ECE421/521 POWER SYSTEM ANALYSIS
Homework #4
total
3.2a
521(100’) 4
421(120’)
3.6a
521(100’) 5
421(120’)
3.2(a)
3.2b
1
3.2c
3
3.4a
5
3.4b
5
3.4c
3
3.5a
5
3.5b
5
3.5c
5
3.5d
5
3.6b
5
3.6c
5
3.6d
1
3.9
5
3.10
8
3.14
14
3.15a
5
3.15b
5
3.17
6
5’
The three-phase apparent power is,
𝑆3𝜙 = 60∠𝑐𝑜𝑠 −1 0.8 = 60∠36.87°𝑀𝑉𝐴 = 48𝑀𝑊 + 𝑗36𝑀𝑣𝑎𝑟
(1’)
The rated voltage per phase is,
𝑉=
69.3
√3
= 40∠0°𝑘𝑉
(1’)
The rated current is,
𝑆∗
3𝜙
𝐼𝑎 = 3𝑉
∗ =
(60∠−36.87°)103
3×40∠0°
= 500∠ − 36.87°𝐴
(1’)
The excitation voltage per phase is,
𝐸 = 40000 + (𝑗15)(500∠ − 36.87°) = 44500 + 𝑗6000𝑉 = 44903∠7.7°𝑉
(1’)
The excitation voltage per phase is 44.9kV and the power angle δ is 7.7°
(b)
1’
The maximum value of three-phase power that the generator can deliver before losing its synchronism
is reached at 𝛿 = 90°,
𝑃3𝜙𝑚𝑎𝑥 = 3
(c)
|𝐸||𝑉|
𝑋𝑠
𝑠𝑖𝑛90° = 3
3’
The power angle is,
36000×40000
15
= 288𝑀𝑊
(1’)
𝛿 = 𝑠𝑖𝑛−1
𝑃3𝜙 𝑋𝑠
3|𝐸||𝑉|
= 𝑠𝑖𝑛−1
48×15
3×46×40
= 7.5°
(1’)
The armature current is,
𝐼𝑎 =
𝐸−𝑉
𝑗𝑋𝑠
=
46000∠7.5°−40000∠0°
𝑗15
= 400 − 𝑗374𝐴 = 548∠ − 43°𝐴
(1’)
The power factor is,
𝑃𝐹 = cos(0° − (−43°)) = 0.73
(1’)
The power factor is lagging.
3.4(a)
5’
(1’)
Based on phasor diagram,
|𝑉|𝑠𝑖𝑛𝛿 = 𝑋𝑞 |𝐼𝑞 |
(1’)
Considering |𝐼𝑞 | = |𝐼𝑎 |cos⁡(𝜃 + 𝛿),
|𝑉|𝑠𝑖𝑛𝛿 = 𝑋𝑞 |𝐼𝑎 |cos⁡(𝜃 + 𝛿)
Applying triangle function,
|𝑉|𝑠𝑖𝑛𝛿 = 𝑋𝑞 |𝐼𝑎 |(𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛿 − 𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛿)
(1’)
Rearrange equation,
|𝑉|𝑠𝑖𝑛𝛿 + 𝑋𝑞 |𝐼𝑎 |𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝛿 = 𝑋𝑞 |𝐼𝑎 |𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝛿
Divide 𝑐𝑜𝑠𝛿 on both sides of equation,
|𝑉|𝑡𝑎𝑛𝛿 + 𝑋𝑞 |𝐼𝑎 |𝑠𝑖𝑛𝜃𝑡𝑎𝑛𝛿 = 𝑋𝑞 |𝐼𝑎 |𝑐𝑜𝑠𝜃
