QUESTION 1
[30]
1.1 With respect to synchronous machines, answer the following questions:
1.1.1 Briefly explain how the voltage regulation of a synchronous generator varies with the type
of loading.
(3)
1.1.2 Briefly explain how you would determine the armature resistance and synchronous reactance
of a given synchronous generator in the laboratory .
(5)
1.1.3 What do you understand by the term “generator capability curve”? What limits the reactive
power capability of a synchronous generator?
(5)
1.1.4 How can the real power sharing between two generators be varied without affecting the
system frequency?
(2)
1.1.5 How can the reactive power sharing between two generators be varied without affecting the
terminal voltage?
(2)
1.2 A 40 MW, 11 kV, 0.8 p.f., steel-making plant is supplied by two synchronous generators of
identical ratings operating in parallel. The internal e.m.f. (line-to-line) of Machine A is
15∠δ A0 kV and the machine delivers 28 MW, the remaining power being supplied by
Machine B. The synchronous reactance for each machine is 2.6 Ω respectively.
Evaluate the performance of this system by determining the following:
1.2.1 The total current supplied to the load.
1.2.2 The load angle δ A of machine A.
(2)
(5)
1.2.3 The current IA of machine A.
1.2.4 The current IB of machine B.
(3)
(3)
Page 1 of 10
Solution:
1.1.1
(a) Resistive load – terminal voltage of generator decreases with increase in load. 
(b) Inductive load – terminal voltage of generator decreases with increase in load. 
(c) Capacitive load – terminal voltage of generator increases with increase in load. 
1.1.2
Procedure for determining the armature resistance, Ra synchronous reactance, XS:
(a) Firstly, carry out DC test to determine the armature resistance Ra
(b) Secondly, carry out Open-circuit test and plot OCC
(c) Thirdly, carry out Short-circuit test and plot SCC
(d) Finally, determine the full-load current ISC and read the corresponding field current from
SCC. For the same field current, read the corresponding open-circuit voltage VOC. The
𝑉𝑉
synchronous impedance, 𝑍𝑍𝑆𝑆 = 𝑂𝑂𝑂𝑂 . Then, the synchronous reactance,
𝐼𝐼
𝑋𝑋𝑆𝑆 =
�𝑍𝑍𝑆𝑆2
−
𝑅𝑅𝑎𝑎2
𝑆𝑆𝑆𝑆

1.1.3 Generator capability curve – this is the P & Q diagram or chart that shows the safe
operating region of the generator. 
Reactive power capability of a synchronous generator is limited by: armature current limit, field
current limit, and end-region heating limits.  
1.1.4 In order to vary the real power sharing between two generators without affecting the
system frequency, increase the governor set points of one generator while decreasing the
governor set points of the other generator simultaneously. The generator with its governor set
points increased will assume more load while the other has its load reduced. 
1.1.5 In order to vary the reactive power sharing between two generators without affecting the
terminal voltage, increase the field current of one generator while decreasing the field current of
the other generator simultaneously. The generator with its field current increased will assume more
reactive power while the other has its reactive power reduced. 
1.2.1 Total load current, 𝐼𝐼𝐿𝐿 =
1.2.2 𝐸𝐸𝐴𝐴(𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎) =
From, 𝑃𝑃 =
√3𝑉𝑉𝑇𝑇
=
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
40×106
√3×11×103 ×0.8
∴ 𝐼𝐼𝐿𝐿 = 2624.32∠−36.87𝑂𝑂 𝐴𝐴 
15
√3
28 =
𝑃𝑃
= 8.66∠𝛿𝛿𝐴𝐴 𝑘𝑘𝑘𝑘 and 𝑉𝑉𝑡𝑡(𝑝𝑝ℎ𝑎𝑎𝑎𝑎𝑎𝑎) =
3𝐸𝐸𝐴𝐴 𝑉𝑉𝑡𝑡
𝑋𝑋𝑠𝑠
sin 𝛿𝛿𝐴𝐴 
3×8.66×6.35
2.6
= 2624.32 𝐴𝐴 
11
√3
= 6.35∠0𝑂𝑂 𝑘𝑘𝑘𝑘 
sin 𝛿𝛿𝐴𝐴 
𝑠𝑠𝑠𝑠𝑠𝑠𝛿𝛿𝐴𝐴 = 0.4412; 𝛿𝛿𝐴𝐴 = 26.2𝑂𝑂 
1.2.3 Current of machine A, 𝐼𝐼𝐴𝐴 = 1000 ×
𝐸𝐸𝐴𝐴 −𝑉𝑉𝑡𝑡
𝑗𝑗𝑋𝑋𝑠𝑠
= 1000 ×
𝑂𝑂
8.66∠26.20𝑂𝑂 −6.35∠0𝑂𝑂
= 1568.73∠ − 20.38 A 
𝑗𝑗2.6
1.2.4. Current of machine B, 𝐼𝐼𝐵𝐵 = 𝐼𝐼𝐿𝐿 − 𝐼𝐼𝐴𝐴 = 2624.32∠ − 36.87𝑂𝑂 − 1568.73∠ − 20.38𝑂𝑂
= 1205.37∠ − 58.55𝑂𝑂 𝐴𝐴 
Page 2 of 10
QUESTION 2
[35]
2.1 A 3-phase, 50 MVA, 2 poles, 22 kV, 50 Hz steam-turbine generator has a synchronous
reactance of 0.8 pu. The excitation voltage 𝐸𝐸𝑎𝑎 is adjusted to 1.2 pu and the machine is
connected to an infinite bus of 22 kV. The torque angle 𝛿𝛿 is 26°.
