ECE421/599 POWER SYSTEM ANALYSIS
Homework #3
total
3.2
421/599(100’) 16
3.2(a)
3.4a
8
3.4b
8
3.4c
8
3.5a
8
3.5b
8
3.5c
8
3.5d
8
3.6
28
8’
The three-phase apparent power is,
𝑆3𝜙 = 60∠𝑐𝑜𝑠 −1 0.8 = 60∠36.87°𝑀𝑉𝐴 = 48𝑀𝑊 + 𝑗36𝑀𝑣𝑎𝑟
(2’)
The rated voltage per phase is,
𝑉=
69.3
√3
= 40∠0°𝑘𝑉
(2’)
The rated current is,
𝑆∗
3𝜙
𝐼𝑎 = 3𝑉
∗ =
(60∠−36.87°)103
3×40∠0°
= 500∠ − 36.87°𝐴
(2’)
The excitation voltage per phase is,
𝐸 = 40000 + (𝑗15)(500∠ − 36.87°) = 44500 + 𝑗6000𝑉 = 44903∠7.7°𝑉
(2’)
The excitation voltage per phase is 44.9kV and the power angle δ is 7.7°
(b)
2’
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
36000×40000
15
= 288𝑀𝑊
6’
The power angle is,
𝑃
𝑋
48×15
3𝜙 𝑠
𝛿 = 𝑠𝑖𝑛−1 3|𝐸||𝑉|
= 𝑠𝑖𝑛−1 3×46×40 = 7.5°
The armature current is,
(2’)
(2’)
𝐼𝑎 =
𝐸−𝑉
𝑗𝑋𝑠
=
46000∠7.5°−40000∠0°
𝑗15
= 400 − 𝑗374𝐴 = 548∠ − 43°𝐴
(2’)
The power factor is,
𝑃𝐹 = cos(0° − (−43°)) = 0.73
(2’)
The power factor is lagging.
3.4(a)
8’
(2’)
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,
𝑋𝑞 |𝐼𝑎 |𝑐𝑜𝑠𝜃
𝑡𝑎𝑛𝛿 = |𝑉|+𝑋
𝑞 |𝐼𝑎 |𝑠𝑖𝑛𝜃
We have,
(2’)
(2’)
𝑋𝑞 |𝐼𝑎 |𝑐𝑜𝑠𝜃
𝛿 = 𝑡𝑎𝑛−1 (|𝑉|+𝑋
𝑞 |𝐼𝑎 |𝑠𝑖𝑛𝜃
(b)
)
8’
The three-phase apparent power is,
𝑆3𝜙 = 60∠𝑐𝑜𝑠 −1 0.8 = 60∠36.87°𝑀𝑉𝐴 = 48𝑀𝑊 + 𝑗36𝑀𝑣𝑎𝑟
(2’)
The rated voltage per phase is,
𝑉=
34.64
√3
= 20∠0°𝑘𝑉
(1’)
The rated current is,
𝐼𝑎 =
∗
𝑆3𝜙
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°
(2’)
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)
8’
Run the following code,
(4’)
%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;
delta = delta*180/pi;
[Pmax,k] = max(P);
dmax = delta(k);
plot(delta,P);
title('salient-pole synchronous generator power angle curve')
(2’)
xlabel('delta, degree');
ylabel('real power, MW');
we have the power angle curve,
(2’)
Read Pmax and dmax from workspace, we have,
(2’)
Steady-state maximum power Pmax=138.71MW
The corresponding power angle dmax=75.05 degree
3.5(a)
8’
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
(2’)
(2’)
(2’)
(2’)
(b)
8’
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
(2’)
The primary voltage,
𝑉1 = 𝑉2′ + 𝐼1 𝑍𝑒1 = 2400 + (50 − 𝑗37.5)(0.4 + 𝑗0.9) = 2453.8 + 𝑗30𝑉 = 2453.9∠0.7°𝑉
(2’)
Voltage regulation,
𝑉𝑅 =
(c)
|𝑉1 |−|𝑉2′ |
|𝑉2′ |
× 100 =
2453.9−2400
×
2400
100 = 2.25%
(2’)
8’
PF=0.8 leading,
𝑆2 = 150 × 0.8 − 𝑗150 × 0.6 = 120 − 𝑗90𝑘𝑉𝐴
𝑆
𝑉2
𝐼1 = ( ′ )∗ = 50 + 𝑗37.5𝐴 = 62.5∠36.87°𝐴
(2')
(2')
The primary voltage,
𝑉1 = 𝑉2′ + 𝐼1 𝑍𝑒1 = 2400 + (50 + 𝑗37.5)(0.4 + 𝑗0.9) = 2386.3 + 𝑗60𝑉 = 2387∠1.4°𝑉
Voltage regulation,
𝑉𝑅 =
(d)
|𝑉1 |−|𝑉2′ |
|𝑉2′ |
× 100 =
2387−2400
×
2400
100 = −0.54%
(2’)
8’
Run trans.m, for the PF=0.8 lagging load, the result is,
Shunt branch ref. to LV side
(2’)
Shunt branch ref. to HV side
(2’)
Rc =
Xm =
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.
