I am trying to design a water-channel that travels through a metal block shaped like a square to cool it down (so it is a 2D problem). One consideration is the wetted surface area, which allows more of the water to be in contact with the metal, thus increasing the cooling effect.

One way to do that is to start at one corner of the square, then form multiple S-shapes on the block to maximize the surface area. However, there will be some sharp U-turns.

I am trying to avoid stagnation points, which slows down the cooling since less fluid travels.

So I did some reading on sharp turns and the term Kutta Condition keeps appearing when describing fluids at sharp points, where there will be a stagnation point at the edge.

So I'm wondering, if I understand it correctly, that is, I should not have any sharp bends or else there will be a stagnation point near(or at) the edge?


1 Answer 1


Yes, sharp bends cause stagnation points, which would reduce heat transfer. More importantly they would increase the resistance to flow, requiring either more pressure to drive the flow or reducing the flow, and thereby reducing the overall heat transfer. If the pump could handle the additional pressure, then it would be better utilized by reducing the channel size which would decrease boundary layer thickness, increasing heat transfer.

  • $\begingroup$ Do you mean at the same flow rate and pipe length, the smaller the diameter, the more the heat transfer? $\endgroup$
    – Naghi
    Sep 7, 2022 at 17:27
  • $\begingroup$ @Alish Well for circular pipe, it's more complicated, because then you're significantly decreasing the surface area. You'd get a higher transfer rate per area, but with the lower area, probably less overall transfer. For flow between plates, keeping a constant average flow velocity, you'd get faster initial heat transfer with the plates closer together, but you'd have a lower total volume of flow, so you might not get as much heat transfer towards the end of the plate. But for constant volumetric flow, you'd definitely get better heat transfer. $\endgroup$
    – Rick
    Sep 7, 2022 at 18:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.