Assignment - 12 : Solution

Q1.Solution

Armature current, Ia = 70A

From the table, at N = 900 rpm and Ia = 70A, Ea = 188V

In regenerative breaking Ea = (Ra +Rse + δRB)× Ia

At maximum speed δ = 0.9. Hence, E ′ a = (0.15 + 0.9× 3)× 70 = 199.5V

E ′ a

Ea = Nmax 900

Nmax = 199.5

188 × 900 = 955 rpm

Q2.Solution

T1 = KfI 2 a1

T2 = KfI 2 a2

Ia2 = Ia1 × √ T2 T1

= 100× √

2 = 141.4A

E1 = KeIa1N1

E2 = KeIa2N2

E2 = E1 × N2 N1 × Ia2 Ia1

= (220− (100× 0.1))× 1000 1500

× 141.4 100

= 210× 1000 1500

× 141.4 100

= 197.96V

E2 = Ia2 × (RB + 0.1)

197.96 = 141.4× (RB + 0.1)

RB = 1.3 Ω

Q3.Solution

When the diverter resistance is not connected, the armature current and field current of the machine are same.

Without field diverter If1 = Ia1 = 45A

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I = 45Aa1

n1 = 1500 rpm

R = 0.06Ωf

R = 0.08Ωa 220V

n2

R = 0.1Ωd

If2

Ia2R = 0.06Ωf

Figure 1:

With field diverter If2 = Rd

Rf +Rd × Ia2

= 0.1

0.06 + 0.1 =

5

8 Ia2

Since, load torque remains the same, T1 = T2

KtI 2 a1 = KtIf2Ia2

452 = If2Ia2

452 = 5

8 I2a2

I2a2 = 3240

Ia2 = 56.92A

If2 = 35.57A

Back emfs, Eb1 = V − Ia1(Ra +Rf )

KgIa1n1 = 220− 45(0.08 + 0.06)

Kg × 45× 1500 = 213.7V

With diverter, Eb2 = V − Ia2 ( Rf ×Rd Rf +Rd

+Ra

) KgIf2n2 = 220− 56.92

( 0.06× 0.1 0.1 + 0.06

+ 0.08

) Kg × 35.57× n2 = 213.3119V Kg × 45× 1500 Kg × 37.57× n2

= 213.7

213.3119

n2 = 1793 rpm

Assignment No : 12 onlinecourses.nptel.ac.in Page 2 / 7

Q4.Solution

Total input Pin = (Vtm + Vfm + Vgm)Iam

= (600 + 42 + 42)× 57 = 38, 988W

Generator output Pout = 520× 38 = 19, 760W

Total losses in the two machines Ploss = 38, 988− 19, 760 = 19, 228W

Total Cu-loss = (572 × 0.2) + (382 × 0.2) + 57(42 + 42) = 5726.6W

No-load rotational loss of both the machines = 19, 228− 5726.6 = 13501.4W

No-load rotational loss of each machine Prot = 13501.4

2 = 6750.7W

Motor input, Pin_m = (600 + 42)× 57 = 36, 594W

Motor losses, Ploss_m = (572 × 0.2) + (57× 42) + 6750.7 = 9794.5W

Efficiency of Motor ηm = (

1− 9794 36, 594

) × 100 = 73.23%

Q5.Solution

Total losses in the generator, Ploss_g = (382 × 0.2) + (57× 42) + 6750.7 = 9433.5W

Generator input, Pin_g = Pout + Ploss_g

= 19, 760 + 9433.5 = 29193.5W

Efficiency of Generator ηg = (

1− 9433.5 29193.5

) × 100 = 67.68%

Q6.Solution

DC series motor cannot be run at no load condition. Hence Swinburne’s test cannot be be conducted on a DC series motor.

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Q7.Solution

Back emf Eb1 when running at 1600rpm at 40A = 240− 30× 0.2 = 234V

φ ∝ Ia φ2 φ1

= Ia2 Ia1

Ia2 = 10A =⇒ φ2 φ1

= 1

3

Speed (N) ∝ Eb φ

=⇒ N2 N1

= Eb2φ1 Eb1φ2

3200

1600 = Eb2 Eb1 × 3

=⇒ Eb2 = Eb1 × 2

3

∴ Eb2 = 234× 2/3

= 156V

∴ Let the external resistance to be added be Rext

Rtot = Rext +Ra +Rse

Where, Ra +Rse = 0.2ohm

Eb2 = V − Ia2Rtot

156 = 240− 10×Rtot

=⇒ Rtot = 240− 156

10 = 8.4ohm

∴ Rext = Rtot −Ra −Rse

Rext = 8.4− 0.2 = 8.2ohm

Q8.Solution

In the given question, data regardingArmature resistance and field winding resistance we not given. Hence the question will be cancelled. But the solution considering Armature resistance=1.1 ohm and resistance of each field winding=0.4 ohm is solved below

