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- 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%, Vs = 2168 Volts]
Figure 2: Equivalent circuit
Solution: For Transformer on LV side: Base kVA = 250; Base MVA = 0.25; Base kV = 2 Base impedance =
(BasekV )^2 BaseMV A
= 16ohm
T ransf ormer p.u. impedance on LV side =
0 .04 + j 0. 125 16
= 0.0025 + j 0. 0078
For Transformer on HV side: Base kVA = 250; Base MVA = 0.25; Base kV = 11
Base impedance =
(BasekV )^2 BaseMV A
= 484ohm
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. 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
For Load: Base kVA = 250; Base MVA = 0.25; Base kV = 2
Base Current =
250 × 1000
= 125amps
p.u. MVA = 1.0; p.u. kV = 1.0; p.u. Current = 1.
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 Vr as the reference, the sending end voltage
Vs = Vr + Ir 6 φr(R^ +^ jX) =^ Vr + (IrCosφr −^ jIrSinφr)(R^ +^ jX)
Vs = 1 + (0. 8 − j 0 .6)(0.0311 + j 0 .0962) = 1.0826 + j 0. 0583 p.u = 1. 08426 3. 0825 Sending end voltage = 2000 × 1 .0842 = 2168.4 Volts
- 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 =
= 231volts
T he impedance per phase = (6 + j9) + (1 + j2) = (7 + j11)ohm
T he line current =
7 + j 11
= 17. 7 amps
P ower input = 3 × 17. 72 × 7 = 6591watts P ower output = 3 × 17. 72 × 6 = 5649watts
When load is delta/mesh connected:
VR =
66 × 1000
= 38105volts
Taking IR as reference, the voltage across the capacitor will be
Vc = (38105 × 0 .8 + 218. 68 × 5) + j(38105 × 0 .6 + 218. 68 × 18 .335) = 31578 + j 26873
T he current Ic = jωCVc = j314(31578 + j26873) × 0. 9959 × 10 −^6 = j 9. 88 − 8. 41 Is = 218.69 + j 9. 88 − 8 .41 = 210.29 + j 9. 88 amps |Is|= 210.52 amps
Vs = Vc + Is
Z
= 31578 + j26873 + (210.29 + j 9 .88)(5 + j 18 .335) = 32448 + j 30778
|Vs| = 44723volts
The no load receiving end voltage will be
|Vs|(−j 3196 .2) 5 + j 17. 55 − j 3196. 2
44723(−j 3196 .2) 5 − j 3178. 65
= 44981volts
% regulation =
× 100 = 18.04%
To determine the efficiency, we evaluate transmission line losses as follows: 3 [218. 692 × 5 + 210. 522 × 5] = 1.3822 MW
% Ef f iciency =
× 100 = 93.54%
- Determine the efficiency and the regulation for above problem using nominal-Π model. [Ans: % Regulation = 18.11%, % Efficiency = 93.51%]
Figure 4:
Solution:
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 IR = 218.69(0.8 - j0.6).
Current Ic 1 = jωCVr = j 314 × 0. 4977 × 10 −^6 × 38105 = j 5. 96 amps Il = IR + Ic 1 = 174. 95 − j 131 .21 + j 5 .96 = 174. 95 − j 125. 25 Vs = VR + IcZ = 38105 + (174. 94 − j 125 .25)(10 + j 36 .67) = 44448 + j 5162. 8 volts
|Vs| = 44746 volts. The no load receiving end voltage will be
44746(−j6398) 10 + j 36. 67 − j 6392. 4
44746(−j6398) 10 − j 6355. 7
= 45005volts
% regulation =
× 100 = 18.11%
The line current Il = 215.17 Loss = 3 × 215. 172 × 10 = 1.389 MW
% ef f iciency =
20 × 100
Same problem can also be solved using generalized circuit constants for a nominal-Π model.
- 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 6 90 = 1.12 × 10 −^3 6 90 mho Z = R+jX = (50+j160)= 167.63 6 72 .65 mho At no-load: Vs = A VR and Is = C VR A and C are calculated as follows:
