SSC JE 2012 Electrical question paper with solution

Iron loss is given as

Pi = Pe + Ph

Pi = Kef² + Khf

Pi/f = Kef + Kh

where Kh = 0.01

Ke = 0.001

Ph = K_hf = 0.01 x 100 = 1W

Pe = K_ef² = 0.001 x (100)2 = 10W

Hysteresis losses  is also known as Iron Loss or Core Loss and it is always constant.

Hysteresis loss is due to the reversal of magnetization of the transformer core whenever it is subjected to the alternating nature of magnetizing force. Whenever the core is subjected to an alternating magnetic field, the domain present in the material will change its orientation after every half cycle. The power consumed by the magnetic domains for changing the orientation after every half cycle is called Hysteresis loss.

Eddy current is given by

Pe = f²B²max

∴ Eddy current losses are directly proportional to the square of the frequency.

Mutual inductance

Properties of the material of a permanent magnet :

(1) It should have high retentivity so that it remains magnetised in the absence of the magnetizing field.

(2) It should have high saturation magnetization.

(3) It should have high coercivity so that it does not get demagnetized easily.

The pole shoe is laminated to reduce the eddy current losses due to any variable air gap reluctance caused by the rotor slots.

In case of open type slot DC machine under teeth position, the air gap is less So reluctance”S, therefore, flux Φ is more

Φ↑ = I_f x N/S↓

Under slot position, Air gap is more Reluctance is more hence flux is less

Φ↓= I_f x N/S↑

So pulsating flux interact with stationary Rotor as a result EMF is induced and eddy current will flow so eddy current loss will be there so pole shoe is laminated to reduce eddy current loss.

The universal motor is so named because it is a type of electric motor that can operate on AC or DC power. It is a commutated series-wound motor where the stator’s field coils are connected in series with the rotor windings through a commutator. It is often referred to as an AC series motor.

The single-phase series motor is a commutator-type motor. If the polarity of the line terminals of a dc series motor is reversed, the motor will continue to run in the same direction. Thus, it might be expected that a dc series motor would operate on alternating current also.

A series motor that is specifically designed for dc operation suffers from low efficiency, poor power factor, and sparking in brushes.

In order to overcome these difficulties, the following modifications are made to a D.C. series motor.

Field core is made up of low hysteresis loss material and is laminated to reduce eddy current loss. The field winding is provided with a small number of turns to reduce iron losses and increase commutation and reduce armature reaction compensating winding are used.

Change in frequency from No load to full load

f = 50 x 0.05 = 2.5Hz

Full load frequency

f_fl1 = 50 – 2.5 = 47.5Hz

Since both alternators have same speed regulation then

f_fl1 = f_fl2

From the above diagram

For machine 1

(50 – f)/(80 – x) = (50 – 47.5)/40

x – 16f = 80 – 16 x 50

x- 16f = -720—–1

For machine 2

(50 – f)/(80 – x) = (50 – 47.5)/60

x – 24f = -1200 ——-2

From equation (1) and (2)

x= 48 MW

f = 60 Hz

So, machine A operates at a load of 48 MW While machine B will operate at a load of

80 – 48 = 32 MW

Power is the energy available at the shaft to drive machine and it is given as

P =2πnT/60

where,

P=Power available at the shaft (W or J/s);

N=Speed of rotation of the shaft (rpm);

T=Torque available at the shaft (N-m)

Hence T = P/602πn

If P2 is the power input to the rotor and Pm is the mechanical output power(including friction losses)

P2 – Pm is the copper losses in the rotor i.e P2 – Pm = I_r²R₂