Question
Question: A 3HP motor requires 2.4 kW to drive it. Its efficiency is about: A. 90% B. 75% C. 60% D. 50...
A 3HP motor requires 2.4 kW to drive it. Its efficiency is about:
A. 90%
B. 75%
C. 60%
D. 50%
Solution
The efficiency of the electrical instruments is the ratio of output electric power to the input electric power. Use 1hp=746W to convert the output power into watts. To determine the percentage efficiency, multiply the final answer by 100.
Formula used:
η=PiPo
Here, Po is the output power and Pi is the input power.
Complete step by step answer:
We know that the efficiency of any electrical instrument is the ratio of output power to input power.
η=PiPo
Here, Po is the output power and Pi is the input power.
We have given that the output power of the motor is 3hp. We also know that 1hp=746W.
Therefore, the output of the electrical motor is,
3×746W=2238W
We have given that the input power to drive the 3hp motor is 2.4 kW. Therefore, we can calculate the efficiency of the motor by substituting 2238W for Po and 2.4 kW for Pi in the above equation for efficiency.
η=2.4kW2238W=2.4×103W2238W
⇒η=0.93
We have to calculate the percentage efficiency of the motor. Therefore, multiply by 100 to the above quantity.
η=0.93×100
⇒η=93%≈90%
So, the correct answer is “Option A”.
Additional Information:
You may wonder if the input of the motor is 2.4 kW and output is 2.238 kW, then where 0.16 kW of input power has gone missing. This is the loss in the input power due to heating of the parts of the electrical motor. For the motors with higher efficiency, this loss is negligible.
Note:
Note that to determine the efficiency of the electrical instruments, both input power and output power must have the same unit. The efficiency is the dimensionless quantity. Hp is a mechanical unit of power used in large industries and it is an abbreviation of horse power given by James Watt. Watt is the unit of power for small electrical components like electric bulbs.