Evaluate the performance of this generator by determining the following:
2.1.1 The synchronous speed.
2.1.2 The active power output.
2.1.3 The armature current.
2.1.4 Draw the phasor diagram.
(2)
(3)
(3)
(2)
2.2 For a 75 kW, 440-V, 0.8-PF-leading, ∆-connected synchronous machine with an armature
resistance of 0.18 Ω and a synchronous reactance of 3.0 Ω, answer the following questions:
2.2.1 If Ea = 470∠ −13.5º V and VΦ = 440∠0º V, is this machine consuming real power from or
supplying real power to the power system? Is it consuming reactive power from or supplying
reactive power to the power system? Explain why?
(3)
2.2.2 Calculate the real power P and reactive power Q supplied or consumed by the machine under
the conditions in part 2.2.1. Is the machine operating within its ratings under these
circumstances?
(5)
2.3 A 100 hp, 440-V 0.8-PF-leading ∆-connected synchronous motor has an armature resistance
of 0.14 Ω and a synchronous reactance of 3.2 Ω. Its efficiency at full load is 95.2%.
2.3.1 What is the input power to the motor at the rated condition?
(2)
2.3.2 What is the line current of the motor at rated conditions? What is the phase current of the
motor at rated conditions?
(2)
2.3.3 What is the reactive power consumed by or supplied by the motor at rated conditions?
(2)
2.3.4 What is the internal generated voltage Ea of this motor at rated conditions?
(2)
2.4
A 100 MVAr, 11 kV, 3000 rpm, 50 Hz synchronous capacitor has a synchronous reactance
of 1.25 Ω and operates at a nominal voltage of 11 kV.
If the machine absorbs 40% of its rated power, determine:
2.4.1
2.4.2
2.4.3
2.4.4
The power factor of the machine.
The induced voltage of the machine.
The torque angle.
Draw the phasor diagram.
(1)
(3)
(1)
(2)
Page 3 of 10
Solution:
2.1.1 Synchronous speed, 𝑛𝑛𝑠𝑠 =
120𝑓𝑓
𝑝𝑝
2.1.2 Active power output, 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 =
=
𝐸𝐸𝑎𝑎 𝑉𝑉𝑡𝑡
𝑋𝑋𝑠𝑠
120×50
2
= 3000 𝑟𝑟𝑟𝑟𝑟𝑟 
1.2×1.0
sin 𝛿𝛿 =
0.8
𝑠𝑠𝑠𝑠𝑠𝑠260 = 0.658 𝑝𝑝𝑝𝑝 
Alternatively, 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 = 0.658 × 50 = 32.88 𝑀𝑀𝑀𝑀 
2.1.3 Armature current, 𝐼𝐼𝑎𝑎 =
Base current, 𝐼𝐼𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 =
𝐸𝐸𝑎𝑎 −𝑉𝑉𝑡𝑡
𝑗𝑗𝑋𝑋𝑠𝑠
𝑆𝑆
√3𝑉𝑉𝑇𝑇
=
=
1.2∠26𝑂𝑂 −1.0∠0𝑂𝑂
𝑗𝑗0.8
50×106
√3×22×103
= 0.665∠−8.5𝑂𝑂 𝑝𝑝𝑝𝑝 
= 1312.16 𝐴𝐴 
Therefore, 𝐼𝐼𝑎𝑎 = 0.665 × 1312.16 = 872.59 𝐴𝐴 
2.1.4 Phasor diagram 
2.2.1 This machine is a motor consuming real power from the power system, because EA is behind
VΦ. It is supplying reactive power because EA cosδ > VΦ. 
