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Violation of first/second law in a Cylic process

  1. Oct 20, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Explain whether or not the first or second law of thermodynamics are violated in the below cyclic process (see attachment)

    2. Relevant equations
    Clausius Inequality, Carnot efficiencies, ΔS ≥ 0 for a cyclic process.


    3. The attempt at a solution
    The first law is satisfied. The fact there are two hot reservoirs instead of one is putting me off a little bit since the K-P version of the second law only applies to heat entering the engine from a single hot reservoir. The generalized Clausius' inequality says that ##\sum_i \frac{\delta Q_i}{T_i} \leq 0 = \frac{600}{600} + \frac{800}{800} - \frac{400}{400} = 1## which is not less than zero. So the second law is violated. Is this all?

    I was wondering if I could replace the two separate hot reservoirs with a single reservoir at 1400K? When I do this, I end up with ##\eta_C = \eta_E## which is not a violation of the second law, so I must have gone wrong in my attempts.

    Many thanks.
     

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  2. jcsd
  3. Oct 20, 2013 #2

    Simon Bridge

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    Sometimes it helps to express the laws in a more general sense - avoids the need for wonderful exotic formulas.

    1st law is conservation of energy.
    ... i.e. does tot energy out = tot energy in?

    2nd law is the entropy law - entropy must increase or stay the same.
    ... what does that mean about the input energy and work?

    Note: if the efficiency of this engine is equal to the carnot efficiency, what does it mean about this engine?
     
  4. Oct 20, 2013 #3

    CAF123

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    This is the case here. Heat in = 800 + 600, while heat out = 1000 + 400.
    When the engine accepts heat it's entropy has increased. When heat is expelled or deposited into the colder reservoir, this tends to decrease it's entropy. The net entropy must be ≥ 0. Since the state of the substance is unchanged in an engine, the entropy of the surroundings has increased, which is accounted for in the fact that the engine does work.

    The engine operates at the maximum possible efficiency. Is it okay to replace the two separate hot reservoirs as a single one?
     
  5. Oct 20, 2013 #4

    Simon Bridge

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    You know - I don't actually know. If the effect of the single reservoir does not change the rate that heat enters the engine, then I'd say yeah.

    ... so are you saying that the second law would be satisfied if all the energy went to work?
     
  6. Oct 20, 2013 #5

    CAF123

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    No, this would violate the K-P formulation of the second law.
     
  7. Oct 20, 2013 #6

    Simon Bridge

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    Never mind specific formalism - that misleads you into focussing on particular equations.
    What does it mean? Like: physics.

    Basically some energy has to go to the cold reservoir, but not all of it.

    It does that in this case - which is a good sign for the 2nd law (and, thus, the engine).

    And you can do all that by inspection. Just about.

    So where did you suspect you'd gone wrong again?
     
  8. Oct 20, 2013 #7

    CAF123

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    Let me explain my attempt using the Clausius' inequality in words:
    Increase in entropy of the engine is by definition Q/T where Q is the heat that enters the engine and T is the temperature of the reservoir supplying this heat. Suppose the engine is in a state ##S_i## initially. Then the increase in entropy is ΔS = Q1/T1 + Q2/T2, Q1=T1=800, Q2=T2=600. Heat 400J flowing out tends to decrease the entropy of magnitude Q3/T3 , Q3=T3=400.

    So the net change in entropy is Q1/T1 + Q2/T2 -Q3/T3 which is not less than zero, so that's why I thought I had a violation of 2nd law.
    However, comparing the efficiencies I see that eff of engine = eff of Carnot which is allowed. Since the formulations are equivalent I have a contradiction here.
     
  9. Oct 20, 2013 #8

    Simon Bridge

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    Doesn't the second law state that the change in entropy has to be bigger than or equal to zero[edit] scrap that - I must be too tired.
    See you tomorrow.

    ... and the net change in entropy has to be zero though.
    Awkward. OTOH: this is where you replaced the reservoirs right?
    Maybe just adding the temperatures does not make it equivalent?

    Discussion on handling a setup with two hot reservoirs:
    https://www.physicsforums.com/showthread.php?t=605332
     
  10. Oct 20, 2013 #9

    CAF123

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    Only for a reversible process, right? I don't think we have this criterion.
    Thanks for the link - I see now that they are not equivalent and I think i understand why: heat flows are additive and so the heat input into the engine is always the 1400J. But this does not mean that the temperature of the reservoir need be 1400K. (for sure it has to be at least this). Since the efficiency of an engine is dependant on the temperature of the reservoirs, the efficiency of the single combined engine is different to that of them separately.
     
