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14/12/2017 Electrical Machines - I - - Unit 10 - Week 9
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Unit 10 - Week 9
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Week 9
Week 10
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Week 12
Lecture 28:Compound DCGenerators
Lecture 29:InterconnectedDC Generators
Lecture 30:Characteristicsof DC ShuntMotors
Quiz : Week 9:Assignment
Week 9 :AssignmentSolution
.
1)
10 points
2)
10 points
3)
10 points
10 points4)
Week 9: Assignment
A compound DC generator has an armature resistance of 0.04Ω, shunt field resistance of 48Ω andseries 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 voltagefor long shunt connection? [Enter only the numerical value. Do not enter the unit]
Accepted Answers:(Type: Range) 219,223
A 230V, 50kW short-shunt compound generator has an armature resistance of Ra = 0.06Ω, serieswinding resistance of 0.04Ω and shunt field winding of resistance 120Ω. Calculate the induced armaturevoltage,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]
Accepted Answers:(Type: Range) 250,255
A 440V dc machine supplies 40A at 400V as a generator. The armature resistance is 0.8Ω. If themachine is now operated as a motor at same terminal voltage and current but with the flux increased by10%, the ratio of motor speed to generator speed is________ . [Enter only the numerical value]
Accepted Answers:(Type: Range) 0.7,0.82
A 10 hp, 230V DC shunt motor has an armature resistance of 0.1Ω and field resistance of180Ω. At no-load and rated voltage, the speed is 1200 rpm and the armature current is 5A. At ful-load
14/12/2017 Electrical Machines - I - - Unit 10 - Week 9
https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=66&assessment=86 2/3
10 points5)
10 points6)
7)
10 points
8)
and rated voltage, the line current is 60A, and the flux is reduced by 6% due to armature reaction effectsfrom its value at no-load. What is the full-load speed?
1197 rpm
1200 rpm
1208 rpm
1247 rpm
Accepted Answers:1247 rpm
A 250V shunt motor is driving a load at 600rpm and the armature draws acurrent of 20A. What resistance should be inserted in series to the shunt field if the speedis to be raised from 600rpm to 800rpm? The armature resistance and shunt fieldresistance are 0.5ohm and 250ohm respectively
88ohm
112ohm
63ohm
43ohm
Accepted Answers:88ohm
A 220V shunt motor with armature resistance of 0.5ohm, runs at 500rpm anddraws 30A at full load. The shunt field is excited to give constant main field. What will bethe speed at full load if a 1ohm resistor is placed in series with the armature circuit?
427rpm
453rpm
557rpm
401rpm
Accepted Answers:427rpm
A 220V shunt motor has armature resistance of 0.5ohm and takes a full load currentof 40A. By what percentage should the main field flux be reduced to raise the speed by50%, 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
Two identical shunt generators are running in parallel, each having a armature resistance of0.02ohm and field resistance of 50ohm. The combined load current is 5000A. The fields are so excitedthat the generated voltages of one machine is 610V and the other is 600V. What will be the terminalvoltage of this parallel combination? Note: Please enter only the numeric value without any units. Decimal approximations can be made
14/12/2017 Electrical Machines - I - - Unit 10 - Week 9
https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=66&assessment=86 3/3
10 points
Accepted Answers:(Type: Range) 550-560
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Assignment - 9 : Solution
Q1.Solution
Ra
= 0.04Ω
