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Air Temp rise across heater in machine
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Source：Internet Author：Unknow Pubdate：2008-04-15
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azrandyh (Mechanical)
28 Jun 07 2:27
I hope somebody can explain how to calculate the following problem to me. Here we go
I have an air duct approx. 2ft by 5ft. the duct has an airflow of 2500CFM. I have a constant 27 degree C air coming into the duct and a 1kW heater element in the middle of the duct. What will be the temperature of the air leaving the duct. If the air entering the duct increases to 32 degrees C, what will be the temp of the air leaving the duct. Thanks for the help
berkshire (Aeronautics)
28 Jun 07 4:49
Is this homework?
B.E.
TADiep (Mechanical)
28 Jun 07 13:11
You are not allowed to post homework questions on the forum. But I will help you out anyway.
Q [BTU/hr] = 1.08 x CFM x (delta Temp)
Delta Temp = Q / (1.08 x CFM) 字串2
Change the electric heater capacity to BTU (1KW = 3412 BTU)
Delta Temp = 3412 / (1.08 x 2500) = 1.26 deg
Entering air temp = 27C / 32C
Leaving air temp = 28.6C / 33.6C ----
A green thought..."We don't inherit the earth from our ancestors, we borrow it from our children." (unknown)
MintJulep (Mechanical)
28 Jun 07 14:23
Q [BTU/hr] = 1.08 x CFM x (delta Temp)
works in F, not C.
TADiep (Mechanical)
28 Jun 07 16:32
MintJulip, thanks for pointing that out.
For some reason I thought delta T applies to any temperature scale.
----
A green thought..."We don't inherit the earth from our ancestors, we borrow it from our children." (unknown)
azrandyh (Mechanical)
29 Jun 07 0:55
Thanks for the help and, no this was not homework. I am determining the root cause of some electronics failures in our equipment and I needed to validate my statement to the company that extreme temperature rise was not the cause. Thanks to your help I feel confident in my statement.
字串3
quark (Mechanical)
29 Jun 07 1:13
1.1 is a better factor and 1.24F is the temperature rise.
If 1.08 is considered then the temperature rise is 1.26F.
TADiep,
Get your hands on Uconeer from www.katmarsoftware.com and you will have even temperature difference conversion in it.
TD of F varies by 9/5times TD of C.
IRstuff (Aerospace)
29 Jun 07 16:06
One caveat I would have is that the analyses presented assume that the heat is evenly distributed into the entire air stream.
As posed in the OP, if the 1-kW source is small compared to the dimensions of the duct opening, the local air temperature might be considerably higher. TTFN
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azrandy (Mechanical)
29 Jun 07 16:58
字串2
The heat source IS equally distributed across the entire duct. I did take that into consideration before posting. Thanks
IRstuff (Aerospace)
29 Jun 07 17:10
Another more intuitive view is to consider that 2500 CFM is equivalent to exchanging the air in a 17 ft x 17 ft x 8 ft room EVERY minute. Now add 5 PCs into the room. What little heat they're generating is trivially removed by the air flow TTFN
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