Heat engine: moles/temp unknown

AI Thread Summary
The discussion focuses on calculating the efficiency of a heat engine using a closed cycle with a monatomic ideal gas. The efficiency formula is derived from work output and heat input, with specific attention to the isothermal process between points 1 and 3. Clarifications are made regarding the heat transfer during different processes, particularly that the heat involved in the transition from 1 to 2 is not included in the heat input for QH. Participants point out missing brackets in the equations and emphasize the importance of correctly interpreting temperature changes and their signs. The conversation concludes with a recognition that the current approach may not be effective in solving the problem.
joe_cool2
Messages
23
Reaction score
0

Homework Statement



A heat engine uses the closed cycle shown in the diagram below. (Figure 1) The working substance is n moles of monatomic ideal gas. Find the efficiency of such a cycle. Use the values for pressure and volume shown in the diagram, and assume that the process between points 1 and 3 is isothermal.

1065003.jpg


Homework Equations



pV=nRT
efficiency = Wout/QH

The Attempt at a Solution



We know that T3 and T1 are on an isotherm, and so are both exactly 3 times T2. So, the absolute value of ΔT is 2 times T2. The reason I refer to the absolute value here is because the sign of ΔT for different processes will be different.

Wout = nRT*ln(V1/V3) - pΔV

So, Wout = nRT*ln(V1/V3) - nRΔT

QH= nRT*ln(V1/V3) + nCvΔT

So the efficiency is:

3nRT2*ln(V1/V3) + 2nRT2 / 3nRT2*ln(V1/V3) + 2nCVT2

Then the idea is to cancel out all the n's and T2's. All the rest are knowns. I get a value of 84.1%. There's definitely something very wrong about my approach.
 
Last edited:
Physics news on Phys.org
Did you consider the heat involved in 1->2?

I don't understand the last, long expression and I think there are brackets missing.
 
The heat involved in 1 -> 2 is actually leaving the system. So, QH is just the heat we put in during the isothermal and isochoric legs. So, the heat associated with the isobaric process, nCPΔT, does not enter into the expression for QH.

Yes, there were brackets missing. Let me clarify as to where this expression comes from.

[3nRT2*ln(V1/V3) + 2nRT2] / [3nRT2*ln(V1/V3) + 2nCVT2]

We know that T3 is three times greater than T2. This should help to explain the first term of the numerator and the first term of the denominator. The former is the heat intake during the isothermal process, and the latter is the work done during the isothermal process. These expressions are the same because in an isothermal process, W = QH.

The expression begins as nRT*ln(V1/V3). Now T can refer to the temeperature at point 1 or point 3, because we are dealing with an isothermal process. And we said in the last paragraph that 3T2 = T3. So this is where the 3T2 comes from in both expressions.

The second terms in the numerator and the denominator contain the factor 2T2. That's because ΔT = T3 - T2, so ΔT = 3T2 - T2.

Note the sign change on the second term in the numerator. That occurs because we lose temperature moving from 1 -> 2, so ΔT is negative. ΔT is not negative in the other place it occurs, moving from 2 -> 3.

EDIT: Let's not get trapped in the paradigm of my thinking. It hasn't gotten me anywhere with this problem.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top