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Old   April 8, 2015, 05:23
Default heat loss in pipe
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I wanted to know how much heat is transfered through insulated pipe with water flowing in it, and then calculate temperature drop. I created a simple matlab script and got results which I am not confident about.
Heat tranfered Q [J] should be equal to Q=(Tfluid-Tambient)/resistance
resistance= (alpha1*2*pi*L*R0)^-1+(alpha2*2*pi*L*R1)^-1+log(R1/R0)/(2*pi*L*lambda)
Therms corresponds to convection inside a tube, conduction through the wall and convection outside the tube.

When I turn off ambient convective thermal resistance I get 1,4*10^3 J lost for 1m tube, when I turn it on I get only 70J/m

heat transfer coefficients for water in pipe and air outside are equal to
2.75*10^3 and 10.21 W/m2/K

It is weird that convection outside a tube insulates the tube so much ;( What do you think about those results?

I am attaching matlab script if anybody wants to see it.
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Old   April 10, 2015, 05:41
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without considering the convection outside the pipe, you can't use Tfluid-Tambient as temperature gradient. T at the outside of the wall is greater than Tambient.
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