Assignment - 11 : Solution

Q1&Q2.Solution

Armature current of the motor, Iam = Iag + Ia = 73 + 17 = 90A

Stray lossen in each machines, Pstray =1

2[V Ia − I2amRam − I2agRag]

=1

2[230× 17− 902 × 0.21− 732 × 0.21] = 544.955W

Pin_motor = V × Iam = 230× 90 = 20, 700W

Ploss_motor = (Pstray + V Ifm) + I2amRam

= (544.955 + 230× 2.3) + 902 × 0.21 = 2774W

The motor efficiency is given by, ηmotor =Pin_motor − Ploss_motor

Pin_motor

=20, 700− 2774

20, 700= 0.865 = 86.5%

Ploss_generator = (Pstray + V Ifg) + I2agRag

= (544.955 + 230× 2.4) + 732 × 0.21 = 2216.045W

Pout_generator = V Iag = 230× 73 = 16, 790W

The efficiency of the generator is given by, ηgenerator =Pout_generator

Pout_generator + Ploss_generator16, 790

16, 790 + 2216.045= 0.8834 = 88.34%

Assignment No : 11 onlinecourses.nptel.ac.in Page 1 / 6

Q3.Solution

From the no-load test

Armature input Parm_ip = 250× 3 = 750W

Rotational losses Prot = Piron + Pmech

= Parm − Parm_Cu − Pbrush

= 750− (32 × 0.22)− (2× 3) = 742.02W

Rotational loss remains unchanged for small variations in speed

On load Prot = 742.02W

Shaft power Psh = 20hp = 14.91kW

Let, Ia be the corresponding armature current

Pin − Parm_Cu − Pbrush − Prot − Pstray = 14.91kW

(250× Ia)− (0.22× I2a)− (2× Ia)− 742.02− (0.01× 14910) = 14910

250Ia − 0.22I2a − 2Ia = 15801

0.22I2a − 248Ia + 15801 = 0

Solving the above quadratic equation Ia = 67.79A

Input power Pin = (250× 67.79) +2502

170= 17315.15W

Output power Pout = 14910W

Hence, efficinecy η =14910

17315.15× 100 = 86.10%

Q4.Solution

Ea = 240− 30(0.3 + 0.2) = 225V

P = EaIa = 225× 30 = 6750W

ωm =1200

60× 2π = 126rad/s

Torqe =6750

126= 53.57Nm

Assignment No : 11 onlinecourses.nptel.ac.in Page 2 / 6

Q5.Solution

The power output Po = 10hp = 10× 746 = 7460W

The power developed Pd = Po + Pr = 7460 + 640 = 8100W

Armature circuit resistance R = Ra +Rse = 1.1Ω

Developed power Pd = EaIa = 8100W

Ea = Vt − IaR

EaIa = (Vt − IaR)Ia

8100 = (230− 1.1Ia)Ia

1.1I2a − 230Ia + 8100 = 0

Ia = 44.83A

Q6.Solution

Power input Pin = VtIa = 230× 44.83 = 10310.9W

Efficiency, η =7460

10310.9× 100 = 72.35%

Q7.Solution

Under no load

Rotational losses = Input Power - Armature resistance loss - Brush Loss

= 250× 13.24− (13.24)2 × 0.06− 2× 13.24

= 3310− 10.52− 26.4 = 3273.08W

Changes in rotational losses due to change in speed from 1515rpm at no load to 1495rpm at full load is neglected

Assignment No : 11 onlinecourses.nptel.ac.in Page 3 / 6

On load

Pshaft = 50kW

Let Armature current be Ia

Input Power = Output Power + Armature Resistance losses + Brush Losses

+ Rotational Losses + Stray Losses

(Note: Input power here does not contain shunt field losses)

Stray losses =50kW × 0.01

250× Ia = 50kW + I2a × 0.06 + 2× Ia + 3273.08 + 0.01× 50kW

Rearranging we get

0.06I2a − 248Ia + 53778.08 = 0

Solving for Ia we get

Ia = 229.6A

Total Input power = 250× 229.6 +2502

50= 58650W

∴ Efficiency =Output Power

Total Input Power× 100 = 85.25%

Q8.Solution

Speed ∝ Eb

φ

Since current is constant, flux remains before and after

=⇒ Speed ∝ Eb

N2

N1=Eb2

Eb1

Eb1 = 200− 15× 1 = 185V

Eb2 = 200− 15× (1 + 5) = 110V

N2 =110

185× 800

= 475.67 rpm

Assignment No : 11 onlinecourses.nptel.ac.in Page 4 / 6

Q9.Solution

Total developed torque =EbIaω

ω =2πN

60

Also Eb =φPNZ

60A

Using Eb in expression for Total developed torque

Total developed torque =φPZIa

2π

=25× 10−3 × 4× 782× 40

2π= 248.9Nm

Q10.Solution

Back EMFEb ∝ φN

Sinceφ ∝ Ia

Eb ∝ IaN

Assignment No : 11 onlinecourses.nptel.ac.in Page 5 / 6

Since losses are neglected Eb = V (Applied Voltage)

∴ Eb2 = Eb1 =⇒ I22N2 = I1N1

I2N2 = 2× 20× 600 = 24000−−−−−−(1)

Torque (T ) ∝ φIa ∝ IfIa (Since φ ∝ If )

