电气机器与系统｜ECS790P Electrical Machines and Systems D

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SOLUTIONS AND MARKING SCHEME
Resit Examination Period 2021 – May
ECS790P Electrical Machines and Systems Duration: 2 hours
This is an open-book exam, which should be completed in approximately 2 hours.
You MUST submit your answers within 24 hours from the exam being released.
Follow instructions on download page.
You can refer to textbooks, notes and online materials to facilitate your working, but
normal referencing and plagiarism rules apply and you must cite any sources used.
You must upload a single PDF document containing your solutions. These can be typed
or handwritten, or a mix of the two. Multiple submissions are not permitted, so be sure
that you check your submission before uploading it.
Calculators are permitted in this examination.
Instructions:
This paper contains FOUR questions. Answer the FIRST question. Then answer ANY TWO
of the REMAINING THREE questions. If you answer all the remaining questions, only the
first two answers (up to the specified number) will be marked.
YOU MUST COMPLETE THE EXAM ON YOUR OWN, WITHOUT CONSULTING OTHERS
Examiners:
Dr Kamyar Mehran and Dr Tijana Timotijevic

c Queen Mary University of London, 2021
Page 2 ECS790P (2021)
Question 1 [50 marks]
Solution: This question is compulsory for all students, and the difficulty of all parts
are designed in such a way that all students are able to answer it with basic revision of
the fundamental subjects. Q1a, b, c, d, e: low-level of difficulty, Q1f,g : mid-level of
difficulty.
Note: All the questions are identical as the UG paper (ECS649U). However, I have
changed the marking weight for Q1 to 50 marks wherein the UG paper is 40 marks. The
main reason for this change is that PG students must have the necessary knowledge
to answer the different parts of Q1 and get the pass mark. Since I designed the parts
Q1a, b, c, d, and e as MCQs; then the students can answer the questions accurately
with high speed. No description is needed for MCQ questions.
Q2, Q3, and Q4 are all have 25 marks compared to the UG paper where they have
30 marks each. PG students must have answer two questions out of Q2, Q3, and Q4.
Since PG students must have more extensive and deeper knowldge, I believe they
must be able to answer fully Q3 and Q4 designed with a low-to-mid level of difficulty.
(a) Which one of the following statements is true for microgrids.
[6 marks]
(i) A microgrid can better integrate renewable energy sources
(ii) A microgrid in islanded mode can provide ancillary services to the main grid like
degraded power quality or shortage.
(iii) The energy management of a microgrid is NOT complex
(iv) The operation of a microgrid is reliable and economical
Solution: option (i) is correct
(b) Which one of the following statements is true in using a nuclear reactor for electricity
generation.
[6 marks]
(i) A nuclear reactor produces damaging and polluting gases
(ii) A nuclear reactors cannot be used in marine vessels
(iii) Storage of radioactive waste is one of the main challenges for a nuclear reactor
(iv) The risk of severe accidents for modern nuclear power stations is zero
Solution:
option (iii) is correct
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ECS790P (2021) Page 3
(c) Which one of the following statements is true for commercial electro-chemical grid
batteries.
[6 marks]
(i) It reduces peak demand charges
(ii) They increase the rate of electricity for residential customers
(iii) They usually can hold power for more than 24 hours
(iv) The residents cannot use them for storing the extra capacity in vehicular batteries
Solution:
option (i) is correct
(d) Which one of the following statements is true when comparing a magnetic circuit with
an electric circuit.
[6 marks]
(i) – Flux always flows in a magnetic circuit.
(ii) – Conductivity in an electric circuit is similar to permeability in a magnetic circuit.
(iii) – A magnetic circuit can be an open circuit or closed circuit.
(iv) – Current density does NOT depend on the conductivity in an electric circuit.
Solution: option (ii) is a correct option.
(e) Which one of the following statements can better explain the Farday law.
[6 marks]
(i) – A time-varying flux density (current) induces an Electro-magnetic flux (EMF).
The magnitude of the EMF is proportional to the rate of change of flux density
and its direction is in opposition to the change.
(ii) – A conductor moved at a variable velocity through a magnetic field has an EMF
induced in it.
(iii) – A current-carrying conductor has a magnetic field around it.
(iv) – A current-carrying conductor in a magnetic field experiences a force. The force
is NOT proportional to the current
Solution: option (i) is a correct option.
(f) A balanced three-phase system with a line voltage of 240V is supplying a balanced
Y-connected load with 1500W and a leading power factor of PF = 0.9.
