Efficiency of a Simple 3-Stage Ideal Gas Cycle: Analyzing Thermal Efficiency η

In summary, the conversation discusses a possible ideal-gas cycle and its efficiency. The cycle consists of three stages where the gas is cooled at constant pressure, heated at constant volume, and expands adiabatically. The thermal efficiency is given by η=1-γ(V2/V1-1)/(p2/p1-1). However, this equation may not be correct as it yields η>1. The correct equation should be η=-1-(γ-1)(V2/V1-1)/(Cvt1((p2/p1-1)V2/V1). It is important to note that the work done by the gas during the cycle equals the total heat added during the cycle. The efficiency can also
  • #1
Toby_phys
26
0
A possible ideal-gas cycle operates as follows:

1. From an initial state (##p_1##, ##V_1##) the gas is cooled at constant pressure to (##p_1##, ##V_2##); Let's call the start and end temperature ##T_1## and ##T_2##

2.The gas is heated at constant volume to (##p_2##, ##V_2##);Lets call the start and end temperature ##T_2## and ##T_3##

3.The gas expands adiabatically back to (##p_1##, ##V_1##). Let's call the start and end temperature ##T_3## and ##T_1##
Assuming constant heat capacities, show that the thermal efficiency η is

$$
\eta=1-\gamma\frac{V_2/V_1 -1}{p_2/p_1-1}
$$Efficiencey is defined as: $$\eta=\frac{W}{Q_h}$$ the work done over the heat entered. The heat enters at stage 2 (and some leaves at stage 1 but that doesn't matter). So I need to find the heat entered at stage 2 and the work done.

**Stage 1:**

From the ideal gas equation we get:
$$
p_1V_1=nRT_1, \ \ \ \ p_2V_2=nRT_2 \implies \frac{T_2}{T_1}=\frac{V_2}{V_1}
$$

The work done is just force times distance which is pressure times change in volume:

$$
\Delta W=-p_1\Delta V=-p_1(V_2-V_1)
$$
**Stage 2:**

It doesn't change in volume and so no work is done. However heat is put into the system, increasing the pressure. We need to find this heat.

##\Delta U= Q_h##

For an ideal gas we have:
$$
\Delta U= C_v\Delta T=C_v(T_3-T_2)
$$

Where ##C_v## is heat capacity at constant volume.

**Stage 3:**

Stage 3 is adiabatic so ##\Delta U=\Delta W=C_v(T_1-T_3)##

We also have, using the ideal gas law:
$$
T_3=\frac{p_2V_2}{p_1V_1}T_1
$$Let us sub this into the efficiency:

$$
\eta=\frac{C_v(T_1-T_3)-p_1(V_2-V_1)}{C_v(T_3-T_2)}
$$

If we get ##T_3## and ##T_2## in terms of ##T_1## and sub these in we get:

$$
\eta=-1-\frac{p_1(V_2-V_1)}{C_vT_1(\frac{p_2V_2}{p_1V_1}-\frac{V_2}{V_1})}
$$
And with the ideal gas law, with ##n=1## for simplicity we get ##T_1=\frac{p_1V_1}{R}##

$$
\implies\eta=-1-\frac{R(V_2/V_1-1)}{C_vT_1(\frac{p_2V_2}{p_1V_1}-\frac{V_2}{V_1})}
$$

##R=C_p-C_v## and ##\gamma=C_p/C_v##

$$
\implies\eta=-1-\frac{(\gamma-1)(V_2/V_1-1)}{C_vT_1(\frac{p_2}{p_1}-1)\frac{V_2}{V_1}}
$$

I have no real clue really if this is right or wrong.
 
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  • #2
Toby_phys said:
Stage 3 is adiabatic so ##\Delta U=\Delta W=C_v(T_1-T_3)##
How did you get this for an adiabatic process?
 
  • #3
kuruman said:
How did you get this for an adiabatic process?

##dq=0## for adiabatic process I think
 
  • #4
Toby_phys said:
Assuming constant heat capacities, show that the thermal efficiency η is

$$
\eta=1-\gamma\frac{V_2/V_1 -1}{p_2/p_1-1}
$$
This can't be correct since it yields ##\eta > 1##. Note ##p_2 > p_1## and ##V_2 < V_1##.
Efficiencey is defined as: $$\eta=\frac{W}{Q_h}$$ the work done over the heat entered.
OK. But ##W## represents the work done by the gas. Later in your work, it looks like you calculate the work done on the gas. This can lead to sign errors.

Let us sub this into the efficiency:

$$
\eta=\frac{C_v(T_1-T_3)-p_1(V_2-V_1)}{C_v(T_3-T_2)}
$$

If we get ##T_3## and ##T_2## in terms of ##T_1## and sub these in we get:

$$
\eta=-1-\frac{p_1(V_2-V_1)}{C_vT_1(\frac{p_2V_2}{p_1V_1}-\frac{V_2}{V_1})}
$$
Looks like maybe you made an error in getting the -1 part of the second equation above. Note that the denominator of the first equation does not cancel the first term of the numerator of the first equation to yield -1.

I think the algebra will be simpler if you use the fact that the work done by the gas during the cycle equals the total heat added during the cycle. (WHY?)

The heat added consists of heat added at constant volume and heat added at constant pressure. So you can write the work as the sum of two terms with the first term involving ##C_V## and the second term involving ##C_P##.

You know that the heat , ##Q_h##, is for the constant volume process and therefore involves ##C_V##. So, when you set up the efficiency, you will find that the ratio of ##C_P## to ##C_V## will pop up in the algebra and that will give you the ##\gamma## factor you need.
 
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Related to Efficiency of a Simple 3-Stage Ideal Gas Cycle: Analyzing Thermal Efficiency η

1. What is a simple 3 stage engine?

A simple 3 stage engine is an internal combustion engine that operates through three stages of the combustion process: intake, compression, and power. It uses a piston-cylinder system to convert fuel into mechanical energy.

2. How does a simple 3 stage engine work?

The first stage is the intake stage, where a mixture of fuel and air is drawn into the cylinder. In the compression stage, the mixture is compressed by the piston, which increases its temperature and pressure. In the power stage, the spark plug ignites the mixture, causing it to expand and push the piston down, creating mechanical energy.

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A simple 3 stage engine is smaller, lighter, and more efficient compared to other types of engines. It also has fewer moving parts, making it easier to maintain and repair. Additionally, it can run on various types of fuels, making it versatile.

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One potential drawback is that a simple 3 stage engine may not be powerful enough for larger vehicles or machinery. It also produces more emissions compared to newer, more advanced engines. Additionally, it may require more frequent maintenance and tuning to keep it running smoothly.

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