PROBLEM SOLUTIONS: Chapter 2

15 PROBLEM SOLUTIONS: Chapter 2 Problem 2.1 At 60 Hz, ω = 120π. primary: (V rms ) max = N 1 ωa c (B rms ) max = 2755 V, rms secondary: (V rms ) max = N 2 ωa c (B rms ) max = 172 V, rms At 50 Hz, ω = 100π. Primary voltage is 2295 V, rms and secondary voltage is 143 V, rms. Problem 2.2 N = 2Vrms ωa c B peak = 167 turns Problem 2.3 75 N = =3 turns 8 Problem 2.4 Resistance seen at primary is R 1 =(N 1 /N 2 ) 2 R 2 =6.25Ω. Thus I 1 = V 1 R 1 =1.6 A and V 2 = ( N2 N 1 ) V 1 =40 V Problem 2.5 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the idealtransformer, equals that of the source (2 kω). Thus the transformer turns ratio N to give maximum power must be Rs N = =6.32 R load Under these conditions, the source voltage will see a total resistance of R tot = 4 kω and the current will thus equal I = V s /R tot =2mA.Thus,thepower delivered to the load will equal P load = I 2 (N 2 R load )=8 mw

16 Here is the desired MATLAB plot: Problem 2.6 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the idealtransformer, equals that of the source (2 kω). Thus the transformer turns ratio N to give maximum power must be Rs N = =6.32 R load Under these conditions, the source voltage will see a total impedance of Z tot = 2+j2 kω whose magnitude is 2 2 kω. The current will thus equal I = V s / Z tot =2 2 ma. Thus, the power delivered to the load will equal P load = I 2 (N 2 R load )=16 mw Here is the desired MATLAB plot:

17 Problem 2.7 ( ) X m V 2 = V 1 = 266 V X l1 + X m Problem 2.8 Referred to the secondary L m,2 = L m,1 = 150 mh N 2 part(b): Referred to the secondary, X m = ωl m,2 =56.7Ω, X l2 and X l1 =69.3 mω. Thus, ( ) X m (i) V 1 = N V 2 = 7960 V X m + X l2 =84.8 mω and Problem 2.9 I 1 = (ii) I sc = V 2 X sc = V 1 X l1 + X m =3.47 A; V 2 = NV 1 V 2 X l2 + X m X l1 = 1730 A ( ) X m = 2398 V X l1 + X m Let X l 2 = X l2 /N 2 and X sc = X l1 + X m (X m + X l 2 ). For I rated = 50 kva/120 V = 417 A V 1 = I rated X sc =23.1 V I 2 = 1 N ( ) X m I rated =15.7 X m + X l2 A Problem 2.10 I L = P load V L =55.5 A and thus I H = I L N =10.6 A; V H = NV L + jx H I H = 2381 9.6 V The power factor is cos (9.6 )=0.986 lagging.

18 Problem 2.11 30 kw Î load = 230 V ejφ =93.8e jφ A where φ is the power-factor angle. Referred to the high voltage side, Î H = 9.38 e jφ A. ˆV H = Z H Î H Thus, (i) for a power factor of 0.85 lagging, V H = 2413 V and (ii) for a power factor of 0.85 leading, V H = 2199 V. part (c):

19 Problem 2.12 Following methodology of Problem 2.11, (i) for a power factor of 0.85 lagging, V H = 4956 V and (ii) for a power factor of 0.85 leading, V H = 4000 V. part (c): Problem 2.13 I load = 160 kw/2340 V = 68.4 Aat =cos 1 (0.89) = 27.1 ˆV t,h = N( ˆV L + Z t I L ) which gives V H =33.7 kv. ˆV send = N( ˆV L +(Z t + Z f )I L )

