Chapter 10: Power System Representation and Equations
10-1.
Sketch the per-phase, per-unit equivalent circuit of the power system in Figure 10-2. (Treat each load on
the systems as a resistance in series with a reactance.) Note that you do not have enough information to
actually calculate the values of components in the equivalent circuit.
SOLUTION The per-phase, per-unit equivalent circuit would be:
T1
Bus 1
T2
Line
Load A
Bus 2
Load B
+
+
-
-
G2
G1
10-2.
A 20,000 kVA, 110/13.8 kV, Y-Δ three phase transformer has a series impedance of 0.02 + j0.08 pu.
Find the per-unit impedance of this transformer in a power system with a base apparent power of 500
MVA and a base voltage on the high side of 120 kV.
SOLUTION The per-unit impedance to the new base would be:
2
per-unit Z new
§V · § S ·
= per-unit Z given ¨ given ¸ ¨ new ¸
© Vnew ¹ © Sgiven ¹
2
(10-8)
§ 110 kV · § 500,000 kVA ·
= 0.42 + j1.68 pu
per-unit Z new = ( 0.02 + j 0.08) ¨
© 120 kV ¸¹ ¨© 20,000 kVA ¸¹
193
10-3.
Find the per-phase equivalent circuit of the power system shown in Figure P10-1.
SOLUTION The per-phase equivalent circuit must be created on some system base voltage and apparent
power. Since this problem has not specified the system base values, we will use the ratings of generator
G1 as the system base values at that point. Therefore, the system base apparent power is Sbase = 30 MVA ,
and the system base voltages in each region are:
Vbase,1 = 13.8 kV
§ 115 kV ·
§ 115 kV ·
Vbase,2 = ¨
Vbase,1 = ¨
(13.8 kV ) = 120 kV
¸
© 13.2 kV ¹
© 13.2 kV ¸¹
§ 12.5 kV ·
§ 12.5 kV ·
Vbase,3 = ¨
Vbase,2 = ¨
(120 kV ) = 12.5 kV
¸
© 120 kV ¹
© 120 kV ¸¹
The base impedance of Region 2 is:
Z base,2
(V
=
LL , base,2
)
2
S3φ , base
=
(120,000 V ) 2
30,000,000 VA
= 480 Ω
The per unit resistance and reactance of G1 are already on the proper base:
Z G1 = 0.1 + j1.0 pu
The per unit resistance and reactance of T1 are:
2
per-unit Z new
§V · § S ·
= per-unit Z given ¨ given ¸ ¨ new ¸
© Vnew ¹ © Sgiven ¹
2
§ 13.2 kV · § 30,000 kVA ·
Z T 1 = ( 0.01 + j 0.10) ¨
= 0.00784 + j 0.0784 pu
© 13.8 kV ¸¹ ¨© 35,000 kVA ¸¹
The per unit resistance and reactance of the transmission line are:
Z line =
Z
5 + j 20 Ω
=
= 0.0104 + j 0.0417 pu
Z base
480 Ω
The per unit resistance and reactance of T2 are already on the right base:
Z T 2 = 0.01 + j 0.08 pu
The per unit resistance and reactance of M1 are:
194
2
§ 12.5 kV · § 30, 000 kVA ·
Z M 1 = ( 0.1 + j1.1) ¨
= 0.15 + j1.65 pu
© 12.5 kV ¸¹ ¨© 20,000 kVA ¸¹
The per unit resistance and reactance of M1 are:
2
ZM 2
§ 12.5 kV · § 30,000 kVA ·
= ( 0.1 + j1.1) ¨
= 0.30 + j 3.30 pu
© 12.5 kV ¹¸ ¨© 10,000 kVA ¸¹
The resulting per-phase equivalent circuit is:
T1
j0.0784
T2
Line
0.00784 j0.0417
0.0104
j0.08
0.1
j1.0
0.01
0.15
0.30
j1.65
j3.30
+
+
G1
-
10-4.
