Concept:

In a DC generator, generated emf is given by

Eg = Vt + IaRa

In a DC motor, back emf is given by

Eb = V – IaRa

Where, Ia is armature current

Ra is armature resistance

Calculation:

Armature resistance (Ra) = 0.8 ohm

Voltage (V) = 220 V

Armature current (Ia) = 40 A

Back emf (Eb) = 200 V

Eb = V – IaRa

= 220 – 40(0.8)

= 188 V

• Residual magnetism is the small magnetic field left in the iron cores of the shunt fields when the generator is at rest
• Without it, the fields would have to be flashed with a DC current in order to start the generator generating; Hence to start the generator, it is necessary for the DC shunt generator
• Once it is destroyed, it can be restored by connecting its field finding to a DC source for a short time to magnetize the poles
• So, we need to connect the field winding to a battery (DC source)

Field’s test:

• This test is applicable to two similar DC series motors of large capacity
• Series motors which are mainly used for traction work are easily available in pairs.
• The two machines are coupled mechanically
• One machine runs normally as a motor and drives generator whose output is wasted in a variable load R
• Iron and friction losses of two machines are made equal by joining the series field winding of the generator in the motor armature circuit so that both machines are equally exciting and by running them at equal speed
• Load resistance R is varied till the motor current reaches its full-load value indicated by ammeter A1
• After this adjustment for full-load current, different ammeter and voltmeter readings are noted

Important Points:

Swinburn’s test and retardation tests are no-load tests. So, these methods of testing are not possible with series motors as series motors have a tendency of attaining dangerous high speed at no load. Hence these tests prefer constant flux machines i.e. shunt (or) compound machines.

Concept:

As load torque increases, the speed of dc differential compound motor decreases it is wrong statement. In the case of dc differential motor if load torque increases, the speed of the dc differential motor increases.

Torque-Speed characteristic for Differential compound motor:

• In the differential compound motor, the series field opposes the shunt field so that the flux is decreased as the load is applied.
• This results in the speed remaining substantially constant or even increasing with the increase of load.
• As we can see by the characteristic of the dc motor if load torque increases, the speed of the dc differential motor increases

• Interpoles are similar to the main field poles and located on the yoke between the main field poles.
• They have windings in series with the armature winding.
• The inter poles are tapering in shape i.e. having a broad base and less pole shoe area to reduce the extra air gap flux under trailing pole tip
• The air gap under inter poles is more than the air gap under the main field poles to avoid the saturation

Interpoles in DC machine has basically two functions:

• Automatic neutralization of cross magnetization due to armature reaction
• To counter and cancel reactance voltage in the coil undergoing commutation

Important Point:

In a DC machine, two kinds of magnetic fluxes are present (armature flux and main field flux). The effect of armature flux on the main field flux is called as armature reaction.

The effect of armature reaction can be reduced by following methods

a) By using compensating winding

b) By using commutation poles or inter-poles

c) By reducing the cross section of pole pieces

• In case of DC shunt motor, flux is practically constant as the field is parallel to the armature.
• If the load is increased, then the speed of the motor will remain almost constant because the field current remains almost constant.
• So that it is also known as constant speed motor.
• Hence, the highest speed attained by the DC shunt motor at rated flux is equal to no-load speed.
• DC shunt motor is used to drive the constant speed line shafting, lathes, blower, and fan, etc.

Note:

In DC series motors, torque increases as the square of armature current. Hence these motors are used in high starting torque and variable speed applications.

Impulse testing:

• Lighting is a common phenomenon in transmission lines because of their tall height. This lightning stroke on the line conductor causes impulse voltage.
• The transformer experiences lightning impulse voltages due to which insulation becomes the most important factor.
• Any weakness in the insulation may cause the failure of the transformer.
• To ensure the effectiveness of the insulation system of a transformer, the dielectric test is used.
• The dielectric test alone cannot determine the dielectric strength of the insulator (for all the frequencies)
• Hence, an impulse test of the transformer performed on it.
• Therefore, Impulse testing of transformers is done to determine the ability of ​insulation to withstand transient voltages.

There are two types of impulse test

1. Lightning impulse test
2. Switching impulse test

Concept:

Hysteresis losses: These are due to the reversal of magnetization in the transformer core whenever it is subjected to alternating nature of magnetizing force.

$W_h=\eta B_{max}^{x}fv$

$B_{max}\propto \frac{V}{f}$

Where

x is the Steinmetz constant

Bm = maximum flux density

f = frequency of magnetization or supply frequency

v = volume of the core

At a constant V/f ratio, hysteresis losses are directly proportional to the frequency.

Wh ∝ f

Eddy current losses: Eddy current loss in the transformer is I2R loss present in the core due to the production of eddy current.

