# Carnot cycle

1. Nov 26, 2014

### toothpaste666

1. The problem statement, all variables and given/known data

A heat engine containing an ideal gas has a reversible cycle which consists of 2 constant
volume segments with V = 1ℓ and V = 3ℓ and two constant pressure segments with P = 1 atm and
P = 2 atm (see figure below). The temperature at point “c” is Tc = 273 K.

A) Is the path of the cycle clockwise or counter-clockwise? Explain.
B) How much work is done by the engine in one cycle?
C) What is the efficiency of the engine? Notice that this is a reversible engine, so recall
the efficiency of a Carnot engine and use the ideal gas law.

2. Relevant equations
W = PdV
W = VdP

3. The attempt at a solution
A) since at point c the engine is at the low temp Tc then since it is a heat engine the cycle must be going counter clockwise to go from high to low temp (please check my understanding on this)
B) this is where I am stuck
W = Wac + Wcd + Wdb + Wba
W = VdeltaP(ac) + PdeltaV(cd) + VdeltaP(db) + PdeltaV(ba)
W = (1L)(1atm - 2atm) + (1atm)(3L-1L) + (3L)(2atm-1atm) + (2atm)(1L-3L)
W = -1atmL + 2atmL + 3atmL - 4atmL
W = 0

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2. Nov 26, 2014

### Simon Bridge

A. an "engine" usually has to do work on it's surroundings - which leg has to do positive work?
B. the work done in a process is the area under the P-V diagram.
... so at constant volume, the work is zero.

Note: this is not a Carnot cycle.

3. Nov 26, 2014

### toothpaste666

im not entirely sure what you mean in A or what you mean by which leg.
but for part B) the cycle has a height of 1 atm and a width of 2L so it has an area of 2atmL which means its does 2 atmL of work?

4. Nov 26, 2014

### Simon Bridge

A. One leg of the cycle would be going from a to b, or from b to a ... it's an analogy from the terms used for a foot race. Perhaps you'd prefer "stage" or maybe "process" since each leg has only one process in it?

B. well done - though the person marking may prefer you put work in Joules.

5. Nov 26, 2014

### toothpaste666

so it would be 203 joules.
for part A) the expansions would do the positive work?
for part c) for a reversible engine
e = 1-Tc/Th
whe have Tc we need Th
using the ideal gas law we find number of moles by using the known temp at point c
PV=nRT
n = PV/RT = (1atm)(1L)/(.0821 atmL/molK)(273K) = .0446 mol
now we can find the temps at the other points
point a) T = PV/nR = (2 atm)(1L)/(.00366atmL/K) = 546 K
point b) T = PV/nR = (2atm)(3L)/(.00366 atmL/K) = 1639 K
point d) T = PV/nR = (1atm)(3L)/(.00366atmL/K) = 820 K
Th = 1639 K
plugging in
e = 1 - Tc/Th = 1 - 273/1639 = .833
e = 83.3%

6. Nov 26, 2014

### Simon Bridge

Well done: - $$\int_{V_1}^{V_2}P(v)\;\text{d}v > 0 : V_2>V_1$$ ... we have a definition for work where work done by the engine is positive.
(Some people rework the equations the other way - but the equation is the same with a different sign.)

C) the carnot effciency is the maximum efficiency.
You can tell which temperatures are lowest and highest by sketching isotherms on the graph - which would have saves you about 2/3 of that work ;)

7. Nov 26, 2014

### toothpaste666

so it goes clockwise because that way V2>V1 and the work is positive?
for C) how would I sketch isotherms on a PV graph?

8. Nov 27, 2014

### Simon Bridge

Look at the PV diagrams for isothermal processes.

9. Nov 27, 2014

### Andrew Mason

A heat engine performs net positive mechanical work. Work is done in only two sections a-b and d-c. Net work done BY the system can only be positive if Wa-b (ie. the area under a-b) is positive. This means the system is expanding from a-b. So it is being compressed from d-c.

I am not sure where you get W = VdP. This is a reversible cycle so $W = \int PdV$ where W is the work done BY the system, P is the pressure of the system and V is its volume. There is ZERO work done from c-a and from b-d.

To determine the net work done per cycle, take the area under a-b (work done by the system in expansion) and subtract the area under d-c (work done on the gas in compression).

AM

10. Nov 27, 2014

### toothpaste666

yeah i knew the W = VdP didnt seem right. So it would still be W = PdV but since dV = 0 then W = 0.
W = Wac + Wcd + Wdb + Wba
W = PdeltaV(ab) + PdeltaV(bd) + PdeltaV(dc) + PdeltaV(ca)
W = PdeltaV(ab) + 0 + PdeltaV(dc) + 0
W = (2atm)(3L-1L) + (1atm)(1L-3L)
W = 4atmL + (-2atmL)
W = 2 atmL

11. Nov 28, 2014

### Andrew Mason

Good. Now express that in Joules.

AM

12. Nov 29, 2014

### toothpaste666

oh right. thats about 203 J