+
M1
M2
-
-
Two 4.16 kV three-phase synchronous motors are connected to the same bus. The motor ratings are:
Motor 1: 5,000 hp, 0.8 PF lagging, 95% efficiency, R = 3%, X S = 90%
Motor 2: 3,000 hp, 1.0 PF, 95% efficiency, R = 3%, X S = 90%
Calculate the per-unit impedances of these motors to a base of 20 MVA, 4.16 kV. (Note: To calculate
these values, you will first have to determine the rated apparent power of each motor considering its rated
output power, efficiency, and power factor.)
SOLUTION The rated input power of Motor 1 is
P1 =
Pout
η
=
§ 746 W ·
© 1 hp ¸¹
= 393 kW
0.95
( 5,000 hp) ¨
The apparent power rating is
S1 =
P1 393 kW
=
= 491 kVA
PF
0.8
The per-unit impedances of Motor 1 are specified to the motor’s own base. The impedances converted to
the specified base are:
2
per-unit Z new
§V · § S ·
= per-unit Z given ¨ given ¸ ¨ new ¸
© Vnew ¹ © Sgiven ¹
2
§ 4.16 kV · § 20,000 kVA ·
Z1 = ( 0.03 + j 0.90) ¨
= 1.22 + j 36.7 pu
© 4.16 kV ¸¹ ¨© 491 kVA ¸¹
The rated input power of Motor 2 is
195
(10-8)
P2 =
Pout
η
=
§ 746 W ·
© 1 hp ¸¹
= 236 kW
0.95
( 3,000 hp) ¨
The apparent power rating is
S2 =
P2 236 kW
=
= 236 kVA
PF
1.0
The per-unit impedances of Motor 2 are specified to the motor’s own base. The impedances converted to
the specified base are:
2
per-unit Z new
§V · § S ·
= per-unit Z given ¨ given ¸ ¨ new ¸
© Vnew ¹ © Sgiven ¹
(10-8)
2
§ 4.16 kV · § 20,000 kVA ·
Z 2 = ( 0.03 + j 0.90) ¨
= 2.54 + j 76.3 pu
© 4.16 kV ¸¹ ¨© 236 kVA ¸¹
10-5.
A Y-connected synchronous generator rated 100 MVA, 13.2 kV has a rated impedance of R = 5% and
X S = 80% per-unit. It is connected to a j10 Ω transmission line through a 13.8/120 kV, 100 MVA, Δ-Y
transformer with a rated impedance of R = 2% and X = 8% per unit. The base for the power system is 200
MVA, 120 kV at the transmission line.
(a) Sketch the one-line diagram of this power system, with symbols labeled appropriately.
(b) Find per-unit impedance of the generator on the system base.
(c) Find per-unit impedance of the transformer on the system base.
(d) Find per-unit impedance of the transmission line on the system base.
(e) Find the per-phase, per-unit equivalent circuit of this power system.
SOLUTION The base quantities for this power system in Region 2 are:
Sbase = 200 MVA
Vbase,2 = 120 kV
Z base,2
(V
=
LL , base,2
)
2
S3φ , base
=
(120,000 V ) 2
200,000,000 VA
= 72 Ω
The base quantities for this power system in Region 1 are:
Sbase = 200 MVA
§ 13.8 kV ·
§ 13.8 kV ·
Vbase,1 = ¨
V
=
(120 kV ) = 13.8 kV
© 120 kV ¸¹ base,2 ¨© 120 kV ¸¹
Z base,1
(V
=
LL , base,1
S3φ, base
)
2
=
(13,800 V ) 2
200,000,000 VA
196
= 0.952 Ω
(a)
The one-line diagram for this power system is shown below:
Region 2
Region 1
T1
G1
L1
1
L1 impedance:
X = j10 Ω
G1 ratings:
100 MVA
13.2 kV
R = 0.05 pu
XS = 0.8 pu
(b)
T1 ratings:
100 MVA
13.8/120 kV
R = 0.02 pu
X = 0.08 pu
Base Values:
200 MVA
120 kV
The per-unit impedance of the generator on the system base is:
2
per-unit Z new
§V · § S ·
= per-unit Z given ¨ given ¸ ¨ new ¸
© Vnew ¹ © Sgiven ¹
(10-8)
2
§ 13.2 kV · § 200 MVA ·
Z G1 = ( 0.05 + j 0.80) ¨
= 0.0915 + j1.464 pu
© 13.8 kV ¸¹ ¨© 100 MVA ¸¹
(c)
The per-unit impedance of the transformer on the system base is:
2
per-unit Z new
§V · § S ·
= per-unit Z given ¨ given ¸ ¨ new ¸
© Vnew ¹ © Sgiven ¹
(10-8)
2
§ 13.8 kV · § 200 MVA ·
Z T 1 = ( 0.02 + j 0.08) ¨
= 0.04 + j 0.16 pu
© 13.8 kV ¸¹ ¨© 100 MVA ¸¹
(d)
The per unit resistance and reactance of the transmission line are:
Z line =
(e)
Z
j10 Ω
=
= j 0.139 pu
Z base
72 Ω
The resulting per-phase equivalent circuit is:
T1
j0.16
Line
0.04
0.0915
j1.464
+
G1
-
197
j0.139
0.00
10-6.