Rearrange it,
(1’)
𝑋𝑞 |𝐼𝑎 |𝑐𝑜𝑠𝜃
𝑡𝑎𝑛𝛿 = |𝑉|+𝑋
𝑞 |𝐼𝑎 |𝑠𝑖𝑛𝜃
(1’)
We have,
𝑋 |𝐼 |𝑐𝑜𝑠𝜃
𝑞 𝑎
𝛿 = 𝑡𝑎𝑛−1 (|𝑉|+𝑋
|𝐼
𝑞 𝑎 |𝑠𝑖𝑛𝜃
(b)
)
5’
The three-phase apparent power is,
𝑆3𝜙 = 60∠𝑐𝑜𝑠 −1 0.8 = 60∠36.87°𝑀𝑉𝐴 = 48𝑀𝑊 + 𝑗36𝑀𝑣𝑎𝑟
(1’)
The rated voltage per phase is,
𝑉=
34.64
√3
= 20∠0°𝑘𝑉
(1’)
The rated current is,
∗
𝑆3𝜙
𝐼𝑎 = 3𝑉 ∗ =
(60∠−36.87°)103
3×20∠0°
= 1000∠ − 36.87°𝐴
(1’)
Power angle,
𝑋 |𝐼 |𝑐𝑜𝑠𝜃
𝑞 𝑎
𝛿 = 𝑡𝑎𝑛−1 (|𝑉|+𝑋
|𝐼
𝑞 𝑎 |𝑠𝑖𝑛𝜃
9.333×1000×0.8
)
20000+9.333×1000×0.6
) = 𝑡𝑎𝑛−1 (
= 16.3°
(1’)
Excitation voltage magnitude is,
|𝐸| = |𝑉|𝑐𝑜𝑠𝛿 + 𝑋𝑑 𝐼𝑑 = 20000 × 𝑐𝑜𝑠16.3° + 13.5 × 1000 × sin(36.87° + 16.3°) = 30𝑘𝑉
Excitation voltage is,
𝐸 = 30∠16.3°𝑘𝑉
(c)
3’
Run the following code,
(1’)
%Problem 3.4(c)
delta = 0:0.05:180;
delta = delta*pi/180;
Xd = 13.5;
Xq = 9.333;
E = 30000;
V = 20000;
P = (3*E*V*sin(delta)/Xd+3*V^2*(Xd-Xq)*sin(2*delta)/(2*Xd*Xq))/10^6;
(1’)
delta = delta*180/pi;
[Pmax,k] = max(P);
dmax = delta(k);
plot(delta,P);
title('salient-pole synchronous generator power angle curve')
xlabel('delta, degree');
ylabel('real power, MW');
we have the power angle curve,
(1’)
Read Pmax and dmax from workspace, we have,
(1’)
Steady-state maximum power Pmax=138.71MW
The corresponding power angle dmax=75.05 degree
3.5(a)
5’
To determine the equivalent circuit to the primary side,
𝑁
2
𝑅𝑒1 = 𝑅1 + (𝑁1 ) 𝑅2 = 0.2 + 102 × 0.002 = 0.4𝛺
2
𝑁
2
𝑋𝑒1 = 𝑋1 + (𝑁1 ) 𝑋2 = 0.45 + 102 × 0.0045 = 0.9𝛺
2
𝑍𝑒1 = 𝑅𝑒1 + 𝑗𝑋𝑒1 = 0.4 + 𝑗0.9𝛺
(1’)
(2’)
(2’)
(b)
5’
The transformer is operating at full load 0.8 power factor lagging and 240V,
𝑉2′ =
𝑁1
𝑉
𝑁2 2
= 10 × 240 = 2400𝑉
(1’)
PF=0.8 lagging,
𝑆2 = 150 × 0.8 + 𝑗150 × 0.6 = 120 + 𝑗90𝑘𝑉𝐴
𝑆
𝐼1 = (𝑉2′ )∗ =
2
120000−𝑗90000
