Gasoline engines operate approximately on the Otto cycle, consisting of two adiabatic and two constant-volume segments. The Otto cycle for a particular engine is shown in figure below. see atchmt
1. Find the maximum temperature in terms of the minimum temperature T_min. (gamma should be part of answer, where gamma = Cp/Cv -->Cp = 5R/2,Cv = 3R/2 --> R = 8.314
2. How does the efficiency compare with that of a Carnot engine operating between the same two temperature extremes? Note: Figure neglects the intake of fuel-air and the exhaust of combustion products, which together involve essentially no net work.
e_otto < e_carnot or e_otto = e_carnot or e_otto > e_carnot?
ideal gas law, PV = nRT
otto cycle efficiency, e = 1 - (Cp/Cv)^-1 where Cp = 5R/2, Cv = 3R/2 where R = 8.314 constant
for otto cycle TV^(gamma -1) = constant where gamma = Cp/Cv
The Attempt at a Solution
for the problem about the temperature i used the TV^(gamma - 1) = constant equation and used T_min for T and thus used the figure to determine the corresponding volume
i used a rough calculation with PV = nRT to get a general idea of temp magnitude
pt3: P = 3P_2, V = V_1/5
pt4: P = 2P_2, V = V_1
pt1: P = P_2/2, V = V_1
pt2: P = P_2, V = V_1/5
and using T = PV/nR and subbing in each pressure and volume, and holding P_2, V_1, n and R = 1, i found pt4 = T = 2, pt3 = T = 3/5, pt2 = T = 1/5, pt1 = T = 1/2, so i assumed T_min is at pt2, and that T_max is at pt4.
how do i use the TV equation using T_min and gamma to describe T_max?
4.5 KB Views: 422