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Filling a tank with steam

  1. Feb 3, 2014 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    A tank made of a material with negligible heat capacity is connected to a pipeline containing steam at 20 kPa and 250°C. Assume the tank is well insulated and that it has a volume of 1 m3. The tank is filled with steam until the pressure equilibrates, then it is disconnected from the pipeline. How much steam is in the tank and what is its temperature if:

    a)the tank is initially evacuated?
    b) the tank initially contains steam at 1 kPa and 125°C?


    2. Relevant equations



    3. The attempt at a solution

    I can't proceed with this problem because I don't know what the conditions are when ''pressure equilibrates''.
     
  2. jcsd
  3. Feb 3, 2014 #2
    The pressure equilibrates when the pressure in the tank reaches 20 kPa.

    This is a tricky problem. Please take some time to carefully dope it out.

    Chet
     
  4. Feb 3, 2014 #3

    Maylis

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    I'm working on part A right now, and this is what I am thinking.

    First and foremost to demonstrate that I know the trick, I know the rate that mass is coming in is not a constant, since as the pressure increases it becomes harder for the gas to enter, so it slows down.

    However for part A, is it not the case that the gas in the tank should be 250 C and 20 kPa at 1 m^3? If so, then I just look for the specific volume at that temperature and pressure on my steam table, then divide volume by specific volume and out comes the mass of the vapor inside the tank. Is this physically correct?
     
  5. Feb 4, 2014 #4

    Maylis

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    Here is my attempt at part a and b. I am being cautious with the mass flow rate in, knowing its a function of time. I am wondering if what I am doing with those integrals is even legitimate math right now, referring to my substitutions for total mass as a function of time with the integral for mass flow rate. The solution for part a is complete (not necessarily correct), and part b is a partial solution.

    I also attached the steam tables used for this problem.
     

    Attached Files:

  6. Feb 4, 2014 #5
    Let's take a step back.

    Have you studied the form of the First Law of Thermo applicable to open systems yet? If not, we can still identify a closed system for this problem that the usual (closed system) form of the First Law can be applied to.

    Chet
     
  7. Feb 4, 2014 #6

    Maylis

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    They're essentially the same thing, right? Meaning the closed system is just a special case of the open system? An open system means mass and energy crosses the control surface, and closed means only energy crosses. Now I realize that I labeled it as being a closed system!! Oops

    Although I mislabeled it as being closed, my mass and energy balance accounts for enthalpy coming into the control volume boundary.
     
  8. Feb 4, 2014 #7
    Please excuse me. I did not look at your solution for part (a) carefully enough. Your final result between the final specific internal energy and the inlet specific enthalpy is correct (IMO). What did you get for the final temperature? (Under the assumption of ideal gas behavior, you would get T=γTin). Your answer should be close to this.

    Another way of arriving at the same result is to envision an imaginary boundary around the steam that eventually gets into the tank. Then treat the contents of the tank plus the steam within this boundary as a closed system. Since the system is adiabatic, the change in internal energy is equal to the work done to push the steam into the tank. This work is just PsteamV = mPsteamvsteam, where V is the volume of steam that enters the tank, vsteam is the specific volume of the steam, and m is the mass of steam that eventually enters the tank. The change in internal energy of the system is just m(u-usteam), where u is the final specific internal energy of the tank contents. This leads to the same result you obtained.

    I haven't checked over what you did at part (b), but I will shortly. However, I have confidence that you probably did that correctly.

    chet
     
  9. Feb 4, 2014 #8
    OK. In part (b), I get a slightly different answer than you:

    (m0+min)u - m0u0=hinmin

    Thoughts?

    Chet
     
  10. Feb 4, 2014 #9

    Maylis

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    I found the specific volume when the internal energy is the same as the enthalpy coming in for part (a), and got the mass of steam to be .0639 kg and its temperature is 405.1 C from linear interpolation.

    By the way, the quiz today made us write an energy balance for a problem exactly like this one, the first one the normal way with the control volume being the tank, then the second way writing the energy balance for the control volume including the steam and being a closed system...wish I paid more attention to your post now :(

    To follow up, I redid the energy balance and got what you got in post #8. The only thing that troubles me right now is that I am doing an integration of d(mU)...and what bothers me about that is that neither m nor U are constant, so I would like to do a product rule on that differential d(mU)/dt rather than integral the product mU since I don't think its constant.

    I'll begin working on part (b) now. We were given a hint that you will probably want to use Excel to guess and check for the answer. Does the pressure in the tank equal 20 kPa when it equilibrates, even though 1 kPa was in it to begin with?
     