  11. Oct 20, 2013 #10

    D H

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    No! You can't do that. That would be a great subject for a troll physics cartoon. If what you propose were allowed, suppose you have a single hot reservoir at temperature T that transfers ΔQ/cycle to the engine. Separate it into two reservoirs, each contributing half of the ΔQ/cycle. Now replace those two reservoirs with a single reservoir at temperature 2T that transfers ΔQ/cycle to the engine. Rinse and repeat. Infinite energy!


    What you can do is replace the two hot reservoirs with an equivalent single reservoir whose temperature is somewhere between 600K and 800K. Take the upper limit: Can an 800K reservoir transfer 1400 J/cycle to a heat engine that produces 1000 J/cycle and transfers 400 J/cycle going to a cold reservoir at 400K?

    Another way to look at this: You could certainly get more out of the system if that 600K hot reservoir was replaced with an 800K hot reservoir. Now you can combine those two 800K reservoirs, but it's not a 1600K reservoir. It's an 800K reservoir that is transferring 1400 J/cycle with 400 J/cycle going to a 400K cold reservoir and 1000 J/cycle going into work. If this system violates the second law, then certainly that 800K / 600K pair of reservoirs violates the second law.

    Yet another way to look at it is to look at the change in entropy, the way you did in the opening post. If entropy is decreasing you have a violation of the second law of thermodynamics. End of story.
     
  12. Oct 20, 2013 #11

    CAF123

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    D H,
    Can you see any error in #7, in particular when I sum all the heat flows into/out of the engine, noting the signs, I get that the second law is violated. Is this correct?
     
  13. Oct 20, 2013 #12

    CAF123

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    How can a reservoir at a temperature less than 1400K supply 1400J of heat?
     
  14. Oct 20, 2013 #13

    D H

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    Huh? Temperature and heat transfer rate are incommensurable quantities. Moreover, temperature is an intrinsic quality while energy is an extrinsic quality. Want more heat transfer from a heat reservoir? Increasing the temperature is one option. Increasing the size of the reservoir is another.
     
  15. Oct 20, 2013 #14

    CAF123

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    Ok thanks. Did you see #11?
     
  16. Oct 20, 2013 #15

    Simon Bridge

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    Oh good - someone awake started helping - good to see.
     
  17. Oct 21, 2013 #16

    CAF123

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    My conclusion at this stage is that the second law is violated, via my attempt in the OP using the Clausius inequality. Is it ok?
     
  18. Oct 21, 2013 #17

    CAF123

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    Why must it be in this range? It is correct to say that it must be > 400 K otherwise no heat would flow (unless we do work, but that is not the setup), but I am not quite sure how you ascertained that range.

    The efficiency of the engine is given by 1-Qout/Qin = 1-400/1400 = 5/7. In your example, if we combine both reservoirs to a single reservoir at 800K, then this gives a Carnot efficiency of 1-400/800=1/2. The engine thus contradicts Carnot's Theorem. Why does a contradiction of Carnot's Theorem necessarily imply a contradiction of the second law?

    Thanks.
     
  19. Oct 21, 2013 #18

    D H

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    Suppose you have three heat engines, each of which involves extracting 1400 J/cycle from a set of hot reservoirs, with some of the heat transferred and the rest used to perform work. One of those heat engines is the one outlined in the problem. The second set comprises a single 800K hot reservoir, the third set comprises a single 600K hot reservoir. The heat engine described in the problem is at most as efficient as the system with an 800K hot reservoir, and at least as efficient as the system with an 600K hot reservoir. In other words, the effective temperature is somewhere in between 600K and 800K.

    Carnot's theorem is one of many, many ways to express the second law of thermodynamics. A violation of Carnot's theorem is a violation of the second law.
     
  20. Oct 21, 2013 #19

    CAF123

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    I'd like to prove this. The maximal efficiency of the 2nd set is 1-400/800 = 1/2. The maximal efficiency of the 3rd set is 1-400/600=1/3. So I need to show that the maximal efficiency of the first set (the one in the problem) is 1/3 < ηcombined < 1/2.

    Let the 800K (=T1) reservoir supply heat Q1 and the 600K(=T2) supply heat Q2. For a reversible Carnot engine, Q1/T1 + Q2/T2 - Q0/T0 = 0. I need to extract Q1+Q2 from this equation to get an expression in terms of the Ti, then sub that into ηcombined = 1- (Q0)/(Q1+Q2)? I am having difficulty obtaining this expression.
     
  21. Oct 21, 2013 #20

    D H

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    Q1/T1 + Q2/T2 - Q0/T0 = 0 is equivalent to Q1/T1 + Q2/T2 = Q0/T0. The equivalent single reservoir Carnot engine is (Q1+Q2)/Ti - Q0/T0 = 0, or (Q1+Q2)/Ti = Q0/T0. Therefore Q1/T1 + Q2/T2 = (Q1+Q2)/Ti. Solve for Ti in terms of those other parameters. You'll find that Ti must lie between T1 and T2 if Q1>0 and Q2>0.
     
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