% the leading currents. I(1,:) = amps.* ( j*0.6); % Lagging I(2,:) = amps.* ( 1.0 ); % Unity I(3,:) = amps.* ( j*0.

Transcription

1 % the leading currents. I(1,:) = amps.* ( 0.8 j*0.6); % Lagging I(,:) = amps.* ( 1.0 ); % Unity I(3,:) = amps.* ( 0.8 j*0.6); % Leading % Calculate VS referred to the primary side % for each current and power factor. avs = V (Req.*I j.*xeq.*i); % Refer the secondary voltages back to the % secondary side using the turns ratio. VS = avs * (00/15); % lot the secondary voltage (in kv!) versus load plot(amps,abs(vs(1,:)/1000),'b','linewidth',.0); hold on; plot(amps,abs(vs(,:)/1000),'k','linewidth',.0); plot(amps,abs(vs(3,:)/1000),'r.','linewidth',.0); title ('\bfsecondary Voltage Versus Load'); xlabel ('\bfload (A)'); ylabel ('\bfsecondary Voltage (kv)'); legend('0.8 F lagging','1.0 F','0.8 F leading'); grid on; hold off; The resulting plot of secondary voltage versus load is shown below: 1. A threephase transformer bank is to handle 600 kva and have a 34.5/13.8kV voltage ratio. Find the rating of each individual transformer in the bank (high voltage, low voltage, turns ratio, and apparent power) if the transformer bank is connected to (a) YY, (b) Y, (c) Y, (d), (e) open, (f) open Y open. SOLUTION For the first four connections, the apparent power rating of each transformer is 1/3 of the total apparent power rating of the threephase transformer. For the open and openy open connections, the apparent power rating is a bit more complicated. The 600 kva must be 86.6% of the total apparent

2 power rating of the two transformers, implying that the apparent power rating of each transformer must be 31 kva. The ratings for each transformer in the bank for each connection are given below: Connection rimary Voltage Secondary Voltage Apparent ower Turns Ratio YY 19.9 kv 7.97 kv 00 kva.50:1 Y 19.9 kv 13.8 kv 00 kva 1.44:1 Y 34.5 kv 7.97 kv 00 kva 4.33: kv 13.8 kv 00 kva.50:1 open 34.5 kv 13.8 kv 346 kva.50:1 openy open 19.9 kv 13.8 kv 346 kva 1.44:1 Note: The openy open answer assumes that the Y is on the highvoltage side; if the Y is on the lowvoltage side, the turns ratio would be 4.33:1, and the apparent power rating would be unchanged.. A 13,800/480 V threephase Y connected transformer bank consists of three identical 100kVA 7967/480V transformers. It is supplied with power directly from a large constantvoltage bus. In the shortcircuit test, the recorded values on the highvoltage side for one of these transformers are V = 560 V I = 1.6 A = 3300 W (a) If this bank delivers a rated load at 0.85 F lagging and rated voltage, what is the linetoline voltage on the primary of the transformer bank? (b) What is the voltage regulation under these conditions? (c) Assume that the primary voltage of this transformer bank is a constant 13.8 kv, and plot the secondary voltage as a function of load current for currents from noload to fullload. Repeat this process for power factors of 0.85 lagging, 1.0, and 0.85 leading. (d) lot the voltage regulation of this transformer as a function of load current for currents from noload to fullload. Repeat this process for power factors of 0.85 lagging, 1.0, and 0.85 leading. SOLUTION From the shortcircuit information, it is possible to determine the perphase impedance of the transformer bank referred to the highvoltage side. The primary side of this transformer is Yconnected, so the shortcircuit phase voltage is V 560 V 33.3 V 3 3 V φ, = = = the shortcircuit phase current is Iφ, = I = 1.6 A and the power per phase is φ, = = 1100 W 3 Thus the perphase impedance is 33.3 V ZEQ = REQ jxeq = = 5.66 Ω 1.6 A W θ = cos = cos = 74.3 VI ( 33.3 V)( 1.6 A) ZEQ = REQ jxeq = Ω= 6.94 j4.7 Ω

3 R EQ = 6.94 Ω XEQ = j4.7 Ω (a) If this Y transformer bank delivers rated kva (300 kva) at 0.85 power factor lagging while the secondary voltage is at rated value, then each transformer delivers 100 kva at a voltage of 480 V and 0.85 F lagging. Referred to the primary side of one of the transformers, the load on each transformer is equivalent to 100 kva at 7967 V and 0.85 F lagging. The equivalent current flowing in the secondary of one transformer referred to the primary side is 100 kva I φ, S = = 1.55 A 7967 V I φ, S = A The voltage on the primary side of a single transformer is thus V φ, = Vφ, S Iφ, S Z EQ, ( )( j ) V φ, = V A Ω = V The linetoline voltage on the primary of the transformer is V LL, V φ, ( ) = 3 = V = 14. kv (b) The voltage regulation of the transformer is VR = 100% = 3.01%

