15PROBLEM SOLUTIONS: Chapter 2 Problem 2.1 At 60 Hz, ω = 120π. primary: (Vrms )max = N1 ωAc (Brms )max = 2755 V, rms secondary: (Vrms )max = N2 ωAc (Brms )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 √ 2Vrms = 167 turns N= ωAc Bpeak Problem 2.3 N= 75 =3 8 turns Problem 2.4 Resistance seen at primary is R1 = (N1 /N2 )2 R2 = 6.25Ω. Thus I1 = and V2 = N2 N1 V1 = 40 V V1 = 1.6 A R1 Problem 2.5 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the ideal transformer, equals that of the source (2 kΩ). Thus the transformer turns ratio N to give maximum power must be N= Rs = 6.32 Rload Under these conditions, the source voltage will see a total resistance of Rtot = 4 kΩ and the current will thus equal I = Vs /Rtot = 2 mA. Thus, the power delivered to the load will equal Pload = I 2 (N 2 Rload ) = 8 mW the source √ voltage will see a total impedance of Ztot = 2 + j2 kΩ whose magnitude is 2 2 kΩ. The current will thus equal I = √ Vs /|Ztot | = 2 2 mA.32 Rload Under these conditions. the power delivered to the load will equal Pload = I 2 (N 2 Rload ) = 16 mW Here is the desired MATLAB plot: . equals that of the source (2 kΩ).16 Here is the desired MATLAB plot: Problem 2.6 The maximum power will be supplied to the load resistor when its impedance. Thus the transformer turns ratio N to give maximum power must be N= Rs = 6. Thus. as reflected to the primary of the ideal transformer. 7 V2 = V1 Xm Xl1 + Xm = 266 V Problem 2.9 part (a): I1 = V1 = 3.3 mΩ.2 = Lm. N VH = N VL + jXH IH = 2381 9.7 Ω.10 IL = and thus IH = IL = 10.47 A.5 A VL 1 N Xm Xm + Xl2 Irated = 15.986 lagging. For Irated = 50 kVA/120 V = 417 A V1 = Irated Xsc = 23.6 A. (i) and (ii) Isc = Problem 2.2 = 56.7 A The power factor is cos (9. Xl1 + Xm V2 = N V1 Xm Xl1 + Xm = 2398 V V2 V2 = = 1730 A Xsc Xl2 + Xm ||Xl1 V1 = N Xm Xm + Xl2 V2 = 7960 V part (b): Let Xl2 = Xl2 /N 2 and Xsc = Xl1 + Xm ||(Xm + Xl2 ).6◦ V Pload = 55.6◦ ) = 0. Xm = ωLm. .8 mΩ and Xl1 = 69.17 Problem 2.1 = 150 mH N2 part(b): Referred to the secondary. Xl2 = 84.8 part (a): Referred to the secondary Lm.1 V I2 = Problem 2. Thus. 38 e A. IH = jφ 9. part (c): . ˆ ˆ VH = ZH IH Thus. VH = 2199 V. Referred to the high voltage side. (i) for a power factor of 0.8 ejφ Iload = 230 V A ˆ where φ is the power-factor angle.85 leading.11 part (a): part (b): 30 kW jφ ˆ e = 93.85 lagging.18 Problem 2. VH = 2413 V and (ii) for a power factor of 0. 89) = 27. part (b): = cos−1 (0. (i) for a power factor of 0.85 leading.12 part (a): part (b): Following methodology of Problem 2.19 Problem 2.H = N (VL + Zt IL ) which gives VH = 33.1◦ ˆ ˆ Vsend = N (VL + (Zt + Zf )IL ) . VH = 4000 V.11.85 lagging.4 A at ˆ ˆ Vt. part (c): Problem 2.13 part (a): Iload = 160 kW/2340 V = 68. VH = 4956 V and (ii) for a power factor of 0.7 kV. L − Poc.L | = Vsc.15 part (a): |Zeq.8 + j108 mΩ part (b): Req.20 which gives Vsend = 33.6.L = 10.L = 4.455 Ω Xeq. efficiency = 98.7 mΩ Zeq.L Soc.8 mΩ Isc.46 mΩ part (c): From the open-circuit test.H = N 2 Req.L = 497 kVA.L = Xeq.14 Following the methodology