(2’)
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.
(2’)
3.6(a)
8’
Open-circuit test
𝑅𝑐2 =
𝐼𝑐 =
𝑉22
𝑃0
𝑉2
𝑅𝑐2
=
=
24002
3456
2400
1666.7
= 1666.7𝛺
= 1.44𝐴
(1’)
(1’)
𝐼𝑚 = √𝐼02 − 𝐼𝑐2 = √2.42 − 1.442 = 1.92𝐴
𝑋𝑚2 =
𝑉2
𝐼𝑚
=
2400
1.92
= 1250𝛺
(1’)
(1’)
Refer to the high voltage side,
𝑁
4800
𝑅𝑐1 = (𝑁1 )2 𝑅𝑐2 = (2400)2 1666.7 = 6666.8𝛺
2
(1’)
𝑁
4800
𝑋𝑐1 = (𝑁1 )2 𝑋𝑚2 = (2400)2 1250 = 5000𝛺
2
(1’)
Short-circuit test
𝑉𝑠𝑐
𝐼𝑠𝑐
𝑍𝑒1 =
=
1250
12.5
𝑃
= 100𝛺
4375
𝑅𝑒1 = (𝐼 𝑠𝑐)2 = 12.52 = 28𝛺
(1’)
𝑠𝑐
2
2
𝑋𝑒1 = √𝑍𝑒1
− 𝑅𝑒1
= √1002 − 282 = 96𝛺
(b)
(1’)
12’
Transformer is operating at full-load, 0.8 PF lagging,
𝑆 = 60 × 0.8 + 60 × 0.6 = 48𝑘𝑊 + 36𝑘𝑣𝑎𝑟 = 60∠36.87°𝑘𝑉𝐴
(1’)
Calculate voltage regulation utilizing the equivalent circuit referred to the primary side,
𝑆
𝐼2 = (𝑉 )∗ =
2
𝑁
60000∠−36.87°
2400
= 25∠ − 36.87°𝐴
2400
(1’)
𝐼2′ = 𝑁2 𝐼2 = 4800 × 25∠ − 36.87° = 12.5∠ − 36.87°𝐴
1
𝑁
𝑉2′ = 𝑁1 𝑉2 = 4800𝑉
2
(1’)
(1’)
𝑉1 = 𝑉2′ + 𝐼2′ 𝑍𝑒1 = 4800 + (12.5∠ − 36.87°)(28 + 𝑗96) = 5800 + 𝑗750𝑉 = 5848.3∠7.37°𝑉
(1’)
𝑉𝑅 =
|𝑉1 |−|𝑉2′ |
×
|𝑉2′ |
100 =
5848.3−4800
× 100
4800
= 21.84%
(1’)
Efficiency,
𝑃𝑐 =
|𝑉2′ |2
𝑅𝑐1
48002
= 6666.8 = 3456𝑊
(2’)
𝑆 = |𝑉2 ||𝐼2 | = 2400 × 25 = 60000𝑉𝐴
𝑃𝑐𝑢 = 𝑅𝑒1 |𝐼2′ |2 = 28 × 12.52 = 4375𝑊
𝑛×𝑆×𝑃𝐹
𝜂 = (𝑛×𝑆×𝑃𝐹)+𝑛2 ×𝑃
𝑐𝑢 +𝑃𝑐
(c)
(2')
(2’)
1×60000×0.8
= (1×60000×0.8)+1×4375+3456 = 85.97%
(2’)
6’
The maximum efficiency occurs when,
𝑃
3456
𝑛 = √𝑃 𝑐 = √4375 = 0.89
𝑐𝑢
(2’)
The load for maximum efficiency is,
|𝑆| = 𝑛60 = 53.4𝑘𝑉𝐴
(2’)
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’)
2’
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%
(2’)