Flow Rate in Venturi Meter: .03825 L/s

AI Thread Summary
The discussion revolves around calculating the flow rate through a Venturi meter using Bernoulli's equation and the principle of pressure drop. The user initially calculated a flow rate of 0.03825 L/s but later found the correct answer to be 12.7 L/s after correcting a conversion factor. There is confusion regarding the use of two different densities (water and mercury) in the pressure drop equation, with a suggestion that both densities must be in compatible units. The importance of consistent units in calculations is emphasized, particularly when dealing with pressure and density. Overall, the conversation highlights the challenges faced in understanding fluid dynamics concepts and the collaborative effort to clarify these points.
LikwidN2
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Homework Statement


  • Known/Given
    • Water flows through a Venturi meter (mercury [density=13.6E3 kg/m^3] in the manometer).
    • Pipe size: 10.0cm diameter
    • Constriction size: 5.6cm diameter
    • Height difference of mercury: 2.3cm
  • Unknown - Flow Rate (L/s)

Homework Equations


  • Bernoulli's - P1 + (1/2)*d*V1^2 = P2 + (1/2)*d*(V2)^2
  • Flow Rate - I = A1*V1 = A2 * V2
  • Pressure drop - the book has this exact example, and says that P1 - P2 = (D_fluid [water] - D_liquid [mercury])*g*H
However, I don't understand why the pressure drop isn't just:
D_mercury*g*H

The Attempt at a Solution


Using the book's guided steps:
  • V2 = (A1/A2)*V1
  • P1 - P2 = (1/2)*D_water*((A1/A2)^2-1)*V1^2
  • (1/2)*D_water*((A1/A2)^2-1)*V1^2 = (D_water - D_mercury)*g*H
  • v1 = 0.7875m/s
  • v1 = 78.75cm/s
  • Flow = 78.75 * pi * (10cm/2)^2 = 6185 cm^3/s
  • 6185 cm^3/s * ([1 L/s] / [1000^3 cm^3/s]) = .03825 L/s

This is off everyone's favorite online homework system, WebAssign, and I have used 5/6 submissions :bugeye: I'm terribly sorry for the awful formatting, but I don't have access to a good typesetting program as of yet (downloading TeX as we speak [well... as I type]).
 
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I agree that you should just use the delta P given in 23mm Hg. See what answer you come up with and we can look that over before No 6 submission.
 
After confirming with a friend, I realized that I just had the conversion factor wrong. Their method worked, and apparently m^3/s * 1000 = L/s. Unfortunately, this leaves me no closer to understand why they used the two densities and not just the mercury density. With a lot of luck I might be able to extract a coherent [say: in English] answer from my TA, but probably not.
 
anychance you can post a picture if given? I get 6.42L/s using just 23mmHg for delta P
 
Picture

13-17.gif


The correct answer turned out to be 12.7 L/s. I get everything except for the use of the change in densities.

Thanks for all your help, by the way, I appreciate it! To think of my 40k / year tuition and I need to find help online from a helpful stranger.
 
LikwidN2 said:
13-17.gif


The correct answer turned out to be 12.7 L/s. I get everything except for the use of the change in densities.

Thanks for all your help, by the way, I appreciate it! To think of my 40k / year tuition and I need to find help online from a helpful stranger.

Thats too funny and sad at the same time. I went to school when many of the TA's still spoke english.

anyhow, all i know is that the delta p and rho need to be in the same units as you can't mix and match say mm Hg with density of water in g/ml etc. Let me know if you get a good explanation.
 
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