Power Stages of Induction Motor MCQ Quiz - Objective Question with Answer for Power Stages of Induction Motor - Download Free PDF

Explanation:

Losses in a 150 kW Electric Motor with 92% Efficiency

Given Data:

• Power Output of the Motor (P_out) = 150 kW
• Efficiency (η) = 92% = 0.92

Formula for Efficiency:

The efficiency of an electric motor is defined as the ratio of the output power to the input power:

η = (P_out ÷ P_in) × 100

Where:

• P_out = Output power (power delivered by the motor)
• P_in = Input power (power required to operate the motor)
• η = Efficiency (expressed as a percentage)

Rearranging the formula to calculate input power:

P_in = P_out ÷ η

Substitute the values:

P_in = 150 ÷ 0.92

P_in ≈ 163.04 kW

Calculating the Losses:

The losses in the motor are the difference between the input power and the output power:

Losses = P_in - P_out

Substitute the values:

Losses = 163.04 - 150

Losses ≈ 13.04 kW

Final Answer:

The losses in the machine are approximately 13 kW, which corresponds to Option 4.

Additional Information

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

Option 1: 92 kW

This option is incorrect because 92 kW does not represent the losses in the machine. Instead, it might be confused with the efficiency percentage (92%), which is unrelated to the actual loss calculation.

Option 2: 150 kW

This option is incorrect because 150 kW represents the output power of the motor, not the losses. The losses are calculated by subtracting the output power from the input power.

Option 3: 163 kW

This option is incorrect because 163 kW represents the input power required to operate the motor at full load, not the losses. Losses are the difference between the input power and the output power.

Option 4: 13 kW

This option is correct because it accurately represents the losses in the machine, calculated as the difference between the input power (163 kW) and the output power (150 kW).

Conclusion:

Understanding the efficiency formula and the relationship between input power, output power, and losses is critical in analyzing the performance of electric motors. In this case, the losses are calculated to be approximately 13 kW, which corresponds to Option 4. This calculation is essential in assessing the energy performance of machines and identifying areas for improvement in their operation.

Explanation:

Calculation of Number of Poles:

To determine the number of poles (P) for a three-phase induction motor, we use the formula that relates the synchronous speed (Ns) in revolutions per minute (rpm) to the frequency (f) of the supply in hertz (Hz) and the number of poles.

The formula is:

Ns = (120 * f) / P

Where:

• Ns = Synchronous speed in rpm
• f = Frequency of the supply in Hz
• P = Number of poles

Given that the supply frequency (f) is 50 Hz and the motor is running at 1460 rpm, we need to find the synchronous speed (Ns) first. However, the 1460 rpm given is the actual speed (N) of the motor, not the synchronous speed. The actual speed is slightly less than the synchronous speed due to slip.

Slip (s) in an induction motor is given by:

s = (Ns - N) / Ns

For small slip values, we can approximate the synchronous speed (Ns) as:

Ns ≈ N / (1 - s)

However, for simplicity, we can assume a small slip value and calculate the synchronous speed directly using common values for slip in induction motors (typically around 2-5%).

Let's assume a slip of around 3% (0.03):

Ns = N / (1 - s)

Substituting the values:

Ns ≈ 1460 / (1 - 0.03) ≈ 1460 / 0.97 ≈ 1505 rpm

For practical purposes, the synchronous speed for a 50 Hz supply is typically one of the standard synchronous speeds: 3000 rpm, 1500 rpm, 1000 rpm, 750 rpm, etc.

Given that 1505 rpm is very close to 1500 rpm, we can assume the synchronous speed Ns to be 1500 rpm.

Now, using the synchronous speed formula:

1500 = (120 * 50) / P

Simplifying, we get:

P = (120 * 50) / 1500

P = 6000 / 1500

P = 4

Therefore, the number of poles of the motor is 4.

The correct answer is option 2.

Additional Information

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

Option 1 (22 poles):

If the motor had 22 poles, the synchronous speed Ns would be calculated as follows:

Ns = (120 * 50) / 22

Ns ≈ 272.7 rpm

This is far from the given running speed of 1460 rpm, so this option is incorrect.

Option 3 (6 poles):

If the motor had 6 poles, the synchronous speed Ns would be:

Ns = (120 * 50) / 6

Ns = 1000 rpm

Since the actual speed (1460 rpm) is much higher than 1000 rpm, this option is incorrect.

Option 4 (8 poles):

If the motor had 8 poles, the synchronous speed Ns would be:

Ns = (120 * 50) / 8

Ns = 750 rpm

The actual speed (1460 rpm) is much higher than 750 rpm, so this option is also incorrect.

Conclusion:

By understanding the relationship between the synchronous speed, frequency, and the number of poles, we can accurately determine the number of poles in a three-phase induction motor. In this case, the correct number of poles is 4, as determined from the given speed and frequency. This type of motor is common in industrial applications where a balance between speed and torque is required.

Explanation:

Mechanical Power Developed in a 3-Phase Induction Motor

Problem Statement: The power input to a 3-phase induction motor is 50 kW. Total losses in the stator are 2 kW. Find the mechanical power developed if the motor is running with a slip of 5%.

