Question

A heat engine that receives heat from a furnace at 1200°C and rejects waste heat to a river at 20°C has a thermal efficiency of 50 percent.

Determine the second-law efficiency of this power plant.

Answer 1

second-law efficiency = 62.42 %

Step-by-step explanation:

given data

temperature T1 = 1200°C = 1473 K

temperature T2 = 20°C = 293 K

thermal efficiency η = 50 percent

solution

as we know that thermal efficiency of reversible heat engine between same temp reservoir

so here

efficiency ( reversible ) η1 = 1 -
`(T2)/(T1)` ............1

efficiency ( reversible ) η1 = 1 -
`(293)/(1473)`

so efficiency ( reversible ) η1 = 0.801

so here second-law efficiency of this power plant is

second-law efficiency =
`(thernal\ efficiency)/(0.801)`

second-law efficiency =
`(50)/(0.801)`

second-law efficiency = 62.42 %