Transcript Per Unit Example, cont`d

ECE 530 – Analysis Techniques for
Large-Scale Electrical Systems
Lecture 2: Power System Modeling
Prof. Hao Zhu
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
[email protected]
8/26/2015
1
Static Power System Analysis
•
•
One of the most common power system analysis tools
is the power flow, which tells how power flows
through a power system in the quasi-steady state time
frame
The power flow can be used to model the full, threephase system, but usually (practically always) for
transmission system analysis the system is assumed to
be balanced. Hence a per-phase equivalent model is
used.
2
Power Flow Problem
•
•
•
Power flow calculations determine the complex
voltages at the nodes and the power flows on the lines
for a specified snapshot of the power system at steady
state with the generation and load schedule at each bus
given
The steady-state timeframe includes common power
system devices, such as transmission lines,
transformers, generators and loads
We will start with the basic power system modeling
3
Power Circuit 101
•
•
•
•
Ohm’s law V = Z×I ⇔ I = Y×V
Complex power S = V×I* = P + jQ (VA)
Real power P (Watts)
Reactive power Q (VAR)
•
•
Balanced 3-phase 𝑆 3𝜙 = 3𝑆1𝜙
Line-to-line vs. line-to-neutral voltage
𝑉 𝐿𝐿 = 3 𝑉 𝐿𝑁
4
Power System Component
Models: Transmission Lines
•
Transmission lines will be modeled using the p circuit
𝐼𝑆
𝐼𝑅
𝑌 1
+
= 2 𝑍
1
𝑍
1
−
𝑉𝑆
𝑍
𝑌 1 𝑉𝑅
− −
2 𝑍
5
Power System Component
Models: Transformers
•
•
Transformer equivalent model
The I/O relation has a similar matrix form, will be
detailed in following lectures
6
Power System Component
Models: Loads
•
•
Operational requirements: 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 = |Vi|2 / Zi
7
Power System Component
Models: Generators
•
•
•
Engineering models depend upon applications
Generators are usually synchronous machines
For generators there are 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
– a short term model treating the generator as a constant voltage
source behind a possibly time-varying reactance
8
Per Phase 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
Solution: perform a normalization of all variables.
This normalization is known as per unit analysis
actual quantity
quantity in per unit 
base value of quantity
9
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,
3.
4.
5.
VB. Voltage bases are related by transformer turns
ratios and ratings. Voltages are line to neutral.
Calculate the impedance base, ZB= (VB)2/SB
Calculate the current base, IB = SB/VB=VB/ZB
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)
10
Per Unit Solution Procedure
1.
2.
3.
Convert to per unit (p.u.) (many problems are already
in per unit)
Solve
Convert back to actual as necessary
11
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 Circuit
12
Per Unit Example, cont’d
Z BLeft
8kV 2

 0.64
100 MVA
Z BMiddle
Z BRight
80kV 2

 64
100 MVA
2
16kV

 2.56
100 MVA
Same circuit, with
values expressed
in per unit.
13
Per Unit Example, cont’d
1.00
I 
 0.22  30.8 p.u. (not amps)
3.91  j 2.327
VL  1.00  0.22  30.8  
    p.u.
2
VL
SL 

 0.189 p.u.
Z
SG  1.00  0.2230.8  30.8 p.u.
VL I L*
14
Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
VLActual  0.859  30.8 16 kV  13.7  30.8 kV
SLActual  0.1890 100 MVA  18.90 MVA
SGActual  0.2230.8 100 MVA  22.030.8 MVA
I Middle
B
100 MVA
 1250 Amps

80 kV
I Actual
Middle  0.22  30.8 Amps  275  30.8 
15
Three Phase Per Unit
Procedure is very similar to 1f except we use a 3f
VA base, and use line to line voltage bases
1.
2.
3.
Pick a 3f VA base for the entire system, S B3f
Pick a voltage base for each different voltage
level, VB. Voltages are line to line.
Calculate the impedance base
ZB 
VB2, LL
S B3f

( 3 VB , LN ) 2
3S 1Bf

VB2, LN
S 1Bf
Exactly the same impedance bases as with single phase!
16
Three Phase Per Unit, cont'd
4.
Calculate the current base, IB
I3Bf
5.
S B3f
3 S 1Bf
S 1Bf