Assume that the power system of the previous problem is connected to a resistive Y-connected load of
200 Ω per phase. If the internal generated voltage of the generator is E A = 1.10∠20° per unit, what is the
terminal voltage of the generator? How much power is being supplied to the load?
SOLUTION The Y-connected load is connected to the end of the transmission line in Region 2, so
Z base,2 = 72 Ω , and the per-unit impedance of the load is
Z load =
Z
200 Ω
=
= 2.78 pu
Z base
72 Ω
The resulting current flow is
EA
Z G1 + Z T 1 + Z line + Z load
1.10∠20°
=
0.0915
j
1.464
0.04
+
+
(
) ( + j0.16) + ( j 0.139) + ( 2.78)
= 0.323∠ − 11.2°
I line =
I line
I line
The per-phase terminal voltage of the generator will be
Vφ = E A − I A RA − jI A X S
Vφ = 1.10∠20° − ( 0.323∠ − 11.2°)( 0.0915) − j ( 0.323∠ − 11.2°)(1.464 )
Vφ = 0.917∠ − 5.14°
Therefore, the terminal voltage will be ( 0.917 )(13.8 kV ) = 12.7 kV . The per-unit power supplied to the
load is
Ppu = I pu 2 R = ( 0.323) ( 2.78) = 0.290
2
Therefore, the total power supplied to the load is ( 0.290)( 200 MVA ) = 58 MW .
10-7.
Figure P10-2 shows a one-line diagram of a three-phase power system. The ratings of the various
components in the system are:
Synchronous Generator 1:
40 MVA, 13.8 kV, R = 3%, X S = 80%
Synchronous Motor 2:
20 MVA, 13.8 kV, R = 3%, X S = 80%
Synchronous Motor 3:
10 MVA, 13.2 kV, R = 3%, X S = 100%
Y-Δ Transformers:
20 MVA, 13.8/138 kV, R = 2%, X = 10%
Line 1:
R = 10 Ω, X = 50 Ω
Line 2:
R = 5 Ω, X = 30 Ω
Line 3:
R = 5 Ω, X = 30 Ω
The per-unit system base for this power system is 40 MVA, 128 kV in transmission line 1. Create the
per-phase, per-unit equivalent circuit for this power system.
