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EEET2263 - Tutorial #5: DC Motors - Solution, Assignments of Electrical Engineering

Acharya Nagarjuna UniversityElectrical Engineering

the documents is filled with full textbook solutions of given document for exam preparation. i assure that it will help you to solve many tough and necessary questions you needed for your preparation

Typology: Assignments

2020/2021

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EEET2263 – Tutorial #5, DC Motors - SOLUTION
1. * A separately excited dc generator turning at 1400 rpm produces an armature voltage of
127V under no load conditions. The armature resistance is 2. If the machine then delivers a
current of 12 A into a load, calculate
(a) The terminal voltage
ANS: 103212127
aaaa RIEV A
(b) The power dissipated in the armature winding
ANS: 2882144
2 aaarm RIP W
(c) The braking torque exerted by the generator on the prime mover.
ANS: 152412127
aamech IEP W 4.10
6014002
1524
mech
mech P
TNm
2. * A separately excited dc generator produces a no load voltage of 127V. What happens to the
output voltage if
(a) the speed is increased by 20%?
ANS: 4.1522.1127
a
VV
(b) the direction of rotation is reversed?
ANS: 127
a
VV (sign reverses)
(c) the exciting current in the field winding is increased by 10%?
ANS: 7.1391.1127
a
VV (assuming linear magnetic field relationship with If)
(d) the polarity of the field winding is reversed?
ANS: 127
a
VV (sign reverses)
3. A 240kW, 500V, 1750 rpm separately excited dc generator has an overall efficiency of 94%.
The shunt field winding has a resistance of 60, and the rated field winding current is 5A.
The resistive loss in the armature winding is 5.52kW at full rated current. Calculate
(a) the rated armature current.
ANS: 480
500
240000
)(
a
a
rateda V
P
IA
(b) the total losses in the machine at full rated current.
ANS: 70205605520 2 faloss PPP W
(c) the resistive losses in the armature when the machine is operating at half rated
current.
ANS: 1380
4
5520
2 aaa IRP W (losses reduce to a quarter for half current)
(d) the torque required from the prime mover to drive the generator at 1750 rpm at half
rated current.
ANS: 9.122500.1380.1120
faomech PPPP kW (assume generator still
supplies field losses despite being separately excited. Then
pf3

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EEET2263 – Tutorial #5, DC Motors - SOLUTION

    • A separately excited dc generator turning at 1400 rpm produces an armature voltage of

127V under no load conditions. The armature resistance is 2. If the machine then delivers a

current of 12 A into a load, calculate

(a) The terminal voltage

ANS: VaEaIaRa  127  12  2  103 A

(b) The power dissipated in the armature winding

ANS: 144 2 288

2 ParmIaRa    W

(c) The braking torque exerted by the generator on the prime mover.

ANS: PmechEaIa  127  12  1524 W 10. 4 2 140060

mech mech

P

T Nm

    • A separately excited dc generator produces a no load voltage of 127V. What happens to the

output voltage if

(a) the speed is increased by 20%?

ANS: Va  127  1. 2  152. 4 V

(b) the direction of rotation is reversed?

ANS: Va  127 V (sign reverses)

(c) the exciting current in the field winding is increased by 10%?

ANS: Va  127  1. 1  139. 7 V (assuming linear magnetic field relationship with I (^) f )

(d) the polarity of the field winding is reversed?

ANS: Va  127 V (sign reverses)

  1. A 240kW, 500V, 1750 rpm separately excited dc generator has an overall efficiency of 94%.

The shunt field winding has a resistance of 60, and the rated field winding current is 5A.

The resistive loss in the armature winding is 5.52kW at full rated current. Calculate

(a) the rated armature current.

ANS: 480

a

a a rated V

P

I A

(b) the total losses in the machine at full rated current.

ANS: 5520 60 5 7020

2 PlossPaPf     W

(c) the resistive losses in the armature when the machine is operating at half rated

current.

ANS: 1380

PaRaIa   W (losses reduce to a quarter for half current)

(d) the torque required from the prime mover to drive the generator at 1750 rpm at half

rated current.

ANS: PmechPoPaPf  120  1. 380  1. 500  122. 9 kW (assume generator still

supplies field losses despite being separately excited. Then

mech mech

P

T Nm

    • A 230V shunt dc motor has a nominal armature current of 60A. If the armature resistance is

0.15, calculate

(a) the internal backemf across the armature

ANS: EaVaRaIa  230  60  0. 15  221 V

(b) the power supplied to the armature

ANS: ParmVaIa  230  60  13. 8 kW

(c) the mechanical output power developed by the motor

ANS: 13800 0. 15 60 13. 26

2 2 PmechParmRaIa     kW

  1. For problem 4, calculate

(a) the initial starting current if the motor is directly connected across the 230V supply

before it starts moving

ANS: 1533

 ^230 

a

a a (^) R

V

I A

(b) the value of a resistor that would need to be connected in series with the armature to

limit the initial current to 115A.

ANS: 2

tot ^ a  

V

R  Rext  2  0. 15  1. 85 

  1. A separately excited dc motor turns at 1200 rpm under no load when the armature is

connected to a 115V dc source. Calculate the armature voltage required to make the motor run

(a) at 1500 rpm under no load

ANS: 143. 8

Va  115   V

(b) at 100 rpm under no load

ANS: 9. 58

Va  115   V

  1. *A 180kW, 230V, 435 rpm dc shunt motor has a nominal full-load current of 860A. Calculate

(a) the total losses and efficiency at full load and rated speed

ANS: PinVaIa  230  860  197. 8 kW 91. 0

  1. 8

 ^180 

in

mech P

P

(b) the approximate shunt field exciting current if the shunt field causes 20% of the total

losses (assume the field winding is directly connected to the 230V supply)

ANS: Ploss  197800  180000  17. 8 kW  Pf  3560 W  15. 5 230

^3560 

I (^) f A

(c) the value of the armature resistance and the backemf, if 50% of the total losses at full-load are attributable to the armature resistance

ANS: Pa  8900 W  0. 012 860

Ra  2    Ea  230  0. 012  860  219. 7 V