    Last edited: Feb 4, 2014
  11. Feb 4, 2014 #10

    Maylis

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    Now that I am at the equation in post #8, I have two unknowns, m_in and U_final, and only one equation, which is that in post #8.

    To find the final internal energy and mass that entered the system, I am basically guessing the two values and waiting to see if the two sides of the equation will be the same value.

    If this is the route I must take, I wonder physically if the internal energy and mass that entered should be less than or more than part (a) so I can make a better guess at the value.
     
  12. Feb 5, 2014 #11
    a) For the first question the answer is 1m^3 steam at 250 degree celcious and 20 Kpa pressure.
    b) For the second question, the problem is not standard, it's need volume flow rate also, if the volume flow rate sufficient enough to provide sufficient time for thermal equillibrium too, then the answer will be same as first one.
     
  13. Feb 5, 2014 #12
    This should help. If V is the volume of the tank, then m0=V/v0, where v0 is the initial specific volume in the tank. You already discovered this. But, (m0+min)=V/v, where v is the final specific volume in the tank. This is the additional equation you need. Actually, V will cancel out of your equations when you substitute into the energy balance. h, u, and v are functions only of the final temperature (since the final pressure is known).

    Chet
     
  14. Feb 5, 2014 #13
    The integral of d(mU)/dt with respect to t is just Δ(mU).

    Chet
     
  15. Feb 5, 2014 #14
    Would the final pressure still be 200 kPa even though there is some initial steam in the tank? If so, why is it the same as in part a, where there is no initial steam? I also don't get how we can get final temperature from final pressure. I know once we get the temperature, we can solve for h and u using the heat capacities of steam.
     
  16. Feb 5, 2014 #15
    The tank is connected to the steam line through a valve. If the pressure in the steam line is higher than the pressure in the tank, steam will continue flowing in through the valve until the two pressures equilibrate. There is essentially an unlimited supply of 200kPa steam in the steam line. The pressure in the tank could not get higher than 200kPa, because, if it did, steam would flow out of the tank back into the steam line until the pressures equilibrated. So, at equilibrium, the pressure in the tank and the pressure in the steam line must match.
    You derived the energy balance equation for part (b):

    (m0+min)u - m0u0=minhin

    where u0 and hin are known. In addition, we have:

    m0=V/v0

    (m0+min)=V/v

    where v0 is the initial specific volume of the tank contents, and v is the final specific volume of the tank contents. If we combine these equations, we get:
    [tex]\frac{u}{v}-\frac{u_0}{v_0}=(\frac{1}{v}-\frac{1}{v_0})h_{in}[/tex]
    Rearranging:
    [tex]\frac{(u-h_{in})}{v}=\frac{(u_0-h_{in})}{v_0}[/tex]
    Everything on the right hand side of this equation is known. At 200 kPa, u and v on the left side of the equation are unique functions of T (in the steam tables). We need to find the value of T that makes good on this equation. The solution for T is going to be somewhere between 125C and 405.1 C. This is a non-linear equation in T. Try solving it using the "half interval method."
     
  17. Feb 5, 2014 #16
    Ahh I think I understand now. Thank you very much! You've been so helpful :) I'll try the problem again later and come back if I have any more questions (which hopefully I won't).
     
  18. Feb 6, 2014 #17

    Maylis

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    Okay, I finally got an answer following chet's suggestion, except I just used the left hand side of unknowns and equated the final specific volume to the final specific internal energy using the steam table, and since the right hand side of the equation is a constant, I just fiddled with the specific volume in excel until the expression on the left with one unknown (since v is now related to U) was equal to the constant on the right.

    My final answer was T= 383.2 C and m_steam = .0661 kg.

    I was wondering if you could, Chet, could comment about the concept why the temperature is less than when it was evacuated, as well as why less mass of steam is present before this thread comes to a close.
     
  19. Feb 6, 2014 #18
    Yes. This approach is what I actually had in mind.
    Less steam is able to enter because there is already steam present in the tank to begin with. The temperature doesn't rise as much because less high enthalpy steam enters. Another way of looking at it is that, if you assume that the entering steam comes through the inlet valve with a pressure drop but no temperature change (zero Joule Thompson effect), the steam that has already entered the tank gets compressed adiabatically back up to its initial pressure by the subsequent steam that enters. If less steam enters, less compression occurs. A third way of looking at it is that, if less steam enters, less work is done to force it into the tank by the steam behind it within the pipe.

    Chet
     
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