EEL303: Power Engineering I - Tutorial 4
1. Determine the voltage at the generating station and the efficiency of the following system
(Figure 1): Both transformers have ratio of 2kV/11kV. The resistance on LV side of both
Figure 1: transformers is 0.04 ohm and that on HV side is 1.3 ohm. Reactance on LV and HV side of both transformers is 0.125 ohm and 4.5 ohm respectively. [Ans: Efficiency = 96.3%,
V s
= 2168 Volts]
Figure 2: Equivalent circuit
Solution: For Transformer on LV side: Base kVA = 250; Base MVA = 0.25; Base kV = 2
( BasekV ) 2 4
Base impedance = = = 16 ohm
BaseMV A 0 .
25
0 .
04 + j 0 .
125
T ransf ormer p.u. impedance on LV side = = 0 .
0025 + j 0 .
0078
16
For Transformer on HV side: Base kVA = 250; Base MVA = 0.25; Base kV = 11
Base impedance =
( BasekV ) 2
BaseMV A
=
121
0 .
25
= 484 ohm
T ransf ormer p.u. impedance on HV side =
1 .
3 + j 4 .
5
484
= 0 .
0027 + j 0 .
0093
Total impedance of Transformer = 0.0052+j0.0171
For Transmission Line: Base kVA = 250; Base MVA = 0.25; Base kV = 11; Base impedance = 484 ohm
T ransmission line p.u. impedance =
10 + j 30
484
= 0 .
0207 + j 0 .
062
Electrical Engineering Dept - IIT Delhi

EEL303: Power Engineering I - Tutorial 4
For Load: Base kVA = 250; Base MVA = 0.25; Base kV = 2
Base Current =
250 × 1000
2000
= 125 amps p.u. MVA = 1.0; p.u. kV = 1.0; p.u. Current = 1.0
P ower Loss = I 2 R = 1 2 × (0 .
0052 + 0 .
0207 + 0 .
0052) = 0 .
0311 p.u.
% η =
Outputrealpower outputrealpower + losses
× 100 =
1 × 0 .
8
1 × 0 .
8 + 0 .
0311
× 100 = 96 .
26%
Taking V r as the reference, the sending end voltage
V s
= V r
+ I r
φ r
( R + jX ) = V r
+ ( I r
Cosφ r
− jI r
Sinφ r
)( R + jX )
V s
= 1 + (0 .
8 − j 0 .
6)(0 .
0311 + j 0 .
0962) = 1 .
0826 + j 0 .
0583 p.u
= 1 .
0842 3 .
0825
Sending end voltage = 2000 × 1 .
0842 = 2168.4 Volts
2. A load of three impedances each (6+j9) is supplied through a line having an impedance of (1+j2) ohm. The line-to-line sending end voltage is 400 volts 50 Hz. Determine the power input and output when the load is
(a) star connected and,
(b) delta connected.
[Ans: (a) 6591W, 5649W (b) 14124.9W, 9416W]
Solution: When load is star connected:
T he line to neutral voltage = √
3
= 231 volts
T he impedance per phase = (6 + j 9) + (1 + j 2) = (7 + j 11) ohm
231
T he line current =
7 + j 11
= 17 .
7 amps
P ower input = 3 × 17 .
7
2
× 7 = 6591 watts
P ower output = 3 × 17 .
7 2 × 6 = 5649 watts
When load is delta/mesh connected:
Electrical Engineering Dept - IIT Delhi

EEL303: Power Engineering I - Tutorial 4
For the same impedance (6+j9), the equivalent star impedance will be
1
3
(6 + j 9) = (2 + j 3) ohm
The impedance per phase = (2+j3) +(1+j2) = (3+j5)
T he line current =
231
3 + j 5
= 39 .
6 amps
P ower input = 3 × 39 .
6 2 × 3 = 14124 .
9 watts
P ower output = 3 × 39 .
6 2 × 2 = 9416 watts
3. Determine the efficiency and the regulation of a 3-phase, 100km, 50Hz transmission line delivering 20MW at a p.f. of 0.8 lagging and line-to-line voltage 66KV to a balanced load. The conductors are of copper, each having resistance 0.1 ohm per km, 1.5 cm outside diameter, spaced equilaterally 2 meters between centres. Neglect leakage and use nominal-T model.
[Ans: % Regulation = 18.04%, % Efficiency = 93.54%]
Figure 3:
Solution: The total resistance of the line = 100 × 0.1 = 10 ohms
T he line inductance of the line = 2 × 10
− 7
200
× 1000 ln
0 .
7788 × 0 .
75
= 11 .
67 × 10
− 2 H
Inductive reactance = 314 × 11 .
67 × 10 − 2 = 36 .
67 ohm z = 10 + j 36 .
67 ohm
T he capacitance/phase =
2Π × 8 .
854 × 10 − 12
200 ln
0 .
75
× 100 × 1000 = 9 .
959 × 10
− 7 = 0 .
9959 µF
The nominal-T circuit for the problem is given in Figure 3.
I
R
= √
20 × 1000
3 × 66 × 0 .
8
= 218 .
69 amps
Electrical Engineering Dept - IIT Delhi