(i) – Determine the line current,
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Page 4 ECS790P (2021)
Solution: Marking: 1 mark
———————————–
phase voltage = 240/
√
3
per-phase power = 1500/3 = 500W
500 =
240
√
3
(IL) (0.9) IL = 4.00A line current
(ii) – Determine the per-phase load impedance,
Solution: Marking: 3 marks (equation 2 marks and correct calculation 1
mark)
———————————–
|ZP| =
VP
IL
=
240
√
3
×
1
4
= 34.6410
PF is 0.9 leading impedance phase angle = 25.84
ZP = 34.641 25.84
(iii) – A balanced 750W lighting load is added (in parallel) to the system. Sketch the
suitable per-phase circuit and describe how the line current will be changed.
Solution: Marking: 8 marks (2 marks for per-phase circuit, and 2 mark for
each equation)
———————————–
The 750W load is assumed to be a balanced load evenly distributed among
the three phases, resulting in an additional 250W consumed by each phase
The amplitude of the lighting current (labed I1) is determined by
250 =
240
√
3
|I1| cos 0 |I1| = 1.804A
In a similar way, the amplitude of the capacitive load current (labeled I2) is
found to be unchanged from its previous ralue, since the voitage across it has
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ECS790P (2021) Page 5
remained the same:
|I2| = 4A
if we assume that the phase with which we are working has a phase voltage
with an angle of 0; then since cos 1
(0.9) = 25.84
, and since the PF is
leading, the angle is negative
I1 = 1.804 0
A, I2 = 4 25.84 A
[12 marks]
(g) A two-legged core is shown in Figure 1. The winding on the left leg of the core (N1)
has 600 turns, and the winding on the right (N2) has 200 turns. If the coils are wound
in the directions as shown in the figure, determine the flux produced by currents
i1 = 0.5A and i2 = 1.0A.
Note: Assume μr = 1200 and constant, and the dimension of the core is as shown in
Figure 1.
[8 marks]
Figure 1: A two-legged core in Question 1-part (g)
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Page 6 ECS790P (2021)
Solution: Marking:
Calculation of FTOT: 2 marks;
Calculation of Rtot: 4 marks;
Calculation of φ: 2 marks.
—————————-
The two coils on this core are wounded so that their magneto-motive forces are
additive, so the total magneto-motive force on this core is
FTOT = N1i1 + N2i2 = (600t)(0.5A) + (200t)(1.00A) = 500A · t
The total reluctance in the core is
Rtot =
l
μrμ0A
=
2.60m
(1200) 4π × 10 7H/m
(0.15m)(0.15m)
= 76.6kA · t/Wb
and the flux in the core is:
φ =
FTOT
RTOT
=
500A · t
76.6kA · t/Wb
= 0.00653Wb
Question 2 [25 marks]
A 460V, 60Hz, four-pole, Y-connected induction motor is rated at 25 hp and has a slip of
0.02. The equivalent circuit parameters of this motor are
R1 = 0.35 R2 = 0.4 XM = 30
X1 = 0.925 X2 = 1.5
PF&W = 600W Pmisc = 200W Pcore = 600W
and the power flow diagram of the motor is shown in the Figure 2.
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ECS790P (2021) Page 7
Figure 2: The power flow diagram of the motor in Question 2
(a) Work out and sketch the equivalent circuit of the motor, and find the line current
(armature current) IL of the motor.
[10 marks]
Solution:
Difficulty level: high level
Marking breakdown:
The equivalent circuit (2 marks),
ZF calculation (5 marks): the Thevenin circuit (2 marks) and the calculation
(3 marks)
IA calculation (3 marks).
———————————-
The easiest way to find the line current or armature current is to get the equivalent
impedance ZF of the rotor circuit in parallel with jXM , and then calculate the current
as of the phase voltage divided the sum of the series impedance, as shown below.
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Page 8 ECS790P (2021)
The equivalent impedance of the rotor circuit in parallel with jXM is as follows with
R2 = R2 + R2((1 s)/s) = 20 and XF = X2 = 1.5ω
ZF =
1
1
jXM
+
1
Z2
=
1
1
j30
+
1
20 + j1.5
= 12.0644 + j0.5439 = 12.0767 0.0451
The phase voltage is
460
√
3
= 266V, so line current is:
IL = IA =
V
R1 + jX1 + RF + jXF
=
266 0
V
0.35 + j0.925 + 12.0644 + j0.5439
=
266 0
V
12.4144 + j1.4689
= 21.1309 j2.5003 = 21.2783 0.1178
(b) Determine the stator power factor, rotor power factor and rotor frequency.
[6 marks]
Solution: Difficulty level: mid to high level
Marking breakdown:
The stator PF (2 marks),
impedance angle (2 marks),
rotor PF (2 marks).
———————————-
The stator power factor
PF = cos (0.1178
) = 0.887 lagging = 0.99
To find the rotor power factor, we must find the impedance angle of the rotor
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ECS790P (2021) Page 9
θR = tan 1 X2
R2/s
= tan 1
1.5
20
= 4.2892
PFR = cos 4.2892
= 0.9972 lagging
The rotor frequency is:
fr = sfs = (0.02)(60 Hz) = 1.2 Hz
(c) Determine the stator copper losses PSCL, the air-gap power PAG, and the electrical
power converted Pconv for this motor.
