Horsepower 81496

The electric motor has an efficiency of 80% (power and input ratio). What is the power input of this motor if its power is 5 horsepower (HP)? Enter the result in kilowatts (kW) and round to one decimal place (1 HP = 3/4 kW).

Correct answer:

P =  4.7 kW

Step-by-step explanation:

P1=5 HP P2=P1 34=5 34=5 34=154=3.75 kW η=80%=80100=45=0.8  η   P = P2  P=P2/η=3.75/0.8=4.7 kWP_{1} = 5 \ \text{HP} \ \\ P_{2} = P_{1} \cdot \ \dfrac{ 3 }{ 4 } = 5 \cdot \ \dfrac{ 3 }{ 4 } = \dfrac{ 5 \cdot \ 3 }{ 4 } = \dfrac{ 15 }{ 4 } = 3.75 \ \text{kW} \ \\ η = 80 \% = \dfrac{ 80 }{ 100 } = \dfrac{ 4 }{ 5 } = 0.8 \ \\ \ \\ η\ \cdot \ \ P\ = \ P_{2} \ \\ \ \\ P = P_{2} / η = 3.75 / 0.8 = 4.7 \ \text{kW}



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