Different answers from equation of continuity and Bernoulli

  • #1
Hi there,
So I was doing the dishes this morning using a sink wand hat can toggle between different flow speeds. The way that I've always thought of this working is using the equation of continuity:

Volume flow rate: = Area*velocity

Pressing a button on the wand decreases the cross-sectional area and correspondingly increases the flow velocity assuming that the flow rate is constant.

The issue that I'm running into is trying to simultaneously explain this with Bernoulli's equation:

P1 + pgy1 + 1/2pv12 = P2 + pgy2 + 1/2pv22 =

I'll take state 1 to be at the water reservoir or wherever the water in the pipes is pressurized and 2 to be at the outlet of the wand. I'm assuming that pressing the button on the wand does not change the pressure at the reservoir (nor at the outlet of the wand, where it is subject to atmospheric pressure).

Based on Bernoulli's equation, this would suggest that pressing the wand button should have NO effect on the flow velocity, which contradicts the result from the equation of motion. So: what am I missing here. I feel like I may be incorrect in my pressure assumption above somehow.

Thanks!

Chris
 

Answers and Replies

  • #2
kuruman
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Let's assume that y1 = y2 to simplify a bit. The Bernoulli equation gives
##v_2^2 - v_1^2 = 2\Delta P/ \rho##, (##\Delta P = P_1-P_2##).
Now ##\Delta P > 0##, therefore ##v_2 > v_1##.
 
  • #3
dRic2
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I'll take state 1 to be at the water reservoir or wherever the water in the pipes is pressurized and 2 to be at the outlet of the wand. I'm assuming that pressing the button on the wand does not change the pressure at the reservoir (nor at the outlet of the wand, where it is subject to atmospheric pressure).
According to your reasoning the geometry of the pipe doesn't play a role so whatever geometry you chose for the pipe you will get the same result. I don't think so. At the outlet of the wand there is a huge pressure-drop so I think you have to take state 2 to be inside the wand where the fluid is still pressurized. Now if you use Bernoulli equation you will see that a change in the section will affect pressure and velocity.
 
  • #4
BvU
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pressing the button on the wand does not change the pressure at the reservoir
Actually, hat doesn't matter: in the reservoir the velocity is near zero. What matters is the pressure before and after the wand opening -- there, where the velocity changes.
 
  • #5
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Hi there,
So I was doing the dishes this morning using a sink wand hat can toggle between different flow speeds. The way that I've always thought of this working is using the equation of continuity:

Volume flow rate: = Area*velocity

Pressing a button on the wand decreases the cross-sectional area and correspondingly increases the flow velocity assuming that the flow rate is constant.
This is the flaw in your logic. Cranking down on the wand reduces the cross sectional area, but does not reduce the velocity at the exit ( at least not for an inviscid fluid, according to the Bernoulli equation). So the exit velocity is unchanged. But the velocity times area (the volumetric flow rate) decreases.
 
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