198
SOLUTION This power system has been divided into regions at the transformers, with the base voltage and
apparent power specified in Region 2 to be 128 kV and 40 MVA. The base quantities for this power
system in all regions are:
Sbase = 40 MVA
§ 13.8 kV ·
§ 13.8 kV ·
Vbase,1 = ¨
Vbase,2 = ¨
(128 kV ) = 12.8 kV
¸
© 138 kV ¹
© 138 kV ¸¹
Z base,1
(V
=
LL , base,1
)
2
=
S3φ, base
(12,800 V ) 2
40,000,000 VA
= 4.096 Ω
Vbase,2 = 128 kV
Z base,2
(V
=
LL , base,2
)
2
=
S3φ , base
(128,000 V ) 2
40,000,000 VA
= 409.6 Ω
§ 13.8 kV ·
§ 13.8 kV ·
Vbase,3 = ¨
Vbase,2 = ¨
(128 kV ) = 12.8 kV
¸
© 138 kV ¹
© 138 kV ¸¹
Z base,3
(V
=
LL , base,3
)
2
=
S3φ, base
(12,800 V ) 2
40,000,000 VA
= 4.096 Ω
§ 138 kV ·
§ 138 kV ·
Vbase,4 = ¨
Vbase,1 = ¨
(12.8 kV ) = 128 kV
¸
© 13.8 kV ¹
© 13.8 kV ¸¹
Z base,4
(V
=
LL , base,3
S3φ , base
)
2
=
(128,000 V ) 2
40,000,000 VA
= 409.6 Ω
199
§ 138 kV ·
§ 138 kV ·
Vbase,5 = ¨
Vbase,3 = ¨
(12.8 kV ) = 128 kV
¸
© 13.8 kV ¹
© 13.8 kV ¸¹
Z base,5
(V
=
LL , base,3
)
2
=
S3φ, base
(128,000 V ) 2
40,000,000 VA
= 409.6 Ω
§ 13.8 kV ·
§ 13.8 kV ·
Vbase,6 = ¨
Vbase,4 = ¨
(128 kV ) = 12.8 kV
¸
© 138 kV ¹
© 138 kV ¸¹
Z base,6
(V
=
LL , base,6
)
2
S3φ, base
=
(12,800 V ) 2
40,000,000 VA
= 4.096 Ω
The base quantities for this power system in Region 1 are:
Sbase = 200 MVA
§ 13.8 kV ·
§ 13.8 kV ·
Vbase,1 = ¨
Vbase,2 = ¨
(128 kV ) = 12.8 kV
¸
© 138 kV ¹
© 138 kV ¸¹
Z base,1
(V
=
LL , base,1
S3φ, base
)
2
=
(128,000 V ) 2
40,000,000 VA
= 409.6 Ω
The per-unit impedances of the various components to the system base are:
2
§V · § S ·
per-unit Z new = per-unit Z given ¨ given ¸ ¨ new ¸
© Vnew ¹ © Sgiven ¹
(10-8)
2
§ 13.8 kV · § 40 MVA ·
Z G1 = ( 0.03 + j 0.80) ¨
= 0.0349 + j 0.930 pu
© 12.8 kV ¸¹ ¨© 40 MVA ¸¹
2
ZM 2
§ 13.8 kV · § 40 MVA ·
= ( 0.03 + j 0.80) ¨
= 0.0697 + j1.860 pu
© 12.8 kV ¸¹ ¨© 20 MVA ¸¹
2
§ 13.8 kV · § 40 MVA ·
Z M 3 = ( 0.03 + j1.00) ¨
= 0.140 + j 4.65 pu
© 12.8 kV ¸¹ ¨© 10 MVA ¸¹
2
§ 13.8 kV · § 40 MVA ·
Z T 1 = ( 0.02 + j 0.10) ¨
= 0.0465 + j 0.233 pu
© 12.8 kV ¸¹ ¨© 20 MVA ¸¹
Z
10 + j50 Ω
Z L1 =
=
= 0.0244 + j 0.1221 pu
Z base
409.6 Ω
Z
5 + j 30 Ω
Z L2 =
=
= 0.0122 + j 0.0732 pu
Z base
409.6 Ω
Z
5 + j 30 Ω
Z L3 =
=
= 0.0122 + j 0.0732 pu
Z base
409.6 Ω
All transformers in this figure are Y-Δ, and the base quantities are the same for all of them, so the per-unit
transformer impedances are:
Z T 1 = Z T 2 = Z T 3 = ZT 4 = Z T 5 = Z T 6 = 0.0465 + j 0.233 pu
The resulting per-phase, per-unit equivalent circuit is shown below:
200
T1
j0.233
T2
L1
0.0465
j0.122
j0.233
0.0244
T3
0.0465
T4
0.0349
0.0697
j0.233
j0.930
0.0465
j0.233
j0.0732
+
0.0465
j1.86
j0.0732
L2
L3
0.0122
0.0122
+
M2
G1
-
-
j0.233
j0.233
T5
T6
0.0465
0.0465
0.140
j4.65
+
M3
-
10-8.
