Heat engine undergoing an elliptical cycle

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
JD_PM
Messages
1,125
Reaction score
156
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).
 
Physics news on Phys.org
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
 
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).
 
  • Like
Likes   Reactions: JD_PM