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An ideal engine operates by taking in steam from a boiler at `327^(@)C` and rejecting heat to a sink at `27^(@)C`. The engine runs at `500 rpm` and heat taken is `600 k cal` in each revolution. Calculate(i) efficiency of engine (ii) work done in each cycle (iii) heat rejected in each revolution and (iv) power output of engine. |
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Answer» Correct Answer - `50% ; 1.26xx10^(6)J ; 300 K cal; 1.05xx10^(4)kW` Here, `327^(@)C= (327+273)K= 600K`, `T_(2)=27^(@)C= (27+273)= 300K` `n=500 rpm= (500)/(60)rps, Q_(1)= 600kcal` (i) `eta= 1-(T_(2))/(T_(1))= 1-(300)/(600)=0.5= 50%` (ii) `eta=W/(Q_(1))` `:. =etaQ_(1)= 0.5xx600= 300Kcal` `= 300xx4.2xx10^(3)J=1.26xx10^(6)J` (iii) `Q_(2)=Q_(1)-W=600 kcal-300kcal` `=300kcal`, (iv) `"Power output" = ("work")/("time")= (500)/(60)xx1.26xx10^(6)J//s` `=1.05xx10^(4)KW` |
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