# The efficiency of a heat engine

• gregorspv
In summary, the problem discussed involves calculating the efficiency of cycles B and C based on the ratio of efficiencies. However, after failing to calculate this ratio, the individual expressions for the efficiency of each cycle were derived. The solution provided for the efficiency of C seems incompatible with this derivation. The conversation then turns to setting up the problem differently and simplifying the math. Lastly, the discussion focuses on comparing the heat flows for each cycle in order to compare their efficiencies.
gregorspv
Homework Statement
An ideal monoatomic gas (ϰ=5/3) is repeating a thermodynamic cycle B, which is made out of two isobaric and two isochoric parts. The same gas is repeating a similar cycle C. Both cycles have the same ΔP and ΔV. Express cycle B's efficiency in terms of the efficiency of cycle A.
Relevant Equations
Q=mcΔT
η=A/Qin
dA=PdV
The first picture was provided along the problem statement. The second has my annotations.

I initially began by calculating the ratio of efficiencies, since the work done is obviously the same and cancels out, but after failing and having seen the form of the solution I saw that that cannot work. Therefore I first derived the expression for the efficiency of cycle B: $$\eta_B=\frac{A}{Q_{in,B}}=\frac{\Delta P\Delta V}{mc_V(T_2-T_1)+mc_P(T_{1'}-T_2)}=\frac{\Delta P\Delta V}{mc_V(T_1\frac{P_2}{P_1}-T_1)+mc_P(T_2\frac{V_2}{V_1}-T_2)}=\frac{\Delta P\Delta V}{mc_VT_1(\frac{P_2}{P_1}-1)+mc_PT_1\frac{P_2}{P_1}(\frac{V_2}{V_1}-1)}=\frac{\Delta P\Delta V}{mT_1}\frac{1}{c_V(\frac{P_2}{P_1}-1)+c_P\frac{P_2}{P_1}(\frac{V_2}{V_1}-1)}$$
and similarly
$$\eta_C=\frac{A}{Q_{in,C}}=\frac{\Delta P\Delta V}{mc_V(T_{2'}-T_{1'})+mc_P(T_{3}-T_{2'})}=\frac{\Delta P\Delta V}{mc_VT_1\frac{P_2}{P_1}\frac{V_2}{V_1}(\frac{P_3}{P_2}-1)+mc_P\frac{P_3}{P_1}\frac{V_2}{V_1}(\frac{V_3}{V_2}-1)}=\frac{1}{mT_1}\frac{1}{c_V\frac{V_2}{V_1}\frac{1}{P_1\Delta V}+c_P\frac{P_2}{P_1}\frac{1}{V_1\Delta P}+c_P\frac{1}{P_1V_1}}$$

I then substituted ##\eta_B## into ##\eta_C## to obtain: $$\eta_C=\eta_B(\frac{1}{P_1\Delta V}+\frac{\kappa}{V_1 \Delta P}+\frac{\kappa}{P_1V_1})\frac{1}{\frac{V_2}{V_1}\frac{1}{P_1\Delta V}+\frac{P_2}{P_1}\frac{\kappa}{V_1 \Delta P}+\frac{\kappa}{P_1V_1}}=\eta_B\frac{1}{1+\frac{\Delta V\Delta P+\kappa\Delta P\Delta V}{V_1\Delta P + \kappa P_1\Delta V + \kappa \Delta V\Delta P}}$$

The solution I was given is: $$\eta_C=\eta_B\frac{1}{1+4\eta_B}$$ but this seems incompatible with my derivation. I should suppose there is a much simpler way of approaching this problem?

#### Attachments

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First of all: what is cycle A?

One has to determine the heat flow in each segment in which there is heat flow into the system. To do that you need to determine the temperatures at the cornerns. Use PV=nRT.

To find the heat flow at constant volume for the isochoric segment use Q = nCvΔT; For the isobaric part use Q = nCpvΔT

AM

Oh, I must have made a typo; the actual problem is to express the efficiency of C in terms of B. Not sure why they're labelled that way though.

I believe the heat flows ##Q_{in}## were determined correctly? I expressed the temperatures in terms of pressure, volume and ##T_1## via Charles's (##\frac{V}{T}=const.##) and Gay-Lussac's laws (##\frac{P}{T}=const.##). Then I substituted those expressions into the two equations containing the specific heat term: this can be seen in the efficiencies' denominators. I did omit a few steps in my post due to tiresome LaTeX. Do you think I have made an arithmetic mistake?

Try setting the problem up a little differently. Call the pressure and volume at point 1' P and V. Then for process B, the coordinates of the corners would be ##(P-\Delta P, V-\Delta V)##, ##(P,V-\Delta V)##, ##(P, V)##, and ##(P-\Delta P, V)##; and the coordinates of the corners in process C would be ##(P+\Delta P, V+\Delta V)##, ##(P,V+\Delta V)##, ##(P, V)##, and ##(P+\Delta P, V)##. This would reduce the number of independent pressure and volume parameters to just three, which should simplify the math.

Also, let $$x=\frac{\Delta P\Delta V}{\left[\frac{C_v}{R}V\Delta P+\frac{C_p}{R}P\Delta V\right]}$$
In terms of this parameter, show that:
$$\eta_B=\frac{x}{\left[1-\frac{C_v}{R}x\right]}$$and
$$\eta_C=\frac{x}{\left[1+\frac{C_p}{R}x\right]}$$

Last edited:
gregorspv said:
Oh, I must have made a typo; the actual problem is to express the efficiency of C in terms of B. Not sure why they're labelled that way though.

I believe the heat flows ##Q_{in}## were determined correctly? I expressed the temperatures in terms of pressure, volume and ##T_1## via Charles's (##\frac{V}{T}=const.##) and Gay-Lussac's laws (##\frac{P}{T}=const.##). Then I substituted those expressions into the two equations containing the specific heat term: this can be seen in the efficiencies' denominators. I did omit a few steps in my post due to tiresome LaTeX. Do you think I have made an arithmetic mistake?
Since you are asked to compare efficiencies and since the work done in each case (Work = the area within the cycle = A) appears to be the same, this is simply a matter of comparing the Qin for each of the two cycles. How do those two heats compare? (hint: for the constant volume heating, ##\Delta T \propto V\Delta P## and for constant pressure expansion ##\Delta T \propto P\Delta V##).

AM

TSny

## What is a heat engine?

A heat engine is a device that converts heat energy into mechanical work. It operates through the transfer of heat from a high temperature source to a low temperature sink, creating a temperature difference that can be used to do work.

## How is the efficiency of a heat engine calculated?

The efficiency of a heat engine is calculated by dividing the work output by the heat input. This ratio is then multiplied by 100 to express the efficiency as a percentage.

## What factors affect the efficiency of a heat engine?

The efficiency of a heat engine is affected by several factors, including the temperature difference between the source and sink, the type of working fluid used, and the design and construction of the engine.

## What is the maximum efficiency that a heat engine can achieve?

According to the Carnot efficiency, the maximum efficiency that a heat engine can achieve is equal to 1 minus the ratio of the absolute temperatures of the source and sink. This means that a heat engine can never have an efficiency of 100%, as some energy will always be lost as heat.

## How can the efficiency of a heat engine be improved?

The efficiency of a heat engine can be improved by increasing the temperature difference between the source and sink, using a more efficient working fluid, and optimizing the design and construction of the engine. However, the efficiency can never exceed the Carnot efficiency limit.

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