[6 marks]
Solution: Difficulty level: mid level
Marking breakdown:
the equivalent circuit (2 marks),
equivalent impedance of the rotor circuit (4 marks),
phase voltage (1 mark),
line current (3 marks).
———————————-
The stator copper losses are:
PSCL = 8IA
2R1 = 3(21.2783A)
2
(0.35) = 475.4044 W
The air-gap power:
PAG = 3I2
2 R2
s
= 3IA
2RF = 16.387 kW
Note: 3I2
2 R2
s
as the power consumed in the original circrit must be the same as
the power consumed by the Therenin equivalent circuit.
Pconverted = (1 s)PAG = (1 0.02)(16.387) = 16.0593 kW
(d) Determine the induced torque τind of the motor.
[3 marks]
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Page 10 ECS790P (2021)
Solution:
Difficulty level: mid level
Marking breakdown:
The equivalent circuit (1 mark),
impedance angle (1 mark),
rotor PF (1 mark).
———————————-
nsync =
120fse
p
=
120 (60H2)
4
= 1800r/min
wsguc = (1800 r/min)
2π
1
rad
r

1 min
60s

= 188.5 rad/s
Therefore the induced torque:
τind =
PAG
wsync
=
16.387 kW
188.5
= 86.9337 N · m
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ECS790P (2021) Page 11
Question 3 [25 marks]
A 50-hp, 250-V, 1200 r/min DC shunt motor with compensating windings has an armature
resistance (including the brushes, compensating windings, and interpoles) of 0.06 . Its
field circuit has a total resistance Radj + RF of 50 , which produces a no-load speed of
1200 r/min. There are 1200 turns per pole on the shunt field winding.
(a) Sketch the circuit diagram of this motor.
[3 marks]
Solution:
Difficulty level: low level
Marking breakdown:
The equivalent circuit (3 marks),
———————————-
(b) Determine the speed of this motor when its input current is 100A.
[6 marks]
Solution:
Difficulty level: mid level
Marking breakdown:
2 marks for each equation, each equation 1 mark and the respective calcu lation 1 mark
———————————-
If IL = 100A, then the armature current in the motor is:
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Page 12 ECS790P (2021)
IA = IL IF = IL
VT
RF
= 100A
250V
50
= 95A
Therefore, EA at this load will be:
EA = VT IARA
= 250V (95A)(0.06 ) = 244.3V
The resulting speed of the motor is:
n2 =
EA2
EA1
n1 =
244.3V
250V
1200 r/min = 1173 r/min
(c) Determine the speed of this motor when its input current is 200A.
[4 marks]
Solution:
Difficulty level: mid level
Marking breakdown:
Each equation 1 mark and the respective calculation 1 mark
———————————-
If IL = 200A,then the armature current in the motor is:
IA = IL IF = IL
VT
RF
= 200A
250V
50
= 195A
Therefore, EA at this load will be:
EA = VT IARA
= 250V (195A)(0.06 ) = 244.3V
The resulting speed of the motor is:
n2 =
EA2
EA1
n1 =
238.3V
250V
1200 r/min = 1173 r/min
(d) Determine the speed of this motor when its input current is 300A.
[4 marks]
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ECS790P (2021) Page 13
Solution:
Difficulty level: mid level
Marking breakdown:
2 marks for each equation, each equation 1 mark and the respective calcu lation 1 mark
———————————-
If IL = 300A,then the armature current in the motor is:
IA = IL IF = IL
VT
RF
= 300A
250V
50
= 295A
Therefore, EA at this load will be:
EA = VT IARA
= 250V (295A)(0.06 ) = 232.3V
The resulting speed of the motor is:
n2 =
EA2
EA1
n1 =
232.3V
250V
1200 r/min = 1115 r/min
(e) Plot the torque-speed characteristic of this motor.
