Solve Carnot Engine: Efficiency (T1-T3)/T1

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Homework Help Overview

The discussion revolves around the efficiency of a two-stage Carnot engine, specifically focusing on the relationship between the temperatures involved and the work done in each stage. The original poster presents a problem statement that includes the absorption and expulsion of heat at different temperatures and seeks to prove the efficiency formula (T1 - T3)/T1.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the efficiency equations for each stage of the Carnot engine and how to combine them. There are attempts to express the overall efficiency in terms of the total work done and total heat flow into the system.

Discussion Status

Some participants have provided guidance on how to relate the efficiencies of the individual stages to the overall efficiency. There is an exploration of the definitions of efficiency and how they can be expressed in terms of the heat flows and temperatures involved. The discussion is ongoing, with participants questioning the setup and seeking clarification on the relationships between the variables.

Contextual Notes

Participants are working under the constraints of the Carnot engine's principles and are discussing the implications of the temperatures T1, T2, and T3 in relation to the efficiency calculations. There is an acknowledgment of potential confusion regarding the terms used in the equations.

LandOfStandar
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[SOLVED] Carnot engine - please help asp

I just typed a whole question and as I typed it I solved it, lol!

Homework Statement



In the 1st stage of a 2-stage Carnot engine, energy is absorbed as heat Q1 at temp T1, work W1 is done, and energy is expelled as heat Q2 at lower temp T2. The 2nd stage absorbs that energy as heat Q2, does work W2 and expels energy as heat Q3 at a still lower temp T3. Prove that the efficiency of the engine is (T1 - T3)/T1

Stage 1 Qin=Q1
Qex=Q2
T2 less then T1
W1

Stage 2 Qin=Q2
Qex=Q3
T3 less then T2
W2

Homework Equations



e =Qex/W = (Qin-Qex)/Qin = (Tin-Tex)/Tin

The Attempt at a Solution



e(stage1) = (T1-T2)/T1
e(stage1) = (T2-T3)/T2

How do you put them together?

or

is this all wrong? how do I approach this?
 
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LandOfStandar said:

The Attempt at a Solution



e(stage1) = (T1-T2)/T1
e(stage1) = (T2-T3)/T2

How do you put them together?

or

is this all wrong? how do I approach this?

Start with the definition of efficiency for each of the cycles and the combined cycle:

E1 = W1/Qh1
E2 = W2/Qh2

What is Ec in terms of W1, W2, Qh1 and Qh2?

What are Qh1 and Qh2 in terms of T1, T2, T3?

AM
 
E1 = W1/Qh1 = (Qh1-Ql1)/Qh1 = (T1 - T2)/T1

what do you mean?

E2 = W2/Qh2 = (Qh2-Ql2)/Qh2 = (T2 - T3)/T2
 
LandOfStandar said:
E1 = W1/Qh1 = (Qh1-Ql1)/Qh1 = (T1 - T2)/T1

what do you mean?

E2 = W2/Qh2 = (Qh2-Ql2)/Qh2 = (T2 - T3)/T2

So what is the overall efficiency in terms of the total work done (W1+W2) and the total heat flow into the system (Qh1+Qh2)?

AM
 
(T1 -T2 + T2 - T3) / (T1 + T2) = (T1 - T3) / (T1 + T2)

the problem is the T2 on the bottom
 
LandOfStandar said:
(T1 -T2 + T2 - T3) / (T1 + T2) = (T1 - T3) / (T1 + T2)

the problem is the T2 on the bottom

Since the heat flow into the system is just Q1 (the heat flow into the second engine is the output heat of the first):

[tex]\eta_{total} = (W_1 + W_2) / Q_1[/tex]

But W1 = Q1-Q2 and W2 = Q2-Q3, so

[tex]\eta_{total} = (Q_1-Q_3) / Q_1 = (T_1-T_3)/T_1[/tex]

AM
 
thank you that makes since

I now understand the question
 

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