EEL303: Power Engineering I - Tutorial 4

Transcription

1 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. ohm. Reactance on LV and HV side of both transformers is ohm and 4.5 ohm respectively. [Ans: Efficiency = 96.%, V s = 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 BaseMVA = = 16ohm Transformer p.u. impedance on LV side = 0.04+j = j For Transformer on HV side: Base kva = 250; Base MVA = 0.25; Base kv = 11 Base impedance = (BasekV)2 BaseMVA = = 484ohm Transformer p.u. impedance on HV side = 1.+j Total impedance of Transformer = j = j0.009 For Transmission Line: Base kva = 250; Base MVA = 0.25; Base kv = 11; Base impedance = 484 ohm Transmission line p.u. impedance = 10+j0 484 = j0.062

2 For Load: Base kva = 250; Base MVA = 0.25; Base kv = 2 Base Current = = 125amps p.u. MVA = 1.0; p.u. kv = 1.0; p.u. Current = 1.0 Power Loss = I 2 R = 1 2 ( ) = 0.011p.u. %η = Outputrealpower outputrealpower+losses 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 j0.6)(0.011+j0.0962) = j0.058p.u = Sending end voltage = = 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) W, 9416W] Solution: When load is star connected: The line to neutral voltage = 400 = 21volts The impedance per phase = (6+j9)+(1+j2) = (7+j11)ohm The line current = 21 7+j11 = 17.7amps Power input = = 6591watts Power output = = 5649watts When load is delta/mesh connected:

3 For the same impedance (6+j9), the equivalent star impedance will be 1 (6+j9) = (2+j)ohm The impedance per phase = (2+j) +(1+j2) = (+j5) The line current = 21 +j5 = 9.6amps Power input = = watts Power output = = 9416watts. Determine the efficiency and the regulation of a -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 = 9.54%] Figure : Solution: The total resistance of the line = = 10 ohms The line inductance of the line = ln = H Inductive reactance = = 6.67ohm z = 10+j6.67ohm Thecapacitance/phase = 2Π ln = = µF 0.75 The nominal-t circuit for the problem is given in Figure. I R = = amps

4 V R = = 8105volts Taking I R as reference, the voltage across the capacitor will be V c = ( )+j( ) = 1578+j2687 The current I c = jωcv c = j14(1578+j2687) = j I s = amps V s = V c +I s Z 2 I s = j = j9.88amps = 1578+j2687+( j9.88)(5+j18.5) = 2448+j0778 The no load receiving end voltage will be V s = 4472volts V s ( j196.2) 5+j17.55 j196.2 = 4472( j196.2) = 44981volts 5 j % regulation = = 18.04% 8105 Todeterminetheefficiency, weevaluatetransmissionlinelossesasfollows: [ ] = MW % Efficiency = = 9.54% Determine the efficiency and the regulation for above problem using nominal-π model. [Ans: % Regulation = 18.11%, % Efficiency = 9.51%] Figure 4: Solution:

5 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 = (0.8 - j0.6). Current I c1 = jωcv r = j = j5.96amps I l = I R +I c1 = j11.21+j5.96 = j V s = V R +I c Z = 8105+( j125.25)(10+j6.67) = j5162.8volts V s = volts. The no load receiving end voltage will be 44746( j698) 10+j6.67 j692.4 = 44746( j698) 10 j655.7 = 45005volts % regulation = = 18.11% The line current I l = Loss = = 1.89 MW % efficiency = = 9.51% 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= ohm/km, x=0.4 ohm/km and y = 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 = KV] Solution: The total line parameters are: R= = 50 ohms X = = 160 ohm Y = = mho Z = R+jX = (50+j160)= mho At no-load: V s = A V R and I s = C V R A and C are calculated as follows:

6 A = A = 1+ YZ 2 = = j0.0 2 C = Y(1+ YZ 4 ) = ( = Now, V R line = 220 A = = KV I s = C V R = = amps It is to be noted that under no-load conditions, the receiving end voltage( kv) is more than the sending end voltage. This phenomenon is known as the Ferranti effect.