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Relief Valve on shell side of heat exchanger

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vancitygd

Mechanical
Dec 23, 2009
6
Hello all,

I have been asked to size a pressure relief valve to be located on the shell side of a heat exchanger. The design conditions are for the tube side 730 gpm of condensate @ 250 F and the shell side 24 gpm of water. The scenario we are concerned with is somebody closing the valves on the inlet and outlet of the shell side while the condensate continues to flow through the heat exchanger while the water boils.

How do I go about calculating the relief capacity required for this valve? Any ideas?
 
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I assume that this is a condensate cooler. Your cooling water flow looks too low. That's a separate question than the one you asked, so I won't say anything other than to suggest re-checking that design, unless it's one that's already built and installed. If you cooling water outlet temp goes above ~120F you're going to get lots of calcium carbonate deposits which will render the exchanger ineffective, and those deposits will be very hard to remove from the shell side.

Back to your question. Assuming the shell side is liquid-full, you'll obviously get an initial overpressure caused by thermal expansion of the shellside water. But once the PSV releases enough liquid to create a vapor pocket, the overpressure scenario is over with, assuming the shellside has an MAWP that is > 15 psig. In other words, your 250F heat source can only generate
a pressure on the shell side that's equal to the vapor pressure of water at 250F. When the shellside water heats to 250F, the resulting pressure is 15.3 psig (see steam table). So, unless the MAWP is lower than 15.3 psig (not likely) then you'll just get an equilibrium that doesn't cause further overpressure. So, size the PSV for thermal expansion of the liquid and your done.

The thermal expansion relief requirement will be set by the initial rate of temperature rise for the water. You know the initial water temp, but need to calculate the rate at which it rises. That can easily be done by first calculating the heat that is transferred from the condensate to the water.

Q=UAdT

Get the tube surface area from the exchanger file. Assume a U value of say 300, and use a conservative dT of (250 - initial water temperature). Knowing the Q and the amount of water in the shell, you can use your steam tables to convert this into X DegF temp rise per hour. The density difference between the starting temp and the final temp is the relief requirement.
 
Also consider the probability of a fire while the Hx shell has both isolation valves closed. If this is a credible Fire Case scenario, you will need to size for a significant ammount of steam. On my water Hx's, this is always bigger than the thermal liquid expansion case.
 
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