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Steam Flow question

  1. Jun 20, 2012 #1
    Hi, here's a problem I've recently come across, and it's bugging me that I can't wrap my mind around it.

    I need to calculate the pressure of steam in a ∅12" pipe. I know that the steam flows at 8700 lb/hr over a 24 minute period, and that the average steam temperature is 176°F. I also know that the steam is not very superheated (if at all), and should be around 10-20 psi.

    My question is: Is there enough information? If so, can someone point me down the right path? I haven't done thermo in a while.

    EDIT: I posted this question before defining my problem. Here it is:-------------------------------------------------------------------------------------------------------

    This is a non-mixing heat exchanger problem.

    I need to calculate the pressure of steam in a ∅12" pipe. I know that the steam flows at 8700 lb/hr over a 24 minute period, and that the average steam temperature is 176°F. I also know that the steam is not very superheated (if at all), and should be around 10-20 psi. I have to find the pressure in the pipe, in addition to the volume of condensate 'released' if the steam gives up 6752 KW, which is the energy it takes to heat the mass of water described here:

    Water flows in the cold side at 800 GPM (also over a 24 minute period) and must be heated from 110° to 170°F.
     
    Last edited: Jun 20, 2012
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  3. Jun 20, 2012 #2

    Q_Goest

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    Hi natedogg, welcome to the board. Do you remember how to use steam tables? Can you determine from the tables at what temperature water boils at atmospheric pressure? How about at 10 or 20 psig? You should find that as you go up in pressure, the temperature at which water boils also increases. So although we all know that water at atmospheric pressure boils at roughly 212 F, if you increase pressure, the water will boil at a higher temperature.

    Turn that around and consider, at what pressure will water boil at 212 F? Obviously, the answer is atmospheric pressure. If you have water boiling then at 176 F, you should be able to look up the pressure it boils at on your steam tables. The answer is less than atmospheric pressure. Water at 176 F and at atmospheric pressure does not boil. It's hot but it has a way to go before it boils.

    Consider also, what does the diameter of the pipe have to do with the temperature or pressure water boils at? And does a flow rate through a pipe change the pressure or temperature water boils at? Note that your steam tables won't reference flow rates or geometry of a container.
     
  4. Jun 20, 2012 #3
    Q_Goest, thanks for replying.

    Interpolating in the steam table gives me a saturation pressure of about 6.8 psi at 176 F. But does that necessarily mean that the pressure in the pipe is 6.8 psi?

    I remembered I could pull the ideal gas trick, and I then calculate a pressure of nearly 6.9 psi - nearly the same. A Tv diagram on this PDF says that I can make the ideal gas assumption for steam under these conditions.

    So I guess the steam is at saturation pressure...

    I should have given my whole problem in my first post: I have to find the pressure in the pipe, in addition to the volume of condensate 'released' if the steam gives up 6752 KW. (This is a heat exchanger problem). So that means I need to find the quality of the steam...but how?
     
  5. Jun 20, 2012 #4

    Q_Goest

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    If this is a homework question, it needs to be moved to the forum for engineering homework here:
    https://www.physicsforums.com/forumdisplay.php?f=158

    Also, when you post in the homework section, a format for your post will appear:
    Anyway, to answer your question, yes, saturated steam at 176 F is at a pressure of approximately 6.8 psia. Note this is absolute pressure, not gage pressure.

    By "released" I believe all they are saying is that the inlet to the heat exchanger is saturated gas and given the flow rate, and given the heat removed at the exchanger, they are asking for the outlet quality. Would you agree?

    This is a question about steam tables. Note that you can draw an open control volume around the heat exchanger and apply the first law of thermo to this to determine the outlet conditions. Do you know how the first law reduces for this situation? If you need help, and this is homework, we can move this thread and I'll help you out in the engineering homework section.
     
  6. Jun 20, 2012 #5
    No, this is a professional question, but it certainly follows like a homework question. If I should move it, let me know. I edited the first post to include the entire problem I need to solve.

    And yes, I do agree that the outlet quality is what I'm after.

