Fluid Mechanics and Venturi meter

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Discussion Overview

The discussion revolves around the calibration of a Venturi meter, focusing on the determination of the coefficient of discharge (Cd) using given flow rates and manometric measurements. Participants explore the theoretical and actual flow rates, the application of Bernoulli's equation, and the implications of these calculations in the context of fluid mechanics.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants assert that the theoretical flow rate is 0.012 m³/s, while others argue that this value represents the actual flow rate measured during calibration.
  • There is a discussion about the need to calculate the actual flow rate using the coefficient of discharge (Cd) and the theoretical flow rate.
  • Participants express confusion regarding the definitions of actual and theoretical flow rates, with some referencing their professor's input on the matter.
  • One participant questions the relevance of calculating flow velocities at the inlet and throat in relation to the theoretical flow rate.
  • Another participant provides equations derived from Bernoulli's principle and attempts to relate them to the problem at hand, indicating a struggle with unknown variables.
  • A later reply clarifies that the actual flow rate is essential for determining Cd and that the 0.012 m³/s value should be used as such.
  • Participants engage in refining their understanding of the equations and variables involved, with some expressing gratitude for assistance in clarifying their confusion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether 0.012 m³/s is the actual or theoretical flow rate initially, leading to confusion. However, there is a later acknowledgment that it is indeed the actual flow rate, suggesting some resolution on that point. The discussion remains unresolved regarding the implications of this distinction for calculating Cd and other variables.

Contextual Notes

Some participants express uncertainty about the definitions of actual and theoretical flow rates, and there are unresolved mathematical steps in deriving the necessary equations for the calibration process. The discussion also highlights dependencies on specific assumptions regarding fluid properties and flow conditions.

Who May Find This Useful

This discussion may be useful for students and practitioners in fluid mechanics, particularly those working with flow measurement devices like Venturi meters and those interested in the calibration processes involved in such measurements.

patep023
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Homework Statement



A Venturi meter is constructed with an inlet diameterof 100 mm and a throat
diameter of 40 mm. The meter must be calibrated prior to leaving the factory.
The meter is installed in a 100 mm diameter water main with a mercury
manometer connected across the inlet and throat of the device. Calibration comprises the determination of the coefficient of discharge (Cd) for the meter at
various flow rates. Cd is defined as the ratio of the actual flow rate to the theoretical flow rate. At one particular flow rate, 0.012 m^3 of water is collected in one second having passed through the meter. The manometric levels differed by 375mm at this flow rate.

Homework Equations


p1/ρg +v1^2/2g+z1=p2/ρg +v2^2/2g+z2
Q=v1A1=v2A2
Cd=mass flow rate actual/mass flow rate theoretical.

The Attempt at a Solution


So the theoretical solution for mass flow rate is 0.012 m^3s^-1
I haven't plugged in the numbers but I have played around with the equations a lot. An image of the question can be accessed from the link. I got as far as to putting a lot of equations together and getting z1-z2=(v2^2-v1^2)/2g - (p1/p2)/ρwg and z1-z2=(p2-p1 - ρwgh +ρmgh)/ρwg where ρw is density of water and ρm is density of mercury. p1 and p2 are pressure at point 1 and 2. The answer is 0.98 but I don't know all the variables to put into get the actual value for the mass flow.
http://i.imgur.com/PgCasr0.png
 
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patep023 said:

Homework Statement



A Venturi meter is constructed with an inlet diameterof 100 mm and a throat
diameter of 40 mm. The meter must be calibrated prior to leaving the factory.
The meter is installed in a 100 mm diameter water main with a mercury
manometer connected across the inlet and throat of the device. Calibration comprises the determination of the coefficient of discharge (Cd) for the meter at
various flow rates. Cd is defined as the ratio of the actual flow rate to the theoretical flow rate. At one particular flow rate, 0.012 m^3 of water is collected in one second having passed through the meter. The manometric levels differed by 375mm at this flow rate.

Homework Equations


p1/ρg +v1^2/2g+z1=p2/ρg +v2^2/2g+z2
Q=v1A1=v2A2
Cd=mass flow rate actual/mass flow rate theoretical.


