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Fluid Mechanics and Venturi meter

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant 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.


    3. 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 in to get the actual value for the mass flow.
    http://i.imgur.com/PgCasr0.png
     
  2. jcsd
  3. Feb 18, 2014 #2
    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?
     
  4. