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A vapour, in a certain process, while condensing at 400°C, transfers heat to water at 200°C. The resulting steam is used in a power cycle which rejects heat at 30°C. 

What is the fraction of the available energy in the heat transferred from the process vapour at 400°C that is lost due to the irreversible heat transfer at 200°C ?

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Temperature of vapour, T1 = 400 + 273 = 673 K 

Temperature of water, T2 = 200 + 273 = 473 K 

Temperature at which heat is rejected, T0 = 30 + 273 = 303 K. 

LMNP  would have been the power cycle, if there was no temperature difference between the vapour condensing and the vapour evaporating, and the area under NP would have been the unavailable energy. RTWP is the power cycle when the vapour condenses at 400°C and the water evaporates at 200°C. The unavailable energy becomes the area under PW. Therefore, the increase in unavailable energy due to irreversible heat transfer is represented by the area under NW.

Now, Q1 = T1∆s = T1′∆s′

\(\cfrac{∆s'}{∆s}\) = \(\cfrac{T_1}{T_1'}\)

W = Work done in cycle LMNP 

= (T1 – T0) ∆s ...per unit mass 

W′ = Work done in cycle RTWP 

= (T1 – T0) ∆s′ ...per unit mass 

The fraction of energy that becomes unavailable due to irreversible heat transfer

Hence the fraction of energy that becomes unavailable = 0.346 or 34.6%.

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