Solution:

To find the mechanical power developed by the 3-phase induction motor, we need to follow these steps:

1. Calculate the Power Input to the Rotor (P_r):

The power input to the rotor can be found by subtracting the stator losses from the total power input to the motor.

Given:

• Total Power Input (P_in) = 50 kW
• Total Stator Losses (P_stator) = 2 kW

Power Input to the Rotor (P_r) = P_in - P_stator

P_r = 50 kW - 2 kW = 48 kW

2. Calculate the Rotor Copper Losses (P_copper):

Rotor copper losses are proportional to the slip of the motor. The formula to calculate rotor copper losses is:

P_copper = Slip × Power Input to the Rotor (P_r)

Given:

• Slip (s) = 5% = 0.05

P_copper = 0.05 × 48 kW = 2.4 kW

3. Calculate the Mechanical Power Developed (P_mech):

The mechanical power developed by the motor can be found by subtracting the rotor copper losses from the power input to the rotor.

P_mech = P_r - P_copper

P_mech = 48 kW - 2.4 kW = 45.6 kW

Therefore, the mechanical power developed by the motor is 45.6 kW.

Correct Option:

The correct option is:

Option 4: 45.6 kW

Additional Information:

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

Option 1: 41.3 kW

This option is incorrect because it does not account for the proper calculation of rotor copper losses and mechanical power developed. The mechanical power developed is higher than this value.

Option 2: 43.5 kW

This option is also incorrect. The calculation of mechanical power developed results in a higher value than this option.

Option 3: 47.2 kW

This option is incorrect because it overestimates the mechanical power developed. The correct calculation shows that the mechanical power developed is lower than this value.

Conclusion:

Understanding the calculation of mechanical power developed in a 3-phase induction motor involves accounting for the stator losses, rotor copper losses, and the slip of the motor. By following the proper steps and formulas, we can accurately determine the mechanical power developed, which is essential for evaluating the performance and efficiency of the motor.

Power flow in Induction Motor

\(P_g:P_c:P_d=1:s:1-s\)

where, P_g = Air gap Power

P_c = Rotor copper loss

P_d = Developed Power 

Calculation

Given, Supply Voltage (V) = 480V (line-to-line)

Frequency (f) = 60Hz

Horsepower = 50 hp

Current (I) = 60 A

Power Factor = 0.85(lagging)

Stator Copper Losses = 2 kW

Rotor Copper Losses = 700 W

Friction and Windage Losses = 600 W

Stator Iron Losses = 1800 W

The input power for a three-phase system is given by:

\(P_{in}=\sqrt{3}V_LI_Lcos\phi\)

\(P_{in}=\sqrt{3}\times 480\times 60\times 0.85=42.4\space kW\)

Air-gap power (P_g) = P_in - Stator Copper Losses - Stator Iron Losses

Air-gap power (Pg) = 42.4 - 2 - 1.8

Air-gap power (Pg) = 38.6 kW

Power flow in the synchronous motor
 

Different types of losses:

1.) Copper losses

• The various windings of the synchronous machine such as armature and field winding are made of copper and have some resistance. When current flows through them, there will be power loss proportional to the square of their respective currents. These power losses are called copper losses.
• Copper losses depend on the load because they are proportional to the square of the current, which increases with load.
   

2.) Core losses

• The losses that occur in the iron parts of the motor are called core losses or magnetic losses.
• Core losses are generally constant and depend on voltage and frequency rather than load. They do not typically increase with the square of the load.
• These losses consist of the following: Hysteresis loss: To minimize this loss, the armature core is made of silicon steel which has a low hysteresis constant.
• Eddy's current loss: To minimize this loss, the armature core is laminated into sheets(0.3-0.5mm).
   

3.) Mechanical losses

• As the field system of a synchronous machine is a rotating part, some power is required to overcome Friction and windage losses.
• The mechanical loss which is the summation of the friction and windage losses is related to the cube of the operating speed.

Power Stages in an Induction Motor:

Stator iron loss (eddy current loss and hysteresis losses) considered as constant loss and it depends on the supply frequency and magnetic flux density in the iron core.

The iron loss of the rotor is negligible because the frequency of rotor currents under normal running conditions is always small.

\( {{P}_{2}}:{{P}_{cu}}:{{P}_{m}}=1:s:\left( 1-s \right)\)

Where,

s is the slip of motor and Pcu is rotor copper loss.

Calculation:

Given,

s = 4% = 0.04

P₂ = 1000 W

From above concept,

P₂ : P_m = 1 : (1 - s)

1000 : P_m = 1 : (1 - 0.04)

1000 : P_m = 1 : 0.96

\(\frac{1000}{P_m}=\frac{1}{0.96}\)

P_m = 1000 × 0.96 = 960 W

Concept:

Power flow in the Induction motor is as shown below.

Slip, \({{s}}=\frac{N_{s}-{{N}_{r}}}{N_s}\)

Stator input power = P

Rotor input or air gap power \({{P}_{in}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}\) = P - Stator loss

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,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

The difference between stator output and rotor gross output is Rotor Cu losses.

