Carnot (car-no) The Carnot cycle can be shown to be the most efficient thermodyn

Question

Carnot (car-no) The Carnot cycle can be shown to be the most efficient thermodynamic cycle for given temperatures. How efficient would a Carnot heat engine be operating between reservoirs at 20degree C and 500degree C? This could be something like a power plant. What would the coefficient of performance be for a Carnot refrigerator operating between reservoirs at 20degree C and 35degree C? This could be something like an air conditioner If the low temperature reservoir is at 20degree C, how high would T H have to be to have a Carnot efficiency of 90%?

Explanation / Answer

a)

Tcold = 20+273 = 293K

Thot = 500+273 = 773K

efficiency = [1 - Tcold/Thot]*100% = [1 - 293/773]*100% = 62.09%

b)

COP(cooling) = Tc/(Th - Tc) = 20/(35 - 20) = 1.334

c)

efficiency = [1 - Tcold/Thot]*100% = 90%

=> [1 - 293/Th] = 0.9

=> 0.1 = 293/Th

=> Th = 2930 K = 2637 0C

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---