A reversible Carnot heat engine has 1.00 lbm of air as the working fluid. Heat i

Question

A reversible Carnot heat engine has 1.00 lbm of air as the working fluid. Heat is received at 740 degrees F and rejected at 40.0 degrees F. At the beginning of the heat addition process, the pressure is 100 psia and during this process the volume triples. Calculate the net cycle work per lbm of air. Assume the air behaves as a constant specific heat ideal gas.

How would I determine Q for this problem? I know that once I have Q, I can determine the net work based off the efficiency, but I'm not sure how to get there.

Explanation / Answer

Heat addition process is isothermal. Hence, Q=W. (Energy change is zero.)
For isothermal process 2-3, W = mRT2 ln(V3/V2)
Here V3/V2 = 3, T2 = 740 deg F = (740+460) deg R = 1200 deg R, m = 1 lbm, R = 53.35 ft-lbf/lbm-R (From Air properties).
W = 70333 ft-lbf.

Next, Efficiency = 1 - T1/T2 with T1 taken as 40 degrees F. Remember that in this formula T1 and T2 to be used are absolute temperatures (Kelvin). Otherwise the result will be wrong.
T1 = 40 deg F = 277.6 K, T2 = 740 deg F = 666.5 K.
Efficiency = 1- 277.6/666.5 = 0.583 or 58.3%.

Finally, net work = Efficiency*Q.
= 0.583*70333 = 41039 ft-lbf.
hope this helps.

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