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Transcript of ECE 333 Renewable Energy Systems Lecture 13: Per Unit, Power Flow Prof. Tom Overbye Dept. of...
![Page 1: ECE 333 Renewable Energy Systems Lecture 13: Per Unit, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois.](https://reader035.fdocuments.net/reader035/viewer/2022062423/56649d825503460f94a67f32/html5/thumbnails/1.jpg)
ECE 333 Renewable Energy Systems
Lecture 13: Per Unit, Power Flow
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
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Announcements
• First exam average: 74.1 • Per unit and power flow is only covered in lecture; not
in the book• No quiz on March 12• HW 6 is posted on the website and is due on Thursday
March 19. HW 6 must be turned in and will count the same as a quiz. There will be no quiz on March 19.
• Start Reading Chapter 4 (The Solar Resource)
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Wind Speed Doubling and Wind Turbine Generator (WTG) Output
• Just about everyone missed problem 3.d, asking about what happens to the output of a WTG if the wind speed doubles; the answer was "It Depends"– Power in wind goes up by 8, but not usually the power output
3Image: http://site.ge-energy.com/prod_serv/products/wind_turbines/en/downloads/GEA14954C15-MW-Broch.pdf
Image shows power output for GE 1.5 and 1.6 MW WTGs;cut-in speed is 3.5 m/s, while cut-out is 25 m/s
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Wind Power and the Power Flow
• The most common power system analysis tool is the power flow (also known sometimes as the load flow)– power flow determines how the power flows in a network– also used to determine all bus voltages and all currents– because of constant power models, power flow is a
nonlinear analysis technique– power flow is a steady-state analysis tool– it can be used as a tool for planning the location of new
generation, including wind
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Simplified Power System Modeling
• Balanced three phase systems can be analyzed using per phase analysis
• A “per unit” normalization is simplify the analysis of systems with different voltage levels.
• To provide an introduction to power flow analysis we need models for the different system devices:– Transformers and Transmission lines, generators and loads
• Transformers and transmission lines are modeled as a series impedances
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Load Models
• Ultimate goal is to supply loads with electricity at constant frequency and voltage
• Electrical characteristics of individual loads matter, but usually they can only be estimated– actual loads are constantly changing, consisting of a large
number of individual devices– only limited network observability of load characteristics
• Aggregate models are typically used for analysis• Two common models
– constant power: Si = Pi + jQi
– constant impedance: Si = |V|2 / Zi
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Generator Models
• Engineering models depend upon application• Generators are usually synchronous machines• For generators we will use two different models:
– a steady-state model, treating the generator as a constant power source operating at a fixed voltage; this model will be used for power flow and economic analysis
– This model works fairly well for type 3 and type 4 wind turbines
– Other models include treating as constant real power with a fixed power factor.
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Per Unit Calculations
• A key problem in analyzing power systems is the large number of transformers. – It would be very difficult to continually have to refer
impedances to the different sides of the transformers
• This problem is avoided by a normalization of all variables.
• This normalization is known as per unit analysis.
actual quantityquantity in per unit
base value of quantity
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Per Unit Conversion Procedure, 1f
1. Pick a 1f VA base for the entire system, SB
2. Pick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.
3. Calculate the impedance base, ZB= (VB)2/SB
4. Calculate the current base, IB = VB/ZB
5. Convert actual values to per unit
Note, per unit conversion on affects magnitudes, not
the angles. Also, per unit quantities no longer have
units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)9
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Per Unit Solution Procedure
1. Convert to per unit (p.u.) (many problems are already in per unit)
2. Solve
3. Convert back to actual as necessary
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Per Unit Example
Solve for the current, load voltage and load power
in the circuit shown below using per unit analysis
with an SB of 100 MVA, and voltage bases of
8 kV, 80 kV and 16 kV.
Original Circuit11
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Per Unit Example, cont’d
2
2
2
80.64
100
8064
100
162.56
100
LeftB
MiddleB
RightB
kVZ
MVA
kVZ
MVA
kVZ
MVA
Same circuit, with
values expressed
in per unit.
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Per Unit Example, cont’d
L
2*
1.0 00.22 30.8 p.u. (not amps)
3.91 2.327
V 1.0 0 0.22 30.8
p.u.
0.189 p.u.
1.0 0 0.22 30.8 30.8 p.u.
LL L L
G
Ij
VS V I
ZS
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Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
LActual
ActualL
ActualG
MiddleB
ActualMiddle
0.859 30.8 16 kV 13.7 30.8 kV
0.189 0 100 MVA 18.9 0 MVA
0.22 30.8 100 MVA 22.0 30.8 MVA
100 MVAI 1250 Amps
80 kV
I 0.22 30.8 Amps 275 30.8
V
S
S
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Three Phase Per Unit
1. Pick a 3f VA base for the entire system,
2. Pick a voltage base for each different voltage level, VB. Voltages are line to line.
3. Calculate the impedance base
Procedure is very similar to 1f except we use a 3f VA base, and use line to line voltage bases
3BSf
2 2 2, , ,3 1 1
( 3 )
3B LL B LN B LN
BB B B
V V VZ
S S Sf f f
Exactly the same impedance bases as with single phase!15
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Three Phase Per Unit, cont'd
4. Calculate the current base, IB
5. Convert actual values to per unit
3 1 13 1B B
, , ,
3I I
3 3 3B B B
B LL B LN B LN
S S S
V V V
f f ff f
Exactly the same current bases as with single phase!
