A 120-V motor has mechanical power output of 3.00 hp. It is 94.0% efficient in converting power...

Question:

A 120-V motor has a mechanical power output of 3.00 hp. It is 94.0% efficient in converting power that it takes in by electrical transmission into mechanical power. Find the energy delivered to the motor by electrical transmission in 2.00 h of operation.

Energy delivered to the motor

If the electrical power input to the motor is {eq}\rm P_i {/eq} and the mechanucal power output of the motor is {eq}\rm P_o {/eq}, then the motor efficiency is given by:

{eq}\rm \eta =\dfrac{P_o}{P_i} {/eq}

Answer and Explanation:

Since the motor is 94% efficient, then the electrical power input to the motor is:

{eq}\rm P_i = \dfrac{P_o}{0.94} {/eq}

{eq}\rm P_i = \dfrac{3 \ hp}{0.94} {/eq}

{eq}\rm P_i = 3.2 \ hp {/eq}

We know 1 hp = 746 W

Hence {eq}\rm P_i = 3.2 \times 746 = 2387.2 \ W = 2387.2 \ J/s {/eq}

Hence electrical energy delivered to the motor in t = 2 hrs:

{eq}\rm E = P_i \times t {/eq}

{eq}\rm E = 2387.2 \times 2 \times 3600 = 17.2 \times 10^6 \ J {/eq}(Because 1 hr = 3600 s)


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