# Engine thermal efficiency and Volume ratios

1. Nov 1, 2013

### sandpants

The question:

A perfect gas undergoes the following cyclic processes:
State 1 to 2 cooling at constant pressure.
State 2 to 3 heating at constant volume.
State 3 to 1 adiabatic expansion.

Deduce an expression for the thermal efficiency of the cycle in terms
of r the volume compression ratio (r=V1/V2) and γ (where γ = ratio of specific heats Cp/Cv)

η = 1 - γ(r-1)/(rγ-1)

My attempt at the solution:
First I tried sketching the cycle

Bare with me as I present you the silly symbol art.

P

3.
^'.
|..|
|...\
|.....'-.
|.........'-._
2<---------':.1 v

I'd like to work in specific terms

As it is a perfect gas
P1v1= RT1
P1v2= RT2
P2v2= RT3

Heats from 1->2, 2->3, 3->1
Q1->2=Cp(T2-T1)
Q2->3=Cv(T3-T2)

Also, polytropic relations
v2/v1 = (P1/P2)1/n
as r = v1/v2⇔ r = (P2/P1)1/n
∴ rn = P2/P1 and
1/rn = P1/P2

Substituting Ideal Gas expressions in terms of Tn
Q1->2=Cp((P1v2-P1v1)/R)
Q2->3=Cv((P2v2-P1v2)/R)

Thermal efficiency
This is what I am unsure off. I begin assuming quite a few things. First I assume that heat in the cooling process is the equivalent of heat escaping to a cold reservoir; coincidentally, heat from the pressurization is the heat INPUT from the hot reservoir. As such:

η = [Q2->3 - Q1->2]/Q2->3

η = 1 - Q1->2/Q2->3

∴ η = 1 - Cp(P1v2-P1v1)/Cv(P2v2-P1v2)

η = 1 - γ((P1v2-P1v1)/(P2v2-P1v2))

From here:

P1v2/P2v2-P1v2 = P1/(P2-P1) = 1/rn-1

And
P1v1/P2v2-P1v2 = r/(rn-1)

My Result

η = γ(1-r)/(rn-1) =/= η = 1 - γ(r-1)/(rγ-1)

Can I assume n=γ in this situation? only 1 process is adiabatic.

Last edited: Nov 1, 2013
2. Nov 1, 2013

### Andrew Mason

Apply the adiabatic condition from 3→1.

AM

3. Nov 1, 2013

### sandpants

Can you be more specific? Apply where?

If it's adiabatic there is no heat - I do not understand how the process can be related to thermal efficiency.

4. Nov 1, 2013

### Andrew Mason

What you need is the relationship between P1 and P2 in terms of V1 and V2. That is determined by the adiabatic condition PVγ = K.

AM

5. Nov 2, 2013

### sandpants

Indeed, the ratios match up and allow you to express them with n=γ. Thermodynamics is always like that - an answer under your nose at all times.

But another issue is that the numerator does not match up.
The expected form is r-1 when I get 1-r despite getting the same denominator.

6. Nov 3, 2013

### Andrew Mason

You are using the absolute values of Qh and Qc so you have to make sure that the original equation reflects that. For example, Qc = |Cp(T2-T1)| = Cp(T1-T2)

AM