DC Motor Efficiency MCQ Quiz in বাংলা - Objective Question with Answer for DC Motor Efficiency - বিনামূল্যে ডাউনলোড করুন [PDF]

Short-circuit current:

• When two or more conductors come in contact with each other then the impedance between them is decreased to a much smaller value, which causes the flow of a very large amount of current known as short-circuit current.
• Due to the flow of a large amount of current, the energy loss will be there in the form of heat.
• The formula for the energy loss is given as I²Rt.

Where I = Current at the time of short-circuit

R = Resistance offered at the time of short-circuiting.

t = Time duration of short-circuiting.

Additional Information

Losses in DC machine:

Armature copper losses:

• Armature copper losses = I_a²R_a, where Ia is armature current and Ra is armature resistance.
• These losses are about 30% of the total full load losses.

Field copper losses:

• Field copper losses = I_sh²R_sh, where Ish is field current and Rsh is field resistance.
• These losses are about 25% theoretically, but practically it is constant.

Magnetic losses: 

• As the iron core of the armature is rotating in the magnetic field, some losses occur in the core which is called core losses or magnetic losses or iron losses.
• Normally, machines are operated with constant speed, so these losses are almost constant. These losses are categorized into Hysteresis losses and Eddy current losses.

Mechanical Losses: 

• The losses associated with mechanical friction of the machine are called mechanical losses.
• These losses occur due to friction in the moving parts of the machine like bearing, brushes etc. and windage losses.
• Windage losses occur due to the air inside the rotating coil of the machine.
• These losses are usually very small about 15% of full load loss.

Out of all these losses, copper losses are the greatest percentage of heat loss in DC machines. 

Shunt resistance = 100 Ω

\({\rm{Shunt\;Current\;}}\left( {{I_{sh}}} \right) = \frac{{240}}{{100}} = 2.4\;A\)

Armature current (I_a) = I_L - I_sh­

= 15 – 2.4 = 12.6 A

At no load, E_b = 240 – (12.6 × 0.25) = 236.86 V

No load losses = (236.85) (12.6) + (12.6)² (0.25) + (240) (2.4) = 3600 W

At load,

I_a = 150 – 2.4 = 147.6 A

P losses = (147.6)² (0.25) + 3600

= 9046.44 W

Input power = 240 × 150 = 36 kW

\(Efficiency\left( \eta \right) = \frac{{36 \times {{10}^3} - 9046.44}}{{36 \times {{10}^3}}} = 74.871\% \)

The armature resistance of the motor is

\({R_A} = \frac{{{V_{A,\;br}}}}{{{I_{A,\;br}}}} = \frac{{16.3}}{{500}} = 0.0326\;{\rm{\Omega }}\) 

Under no-load conditions, the core and mechanical losses taken together (i.e. the rotational losses) of this motor are equal to the product of the internal generated voltage E_A and the armature current I_A, since this is no output power from the motor at no-load conditions.

Therefore, the rotational losses at rated speed can be found as

E_A = V_A - V_br - I_A R_A

= 442 – 2 – (23.1) (0.0326) = 439.25 V

P_rot = E_AI_A = (439.25) (23.1) = 10.15 kW

The output from the motor at full load,

P_out = P_in - P_cu - P_rot - P_br - P_stray

\({P_{cu}} = I_A^2{R_A} + {I_F}{V_F}\) 

= (560)² (0.0326) + (440) (8.86) = 14.1 kW

P_br = V_br I_A = (2) (560) = 1.12 kW

P_in = V_T I_L = 440 × 560 = 246.4 kW

P_out = 264.4 – 14.1 – 10.15 – 1.12 = 2.46

= 218.6 kW

The induced emf, E_b = V_t - I_a R_a = 240 - (100) (0.1) = 230 V

Output power at full load = 30 hp = 30 × 746 = 22380 W

\({\rm{The\;load\;current}} = {{\rm{I}}_{\rm{L}}} = {{\rm{I}}_{\rm{a}}} + {{\rm{I}}_{{\rm{sh}}}} = 100 + \frac{{240}}{{200}} = 101.2{\rm{A}}\)

The input Power P_in = V_LI_L = 240 × 101.2 = 24288 W

\({\rm{Efficiency}}\left( {\rm{\eta }} \right) = \frac{{{\rm{output\;power}}}}{{{\rm{input\;power}}}} = \frac{{22380}}{{24288}} \times 100 = 92.14\)

For no load,

r_a = 1.2 Ω, V = 100, I_a = 2 A, N = 1200 rpm

No load losses = E × I

= (100 – 2 × 1.2) 2

= 195.2 W

At load conditions,

Variable loss = I²r_a = 5² × 1.2

= 30 W

Total loss = 225.2

Total power = 100 × 5 = 500

Efficiency, \(\eta = \frac{{500 - 225.2}}{{500}}\)

= 54.96%

Explanation:

Direct Testing of DC Machines

Definition: Direct testing of a DC machine involves operating the machine under actual working conditions and measuring its performance directly. This method provides accurate results as the machine is tested in real-time under actual load conditions. Among the various testing methods available for DC machines, the Brake Test is one of the most common direct testing methods.

Correct Option Analysis:

The correct option is:

Option 4: Brake Test

The Brake Test is a direct testing method where the DC machine is tested under load conditions, and its performance is evaluated. In this method, the machine is loaded mechanically using a brake drum, and the torque produced by the machine is measured. The output power is calculated based on the torque and speed. The input power is measured using electrical instruments to evaluate the efficiency of the machine.

