As shown in the diagram, two reversible cycles arranged in series both produce t

ID: 1862398 • Letter: A

Question

As shown in the diagram, two reversible cycles arranged in

series both produce the same net work, Wout. The first cycle

receives energy QH by heat transfer from a hot reservoir at

1000 R and rejects energy Q by heat transfer to a reservoir at an

intermediate temperature, T. The second cycle receives energy

Q by heat transfer from the reservoir at temperature T and

rejects energy QL by heat transfer to a cold reservoir at 400 R.

Determine the following:

(a) the intermediate temperature T (R) and the thermal

efficiency (%) for each of the two power cycles.

(b) the thermal efficiency (%) of a single reversible power

cycle operating between the same hot and cold

reservoirs at 1000 R and 400 R, respectively.

in terms of the net work produced by each of the two

cycles (i.e., Wout) in the original problem.

QH = Th*dS

Q = T*dS

QL = Tl*dS

QH = Q + Wout-------1

Q = QL + Wout----------2

Therefore

QH - Q = Q - QL

2Q = QH +QL

2*T = Th+Tl

T = 700 R = 388.89 K

Th = 1000*5/9 = 555.56 K

Tl = 222.22 K

Efficiency of cycle 1 = 1-(T/Th) = 1-(388.89/555.56) = 0.3 = 30%

Efficiency of cycle 2 = 1-(Tl/T) = 1- (222.22/388.89) = 0.4385 = 43.85 %

Efficiency of single cycle = 1-(Tl/Th) = 1-(222.22/555.56) = 0.6 = 60%

QH = QL + W

W = QH-QL = (Th-Tl)*dS

Adding eq 1 and 2

QH - QL = 2*Wout

W = 2*Wout

Work produced by single cycle is twice the work produced by individual cycle

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