$W_e=K{f}^2{B}_m^2{t}^{2}V$

$B_{max}\propto \frac{V}{f}$

Where,

K - coefficient of eddy current. Its value depends upon the nature of magnetic material

Bm - Maximum value of flux density in Wb/m2

t - Thickness of lamination in meters

f - Frequency of reversal of the magnetic field in Hz

V - Volume of magnetic material in m3

At a constant V/f ratio, eddy current losses are directly proportional to the square of the frequency.

We ∝ f2

Iron losses or core losses or constant losses are the sum of both hysteresis and eddy current losses.

Wi = Wh­ + We

At constant V/f ratio, Wi = Af + Bf2

Calculation:

In the given question, voltage kept constant and frequency is varied.

The V/f ratio is not constant.

Hysteresis losses, W_h ∝ V^1.6 f^-0.6

As the frequency is decreased to 40 Hz from 50 Hz, the hysteresis losses will increase.

Eddy current losses, W_e ∝ V²

As eddy current losses are independent of frequency, the eddy current losses remain the same.

Transformer: An electrical device that is used to transfer electrical energy from one electrical circuit to another is called a transformer.

There are two types of transformer:

1. Step-up transformer:

• If the number of turns in the primary coil is less than the turns in the secondary coil then it is called a Step-up transformer.
• A step-up transformer gives voltage greater than the input voltage 
• The step-up transformer increases the voltage and reduces the current in the output or secondary side. 

2.Step-down transformer:

• If the number of turns in the primary coil is greater than the turns in the secondary coil then it is called Step down transformer.
• A step-down transformer gives output voltage less than the input voltage.
• The step-down transformer reduces the voltage and increases the current in output or secondary side. 

Concept:

A transformer has mainly two types of losses

• core losses
• copper losses.

Core loss, which is also referred to as iron loss, consists of hysteresis loss and eddy current loss.

These two losses are constant when the transformer is charged. That means the amount of these losses does not depend upon the condition of the secondary load of the transformer. In all loading conditions, these are fixed.

Copper losses are directly proportional to the square of the load on the transformer.

$W_{cu}=x^2{W}_{cufl}$

Here, x is the percentage of the full load of the transformer.

Wcufl­ is the copper losses at the full load.

The maximum efficiency in a transformer that occurs at copper losses is equal to iron losses.

Efficiency is maximum at some fraction x of full load.

$x=\sqrt{\frac{W_i}{W_{cu}}}$

Where Wi = iron losses

Wcu = copper losses

kVA at maximum efficiency is given by,

kVA at ηmax = full load kVA × $\sqrt{\frac{W_i}{W_{cu}}}$

Calculation:

Given that, full load kVA = 10 kVA

Iron losses (Wi) = 1 kW

Copper losses (Wcu) = 2 kW

Load kVA corresponds to maximum efficiency $=10\times \sqrt{\frac{1}{2}}=7.07kVA$

• Silica gel is used to absorb moisture and prevent entering the oil tank while breathing. It is used in an oil transformer breather.
• The Colour of fresh silica gel is blue. The moist silica gel became pink in color.
• A transformer consists of Breather, Conservator and Buchholz relay, etc.
• The breather is used in the transformer to filter out the moisture from the air. Breather consists of silica gel which absorbs moisture from the air.
• Conservator tank present at the top of the transformer which allows adequate space for expansion of oil. Therefore during an overloading condition, the oil moves to the conservator tank.
• Buchholz relay is used for the protection of transformers from the faults occurring inside the transformer.
• Whenever the transformer is loaded, the temperature of the transformer insulating oil increases. Consequently, the volume of oil is increased.
• As the volume of the oil is increased, the air above the oil level in conservator will come out. At low oil temperature, the volume of the oil is decreased, causes air to enter into a conservator tank.
• The air consists of moisture in it and this moisture can be mixed up with oil. Hence to filter the air from a moisture silica gel breather is used.

Concept:

• From the below diagram we can observe that the maximum torque of an induction motor is independent of rotor resistance 
• Slip at which maximum torque occurs depends on rotor resistance and they change on adding the additional resistance to the rotor circuit.

The maximum torque of an induction motor is given by

$T_{max}=\frac{k{E}_{20}^2}{2X_{20}}$

∴ Maximum torque is directly proportional to supply voltage & maximum torque is inversely proportional to rotor reactance.

Hence, the maximum torque is dependent on the supply voltage & reactance of the rotor and is independent of the rotor resistance.

Sm the value of slip corresponding to the maximum torque is

$S_m=\frac{R_2}{X_{20}}$

Explanation:

$S_m=\frac{R_2}{X_{20}}$

R₂ = S_m X₂₀

Therefore, the rotor resistance per phase is equal to slip times the rotor reactance per phase.

Concept:

In a three-phase induction motor,

• The stator is stationary, and the rotor is a rotating part
• Both the stator field and rotor field rotate with synchronous speed
• The rotor rotates with a speed less than synchronous speed
• The relative speed between the rotor field and a stator is synchronous speed
• The relative speed between the rotor field and stator field is zero
• The relative speed between stator filed and stator is synchronous speed

The speed of rotor field flux is synchronous speed (Ns) and the speed of the stator is zero as the stator doesn't rotate.