EEL303: Power Engineering I - Tutorial 4
V
R
=
66 × 1000
3
= 38105 volts
Taking I
R as reference, the voltage across the capacitor will be
V c
= (38105 × 0 .
8 + 218 .
68 × 5) + j (38105 × 0 .
6 + 218 .
68 × 18 .
335) = 31578 + j 26873
T he current I c
= jωCV c
= j 314(31578 + j 26873) × 0 .
9959 × 10 − 6 = j 9 .
88 − 8 .
41
| I s
| = 210.52 amps
I s
= 218 .
69 + j 9 .
88 − 8 .
41 = 210 .
29 + j 9 .
88 amps
V s
= V c
+ I s
Z
2
= 31578 + j 26873 + (210 .
29 + j 9 .
88)(5 + j 18 .
335) = 32448 + j 30778
| V s
| = 44723 volts
The no load receiving end voltage will be
| V s
| ( − j 3196 .
2)
5 + j 17 .
55 − j 3196 .
2
% regulation =
=
44723( − j 3196 .
2)
5 − j 3178 .
65
44981 − 38105
38105
×
= 44981
100 = 18 .
volts
04%
To determine the efficiency, we evaluate transmission line losses as follows: 3 [218 .
69 2 ×
5 + 210 .
52 2 × 5] = 1.3822 MW
% Ef f iciency =
20
20 + 1 .
3822
× 100 = 93 .
54%
4. Determine the efficiency and the regulation for above problem using nominal-Π model.
[Ans: % Regulation = 18.11%, % Efficiency = 93.51%]
Figure 4:
Solution:
Electrical Engineering Dept - IIT Delhi

EEL303: Power Engineering I - Tutorial 4
The nominal-Π circuit for the above problem is shown in Figure 4.
For nominal-Π it is preferable to take receiving end voltage as the reference phasor.
The current I
R
= 218.69(0.8 - j0.6).
Current I c 1
= jωCV r
= j 314 × 0 .
4977 × 10
− 6 × 38105 = j 5 .
96 amps
I l
= I
R
+ I c 1
= 174 .
95 − j 131 .
21 + j 5 .
96 = 174 .
95 − j 125 .
25
V s
= V
R
+ I c
Z = 38105 + (174 .
94 − j 125 .
25)(10 + j 36 .
67) = 44448 + j 5162 .
8 volts
| V s
| = 44746 volts.
The no load receiving end voltage will be
44746( − j 6398)
10 + j 36 .
67 − j 6392 .
4
% regulation =
=
44746( − j 6398)
10 − j 6355 .
7
45005 − 38105
38105
×
= 45005
100 = 18 .
volts
11%
The line current I l
= 215.17 Loss = 3 × 215 .
17 2 × 10 = 1.389 MW
% ef f iciency =
20 × 100
21 .
389
= 93 .
51%
Same problem can also be solved using generalized circuit constants for a nominal-Π model.
5. A three phase 50 Hz transmission line is 400 km long. The voltage at the sending end is
220 kV. The line parameters are r=0.0125 ohm/km, x=0.4 ohm/km and y = 2.8
× 10 − 6 mho/km. Find the sending end current and receiving end voltage when there is no-load on the line. Assume the line to be a medium length line. [Ans: Sending end current =
152A, Receiving end voltage = 241.7582 KV]
Solution: The total line parameters are:
R=0.125
× 400 = 50 ohms
X = 0.4
× 400 = 160 ohm
Y = 2.8
× 10 − 6 × 400 90 = 1.12
× 10 − 3 90 mho
Z = R+jX = (50+j160)= 167.63
72 .
65 mho
At no-load:
V s
= A V
R and I s
= C V
R
A and C are calculated as follows:
Electrical Engineering Dept - IIT Delhi

EEL303: Power Engineering I - Tutorial 4
A = 1 +
Y Z
2
= 1 +
0 .
1875 162 .
65
2
= 0 .
904 + j 0 .
03
| A | = 0.9045
C = Y (1 +
Y Z
4
) = 1 .
12 × 10
− 3 90(1 +
0 .
1875 162 .
65
4
= 1 .
1 × 10
− 3 90 .
8367
Now,
| V
R
| line
=
220
| A |
=
0 .
220
9045
= 243 .
2295 KV
| I s
| = | C || V
R
| = 1 .
1 × 10
− 3 ×
243 .
2295
3
× 10 3 = 154 .
4989 amps
It is to be noted that under no-load conditions, the receiving end voltage (243.2295KV) is more than the sending end voltage. This phenomenon is known as the Ferranti effect.
Electrical Engineering Dept - IIT Delhi