InterviewSolution
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InterviewSolution
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Form the given T-S diagram of a reversible carnot engine, find (i) work delivered by engine in one cycle (ii) heat taken from the source in each cycle . (iii) `DeltaS_(sink)` in each cycle . |
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Answer» (i) `W_(AB) = - nRT ln.(V_(2))/(V_(1))` `DeltaS =(q_(rev))/(T) = - (W_(AB))/(T)` `rArr - W_(AB) = TDeltaS = 600 xx100` `" "- W_(BC) = -.^(n)C_(v)(T_(1) - T_(2))` `" "- W_(CD) = TDeltaS = 300 xx(-100)` `" "- W_(DA)= - .^(n)C_(v) (T_(2) - T_(1))` net work delivered during one cycle `= - W_(AB) - W_(BC) - W_(CD) - W_(DA) = 300 xx 100 = 30 kJ` Note: Net work done = area of the rectangle (ii) `(W_("net"))/(q) = eta" "and " " eta= (600-300)/(600) =(1)/(2)` `rArr q=` "heta taken from the the source" `(-W_("net"))/(1//2) = (30kJ)/( 1//2) = + 60 kJ` (iii) `DeltaS_("sink") = - (Q_("sink"))/(T) " "also" "(q_("source") + q_("sink"))= 30` `" "q_("source") = 600 " " rArr " "q_("sink") = - 30 kJ` `rArr " "DeltaS_("sink") = (-q_("sin K"))/(T) = (_(-30000J))/(100) = 100 J//K` |
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