To solve the problem, we will find the efficiency of the given three-phase induction motor based on the provided specifications.
Step 1: Calculate the input power (P_input)
For a three-phase motor, the input power can be calculated using the following formula:
\[ P_{\text{input}} = \sqrt{3} \cdot V_L \cdot I_L \cdot PF \]
Where:
- \(V_L\) = Line voltage = 220 V
- \(I_L\) = Line current = 77 A
- \(PF\) = Power factor = 0.88
Now, calculate \(P_{\text{input}}\):
\[ P_{\text{input}} = \sqrt{3} \cdot 220 \cdot 77 \cdot 0.88 \]
Calculating \( \sqrt{3} \):
\[ \sqrt{3} \approx 1.732 \]
Now substituting it back into the equation:
\[ P_{\text{input}} = 1.732 \cdot 220 \cdot 77 \cdot 0.88 \]
Calculating that step by step:
- Calculate \(1.732 \cdot 220 \approx 381.04\)
- Calculate \(381.04 \cdot 77 \approx 29320.08\)
- \(29320.08 \cdot 0.88 \approx 25745.27\) W
Thus,
\[ P_{\text{input}} \approx 25745.27 \text{ W} \quad \text{(or 25.75 kW)} \]
Step 2: Determine the output power (P_output)
The output power of the motor can be calculated from its rated horsepower (hp). The conversion from horsepower to watts is:
\[ 1 \text{ hp} = 746 \text{ W} \]
Thus, for a 30 hp motor:
\[ P_{\text{output}} = 30 \cdot 746 = 22380 \text{ W} \quad \text{(or 22.38 kW)} \]
Step 3: Calculate the efficiency (η)
The efficiency (\( \eta \)) of the motor can be calculated using the following formula:
\[ \eta = \frac{P_{\text{output}}}{P_{\text{input}}} \cdot 100% \]
Substituting the values we have:
\[ \eta = \frac{22380}{25745.27} \cdot 100% \]
Calculating:
\[ \eta \approx 0.868 \cdot 100% \approx 86.8% \]
Conclusion
The efficiency of the three-phase, four-pole, 30-hp, 220-V, Y-connected induction motor, under the given operating conditions, is approximately 86.8%.