14/12/2017
Electrical Machines - I - - Unit 10 - Week 9
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Unit 10 - Week 9
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Week 9: Assignment
.
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Week1
Week 2
Week 3
1) A compound DC generator has an armature resistance of 0.04Ω, shunt field resistance of 48Ω and
series field resistance of 0.02Ω. The field excitation is adjusted to get a rated terminal voltage of 220V.
Find out the internal generated voltage, E(in Volts), when the motor is supplying 4.4kW at rated voltage
for long shunt connection?
[Enter only the numerical value. Do not enter the unit]
Week 4
Week 5
Week 6
Accepted Answers:
(Type: Range) 219,223
Week 7
Week 8
Week 9
10 points
2) A 230V, 50kW short-shunt compound generator has an armature resistance of Ra = 0.06Ω, series
winding resistance of 0.04Ω and shunt field winding of resistance 120Ω. Calculate the induced armature
voltage,E (in Volts) at rated load and terminal voltage. The total brush contact drop is 2V?
[Enter only the numerical value. Do not enter any units]
Lecture 28:
Compound DC
Generators
Lecture 29:
Interconnected
DC Generators
Lecture 30:
Characteristics
of DC Shunt
Motors
Quiz : Week 9:
Assignment
Week 9 :
Assignment
Solution
Accepted Answers:
(Type: Range) 250,255
10 points
3) A 440V dc machine supplies 40A at 400V as a generator. The armature resistance is 0.8Ω. If the
machine is now operated as a motor at same terminal voltage and current but with the flux increased by
10%, the ratio of motor speed to generator speed is________ .
[Enter only the numerical value]
Week 10
Week 11
Accepted Answers:
(Type: Range) 0.7,0.82
Week 12
10 points
4) A 10 hp, 230V DC shunt motor has an armature resistance of 0.1Ω and field resistance of 10 points
180Ω. At no-load and rated voltage, the speed is 1200 rpm and the armature current is 5A. At ful-load
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14/12/2017
Electrical Machines - I - - Unit 10 - Week 9
and rated voltage, the line current is 60A, and the flux is reduced by 6% due to armature reaction effects
from its value at no-load. What is the full-load speed?
1197 rpm
1200 rpm
1208 rpm
1247 rpm
Accepted Answers:
1247 rpm
5) A 250V shunt motor is driving a load at 600rpm and the armature draws a
10 points
current of 20A. What resistance should be inserted in series to the shunt field if the speed
is to be raised from 600rpm to 800rpm? The armature resistance and shunt field
resistance are 0.5ohm and 250ohm respectively
88ohm
112ohm
63ohm
43ohm
Accepted Answers:
88ohm
6) A 220V shunt motor with armature resistance of 0.5ohm, runs at 500rpm and 10 points
draws 30A at full load. The shunt field is excited to give constant main field. What will be
the speed at full load if a 1ohm resistor is placed in series with the armature circuit?
427rpm
453rpm
557rpm
401rpm
Accepted Answers:
427rpm
7) A 220V shunt motor has armature resistance of 0.5ohm and takes a full load current
of 40A. By what percentage should the main field flux be reduced to raise the speed by
50%, if the developed torque remains constant?
Note: Please enter only the numeric value without any units. Decimal approximations can be made
Accepted Answers:
(Type: Range) 35-40
10 points
8) Two identical shunt generators are running in parallel, each having a armature resistance of
0.02ohm and field resistance of 50ohm. The combined load current is 5000A. The fields are so excited
that the generated voltages of one machine is 610V and the other is 600V. What will be the terminal
voltage of this parallel combination?
Note: Please enter only the numeric value without any units. Decimal approximations can be made
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14/12/2017
Electrical Machines - I - - Unit 10 - Week 9
Accepted Answers:
(Type: Range) 550-560
10 points
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Assignment - 9 : Solution
Q1.Solution
Rse = 0.02Ω
Ra = 0.04Ω
Rsh = 48Ω
4.4kW
220V
Figure 1:
4400
= 20A
220
220
= 4.58A
The shunt-field current , If _sh =
48
Armature current, Ia = IL + If _sh = 24.58A
The load current
IL =
