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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. None

11 months ago Give Answer
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Answer:second-law efficiency  = 62.42 %Explanation:given data temperature T1 = 1200°C = 1473 Ktemperature T2 = 20°C  =  293 K thermal efficiency η = 50 percentsolutionas we know that thermal efficiency of reversible heat engine between same  temp reservoir so here efficiency ( reversible ) η1 = 1 - 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.      ............1efficiency ( reversible ) η1  = 1 - 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.   so efficiency ( reversible ) η1  = 0.801so here second-law efficiency of this power plant is second-law efficiency = 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. second-law efficiency = 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.  second-law efficiency  = 62.42 %

Answerd by rbwatson
11 months ago 10 5.0