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

Answer:

second-law efficiency  = 62.42 %

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 - $\frac{T2}{T1}$      ............1

efficiency ( reversible ) η1  = 1 - $\frac{293}{1473}$  

so efficiency ( reversible ) η1  = 0.801

so here second-law efficiency of this power plant is

second-law efficiency = $\frac{thernal\ efficiency}{0.801}$

second-law efficiency = $\frac{50}{0.801}$  

second-law efficiency  = 62.42 %