[8 marks]
Solution:
Difficulty level: mid level
Marking breakdown:
1 marks for each induced torque τind equation , and 2 mark for description,
and 1 mark for the torque-speed characteristic
———————————-
To plot the output characteristic of this motor, it is necessary to find the torque
corresponding to each value of speed. At no load, the induced torque is clearly
zero. The induced torque τind for any other load can be found from the fact that
power converted in a dc motor is:
Pconv = EAIA = τind ω
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Page 14 ECS790P (2021)
From this equation, the induced torque in a motor is:
τind =
EAIA
ω
Therefore, the induced torque when IL = 100A is:
Therefore, the induced torque when IL = 100A is
τind =
(244.3V)(95A)
(1173r/min)(1min/60s)(2πrad/r)
= 190N m
The induced torque when IL = 200A is
τind =
(238.3V)(95A)
(1144r/min)(1 min(60s)(2πrad/r)
= 388N m
The induced torque when IL = 300A is
τind =
(232.3V)(295A)
(1115r/min)(1 min /60s)(2πrad/r)
= 587N m
The resulting torque-speed characteristic for this motor is plotted as below:
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ECS790P (2021) Page 15
Question 4 [25 marks]
A 13.8-kV, 50-MVA, 60-Hz, four-pole Y-connected synchronous generator has a syn chronous reactance of 2.5 , an armature resistance of 0.2 , and a power factor of 0.9
(lagging). At 60 Hz, its friction and windage losses are 1 MW, and its core losses are 1.5
MW. The field circuit has a dc voltage of 120 V, and the maximum IF is 10A. The current of
the field circuit is adjustable over the range from 0 to 10A. The open-circuit characteristics
(OCC) of this generator is shown in the Figure 3.
Figure 3: open-circuit characteristics of the synchronous generator in Question 4. The horizen tal axis represents ”Field Current (A)”, and the vertical axis represents ”Open-circuit voltage
(kV)”.
(a) Determine the field current required to make the terminal voltage Vt (or line voltage
VL) equal to 13.8 kV when the generator is running at no load
[2 marks]
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Page 16 ECS790P (2021)
Solution:
Difficulty level: low ot mid level
Marking breakdown:
The readings from OCC (2 marks),
———————————-
If the no-load terminal voltage is 13.8 kV, the required field current can be read
directly from the OCC, which is 3.50 A
(b) Determine the internal generated voltage EA of this machine at the rated conditions
[8 marks]
Solution:
Difficulty level: mid to high level
Marking breakdown:
The description that IL = IA (2 marks)
The calculation of IA (2 mark)
The calculation of V (2 marks)
The calculation of EA (2 marks)
———————————-
This generator is Y-connected, so IL = IA. At rated conditions, the line and phase
current in this generator is:
IA = LL =
P
√
3VL
=
50 MVA
√
3(13800 V)
= 2092 at and aingle of 25.8
The phase voltage:
V = Vt/
√
3 = 7967V
The internal voltage generated:
EA = V + RAIA + jXSIA
EA = 7967 0
+ (0.20)(2092 25.8 A) + j(2.5)(2092 25.8 A)
EA = 11544 23.1 V
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ECS790P (2021) Page 17
(c) Determine the phase voltage of this generator at the rated conditions
[3 marks]
Solution:
Difficulty level: low to mid level
Marking breakdown:
The calculation of V (2 mark)
OCC reading (1 mark)
———————————-
The phase voltage of the machine at rated coudition
V = Vt/
√
3 = 7967V
From OCC, the required field cirreut is 10 A
(d) Determine the field current required to make the terminal voltage Vt equal to 13.8 kV
when the generator is running at the rated conditions
[3 marks]
Solution:
Difficulty level: low to mid level
Marking breakdown:
The calculation of VT, OC (2 mark)
OCC reading (1 mark)
———————————-
The equivalent open-circiut terminal voltage corresponding to an EA of 11544V is
VT,OC =
√
3(11544V) = 20kV
From the OCC, the required field corrent is 10 A
(e) Suppose that this generator is running at the rated conditions, and then the load is
removed without changing the field current. Determine the terminal voltage of the
generator.
[3 marks]
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Page 18 ECS790P (2021)
Solution:
Difficulty level: low to mid level
Marking breakdown:
The calculation of V (2 mark)
OCC reading (1 mark)
———————————-
if the load is removed without changing the field current,
V = EA = 11544V
From the OCC, the required field corrent is 10 A Terminal voltage Vt = 20kV
(f) Determine the steady-state power and torque that the generator’s prime mover must
be capable of supplying to handle the rated conditions.
[6 marks]
Solution:
Difficulty level: mid to high level
Marking breakdown:
POUT calculation (1 mark)
PIN calculation (3 mark)
τAPP calculation (2 mark)
———————————-
The input power to this generator is equal to the output power plus losses.
The rated output power: POUT = (50 MVA)(0.9) = 45 MW
PCU = 3IA
2RA = 3(2092A)
2
(0.2 ) = 2.6 MW
PF&W = 1 MW
Pcore = 1.5 MW
Pstray = ( assumed 0)
PIN = POUT + PCU = PF&W + Pcore + Pstray = 50.1 MW
Therefore the prime mover must be capable of supplying 50.1MW. The generator
is a four-pole 60Hz machine that must be timing at 1800 r/min. The required
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ECS790P (2021) Page 19
torque is:
τAPP =
PIN
wm
=
50.1 MW
(1800r/min)
1min
60s
2π
1
rad
r

= 265, 800 N · m
End of questions

电气机器与系统｜ECS790P Electrical Machines and Systems D最先出现在KJESSAY历史案例。