In first case If = I1 = 20A

and in second case If = I2/2

But, the torque is said to be proportional to speed square i.e. T ∝ N2

=⇒ IfIa ∝ N2

∴If2I2

N22

=If1I1

N21

I2/2× I2N2

2

=20× 20

600× 600

I2N2

=

√2

30−−−−−−(2)

From (2) and (1)

I22 = 800√

2 =⇒ I2 = 33.5A

Again from (1)N2 =40× 600

33.5= 714rpm

Assignment No : 11 onlinecourses.nptel.ac.in Page 6 / 6

14/12/2017 Electrical Machines - I - - Unit 12 - Week 11

https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=68&assessment=90 1/3

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Unit 12 - Week 11

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Week 12

Lecture 35:Testing of DCShunt Motors

Lecture 36:Characteristicsof DC SeriesMotors

Lecture 37:Starting andBraking of DCSeries Motors

Quiz : Week 11:Assignment

Week 11 :AssignmentSolution

Due on 2017-10-14, 23:59 IST.

1)

10 points

2)

10 points

3)

10 points

Week 11: AssignmentThe due date for submitting this assignment has passed.

Submitted assignment

Hopkinson’s test was performed on two similar dc shunt machines and the following results areobtained. Supply voltage : 230V Field current of motor : 2.3A Field current of generator : 2.4A Armature current of generator : 73A Current taken by the two armatures from supply : 17A Resistance of each armature circuit : 0.21 Ω Calculate the efficiency of the motor in percentage under these conditions of load? [Enter only the numerical value, do not put any symbols]

No, the answer is incorrect. Score: 0

Accepted Answers:(Type: Range) 85,88

For the data given in problem-1, find the efficiency of the generator? [Enter only the numerical value. Do not put any symbols]

No, the answer is incorrect. Score: 0

Accepted Answers:(Type: Range) 86.5,89.5

A 20hp, 250V DC shunt motor has an armature resistance of 0.22Ω and a field resistance of 170Ω.At no-load and rated voltage the speed of the motor is 1222 rpm and the armature current is 3A. Whatwill be the efficiency of the motor when the shaft power is 20hp and the motor is running at 1190 rpm.The stray load losses is 1% of the output. The brush contact drop is 2V. [Enter only the numerical value. Do not put any symbols]

No, the answer is incorrect. Score: 0

Accepted Answers:(Type: Range) 84,88

14/12/2017 Electrical Machines - I - - Unit 12 - Week 11

https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=68&assessment=90 2/3

10 points4)

10 points5)

6)

10 points

7)

10 points

8)

A dc series motor (240V, 12hp, 1200 rpm) is connected to a 240V supply, draws a currentof 30A and rotates at 1200 rpm. Ra = 0.3Ω and series resistance Rsr = 0.2Ω. Determine the torquedeveloped at rated speed.Assume magnetic linearity.

45 Nm

36 Nm

72 Nm

54 Nm

No, the answer is incorrect. Score: 0

Accepted Answers:54 Nm

A 10hp, 230V,1200 rpm DC series motor has the total armature circuit resistance Ra +Rse = 1.1Ω. The rotational losses of the machine is 640W. What is the armature current when the motordelivers its rated load at 1200rpm when connected to a rated supply voltage?

24 A

16 A

45 A

60A

No, the answer is incorrect. Score: 0

Accepted Answers:45 A

What is the efficiency in percentage of the motor given in problem-5, at full load and rated speed? [Enter only the numerical value. Do not put any symbols]

No, the answer is incorrect. Score: 0

Accepted Answers:(Type: Range) 70,74

The following test results were obtained when Swinburne's test wasconducted on a 250V, 50kW, 1500rpm, DC shunt motor.No load- Terminal voltage (Vt)=250V, Armature Current (Ia)=13.24A.Speed(N)=1515rpm. Armature Resistance and shunt field resistanceare 0.06ohm and 50ohm respectively. Take overall brush drop as 2V.Find the efficiency of the machine when it delivers a load of 50kW at1495rpm. Assume that the stray load losses is 1% of output

No, the answer is incorrect. Score: 0

Accepted Answers:(Type: Range) 84-86

A DC series motor is running at 800rpm at 200V with a current of15A. Total resistance of the motor is 1ohm. If a 5ohm resistor isconnected in series with the motor and if it draws the same current asbefore at 200V, what will be the new speed in rpm?

14/12/2017 Electrical Machines - I - - Unit 12 - Week 11

https://onlinecourses.nptel.ac.in/noc17_ec10/unit?unit=68&assessment=90 3/3

10 points

9)

10 points

10 points10)

No, the answer is incorrect. Score: 0

Accepted Answers:(Type: Range) 470-480

What will be the total torque developed in Nm for a 250V, 4pole DCseries motor, if it delivers 11bhp at 40 A, with flux per pole of 25mWb?Take number of conductors as 782 (wave connected) and motorresistance as 0.75ohm.

No, the answer is incorrect. Score: 0

Accepted Answers:(Type: Range) 245-255

A 4 pole series motor draws 20A at 250V running at 600rpm. All the field coils areconnected in series. What will be the speed of the motor if the field coils are reconnected to have toparallel groups instead of having them all in series? Assume that the torque is proportional to square ofspeed. Also ignore the losses in the machine.

827rpm

714rpm

553rpm

630rpm

No, the answer is incorrect. Score: 0

Accepted Answers:714rpm

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