1 Answer

Optiq · Answer 1 · 2020-03-22T19:16:22+0000

Answer:

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

Step-by-step 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

$\eta=(Work\ Output)/(Work\ input)$

$Work\ Input =(Work\ output)/(\eta)$

$Work\ Input =(75*746)/(0.91)$

$Work\ Input =61483.51\ W$

$Work\ Input = 61.48 kW$

We need to calculate the electric input

For, heigh efficiency

$Work\ Input_(inc) =(75*746)/(0.954)$

$Work\ Input_(int) =58647.7\ W$

$Work\ Input_(int) = 58.64\ kW$

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.

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