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

In summary: R/2 / 3R/2 = 5/3In summary, the discussion focused on the operating cycle of gasoline engines, specifically the Otto cycle which consists of two adiabatic and two constant-volume segments. The maximum temperature of the cycle was found using the ideal gas law and the Otto cycle efficiency equation, with the minimum temperature and specific heat ratios given. The efficiency of the Otto cycle was also compared to that of a Carnot engine, with the conclusion that it is less efficient. The conversation then turned to determining the maximum temperature in terms of the minimum temperature, with calculations using the ideal gas law and the given pressure and volume values. The final result was Tmax = (5/12)Tmin
  • #1
gnarkil
26
0

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?
 

Attachments

  • otto.JPG
    otto.JPG
    4.5 KB · Views: 540
Physics news on Phys.org
  • #2
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
 
Last edited:
  • #3




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.
 

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

1. What is the Otto cycle and how does it relate to temperature?

The Otto cycle is a thermodynamic cycle that describes the operation of a typical spark-ignition internal combustion engine. Temperature plays a critical role in this cycle, as it determines the efficiency and power output of the engine.

2. What is the maximum temperature reached in an Otto cycle?

The maximum temperature reached in an Otto cycle is the temperature at the end of the adiabatic compression process, also known as the compression ratio. This temperature can reach up to 2000°C in a typical gasoline engine.

3. How does the air-fuel ratio affect the temperature in an Otto cycle?

The air-fuel ratio is a crucial factor in determining the temperature in an Otto cycle. A lean air-fuel mixture results in a lower temperature, while a rich mixture leads to a higher temperature. This is due to the amount of oxygen available for combustion.

4. What is the role of the cooling system in regulating the temperature in an Otto cycle?

The cooling system is responsible for maintaining the temperature within a safe operating range in an Otto cycle. It does this by circulating coolant through the engine to absorb excess heat and dissipate it through the radiator.

5. How does ignition timing affect the temperature in an Otto cycle?

Ignition timing refers to the precise moment when the spark plug ignites the air-fuel mixture in the engine. A delayed ignition timing can result in a higher temperature due to incomplete combustion, while an advanced timing can lead to a lower temperature due to early combustion.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
230
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
7K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Replies
1
Views
980
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
4K
Back
Top