2.2.2 This machine is acting as a motor, and the current flow in these conditions:
E A − Vφ
430∠13.5 − 440∠0 
Ia =
=
= 36.94∠12.25 A 
0.18 + j 3.0
Ra + jX S
The real power consumed by this machine is
P = 3Vφ I a cosθ = 3× 440 × 36.94 × cos(12.25 ) = 47.65 kW 
The reactive power supplied by this machine is
Q = 3Vφ I a sin θ = 3× 440 × 36.94 × sin 12.25 = 10.35 kVAr 
kVA rating of the generator, 𝑆𝑆𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 =
𝑃𝑃
=
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
75
0.8
= 93.75 𝑘𝑘𝑘𝑘𝑘𝑘 (1/2)
Apparent power produced by generator, 𝑆𝑆𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = �𝑃𝑃2 + 𝑄𝑄 2 = 48.76 𝑘𝑘𝑘𝑘𝑘𝑘 (1/2)
Therefore, the generator is operating within its rating since 𝑆𝑆𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 > 𝑆𝑆𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 . 
2.3.1 𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜 = 0.746 × 100 = 74.6 𝑘𝑘𝑘𝑘 
𝑃𝑃𝑖𝑖𝑖𝑖 =
2.3.2 I L =
𝑃𝑃𝑜𝑜𝑜𝑜𝑜𝑜
η
Pin
74.6
= 0.952 = 78.36 𝑘𝑘𝑘𝑘
3VT PF
=
128.5
78.36 kW
= 128.5 A , I a =
= 74.2 A 
3
3 × 440 X 0.8
Page 4 of 10
2.3.3 Qrated = 3Vφ I a sin θ = 3× 440 × 74.2 × sin 36.87  = 58.77 kVAr 
2.3.4 The internal generated voltage of the machine:
𝐸𝐸𝑎𝑎 = 𝑉𝑉𝑇𝑇 − 𝐼𝐼𝑎𝑎 (𝑅𝑅𝑎𝑎 + 𝑗𝑗𝑋𝑋𝑆𝑆 )
= 440∠0𝑜𝑜 − (74.2∠−36.87𝑂𝑂 )(0.14 + 𝑗𝑗3.2) = 342.64 ∠ − 32.42𝑂𝑂 𝑉𝑉
2.4.1 Power factor, cos 𝜃𝜃 = cos 900 = 0 
2.4.2 Load current, 𝐼𝐼𝑎𝑎 =
𝑄𝑄
√3𝑉𝑉𝑇𝑇
=
0.4×100×106
√3×11×103
= 2 099.46 𝐴𝐴 
The excitation voltage of the machine is:
𝐸𝐸𝑎𝑎 = 𝑉𝑉𝑇𝑇 − 𝑗𝑗𝐼𝐼𝑎𝑎 𝑋𝑋𝑆𝑆 =
11000
√3
∠0𝑂𝑂 − (2099.46∠−90𝑂𝑂 )(𝑗𝑗1.25)
= 3 726.5∠0𝑂𝑂 𝑉𝑉
2.4.3 Torque angle, 𝛿𝛿 = 00 
2.4.4 Phasor diagram 
Page 5 of 10
QUESTION 3
[35]
3.1 A 48 kW, 400 V, 1000 rpm dc motor with an armature resistance of 0.25 Ω and a rated armature
current of 120 A is driven by an AC-DC converter as shown in Figure Q3.1. The converter is
supplied by a three-phase 360V, 50 Hz system.
Evaluate the performance of this dc drive by determining the following:
3.1.1
3.1.2
3.1.3
3.1.4
3.1.5
3.1.6
The firing angle to obtain rated motor voltage.
The speed of the dc motor at rated converter voltage when 𝛼𝛼 = 68°.
The firing angle in order for the dc motor to develop its rated torque at 200 rpm.
The dc power supplied to the dc motor at its rated torque at 200 rpm.
The dc power developed by the dc motor at its rated torque at 200 rpm.
The reactive power absorbed by the converter at its rated torque at 200 rpm.
Figure Q3.1
Page 6 of 10
(3)
(5)
(4)
(2)
(2)
(3)
3.2 A 30 hp, 208 V, 3-phase, 2850 r/min, 50 Hz, wound-rotor induction motor is to be used for a
mining application. The machine produces an open-circuit rotor line voltage of 250 V.
