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ArbitrLikvidat [17]
3 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.
Engineering
1 answer:
viktelen [127]3 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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