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How do you calculate (or otherwise determine) $h_\text{inner}$ and $h_\text{outer}$ (convective heat transfer coefficients) for a heat exchanger?

For reference, the standard equation is: $$ \text{UA}~=~\frac{n}{\frac{1}{h_\text{inner} A_\text{inner}} + \frac{1}{2 \pi kL} \ln{\left(\frac{r_\text{outer}}{r_\text{inner}}\right)} + \frac{1}{h_\text{outer} A_\text{outer}}} \,. $$

My specific situation is water and a stainless steel pipe, but I am curious about both the specific and general case.

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  • $\begingroup$ An older question now, but this probably would've been better at SE.Engineering, as it's a routine calculation done during typical engineering design processes but not typically seen by physicists. $\endgroup$
    – Nat
    Commented Apr 20, 2019 at 2:31

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Heat transfer coefficients are a bit tricky, for good reason. You need to know something about the conditions that exist on both sides of the pipe, and you only stated that you have water in a stainless steel pipe. For heat transfer to occur, you need a second fluid on the outside of the pipe that you can transfer heat with.

In addition to the above, there are thin boundary layers involved with the fluid on both sides of the pipe. These boundary layers impede heat transfer, and their thickness is strongly dependent on fluid viscosity, fluid velocity, whether fluid flow is laminar or turbulent, whether your heat transfer conditions are for counter-current flow, co-current flow, or cross-flow, etc. Due to this, your exact geometry and heat transfer conditions must be known, such that you can find the published empirical equation that matches the equipment and conditions that you want to model.

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  • $\begingroup$ Thanks for your insight. I believe I am looking for something specific to the pipe and the water flowing though it (I am new to this). "k" in the equation above is the thermal conductivity of the pipe material (steel) so that is already accounted for. I have water flowing on both sides of the pipe: laminar flow, low pressure, temperature between 5 degrees C and 40 degrees C. Does this help narrow things down? $\endgroup$
    – Tom Bombadil
    Commented Apr 25, 2016 at 2:57
  • $\begingroup$ Is the flow co-current or counter-current? $\endgroup$
    – David White
    Commented Apr 25, 2016 at 11:03
  • $\begingroup$ It would take too long to explain in detail here. The heat transfer coefficient inside the pipe is a function of the Reynolds number for the flow and the Prantdl number of the fluid. Look up Dietus Boelter equation. $\endgroup$
    – Chet Miller
    Commented Apr 25, 2016 at 11:15

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