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tank pressure after water is made into steam

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wildcatrod

Mechanical
Nov 15, 2005
2
I am looking for a way to figure what size of vent I would need to keep the pressure of a 30,000 gallon tank from increasing 2 PSI if the water in it is turned into steam. The amount of water is between 100 and 400 gallons.
 
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What are the thermodynamic conditions in the tank? ie Pressure, temp,
What is amount of water in the tank? Is air over the water?
What is the flow rate into/outof the tank?
Is the energy added to tank a constant or does exit flow increase/decrease?
 
This looks like a "credible scenario" discussion. You need to list the things that can happen (e.g., rapid introduction of heat, rapid introduction of fluid, blocked-in tank with a solar temperature gain, etc.), and then define the magnitude of pressure change associated with each. That magnitude and the rate of pressure change will let you determine the minimum vent size.

One example: a liquid-full vessel will change pressure approximately 100 psi for every degree F change in temperature. That is a really big pressure change for a very small temperature change, but the mass of fluid required to be released/added to prevent catastrophe is very small. So for the credible scenario of a liquid-full vessel facing a temperature change you could size a tiny bi-directional relieving device that allowed flow in or out.

It is the same way with your 30,000 gallon tank. "Just" determine what is a credible source for the steam you are concerned about, determine the magnitude, rate, and duration of that event and size the vent to handle that magnitude of mass flow.

David Simpson, PE
MuleShoe Engineering
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Wildcat:

Before you can size or engineer a relief for a vessel or a process, you must identify a rate of flow or establish the basis of a rate for the relief. That is what determines the “size” of the vent (the diameter, the length, the configuration, etc.).

It doesn’t matter if you have 100 or 400 gallons of water in the tank. The conventional way of determining the rate of vent flow required is to identify the energy or heat source that may cause the phase change (convert the water into steam). That energy or heat source has a rate dimension to it and it establishes the rate at which steam is being formed at the determined pressure. To do that, the tank must be at the temperature and pressure where the heat/energy added is sufficient to contribute the latent heat of vaporization of water at the stated process conditions.

Knowing how fast you’re creating steam inside the tank, you automatically know how fast you must vent it or get rid of it in order to prevent a pressure build-up.
 
The material in the tank is an emulsified asphalt, which has 30% water in it. We can only empty the tank to approx. 400 gallons. We then add approx. 1000 gallons of hot asphalt to boil off the water if we no longer need the tank to store emulsified asphalt. We pump the asphalt in at a rate of 350 gallons per minute. The temperature of the emulsion is 190 F and the temperature of the asphalt is 275 F. I am being asked to figure out if we have adequate venting in the tanks. It has been a long time since I have worked any heat transfer problems and I am needing some help.

Thanks,

 
What is the diameter and fill height of your tank?
Is your tank a pressure vessel, cone roof, floating roof or other?
If a pressure vessel, what is the maximum allowable working pressure MAWP?
Density of your asphalt material?
Tank Elevation?
Ambient temperature range?
What is the down stream pressure of your tank?
What is the pressure rating on your infeed pump?
 
How I'd approach it (approximation):
You have the flow rate of asphalt. You need internal energy for the asphalt at boiling temp and input temp. Take mass flow rate times difference in internal energy as the heat input. From that heat input and steam tables, figure what rate of steam generation you can get, going from 190 degree water to 212 degree saturated steam. Figure volume of that steam. That should be an approximate flow rate that you'll need. Size it generously.
 
400 gl at 30% say you have 120Gl of water right, x 8.3 #/GL
= about 1000Lb of water = 1000 steam when boiled, then you will figure out the time taking in boiling that 120GL and I figure it will be very fast 5 minutes perhaps,
then 1000LB x 12 becomes 12000LB of steam per hour (safety valves for steam are rated in PPH oriffice capacity as to be calculated,
at 2psi I imaging you will be in the approx 12" opening...
genb
 
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