You are part of a team that is completing a mechanical-engineering project. Your

Question

You are part of a team that is completing a mechanical-engineering project. Your team built a steam engine that takes in superheated steam at 270 ºC and discharges condensed steam from its cylinder at 50 ºC. Your team has measured its efficiency to be 30.0%.

(a) How does this efficiency compare with the maximum possible efficiency for your engine?
esteam engine/emax = ______ %



(b) If the useful power output of the engine is known to be 200 kW, how much heat does the engine release to its surroundings in 1.00 h?
______ GJ

Explanation / Answer

a) emax = 1-(Tlow/Thigh)

Tlow = 50oC = 50 + 273.15 K = 323.15K

Thigh = 270oC = 270 + 273.15 K = 543.15K

Tlow / Thigh = 0.595 = 59.5%

Therefore emax = 1 - 0.595 = 0.405

esteam engine/emax = 0.3/0.405 = 0.7407 = 74.07%

b)Given useful power = 200kW

=> dW/dt = 200 KW

=>dW = 200*103*(1) watt-hour = 2*105*60*60 Joules = 72*107 Joules

But dW/dQ = 0.3

=>dQ = dW/0.3 = (72*107)/(0.3) J= (72*107*10)/(0.3*10) J= (72*108)/(3) J= 24*108 J= 2.4*109 J= 2.4GJ

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