Concept:

Power Stage in Induction Motor:

The power stages diagram is an important tool for power flow analysis in any machine.

For an induction motor, the power stages diagram is given below

Calculation:

P_in = √3 VI cos ϕ

P_in = √3 × 400 × 50 × 0.8

= 27.71 kW

P (air gap) = P_in - stator copper loss - stator core loss

P (air gap) = 27.71 - 1.5 - 1.2

P = 25.01 kW

Concept:

Power flow in the Induction motor is as shown below.

Rotor input or air gap power, \({{P}_{ag}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}\)

Rotor copper losses, \({{P}_{cu}}=s\times {{P}_{ag}}=3I_{2}^{2}{{R}_{2}}\)

Mechanical power developed or shaft power output is\({{P}_{o}}={{P}_{ag}}-{{P}_{cu}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}-3I_{2}^{2}{{R}_{2}}=3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

Now mechanical power developed (P_o) in terms of the air gap (Pag) will be (1-s) Pag

∴ P_o = (1 - s) Pag

Where, s = slip

Additional Information

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

\({{P}_{ag}}:{{P}_{cu}}:{{P}_{o}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{ag}}:{{P}_{cu}}:{{P}_{o}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Concept:

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 = P_in × s = \(3I_{2}^{2}{{R}_{2}}\)

Gross mechanical power output P_d = P_in - P_cu

Application:

Given: P_in = 2000 W, s = 5 % = 0.05

P_cu = P_in × s = 2000 × 0.05 = 100 W

P_d = P_in - P_cu = 2000 - 100 = 1900 W

Important Points

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

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\dfrac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \dfrac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\dfrac{1}{s}:1:\left( \dfrac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Concept:

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,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Calculation:

Given that, number of poles (P) = 6

Frequency (f) = 50 Hz

Rotor speed (N_r) = 950 rpm

Synchronous speed \( = \frac{{120 \times 50}}{6} = 1000\;rpm\)

Slip, \(s = \frac{{1000 - 950}}{{1000}} = 0.05\)

Copper losses = 5 kW

Rotor input = 5/0.05 = 100 kW

Concept:

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,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

Calculation:

Frequency of rotor emf (f_r) = 100 oscillations/minute

= 100/60 oscillations/second

= 1.66 Hz

Supply frequency (f) = 50 Hz

Slip (s) = f_r/f = 1.66/50 = 0.033

Motor input = 50 kW

Stator losses = 2 kW

Rotor input = stator output = 50 – 2 = 48 kW

Rotor copper losses = 0.033 × 48 = 1.6 kW

Concept:

Power Stages in an Induction Motor:

Stator iron loss (eddy current loss and hysteresis losses) considered as constant loss and it depends on the supply frequency and magnetic flux density in the iron core.

The iron loss of the rotor is negligible because the frequency of rotor currents under normal running conditions is always small.

\( {{P}_{2}}:{{P}_{cu}}:{{P}_{m}}=1:s:\left( 1-s \right)\)

Where,

s is the slip of motor and P_cu is rotor copper loss.

Calculation:

Given,

Stator input (P₁) = 40 kW

Stator loss = 1 kW

Rotor input (P₂) = 40 kW - 1 kW = 39 kW

f = 50 Hz

P = 6

\({N_s} = \frac{{120f}}{P} = \frac{{120 × 50}}{6} = 1000\;RPM\)

N_r = 975 RPM

\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}} = \frac{{1000 - 975}}{{1000}} = 0.025\)

∴ P_m = (1 - s) × P₂ = (1 - 0.025) × 39 = 38.025 kW

 Total friction and windage losses = 2.025 kW

From power flow diagram,

Rotor output (P_o) = (P_m) - (windage friction loss)

P_o = 38.025 kW - 2.025 kW = 36 kW

\(\eta = \frac{{{P_o}}}{{{P_1}}} = \frac{{36}}{{40}} = 0.9\)

Hence, the efficiency of the motor is 90%.

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,

\({{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{3I_{2}^{2}{{R}_{2}}}{s}:3I_{2}^{2}{{R}_{2}}:3I_{2}^{2}{{R}_{2}}\left( \frac{1-s}{s} \right)\)

\(\Rightarrow {{P}_{in}}:{{P}_{cu}}:{{P}_{g}}=\frac{1}{s}:1:\left( \frac{1-s}{s} \right)=1:s:\left( 1-s \right)\)

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

Concept:

Power Stages in an Induction Motor:

• Stator iron loss (eddy current loss and hysteresis losses) is considered a constant loss and it depends on the supply frequency and magnetic flux density in the iron core.
• The iron loss of the rotor is negligible because the frequency of rotor currents under normal running conditions is always small.

\( {{P}_{2}}:{{P}_{cu}}:{{P}_{m}}=1:s:\left( 1-s \right)\)

Where,

s is the slip of motor and Pcu is rotor copper loss.

Calculation:

Given,

s = 0.04

P_cu = 625 W

From the above concept,

Motor output power (P_m) = \(\frac{P_{cu}}{s}\times (1-s)=\frac{625}{0.04}\times (1-0.04)=15000\ W\) = 15 kW