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Heating water in a long container

  1. Aug 4, 2009 #1
    This is an oversimplification of a practical problem I am currently trying to design for. Suppose, we have a very long cylinder closed at both ends which contains water at a given pressure P1 and temperature T1.

    P1 and T1 are such that water is in a compressed state (subcooled). One end of the vessel is heated slowly. The cylinder is sufficiently long so that there is temperature gradient within the volume of water inside the cylinder. As one can imagine the pressure of water increases.

    1. How to determine the increase in pressure?

    Additionally, a relief valve is provided. The valve opens to relieve pressure once the pressure reaches P2.

    2. How much volume will need to be vented to relieve the built up pressure?
  2. jcsd
  3. Aug 5, 2009 #2
    Well, assuming you don't heat the water past it's boiling point, then the additional pressure inside of the tube is due to the expansion of an incompressible material. In this case water is the material. Just like heating up a metal to make it expand, if you heat up a liquid, it will have a slight change in it volume.

    The equation governing this is:

    [tex]\Delta V = \beta V_0 \Delta T[/tex]

    where [tex]\Delta T[/tex] is the change in temperature, [tex]\Delta V[/tex] is the change in volume, [tex]V_0[/tex] is the initial volume, and [tex]\beta[/tex] is the thermal expansion coefficient of the system.

    The thermal expansion coefficient for water is [tex]\beta = 207*10^{-6} m^3/K[/tex].
  4. Aug 5, 2009 #3
    Thank you, jpreed.

    The water is already pressurized to supercritical pressures. That is, no amount of heat addition or temperature rise will boil it off.

    How should I determine the volume of water to be vented?
  5. Aug 5, 2009 #4
    Assuming that the cylinder itself (ie, the pipe) doesn't stretch, the volume is constant. If it doesn't leak, the mass is also constant. So, you know the density. If you know how much energy is added to the water, you also know its enthalpy. Then you can use the steam tables to find the pressure.

    To refine the calc you could then determine the actual stretch in the pipe (based on its material properties, and dimensions), and then redo the density calc. Iterate to a suitably accurate result.

    You need to vent off enough water to bring the pressure back to its initial value. You know this value, and you know the enthalpy (see above). Use the steam tables again, this time to find the density. From the pipe volume (constant as stated above) you can find the mass. The difference is the mass which must be relieved.

    In practice, thermal relief valves are small (like 5 gpm capacity). That's enough for 'normal size' piping systems. If you're working with really big systems (lots of volume) you might want to do a check calc as outlined above.

    If you are concerned about the 'gradient' - really long pipe - then you can break it up into a number of segments, find the temp or enthalpy of each segment, do the calc on each segment, then sum the results. In that case, you would need to do the 'thermal' problem first, to get the 'gradient.'

    Last edited: Aug 5, 2009
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