Calculate the bus admittance matrix Ybus and the bus impedance matrix Z bus for the power system shown
in Figure P10-2.
SOLUTION The voltage sources can be converted to current sources, and the series impedances between
each bus can be replaced by a single admittance, resulting in the system shown below. Note that we have
labeled each bus with a number.
201
Ya
1
I1
2
Yd
I1
Yf
Yc
Yb
I2
3
I3
Ye
n
The admittances in this circuit are:
1
1
=
ZT 1 + Z L1 + Z T 2 ( 0.0465 + j 0.233 pu ) + ( 0.0244 + j 0.122 ) + ( 0.0465 + j 0.233 pu )
= 0.3265 − j1.6355 pu
1
1
=
=
ZT 3 + Z L 2 + Z T 5 ( 0.0465 + j 0.233 pu ) + ( 0.0122 + j 0.0732 ) + ( 0.0465 + j 0.233 pu )
= 0.3486 − j1.7866 pu
1
1
=
=
Z T 4 + Z L 3 + ZT 6 ( 0.0465 + j 0.233 pu ) + ( 0.0122 + j 0.0732 ) + ( 0.0465 + j 0.233 pu )
= 0.3486 − j1.7866 pu
1
1
=
=
= 0.3537 − j 0.9425 pu
Z G1 0.0349 + j 0.930
1
1
=
=
= 0.0064 − j 0.2149 pu
Z M 3 0.140 + j 4.65
1
1
=
=
= 0.0201 − j 0.5369 pu
Z M 2 0.0697 + j1.860
Ya =
Ya
Yb
Yb
Yc
Yc
Yd
Ye
Yf
The bus admittance matrix Ybus is:
Ybus
Ybus
−Ya
−Yb
ªYa + Yb + Yd
º
«
»
=«
−Ya
Ya + Yc + Y f
−Yc
»
«¬
Yb + Yc + Ye »¼
−Yb
−Yc
ª 1.0288 − j 4.3646 −0.3265 + j1.6355 −0.3486 + j1.7866º
= «« −0.3265 + j1.6355 0.6952 − j 3.9590 −0.3486 + j1.7866»»
«¬ −0.3486 + j1.7866 −0.3486 + j1.7866 0.7036 − j 3.7881 »¼
The bus impedance matrix Z bus is:
202
Z bus = Ybus −1
Z bus
10-9.
ª0.1532 + j 0.6048 0.1068 + j 0.4875 0.1181 + j 0.5172 º
= «« 0.1068 + j 0.4875 0.1251 + j 0.7046 0.1045 + j 0.5642 »»
«¬ 0.1181 + j 0.5172 0.1045 + j 0.5642 0.1479 + j 0.7670»¼
Assume that internal generated voltages of the generators and motors in the per-unit equivalent circuit of
the previous problem have the following values:
E A1 = 1.15∠22°
E A2 = 1.00∠ − 20°
(a)
(b)
(c)
(d)
(e)
E A3 = 0.95∠ − 15°
Find the per-unit voltages on each bus in the power system.
Find the actual voltages on each bus in the power system.
Find the current flowing in each transmission line in the power system.
Determine the magnitude and direction of the real and reactive power flowing in each transmission
line.
Are any of the components in the power system overloaded?