The Attempt at a Solution


So the theoretical solution for mass flow rate is 0.012 m^3s^-1
I haven't plugged in the numbers but I have played around with the equations a lot. An image of the question can be accessed from the link. I got as far as to putting a lot of equations together and getting z1-z2=(v2^2-v1^2)/2g - (p1/p2)/ρwg and z1-z2=(p2-p1 - ρwgh +ρmgh)/ρwg where ρw is density of water and ρm is density of mercury. p1 and p2 are pressure at point 1 and 2. The answer is 0.98 but I don't know all the variables to put into get the actual value for the mass flow.
http://i.imgur.com/PgCasr0.png
Let Q be the theoretical flow rate. What would the mercury height difference in the manometer be if the volumetric flow rate was zero? In terms of Q, what is the flow velocity at the inlet? In terms of Q, what is the flow velocity at the throat? From Bernoulli's equation, what would the theoretical flow rate Q be if the height difference of mercury in the manometer were 375 mm? How does this compare with the observed value?
 
How does calculating the flow velocity at the inlet and throat help in terms of Q as Q is only theoretical and I am trying to work out practical?
 
patep023 said:
How does calculating the flow velocity at the inlet and throat help in terms of Q as Q is only theoretical and I am trying to work out practical?
You said the Cd is the ratio of the actual to the theoretical flow rate. So, you need to determine the theoretical flow rate at the measured 375 mm Hg in order to calculate the Cd.
 
Theoretical flow rate is already given 0.012 m^3s^-1. I need to calculate the actual flow rate
 
patep023 said:
Theoretical flow rate is already given 0.012 m^3s^-1. I need to calculate the actual flow rate
No. The problem statement says that 0.012 is the actual measured flow rate. You are trying to use this measurement to calibrate the venturi meter by determining the Cd. You can then use this Cd to determine the actual flow rates for other sets of flow conditions.

Chet
 
I am really confused now, my professor said that was the theoretical flow rate
 
patep023 said:
I am really confused now, my professor said that was the theoretical flow rate
What does the following mean to you: At one particular flow rate, 0.012 m^3 of water is collected in one second having passed through the meter.

This is the only piece of data you have to calibrate the flow meter (i.e., determine the Cd). With all due respect to your professor, it has to be the actual flow rate. Try it, and see if you get your 0.98 value.

Chet
 
Assuming I am only expected to calculate the flow rate and let's imagine that 0.012m^3 and Cd doesn't exist. How would I do it? I got as far as finding the equation so that I have the variables P2 and v2 which I don't know and am really stuck on that part. I did the following, equation 1=p2+Ro(w)g(z2-h)+Ro(m)gh=P1+Ro(w)g(z1) which I got by making Pa = Pb. My second equation was by using Bernoullies equation and making that equal to z1-z2 and then I put both equation together with 3 variables I didn't know. v1,v2,p2. I replaced v1 with v2A2/A1 using v1a1=v2A2.
 
  • #10
patep023 said:
Assuming I am only expected to calculate the flow rate and let's imagine that 0.012m^3 and Cd doesn't exist. How would I do it? I got as far as finding the equation so that I have the variables P2 and v2 which I don't know and am really stuck on that part. I did the following, equation 1=p2+Ro(w)g(z2-h)+Ro(m)gh=P1+Ro(w)g(z1) which I got by making Pa = Pb. My second equation was by using Bernoullies equation and making that equal to z1-z2 and then I put both equation together with 3 variables I didn't know. v1,v2,p2. I replaced v1 with v2A2/A1 using v1a1=v2A2.

This is a really good start. Your first equation, when rearranged, gives:

(Ro(m)-Ro(w))gh=(p1-p2)-Ro(w)g(z1-z2)

What does your Bernoulli equation give for the right hand side of this equation?

Also, don't forget that v1a1=v2A2=Q. So, v1 = Q/a1, and v2 = Q/a2.

Chet
 
  • #11
I got it. Thank you for the help and yes you were right, 0.012 is the actual value, turns out i just heard him wrong. sorry for the inconvenience
 

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