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The Carnot Engine

  1. Aug 11, 2008 #1
    A firebox is at 750K, and the ambient temperature is 300K. The efficiency of a Carnot engine doing 150J of work as it transports energy between these constant-temperature baths is 60.0%. The Carnot engine must take in energy 150J/.600 = 250J from the hot reservoir and must put out 100J of energy by heat into the environment. To follow Carnot's reasoning, suppose that some other heat engine S could have efficiency 70.0%.

    (a) Find the energy input and wasted energy output of energy S as it does 150J of work.
    (b) Let engine S operate as in part (a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statement of the second law of thermodynamics is violated.

    Part (a) was done relatively easily but I have no idea what situation part (b) is bringing up (its a problem in conceptualizing what they're asking). How is engine S operating with the Carnot engine?
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  3. Aug 12, 2008 #2


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    I believe it means to operate the Carnot engine as a heat pump, letting the output of work by engine S be the input of work to the Carnot heat pump. After following the process, do you see how Clausius's statement is violated?
  4. Aug 12, 2008 #3
    Somewhat...the work being done on the heat pump is internal to the system, whereas the Clausius statement does not allow for such cyclic processes to continue constantly without external work being applied to the system. However, if the work done by engine S is applied to the heat pump to bring the exhaust energy back to the hot reservoir (while ignoring friction, conduction, or any unnecessary dissipation of energy), then how and where in the process is the limitation mentioned in the Clausius statement produced?
  5. Aug 12, 2008 #4


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    Let the engine S produce 150 J of work, and the Carnot engine use 150 J of work. What's the total (net) change in this closed system? In other words, what has been accomplished?
  6. Aug 12, 2008 #5
    Hmmm...if I'm thinking in the right direction...then there is no net work done throughout the process...I see...and that violates the Clausius statement...heh, well thats something.

    Also, if its not too much of a bother...in part (b)...whats the difference between the first find and the second find. Both of them are -35J (according to the answer given). I know the second is exhaust energy...what is the first one supposed to be?
  7. Aug 12, 2008 #6


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    Well, I think you need a bit more. No work has been done, true, but what has been accomplished? What has occured due to the two of them running?

    What is the Clausius statement? If you consider what it states, it might be easier to focus on the relevant happening.

    I'm a bit confused by the way you're using the word "find" here, but from the number you give, it appears that you are looking at the input/output effects of the total combined machine. This relates to answering my questions above.
  8. Aug 12, 2008 #7
    Would this fulfill it?: There is a cyclical process that is now continuous and working by itself...and so the entropy change for one cycle is 0. The Clausius statement indicates that such an accomplishment is not possible.
    To stray a little...why is the Clausius statement true? I'm not denying it...I'm just asking how it is true. Apparently, according to the statement, even if a system did not hold any friction or any other dissipation of energy, a limitation will still exist and external work is needed. What is the source of the limitation?
  9. Aug 12, 2008 #8


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    What is the Clausius statement? For example the Kelvin statement of the second law goes something like "No process can convert heat completely into work." What is the Clausius statement of the second law?

    (I'd rather not just write it out here, because it kind of gives away this problem. If you can't find it in your text you can find it on the web.)

    Also, why do you say the entropy change is zero? Try calculating it for a single cycle. What do you get?
  10. Aug 12, 2008 #9
    According to text, the Clausius statement states that "it is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work." Wouldn't that be somewhat similar to my explanation...or am I still missing something?

    The reason why I stated that the entropy of one cycle would be zero would be that I was assuming that the entire process would be reversible...although, looking back at things, my reasoning in that was off.
  11. Aug 12, 2008 #10


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    When the problem directly references the Clausius statement, I think you would need to constuct your answer in the language of the Clausius statement. You have found that the total effect of the combined device is to transfer 35 J (or something like that) of heat from the cold reservoir to the hot reservoir, and there is no other effect. That specifically is what the Clausius statement says is impossible.

    You could also think about the problem in terms of entropy, but in my opinion the problem was specifically wanting you to think about what Clausius had to say.

    What did you get for the change in entropy?
  12. Aug 14, 2008 #11
    Sorry it took so long to respond...it took me a while to get the question back in front of me. I understand the problem now. I wasn't conceptualizing the process right. Apparently there is a net transfer of heat energy of 35.7J from the cold reservoir to the hot reservoir without any net external work, which violates the Clausius statement.

    Since you brought up the matter, wouldn't I need to know the medium I'm using (the type of gas mixture, amount, etc...) to solve for the change in entropy (considering that the heat energy shifted from a colder temperature to a higher temperature)?
    Last edited: Aug 14, 2008
  13. Aug 14, 2008 #12


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    Since you know that the temperatures of the reservoirs is constant, and you know the heat transfer to and from each reservoir, you can simplify

    \Delta S = \int \frac{dQ}{T}

    directly for each reservoir.
  14. Aug 14, 2008 #13
    Oh...I see...and the sum would give total change in entropy for the system of the 2 reservoirs, right?
  15. Aug 14, 2008 #14


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    That sounds right; and once you calculate the number, it should give more evidence as to why that combined device is impossible.
  16. Aug 14, 2008 #15
    I see, I see...the entropy is apparently decreasing for this kind of device, making it impractical...thanks for the help.
  17. Aug 15, 2008 #16


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    Just wanted to point out that the entropy change of a system may very well decrease due to heat transfer from the system for a reversible process. It is the entropy generated in the system that may never decrease. It can only remain zero in the case of a reversible process. Hence, if the entropy transfer is negative (i.e. out of the system), the system entropy will decrease.

