A heat engine receives 500 BTU of heat per cycle from a reservoir at 540oF and rejects heat to a sink at 40oF in a hypothetical amounts of (a) 375 BTU per cycle (b) 250 BTU per cycle and (c) 150 BTU per cycle. Which of these respective cases represent a reversible cycle, an irreversible cycle and an impossible cycle?
Calculations:
Maximum efficiency of a heat engine = efficiency of Carnot engine
Efficiency of Carnot engine η max = 1 - T2/T1
T1 = temperature of source = 540oF = 540 + 460 = 1000oR
T2 = temperature of sink = 40oF = 40 + 460 = 500oR
ηmax = 1 - 500/1000 = 0.5
(a) Heat rejected to the sink = 375 BTU
ηa = 1 - Q2/Q1 = 1 - 375/500 = 0.25
(b) Heat rejected to the sink = 250 BTU
ηb = 1 - 250/500 = 0.5
(c) Heat rejected to the sink = 150 BTU
ηc = 1 - 150/500 = 0.7
From the above calculations we can see that the case 'c' is an impossible one (since the efficiency of a heat engine can not be more than that of Carnot cycle efficiency).
Summary:
Case: a - irreversible cycle
Case: b - reversible cycle
Case: c - impossible cycle
Last Modified on: 04-Feb-2022
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