2400
(1’)
= 50 − 𝑗37.5𝐴 = 62.5∠ − 36.87°𝐴
(1’)
The primary voltage,
𝑉1 = 𝑉2′ + 𝐼1 𝑍𝑒1 = 2400 + (50 − 𝑗37.5)(0.4 + 𝑗0.9) = 2453.8 + 𝑗30𝑉 = 2453.9∠0.7°𝑉
(1’)
Voltage regulation,
𝑉𝑅 =
(c)
|𝑉1 |−|𝑉2′ |
|𝑉2′ |
× 100 =
2453.9−2400
×
2400
100 = 2.25%
(1’)
5’
PF=0.8 leading,
𝑆2 = 150 × 0.8 − 𝑗150 × 0.6 = 120 − 𝑗90𝑘𝑉𝐴
𝑆
𝐼1 = (𝑉 ′ )∗ = 50 + 𝑗37.5𝐴 = 62.5∠36.87°𝐴
2
(0.5’)
(0.5’)
The primary voltage,
𝑉1 = 𝑉2′ + 𝐼1 𝑍𝑒1 = 2400 + (50 + 𝑗37.5)(0.4 + 𝑗0.9) = 2386.3 + 𝑗60𝑉 = 2387∠1.4°𝑉
Voltage regulation,
(2’)
𝑉𝑅 =
(d)
|𝑉1 |−|𝑉2′ |
|𝑉2′ |
× 100 =
2387−2400
×
2400
100 = −0.54%
(2’)
5’
Run trans.m, for the PF=0.8 lagging load, the result is,
Shunt branch ref. to LV side
Rc =
Xm =
(2’)
Shunt branch ref. to HV side
10.000 ohm
Rc =
15.000 ohm
1000.000 ohm
Xm =
Series branch ref. to LV side
1500.000 ohm
Series branch ref. to HV side
Ze = 0.004000 + j 0.009000 ohm
Ze = 0.400000 + j 0.900000 ohm
Enter load kVA, S2 = 150
Enter load power factor, pf = 0.8
Enter 'lg' within quotes for lagging pf
or 'ld' within quotes for leading pf -> 'lg'
Enter load terminal voltage in volt, V2 = 240
Secondary load voltage = 240.000 V
Secondary load current = 625.000 A at -36.87 degrees
Current ref. to primary =
Primary no-load current =
Primary input current =
62.500 A at -36.87 degrees
2.949 A at -33.69 degrees
65.445 A at -36.73 degrees
Primary input voltage = 2453.933 V at 0.70 degrees
Voltage regulation
=
Transformer efficiency =
2.247 percent
94.249 percent
Maximum efficiency is 95.238 percent, occurs at 288.000 kVA with 0.80 pf
The result agrees to the calculation.
(1’)
for the PF=0.8 leading load, the result is,
(2’)
