1. A dc shunt motor has the following characteristics, Ra = 0.5 Ω, Rf = 200 Ω, base speed = 1000 rpm, rated voltage = 250 V. On no-load, it draws a current = 5 A. At what speed will this run while delivering a torque of 150 N.m?
881 rpm
920 rpm
950 rpm
990 rpm
Answer: 1. 881 rpm
Explanation:
In motor back emf is given by
Eb = V – Iara
And
Eb = KnϕN
Where N = speed in rpm
Kn in V/rpm
ϕ = flux per pole
Torque T = KaϕIa
Ka=30πKn
Calculation:
Given V = 250 volt
Ia = 5 A
If = 250/200 = 1.25 A
Ia = 5 – 1.25 = 3.75 A
Eb = 250 – 3.75 × 0.5 = 248.125 volt
Knφ = 248.125/1000 = 0.248125
$T = \frac{{30}}{\pi }{K_n}\phi {I_a}$
Ia = (150 × π)/(0.248 × 30) = 63.33 A
Eb = 250 – 63.33 × 0.5 = 218.335
N = 218.335/0.248125 = 881 RPM
2. A 4-pole dc motor takes a 50 A armature current. The armature has lap-connected 480 conductors. The flux per pole is 20 mWb. Calculate the gross torque developed by the armature of the motor.
3. With an armature voltage of 100 V and rated field winding voltage, the speed of a separately excited DC motor driving a fan is 1000 rpm, and its armature current is 10 A. The armature resistance is 1 Ω. The load torque of the fan load is proportional to the square of the rotor. Neglecting rotational losses, the value of the armature voltage (in Volt) which will reduce the rotor speed to 500 rpm is
50 V
45.5 V
30 V
47.5 V
Answer:4. 47.5 V
Explanation:
Let,
Torque T1 at N1 = 1000 rpm
Torque T2 = at N2 = 500 rpm
∴ T ∝ N2 {Given}
T2/T1 = (N2/N1)2
T2/T1 = (500/1000)2
T2/T1 = 1/4
Load torque can be expressed as,
T = k ϕ. Ia
T ∝ ϕ Ia
T ∝ Ia {∴ ϕ = constant}
T2/T1 = Ia2/Ia1
Ia2/Ia1 = 1/4
Ia2 = 10/4 = 2.5 A
Therefore,
Eb1 = V − Ia1Ra = (100 − 10) = 90 V
${E_b} = \frac{{NP\phi Z}}{{60{\rm{\;}}A}}$
Eb1/Eb2 = N1/N2
Eb2 = 90 × 500/1000
Eb2 = 45 V
Therefore,
V2 = Eb2 + Ia2Ra
⇒ V2 = 45 + 2.5 × 1
⇒ V2 = 47.5 V
Hence, The armature voltage is 47.5 V
4. A DC shunt machine develops an AC EMF of 250 V at 1500 rpm. Find the torque developed for an armature current of 50 A.
59.6 N-m
79.6 N-m
69.6 N-m
49.6 N-m
Answer: 2. 79.6 N-m
Explanation:
Speed is in radians per second (ω).
ω = (2 × π × N)/60
ω = (2 × π × 1500)/60
= 157.08 radians/sec.
Power = generated emf × armature current
= 250 × 50
= 12500 watts.
Torque = power/speed.
T = 12500/157.08
T = 79.577 Nm
5. An Electric motor, developing a starting torque of 15 Nm, starts with a load torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/s2, the moment of inertia of the system is (neglecting viscous and Coulomb friction)
0.25 kg m2
0.25 Nm2
4 kg m2
4 Nm2
Answer: 3. 4 kg m2
Explanation:
Accelerating torque in an electric motor is given as
Ta = Tm – Te
Ta = J α
Where, Tm = Mechanical torque or starting torque
Te = Electrical torque or load torque
J = Moment of inertia
α = Angular acceleration
Calculation:
Given-
Tm = 15 Nm, Te = 7 Nm,α = 2 rad /sec2
J = (15 – 7) / 2
J = 4 Nm2
6. Neglecting saturation, if the current taken by a series motor is increased from 10 A to 12 A, the percentage increase in its torque will be:
44%
40%
20%
16%
Answer: 1. 44%
Explanation:
In dc series motors, flux is directly proportional to armature current. Hence torque is directly proportional to square of the armature current.
Ta ∝ Ia2 (∵ Flux saturation is neglected)
In dc series motor at saturation condition, flux is constant. Hence torque is proportional to armature current.
T ∝ Ia
Where, T = torque, Ia = armature current, ϕ = field flux
Calculation:
Let at I = 10 A, the torque is T1 and at I = 12 A torque is T2
T1/T2 = (10/12)2
T2 = 1.44 T1
i.e. T2 increases by 44%
7. A DC motor has the following specifications: 10 hp, 37.5 A, 230 V; flux / pole = 0.01 Wb, number of poles = 4, number of conductors = 666, number of parallel paths = 2. Armature resistance = 0.267Ω. The armature reaction is negligible and rotational losses are 600W. The motor operates from a 230V DC supply. If the motor runs at 1000 rpm, the output torque produced in (in Nm) is _______
60.34 Nm
57.78 Nm
51.34 Nm
40.23 Nm
Answer: 2. 57.78 Nm
Explanation:
The back emf induced in the dc motor can be given by,
8. A DC Motor develops a torque of 150 N-m. A 10 percent reduction in the field flux causes a 50 percent increase in armature current. The new value of torque is:
102.5 N-m
202.5 N-m
172.5 N-m
232.5 N-m
Answer: 2. 202.5 N-m
Explanation:
The back emf induced in the dc motor can be given by,
9. A shunt–connected DC motor operates at its rated terminal voltage. Its no-load speed is 200 radians/second. At its rated torque of 500 Nm, its speed is 180 radian/second, The motor is used to directly drive a load whose load torque TL depends on its rotational speed (in radian/second), such that TL = 2.78 × ωT. Neglecting rotational losses, the steady-state speed (in radian/second) of the motor, when it drives this load is __________
180 rad/sec
260 rad/sec
360 rad/sec
90 rad/sec
Answer: 1. 180 rad/sec
Explanation:
Given,
no load speed (ω0) = 200 rad/sec
Rated torque (Tr) = 500 Nm
Rated Speed (ωr) = 180 rad/sec
Load torque (TL) = 2.78 ωr
Developed torque can be expressed as,
T = (200 − ωr) × 500/20
⇒ ωr = 180 rad/sec
Therefore, steady state speed is 180 rad/sec
10. Which of the given options is correctly represented by the following expression? 9.55 (output / N), where N is in rpm?
Back EMF
Maximum efficiency
Armature torque
Shaft torque
Answer:4. Shaft torque
Explanation:
The torque that is available at the motor shaft for doing useful work is known as shaft torque. It is represented by Tsh.
The total or gross torque (Ta) developed in the armature of a motor is not available at the shaft because a part of it is lost in overcoming the iron and frictional losses in the motor.
Therefore, shaft torque Tsh is less than the armature torque Ta.
The difference Ta – Tsh is called lost torque.
We know that, Power (P) = Torque (T) × Angular Speed (ω)