Assignment No : 12 onlinecourses.nptel.ac.in Page 4 / 7

Ra1 = 1.1 + 2× 0.4 = 1.9ohm

Ra2 = 1.1 + 2× 0.4/2 = 1.3ohm

Eb1 = 250− Ia1Ra1 = 250− 20× 1.9 = 212V

Eb2 = 250− Ia2Ra2

= 250− 1.3Ia2 −−−−−−(1)

Since ouput for both motors are same

Eb1Ia1 = Eb2Ia2

Eb2Ia2 = 250× 20−−−−−−− (2)

Using (2) in (1)

(250− 1.3Ia2)Ia2 = 4240

Reaaranging, we get

1.3I2a2 − 250Ia2 + 4240 = 0

Ia2 = 18.8A

=⇒ Eb2 = 4240

18.8 = 225.5V

Field current in the second case = 18.8

2 = 9.4A

N2 N1

= Eb2φ1 Eb1φ2

= 225.5× 20 212× 9.4

N2 = 2263.467rpm

Assignment No : 12 onlinecourses.nptel.ac.in Page 5 / 7

Q9.Solution

For any DC motor T ∝ φIa

For Shunt motor

==================

For a shunt motor flux (φ)is constant and hence torques T rises linearly with Ia

Speed for a shunt motor is nearly constant with slight reduction in speed due to IaRa drop

Hence the Torque-speed charecteristics will be a straight line almost parallel to x axis

For Series motor

==================

In series motor T ∝ I2a Also flux φ ∝ Ia

Hence as load current increases Speed decreases

Hence we get an inverse relationship between Torque and speed

For Compound motor

==================

A compound motor combines the charecteristics of a shunt and series motor.

Hence we expect the characteristics of it to lie between that of shunt and series motor

Ans = Shunt motor, Compound motor and Series motor

Assignment No : 12 onlinecourses.nptel.ac.in Page 6 / 7

Q10.Solution

The load-speed charecteristics of a Differential compund motor is shown in figure

It can be seen that at some point in the characteristics crosses the no load speed represented by dashed lines

At this point the speed regulation is zero

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14/12/2017 Electrical Machines - I - - Unit 13 - Week 12

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Lecture 38 : Speed Control of DC Series Motors

Lecture 39 : Testing of DC Series Motors

Lecture 40 : Characteristics of Compound DC Series Motors

Quiz : Week 12: Assignment

Additional Questions on Speed Control of DC Motor

Due on 2017-10-18, 23:59 IST.

10 points1)

10 points2)

3)

Week 12: Assignment The due date for submitting this assignment has passed.

Submitted assignment

A 240V,70A dc series motor has a combined armature circuit resistance Ra+Rse = 0.15Ω. The magnetization curve expressed in terms of Ea vesus Ia at 900 rpm is given by the following table.

EA,V 95 150 188 212 229 243 Ia, A 30 52 70 78 85 90

A chopper whose duty ratio can be changed from 0.1 to 0.9 is used to brake the motor dynamically. What is the maximum speed the motor can achieve when the armature current is 70A and the breaking resistance is 3Ω?

850 rpm

1045 rpm

894 rpm

955 rpm

No, the answer is incorrect. Score: 0

Accepted Answers: 955 rpm

The armature winding and field winding resistance of a 220V, 100A, 1500 rpm dc series motor are 0.05Ω each. The motor is operated under dynamic breaking at twice the rated torque and at 1000 rpm. Calculate the value of braking resistance. Assume linear magnetic circuit.

5.1Ω

2.6Ω

1.3Ω

4Ω

No, the answer is incorrect. Score: 0

Accepted Answers: 1.3Ω

Field test is performed on two identical DC series machines, one acting as a motor and the other one as a generator. The readings obtained are given below: For Motor: Armature current = 57 A Armature voltage = 600 V Voltage drop across field winding = 42 V For Generator Armature current = 38 A Armature voltage = 520 V

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14/12/2017 Electrical Machines - I - - Unit 13 - Week 12

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Week 12 : Assignment Solution

10 points

4)

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6)

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10 points7)

Voltage drop