Would you like the analysis for another load? Enter 'Y' or 'N' within quotes -> 'Y'
Enter load kVA, S2 = 150
Enter load power factor, pf = 0.8
Enter 'lg' within quotes for lagging pf
or 'ld' within quotes for leading pf -> 'ld'
Enter load terminal voltage in volt, V2 = 240
Secondary load voltage = 240.000 V
Secondary load current = 625.000 A at 36.87 degrees
Current ref. to primary =
Primary no-load current =
Primary input current =
62.500 A at 36.87 degrees
2.869 A at -33.69 degrees
63.512 A at 34.43 degrees
Primary input voltage = 2387.004 V at 1.44 degrees
Voltage regulation
=
Transformer efficiency =
-0.541 percent
94.249 percent
Maximum efficiency is 95.238 percent, occurs at 288.000 kVA with 0.80 pf
The result agrees to the calculation.
(1’)
3.6(a)
5’
Open-circuit test
𝑉22
𝑃0
𝑅𝑐2 =
𝑉
=
24002
3456
2400
= 1666.7𝛺
𝐼𝑐 = 𝑅 2 = 1666.7 = 1.44𝐴
𝑐2
(0.5’)
(0.5’)
𝐼𝑚 = √𝐼02 − 𝐼𝑐2 = √2.42 − 1.442 = 1.92𝐴
(0.5’)
𝑉
2400
1.92
𝑋𝑚2 = 𝐼 2 =
𝑚
= 1250𝛺
(0.5’)
Refer to the high voltage side,
𝑁
4800
𝑅𝑐1 = (𝑁1 )2 𝑅𝑐2 = (2400)2 1666.7 = 6666.8𝛺
2
𝑁
𝑁2
4800 2
) 1250
2400
𝑋𝑐1 = ( 1 )2 𝑋𝑚2 = (
= 5000𝛺
(0.5’)
(0.5’)
Short-circuit test
𝑍𝑒1 =
𝑉𝑠𝑐
𝐼𝑠𝑐
𝑅𝑒1 =
𝑃𝑠𝑐
(𝐼𝑠𝑐 )2
=
1250
12.5
=
= 100𝛺
4375
12.52
= 28𝛺
(1’)
2
2
𝑋𝑒1 = √𝑍𝑒1
− 𝑅𝑒1
= √1002 − 282 = 96𝛺
(b)
(1’)
5’
Transformer is operating at full-load, 0.8 PF lagging,
𝑆 = 60 × 0.8 + 60 × 0.6 = 48𝑘𝑊 + 36𝑘𝑣𝑎𝑟 = 60∠36.87°𝑘𝑉𝐴
(0.5’)
Calculate voltage regulation utilizing the equivalent circuit referred to the primary side,
𝑆
𝐼2 = (𝑉 )∗ =
2
60000∠−36.87°
2400
= 25∠ − 36.87°𝐴
(0.5’)
𝑁
2400
𝐼2′ = 𝑁2 𝐼2 = 4800 × 25∠ − 36.87° = 12.5∠ − 36.87°𝐴
1
𝑁
𝑉2′ = 𝑁1 𝑉2 = 4800𝑉
2
(0.5’)
(0.5’)
𝑉1 = 𝑉2′ + 𝐼2′ 𝑍𝑒1 = 4800 + (12.5∠ − 36.87°)(28 + 𝑗96) = 5800 + 𝑗750𝑉 = 5848.3∠7.37°𝑉