The internal generated voltage, E
= VL + Ia (Rse + Ra )
= 220 + 24.58(0.02 + 0.04) = 221.47V
Q2.Solution
Rse = 0.04Ω
Ra=0.06Ω
Rsh = 120Ω
50kW
230V
Figure 2:
Assignment No :9
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50, 000
= 217.4A
230
= 217.4 × 0.04 = 8.696V
The load current IL =
IL Rse
Voltage across shunt-field winding, Vf = 230 + 8.696 = 238.696V
238.696
Shunt field current, If =
= 1.99A
120
Armature current, Ia = IL + If = 217.4 + 1.99 = 219.39A
Induced armature voltage, E = Vf + Ia Ra + eb
= 238.696 + (219.39 × 0.06) + 2 = 253.86V
Q3. Solution
Rse = 0.01Ω
Ra = 0.1Ω
Rsh = 63Ω
4.4kW
110V
Figure 3:
Rated load
= 4.4kW
No-load terminal voltage VN L = 110V = VF L
4400
Full load current, IL_F L =
= 40A
110
110
Shunt-field current at full-load Ish_F L =
= 1.746A
63
110
Shunt-field current at No-load Ish_N L =
= 1.746A
63
Assignment No :9
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Armature current at full load, Ia = IL_F L + Ish_F L = 40 + 1.746 = 41.746A
Internal generated voltage at full-load EF L = VF L + Ia (Ra + Rse ) + Eb
110 + 41.746(0.1 + 0.02) + 2 = 117V
From the magnetization data, the shunt-field current for E = 117V
For a long-shunt machine Ish_ef f = Ish +
= 2.1A
Nse
Ia
Nsh
Nse
× 41.746
1400
= 11.87 ≈ 12 turns
2.1 = 1.746 +
Nse
Q4.Solution
T = 180 = ka φIa
Ea = 440 − Ia × 0.3 = ka φωm
1500 × 2π
= 157.079rad/s
ωm =
60
180
440 − Ia × 0.3
=
Ia
157.079
Solving the above equaltion, we get; Ia = 67.39A
or
1399.26A
Q5.Solution
For generator
E = V + Ia Ra
= 400 + (40 × 0.8) = 432V
For motor
V = E − Ia Ra
E = 400 − (40 × 0.8) = 368V
E1
N1 φ1
=
×
E2
N2 φ2
432
N1
1
=
×
368
N2 1.1
N2
368
=
= 0.77
N1
(432 × 1.1)
Hence
Assignment No :9
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Q6. Solution
EN L = 230 − (5 × 0.1) = 229.5V
230
Field current, If =
= 1.2778A
180
EF L = 230 − (60 − 1.2778) × 0.1 = 224.13V
E ∝ φωm
E1
N1 φ1
=
×
E2
N2 φ2
E2 φ1
N2 = N1 ×
×
E1 φ2
224.13
1
= 1200 ×
×
= 1246.73rpm
229.5
1 − 0.06
Q7. Solution
Torque(T) ∝ flux(φ) × Armature current
T ∝ φI
Here the torque remains constant and hence
φ2 I2 = φ1 I1
Also the flus is proportional to the shunt field current(assuming linear magnetization curve)
∴ Ish2 I2 = Ish1 I1
250
=1
Ish1 I2 =
250
=⇒ Ish2 I2 = 20 (Since armature current before change is 20A)
Eb1 = 250 − 0.5 × 20 = 240V
Eb2 = 250 − 0.5I2
Eb
=⇒
φ
250 − 0.5I2
∴
=
240
Speed (N) ∝
Eb2
Ish2 N2
=
Eb1
Ish1 N1
Ish2 × 800
1 × 600
But, Ish1 I2 = 20
Combining the above two equations, we get
32Ish2 2 − 25Ish2 − 1 = 0
Solving the above equation for Ish1 gives
Ish1 = 0.7389A
250
= 338ohm
0.7389
∴ Resistance to be added = 338 − 250 = 88ohm
∴ Corresponding shunt resistance =
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Q8. Solution
T ∝ φIa ∝ Ia (Since flux is constant)
Since torques is constant, armature current remains constant
Eb1 = 220 − 0.5 × 30 = 205V
Eb2 = 220 − 1.5 × 30 = 175V
Speed ∝
∴
Eb
φ
N2
Eb2
=
(Since flux is constant)
N1
Eb1
175
N2 =
× 500 = 427rpm
205
Q9. Solution
Torque (T) remains constant, which implies T ∝ φIa
φ1 Ia1
=1
φ2 Ia2
∴
Also the speed (N) is raised by 150 %
N∝
=⇒
Eb
φ
Eb2 φ1
N2
= 1.5 =
N1
Eb1 φ2
φ2
Eb2 = 300
φ1
Eb1 = 220 − 0.5 × 40 = 200V
Eb2 = 220 − 0.5 × Ia2
But Ia2 =
40 × φ1
φ1 Ia1
=
φ2
φ2
Substituting in equation for Eb2
Eb2 = 220 − 0.5 ×
40 × φ1
φ2
φ1
=⇒ 300
= 220 − 20
φ2
φ1
φ2
Let φ2 /φ1 be equal to ’x’
20
∴ 300x = 220 −
x
2
Rearranging, we get 300x − 220x + 20 = 0
Solving for x, we get x = 0.625
=⇒ φ2 is 37.5% less than φ1
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Q10. Solution
Eb1 = 600V, Ra1 = 0.02ohm
Eb2 = 610V, Ra2 = 0.02ohm
Converting voltage sources in series with resistance into current sources in paralell to resistance
∴ I1 = 600/0.02 = 30000A and I2 = 610/0.02 = 30500A
The equivalent resistance (Req ) of Rsh1 , Rsh2 , Ra1 and Ra2 = 50||50||0.02||0.02 = 1/100.4ohm
The two parallel current sources can be combined to single current source (I) of value = 30000 + 30500 = 60500A
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∴ Current through Req = 60500 − 5000 = 55500A
1
∴ Voltage across the load = 55500 × Req = 55500 ×
= 554.778V
100.04
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