Efficiency of an Ideal-Gas Cycle: How is it Derived?

In summary, the homework statement is that the thermal efficiency for a gas cycle operating as follows is 1-ɣ[((V1/V2) - 1) / ((p2/p1) -1)]
  • #1
JesseCoffey
6
0

Homework Statement


A possible ideal-gas cycle operates as follows:
(i) from an initial state (p1,V1) the gas is cooled at constant pressure to (p1,V2);
(ii) the gas is heated at constant volume to (p2,V2);
(iii) the gas expands adiabatically back to (p1,V1).

Assuming constant heat capacities, show that the thermal eficiency is:

1-ɣ[((V1/V2) - 1) / ((p2/p1) -1)]


Homework Equations



I used first law of thermal dynamics, n = W/Qh, differential forms of Cv and Cp.

change in energy for the complete cycle = 0, therefor W = Qh + Ql.

The Attempt at a Solution



Using efficiency = W / heat absorbed.

I attempted to find the W and Q of the 3 processes separately.
I was using for Isobaric: dW = -Pdv but I am pretty sure this is incorrect because no where does it state it is a reversible process. But the book for the class only shows reversible processes for examples and does not actually show a isobaric or isochoric process in any example.

I tried using the differential form of Cv and Cp and rearranging them to find dQ, but failed.

I have used 4 or 5 sheets of paper, trying every way i could think of to work this problem out. I know I am missing something and if someone could just point me in the right direction I would gladly appreciate it and work out the problem on my own.

Thanks
 
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  • #2
Welcome to Physics Forums.

JesseCoffey said:

The Attempt at a Solution



Using efficiency = W / heat absorbed.

I attempted to find the W and Q of the 3 processes separately.
I was using for Isobaric: dW = -Pdv but I am pretty sure this is incorrect because no where does it state it is a reversible process. But the book for the class only shows reversible processes for examples and does not actually show a isobaric or isochoric process in any example.
Actually, we would assume a reversible process here, so dW=-PdV as you said.

I tried using the differential form of Cv and Cp and rearranging them to find dQ, but failed.

I have used 4 or 5 sheets of paper, trying every way i could think of to work this problem out. I know I am missing something and if someone could just point me in the right direction I would gladly appreciate it and work out the problem on my own.

Thanks
You'll need W for the complete cycle, the tricky part is the adiabatic leg. Can you set up the integral for that part and show what you get? Here is an integral sign you can copy-and-paste: ∫

You'll also need Q for one leg of the cycle (do you know which leg that is?).
 
  • #3
Using Qh for heat absorbed and Ql for heat lost I was using the equation as:

n = w/Qh = (Qh +Ql)/Qh = 1 + Ql/Qh.

Ql being from (i) and Qh being from (ii). Q = 0 from (iii).

Qh = ∫Cv dT
Ql = ∫Cp dt + ∫pdV

For adiabatic ΔQ = 0 so ΔU = W.

Since, you are ending where you started ΔU for the complete cycle should = 0 correct?

so ΔU for adiabatic should equal ΔU for isochoric + isobaric.

Isobaric: ΔU = Ql + pdV
Isochoric: ΔU = Qh

ΔU for adiabatic = Ql + Qh - ∫pdV.

ΔU = W so W = Ql + Qh - ∫pdV

and now i lost myself. Maybe i just need some sleep.

Thanks for trying.
 
  • #4
Redbelly98 said:
You'll need W for the complete cycle, the tricky part is the adiabatic leg. Can you set up the integral for that part and show what you get? Here is an integral sign you can copy-and-paste: ∫

You'll also need Q for one leg of the cycle (do you know which leg that is?).

Ok, so I tried 3 different ways and started to post my questions on here, and while posting one of my questions, I think I answered it myself.

I stated that I wasnt sure how to relate the change in T to p and V, but I forgot about the ideal gas equation.

So I just solved for Q's like before and put in pV/R in place of T. Doing that I ended up with

n = Qin - Qout / Qin

dQin = Cv/R(Vdp + pdV) and pdV = 0
dQout = Cp/R(Vdp + pdV) and Vdp = 0

n = Cv(V2(p2-p1)) - Cp(p1(V2-V1)) / Cv(V2(p2-p1)) because R's cancel out.

n = 1 + (-Cp/Cv) *((V2-V1)/V2) * (p1(p2-p1)) sub in ɣ for Cp/Cv and clean up the rest

and that shows that n = 1+ɣ[((V1/V2) - 1) / ((p2/p1) -1)]

It is very close, only missing a - sign. I am not sure If i did a calculation wrong or that it is purely coincidental that I got this close.

