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ArbitrLikvidat [17]

2 years ago

14

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.

1 answer:

viktelen [127]2 years ago

3 0

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 %

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