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PROBLEM SOLUTIONS: Chapter 2
Problem 2. At 60 Hz, ω = 120π.
primary: (Vrms)max = N 1 ωAc(Brms)max = 2755 V, rms
secondary: (Vrms)max = N 2 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.
N =
2 Vrms ωAcBpeak
= 167 turns
Problem 2.
N =
= 3 turns
Problem 2. 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 =
N 2
N 1
V 1 = 40 V
Problem 2. The maximum power will be supplied to the load resistor when its im- pedance, 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
N =
Rs 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
Here is the desired MATLAB plot:
Problem 2. The maximum power will be supplied to the load resistor when its im- pedance, 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
N =
Rs Rload
Under these conditions, 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. Thus, the power delivered to the load will equal
Pload = I^2 (N 2 Rload) = 16 mW
Here is the desired MATLAB plot:
Problem 2. part (a):
part (b):
Iˆload = 30 kW 230 V
ejφ^ = 93. 8 ejφ^ A
where φ is the power-factor angle. Referred to the high voltage side, IˆH =
- 38 ejφA.
VˆH = ZH IˆH
Thus, (i) for a power factor of 0.85 lagging, VH = 2413 V and (ii) for a power factor of 0.85 leading, VH = 2199 V. part (c):
Problem 2. part (a):
part (b): Following methodology of Problem 2.11, (i) for a power factor of 0.85 lagging, VH = 4956 V and (ii) for a power factor of 0.85 leading, VH = 4000 V. part (c):
Problem 2. part (a): Iload = 160 kW/2340 V = 68.4 A at = cos−^1 (0.89) = 27. 1 ◦
V^ ˆt,H = N ( VˆL + ZtIL)
which gives VH = 33.7 kV. part (b):
Vˆsend = N ( VˆL + (Zt + Zf )IL)
Xm,L =^
V (^) oc^2 ,L Qoc,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 high- voltage side but referred to the low-voltage side. The rated low-voltage current is IL = 50 MVA/8 kV = 6.25 kA. Assume the load is at rated voltage. Thus the low-voltage terminal voltage is
VL = |Vload + Zeq,LIL| = 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 =
Pload Pin
= 0.992 = 99.2 percent
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, IˆL = 6. 25 25. 8 ◦^ kA. Assume the load is at rated voltage. Thus the low-voltage terminal voltage is
VL = |Vload + Zeq,L IˆL| = 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. 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
The magnetizing inductance is proportionalto the area and inversely pro- portionalto 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
Ino−load =
2 4.93 = 6. 97 A
Problem 2. part (a): Rated current at the high-voltage side is 20 kVA/2.4 kV = 8.33 A. Thus the total loss will be Ploss = 122 + 257 = 379 W. The load power is equal to 0. 8 × 20 = 16 kW. Thus the efficiency is
η =
= 0.977 = 97.7 percent
part (b): First calculate the series impedance (Zeq,H = Req,H + jXeq,H) of the transformer from the short-circuit test data.
Req,H =
Psc,H Isc^2 ,H
Ssc,H = Vsc,HIsc,H = 61. 3 × 8 .33 = 511 kV A
Thus Qsc,H =
S^2 sc,H − P (^) sc^2 ,H = 442 VAR and hence
Xeq,H =
Qsc,H Isc^2 ,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| = 61 .2 V and the regulation will equal 61.2 V/2.4 kV = 0.026 = 2.6 percent.
Problem 2. For a power factor of 0.87 leading, the efficiency is 98.4 percent and the regulation will equal -3.48 percent.
Problem 2. part (a): The voltage rating is 2400 V:2640 V. part (b): The rated current of the high voltage terminal is equal to that of the 240-V winding, Irated = 30 × 103 /240 = 125 A. Hence the kVA rating of the transformer is 2640 × 125 = 330 kVA.
part (b):
(i) 23.9 kV:66.4 kV, 300 MVA (ii) Zeq = 0.0045 + j 0 .19 Ω (iii) Zeq = 0.0347 + j 1 .47 Ω
Problem 2. Following the methodology of Example 2.8, Vload = 236 V, line-to-line.
