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I've recently been learning about how to use Reynolds number to estimate if a fluid flow is laminar or turbulent while flowing through a pipe, and it has got me wondering if there is a similar way to tell if a flow is laminar or turbulent in a partially filled pipe.

I've tried googling a way to calculate this, but have not found anything helpful.

I'm not sure if this makes a difference, but I'm assuming the pipe is half filled and cylindrical.

Update:

Reynolds number for a full pipe is $\frac{\rho{V}{d}}{\mu}$, where $\rho$ is density, $V$ is velocity of the fluid, $d$ is diameter of the pipe, and $\mu$ is viscosity. How would this formula change for a half pipe?

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  • $\begingroup$ Sure, if there is geometric similarity (pipes of the same cross-section shape, filled to the same fraction of the diameter) then transition to turbulence will occur at the same Reynolds number. $\endgroup$ Nov 24, 2019 at 21:41
  • $\begingroup$ @MaximUmansky, how would the Reynolds number be calculated for a partially filled pipe? Would the diameter be the diameter of the full pipe, the average diameter of the half pipe, or something else? $\endgroup$
    – 63677
    Nov 24, 2019 at 23:13
  • $\begingroup$ It does not really matter, as long as a consistent definition is used. You can use the full pipe diameter, that's convenient. $\endgroup$ Nov 25, 2019 at 19:53

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Interesting question. I briefly skimmed through this paper which described this phenomenon.

They were able to model the flow by introducing a quantity known as the equivalent diameter, which was the diameter of a tube that would have the same cross sectional area as the "partially filled tube", given by:

$$D_\text{eq} = 2\sqrt{\frac{A}{\pi}}$$

where $A$ was the cross sectional area of the flow. Substituting in $A=\frac{1}{2}\pi R^2$ gives:

$$D_\text{eq} = \frac{d}{\sqrt{2}}$$

so we can say the Reynolds number is:

$$\text{Re} = \frac{\rho V d}{\sqrt{2}\mu}$$

for a half filled pipe.

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