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Thread: Thermodynamics: Required work being done, in spite of irreversibility and no additional force being spent

  1. #1 Thermodynamics: Required work being done, in spite of irreversibility and no additional force being spent 
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    Hello. I'm new here. I'd like to ask a question which has been puzzling me for quite some time. The illustration is attached below the text.

    Imagine a tall, airtight container C1 having a certain amount of liquid (say water) at an air pressure of P1. An airtight valve V1 separates C1 from another container C2 containing air at a pressure of P2. C2 is at an elevation with respect to C1. Pressure P1 is greater P2. When this valve is opened, the water is forced up through the tube into C2. The height through which it rises h, is given by P1'-P2'/ (rho g), where P1' and P2' are the final pressures in containers 1 and 2 respectively.

    The time taken to reach equilibrium condition is t1.

    Now, supposing there is a friction loss 4flv2/2gD, where f is Fanning friction factor, v is the varying velocity of the flow and D is the diameter of the pipe. As a result of this, there is a kinetic energy loss, and therefore the flow is slower. The time taken to reach equilibrium condition is now t2.
    In either case (with or without friction loss), the work done is the same- the same mass of water is moved through the same height, and the final pressures are also the same.

    In the first case, only PV work is done and there is no change in internal energy (adiabatic). The process is also reversible- if the same amount of energy is given to a pump/blower operating between the two air chambers, the initial pressures P1 and P2 are reached in container C1 and C2 respectively.

    So how is it that in spite of the irreversibility occurring in the second case, the work done is the same (though it takes a longer time to be done)?

    Please enlighten me.

    state1.JPGstate3.JPGstate2.JPG
    Last edited by radi1113; 11-16-2013 at 09:04 AM.
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  2. #2  
    KJW
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    In the frictionless case, you'll have oscillations of the water flowing in and out of the two tanks. At the equilibrium point, the kinetic energy will be at its maximum. In the friction case, the loss due to friction will be of this kinetic energy and the system will eventually become stationary at the equilibrium.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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  3. #3  
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    Yes, in the frictionless case, water will have higher kinetic energy at the equilibrium point.

    But work done in both the cases is the same is it not?
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  4. #4  
    KJW
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    Quote Originally Posted by radi1113 View Post
    But work done in both the cases is the same is it not?
    There's no overall work being done.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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  5. #5  
    KJW
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    The system starts out at maximum potential energy. The equilibrium point is at minimum potential energy. In the frictionless case, the kinetic energy at the equilibrium point will be the difference between these two potential energies, with the kinetic energy converting to potential energy away from the equilibrium point. In the friction case, the energy loss will be the difference between the two potential energies. If a turbine is placed between the two tanks, then the difference in potential energy will be shared between the turbine and friction.
    A tensor equation that is valid in any coordinate system is valid in every coordinate system.
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  6. #6  
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    Thank you... No overall work is being done...Got it..
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