of Example 2. part (c): ˆ∗ ˆ Ssend = Psend + jQsend = Vsend Isend = 164 kW − j64.L Psc.L = 2 2 Soc.4 percent and regulation = 1.H = N 2 Xeq.L = 311 Ω Poc.24 Ω Zeq.L Req. and thus Qoc.5 kVAR.2 kVAR .L = and thus 2 |Zeq.25 percent. the core-loss resistance and the magnetizing reactance as referred to the low-voltage side can be found: Rc.L = Voc.L = 107.L Ioc.L = 0.L = 2 Voc.L |2 − Req.5 kVAR Thus Psend = 164 kW and Qsend = −64.L = 45. Problem 2.H = 10.78 mΩ 2 Isc.3 + j0. Problem 2.4 kV.L = 107.L = 4. Hence.0302 = -3. Thus the low-voltage terminal voltage is ˆ VL = |Vload + Zeq. Assume the load is at rated voltage. The rated low-voltage current is IL = 50 MVA/8 kV = 6. Since the voltage applied to the second transformer is twice that of the first. Thus the efficiency is given by η= 50.L = 2 Voc.053-8)/8 = 0.0 Pload = 0.67 kW .72 percent.758 kV and thus the regulation is given by (7.21 Xm.2 percent = Pin 50. The efficiency is the same as that found in part (d).L = 141 Ω Qoc. and its volume is √ 2 2 times that of the first.02 percent.25 25. η = 99. the flux densitities will be the same.992 = 99. Assume the load is at rated voltage.2 percent. Now.8◦ kA. the core loss will be proportional to the volume and √ Coreloss = 2 23420 = 9.0072 = 0.25 kA. IL = 6.058 kV and thus the regulation is given by (8.L 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. Thus the low-voltage terminal voltage is VL = |Vload + Zeq.758-8)/8 = -0.39 part (e): We will again solve this problem with the load connected to the ˆ high-voltage side but referred to the low-voltage side. The total loss is approximately equal to the sum of the open-circuit loss and the short-circuit loss (393 kW). its core area of the second transformer is twice that of the first.16 √ The core length of the second transformer is is 2 times that of the first. Problem 2.L IL | = 8.L IL | = 7. 97 A Problem 2. Req.026 = 2.H = Req. Problem 2.3 × 8.977 = 97. Thus the no-load magnetizing current will be 2 times larger in the second transformer or √ Ino−load = 2 4. .4 kV = 8.87 leading.H = 6.H = 3.33 = 511 kV A Thus Qsc. part (b): The rated current of the high voltage terminal is equal to that of the 240-V winding.H = 2 2 Ssc.H = 61.35 Ω 2 Isc. Irated = 30 × 103/240 = 125 A.69 Ω 2 Isc. The load power is equal to 0. Thus the total loss will be Ploss = 122 + 257 = 379 W.33 A.H = Psc. the efficiency is 98.7 percent 16.18 For a power factor of 0.H = Qsc.H = Vsc.2 V and the regulation will equal 61.H 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 = |Iload ||Zeq.H − Psc.6 percent.22 The magnetizing inductance is proportional to the area and inversely pro√ portional to the core length√ and hence is 2 times larger.H + jXeq.H | = 61.17 part (a): Rated current at the high-voltage side is 20 kVA/2.379 part (b): First calculate the series impedance (Zeq.H = 442 VAR and hence Xeq.8 × 20 = 16 kW.H ) of the transformer from the short-circuit test data.H Ssc.93 = 6. Problem 2.4 kV = 0. Thus the efficiency is η= 16 = 0.19 part (a): The voltage rating is 2400 V:2640 V.48 percent. Hence the kVA rating of the transformer is 2640 × 125 = 330 kVA.2 V/2.4 percent and the regulation will equal -3.H Isc. 33 kV. 1500 kVA part (b): 13. 