Tank Freeze Protection Heat Losses/NFPA help
Tank Freeze Protection Heat Losses/NFPA help
(OP)
Hi,
I have received quotes for electric heaters to freeze protect a large water holding tank. One suggetsts a 25 KW (approx 85000 BTU/hr loss) heating requirement & one a 90 KW (approx 307000 BTU/hr loss)heating requirement. it is a 36' diameter x 40' high tank on a concrete slab w/ R-10.2 insulation on the walls and an uninsulated roof. from what i can gather, the ground losses can be neglected at this temperature/tank size. The losses through the walls will be calculated using the following:
(SA(walls) x k x delta T)/3412 BTUh/KWh =
(4524ft^2 x .098 Btu/Hr/ft^2/°F/in(for R10.2) x 42F)/3412 = 5.5 KWh
From the Chromalox catalog (beginning @ page 19):
htt p://www.ch romalox.co m/content/ training-m anuals/Tra iningTankH eating.pdf
I can see that the losses from uninsulated steel walls would be calculated using the following:
U x delta T x SA(roof) =
(5.1 BTU/Hr.Sq.Ft.°F (for 15 mph wind) x 42 x PI*18^2)/3412 = 64 KWh
now it would appear that the 90 KW quote is more correct based an including a safety factor & some ground loss value, however if the water is never in direct contact with the roof of the tank, would the air at the top act as insulation(hence the 25 KW requirement)? Obviously there is a large difference in product/operation costs of these units so I would like to know if 25 KW is sufficient or if i should go with the 90 KW to be safe as this is a fire protection tank. All help is appreciated - Thanks!
I have received quotes for electric heaters to freeze protect a large water holding tank. One suggetsts a 25 KW (approx 85000 BTU/hr loss) heating requirement & one a 90 KW (approx 307000 BTU/hr loss)heating requirement. it is a 36' diameter x 40' high tank on a concrete slab w/ R-10.2 insulation on the walls and an uninsulated roof. from what i can gather, the ground losses can be neglected at this temperature/tank size. The losses through the walls will be calculated using the following:
(SA(walls) x k x delta T)/3412 BTUh/KWh =
(4524ft^2 x .098 Btu/Hr/ft^2/°F/in(for R10.2) x 42F)/3412 = 5.5 KWh
From the Chromalox catalog (beginning @ page 19):
htt
I can see that the losses from uninsulated steel walls would be calculated using the following:
U x delta T x SA(roof) =
(5.1 BTU/Hr.Sq.Ft.°F (for 15 mph wind) x 42 x PI*18^2)/3412 = 64 KWh
now it would appear that the 90 KW quote is more correct based an including a safety factor & some ground loss value, however if the water is never in direct contact with the roof of the tank, would the air at the top act as insulation(hence the 25 KW requirement)? Obviously there is a large difference in product/operation costs of these units so I would like to know if 25 KW is sufficient or if i should go with the 90 KW to be safe as this is a fire protection tank. All help is appreciated - Thanks!





RE: Tank Freeze Protection Heat Losses/NFPA help
think of how ice is made in an ice tray in your freezer.
think of how ice is formed in a puddle of water when cold air is present.
although the ideas are different processes, these ideas will make you think a little bit.
lastly, fire protection, a prudent engineer/designer should take measures to ensure the fire protection system will work in accordance with design conditions.
hope this helps.
good luck!
-pmover
RE: Tank Freeze Protection Heat Losses/NFPA help
Looking at your equation, it appears your outside design conditions are -10 degF with 15 mph wind. How long do you anticipate the design conditions to last? Days, Weeks, Months??
RE: Tank Freeze Protection Heat Losses/NFPA help
RE: Tank Freeze Protection Heat Losses/NFPA help
I believe that you are grossly underestimating the U value of the air gap on top of the tank (there is no 15 mph wind on the inside of the tank). I would conservatively estimate the heat loss off the top of the tank be around 65 BTU/ft^2 for your 42 degF delta T. Or around 66,000 BTU/hr for the roof with minimal air gap. All together, you have 66,000 BTU/hr roof + 19,000 BTU/hr wall (I think your wall value is a little high but I still used it) would yield 85,000 BTU/hr. There is some software from 3EPlus that is available at no cost that you can model your tank with http://www.pipeinsulation.org/ .
Think about what pmover stated about how ice will form. I would think that the top of the tank will freeze first around the edges on top and move inwards towards the center. As the ice forms, it provides additional insulation in the local freeze areas.
Lastly why would operation costs be grossly higher for a larger heater? The first cost would be greater; however, I suspect that you would regulate the temperature in the tank so there should be no significant difference in the operating cost between the two heaters.
RE: Tank Freeze Protection Heat Losses/NFPA help
I think they assume heat loss through the roof is the same as through the walls. I suppose the motivation being that it is 100% saturated condition, so cooling the roof condenses water vapor on it and drips it back into the tank.