Question

A single phase 150 V electric motor absorbs 8.0 A while running at 1500 rev/min and developing 2.8 N-m of torque. The phase angle between voltage and current is \(60^\circ\). What is power efficiency of the motor?

• A. 53%
• B. 63%
• C. 73%
• D. 83%

Hint:

To make calculations faster during exams, use the approximation \(\pi \approx \frac{22}{7}\):
\[ \omega = \frac{2 \times \frac{22}{7} \times 1500}{60} = \frac{1100}{7} \text{ rad/s} \] \[ P_{\text{out}} = 2.8 \times \frac{1100}{7} = 0.4 \times 1100 = 440 \text{ W} \] \[ \text{Efficiency} = \frac{440}{600} \approx 73.33% \] This simple approximation avoids complex decimal division.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The power efficiency (\(\eta\)) of any electric motor is the ratio of the mechanical power output to the electrical power input, expressed as a percentage:
\[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100% \] Key Formula or Approach:
1. Electrical Power Input (\(P_{\text{in}}\)) for a single-phase AC motor is:
\[ P_{\text{in}} = V \cdot I \cdot \cos\theta \] Where \(\cos\theta\) is the power factor, and \(\theta\) is the phase angle.
2. Mechanical Power Output (\(P_{\text{out}}\)) is:
\[ P_{\text{out}} = T \cdot \omega = T \cdot \left( \frac{2\pi N}{60} \right) \] Where \(T\) is torque and \(N\) is rotational speed in rev/min.

Step 2: Detailed Explanation:
Let us calculate the input and output power values step-by-step:
1. Calculate \(P_{\text{in}}\):
Given: - \(V = 150 \text{ V}\)
- \(I = 8.0 \text{ A}\)
- \(\theta = 60^\circ \implies \cos(60^\circ) = 0.5\)
\[ P_{\text{in}} = 150 \times 8.0 \times 0.5 = 600 \text{ W} \] 2. Calculate \(P_{\text{out}}\):
Given: - \(T = 2.8 \text{ N-m}\)
- \(N = 1500 \text{ rev/min}\)
\[ \omega = \frac{2 \cdot \pi \cdot 1500}{60} = 50\pi \approx 50 \times 3.14159 = 157.08 \text{ rad/s} \] Now, calculate the mechanical power output:
\[ P_{\text{out}} = 2.8 \times 157.08 \approx 439.82 \text{ W} \] 3. Calculate Efficiency (\(\eta\)):
\[ \eta = \frac{439.82 \text{ W}}{600 \text{ W}} \times 100% \approx 73.3% \] The power efficiency is approximately 73%.

Step 3: Final Answer:
The correct option is (C).