Question

A 75-hp (shaft output) motor that has an efficiency of 91% is worn out and is replaced by a high efficiency 75-hp motor that has an efficiency of 95.4%. Determine the reduction in the heat gain of the room due to higher efficiency under full-load conditions.

Answer 1

Answer:

The heat gain of the room due to higher efficiency is 2.84 kW.

Explanation:

Given that,

Output power of shaft = 75 hp

Efficiency = 91%

High efficiency = 95.4%

We need to calculate the electric input given to motor

Using formula of efficiency

$\eta=\dfrac{Work\ Output}{Work\ input}$

$Work\ Input =\dfrac{Work\ output}{\eta}$

$Work\ Input =\dfrac{75\times746}{0.91}$

$Work\ Input =61483.51\ W$

$Work\ Input = 61.48 kW$

We need to calculate the electric input

For, heigh efficiency

$Work\ Input_{inc} =\dfrac{75\times746}{0.954}$

$Work\ Input_{int} =58647.7\ W$

$Work\ Input_{int} = 58.64\ kW$

The reduction of the heat gain of the room due to higher efficiency is

$Q=Work\ Input-Work\ Input_{int}$

Put the value into the formula

$Q=61.48 -58.64$

$Q=2.84\ kW$

Hence, The heat gain of the room due to higher efficiency is 2.84 kW.