
L E D
1
2
C 4
D 1 2
D 1 0
R 9
R 2
D 1
R 5
D 3
R 3
D 1 5C 5
R 3 0
R 2 4
D 4
D 2
D 1 3
R 2 2
R 1 9
R 7
R 8
B a t t e r y L o w
8K 21 %
(1 N4 1 4 8 )
1 N 5 8 2 2
3 V3
4K 7
47 K
1 N 5 8 2 2
2K 2
1 2K
( 8 K 2 )
1 002 5V
D 7 t o D 1 4M T P 2 9 5 5 E
C h a rg e C o n t r o l l e r f o r C F L 5 W / 7 W b a s e dL a n t e r n
N o t e :
4 V7
47 K
1 0K
5V
0 .1
4K 7
1 K
2K 2
1M
To L 2
0 . 0 12 K7
5 K63 9K1 %
22 K2 K 2 )
B+
S W1 B
R ED
5 K6
3 9K1 %
10 K
2K 2
PV +
8 * 1 N 4 1 4 8
0 . 0 1
2 7 0 K
2 K7
L M 3 8 5 - 2 . 5 V
1 N 4 1 4 8
8K 2B-
22 K
27 0
PV -
0 .1
1 N
4 1 4 8
T O C T O FT R A N S F O R M E RB A S EW I N D I N G
8K 21 %
1 N 4 0 0 7
S W1 A
F l o a t C h a r g i n g = 1 4 . 5 VL o a d d i s c o n n . = 11 . 5 VL o a d R e - c o n n . = 1 2 V
0 . 0 1
1 N 5 8 2 2
Te m p . C o m p . a t B + = 1 4 . 5 6 m V / C ( 1 6 . 1 3 m V / C )
1 K8
G r e e n
1 8 K / 1 %
1K 8
3 V3
1 8 K / 1 %
( C h a rg i n g )
2 M2
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A Lead Acid Battery
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Chemical EquationsFor charging: the cell the positiveterminal of a DC voltage, higher thanthat of the cell, is applied to the anodewith the negative end attached to itscathode. The governing chemicalequations are:
1. PbSO4 + 2H2O PbO2 + 4H+ +SO42- + 2e- At the anode, and
2. PbSO4 + 2e- Pb + SO42-At the Cathode
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Discharging:
The equations at the anodeand the cathode become:PbO 2 + 4H + + SO 42- + 2e - PbSO 4 +2H 2O
And, Pb + SO 42- PbSO 4 + 2e -
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Charge Versus Rate of
Discharge
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Life Cycles Versus
Discharge
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Alone Photovoltaic Power
Supply1. Average power output = 1 kW into a DC load at a
DC voltage of 108 Volts2. Duration of operation = 24 hours/day
3. Average time of sunlight available = 8 hours/day4. Number of sunless days = 2/week5. Peak value of insolation in Delhi = 900
Watts/meters26. Maximum depth of discharge of battery = 50 %7. Array should have a fixed tilt of 28 Deg. For Delhi
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Block diagram of the 1 kWPV power supply
For DC loads only
DC LOADPV ARRAY
DIODE
BATTERY
FANS,
LAMPS
etc.
MPPT
CHARGE/DISCHARGECONTROLLER
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Assumptions The following assumptions have been made: The electrical efficiency of the circuit of the MPPT= 90 %
The charge/discharge cycle efficiency of thebattery (assuming new ones) = 90 %
The diode is usually a built-in part of the MPPT andtherefore neglected. However it is necessary tosave the circuit from accidental input voltageinversion.Wiring and cabling will introduce another 5 % loss.
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Calculations The energy requirement for 7 days will becalculated below.Power required = 1000 Watts
Therefore, energy needed for 7 days = 1000W X 24 Hr X 7 days = 168,000 Watt hours.For an 8 hour sunlit day the energy given
directly to the load is:1000 W X 8 Hr X 5 Days = 40,000 W Hr-------------------- A1Since sunlight is available for only 5 days.
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Calculations (Continued)Hence, the rest of the energy must bestored in and supplied by the batterybank.
This energy is:168,000 40,000 = 128,000 W Hr--------A2Again, the charge discharge efficiency
of the battery bank is 90 %. Therefore,the energy supplied to the battery is:128,000/0.9 = 142,222 W Hr-------------------A3
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Battery Sizing (Continued)1. For 108 volts a string of 9 batteries, of 12
volts each, must be used.
2. The charge capacity of each battery must be:3. 2633.75 A - Hr/9 = 292.6 A Hr.-----------B34. In case 300 A - Hr batteries, which are rare,
are not available then two strings of 9, 150 A
Hr ones may be connected in parallel
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Diodes D1 , D2 ,D3 and D4
1. At 108 volts the load current is:2. 1000 W/108 V = 9.26 Amperes -----B4
3. Nominal voltage, during conduction,across them is 0.7 Volt.
4. Therefore both the diode will dissipate9.26 A X 0.7 V = 6.8 Watts -----B5
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Dissipation in Diodes1. Each diode will conduct half the
current of 9.26 Amperes, that is, 4.63
Amperes2. Each diode, with 100 % overrating,should be 10 amperes, 200 volts
3. Energy Consumed by Diodes:6.8 W X 24 hrs X 7 days = 1,142.4 W Hr -------------- B6
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Power loss in D3 & D4(Continued)1. The voltage drop across the diodes is 0.7
Volts.
