How Is the Maximum Temperature of an Otto Cycle Engine Calculated?

[gnarkil]

Homework Statement

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?

Homework Equations

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?

 

[gnarkil]

no ideas?

i had another look and Tmax is at 3, and Tmin is at 1

using PV= nRT:
Tmax = PV/nR = [3P_2(V_1/5)]/nR assume n, R, P_2 and V_1 = 1
Tmin = PV/nR = [P_2/4(V_1)]/nR

so assuming n, R, P_2, V_1 = 1, then Tmax = 3/5 and Tmin = 1/4

so Tmax in terms of Tmin ---> (3/5)x = 1/4 --> x = 5/12 so Tmax = (5/12)Tmin

i'm not sure if i am correct because i need gamma in the answer, gamma = Cp/Cv

 

[baba89]

Based on your calculations, it seems that T_max is indeed at pt4, where the volume is the same as the initial volume V_1 and the pressure is 2 times the initial pressure P_2. To find the maximum temperature in terms of T_min, we can use the TV^(gamma - 1) = constant equation and substitute the values at pt4:

T_max = (P_2 * V_1^(gamma - 1)) / (P_2/2)^(gamma - 1)

Since gamma = Cp/Cv and we know the values for Cp and Cv, we can substitute them in the equation to get:

T_max = (2 * V_1^(gamma - 1)) / (1/2)^(gamma - 1)

Simplifying, we get:

T_max = 4 * V_1^(gamma - 1)

Therefore, the maximum temperature in terms of T_min is 4 times T_min.

In terms of efficiency, the Otto cycle is less efficient than a Carnot engine operating between the same two temperature extremes. This is because the Otto cycle involves constant volume processes, which are less efficient than constant pressure processes. Additionally, the figure provided neglects the intake and exhaust processes, which also contribute to a decrease in efficiency. Therefore, e_otto < e_carnot.