Rse
=
Rsh
= 48Ω4.4kW
220V
0.02Ω
Figure 1:
The load current IL =4400
220= 20A
The shunt-field current , If_sh =220
48= 4.58A
Armature current, Ia = IL + If_sh = 24.58A
The internal generated voltage, E = VL + Ia(Rse +Ra)
= 220 + 24.58(0.02 + 0.04) = 221.47V
Q2.Solution
R =0.06Ωa
R = 0.04Ωse
R = 120Ωsh
50kW
230V
Figure 2:
Assignment No :9 onlinecourses.nptel.ac.in Page 1 / 7
The load current IL =50, 000
230= 217.4A
ILRse = 217.4× 0.04 = 8.696V
Voltage across shunt-field winding, Vf = 230 + 8.696 = 238.696V
Shunt field current, If =238.696
120= 1.99A
Armature current, Ia = IL + If = 217.4 + 1.99 = 219.39A
Induced armature voltage, E = Vf + IaRa + eb
= 238.696 + (219.39× 0.06) + 2 = 253.86V
Q3. Solution
Ra
= 0.1Ω
Rse
=
Rsh
= 63Ω4.4kW
110V
0.01Ω
Figure 3:
Rated load = 4.4kW
No-load terminal voltage VNL = 110V = VFL
Full load current, IL_FL =4400
110= 40A
Shunt-field current at full-load Ish_FL =110
63= 1.746A
Shunt-field current at No-load Ish_NL =110
63= 1.746A
Assignment No :9 onlinecourses.nptel.ac.in Page 2 / 7
Armature current at full load, Ia = IL_FL + Ish_FL = 40 + 1.746 = 41.746A
Internal generated voltage at full-load EFL = VFL + Ia(Ra +Rse) + Eb
110 + 41.746(0.1 + 0.02) + 2 = 117V
From the magnetization data, the shunt-field current for E = 117V = 2.1A
For a long-shunt machine Ish_eff = Ish +Nse
NshIa
2.1 = 1.746 +Nse
1400× 41.746
Nse = 11.87 ≈ 12 turns
Q4.Solution
T = 180 = kaφIa
Ea = 440− Ia × 0.3 = kaφωm
ωm =1500× 2π
60= 157.079rad/s
180
Ia=
440− Ia × 0.3
157.079
Solving the above equaltion, we get; Ia = 67.39A or 1399.26A
Q5.Solution
For generator E = V + IaRa
= 400 + (40× 0.8) = 432V
For motor V = E − IaRa
E = 400− (40× 0.8) = 368V
HenceE1
E2=N1
N2× φ1φ2
432
368=N1
N2× 1
1.1N2
N1=
368
(432× 1.1)= 0.77
Assignment No :9 onlinecourses.nptel.ac.in Page 3 / 7
Q6. Solution
ENL = 230− (5× 0.1) = 229.5V
Field current, If =230
180= 1.2778A
EFL = 230− (60− 1.2778)× 0.1 = 224.13V
E ∝ φωm
E1
E2=N1
N2× φ1φ2
N2 = N1 ×E2
E1× φ1φ2
= 1200× 224.13
229.5× 1
1− 0.06= 1246.73rpm
Q7. Solution
Torque(T) ∝ flux(φ)× Armature current
T ∝ φI
Here the torque remains constant and hence
φ2I2 = φ1I1
Also the flus is proportional to the shunt field current(assuming linear magnetization curve)
∴ Ish2I2 = Ish1I1
Ish1I2 =250
250= 1
=⇒ Ish2I2 = 20 (Since armature current before change is 20A)
Eb1 = 250− 0.5× 20 = 240V
Eb2 = 250− 0.5I2
Speed (N) ∝ Eb
φ=⇒ Eb2
Eb1=Ish2N2
Ish1N1
∴250− 0.5I2
240=Ish2 × 800
1× 600
But, Ish1I2 = 20
Combining the above two equations, we get
32Ish22 − 25Ish2 − 1 = 0
Solving the above equation for Ish1 gives
Ish1 = 0.7389A
∴ Corresponding shunt resistance =250
0.7389= 338ohm
∴ Resistance to be added = 338− 250 = 88ohm
Assignment No :9 onlinecourses.nptel.ac.in Page 4 / 7
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
N1=Eb2
Eb1(Since flux is constant)
N2 =175
205× 500 = 427rpm
Q9. Solution
Torque (T) remains constant, which implies T ∝ φIa
∴φ1Ia1φ2Ia2
= 1
Also the speed (N) is raised by 150 %
N ∝ Eb
φ
=⇒ N2
N1= 1.5 =
Eb2φ1Eb1φ2
Eb2 = 300φ2φ1
Eb1 = 220− 0.5× 40 = 200V
Eb2 = 220− 0.5× Ia2
But Ia2 =φ1Ia1φ2
=40× φ1φ2
Substituting in equation for Eb2
Eb2 = 220− 0.5× 40× φ1φ2
=⇒ 300φ2φ1
= 220− 20φ1φ2
Let φ2/φ1be equal to ’x’
∴ 300x = 220− 20
x
Rearranging, we get 300x2 − 220x+ 20 = 0
Solving for x, we get x = 0.625
=⇒ φ2 is 37.5% less than φ1
Assignment No :9 onlinecourses.nptel.ac.in Page 5 / 7
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
Assignment No :9 onlinecourses.nptel.ac.in Page 6 / 7
∴ Current through Req = 60500− 5000 = 55500A
∴ Voltage across the load = 55500×Req = 55500× 1
100.04= 554.778V
Assignment No :9 onlinecourses.nptel.ac.in Page 7 / 7