Heat engine undergoing an elliptical cycle

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SUMMARY

An ideal diatomic gas undergoes an elliptical cyclic process in a heat engine, characterized by specific points on a PV diagram. The efficiency of this engine is defined as the ratio of work done (W) to the heat added (Qh), expressed as e = W/Qh. The work done by the engine is calculated using the formula W = π(P2 - P1)(V2 - V1), where P1 = 1 and P2 = 2P1. The discussion highlights the need for a correct calculation of Qh and emphasizes the importance of accurately determining the area under the PV graph for the elliptical cycle.

PREREQUISITES
  • Understanding of thermodynamic cycles, specifically elliptical cycles
  • Familiarity with the PV diagram and its significance in heat engines
  • Knowledge of work and efficiency calculations in thermodynamics
  • Basic concepts of ideal gases and their properties
NEXT STEPS
  • Research the calculation of heat transfer (Qh) in elliptical cycles
  • Learn about the efficiency of different thermodynamic cycles, including rectangular cycles
  • Explore the implications of compression ratios on work done in heat engines
  • Study the relationship between area under the PV curve and work done in thermodynamic processes
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Students and professionals in thermodynamics, mechanical engineers, and anyone interested in the efficiency of heat engines and cyclic processes.

JD_PM
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Homework Statement
Evaluate the efficiency of this engine
Relevant Equations
The equation for an ellipse:

##\frac{(x-x_o)^2}{a^2} + \frac{(y-y_o)^2}{b^2} = 1##
An ideal diatomic gas undergoes an elliptic cyclic process characterized by the following points in a ##PV## diagram:

$$(3/2P_1, V1)$$
$$(2P_1, (V1+V2)/2)$$
$$(3/2P_1, V2)$$
$$(P_1, (V1+V2)/2)$$This system is used as a heat engine (converting the added heat into mechanical work).

Evaluate the efficiency of this engine setting ##P_1=1## and ##P_2= 2P_1##

We know that the efficiency is defined as the benefit/cost ratio:

$$e = \frac{W}{Q_h}$$

Let's focus first on the work done by the engine; the work done by the working substance is the area under the ##PV## graph. Then:

$$W = \pi (P_2 - P_1)(V_2 - V_1)$$

$$W = \pi P_1(V_2 - V_1)$$

My problems come when calculating ##Q_h##; I have been told an analytic method: https://chemistry.stackexchange.com...at-engine-which-undergoes-an-elliptical-cycle . But I am convinced there has to be an easier one...

I have been thinking I have been thinking about how I could make an analogy with the same problem but with a rectangular shape (which is much easier to solve).
 
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The rectangle that circumscribes the ellipse has area ##\Delta P \Delta V##. Your expression for ##W## is ##\pi## times that. That can't possibly be correct.
 
vela said:
The rectangle that circumscribes the ellipse has area ##\Delta P \Delta V##. Your expression for ##W## is ##\pi## times that. That can't possibly be correct.

$$W = \pi (P_2 - P_1)(V_2 - V_1)$$

It is for the ellipse (my question). I have just mentioned the rectangle as an example. However, the exercise is about an ellipse cycle
 
You missed the point.
 
Sorry I realized that:

$$W = \pi (P_2 - P_1)(V_2 - V_1)$$

It 's wrong. Instead we have:

$$W=\text{Area}=\pi\left(\frac{s-1}2V_1\right)\left(\frac{r-1}2P_1\right)$$

Where ##s## and ##r ## are the volume and pressure compression ratios respectively.
 
JD_PM said:
I have been thinking I have been thinking about how I could make an analogy with the same problem but with a rectangular shape (which is much easier to solve).
Once you solve the problem properly for the elliptical cycle, it will be possible to identify what would be considered an equivalent rectangular cycle with the same efficiency and the same center (for whatever that might be worth).
 
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