𝑉𝑅 =
|𝑉1 |−|𝑉2′ |
×
|𝑉2′ |
100 =
5848.3−4800
× 100
4800
= 21.84%
(0.5’)
Efficiency,
𝑃𝑐 =
|𝑉2′ |2
𝑅𝑐1
=
48002
6666.8
= 3456𝑊
(0.5’)
𝑆 = |𝑉2 ||𝐼2 | = 2400 × 25 = 60000𝑉𝐴
𝑃𝑐𝑢 = 𝑅𝑒1 |𝐼2′ |2 = 28 × 12.52 = 4375𝑊
𝑛×𝑆×𝑃𝐹
(0.5’)
1×60000×0.8
𝜂 = (𝑛×𝑆×𝑃𝐹)+𝑛2 ×𝑃
𝑐𝑢 +𝑃𝑐
(c)
(0.5’)
= (1×60000×0.8)+1×4375+3456 = 85.97%
(0.5’)
5’
The maximum efficiency occurs when,
𝑃𝑐
𝑃𝑐𝑢
𝑛=√
3456
4375
=√
= 0.89
(2’)
The load for maximum efficiency is,
|𝑆| = 𝑛60 = 53.4𝑘𝑉𝐴
(1’)
The maximum efficiency at 0.8 PF lagging is,
𝑛×𝑆×𝑃𝐹
𝜂 = (𝑛×𝑆×𝑃𝐹)+𝑛2 ×𝑃
𝑐𝑢 +𝑃𝑐
(d)
0.89×60000×0.8
= (0.89×60000×0.8)+0.892 ×4375+3456 = 86.06%
(2’)
1’
Efficiency for 0.75 full-load, 0.8 PF lagging is,
𝑛×𝑆×𝑃𝐹
𝜂 = (𝑛×𝑆×𝑃𝐹)+𝑛2 ×𝑃
𝑐𝑢 +𝑃𝑐
0.75×60000×0.8
= (0.75×60000×0.8)+0.752 ×4375+3456 = 85.88%
(1’)
(0.5’)
3.9
5’
𝑉2𝐿 , line voltage of low voltage side
𝑉1𝐿 , line voltage of high voltage side
𝑉2𝑃 , phase voltage of low voltage side
𝑉1𝑃 , phase voltage of high voltage side
Low side is Δ connected, take 𝑉2𝐿 as reference
𝑉2𝑃 = 𝑉2𝐿 = 24∠0°𝑘𝑉
(0.5’)
′
𝑉2𝐿
= 𝑎𝑉2𝐿 = 10 × 24 = 240∠0°𝑘𝑉
(0.5’)
The complex power of low side,
𝑆3𝜙 = 400∠36.87°𝑀𝑉𝐴
′
𝐼1𝑃 = 𝐼2𝑃
=(
𝑆3𝜙
′
√3𝑉2𝐿
)∗ =
′
𝑉1𝑃 = 𝑉2𝑃
+ 𝑍𝑒1 𝐼1𝑃 =
(0.5’)
400∠−36.87°
×
√3×240
240
+
√3
1000 = 962.3∠ − 36.87°𝐴
(0.5’)
(1.2 + 𝑗6) × 0.9623∠ − 36.87° = 142.96 + 𝑗3.93𝑘𝑉 = 143.01∠1.57°𝑘𝑉
(0.5’)
The line voltage at the high voltage side is,
𝑉1𝐿 = √3∠30°𝑉1𝑃 = 247.69∠31.57°𝑘𝑉
(1’)
The voltage of the sending end of the feeder is,
𝑉𝑓𝑒𝑒𝑑𝑒𝑟 = 𝑉1𝑃 + 𝑍𝐼1 = 143.01∠1.57° + (0.6 + 𝑗1.2) × 0.9623∠ − 36.87° = 144.18∠1.79°𝑘𝑉
𝑉𝑓𝑒𝑒𝑑𝑒𝑟𝐿 = √3∠30°𝑉1𝑃 = 249.72∠31.79°𝑘𝑉
3.10
(1’)
8’
Take the transformer rated value 400MVA, 240kV as base quantities,
𝑆𝐵1 = 400𝑀𝑉𝐴
