Reynolds Number Question - Tube pinched slightly (picture inside)

[KWilliston]

Hello everyone,

I have a question regarding the Reynolds Number (Re) of water in a tube that is slightly pinched in the center (see attached image). I'm trying to calculate the Re at the point of the greatest deformation in the center. From what I have found Re for this case could be calculated from:

Re = [(velocity)(density)(characteristic length)]/(dynamic viscosity)

where characteristic length = (4*wettedarea)/perimeter)

I am unsure if I am headed in the correct direction with that equation and if it is correct, how would I apply it to my situation (how to calculated wetted area for example).

Thank you for any assistance, please let me know if more details are required.

Tube Picture

Edit: added alternate imgur link

 

[moltenmetal]

You use the ACTUAL, calculated velocity, i.e. volumetric flowrate divided by the actual cross sectional area available for flow, but you use the hydraulic RADIUS in place of the diameter in the Reynolds number calculation.

 

[bimr]

The Reynolds number is a dimensionless number expressed as (Chow, 1959):

Re = ρVL / μ

The Reynolds number can also be expressed in terms of kinematic viscosity (ν):

Re = VL / ν

The parameter L is a characteristic length. In closed-conduit flow, L is interpreted as the pipe diameter D, such that:

Re = VD / ν

In open-channel flow, L is interpreted as the hydraulic radius R. Thus, the Reynolds number is:

Re = VR / ν (


For a circular pipe: R = D/4. Therefore, the Reynolds number for pipe flow in terms of hydraulic radius is:


Re = 4VR / ν

For irregular shapes, substitute the equivalent diameter which is equal to 4* cross-sectional area/ wetted perimeter.

 

[moltenmetal]

Sorry, I replied too quickly without references handy! I believe that bimr has explained it correctly. Just be sure to use the real velocity, not a fictitious velocity calculated using the hydraulic diameter/radius.

 

[btrueblood]

You can certainly calculate the Re that way. What you would do with that information is another story...

 

[georgeverghese]

This procedure would give you the plain head loss due to straight flow - we also have to account for the head loss due to the eddies caused in this deformation - but I cant imagine what fitting would have a shape like this to get the corresponding K value or Le/D value to enable calculating the total head loss??

 

[racookpe1978]

You're removing a large part of the flow area - just under 1/3 of the original simple circle of the pipe, but you're removing with a goofy shape that - frankly - will NOT result in the assumed even-cross-section, uniformly-turbulent flat speed profile that forms the basic flow-loss calculations.

But it (the xtra friction losses caused by this deformed pipe wall) will be slightly less than the sharp-edge losses through a round hole that are calculated by a orifice plate. Use the orifice flow loss as your limiting condition.

Any chance of trapped residue or eventual plugging at this point?

 

[KWilliston]

Thank you for the help so far everyone.

Let me provide some more context so you know what exactly the scenario is. The tube I showed above is actually a compressible plastic tube. At the point of deformation the tube is actually being pinched by a plunger (this is a tube for a machine that dispenses water). The shape I made in my drawing is not an accurate representation of what it looks in the machine, it was just a rough idea so that I could wrap my head around the calculation side of things, and be able to apply it to a variety of shapes. Even when plunger is retracted (to allow flow) the tube cross section is reduced is reduced by quite a bit (because of the shape of the plunger housing). I did a rough calculation and determined that the cross sectional area when the plunger is retracted is 30-40% of that when it is not deformed.

I hope this information helps everyone picture the problem a bit better. So far I have used the information Bimr provided and calculated Re at the point of the smallest cross section. From these calculations the Re goes from around ~1500 in the non deformed part and upwards of 6000 in the area of smallest cross sectional area. My ultimate goal with this is to determine whether turbulence occurs at that point (it seems it does) and if that turbulence effects the amount that dispenses out of the other end.

 

[georgeverghese]

A concentric orifice may be one approximation to get the total head loss for this deformation, another may be two reducers directly in series?