Working Principle of Brake Test:

• A belt or rope is wound around the brake drum, which is mounted on the shaft of the DC machine.
• The ends of the belt or rope are connected to a spring balance and a fixed weight. The tension in the rope creates a braking torque on the drum.
• The torque produced by the machine is measured by the difference in readings of the spring balance on either side of the drum.
• The rotational speed of the shaft is measured using a tachometer.
• Electrical instruments are used to measure the input power supplied to the machine.
• The output power is calculated using the formula:

Output Power, P_out = 2 × π × N × T

• Where:
• N = Rotational speed of the shaft in revolutions per second (rps)
• T = Torque produced by the machine (measured using the spring balance)
• The efficiency of the DC machine can be calculated using the formula:

Efficiency, η = (P_out / P_in) × 100%

Advantages of Brake Test:

• Provides accurate results as the machine is tested under actual working conditions.
• Simple and straightforward method for performance evaluation.

Disadvantages of Brake Test:

• Cannot be used for large machines due to the difficulty in dissipating the heat generated during testing.
• Requires a mechanical setup, which may involve additional costs and effort.

Applications:

• Commonly used for small and medium-sized DC machines to evaluate their performance.
• Widely used in laboratories for educational and research purposes.

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: Swinburne's Test

Swinburne's Test is an indirect testing method used for DC machines. It involves measuring the no-load losses of the machine and estimating its performance under load conditions. This method is not a direct test as the machine is not tested under actual working conditions. It is suitable for machines where loading is difficult or impractical, such as large DC machines.

Option 2: Field Test

The Field Test is a specific method used for testing two identical DC machines connected back-to-back, where one machine acts as a motor and the other as a generator. This method is not a direct testing method, as it involves testing two machines simultaneously under controlled conditions.

Option 3: Hopkinson’s Test

Hopkinson’s Test, also known as the Regenerative Test, is another indirect testing method. It involves connecting two identical DC machines back-to-back, with one operating as a motor and the other as a generator. The power circulating between the machines reduces the external power requirement. This method is suitable for large machines but is not considered a direct testing method.

Conclusion:

The Brake Test is the correct answer as it is a direct testing method where the DC machine is tested under actual load conditions. While other methods like Swinburne's Test, Field Test, and Hopkinson’s Test are valuable for specific applications, they are not direct testing methods. The Brake Test provides accurate results and is widely used for evaluating the performance of small and medium-sized DC machines.

Concept

The efficiency of the motor is given by:

% η \(={P_{out}\over P_{in}}× 100\)

Calculation

Given, P_out = 10 × 735.5 = 7355 W

P_in = 500 × 18 = 9000 W

% η \(={P_{out}\over P_{in}}× 100\)

% η \(={7355\over 9000}× 100\)

% η = 81.7%

Short-circuit current:

• When two or more conductors come in contact with each other then the impedance between them is decreased to a much smaller value, which causes the flow of a very large amount of current known as short-circuit current.
• Due to the flow of a large amount of current, the energy loss will be there in the form of heat.
• The formula for the energy loss is given as I²Rt.

Where I = Current at the time of short-circuit

R = Resistance offered at the time of short-circuiting.

t = Time duration of short-circuiting.

Additional Information

Losses in DC machine:

Armature copper losses:

• Armature copper losses = I_a²R_a, where Ia is armature current and Ra is armature resistance.
• These losses are about 30% of the total full load losses.

Field copper losses:

• Field copper losses = I_sh²R_sh, where Ish is field current and Rsh is field resistance.
• These losses are about 25% theoretically, but practically it is constant.

Magnetic losses: 

• As the iron core of the armature is rotating in the magnetic field, some losses occur in the core which is called core losses or magnetic losses or iron losses.
• Normally, machines are operated with constant speed, so these losses are almost constant. These losses are categorized into Hysteresis losses and Eddy current losses.

Mechanical Losses: 

• The losses associated with mechanical friction of the machine are called mechanical losses.
• These losses occur due to friction in the moving parts of the machine like bearing, brushes etc. and windage losses.
• Windage losses occur due to the air inside the rotating coil of the machine.
• These losses are usually very small about 15% of full load loss.

Out of all these losses, copper losses are the greatest percentage of heat loss in DC machines. 

Concept:

Efficiency (η) = \({Input\;power- losses \over Input\;power}\times 100%\)%

Calculation:

Given,

Armature resistance = 0.8 Ω

No-load Current = 1.5 A

No-load power developed by motor P = E_b × I_a

E_b = 25 - (1.5 × 0.8) = 23.8 V

No-load power = 23.8 × 1.5A = 35.7 W (Constant losses)

This Power is used to overcome frictional and windage losses

Under loaded condition (at 1500 rpm)

 The loaded Current = 3.5 A

The Copper losses at this speed = 3.5²× 0.8 = 9.8 W

Total losses at 1500rpm = constant losses + copper losses

= 35.7 + 9.8 = 45.5 W

Input power (P) = 25 * 3.5 = 87.5 W

Efficiency (η) = \({87.5-45.5 \over 87.5}\times 100 = 48%\) %

Maximum efficiency will be obtained only when, variable losses are equal to constant losses.

Total losses = Input - output = 20 – 17 = 3 kW

 But total losses = copper losses + constant losses

\({I_{sh}} = \frac{V}{{{R_{sh}}}} = \frac{{220}}{{50}} = 4.4\;A\)

Output, \(V{I_L} = 20\;kW\)

\(\begin{array}{l} \Rightarrow 220{I_L} = 20 \times {10^3}\\ \Rightarrow {I_L} = 90.90\;A \end{array}\)

For shunt motor, \({I_a} = {I_L} - {I_{sh}} = 90.90 - 4.4 = 86.6\;A\)

Copper losses\( = I_a^2{R_a} = {86.6^2} \times 0.04 = 300\)

Constant losses = total losses – copper losses = 3000 – 300 = 2700 W

When the total constant loss equals to copper loss, then the maximum efficiency occurs.