    From the standpoint of the first law, conservation of mass yields:

    m1 = steam in
    m2 = cold water in
    m3 = steam out
    m4 = condensate out
    m5 = hot water out

    m1 = m3 + m4
    m2 = m5

    Conservation of energy yields:

    m1*h1 + m2*h2 = m3*h3 + m4*h4 + m5*h5

    So now,

    m1*h1 + (m1-m4)*h3 + m4*h4 = m2*(h5-h4)

    At 800 GPM over a 24 minute period and with 1 gal = 0.13368 ft3, and then looking up 110°F water to be 8 lb/gal, I find that 153,000 lb of water goes through my heat exchanger.

    Subbing values,

    (3480 lb)(1135 Btu/lb) + (3480 lb - m4)*(1133.9 Btu/lb) + m4*h4 = (153,000 lb)(138.02-78.02)(Btu/lb)

    to solve for m4 and h4. Can't quite solve yet - I need another equation (or I've missed something). Hints?
     
  7. Jun 20, 2012 #6

    Q_Goest

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    The way you're setting this up isn't incorrect, but it fails to capture a key bit of information. The temperature of m3 and m4 is the same temperature and pressure. So when you split it up like this, you are no longer able to properly use the steam tables to determine quality of the mixture of m3 and m4.

    Quality is defined as the ratio of the mass of vapor divided by the total mass, so using your notation, the quality of the water coming out of the heat exchanger that you denote by m3 + m4, can be rewritten as:

    x = m3 / (m3 + m4)
    or
    x = m3 / m1

    Also, since m1 is your steam, and conservation of mass says what goes in also comes out, you can just keep referring to m1 as the flow on that particular side of your heat exchanger that comes from the steam. Then use m2 to signify the water since again, there is no change in flow in and out for this water.

    Once you do this, you can eliminate m3, m4 and m5 and rewrite your conservation of energy equation above:
    m1*h1i + m2*h2i = m1*h1o + m2*h2o
    Where subscript i = inlet condition and subscript o = outlet condition.

    Since you have the inlet (110 F) and outlet (170 F) temperature of m2, and assuming the water leg is approximately at atmospheric pressure (or insert actual pressure) and doesn't change phase, this determines the inlet and outlet enthalpy. Now you can solve for the outlet enthalpy of m1, h1o, since you have all the other values which you can take from your steam tables. For example, h1i is simply the enthalpy of saturated water vapor at 176 F.

    Once you solve for h1o, you can plug that in to the definition of vapor quality:
    x = (h1o - hf ) / hfg
    Where hf is the enthalpy of saturated liquid and hfg is the enthalpy of the saturated gas minus the enthalpy of the saturated liquid. See the definition of vapor quality: http://en.wikipedia.org/wiki/Vapor_quality

    Note that you're also given the total energy exchanged between the steam and water, so there is a second way to do this that should give you the same answer. You can apply the first law and show that the change in enthalpy of either flow stream is equal to your energy exchanged which you give as 6752 kW.
     
  8. Jun 21, 2012 #7
    Thank you for guiding me through this. Trust me, I don't plan on dragging you through this much longer.

    However, I get a negative enthalpy when I solve for h1o.

    From the steam tables,
    h1i = 1135 (Btu/lb) [saturated steam]
    h2i = 78 [110 F saturated water]
    h2o = 138 [170 F saturated water]

    With masses of m1 = 3480 lb and m2 = 153,600 lb, h1o = -1513 Btu/lb, a nonsensical value.
     
  9. Jun 21, 2012 #8

    Q_Goest

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    A negative value for enthalpy isn't a problem. Each database for steam can differ significantly. As long as you use the same database for a given problem, you will be fine because it's only the difference between the two values for enthalpy that matters. For example, my database has a value of -5733 Btu/lbm for saturated steam at 176 F.

    Let's go over the math then. Please verify these are the values you're using:
    m1 = 3480 lbm
    m2 = 154,000 lbm
    h2i - h2o = -40 Btu/lbm
    h1fg = 996 Btu/lbm

    When I do the calculation to solve for h1o, I get a very large, negative value which put the steam well below the solidification temperature. In other words, the steam comes out frozen. So there's a problem with the set up. Either there's not enough steam, too much water, or the dT of the water is too high. Check those three values and see if one of them is incorrect.

    Note that you can do a quick check just to verify, the absolute value of energy difference of the water is:
    m*dh = 6,160,000 Btu
    while the energy difference for the steam assuming it all condenses and becomes saturated water is: 3,466,000 Btu

    So the condensation of the steam is roughly half the energy needed to warm the liquid water. Clearly, there's a problem with one of the assumptions or with the masses.
     
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