Question

A steam power plant receives heat from a furnace at a rate of 280 GJ/h. (that is

‘giga’ joules.) Heat losses to the surrounding air from the steam as it passes through the pipes and other components are estimated to be about 8 GJ/h. If the waste heat is transferred to the cooling water at a rate of 145 GJ/h, determine (a) the net power output and (b) the thermal efficiency of this power plant.

Answer 1

a)
`\dot W_(out) = 35.278\,MW`, b)
`\eta_(th) = 45.357\,\%`

Step-by-step explanation:

a) The net power output is:

`\dot W_(out) = \dot Q_(in) - \dot Q_(out) - \dot Q_(loss)`

`\dot W_(out) = \left(280\,(GJ)/(h) - 145\,(GJ)/(h) - 8\,(GJ)/(h)\right)\cdot \left((1000\,MJ)/(1\,GJ)\right)\cdot \left((1\,h)/(3600\,s) \right)`

`\dot W_(out) = 35.278\,MW`

b) The thermal efficiency of the power plant is:

`\eta_(th) = (\dot Q_(in)-\dot Q_(out) - \dot Q_(loss))/(\dot Q_(in))* 100\,\%`

`\eta_(th) = (280\,(GJ)/(h)-145\,(GJ)/(h)-8\,(GJ)/(h) )/(280\,(GJ)/(h) ) * 100\,\%`

`\eta_(th) = 45.357\,\%`