Diborane
Chemical
- May 3, 2005
- 16
Could you all discuss vacuum relief due to steam 'collapsing' (condensing) in a tank not protected for full vacuum, which does not have an open connection to atmosphere?
Vacuum protection from pump out, drain out, thermal vacuum, all seem to have relatively small vacuum relief flows compared to steam collapsing in the vessel.
For the steam collapsing case there is a credible scenario in which the tank could be filled with the steam. There is also a credible scenario where cold water could flow into the vessel.
Lets assume a vessel is filled with steam at 212 deg F and there is a flow of 250 gpm of water a 60 deg F into the vessel.
The most conservative method to calculate this is 250 gpm * 8.34 lb/gal* (212 F - 60 F) = 316920 BTU removed/min.
316920 BTU/min / 970 BTU/lb = 326 lb/min of steam * 26 ft3/lb = 8500 ft3/min of air through vacuum protection.
If your tank is rated for only 4" of water column you are talking about many vacuum breakers.
The problem I see with this method is it assumes that heat is transferred instantly to the water as it flow in and the temperatures reach equilibrium. I would think the rate of steam collapse/heat transfer is dependent on the amount of surface area that the cold water is in contact with the steam. A fine spray of water would condense more steam than a stream of water.
The second problem is that with that air inflow rate even if your tank is 100,000 gallons the steam would totally be replaced < 2 minutes.
I know enough to diagnose my problem that I am incorrectly calculating the rate at which the steam could possibly condense, however I do not know enough to accurately calculate what the real rate would be.
Vacuum protection from pump out, drain out, thermal vacuum, all seem to have relatively small vacuum relief flows compared to steam collapsing in the vessel.
For the steam collapsing case there is a credible scenario in which the tank could be filled with the steam. There is also a credible scenario where cold water could flow into the vessel.
Lets assume a vessel is filled with steam at 212 deg F and there is a flow of 250 gpm of water a 60 deg F into the vessel.
The most conservative method to calculate this is 250 gpm * 8.34 lb/gal* (212 F - 60 F) = 316920 BTU removed/min.
316920 BTU/min / 970 BTU/lb = 326 lb/min of steam * 26 ft3/lb = 8500 ft3/min of air through vacuum protection.
If your tank is rated for only 4" of water column you are talking about many vacuum breakers.
The problem I see with this method is it assumes that heat is transferred instantly to the water as it flow in and the temperatures reach equilibrium. I would think the rate of steam collapse/heat transfer is dependent on the amount of surface area that the cold water is in contact with the steam. A fine spray of water would condense more steam than a stream of water.
The second problem is that with that air inflow rate even if your tank is 100,000 gallons the steam would totally be replaced < 2 minutes.
I know enough to diagnose my problem that I am incorrectly calculating the rate at which the steam could possibly condense, however I do not know enough to accurately calculate what the real rate would be.
