Question

A system undergoing a power cycle develops a steady power output of 0.3 kW while receiving energy input by heat transfer at a rate of 2400 Btu/h. Determine the thermal efficiency and the total amount of energy developed by work, kW · h, for 1 full year of operation.

Answer 1

Answer:

The thermal efficiency is 43%.

The total amount of energy developed by work is 397 kW in one year.

Explanation:

Given that,

Power output = 0.3 kW

Heat energy = 2400 Btu/h = 0.703 kW

We need to calculate the thermal efficiency

Using formula of efficiency

$\eta=\dfrac{out-put}{in-put}$

Put the value into the formula

$\eta=\dfrac{0.3}{0.703}$

$\eta=0.43\times100$

$\eta=43\%$

The thermal efficiency is 43%.

(b). We need to calculate the total amount of energy

Using formula of energy

$E=0.43\times0.703\times60\times60\times365$

$E=397209.06\ J$

$E=397\ kW$

Hence, The thermal efficiency is 43%.

The total amount of energy developed by work is 397 kW in one year.