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Three Phase Per Unit Example
Solve for the current, load voltage and load power in the previous circuit, assuming a 3f power base of300 MVA, and line to line voltage bases of 13.8 kV,138 kV and 27.6 kV (square root of 3 larger than the 1f example voltages). Also assume the generator is Y-connected so its line to line voltage is 13.8 kV.
Convert to per unit
as before. Note the
system is exactly the
same!
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3f Per Unit Example, cont'd
L
2*
1.0 00.22 30.8 p.u. (not amps)
3.91 2.327
V 1.0 0 0.22 30.8
p.u.
0.189 p.u.
1.0 0 0.22 30.8 30.8 p.u.
LL L L
G
Ij
VS V I
ZS
Again, analysis is exactly the same!
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3f Per Unit Example, cont'd
LActual
ActualL
ActualG
MiddleB
ActualMiddle
0.859 30.8 27.6 kV 23.8 30.8 kV
0.189 0 300 MVA 56.7 0 MVA
0.22 30.8 300 MVA 66.0 30.8 MVA
300 MVAI 125 (same cur0 Amps
3138 kV
I 0.22 30.
rent!)
8
V
S
S
Amps 275 30.8
Differences appear when we convert back to actual values
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3f Per Unit Example 2
•Assume a 3f load of 100+j50 MVA with VLL of 69 kV is connected to a source through the below network:
What is the supply current and complex power?
Answer: I=467 amps, S = 103.3 + j76.0 MVA
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Power Flow Analysis
• We now have the necessary models to start to develop the power system analysis tools
• The most common power system analysis tool is the power flow (also known sometimes as the load flow)– power flow determines how the power flows in a network– also used to determine all bus voltages and all currents– because of constant power models, power flow is a
nonlinear analysis technique– power flow is a steady-state analysis tool
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Linear Power System Elements
Resistors, inductors, capacitors, independent
voltage sources and current sources are linear
circuit elements
1V = R I V = V =
Such systems may be analyzed by superposition
j L I Ij C
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Nonlinear Power System Elements
•Constant power loads and generator injections are nonlinear and hence systems with these elements can not be analyzed by superposition
Nonlinear problems can be very difficult to solve,
and usually require an iterative approach
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Nonlinear Systems May Have Multiple Solutions or No Solution
•Example 1: x2 - 2 = 0 has solutions x = 1.414…•Example 2: x2 + 2 = 0 has no real solution
f(x) = x2 - 2 f(x) = x2 + 2
two solutions where f(x) = 0 no solution f(x) = 0
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Multiple Solution Example 3
• The dc system shown below has two solutions:
where the 18 watt
load is a resistive
load
22
Load
Load
Load
The equation we're solving is
9 voltsI 18 watts
1 +R
One solution is R 2
Other solution is R 0.5
Load LoadR R
What is the
maximum
PLoad?
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Adding New Generation
• Adding (siting) large amounts of new generation, such as for a wind farm, often require changes to the transmission system– Developers try to site where existing capacity is available
• Getting new right-of-ways for transmission lines can be quite difficult (i.e., not in my backyard [NIMBY]) even when the transmission is associated with "green" energy
• High voltage (e.g., 345 kV) overhead transmission lines can cost $2 million per mile, whereas burying the conductors can increase costs by a factor of 10– AC lines cannot be buried for long distances because of the
high capacitance associated with underground lines26
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Adding New Generation and Transmission
• Power flow studies are done to determine how much new transmission is required, and which would be the best right-of-ways– Often there is no single best answer
• Example: Ameren Illinois Rivers Project
27Project Website: http://www.ilriverstransmission.com/maps
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Ameren Illinois Rivers Project: Sidney to Rising 345 kV Line
• One portion of project is adding a new 28 mile 345 kV transmission line in Champaign County – Estimated in-service Nov 2016, cost $66 million
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Bus Admittance Matrix or Ybus
• First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus.
• The Ybus gives the relationships between all the bus current injections, I, and all the bus voltages, V,
I = Ybus V
• The Ybus is developed by applying KCL at each bus in the system to relate the bus current injections, the bus voltages, and the branch impedances and admittances
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Ybus Example
Determine the bus admittance matrix for the network
shown below, assuming the current injection at each
bus i is Ii = IGi - IDi where IGi is the current injection into the bus from the generator and IDi is the current flowing into the load
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Ybus Example, cont’d
1 1 1
1 2 1 31 12 13
1 1 2 1 3 j
1 2 3
2 21 23 24
1 2 3 4
By KCL at bus 1 we have
1( ) ( ) (with Y )
( )
Similarly
( )
G D
A B
A Bj
A B A B
A A C D C D
I I I
V V V VI I I
Z Z
I V V Y V V YZ
Y Y V Y V Y V
I I I I
Y V Y Y Y V Y V Y V
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Ybus Example, cont’d
1 1
2 2
3 3
4 4
We can get similar relationships for buses 3 and 4
The results can then be expressed in matrix form
0
0
0 0
bus
A B A B
A A C D C D
B C B C
D D
I Y Y Y Y V
I Y Y Y Y Y Y V
I Y Y Y Y V
I Y Y V
I Y V
For a system with n buses, Ybus is an n by n
symmetric matrix (i.e., one where Aij = Aji)
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