20 which gives V send =33.4 kv. part (c): S send = P send + jq send = ˆV send Î send = 164 kw j64.5 kvar Thus P send = 164 kw and Q send = 64.5 kvar. Problem 2.14 Following the methodology of Example 2.6, efficiency = 98.4 percent and regulation = 1.25 percent. Problem 2.15 Z eq,l = V sc,l I sc,l = 107.8 mω and thus X eq,l = R eq,l = P sc,l I 2 sc,l =4.78 mω Z eq,l 2 Req,L 2 = 107.7 mω Z eq,l =4.8+j108 mω R eq,h = N 2 R eq,l =0.455 Ω X eq,h = N 2 X eq,l =10.24 Ω Z eq,h =10.3+j0.46 mω part (c): From the open-circuit test, the core-loss resistance and the magnetizing reactance as referred to the low-voltage side can be found: and thus R c,l = V 2 oc,l P oc,l = 311 Ω S oc,l = V oc,l I oc,l = 497 kva; Q oc,l = Soc,L 2 P oc,l 2 =45.2 kvar

21 X m,l = V 2 oc,l Q oc,l = 141 Ω The equivalent-t circuit for the transformer from the low-voltage side is thus: part (d): We will solve this problem with the load connected to the highvoltage side but referred to the low-voltage side. The rated low-voltage current is I L =50MVA/8 kv=6.25 ka. Assume the load is at rated voltage. Thus the low-voltage terminal voltage is V L = V load + Z eq,l I L =8.058 kv and thus the regulation is given by (8.053-8)/8 = 0.0072 = 0.72 percent. The total loss is approximately equal to the sum of the open-circuit loss and the short-circuit loss (393 kw). Thus the efficiency is given by η = P load = 50.0 =0.992 = 99.2 percent P in 50.39 part (e): We will again solve this problem with the load connected to the high-voltage side but referred to the low-voltage side. Now, ÎL =6.25 25.8 ka. Assume the load is at rated voltage. Thus the low-voltage terminal voltage is V L = V load + Z eq,l Î L =7.758 kv and thus the regulation is given by (7.758-8)/8 = -0.0302 = -3.02 percent. The efficiency is the same as that found in part (d), η =99.2 percent. Problem 2.16 The core length of the second transformer is is 2 times that of the first, its core area of the second transformer is twice that of the first, and its volume is 2 2 times that of the first. Since the voltage applied to the second transformer is twice that of the first, the flux densitities will be the same. Hence, the core loss will be proportional to the volume and Coreloss = 2 23420 = 9.67 kw

22 The magnetizing inductance is proportionalto the area and inversely proportionalto the core length and hence is 2 times larger. Thus the no-load magnetizing current will be 2 times larger in the second transformer or I no load = 24.93 = 6.97 A Problem 2.17 Rated current at the high-voltage side is 20 kva/2.4 kv = 8.33 A. Thus the total loss will be P loss = 122 + 257 = 379 W. The load power is equal to 0.8 20 = 16 kw. Thus the efficiency is η = 16 =0.977 = 97.7 percent 16.379 First calculate the series impedance (Z eq,h = R eq,h + jx eq,h )of the transformer from the short-circuit test data. R eq,h = P sc,h I 2 sc,h =3.69 Ω S sc,h = V sc,h I sc,h =61.3 8.33 = 511 kv A Thus Q sc,h = Ssc,H 2 P sc,h 2 = 442 VAR and hence X eq,h = Q sc,h I 2 sc,h =6.35 Ω The regulation will be greatest when the primary and secondary voltages of the transformer are in phase as shown in the following phasor diagram Thus the voltage drop across the transformer will be equal to V = I load Z eq,h = 61.2 V and the regulation will equal 61.2 V/2.4 kv = 0.026 = 2.6 percent. Problem 2.18 For a power factor of 0.87 leading, the efficiency is 98.4 percent and the regulation will equal -3.48 percent. Problem 2.19 The voltage rating is 2400 V:2640 V. The rated current of the high voltage terminal is equal to that of the 240-V winding, I rated =30 10 3 /240 = 125 A. Hence the kva rating of the transformer is 2640 125 = 330 kva.

