Engine thermal efficiency and Volume ratios

AI Thread Summary
The discussion centers on deriving the thermal efficiency of a cyclic process involving a perfect gas, characterized by three distinct stages: cooling at constant pressure, heating at constant volume, and adiabatic expansion. The proposed expression for thermal efficiency is η = 1 - γ(r-1)/(rγ-1), but the calculations presented lead to a different result, prompting confusion about the assumptions made, particularly regarding the adiabatic process. Participants emphasize the importance of using the adiabatic condition, PVγ = K, to relate pressures and volumes effectively. There is also a focus on ensuring that heat exchanges are correctly represented in terms of absolute values to align with the expected efficiency formula. The conversation highlights the complexities of thermodynamic calculations and the necessity of careful assumptions in deriving accurate results.
sandpants
Messages
21
Reaction score
0
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)
Q3->1 = 0 ; adiabatic.

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:
Science news on Phys.org
Apply the adiabatic condition from 3→1.

AM
 
Andrew Mason said:
Apply the adiabatic condition from 3→1.

AM

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.
 
sandpants said:
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.
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
 
Andrew Mason said:
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

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.
 
sandpants said:
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.
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
 
Thread 'Thermo Hydrodynamic Effect'
Vídeo: The footage was filmed in real time. The rotor takes advantage of the thermal agitation of the water. The agitation is uniform, so the resultant is zero. When the aluminum cylinders containing frozen water are immersed in the water, about 30% of their surface is in contact with the water, and the rest is thermally insulated by styrofoam. This creates an imbalance in the agitation: the cold side of the water "shrinks," so that the hot side pushes the cylinders toward the cold...
Back
Top