𝑉𝐵1 = 240𝑘𝑉
(0.5’)
(0.5’)
24
𝑉𝐵2 = 240 (240) = 24𝑘𝑉
𝑍𝑇 = 1.2 + 𝑗6𝛺
(0.5’)
(0.5’)
(0.5’)
𝑉2
𝑍𝐵1 = 𝑆𝐵1 =
𝐵1
2402
400
= 144𝛺
𝑍𝑇𝑝𝑢 =
𝑍𝑇
𝑍𝐵1
=
𝑉2𝑝𝑢 =
24
𝑉𝐵2
= 1𝑝𝑢
𝑆𝑝𝑢 =
1.2+𝑗6
144
400∠36.87°
𝑆𝐵1
(0.5’)
= 0.0083 + 𝑗0.0417𝑝𝑢
(0.5’)
(0.5’)
= 1∠36.87°𝑝𝑢
(0.5’)
𝐼𝑝𝑢 = (𝑉 𝑝𝑢 )∗ = 1∠ − 36.87°𝑝𝑢
(0.5’)
𝑆
2𝑝𝑢
𝑉1𝑝𝑢 = 𝑉2𝑝𝑢 + 𝑍𝑇𝑝𝑢 𝐼𝑝𝑢 = 1 + (0.0083 + 𝑗0.0417)(1∠ − 36.87°) = 1.0317 + 𝑗0.0284 =
1.0321∠1.58°𝑝𝑢 (1’)
High-voltage side line voltage is,
𝑉1𝐿 = 𝑉𝐵1 𝑉1𝑝𝑢 = 247.69∠1.58𝑘𝑉
(0.5’)
Feeder voltage,
𝑍
𝑍𝑓𝑝𝑢 = 𝑍 𝑓 =
𝐵1
0.6+𝑗1.2
144
= 0.0042 + 𝑗0.083𝑝𝑢
(0.5’)
𝑉𝑓𝑝𝑢 = 𝑉1𝑝𝑢 + 𝑍𝑓𝑝𝑢 𝐼1𝑝𝑢 = 1.0317∠1.58° + 𝑗0.0284 + (0.0042 + 𝑗0.0083)(1∠ − 36.87°) = 1.04 +
𝑗0.03𝑝𝑢 = 1.04∠1.79°𝑝𝑢 (1’)
𝑉𝑓 = 𝑉𝑓𝑝𝑢 𝑉𝐵1 = 249.72∠1.79°𝑘𝑉
3.14
(0.5’)
14’
𝑉𝐵1 = 22𝑘𝑉
(0.5’)
𝑆𝐵 = 100𝑀𝑉𝐴
(0.5’)
220
𝑉𝐵2 = 𝑉𝐵3 = 22 ( 22 ) = 220𝑘𝑉
22
𝑉𝐵4 = 220 (220) = 22𝑘𝑉
(0.5’)
110
𝑉𝐵5 = 𝑉𝐵6 = 22 ( 22 ) 110𝑘𝑉
22
𝑉𝐵𝑚 = 110 (110) = 22𝑘𝑉
4
𝑉𝐵𝐿 = 110 (110) = 4𝑘𝑉
(0.5’)
(0.5’)
(0.5’)
(0.5’)
𝑍𝐵2 =
2
𝑉𝐵2
𝑆𝐵
𝑍𝑙𝑖𝑛𝑒1 =
𝑍𝐵5 =
2202
100
𝑗121
484
2
𝑉𝐵5
𝑆𝐵
𝑍𝑙𝑖𝑛𝑒2 =
=
=
= 484𝛺
= 𝑗0.25𝑝𝑢
1102
100
𝑗42.35
121
(0.5’)
(0.5’)
= 121𝛺
(0.5’)
= 𝑗0.35𝑝𝑢
(0.5’)
20 2
100
𝑋𝑚 = 0.225 (68.85) (22) = 0.27𝑝𝑢
(0.5’)
Load is Δ-connected capacitors,
𝑋𝑙𝑜𝑎𝑑 =
42
10
𝑋𝐵𝑙𝑜𝑎𝑑 =
𝑋𝑙𝑜𝑎𝑑 =
= 1.6𝛺
𝑉72
𝑆
42
10
(0.5’)
= 0.16𝛺
= 10𝑝𝑢
100
(0.5’)
(0.5’)
22
𝑋𝐺 = 0.24( 80 )(22)2 = 0.3𝑝𝑢
100 22 2
)( )
50 22
𝑋𝑇1 = 0.1(
100
= 0.2𝑝𝑢
(0.5’)
(0.5’)
220
𝑋𝑇2 = 0.06( 40 )(220)2 = 0.15𝑝𝑢
100
(0.5’)
22
𝑋𝑇3 = 0.064( 40 )(22)2 = 0.16𝑝𝑢
(0.5’)
(0.5’)
100 110 2
)( )
40 110
𝑋𝑝𝑠 = 0.096(
= 0.24𝑝𝑢
(0.5’)
𝑋𝑝𝑡 = 0.072( 40 )(110)2 = 0.18𝑝𝑢
(0.5’)
100
100
110
22
𝑋𝑠𝑡 = 0.12( 40 )(22)2 = 0.3𝑝𝑢
𝑍𝑝 =
𝑍𝑝𝑠 +𝑍𝑝𝑡 −𝑍𝑠𝑡
2
=
(0.5’)
𝑗0.24+𝑗0.18−𝑗0.3
2
= 𝑗0.06𝑝𝑢
(0.5’)
𝑍𝑠 =
𝑍𝑡 =