3.2.1 Design a concept of the drive based on 3-phase bridge rectifier composed of six diodes
connected to the three slip-rings. As part of the conditions of the design, it is to limit the lockedrotor torque to a maximum value of 35 Nm to ensure a small starting current. A single manual
rheostat is to be connected across the dc output of the rectifier as part of power dissipating
circuit. Represent your design concept by a circuit diagram of the motor and the converter.
(4)
3.2.2 Estimate the power dissipated in the rotor circuit under blocked-rotor conditions, converter
dc voltage, and resistance of dc manual rheostat.
(6)
3.2.3 How would you modify this design so that rotor power which is being dissipated in dc
manual rheostat can be recovered, i.e. fed back to the supply. Draw your new design and
explain how it works.
(6)
Page 7 of 10
Solution:
3.1.1 𝐸𝐸𝑑𝑑 = 1.35 × 𝐸𝐸 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼
∴ 400 = 1.35 × 360 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼
400
𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼 = 1.35 ×360 = 0.823
∴ 𝛼𝛼 = 31.61𝑜𝑜 
3.1.2 At 1000rpm, 𝐸𝐸𝑑𝑑1 = 𝐼𝐼𝑎𝑎 𝑅𝑅𝑎𝑎 + 𝐸𝐸01
∴ 𝐸𝐸01 = 400 − 120 × 0.25 = 370 𝑉𝑉 
With 𝛼𝛼 = 68𝑜𝑜 ; 𝐸𝐸𝑑𝑑2 = 1.35 × 360 𝑐𝑐𝑐𝑐𝑐𝑐 68 = 182.06 𝑉𝑉 
𝐸𝐸𝑑𝑑2 = 𝐼𝐼𝑎𝑎 𝑅𝑅𝑎𝑎 × 𝐸𝐸02
∴ 𝐸𝐸02 = 182.06 − 30 = 152.06 𝑉𝑉 
But 𝐸𝐸0 = 𝑘𝑘∅𝑛𝑛, so
𝐸𝐸01
𝐸𝐸02
152.06
∴ n2 = �
370
𝑛𝑛
= 𝑛𝑛1 
2
� (1000) = 410.97 rpm 
200
3.1.3 At 200 rpm; E0 = �1000� (370) = 74 V 
∴ 𝐸𝐸𝑑𝑑 = 𝐼𝐼𝑎𝑎 𝑅𝑅𝑎𝑎 + 𝐸𝐸0 = 30 + 74 = 104 𝑉𝑉 
Using 𝐸𝐸𝑑𝑑 = 1.35 × 𝐸𝐸 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼
∴ 104 = 1.35 × 360 𝑐𝑐𝑐𝑐𝑐𝑐 𝛼𝛼
∴ 𝛼𝛼 = 77.64𝑜𝑜 
3.1.4 Power supplied to dc motor, 𝑃𝑃𝑖𝑖𝑖𝑖 = 𝐸𝐸𝑑𝑑 𝐼𝐼𝑎𝑎 = 104 × 120 = 12.48 𝑘𝑘𝑘𝑘 
3.1.5 Power developed by dc motor, 𝑃𝑃𝑑𝑑𝑑𝑑𝑑𝑑 = 𝐸𝐸0 𝐼𝐼𝑎𝑎 = 74 × 120 = 8.88 𝑘𝑘𝑘𝑘 
3.1.6 Reactive power absorbed by converter, 𝑄𝑄𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑃𝑃𝑖𝑖𝑖𝑖 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡
= 12.48 tan 77.64𝑜𝑜 = 56.95 𝑘𝑘𝑘𝑘𝑘𝑘𝑘𝑘 
Page 8 of 10
3.2.1 Design concept 
3.2.2 Airgap power, 𝑃𝑃𝑎𝑎𝑎𝑎 =
𝑇𝑇𝑚𝑚 ×𝑛𝑛𝑠𝑠
9.55
=
35×3000
9.55
= 10.995 𝑘𝑘𝑘𝑘
Power dissipated in the rotor, Pcur = sPag =1×10.995 = 10.995 kW 
Rotor voltage: E = sEOC =1× 250 = 250V 
Converter dc voltage: Ed =1.35 E =1.35× 250 = 337.5 V 
Resistance of dc manual rheostat, 𝑅𝑅𝑟𝑟 =
𝐸𝐸𝑑𝑑2
𝑃𝑃𝑐𝑐𝑐𝑐𝑐𝑐
=
337.52
10995
= 10.36 Ω
3.2.3 Use the power recovery system
•
Instead of dissipating rotor power in a resistor, we use a 3-phase inverter that returns the
power to the ac source. The line-commutated inverter is connected to the same feeder
that supplies power to the stator of the motor through a transformer T. 
Page 9 of 10

Page 10 of 10