pbmechtech
Mechanical
- Dec 12, 2014
- 4
I am trying to design a heating coil for a snowmelt pit in a hockey rink. Using the Zamboni specification, it says that one Zamboni can hold 100 cu.ft. of snow. We have two rinks and the Zamboni will clear each rink once an hour, so that gives me 200 cu. Ft of snow I have to melt every hour.
1 cu. ft. foot of water weighs 62 lbs, so let us say 1 cu. Ft. of compacted dirty snow weighs 50 lbs.
It takes 144 btu to melt one pound of snow @ 32 F to water a @32F
200 cu. Ft of snow X 50 lbs = 10000 lbs of snow that needs to be melted every hour
144 BTU X 10000 lbs of snow =1440 MBH to change the snow to water. Once the snow has turned to water, I would only need 1 btuh per lb of water to heat the water to melt the snow faster?
Am I correct so far?
I need to transfer this amount of heat through a heating coil in the bottom of a pit that is 120 sq. ft. in area and 36” high. (360 cu. Ft)
I want to use ¾” copper to make a coil(s) to place in the bottom of the pit. How do I calculate the length of copper tubing to transfer this much heat?
Thanks
1 cu. ft. foot of water weighs 62 lbs, so let us say 1 cu. Ft. of compacted dirty snow weighs 50 lbs.
It takes 144 btu to melt one pound of snow @ 32 F to water a @32F
200 cu. Ft of snow X 50 lbs = 10000 lbs of snow that needs to be melted every hour
144 BTU X 10000 lbs of snow =1440 MBH to change the snow to water. Once the snow has turned to water, I would only need 1 btuh per lb of water to heat the water to melt the snow faster?
Am I correct so far?
I need to transfer this amount of heat through a heating coil in the bottom of a pit that is 120 sq. ft. in area and 36” high. (360 cu. Ft)
I want to use ¾” copper to make a coil(s) to place in the bottom of the pit. How do I calculate the length of copper tubing to transfer this much heat?
Thanks