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Making use of the Clausius inequality, demonstrate that all cycles having the same maximum temperature T_(max) and the same minimum temperature T_(min) are less efficient compared to the Carnot cycle with the same T_(max) and T_(min). |
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Answer» Solution :We WRITE Claussius inequality in the form `int(₫_1 Q)/(T) - int (₫ Q)/(T) le 0` where `₫ Q` is the HEAT transferred to the system but `₫_2 Q` is heat rejected by the system, both are `+ve` and this axplains the minus sign before `₫_2 Q`, In this inequality `T_(max) gt T gt T_(min)` and we can wire `int (₫ _1 Q)/(T_(max)) - int (₫ _2 Q)/(T_(min)) lt 0` Thus `(Q_1)/(T_(max)) lt (Q'_2)/(T_(min))` or `(T_(min))/(T_(max)) lt (Q'_2)/(Q_1)` `eta = 1 -(Q'_2)/(Q_1) lt 1 -(T_(min))/(T_(max)) = eta_(carnol)`. |
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