Thermal efficiency in a reversible cycle for an ideal gas pV diagram

In summary, a heat engine takes 4.0 moles of an ideal gas through a reversible cycle, with an isothermal process at 600 K and volumes of 0.04 m^3 and 0.10 m^3 at points a and c, respectively. The molar heat capacity at constant volume for the gas is 30 J/mol*K. Using the equations P = nRT, Q = nC_{v}\DeltaT, Q = nC_{p}\DeltaT, W = P\DeltaV, and W = nRT ln \frac{V_{f}}{V_{i}}, the thermal efficiency of the engine is calculated to be closest to 0.10 when using the formula e
  • #1
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Homework Statement



A heat engine takes 4.0 moles of an ideal gas through the reversible cycle abca, on the pV diagram, as shown. The path bc is an isothermal process. The temperature at c is 600 K and the volumes at a and c are 0.04 m^3 and 0.10m^3, respectively. The molar heat capacity at constant volume, of the gas, is 30 J/mol*K. In the figure, the thermal efficiency of the engine is closest to:
a) 0.07
b) 0.10
c) 0.15
d) 0.22
e) 0.30

Homework Equations



[tex]P = nRT[/tex]
[tex]Q = nC_{v}\Delta[/tex]T
[tex]Q = nC_{p}\Delta[/tex]T
W = P[tex]\Delta[/tex]V
[tex]W = nRT ln \frac{V_{f}}{V_{i}}[/tex]
[tex]e = \frac{W}{Q_{h}} = 1 - \frac{Q_{c}}{Q_{h}}[/tex]

The Attempt at a Solution



First, I calculated the temperature and pressures at points a, b, and c using PV = nRT.

Point A:
V = 0.04 m^3
T = 600 K
P = (4)(8.31)(600) / 0.04 = 498000 Pa

Point B:
V = 0.04 m^3
T = PV / NR = (199440)(0.04) / (4)(8.31) = 240 K
P = 199440 Pa

Point C:
V = 0.10 m^3
T = 600 K
P = nRT / v = (4)(8.31)(600) / .10 = 199440

Then I calculated the heat absorbed on the path ab in kJ:
[tex]Q = nC_{v}\Delta[/tex]T
Q = (4)(30)(600-240) = 43200 ~ 43

Then I calculated the heat absorbed on the path ca in kJ:
[tex]Q = nC_{p}\Delta[/tex]T
Q = (4)(30 + 8.31)(-360) = -55166.4 ~ -55

Then I calculated the work done on the path bc in kJ:
[tex]W = nRT ln \frac{V_{f}}{V_{i}}[/tex]
(4)(8.31)(600)ln (.10/.04) = 18274.5 ~ 18
[/tex]
Then I calculated the work done on the path ca in kJ:
W = P[tex]\Delta[/tex]V
(199440)(.04-.10) = -11966.4 ~ -12To calculate the thermal efficiency, I tried this:

[tex]e = \frac{W}{Q_{h}} = 1 - \frac{Q_{c}}{Q_{h}}[/tex]

[tex]e = \frac{W}{Q_{h}}[/tex] = (18-12) / (43 -55) = -0.5

Obviously that cannot be correct.

The correct answer is B. 0.10.

How should I calculate the efficiency?
 

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  • #2
Please let me know, this problem has been plaguing me forever!
 
  • #3
Any help?
 
  • #4
Please? I hate bump posts, but I'm kind of desperate on this one.
 
  • #5
You need to include Q for path bc.
 
  • #6
Also, looking again at your derivation:

I don't think Qh would include the -55 kJ of path ac; wouldn't that be part of Qc instead?
 
  • #7
Since there is no change in temperature from bc, wouldn't there be no heat for bc?
 
  • #8
That's a common misconception.

If no heat were added, an expanding gas will cool because it is doing work (i.e. losing energy) to the environment. To maintain temperature, it is necessary to add heat equal to the amount of work done in path bc.

Has your class covered the Carnot Cycle? Note the distinction between constant temperature and adiabatic (no heat exchange) paths in that cycle.

Hope that helps.
 
  • #9
Oh I see, I'm not sure how to calculate the amount bc? Would it just remain the same as in ab?
 
  • #10
No (or at least I don't see why it would).

If you can figure out both the work done and the energy change along bc, then you can use the 1st law to calculate the heat.
 
  • #11
So the heat would be just equal to the work done right, because there is no change in internal energy?
 
  • #13
Using that info, here's what I tried:

[tex]Q_{ab} = +43[/tex]
[tex]Q_{bc} = +18[/tex]
[tex]Q_{ca} = -55[/tex]

[tex]W_{ab} = 0[/tex]
[tex]W_{bc} = +18[/tex]
[tex]W_{ca} = -12[/tex]

So first, I tried adding all the heat up, and got a value of 6. Then I added up all the work and got 6. 6/6 = 1 Obviously wasn't going to work.

I guessed that because on the path ca, the heat is negative, that wouldn't be Qh because it would be the heat expelled.

So then what I did was just took 43+18 for Q, which comes out to be 61, and then took the net work of 6, and did 6/61 = .098361, which I think matches to the answer of .1.

I just want to make sure my logic was correct on this.
 
  • #14
Yes, you've got it. :smile:

Efficiency is (total work) / Qh

Qh refers to any heat absorbed. Presumably, this is the energy cost to operate the heat pump, whether it's from burning coal or gasoline or nuclear fuel, etc.

Qc is any heat that is given off, and is not a part of the "energy cost".
 
  • #15
Thank you very much! :D
 

What is thermal efficiency?

Thermal efficiency is a measure of how well a system converts heat energy into mechanical work. It is expressed as a percentage and is calculated by dividing the work output by the heat input.

What is a reversible cycle?

A reversible cycle is a thermodynamic process that can be reversed without leaving any trace on the surroundings. This means that the system will return to its initial state after completing a full cycle.

What is an ideal gas?

An ideal gas is a theoretical gas composed of randomly moving particles that do not interact with each other. It follows the ideal gas law, which describes its behavior in terms of pressure, volume, and temperature.

How is thermal efficiency calculated in a reversible cycle for an ideal gas pV diagram?

In a reversible cycle for an ideal gas pV diagram, thermal efficiency is the ratio of the area enclosed by the cycle to the area under the isothermal curve. This can be calculated using the formula: efficiency = (work output / heat input) x 100%

What factors affect thermal efficiency in a reversible cycle for an ideal gas pV diagram?

The thermal efficiency in a reversible cycle for an ideal gas pV diagram is affected by the temperature, pressure, and volume of the gas, as well as the type of process (isothermal, adiabatic, etc.). It is also influenced by the efficiency of the heat engine or device used to convert heat energy into work.

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