Thermal Machine: Solving for Q23 to Complete the Otto Cycle

In summary, the conversation involves a series of equations that describe the relationship between pressure, volume, and temperature for different processes (adiabatic and isochoric). The equations ultimately lead to the question of how to find the missing data (P3, P4, T3, T4) in order to close the loop at point 1. The solution involves expressing everything in terms of Q23 and solving for the value of Q23 that closes the loop at point 1.
  • #1
A13235378
50
10
Homework Statement
Consider an Otto Cycle:
1-->2 (adiabatic)
2-->3 (isochoric)
3-->4 (adiabatic)
4-->1 (isochoric)

Knowing that:

P1 = 10^5 Pa
V1= 0,8 m^3
T1 = 290K
V3= 0,1m^3
cv= 0,171 Kcal/kg . K
Gas mass = 1 kg
gamma (coefficient of adiabatic expansion) = 1,4
V1=V4
V2=V3
Q23 = heat received
Q41 = heat expelled

Find T2,T3,T4 and P2,P3,P4
Relevant Equations
pv/t = constant
p(v)^gamma= constant
t v^(1-gamma) = constant
Q = m. cv . (T'-T)
1-->2 (adiabatic)

$$P_1V_1^γ=P_2V_2^γ$$

$$P_2=18,4.10^5Pa$$

$$T_1V_1^{1-\gamma}=T_2V_2^{1-\gamma}$$

$$T_2=429,32K$$

2-->3 (isochoric)

$$\frac{P_2}{T_2}=\frac{P_3}{T_3}$$

$$\frac{18,4.10^5}{429,32}=\frac{P_3}{T_3}$$

3--> 4 (adiabatic)

$$P_3V_3^{\gamma}=P_4V_4^{\gamma}$$

$$T_3V_3^{\gamma-1}=T_4V_4^{\gamma-1}$$

4--> 1 (isochoric)

$$\frac{P_1}{T_1}=\frac{P_4}{T_4}$$

From here, I can no longer find the values of P3, P4, T3, T4, as it is as if each equation is an association of the other. So I was wondering if there is a way to resolve this issue, because for me, I believe that data is missing.
 
Physics news on Phys.org
  • #2
Data is not missing. Express everything for points 3 and 4 in terms of Q23. Solve for the value of Q23 that closes the loop at point 1.
 

1. What is an Otto Cycle?

The Otto Cycle is a thermodynamic cycle that describes the process of converting heat into work in an internal combustion engine. It is named after its inventor, Nikolaus Otto, and is used in most gasoline-powered engines.

2. How does the Otto Cycle work?

The Otto Cycle consists of four processes: intake, compression, power, and exhaust. During the intake process, a mixture of fuel and air is drawn into the engine. In the compression process, the mixture is compressed by a piston, increasing its temperature and pressure. The power process involves igniting the compressed mixture, causing it to expand rapidly and push the piston down. Finally, the exhaust process removes the burned gases from the engine.

3. What is the efficiency of the Otto Cycle?

The efficiency of the Otto Cycle is determined by the compression ratio, which is the ratio of the volume of the cylinder when the piston is at the bottom of its stroke to the volume when the piston is at the top of its stroke. The higher the compression ratio, the more efficient the cycle will be.

4. What are the advantages of the Otto Cycle?

The Otto Cycle is advantageous because it is a relatively simple and efficient way to convert heat into work. It also produces less pollution compared to other types of engines, making it a popular choice for automobiles.

5. What are the limitations of the Otto Cycle?

One limitation of the Otto Cycle is that it is not as efficient as other types of engines, such as the Diesel Cycle. It also requires a constant supply of fuel and air, making it less suitable for applications where these resources are limited. Additionally, the combustion process can produce harmful emissions, which can contribute to air pollution.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
134
Replies
3
Views
1K
  • Classical Physics
Replies
1
Views
922
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
7K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
4K
Back
Top