What is the thermal efficiency of an ideal gas cycle?

In summary: I think I understand what you are trying to do now. In summary, the gas is cooled at constant pressure to (p_1,V_2); the gas is heated at constant volume to (p_2,V_2); the gas expands adiabatically back to (p_1,V_1). Assuming constant heat capacities, the thermal efficiency is 1 - \gamma\frac{(V_1/V_2)-1}{(p_2/p_1) - 1}
  • #1
tjackson3
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Homework Statement



A possible ideal gas cycle operates as follows:

(i) From an initial state [tex](p_1,V_1)[/tex], the gas is cooled at constant pressure to [tex](p_1,V_2)[/tex];

(ii) the gas is heated at constant volume to [tex](p_2,V_2)[/tex];

(iii) the gas expands adiabatically back to [tex](p_1,V_1)[/tex].

Assuming constant heat capacities, show that the thermal efficiency is

[tex]1 - \gamma\frac{(V_1/V_2)-1}{(p_2/p_1) - 1}[/tex]

where [tex]\gamma = c_p/c_v[/tex]

Homework Equations



Carnot efficiency: [tex]\nu = \frac{W}{Q_H} = 1 - \frac{T_l}{T_h}[/tex]

In an adiabatic process, [tex]pV^{\gamma},TV^{\gamma-1},p^{1-\gamma}T^{\gamma}[/tex] are all constant.

The Attempt at a Solution



I've been spinning my wheels a lot with this one, and I think the issue may be algebraic. My first thought was that I'd calculate the work done in the cycle. For part (i), it's just [tex]p(V_1-V_2)[/tex]. For part (ii), it's zero, since dV = 0. For part (iii), I used the fact that since it's adiabatic, [tex]pV^{\gamma}[/tex] is constant, which I'll call k. Then the work becomes

[tex]W_3 = \frac{k}{\gamma-1}(V_2^{1-\gamma} - V_1^{1-\gamma})[/tex]

You can set [tex]k = p_1V_1^{\gamma} = p_2V_2^{\gamma}[/tex], but there doesn't seem to be a preferable way. Adding these together gives you

[tex]W = p(V_1-V_2) + \frac{p_iV_i^{\gamma}}{\gamma-1}(V_2^{1-\gamma} - V_1^{1-\gamma})[/tex]

Since none of these processes are isothermal, I can't figure out a meaningful expression for [tex]Q_H[/tex], so I'm stuck here. Any help is very much appreciated!
 
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  • #2
This is not a Carnot engine, so Carnot efficiency is not applicable.

Heat flow is into the system in during the isochoric part and out on the isobaric part. Calculate the heat flow in and the heat flow out. Apply the first law to determine the work done in terms of heat in and heat out. From that, determine the efficiency.

AM
 
  • #3
Thank you very much, Andrew!
 

Related to What is the thermal efficiency of an ideal gas cycle?

What is a Carnot engine?

A Carnot engine is a theoretical thermodynamic engine that operates on a reversible cycle, meaning it can be run in both forward and reverse directions without any loss of energy. It serves as a model for the most efficient engine possible, known as the Carnot cycle.

How does a Carnot engine work?

A Carnot engine works by using heat energy to do work. It takes in heat from a high-temperature source, uses some of that energy to do work, and then releases the remaining heat to a low-temperature sink. The cycle then repeats, with the engine converting heat energy into mechanical work.

What is Carnot engine efficiency?

Carnot engine efficiency is the maximum possible efficiency that can be achieved by a heat engine operating between two temperature reservoirs. It is defined as the ratio of the work output to the heat input, and is represented by the equation: efficiency = 1 - (Tlow/Thigh), where Thigh is the temperature of the high-temperature reservoir and Tlow is the temperature of the low-temperature reservoir.

What factors affect the efficiency of a Carnot engine?

The efficiency of a Carnot engine is affected by the temperatures of the high-temperature and low-temperature reservoirs. The higher the temperature difference between the two, the higher the efficiency. Additionally, the type of working substance used and the design of the engine can also affect its efficiency.

Can a Carnot engine have 100% efficiency?

No, according to the Carnot efficiency formula, the maximum efficiency a Carnot engine can have is 1 - (Tlow/Thigh). This means that even with perfect design and a large temperature difference between the two reservoirs, the efficiency of a Carnot engine cannot reach 100%. This is due to the second law of thermodynamics, which states that no heat engine can have 100% efficiency.

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