2. Therefore the power dissipated in D3 andD4 is: 42 A X 0.7 V = 29.53 Watts3. Therefore energy consumed is:
29.53 W X (8 Hours X 5 days) = 1181 W Hr --------------- B11
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Total Array Energy1. Thus the energy consumed by the four
diodes is:1,142.4 W Hr + 1,181 W Hr = 2,323.4 W Hr ------------------ B12
2. This must be supplied by the PV array. Therefore the total array energy rises to:
40,000 W Hr +1, 42,222 W Hr + 2,323.4 W- Hr = 1, 84,545.4 Watt hours -----B13
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Total Array Energy(Contd.)
1. This energy is given by the MPPTwhich itself has an efficiency of 90%. Hence the energy delivered atthe input of the MPPT from thearray is:
1,84,545.4/0.9 = 2, 05,050.44Watt hours -------------- B14
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Energy given to MPPTInput
1. As assumed earlier there is a 5 %loss in wiring and cabling, hence
the output of the array should be:2, 05,050.44 + (5 X 2,05,050.44)/100 = 2, 05,050.44 +
10,252.52 = 2, 15,302.96 W Hr----B15
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. Array Size
This energy of 2, 15,302.96 W Hris to be generated by the array in 5days with 8 hours of sunlight oneach day. Hence the power of thearray becomes:
2, 05,050.44 W Hr/ (8 hr X 5days) = 5126.25 Watts------------------ C1
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Daily Variation of Insolation
Modules are Rated at 1000 W/m2
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Electrical Parameters PM150
1. Maximum Power Rating Pmax. (Wp)* 150.02. Minimum Power Rating Pmin (Wp)* 180.03. Rated Current IMPP (A) 4.804. Rated Voltage VMPP (V) 34.05. Short Circuit Current Isc (A) 5.0
6. Open Circuit Voltage Voc (V) 42.8
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Number of Modules1. Since these modules are calibrated at
1000 W/m2 their MPP value will reduce at900 W/m2. The real MPP voltage will be
then:34 X 9/10 = 30.6 Volts -------------------- C72. Therefore 3 modules in series will yield
91.8 Volts and the total number of modulesrequired for the array:8977.67 Watts/150 Watts (Wp) = 59.85 =60 ----- C8
3. Number of strings, with each stringscontaining 3 modules is 60/3 = 20
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Costing The cost of the components can be tabulated
below:Solar module @ Rs. 200.00 per Watt = 8977.67Watts X 200 = Rs. 17,95,534.0018 batteries, 150 A Hr, 12 Volts each @ Rs.10,000.00 each = Rs. 1,80,000.00MPPT and Charge/Discharge controller
= Rs. 50,000.00_
Total = Rs.20,25,534.00
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ComparisionIf a life time of 10 years is taken for thearray and 5 years for the battery bankthen the cost amounts to: Rs. 22,05,534.00. In ten years the electricityproduced is equal to:1 X 24 hours X 365 days X 10 years =87,600 kW-Hr. ------------- D1
Therefore the cost of this energy is:Rs. 22, 05,534/87,600 = Rs. 25.17 perkW Hr --------------D2
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Comparison (Continued)1. At present the cost of domestic
electricity from the grid is Rs. 4.60 per
kW Hr2. Cost of electricity from Diesel = Rs.
12.50/unit3. If the life span of the PV array is taken
to be 20 years, the PV generatedpower will compete with dieselgenerated power
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The Additional Benefits of PV Power
1. Carbon credits, and2. Lack of emission of polluting
gases.
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Design of Solar Pump1. The pump horsepower
HP = (4.19 X 10-6) (GPD)(h) ---------- E1(PT)(PTF)()
GPD is the gallons per day to be
pumped,PT is the pumping time,
PTE is the pumping time factor,h is the effective height and is the wire-to-water efficiency of thepump-motor combination.
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Horsepower in MKS Units
In MKS, the horsepower is given byHP = (3.658 X 10-6) (LPD)(h) ----- E2
(PT)(PTF)()Where now LPD is the pumpingrequirement in liters per day
h is the effective pumping height inmeters.
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Pumping Time Factor
Use of an MPPT in the systemnormally increases the daily
volume pumped by an additional20%. Hence, a reasonable defaultvalue for PTF when a MPPT is used
is 1.2 if the pump is connecteddirectly to the PV array, then thePTF will be 1.0.
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Pump Efficiency
The wire-to-water efficiency, , willbe specified by the pump
manufacturer. For fractionalhorsepower pumps, it is typicallyabout 25% while larger pumps will
be more efficient.
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NumericalNumerical:Specification for pumping system:Volume of water to be lifted = 2000gallons/day.Water reservoir = 200 ft. undergroundWorst case peak Sun day = 6 hrs.PTF = 1Peak Sun = 6 hrs.Assume pump efficiency = 25%Piping friction losses = 5%
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Calculations Therefore effective height = 200X 1.05= 210 ft.
Substituting in equation E1, pump HP =1.17. However, the service factor is 25%for a 1HP motor which means that a 1HP motor can operate at 1.25 HPwithout any damage to itself.1.17 HP = 1.17 X 746W = 872.82 W.Pump Operating DC Voltage = 96 V
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