Question

A 45kW three-phase motor operating at a full-load efficiency of 95% has its total power drawn from the supply measured by using two-wattmeter method. Calculate the readings on the two wattmeter if thr power factor of the motor is 0,75 lagging.

Answer 1

To find the readings on the two wattmeters when measuring the power drawn by a three-phase motor using the two-wattmeter method, we follow these steps:

1. **Calculate the Input Power (P)**:
The output power (useful power, \( P_{out} \)) of the motor at full load is given as 45 kW. Since the efficiency (\( \eta \)) is 95%, we can find the input power using the formula:
\[
P_{in} = \frac{P_{out}}{\eta} = \frac{45 \text{ kW}}{0.95} = 47.368 \text{ kW}
\]
Therefore, the input power is approximately 47.37 kW.

2. **Determine the Power Factor (pf)**:
The problem states that the power factor is 0.75 lagging. For a three-phase system, the total power is related to the individual wattmeter readings by:
\[
P = V_{LL} I_{ph} \cdot \sqrt{3} \cdot \text{pf}
\]
However, we can also find the readings for the two wattmeters directly.

3. **Calculate the Wattmeter Readings**:
For a balanced three-phase load, the two wattmeter readings \( W_1 \) and \( W_2 \) can be calculated using the formula:
\[
W_1 = P \cdot \frac{1 + \cos \phi}{2}
\]
\[
W_2 = P \cdot \frac{1 - \cos \phi}{2}
\]
First, we need to calculate \( \cos \phi \):
\[
\cos \phi = 0.75
\]
Now, substituting into the equations:
\[
W_1 = 47.37 \text{ kW} \cdot \frac{1 + 0.75}{2} = 47.37 \text{ kW} \cdot \frac{1.75}{2} = 47.37 \text{ kW} \cdot 0.875 = 41.5 \text{ kW}
\]

\[
W_2 = 47.37 \text{ kW} \cdot \frac{1 - 0.75}{2} = 47.37 \text{ kW} \cdot \frac{0.25}{2} = 47.37 \text{ kW} \cdot 0.125 = 5.92 \text{ kW}
\]

4. **Final Results**:
The readings on the two wattmeters will be:
\[
W_1 \approx 41.5 \text{ kW}
\]
\[
W_2 \approx 5.92 \text{ kW}
\]

So, the readings on the two wattmeters are approximately \( W_1 = 41.5 \text{ kW} \) and \( W_2 = 5.92 \text{ kW} \).