14/12/2017 Electrical Machines - I - - Unit 10 - Week 9

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Unit 10 - Week 9

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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