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

AI Thread Summary
The discussion focuses on solving the efficiency of a two-stage Carnot engine, where heat is absorbed and expelled at different temperatures. The efficiency for each stage is expressed as e(stage1) = (T1 - T2)/T1 and e(stage2) = (T2 - T3)/T2. Participants clarify how to combine these efficiencies to find the overall efficiency of the system. The final conclusion reached is that the total efficiency can be expressed as (T1 - T3)/T1, confirming the initial problem statement. The discussion ultimately resolves the question and enhances understanding of Carnot engine efficiency calculations.
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):

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

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

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

AM
 
thank you that makes since

I now understand the question
 

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