Thermodynamics Of Coupled Cycles

If two cycles are coupled in series where heat lost by one is absorbed by the other, as in the mercury-steam binary cycle, let n1 and n2 be the efficiencies of the topping and bottom cycles respectively, and N be the overall efficiency of the combined cycle.

N1 = 1 - (Q2 /Q1) and N2 = 1 - (Q3 / Q2)

or Q2 = Q1(1-N1) and Q3 = Q2 (1-N2)

Now N= 1-(Q3 / Q1) = 1 - (Q2(1-N2) / Q1)

= 1- (Q(1-N1)(1-N2) / Q1)

= 1 - (1-N1) (1-N2)

If There are N cycles coupled in series, the overall efficiency would be given by

N = 1- ((1-N1)(1-N2)(1-N3)........(1-Nn))

1-N = (1-N1)(1-N2)(1-N3).........(1-Nn)

Total loss = Product of losses in all the cycles.

For two cycles coupled in series.

N = 1 - (1-N1) (1-N2)

= 1 - (1-N1-N2 +N1N2)

= N1 + N2 - N1N2

N = N1 + N2 - N1N2

This shows that the overall efficiency of two cycles coupled in series equals the sum of the individual efficiencies minus their product.

By combining two cycles in series, even if the individual efficiencies are low, it is possible to have fairly high combined efficiency , which cannot be attained by a single cycle.

For example ,

N1 = 0.50 and N2 = 0.40

N = 0.5 + 0.4 - 0.5(0.4) = 0.70

It is almost impossible to achieve such an inefficiency in a single cycle.

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