23 Problem 2.20 The rated current of the high voltage terminal is equal to that of the 120-V winding, I rated =10 4 /120 = 83.3 A. Hence the kva rating of the transformer is 600 83.3 =50kVA. part (c): The full load loss is equal to that of the transformer in the conventionalconnection, P loss =(1 0.979) 10 kw = 210 W. Hence as an autotransformer operating with a load at 0.85 power factor (P load =0.85 50 kw = 42.5 kw), the efficiency will be 42.5 kw η = =0.995 = 99.5 percent 42.71 kw Problem 2.21 The voltage rating is 78 kv:86 kv. The rated current of the high voltage terminal is equal to that of the 8-kV winding, I rated =50 10 6 /8000 = 6.25 ka. Hence the kva rating of the transformer is 86 kv 6.25 ka = 537.5 MVA. The loss at rated voltage and current is equal to 393 kw and hence the efficiency will be 537.5 MW η = =0.9993 = 99.93 percent 538.1 MW Problem 2.22 No numerical result required for this problem. Problem 2.23 7.97 kv:2.3 kv; 191 A:651 A; 1500 kva 13.8 kv:1.33 kv; 109 A:1130 A; 1500 kva part (c): 7.97 kv:1.33 kv; 191 A:1130 A; 1500 kva part (d): 13.8 kv:2.3 kv; 109 A:651 A; 1500 kva Problem 2.24 (i) 23.9 kv:115 kv, 300 MVA (ii) Z eq =0.0045 + j0.19 Ω (iii) Z eq =0.104 + j4.30 Ω

24 (i) 23.9 kv:66.4 kv, 300 MVA (ii) Z eq =0.0045 + j0.19 Ω (iii) Z eq =0.0347 + j1.47 Ω Problem 2.25 Following the methodology of Example 2.8, V load = 236 V, line-to-line. Problem 2.26 The totalseries impedance is Z tot = Z f + Z t = j11.7 +0.11 + j2.2 Ω = 0.11 + j13.9 Ω. The transformer turns ratio is N =9.375. The load current, as referred to the transformer high-voltage side will be ( ) 325 MVA I load = N 2 e jφ =7.81e jφ ka 324kV where φ = cos 1 0.93 = 21.6. The line-to-neutral load voltage is V load = 24 3kV. At the transformer high-voltage terminal V = 3 NV load + I load Z t = 231.7 kv, line-to-line At the sending end V = 3 NV load + I load Z tot = 233.3 kv, line-to-line Problem 2.27 Problem 2.28 First calculate the series impedance (Z eq,h = R eq,h + jx eq,h )ofthetransformer from the short-circuit test data. Z eq,h =0.48 = j1.18 Ω

25 The totalimedance between the load and the sending end of the feeder is Z tot = Z f + Z eq,h =0.544+j2.058Ω. The transformer turns ration is N = 2400:120 3 = 11.6. The referred load voltage V load and current I load will be in phase and can be assumed to be the phase reference. Thus we can write the phasor equation for the sending-end voltage as: ˆV s = V load + I load Z tot We know that V s = 2400/sqrt3 = 1386 V and that I load = 100 kva/( 32.4kV). Taking the magnitude of both sides of the above equation gives a quadradic equation in V load whichcanbesolvedforv load Vload 2 +2R tot I load V load + Z tot 2 Iload 2 Vs 2 V load = R tot I load + V 2 s (X toti load ) 2 =1.338 kv Referred to the low-voltage side, this corresponds to a load voltage of 1.338 kv/n = 116 V, line-to-neutral or 201 V, line-to-line. Feeder current = 2400 = 651 A 3Ztot HV winding current = 651 3 = 376 A LV winding current = 651N = 7.52 ka Problem 2.29 The transformer turns ratio is N = 7970/120 = 66.4. The secondary voltage will thus be ˆV 2 = V 1 N ( ) jx m = 119.74 0.101 R 1 + jx 1 + jx m Defining R L = N 2 R L = N 2 1kΩ=4.41 MΩ and Z eq = jx m (R 2 + R L + jx 2) = 134.3+j758.1 kω the primary current will equal Î 1 = 7970 R 1 + jx 1 + Z eq =10.3 79.87 ma

