Difference in height in a U-shaped tube

AI Thread Summary
The discussion focuses on calculating the height difference between mercury columns in a U-shaped tube connected to a horizontal pipe with varying cross-sectional areas. Participants emphasize the importance of applying Bernoulli's Principle correctly, noting that velocity differences due to the pipe's constriction must be accounted for using the continuity equation (A1v1 = A2v2). Initial calculations yielded an incorrect height difference of 0.127 m, prompting a reevaluation of units and flow rates. It is highlighted that the flow rate should be converted to meters per second for accurate velocity calculations. The correct approach involves determining velocities at both sections of the pipe before applying Bernoulli's equation.
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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. :(
 
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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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