Question

Problem 8: Consider the reversible Carnot's cycle of an ideal monatomic gas in the cold cylinder of 290 K corresponding to the isothermal compression step. Then the volume of the gas is further compressed by a factor of 7.5 in the adiabatic compression step. a) Find the temperature at the final step of the adiabatic compression. b) What is Thot for the isothermal expansion step? c) What is the maximum thermodynamic efficiency for this engine? d) How much would the efficiency increase if the adiabatic compression factor was 25? e) What is a problem in compressing the gas by a factor of 25 in the adiabatic compression step of an internal combustion engine? f) Plot the P-V diagram for the given Carnot's cycle (Vi/Vi-7.5), assuming an ideal monatomic gas in the cylinder.

Answer 1

as per HomeworkLib policy i had done first four parts

AB rightarrow isothermal expansion CD rightarrow isothermal compression BC rightarrow adiabatic expansion DA rightarrow adiabatic compression ideal non atomic gas T_2 = 290 K v_2 = v v_1 = v/7.5 at final step of adiabatic compression i.e at point A. tempt = T_1 Using P V^r = Constant also T V^r - 1 = constant r = 5/3 - for non atomic gas = > T_1 V^r - 1_1 = T_2 V^r - 1_2 T_1 = T_2 (V_2/V_1)^2/3 T_1 = T_2 (7.5)^2/3 = 1111.15 K T_hot = T_1 = 1111.15 K for max efficiency eta = 1 - T_2/T_1 = 1 - 290/1111.15 eta = 0.74 = > eta = 74 % If V_1 = V/25 T_1 = T_2 (25)^2/3 = 2479.47 K eta prime = 0.88 = eta prime = 88 % \r\n Therefore increase in efficiency = eta prime - eta = 0.14 = > by 14 %