What is the Efficiency of an Ideal Gas Engine?

[SonOfOle]

Homework Statement

An ideal gas engine that works according to the following cycle.

Find the efficiency of this engine assuming that the heat capacities of the gas may be taken to be constant. Recall that the efficiency may be defined as:
[tex] \eta = \frac{Net work done over full cycle}{Heat absorbed along an isotherm} [/tex]

Express your answer in terms of the volumes [tex]V_1[/tex] and [tex]V_2[/tex] and the pressures [tex]P_1[/tex] and [tex]P_2[/tex] and the heat capacities at constant pressure and volume [tex]C_p[/tex] and [tex]C_v[/tex].

Homework Equations

[tex]W = \int P dV [/tex]
[tex]P V = N kB T [/tex]
[tex]C_v = 3N k_b /2 [/tex]
[tex]C_p = 5N k_b /2 [/tex]

The Attempt at a Solution

[tex] \eta = \frac{W}{Q} = \frac {W_a + W_b + W_c}{Q_c} [/tex]
[tex] W_a = P_2 (V_2 - V_1) [/tex]
[tex] W_b = 0 [/tex]
[tex] W_c = \int^{V_1}_{V_2} P dV = \int^{V_1}_{V_2} \frac{N k_b T}{V} dV = N k_b T ln (V_1 / V_2)[/tex]
To find [tex]Q_c[/tex], recall [tex]U = Q + W[/tex] and [tex]U=(3/2)N k_b T[/tex] which is constant on an isotherm. So [tex]Q_c=-W_c= -N k_b T ln (V_1 / V_2) [/tex]

Thus, by the equation of the efficiency,

[tex]\eta = \frac{N k_b T ln (V_1 / V_2)+P_2 (V_2 - V_1)}{-P_2 (V_2 - V_1)}[/tex]

However, I'm not sure if all my assumptions above are valid, and I don't see how the above could be put in terms of [tex]C_v[/tex] and [tex]C_p[/tex], since there's no relation between them and the temperature.

Any help is appreciated.

(So you know, I'm studying for a qualification exam this fall in graduate school. These problems are from past exams that they've given us to help study. I've been out of the physics world for a year and a half, so this stuff comes back slowly sometimes. I appreciate everyone who has responded so far to my questions.)

 

[Andrew Mason]

Homework Statement

An ideal gas engine that works according to the following cycle.

Find the efficiency of this engine assuming that the heat capacities of the gas may be taken to be constant. Recall that the efficiency may be defined as:
[tex] \eta = \frac{Net work done over full cycle}{Heat absorbed along an isotherm} [/tex]

Where did you get this definition from? Efficiency is:

[tex]\eta = \frac{W}{Q_h}[/tex]

where W is the net work done per cycle and [itex]Q_h[/itex] is the heat absorbed from the hot reservoir. Since [itex]W = Q_h - Q_c[/itex] as there is no change in U in one complete cycle, you can replace the denominator with [itex]W + Q_c[/itex]. In this case, heat is absorbed in parts b and c of the cycle and released to the cold reservoir in a.

The Attempt at a Solution

[tex] \eta = \frac{W}{Q} = \frac {W_a + W_b + W_c}{Q_c} [/tex]

This should be:

[tex] \eta = \frac{W}{Q_h} = \frac {|W_b| + |W_c| - |W_a| }{Q_h} [/tex]

[tex] W_a = P_2 (V_2 - V_1) [/tex]

This is a negative value since work is being done ON the system (ie should be:

[tex] W_a = P_2 (V_1 - V_2) [/tex]

[tex] W_b = 0 [/tex]
[tex] W_c = \int^{V_1}_{V_2} P dV = \int^{V_1}_{V_2} \frac{N k_b T}{V} dV = N k_b T ln (V_1 / V_2)[/tex]

Correct.

To work out the heat flow into the cold reservoir (part b) you should recognize that heat flows at constant pressure so:

[tex]Q_c = nC_p\Delta T[/tex]

You will have to work out the temperature change, which is simple to do.

AM

 

[SonOfOle]

Where did you get this definition from? Efficiency is:

[tex]\eta = \frac{W}{Q_h}[/tex]

where W is the net work done per cycle and [itex]Q_h[/itex] is the heat absorbed from the hot reservoir.

Good question Andrew. Interestingly enough, it's verbatim on the graduate entrance exam for a state university. I was wondering about it myself, but that's why I asked here. Thanks, I think with your help I can solve the problem.

 

[Andrew Mason]

Good question Andrew. Interestingly enough, it's verbatim on the graduate entrance exam for a state university. I was wondering about it myself, but that's why I asked here. Thanks, I think with your help I can solve the problem.

Someone may be confusing coefficient of performance of a refrigerator with efficiency of a heat engine. They are quite different. COP of a refrigerator is W(input)/Qc

Incidentally, I had the volumes backward in my mind so ignore my comment that Wa = P2(V1-V2). You had it right.

AM

 

[SonOfOle]

Incidentally, I had the volumes backward in my mind so ignore my comment that Wa = P2(V1-V2). You had it right.

I had to double check my work, but I had convinced myself that you were mistaken. Thanks for letting me know though.

Here's the solution I worked out.

[tex] \eta = \frac{W_{total}}{Q_{hot}} = \frac{W_a+W_b + W_c}{W_{total}+Q_{cold}} = \frac{-P_2 (V_1 -V_2)+ 0 + nRT ln(V_1 / V_2)}{-P_2 (V_1 -V_2)+ 0 + nRT ln(V_1 / V_2)+Q_{a}} [/tex]

where [tex] Q_a = n C_p \Delta T = n C_p (T-P_2 V_2 /(nR)) = C_p (n T - P_2 V_2 /R) [/tex]

so [tex] \eta = \frac{-P_2 (V_1 -V_2) + nRT ln(V_1 / V_2)}{-P_2 (V_1 -V_2) + nRT ln(V_1 / V_2)+C_p (n T - P_2 V_2 /R)} [/tex].

Which can be re-written

[tex] \eta = \frac{- (V_1 -V_2) + ( V_1) ln(V_1 / V_2)}{( V_1) ln(V_1 / V_2)+(C_p/R - 1) (V_1- V_2 )} [/tex]

Howabout that... no explicit P2 dependence. That answer, though, doesn't fit well with what the problem asked for. It may just be a poorly asked question (as the [tex]\eta[/tex] definition they gave already showed).