 I1Bf
3 VB , LL
3 3 VB , LN VB , LN
Exactly the same current bases as with single phase!
Convert actual values to per unit
17
Three Phase Per Unit Example
Solve for the current, load voltage and load power
in the previous circuit, assuming a 3f power base of
300 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!
18
3f Per Unit Example, cont'd
1.00
I 
 0.22  30.8 p.u. (not amps)
3.91  j 2.327
VL  1.00  0.22  30.8  
    p.u.
2
VL
SL 

 0.189 p.u.
Z
SG  1.00  0.2230.8  30.8 p.u.
*
VL I L
Again, analysis is exactly the same!
19
3f Per Unit Example, cont'd
Differences appear when we convert back to actual values
Actual
VL
 0.859  30.8  27.6 kV  23.8  30.8 kV
S LActual  0.1890  300 MVA  56.70 MVA
SGActual  0.2230.8  300 MVA  66.030.8 MVA
I Middle
B
300 MVA

 1250 Amps (same current!)
3 138 kV
I Actual
Middle  0.22  30.8   Amps  275  30.8 
20
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
21
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
22
Ybus Example, cont’d
By KCL at bus 1 we have
I1 = I G1  I D1
I1  I12  I13
V1  V2 V1  V3


ZA
ZB
I1  (V1  V2 )YA  (V1  V3 )YB
1
(with Yj  )
Zj
 (YA  YB )V1  YA V2  YB V3
Similarly
I 2  I 21  I 23  I 24
 YA V1  (YA  YC  YD )V2  YC V3  YD V4
23
Ybus Example, cont’d
We can get similar relationships for buses 3 and 4
The results can then be expressed in matrix form
I  Ybus V
YA
YB
 I1  YA  YB
 I   Y
YA  YC  YD
YC
2
A
  
YC
YB  YC
 I 3   YB
I   0
YD
0
 4 
0  V1 
YD  V2 
 
0  V3 
YD  V4 
For a system with n buses, Ybus is an n by n symmetric
matrix (i.e., one where Aij = Aji); however this will not be
true if we consider phase shifting transformers
24
Ybus General Form
•The diagonal terms, Yii, are the self admittance
terms, equal to the sum of the admittances of all
devices incident to bus i.
•The off-diagonal terms, Yij, are equal to the
negative of the sum of the admittances joining the
two buses.
•With large systems Ybus is a sparse matrix (that is,
most entries are zero)
•Shunt terms, such as with the p line model, only
affect the diagonal terms.
25
Modeling Shunts in the Ybus
Ykc
Since I ij  (Vi  V j )Yk  Vi
2
Ykc
Yii 
 Yk 
2
Rk  jX k Rk  jX k
1
1
Note Yk 

 2
Z k Rk  jX k Rk  jX k Rk  X k2
Yiifrom other lines
26
Two Bus System Example
Yc
(V1  V2 )
I1 
 V1
Z
2
1
 12  j16
0.03  j 0.04
 I1 
12  j15.9 12  j16  V1 
 I    12  j16 12  j15.9  V 

 2
 2
27
Using the Ybus
If the voltages are known then we can solve for
the current injections:
Ybus V  I
If the current injections are known then we can
solve for the voltages:
1
Ybus
I  V  Zbus I
where Z bus is the bus impedance matrix
However, this requires that Ybus not be singular; note it
will be singular if there are no shunt connections!
28
Solving for Bus Currents
For example, in previous case assume
 1.0 
V

0.8

j
0.2


Then
12  j15.9 12  j16   1.0   5.60  j 0.70 
 12  j16 12  j15.9  0.8  j 0.2    5.58  j 0.88


 

Therefore the power injected at bus 1 is
S1  V1I1*  1.0  (5.60  j 0.70)  5.60  j 0.70
*
S2  V2 I 2
 (0.8  j 0.2)  (5.58  j 0.88)  4.64  j 0.41
29
Solving for Bus Voltages
For example, in previous case assume
 5.0 
I


4.8


Then
1
12  j15.9 12  j16   5.0   0.0738  j 0.902 
 12  j16 12  j15.9   4.8   0.0738  j1.098

 
 

Therefore the power injected is
S1  V1I1*  (0.0738  j 0.902)  5  0.37  j 4.51
S2  V2 I 2*  (0.0738  j1.098)  (4.8)  0.35  j 5.27
30