Now that part that scares me and why I didnt do this in the beginning is that I complete ignore the adiabatic leg, and all work done. I just simple used heat.

It is close to the right answer but I am not convinced I did it correctly.

Thanks for your help so far.
 
Last edited:
  • #5
JesseCoffey said:
...
So I just solved for Q's like before and put in pV/R in place of T. Doing that I ended up with

n = Qin - Qout / Qin

dQin = Cv/R(Vdp + pdV) and pdV = 0
dQout = Cp/R(Vdp + pdV) and Vdp = 0

n = Cv(V2(p2-p1)) - Cp(p1(V2-V1)) / Cv(V2(p2-p1)) because R's cancel out.

n = 1 + (-Cp/Cv) *((V2-V1)/V2) * (p1(p2-p1)) sub in ɣ for Cp/Cv and clean up the rest

and that shows that n = 1+ɣ[((V1/V2) - 1) / ((p2/p1) -1)]

It is very close, only missing a - sign. I am not sure If i did a calculation wrong or that it is purely coincidental that I got this close.
The work done is the difference in heat in and heat out. It is also the work done in the adiabatic expansion less the work done in the isobaric compression. So you can approach it either by calculating 1) heat in and heat out (as you have done) or 2) heat in and work.

If you are using 1 - Qout/Qin for efficiency you have to take the absolute value of Qout:

[tex]|Q_{out}| = C_pP_1(V_1 - V_2)[/tex]

That is where you got the signs mixed up.

AM
 
  • #6
thanks, that makes me feel better.

can you explain to me exactly why we take the absolute value? Just so i can understand it and apply it to future problems.

thanks again.

I think I understand. Work is actually equal to the sum of the heat. By me stating Qin - Qout I am taking the negative out of Qout already. To do it the way I did I should have used
n = (Qin + Qout)/ Qin
 
Last edited:
  • #7
JesseCoffey said:
thanks, that makes me feel better.

can you explain to me exactly why we take the absolute value? Just so i can understand it and apply it to future problems.

thanks again.

I think I understand. Work is actually equal to the sum of the heat. By me stating Qin - Qout I am taking the negative out of Qout already. To do it the way I did I should have used
n = (Qin + Qout)/ Qin
Right. If you define the net heat flow (=work) being the sum of all heat flows then you have to stick with the signs (+ being into and - being out of the system). If you define it as heat in MINUS heat out, you are implicitly using positive quantities for all heat flows.

AM
 

Related to Efficiency of an Ideal-Gas Cycle: How is it Derived?

What is an ideal-gas cycle?

An ideal-gas cycle is a thermodynamic process in which a gas undergoes a series of changes in pressure, volume, and temperature that can be described using the ideal gas law. This cycle is often used as a model for real-world processes, such as internal combustion engines and refrigeration systems.

What is the efficiency of an ideal-gas cycle?

The efficiency of an ideal-gas cycle is a measure of how much of the energy put into the system is converted into useful work. It is calculated by dividing the work output by the heat input. In an ideal-gas cycle, the efficiency is determined by the temperature at which the gas expands and contracts, and can approach 100% under certain conditions.

How does the efficiency of an ideal-gas cycle compare to real-world processes?

In reality, no process can achieve 100% efficiency. The efficiency of an ideal-gas cycle is a theoretical limit that real-world processes can approach, but not exceed. Many factors, such as friction and heat loss, contribute to the lower efficiency of real-world processes.

What factors affect the efficiency of an ideal-gas cycle?

The efficiency of an ideal-gas cycle is affected by the temperature at which the gas expands and contracts, the pressure and volume of the gas, and the type of gas used. In addition, any heat loss or work required to overcome friction will decrease the efficiency of the cycle.

How is the efficiency of an ideal-gas cycle improved?

To improve the efficiency of an ideal-gas cycle, the gas must expand and contract at higher temperatures, and any sources of heat loss or friction must be minimized. This can be achieved through careful design and engineering of the system, as well as using more efficient gases or processes.

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