heat transfer in copper pipe

[dcl0125]

I am first time here.
I like to know how to calculate BTU generated from 1 inch copper pipe with the length of 1 foot through which 176F water is passing.

Thanks in advance

 

[imok2]

dcl0125
If you pump 1 gpm/ft (8.33#)of water through 1" pipe and the ambient temperature is 76*F then you would multiply 500 x 1 gpm/ft x (176*F- 76*F)= 500x1x100*F= 50,000BTU/HR or you can do this: It takes (1 btu to raise lb of water 1*F), so if you multiply 1 gpm x 60 min = 60 gallons/hr x 8.33#/gal = 500# of water x 100*F = 50,000btu/hr. hope this helps

 

[IRstuff]

You don't give the actual flow rate, so the problem depends on whether the flow rate is slow or fast. If it's really slow, then the temperature of the water will substantially change while it's in the 1 ft section.

If it's fast, then the water essentially looks like an infinite source.

flow_rate/vol_of_pipe=number of exchanges/unit time

What's the orientation of the pipe? You'll need to figure out the convection coefficient, assuming natural convection. Figure out the infinite source heat flow. Using the same values as the previous posting, the heat flow with hc=4W/m^2°C is 18 BTU/hr

Figure out the amount of latent heat as a function of the volume flow. As indicated above, 1gpm would be equivalent of 50,000 BTU/hr of available power.

This means that the heat loss assuming constant temperature is less than .04% of the available heat in the water flow.

End result is that the heat flow is 18 BTU/hr and the temperature drop in the water is negligible.



TTFN

 

[MortenA]

another thing that is not mentioned is the medium on the outside - what is it (air, water,...?) and what is the temperature.

Best regards

Morten

 

[imok2]

Sorry fellas, But in order to answer his question I had to make some assumptions in hopes he could get some understanding

 

[jdainis]

You may find the information on this site:

http://www.pmengineer.com/CDA/ArticleInformation/features/BNP__Features__Item/0,2732,88674,00.html