Physics Asked by user36510 on May 19, 2021
Good day everyone,
I am new on this site and I hope to find here help, since I am not going anywhere with the literature I have found.
I try to calculate realistically the heat loss of a hot, uninsulated pipe. Let’s say,
it is $170,,C$, 1 meter long, $8″ (=0,203,,m)$ of diameter. I want to calculate the whole losses for the situation of no wind (only normal convection) and $20C$ of outside temperature.
The $170C$ is the temperature at the outside surface of the pipe, the surface is known, I also know the Stefan–Boltzmann constant ($sigma$) and I take $alpha=5$ for the convective coefficient of air at the surface of the pipe as well as $epsilon=0.85$ for the steel pipe emissivity.
My problem is, that in literature, some parameter is always being rejected as negligible due to the fact, that most of the times the example is some kind of exercise of a heat transfer class. I want to calculate the real thing and then decide what is negligible and what is not.
So my main question is: do I just add the radiation losses to the convective losses of the pipe ?
$$Q_{loss} = Q_{conv}+ Q_{rad}$$
$$= (alpha*Α*Delta T) + epsilon*Α*sigma*((T_{pipe})^4-(T_{air})^4) tag{with T in K}$$
According the above, I get $Q_{conv} = 479,,W$ and $Q_{rad} = 958,,W$.
Is there any mistake in my way of thinking or is it truly that simple ?
Thanks in advance.
Marcus
This should really only be a comment but I can't comment without 50 reputation points..
Your calculations are very simplified. The general procedure for the convection part for this type of pipe through room problem is: 1 - Calc. Coefficient of thermal expansion 2 - calc. Rayleigh number 3 - calc. Nusselt number 4 - calc. convective heat transfer coefficient 5 - calc. convective heat transfer
As for your main question, yes you do just add the heat transfer from the different modes to get the overall heat transfer.
As for whether your transfer via radiation value is correct, that depends mostly of if the assumption that all surfaces within view are the same temp. as the air is correct.
Answered by Fergus on May 19, 2021
According to http://www.engineeringtoolbox.com/steel-pipes-heat-loss-d_53.html , the heat losses are close to what you get. I'd note that the convective coefficient of air is rather tricky: it depends even on the orientation of the pipe (horisontal/vertical).
Answered by akhmeteli on May 19, 2021
Other small effects that you may be ignoring are heat conduction to the air and temperature drop along the pipe.
If the fluid is a gas, its temperature may drop by some small amount due to pressure drop along the pipe. The fluid temperature will also drop due to the heat transfer that you calculate. An old reference for these effects is Chapter 8 of Compressible Fluid Flow by Ascher Shapiro (1953).
Some practical data on heat transfer may be found in Marks' Standard Handbook for Mechanical Engineers. In the 2007 edition, Eqn 4.4.15 is applicable to practical calculations. This equation uses a combined convective and radiative heat transfer coefficient approach. A table of combined coefficients is also given.
Answered by LouisB on May 19, 2021
The natural convective heat transfer coefficient can be estimated through correlations and depends on geometry and its orientation.
There is a calculator which does this for a horizontal pipe here: https://www.poppyi.com/app_design_form/public_render/free%20convection-%20horizontal%20cylinder
Hope this helps.
Answered by kitt91 on May 19, 2021
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