The speed of rotor field flux with respect to the stator = Ns - 0 = Ns

Calculation:

Given that, number of poles (P) = 4

Frequency (f) = 50 Hz

Synchronous speed $=\frac{120\times 50}{4}=1500rpm$

The relative speed between the rotor field and stator is synchronous speed and it is 1500 rpm.

Power flow in the Induction motor is as shown below.

Rotor input or air gap power $P_{in}=\frac{3I_2^2{R}_2}{s}$

Rotor copper losses $P_{cu}=s\times P_{in}=3I_2^2{R}_2$

Gross mechanical power output $P_g=P_{in}-P_{cu}=\frac{3I_2^2{R}_2}{s}-3I_2^2{R}_2=3I_2^2{R}_2\left(\frac{1-s}{s}\right)$

The relation between the rotor air gap power, rotor copper losses and gross mechanical power output is,

$P_{in}:P_{cu}:P_g=\frac{3I_2^2{R}_2}{s}:3I_2^2{R}_2:3I_2^2{R}_2\left(\frac{1-s}{s}\right)$

$\Rightarrow P_{in}:P_{cu}:P_g=\frac{1}{s}:1:\left(\frac{1-s}{s}\right)=1:s:(1-s)$

Therefore, the electrical rotor losses are proportional to the slip.

• If the two terminals of a rotor windings are kept open, because of the rotating magnetic field produced by currents in the three-phase stator windings, there will be induced a voltage in stator coils
• But this open circuit voltage will not be able to make any flow of currents in the rotor windings since these windings are kept open; The motor will not start at this condition
• When the two terminals are short-circuited, the induced voltage in this coil will generate the flow of current which is known as induced current
• This induced current will generate its own rotating magnetic field in the same direction of rotor rotation
• Due to the interaction of stator produced and rotor produced rotating magnetic fields, the rotor conductors experience a torque due to which the rotor starts rotating

Concept:

In a three-phase induction motor,

• The stator is stationary, and the rotor is a rotating part
• Both the stator field and rotor field rotate with synchronous speed
• The rotor rotates with a speed less than synchronous speed

Application:

• The relative speed between the rotor field and a stator is synchronous speed
• The relative speed between the rotor field and stator field is zero
• The relative speed between stator filed and stator is synchronous speed

• A single-phase induction motor is not a self-starting motor.
• By splitting the single phase to two phases 90° to each other rotating magnetic field is produced.
• This concept is used to start the motor known as a split-phase technique.
• The starting winding also called the auxiliary or temporary winding is connected in parallel to the main winding through a centrifugal switch in order to get isolated under running conditions.
• Both main winding and auxiliary windings are placed on the stator.

Important Points:

For a single-phase capacitor start induction motor, the direction of rotation can be changed by reversing main field terminals but not supply terminals.

For this motor starting performance is good but efficiency and power factor are poor.

• The shaded pole motor is simply a self-starting single-phase induction motor whose one of the pole is shaded by the copper ring.
• The copper ring is also called the shaded ring. This copper ring acts as a secondary winding for the motor.
• The shaded pole motor rotates only in one particular direction, and the reverse rotation of the motor is not possible.
• The stator of the shaded pole single phase induction motor has salient or projected poles.
• These poles are shaded by copper band or ring which is inductive in nature.
• The poles are divided into two unequal halves. The smaller portion carries the copper band and is called a shaded portion of the pole.

Advantages:

• Very economical and reliable.
• Construction is simple and robust because there is no centrifugal switch.

Disadvantages:

• Low power factor.
• The starting torque is very poor.
• The efficiency is very low as the copper losses are high due to presence of copper band. It is around 5 to 35%.
• The speed reversal is also difficult and expensive as it requires another set of copper rings.
• ​Very little overload capacity

Applications:

• They are suitable for small devices like relays and fans because of their low cost and easy starting.
• Used in exhaust fans, hairdryers and also in table fans.
• Used in air conditioning and refrigeration equipment and cooling fans.
• Used in record players, tape recorders, projectors, photocopying machines.
• Used for starting electronic clocks and single-phase synchronous timing motors.

Split phase induction motor:

• A split-phase motor has no capacitance in the auxiliary circuit.
• A phase shift with respect to the main current is achieved by using narrow conductors to achieve high resistance to reactance ratio.
• Increasing the resistance means that the auxiliary winding can only be used during starting, otherwise, it would overheat.
• A split-phase motor has significantly Moderate starting current and lower torque at starting than any of the capacitor motors due to the reduced phase angle between main and starting winding currents.
• Torques developed by a split-phase motor is proportional to the sine of the angle between Im and Is

Applications:

• ​These motors are used in fans, blowers, centrifugal pumps, washing machines, grinders, lathes, air conditioning fans, etc.
• These motors are available in the size ranging from 0.05 kW to 0.5 kW