𝑍𝑝𝑠 +𝑍𝑠𝑡 −𝑍𝑝𝑡
2
𝑍𝑝𝑡 +𝑍𝑠𝑡 −𝑍𝑝𝑠
2
=
𝑗0.24+𝑗0.3−𝑗0.18
2
= 𝑗0.18𝑝𝑢
(0.5’)
=
𝑗0.18+𝑗0.3−𝑗0.24
2
= 𝑗0.12𝑝𝑢
(0.5’)
(1’)
3.15(a)
5’
𝑉𝐵𝑔 = 20𝑘𝑉
(0.5’)
200
)
20
𝑉𝐵1 = 𝑉𝐵2 = 20 (
= 200𝑘𝑉
20
𝑉𝐵𝑚 = 200 (200) = 20𝑘𝑉
100
(0.5’)
20
𝑋𝐺 = 0.09( 60 )(20)2 = 0.15𝑝𝑢
100 20 2
)( )
50 20
𝑋𝑇1 = 0.1(
= 0.2𝑝𝑢
100 200 2
)( )
50 200
𝑋𝑇2 = 0.1(
100
(0.5’)
𝑋𝑚 = 0.08(43.2)(20)2 = 0.15𝑝𝑢
𝑍𝐵𝑙𝑖𝑛𝑒 =
2
𝑉𝐵1
𝑆𝐵
=
𝑍
2002
100
𝑍𝑙𝑖𝑛𝑒 = 𝑍 𝑙𝑖𝑛𝑒 =
𝐵𝑙𝑖𝑛𝑒
= 400𝛺
120+𝑗200
400
(0.5’)
(0.5’)
= 0.2𝑝𝑢
18
(0.5’)
(0.5’)
(0.5’)
= 0.3 + 𝑗0.5𝑝𝑢
(0.5’)
(0.5’)
(b)
5’
18
𝑉𝑚 = 20 = 0.9𝑝𝑢
(0.5’)
45
|𝑆𝑚 | = 100 = 0.45𝑝𝑢
(0.5’)
𝑆𝑚 = 0.45 × 0.8 + 𝑗0.45 × 0.6 = 0.36 + 𝑗0.27𝑝𝑢
𝑆
𝐼 = (𝑉𝑚 )∗ =
𝑚
0.36−𝑗0.27
0.9
= 0.4 − 𝑗0.3𝑝𝑢
(0.5’)
(0.5’)
Terminal voltage of generator is,
𝑉𝑔 = 𝑉𝑚 + 𝐼(𝑗0.2 + 𝑗0.2 + 0.3 + 𝑗0.5) = 0.9 + (0.4 − 𝑗0.3)(0.3 + 𝑗0.9) = 1.29 + 𝑗0.27𝑝𝑢 =
1.318∠11.8°𝑝𝑢 (1’)
𝑉𝑔 = 1.318∠11.8° × 20 = 26.36∠11.8°𝑘𝑉
(0.5’)
𝐸𝑔 = 𝑉𝑔 + 𝐼(𝑗0.15) = 1.29 + 𝑗0.27 + (0.4 − 𝑗0.3)(𝑗0.15) = 1.335 + 𝑗0.33𝑝𝑢 = 1.38∠13.9°𝑝𝑢
𝑉𝑔 = 1.38∠13.9° × 20 = 27.5∠13.9°𝑘𝑉
3.17
(0.5’)
6’
Choose 𝑉𝐵1 = 23𝑘𝑉, 𝑆𝐵 = 100𝑀𝑉𝐴 as common base,
𝑋𝑇 = 0.2𝑝𝑢
115
)
23
𝑉𝐵2 = 𝑉𝐵3 = 23 (
𝑍𝐵𝑙𝑖𝑛𝑒 =
2
𝑉𝐵2
𝑆𝐵
=
1152
100
= 115𝑘𝑉
= 132.25𝛺
(0.5’)
(1’)
𝑍𝑙𝑖𝑛𝑒 =
𝑗66.125
132.25
𝑉3 =
115
𝑉𝐵3
𝑆2 =
184.8+𝑗6.6
𝑆𝐵
𝑆3 =
𝑗20
𝑆𝐵
= 𝑗0.5𝑝𝑢
= 1𝑝𝑢
(0.5’)
= 1.848 + 𝑗0.066𝑝𝑢
= 𝑗0.2𝑝𝑢
𝑆
𝐼3 = (𝑉3 )∗ =
3
(0.5’)
−𝑗0.2
1
(0.5’)
(0.5’)
= −𝑗0.2𝑝𝑢
(0.5’)
𝑉2 = 𝑉3 + 𝐼3 𝑍𝑙𝑖𝑛𝑒 = 1 + (−𝑗0.2)(𝑗0.5) = 1.1𝑝𝑢
𝑆
𝐼2 = (𝑉2 )∗ =
2
1.848−𝑗0.066
1.1
= 1.68 − 𝑗0.06𝑝𝑢
(0.5’)
(0.5’)
𝐼1 = 𝐼2 + 𝐼3 = −𝑗0.2 + 1.68 − 𝑗0.06 = 1.68 − 𝑗0.26𝑝𝑢
(0.5’)
𝑉1 = 𝑉2 + 𝐼1 𝑍𝑇 = 1.1 + (1.68 − 𝑗0.26)(𝑗0.2) = 1.15 + 𝑗0.34𝑝𝑢 = 1.2∠16.26°𝑝𝑢
In kV,
𝑉1 = 1.2𝑉𝐵1 = 27.6𝑘𝑉
𝑉2 = 1.1𝑉𝐵2 = 126.5𝑘𝑉
(0.5’)
(0.5’)
(0.5’)