Question

In a steam power plant, 1 MW is added in the boiler, 0.58 MW is taken out in the condenser and the pump work is 0.02 MW.

Find plant thermal efficiency. If everything could be reversed find the coefficient of performance as a refrigerator.

Answer 1

a)
`\eta = 42\,\%`, b)
`COP_(R) = 29`

Step-by-step explanation:

a) The thermal efficiency is:

`\eta = (\dot Q_(in) - \dot Q_(out))/(\dot Q_(in))* 100\,\%`

`\eta = (1\,MW-0.58\,MW)/(1\,MW) \,* 100\,\%`

`\eta = 42\,\%`

b) The coefficient of performance is:

`COP_(R) = (\dot Q_(L))/(\dot W)`

`COP_(R) = (0.58\,MW)/(0.02\,MW)`

`COP_(R) = 29`