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I and a friend have made a venturi tube trough which water flows at a low speed. The flow is therefore laminar. We measured the static pressure using some piëzometers at the points 1,2 and 3. We know the average velocity with which the water flows at the end of reservoir 3.

My friend suggests calculating the speeds at other resvoirs(1,2) of the venturi-tube by using the volumetric flow rate. We namely know the cross-sectional area of the venturi tube at all places.

I however don't think this volumetric flow rate will stay the same because of frictional losses, and so prefer to calculating the speed by using a bernoulli equation with a head loss incorporated in it. What's best to calculate the other speeds in this situation? Or are we both completely wrong?!

Diameter of the venturi tube at different points:

  • Point 1: 17 mm
  • Point 2: 13 mm
  • Point 3: 10 mm

Arrows show length of venturi tubes reservoirs

Measured static pressure at points:

  • Point 1: 106503 Pa
  • Point 2: 105375 Pa
  • Point 3: 102903 Pa

Measured velocity at the end of reservoir 3:

1.8 m/s

Differences between methods:

  • Calculated head loss between the end of reservoir 2 and the end of reservoir 3 : 0.1208901666 Pa

  • Calculated velocity at the end of reservoir 2 using bermoulli with head loss: 0.81 m/s

  • Calculated velocity at the end of reservoir 2 using constant volumetric flow rate and area changes: 1.06 m/s

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  • $\begingroup$ Can you put a value on the frictional losses? $\endgroup$
    – Kyle Kanos
    Commented Jun 16, 2015 at 13:29
  • $\begingroup$ What pressures are you measuring? You can't be measuring the Bernoulli pressure because that falls with velocity so it should be lower at point 1 than point 3. If you're measuring the static pressure then how is that related to the flow rate? The Hagen-Poiseuille equation? $\endgroup$
    – John Rennie
    Commented Jun 16, 2015 at 15:34
  • $\begingroup$ Now that you edited the question, $1.8*(10/13)^2 = 1.06$ should be the velocity at point 2, based on volumetric flow rate. $\endgroup$
    – Mike Dunlavey
    Commented Jun 17, 2015 at 17:25
  • $\begingroup$ @MikeDunlavey You're correct. I have been using the wrong cross-sectional area for calculating the speed at the end of the second reservoir, this is why my original answer was so strange. Thanks for your help. $\endgroup$
    – Ruben23630
    Commented Jul 8, 2015 at 16:24

1 Answer 1

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The volumetric flow rate has to stay the same, because the same amount of water flows out of the tube as flows into it. The only way for the volumetric flow rate to change along the tube would be if there was a leak. Friction (i.e. viscous drag) can't change the volumetric flow rate, all it can do is change the pressure gradient along the tube.

So you are safe to use your friend's suggestion for calculating the average velocity. Note that this will be an average because the flow velocity is zero at the walls and rises to a maximum at the centre of the tube.

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  • $\begingroup$ Why is there then such a big difference between the two calculated values? $\endgroup$
    – Ruben23630
    Commented Jun 16, 2015 at 15:07
  • $\begingroup$ @Ruben23630: Your diagram says the areas are getting smaller, but your numbers say they are getting bigger. Could that be the problem? $\endgroup$
    – Mike Dunlavey
    Commented Jun 16, 2015 at 15:24
  • $\begingroup$ @Ruben23630: see my comment to your question. It doesn't help that your diagram has the tubes labelled incorrectly. $\endgroup$
    – John Rennie
    Commented Jun 16, 2015 at 15:35
  • $\begingroup$ @JohnRennie The calculated values differed from eachother as a result of me making small mistakes in my calculation of the speed at the end of reservoir 2 (missing squared signs etc.) and me using the wrong cross-sectional area in the calculation of the speed using the volumetric flow rate. Thanks for your help. $\endgroup$
    – Ruben23630
    Commented Jul 8, 2015 at 16:27

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