Problem 2. The totalseries impedance is Ztot = Zf + Zt = j 11 .7 + 0.11 + j 2 .2 Ω = 0 .11 + j 13 .9 Ω. The transformer turns ratio is N = 9.375. The load current, as referred to the transformer high-voltage side will be
Iload = N 2
325 MVA
3 24 kV
ejφ^ = 7. 81 ejφ^ kA
where φ = − cos−^1 0 .93 = − 21. 6 ◦. The line-to-neutral load voltage is Vload = 24
3 kV. part (a): At the transformer high-voltage terminal
V =
3 |N Vload + IloadZt| = 231. 7 kV, line-to-line
part (b): At the sending end
V =
3 |N Vload + IloadZtot| = 233. 3 kV, line-to-line
Problem 2.
Problem 2. First calculate the series impedance (Zeq,H = Req,H + jXeq,H) of the trans- former from the short-circuit test data.
Zeq,H = 0.48 = j 1 .18 Ω
The totalimedance between the load and the sending end of the feeder is Ztot = Zf + Zeq,H = 0.544 + j 2 .058 Ω. The transformer turns ration is N = 2400:
part (a): The referred load voltage Vload and current Iload 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:
Vˆs = Vload + IloadZtot
We know that Vs = 2400/sqrt3 = 1386 V and that Iload = 100 kVA/(
32 .4 kV). Taking the magnitude of both sides of the above equation gives a quadradic equation in Vload
V (^) load^2 + 2RtotIloadVload + |Ztot|^2 Iload^2 − V (^) s^2
which can be sol ved forVload
Vload = −RtotIload +
V (^) s^2 − (XtotIload)^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. part (b):
Feeder current =
3 Ztot
∣ = 651^ A
HV winding current =
= 376 A
LV winding current = 651N = 7. 52 kA
Problem 2. part (a): The transformer turns ratio is N = 7970/120 = 66.4. The sec- ondary voltage will thus be
Vˆ 2 = V^1
N
jXm R 1 + jX 1 + jXm
part (b): Defining R′ L = N 2 RL = N 2 1 kΩ = 4.41 MΩ and
Zeq = jXm||(R′ 2 + R′ L + jX′ 2 ) = 134.3 + j 758. 1 kΩ
the primary current will equal
Iˆ 1 = 7970
R 1 + jX 1 + Zeq
= 10. 3 − 79. 87 ◦^ mA
Problem 2. part (a): The transformer turns ratio N = 200/5 = 40. For I 1 = 200 A
I 2 =
I 1
N
jXm R′ 2 + j(Xm + X′ 2 )
part (b): Defining R′ L = N 2250 μΩ = 0.4 Ω
I 2 =
I 1
N
jXm R′ 2 + R′ L + j(Xm + X 2 ′)
Problem 2. part (a):
part (b):
Problem 2.
Zbase,L =
V (^) base^2 ,L Pbase
Zbase,H =
V (^) base^2 ,H Pbase
Thus
R 1 = 0. 0095 Zbase,L = 17.1 mΩ; X 1 = 0. 063 Zbase,L = 113 mΩ
Xm = 148Zbase,L = 266 Ω
R 2 = 0. 0095 Zbase,H = 2.33 Ω; X 2 = 0. 063 Zbase,H = 15.4 Ω
Problem 2. part (a):
(i) Zbase,L =
(7. 97 × 103 )^2
75 × 103
= 0.940 Ω; XL = 0. 12 Zbase,L = 0.113 Ω
(ii) Zbase,H =
(7970)^2
75 × 103
= 847 Ω; XH = 0. 12 Zbase,H = 102 Ω
part (b):
(i) 797 V:13.8 kV, 225 kVA (ii) Xpu = 0. 12 (iii) XH = 102 Ω (iv) XL = 0.339 Ω
part (c):
(i) 460 V:13.8 kV, 225 kVA (ii) Xpu = 0. 12 (iii) XH = 102 Ω (iv) XL = 0.113 Ω
Problem 2. part (a): In each case, Ipu = 1/ 0 .12 = 8.33 pu.
(i) Ibase,L = Pbase/(
3 Vbase,L) = 225 kVA/(
3 797 V) = 163 A
IL = IpuIbase,L = 1359 A (ii) Ibase,H = Pbase/(
3 Vbase,H) = 225 kVA/(
3 13.8 kV) = 9.4 A IH = IpuIbase,H = 78.4 A
part (b): In each case, Ipu = 1/ 0 .12 = 8.33 pu.
(i) Ibase,L = Pbase/(
3 Vbase,L) = 225 kVA/(
3 460 V) = 282 A
IL = IpuIbase,L = 2353 A (ii) Ibase,H = Pbase/(
3 Vbase,H) = 225 kVA/(
3 13.8 kV) = 9.4 A IH = IpuIbase,H = 78.4 A