4 3. Three 5kVA 4,000/77V distribution transformers are connected in Y. The opencircuit test was performed on the lowvoltage side of this transformer bank, and the following data were recorded: V = 480 V I line, = 4.10 A 3 φ,= 945 W line, The shortcircuit test was performed on the highvoltage side of this transformer bank, and the following data were recorded: V = 1600 V I line, =.00 A 3 φ,= 1150 W line, (a) Find the perunit equivalent circuit of this transformer bank. (b) Find the voltage regulation of this transformer bank at the rated load and 0.90 F lagging. (c) What is the transformer bank s efficiency under these conditions? SOLUTION (a) The equivalent of this threephase transformer bank can be found just like the equivalent circuit of a singlephase transformer if we work on a perphase bases. The opencircuit test data on the lowvoltage side can be used to find the excitation branch impedances referred to the secondary side of the transformer bank. Since the lowvoltage side of the transformer is Yconnected, the perphase opencircuit quantities are: V φ = 77 V I φ, = 4.10 A φ, = 315 W, The excitation admittance is given by Iφ, 4.10 A YEX = = = S V 77 V φ, The admittance angle is Therefore, 315 W θ = = = 1 φ, cos cos Vφ, Iφ, ( 77 V)( 4.10 A) YEX = GC jb = = j0.014 RC = 1/ GC = 44 Ω X = 1/ B = 70.3 Ω The base impedance for a single transformer referred to the lowvoltage side is Z base, S ( Vφ, S ) ( 77 V) = = = Ω S 5 kva φ so the excitation branch elements can be expressed in perunit as 44 Ω 70.3 Ω = = 79.5 pu X.9 pu Ω = = Ω 4

5 The shortcircuit test data can be used to find the series impedances referred to the highvoltage side, since the shortcircuit test data was taken on the highvoltage side. Note that the highvoltage is connected, so Vφ, = V = 1600 V, I I / A φ, = =, and / W φ, = =. Z EQ V φ, = = = Ω I φ, 1600 V A 1 φ, 383 W θ = cos = cos 1 = 78.0 V I φ, φ, ( 1600 V)( A) Z = R jx = = 88 j1355 Ω EQ EQ EQ The base impedance referred to the highvoltage side is Z base, ( Vφ, ) ( 4,000 V) = = = 3,040 Ω S 5 kva φ The resulting perunit impedances are 88 Ω = = pu 3,040 Ω X EQ 1355 Ω = = pu 3,040 Ω The perunit, perphase equivalent circuit of the transformer bank is shown below: I I S jx EQ j V jx V S 79.5 j.9 (b) If this transformer is operating at rated load and 0.90 F lagging, then current flow will be at an cos 1 0.9, or 5.8. The perunit voltage at the primary side of the transformer will be angle of ( ) ( )( ) V = V S I Z S EQ = j = The voltage regulation of this transformer bank is VR = 100% = 3.8% 1.0 (c) The output power of this transformer bank is ( )( )( ) OUT = V I cos θ = = 0.9 pu The copper losses are S S I R ( ) ( ) CU = S EQ = = pu 5

6 The core losses are core ( 1.038) V = = = pu R 79.5 C Therefore, the total input power to the transformer bank is IN = OUT CU core = = 0.96 and the efficiency of the transformer bank is OUT 0.9 η = 100% = 100% = 97.% 0.96 IN 4. A 0kVA 0,000/480V 60Hz distribution transformer is tested with the following results: Opencircuit test Shortcircuit test (measured from secondary side) (measured from primary side) V = 480 V V = 1130 V I = 1.60 A I = 1.00 A V = 305 W = 60 W (a) Find the perunit equivalent circuit for this transformer at 60 Hz. (b) What would the rating of this transformer be if it were operated on a 50Hz power system? (c) Sketch the equivalent circuit of this transformer referred to the primary side if it is operating at 50 Hz. SOLUTION (a) The base impedance of this transformer referred to the primary side is ( V ) ( 0,000 V) Zbase, = = = 0 kω S 0 kva The base impedance of this transformer referred to the secondary side is ( V ) ( 480 V) S Z base, S = = = 11.5 Ω S 0 kva The open circuit test yields the values for the excitation branch (referred to the secondary side): Iφ, 1.60 A YEX = = = S Vφ, 480 V W θ = cos = cos = 66.6 V I 480 V 1.60 A ( )( ) Y = G jb = = j R X EX C C = 1/ G = 757 Ω C = 1/ B = 37 Ω The excitation branch elements can be expressed in perunit as 757 Ω 37 Ω = = 65.7 pu X 8.4 pu 11.5 Ω = = 11.5 Ω The short circuit test yields the values for the series impedances (referred to the primary side): 6

7 Z EQ V 1130 V = = = 1130 Ω I 1.00 A W θ = cos = cos = 76.7 V I ( 1130 V)( 1.00 A) Z = R jx = = 60 j1100 Ω EQ EQ EQ The resulting perunit impedances are 60 Ω = = pu 0,000 Ω The perunit equivalent circuit is X EQ 1100 Ω = = pu 0,000 Ω I I S jx EQ j0.055 V jx V S 65.7 j8.4 (b) If this transformer were operated at 50 Hz, both the voltage and apparent power would have to be derated by a factor of 50/60, so its ratings would be kva, 16,667/400 V, and 50 Hz. (c) The transformer parameters referred to the primary side at 60 Hz are: RC = Zbase RC,pu = ( 0 kω )( 65.7) = 1.31 Ω X = Zbase X,pu = ( 0 kω )( 8.4) = 568 kω REQ = ZbaseRE Q,pu = ( 0 kω )( 0.013) = 60 Ω XEQ = Zbase X E Q,pu = ( 0 kω )( 0.055) = 1100 Ω At 50 Hz, the resistance will be unaffected but the reactances are reduced in direct proportion to the decrease in frequency. At 50 Hz, the reactances are 50 Hz X = ( 568 kω ) = 473 kω 60 Hz 50 Hz X EQ = ( 1100 Ω ) = 917 Ω 60 Hz 7

8 The resulting equivalent circuit referred to the primary at 50 Hz is shown below: I I S jx EQ 60 Ω j917 Ω V jx V S 1.31 Ω j473 kω 8