26 The secondary current will be equal to ( ) jx m Î 2 = NÎ1 R 2 + R L + j(x = 119.7 m + X 2 ) 0.054 ma and thus ˆV 2 = R L Î 2 = 119.7 0.054 V part (c): Following the methodology of part (b) ˆV 2 = 119.6 0.139 V Problem 2.30 This problem can be solved iteratively using MATLAB. The minimum reactance is 291 Ω. Problem 2.31

27 Problem 2.32 The transformer turns ratio N = 200/5 = 40. For I 1 = 200 A I 2 = I ( ) 1 jx m N R 2 + j(x m + X 2 ) =4.987 0.024 Defining R L = N 2 250µΩ =0.4Ω I 2 = I ( ) 1 jx m N R 2 + R L + j(x m + X 2 ) =4.987 0.210 Problem 2.33 Problem 2.34 Z base,l = V base,l 2 =1.80 Ω P base Z base,h = V base,h 2 = 245 Ω P base

28 Thus R 1 =0.0095Z base,l =17.1 mω; X 1 =0.063Z base,l = 113 mω X m = 148Z base,l = 266 Ω R 2 =0.0095Z base,h =2.33 Ω; X 2 =0.063Z base,h =15.4 Ω Problem 2.35 (i) Z base,l = (7.97 103 ) 2 75 10 3 =0.940 Ω; X L =0.12Z base,l =0.113 Ω (ii) Z base,h = (7970)2 75 10 3 = 847 Ω; X H =0.12Z base,h = 102 Ω (i) 797 V:13.8 kv, 225 kva (ii) X pu =0.12 (iii) X H = 102 Ω (iv) X L =0.339 Ω part (c): (i) 460 V:13.8 kv, 225 kva (ii) X pu =0.12 (iii) X H = 102 Ω (iv) X L =0.113 Ω Problem 2.36 In each case, I pu =1/0.12 = 8.33 pu. (i) I base,l = P base /( 3 V base,l ) = 225 kva/( 3 797 V) = 163 A I L = I pu I base,l = 1359 A (ii) I base,h = P base /( 3 V base,h ) = 225 kva/( 313.8 kv)=9.4 A I H = I pu I base,h =78.4 A In each case, I pu =1/0.12 = 8.33 pu. (i) I base,l = P base /( 3 V base,l ) = 225 kva/( 3 460 V) = 282 A I L = I pu I base,l = 2353 A (ii) I base,h = P base /( 3 V base,h ) = 225 kva/( 313.8 kv)=9.4 A I H = I pu I base,h =78.4 A

29 Problem 2.37 On the transformer base ( ) ( ) Pbase,t 800 MVA X gen = 1.57 = 1.57 = 1.27 pu 850 MVA P base,g On the transformer base, the power supplied to the system is P out = 700/850 = 0.824 pu and the totalpower is S out = P out /pf =0.825/0.95 = 0.868 pu. Thus, the per unit current is Î =0.868 φ, whereφ = cos 1 0.95 = 18.2. (i) The generator terminalvoltage is thus ˆV t =1.0+ÎZ t =1.03 3.94 pu = 26.8 3.94 kv and the generator internalvoltage is ˆV gen =1.0+Î(Z t + Z gen )=2.07 44.3 pu = 53.7 44.3 kv (ii) The totaloutput of the generator is given by S gen = ˆV t Î =0.8262 + 0.3361. Thus, the generator output power is P gen =0.8262 850 = 702.2 MW. The correspoinding power factor is P gen / S gen = 0.926 lagging.