Difference in height in a U-shaped tube

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SUMMARY

The discussion centers on calculating the height difference between mercury columns in a U-shaped tube using Bernoulli's Principle and the continuity equation. The problem involves a horizontal pipe with a cross-sectional area of 40.0 cm² and 10.0 cm², discharging water at a rate of 6.00 L/s. Participants emphasize the need to incorporate velocity terms into the Bernoulli equation due to the varying cross-sectional areas, leading to the conclusion that the correct approach requires calculating velocities using the formula A₁v₁ = A₂v₂.

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  • Knowledge of fluid dynamics, specifically the continuity equation
  • Familiarity with unit conversions, particularly between liters and cubic meters
  • Basic skills in algebra for manipulating equations
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kottur
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Homework Statement



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

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.

P_{1}+\rho*g*H_{1}=P_{2}+\rho*g*H_{2}

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:
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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.
 
You have the density of water and the density of mercury as roughly equal.
 
Sorry! I've put in the picture and the correct density of water. :)
 
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.
 
Should I use A_{1}v_{1}=A_{2}v_{2}? A and v are the areas and velocities. I can use A_{1} and v_{1} for the wide part... then I can find v_{2}. Does that help?
 
Okay my calculations are A_{1}v_{1}=40 cm^{2}*6,00\frac{L}{s}=240\frac{L*cm^{2}}{s}. Then I get A_{2}v_{2}=10 cm^{2}*v_{2}=240\frac{L*cm^{2}}{s} so v_{2}=24\frac{L}{s}.
 
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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