SOLUTION The Norton equivalent currents associated with each voltage source are:
E A1
1.15∠22°
=
= 0.7832 − j 0.8526 = 1.158∠ − 47.4°
Z G1 0.0349 + j 0.930
1.00∠ − 20°
E
I 2 = A2 =
= −0.1647 − j 0.5114 = 0.5373∠ − 107.9°
Z M 2 0.0697 + j1.86
E
0.95∠ − 15°
I 3 = A3 =
= −0.0469 − j 0.1988 = 0.204∠ − 103.2°
Z M 3 0.14 + j 4.65
I1 =
The nodal equations for this power system are:
Ybus V = I
ª 1.0288 − j 4.3646 −0.3265 + j1.6355 −0.3486 + j1.7866º
V = «« −0.3265 + j1.6355 0.6952 − j 3.9590 −0.3486 + j1.7866»»
«¬ −0.3486 + j1.7866 −0.3486 + j1.7866 0.7036 − j 3.7881 »¼
ª0.9646 + j 0.1604 º ª0.978∠9.4°º
V = «« 0.9463 + j 0.0635 »» = ««0.948∠3.8°»»
«¬ 0.9503 + j 0.0926 »¼ «¬ 0.955∠5.6° »¼
(a)
The per-unit voltages at each bus in the power system are:
(b)
The actual voltages at each bus in the power system are:
V1 = 0.978∠9.4°
V2 = 0.948∠3.8°
V3 = 0.955∠5.6°
V1 = V1,puVbase = ( 0.978)(12.8 kV ) = 12.5 kV
V2 = V2,puVbase = ( 0.948)(12.8 kV ) = 12.1 kV
V3 = V3,puVbase = ( 0.955)(12.8 kV ) = 12.2 kV
203
−1
ª 0.7832 − j 0.8526 º
« −0.1647 − j 0.5114 »
«
»
«¬ −0.0469 − j 0.1988»¼
(c)
The current flowing in Line 1 is:
(start here)
I L1 = Ya ( V1 − V2 ) = ( 0.3265 − j1.6355)( 0.978∠9.4° − 0.948∠3.8°)
I L1 = 0.1645 + j 0.0017 = 0.165∠0.6°
The current flowing in Line 2 is:
I L 2 = Yb ( V1 − V3 ) = ( 0.3486 − j1.7866)( 0.978∠9.4° − 0.955∠5.6°)
I L 2 = 0.1263 − j 0.0029 = 0.1269∠ − 0.9°
The current flowing in Line 3 is:
I L 3 = Yc ( V2 − V3 ) = ( 0.3486 − j1.7866)( 0.948∠3.8° − 0.955∠5.6°)
I L 3 = −0.0533 + j 0.0029 = 0.0534∠ − 177°
(d)
The real and reactive power flowing from Bus 1 to Bus 2 in Line 1 is:
S L1 = V1I L1* = ( 0.978∠9.4°)( 0.165∠0.6°)
S L1 = 0.159 + j 0.0248 = 0.161∠8.9°
*
PL1 = Sbase PL1,pu = ( 40 MVA )( 0.159 ) = 6.4 MW
QL1 = S baseQL1,pu = ( 40 MVA )( 0.0248) = 1.0 MVAR
The real and reactive power flowing from Bus 1 to Bus 3 in Line 2 is:
S L 2 = V1I L 2* = ( 0.978∠9.4°)( 0.1269∠ − 0.9°)
S L 2 = 0.1215 + j 0.0221 = 0.124∠10.3°
*
PL 2 = Sbase PL 2,pu = ( 40 MVA )( 0.1215) = 4.86 MW
QL 2 = SbaseQL 2,pu = ( 40 MVA )( 0.0221) = 0.88 MVAR
The real and reactive power flowing from Bus 2 to Bus 3 in Line 3 is:
S L 3 = V3I L 3* = ( 0.955∠5.6°)( 0.0534∠ − 177°)
S L 3 = −0.0509 − j 0.0021 = 0.051∠ − 178°
*
PL 3 = Sbase PL 3,pu = ( 40 MVA )( −0.0509 ) = −2.04 MW
QL 3 = S baseQL 3,pu = ( 40 MVA )( −0.0021) = −0.09 MVAR
The negative power here means that the power is really flowing from Bus 3 to Bus 2.
(e)
None of the components in the power system are even close to being overloaded.
204