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

The thermal efficiency is 43%.

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

Step-by-step 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=(out-put)/(in-put)`

Put the value into the formula

`\eta=(0.3)/(0.703)`

`\eta=0.43*100`

`\eta=43\%`

The thermal efficiency is 43%.

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

Using formula of energy

`E=0.43*0.703*60*60*365`

`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.