Difference in height in a U-shaped tube

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Homework Help Overview

The problem involves fluid dynamics within a U-shaped tube, specifically examining the difference in height between mercury columns due to water flow in a horizontal pipe with varying cross-sectional areas. The context includes the application of Bernoulli's Principle and the relationship between flow rate, velocity, and pressure changes.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Bernoulli's Principle and the need to account for velocity changes due to differing cross-sectional areas. There are questions regarding the correct use of units and the setup of the problem, including the necessity of including velocity terms in the equations.

Discussion Status

The discussion is ongoing, with participants providing guidance on the need to compute velocities from the flow rate and cross-sectional areas. There is a recognition of the importance of unit consistency, and some participants are exploring the implications of their calculations and assumptions.

Contextual Notes

There is a mention of a missing diagram initially, which may have contributed to misunderstandings about the problem setup. Participants are also addressing the correct densities of water and mercury in their calculations.

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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