1500 kVA part (c): 7.97 kV:1.8 kV:1.23 part (a): 7.20 part (a): part (b): The rated current of the high voltage terminal is equal to that of the 120-V winding.5 percent 42. Problem 2.97 kV:2. 191 A:1130 A.3 = 50 kVA. Hence as an autotransformer operating with a load at 0.30 Ω .24 part (a): (i) 23.93 percent 538.0045 + j0.104 + j4. 1500 kVA Problem 2.23 Problem 2. Hence the kVA rating of the transformer is 600 × 83.995 = 99.3 A.85 × 50 kW = 42. 1500 kVA part (d): 13.5 MW = 0.5 kW = 0.22 No numerical result required for this problem. Irated = 50 × 106 /8000 = 6. Hence the kVA rating of the transformer is 86 kV × 6.25 kA. part (c): The full load loss is equal to that of the transformer in the conventional connection. Ploss = (1 − 0.3 kV. the efficiency will be η= 42.9993 = 99. part (b): The loss at rated voltage and current is equal to 393 kW and hence the efficiency will be η= 537.19 Ω (iii) Zeq = 0.3 kV.8 kV:2. 191 A:651 A. Irated = 104 /120 = 83.85 power factor (Pload = 0.9 kV:115 kV. 300 MVA (ii) Zeq = 0.71 kW Problem 2.979) 10 kW = 210 W. 109 A:651 A. 109 A:1130 A.5 kW).25 kA = 537. The rated current of the high voltage terminal is equal to that of the 8-kV winding.1 MW Problem 2.33 kV.5 MVA.21 part (a): The voltage rating is 78 kV:86 kV. line-to-line.47 Ω Problem 2. The line-to-neutral load voltage is Vload = √ 24 3 kV. line-to-line Problem 2.9 kV:66.7 kV.3 kV.375. The load current.48 = j1. 300 MVA (ii) Zeq = 0.9 Ω.24 part (b): (i) 23.26 The total series impedance is Ztot = Zf + Zt = j11. as referred to the transformer high-voltage side will be Iload = N 2 325 MVA √ 3 24 kV ejφ = 7.8.H + jXeq.27 √ 3 |N Vload + Iload Ztot | = 233.81ejφ kA where φ = − cos−1 0.28 First calculate the series impedance (Zeq.11 + j2. The transformer turns ratio is N = 9.25 Following the methodology of Example 2. line-to-line part (b): At the sending end V = Problem 2.18 Ω . Problem 2.11 + j13.7 + 0.6◦.19 Ω (iii) Zeq = 0.H ) of the transformer from the short-circuit test data.H = Req.0045 + j0.H = 0. Zeq. Vload = 236 V.93 = −21. part (a): At the transformer high-voltage terminal V = √ 3 |N Vload + Iload Zt | = 231.4 kV.0347 + j1.2 Ω = 0. part (a): The referred load voltage Vload and current Iload will be in phase and can be assumed to be the phase reference.41 MΩ and Zeq = jXm ||(R2 + RL + jX2 ) = 134. line-to-neutral or 201 V. part (b): 2400 = 651 A Feeder current = √ 3Ztot 651 HV winding current = √ = 376 A 3 LV winding current = 651N = 7.87◦ mA R1 + jX1 + Zeq .H = 0.058 Ω.6. line-to-line.338 kV Referred to the low-voltage side.3 − 79.4 kV).544 + j2.3 + j758.29 part (a): The transformer turns ratio is N = 7970/120 = 66.4. this corresponds to a load voltage of 1.74 0.25 = The total imedance between the load and the sending end of the feeder is Ztot√ Zf + Zeq. Thus we can write the phasor equation for the sending-end