Difference in height in a U-shaped tube

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Homework Statement



The horizontal pipe, shown in the figure, has a cross-sectional area of [itex]40,0 cm^{2}[/itex] at the wider portions and [itex]10,0 cm^{2}[/itex] at the constriction. Water is flowing in the pipe, and the discharge from the pipe is [itex]6,00*10^{-3}\frac{m^{3}}{s} (6,00\frac{L}{s})[/itex]. The density of mercury is [itex]\rho_{Hg}=13,6*10^{3}\frac{kg}{m^{3}}[/itex] and the density of water is [itex]\rho_{w}=1,00*10^{3}\frac{kg}{m^{3}}[/itex].

What is the difference in height between the mercury columns in the U-shaped tube?

YF-14-45.jpg


Homework Equations



I've been using Bernoulli's Principle.

[itex]P_{1}+\rho*g*H_{1}=P_{2}+\rho*g*H_{2}[/itex]

The Attempt at a Solution



When I use the formula above I get 0.127 m which is incorrect. I also found this problem on the internet where they get the same answer so I was pretty sure but apparently it's wrong. :(
 
Last edited:

Answers and Replies

  • #2
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Where is your picture? I suspect one end of the manometer attached to the smaller diameter while the other is sampling the larger diameter. If so there is a difference in velocity so the pressure must change. You'll probably need some V^2 terms in your equation to account for velocity differences. We need a picture of your problem.
 
  • #3
SteamKing
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You have the density of water and the density of mercury as roughly equal.
 
  • #4
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Sorry! I've put in the picture and the correct density of water. :)
 
  • #5
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OK I see your picture now. Don't know where it was before because it was not showing up on my screen. Because the velocity changes you must have a V^2 on each side of the equation. You compute the velocities from the cross sectional areas and flow rate.
 
  • #6
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Should I use [itex]A_{1}v_{1}=A_{2}v_{2}[/itex]? A and v are the areas and velocities. I can use [itex]A_{1}[/itex] and [itex]v_{1}[/itex] for the wide part... then I can find [itex]v_{2}[/itex]. Does that help?
 
  • #7
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Okay my calculations are [itex]A_{1}v_{1}=40 cm^{2}*6,00\frac{L}{s}=240\frac{L*cm^{2}}{s}[/itex]. Then I get [itex]A_{2}v_{2}=10 cm^{2}*v_{2}=240\frac{L*cm^{2}}{s}[/itex] so [itex]v_{2}=24\frac{L}{s}[/itex].
 
  • #8
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Your units are liters per second. That is not the unit of velocity. You need meters/second. V1 is not 6 liters/second.

Always check to ensure your units describe the variable you seek.
 
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