  18. Aug 15, 2008 #17

    Andrew Mason

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    This is a bit confusing. It is not clear what you mean by "system".

    A consequence of the second law is that the entropy of the universe cannot decrease in any process. If the system includes the reservoirs, then it is not correct to say that the entropy of the system can decrease (since this is a closed system with no external work being supplied).

    It is also unclear what you mean by "entropy generated in the system". Work is generated and heat flows in these thermodynamic processes. As a result of heat flowing over one full cycle, the entropy of each of the reservoirs changes (by dQ/T where dQ is the heat flow into/out of the reservoir and T is the temperature of the reservoir). The sum total entropy of those changes cannot be less than 0 if no external work is being supplied.

  19. Aug 15, 2008 #18


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    The increase of entropy principle does not imply that entropy of a system cannot decrease. The entropy change of a system can be negative during a process, but entropy generation cannot. If the process were reversible, then the entropy generation would be zero. However, since all processes in the real world are irreversible, it is said that the entropy of the universe is always increasing. That was my point to the OP.

    Here's an example: Take rigid tank and some refrigerant as the system. If heat is transferred out of the system, the entropy change of the refrigerant will be negative.

    By entropy generation I mean irreversibilities such as friction, mixing, chemical reactions, expansion, etc.

    Entropy generation and entropy transfer are not the same. Entropy transfer can occur by two mechanisms: heat transfer and mass flow. There is no entropy transfer associated with energy transfer as work.

  20. Aug 17, 2008 #19

    Andrew Mason

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    Entropy describes a thermodynamic state, not a quantity of something physical. So it is very confusing to speak about entropy generation. One does not speak about generating temperature, for example.

    This is not exactly a thermodynamic system. You need some place to deliver the heat to. The system has to include the hot reservoir.

    The entropy change of the refrigerant is always 0 after one complete cycle because it returns to its initial thermodynamic state. The entropy of the cold reservoir is negative in a refrigeration cycle simply because heat flows out of it (dQ<0 so dQ/T < 0). This is the case whether the cycle is reversible or irreversible.

    This is, again, confusing. If I take a Carnot engine system (engine + reservoirs) and I do work that is dissipated by friction externally to the system, is the change in entropy of the system not still zero?

    If entropy is a state function, how is it transferred or generated?

    There can be an entropy change associated with the transfer of work. Consider work done against friction force where the heat generated is transferred to a reservoir: the entropy of the reservoir will increase (by dQ/T = W/T).

  21. Aug 18, 2008 #20


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    Entropy generation is a well known term used in thermodynamic texts. It is not a term I coined by any means.

    Of course is it a thermodynamic system. Thermodynamics is the study of energy. Let’s put some numbers in and see what happens to the entropy change of this system. Say the tank has 5kg of R134a initially at 20 °C and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa.

    Since the tank is rigid [tex] v_2 = v_1 [/tex]. It is a closed system since no mass crosses the system boundary during the process. The entropy change of a substance is simply the difference between the entropy values at the final and initial states. Since the specific volume remains constant during this process the properties of the refrigerant at both states are:

    State 1:

    [tex] P_1 = 140 [/tex] kPa
    [tex] T_1 = 20 [/tex] °C

    Given these two intensive independent properties, one can see from a property table for R134a the following:

    [tex] s_1 = 1.0624 \frac{kJ}{kg \cdot K} [/tex]
    [tex] v_1 = 0.16544 \frac{m^3}{kg} [/tex]

    State 2:

    [tex] P_2 = 100 [/tex] kPa
    [tex] v_2 = v_1 [/tex]

    Using the property table again one finds:

    [tex] V_f = 0.0007259 \frac{m^3}{kg} [/tex]
    [tex] V_g = 0.19252 \frac{m^3}{kg} [/tex]

    From these the quality at state 2 is found to be 0.859.


    [tex] s_2 = s_f + x_2 \cdot s_{fg} = 0.07188 + (0.89)(0.87995) = 0.8278 \frac{kJ}{kg \cdot K} [/tex]

    The entropy change of the refrigerant during this process is:

    [tex] \Delta{S} = m({s_2} - {s_1}) = (5)(0.8278 – 1.0624) \frac{kJ}{kg \cdot K} [/tex]
    EDIT: For some reason latex is not being displayed properly here. Delta S should equal m(s_2 - s_1).

    = -1.173 kJ/K

    Notice the negative sign. This indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however, since it is the entropy generation that cannot be negative. This is why one must know the difference between the general term “entropy” and entropy generation. They are not the same. Equating the terms is more confusing, not less.

    BTW, this is not a cyclic device as you seem to think (in my example).

    This was my point all along which you seemed to have been saying wasn’t possible. The stipulation about being reversible simply points out that the entropy generation is zero for a reversible process.

    Friction is an irreversibility of a system. If friction is involved, the entropy generation is positive, not zero.

    Entropy generation is a measure of the magnitudes of the irreversibilities present during a process.

    Being an organized form of energy, work is free of disorder or randomness and thus free of entropy. The work done against a frictional force results in an entropy change due to the irreversibility (i.e. the friction). It is not a result of the work itself, rather it is the fact that the organized energy (work) is being degraded to a less useful form (heat). Hence, the entropy generation is increased.

    Last edited: Aug 18, 2008
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