voltage as: ˆ Vs = Vload + Iload Ztot √ We know that Vs = 2400/sqrt3 = 1386 V and that Iload = 100 kVA/( 32.338 kV/N = 116 V.1 kΩ the primary current will equal ˆ I1 = 7970 = 10. The secondary voltage will thus be V1 ˆ V2 = N jXm R1 + jX1 + jXm = 119.52 kA Problem 2.101◦ part (b): Defining RL = N 2 RL = N 2 1 kΩ = 4. Taking the magnitude of both sides of the above equation gives a quadradic equation in Vload 2 2 Vload + 2Rtot Iload Vload + |Ztot |2 Iload − Vs2 which can be solved for Vload Vload = −Rtot Iload + Vs2 − (Xtot Iload )2 = 1. The transformer turns ration is N = 2400:120 3 = 11. The minimum reactance is 291 Ω. Problem 2.139◦ V V jXm R2 + RL + j(Xm + X2 ) = 119.7 0.054◦ part (c): Following the methodology of part (b) ˆ V2 = 119.31 part (a): part (b): .7 0.6 0.26 The secondary current will be equal to ˆ ˆ I2 = N I1 and thus ˆ ˆ V2 = RL I2 = 119.30 This problem can be solved iteratively using MATLAB.054◦ mA Problem 2. L = 2 Vbase.024◦ part (b): Defining RL = N 2 250µΩ = 0.210◦ part (b): Problem 2.27 Problem 2.34 Zbase.H = Pbase = 245 Ω .33 part (a): I1 N jXm R2 + RL + j(Xm + X2 ) = 4.L Pbase 2 Vbase.987 0.987 0.80 Ω Zbase.4 Ω I2 = Problem 2.H = 1. For I1 = 200 A I2 = I1 N jXm R2 + j(Xm + X2 ) = 4.32 part (a): The transformer turns ratio N = 200/5 = 40. H = Pbase /( 3 Vbase. 75 × 103 XL = 0.33 pu.H = Pbase /( 3 Vbase.0095Zbase.L = Pbase /( 3 Vbase.4 A part (b): In each case.33 Ω.113 Ω X2 = 0. 225 kVA (ii) Xpu = 0.L ) = 225 kVA/( 3 460 V) = 282 A IL = Ipu Ibase.L = Pbase /( 3 Vbase.L = 17.8 kV.8 kV) = 9.8 kV.339 Ω part (c): (i) 460 V:13.H = 2.H = 15.L = 266 Ω R2 = 0. √ √ (i) Ibase.4 A IH = Ipu Ibase.L = 2353 A √ √ (ii) Ibase. Problem 2.4 A IH = Ipu Ibase. Ipu = 1/0.H = part (b): XH = 0.063Zbase. 75 × 103 (7970)2 = 847 Ω.12 (iii) XH = 102 Ω (iv) XL = 0.940 Ω.4 Ω (ii) Zbase.L ) = 225 kVA/( 3 797 V) = 163 A IL = Ipu Ibase.33 pu.0095Zbase.L = 0. Ipu = 1/0.4 A .L = 1359 A √ √ (ii) Ibase.35 part (a): (i) Zbase.L = (7.H ) = 225 kVA/( 3 13.063Zbase. X1 = 0.97 × 103 )2 = 0.28 Thus R1 = 0.H = 78.12 = 8.H ) = 225 kVA/( 3 13.113 Ω Problem 2.1 mΩ.12Zbase.12 = 8.L = 113 mΩ Xm = 148Zbase.12 (iii) XH = 102 Ω (iv) XL = 0.H = 78.36 part (a): In each case. √ √ (i) Ibase.12Zbase. 225 kVA (ii) Xpu = 0.H = 102 Ω (i) 797 V:13.8 kV) = 9. 7 44.57 = 1.t Pbase.3◦ pu = 53.95 = ◦ −18. The correspoinding power factor is Pgen /|Sgen | = 0.29 Problem 2. Thus.8262 + 0. where φ = − cos−1 0. the per unit current is I = 0.27 pu part (b): On the transformer base.8262 × 850 = 702.3361.0 + I(Zt + Zgen ) = 2. .03 3.2 MW.926 lagging.3◦ kV ˆ ˆ (ii) The total output of the generator is given by Sgen = Vt I ∗ = 0. Thus.2 . the generator output power is Pgen = 0.95 = ˆ 0.868 φ.825/0.57 = 800 MVA 850 MVA 1.94◦ pu = 26. (i) The generator terminal voltage is thus ˆ ˆ Vt = 1.94◦ kV and the generator internal voltage is ˆ ˆ Vgen = 1.07 44.g 1.0 + IZt = 1.868 pu.824 pu and the total power is Sout = Pout /pf = 0.8 3. the power supplied to the system is Pout = 700/850 = 0.37 part (a): On the transformer base Xgen = Pbase.