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Figure depicts an idealized cycle of a petrol internal combustion engine. The segment 1-2 corresponds to the adiabatic compression of the combustible mixture, segment 2-3, to the isochoric combustion of fuel in the course of which the working fluid receives an amount of heat Q, segment 3-4 corresponds to the adiabatic expansion of the working fluid, segment 4-1, to the isochoric exhaust of spent gases. Express the engine's efficiency in terms of the gas compression ratio x = V_2//V_1. |
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Answer» `W = m/M C_(mV) (T_3 - T_4)`- `- m/M C_(mV) (T_2 - T_1) = ` `= m/M C_(mV) (T_3 - T_2 + T_1 - T_4)` The WORKING MEDIUM receives heat in the process of isochoric combustion of fuel : `Q = m/M C_(mV) (T_3 - T_2)` Hence `eta = 1 - (T_4- T_1)/(T_3 - T_2)` Making use of the result, express the temperatures in terms of the volumes. We have `V_2^(gamma) T_2 = V_1^(gamma - 1) T_1 and V_(2)^(gamma-1) T_3 = V_(1)^(gamma -1) T_4`. Dividing the first equality by the second, we obtain `T_2//T_3 = T_1//T_4`. Transform the expression for the efficiency and reduce it to the FOLLOWING form: `eta = 1 - (T_4)/(T_3) cdot (1 - (T_1//T_4))/(1- (T_2//T_3))` But the second fraction is , evidently , unity and the first fraction `T_4//T_3 = (V_2//V_1)^(gamma-1) = x^(gamma -1)`. Hence `eta= 1 - x^(gamma -1)`. 
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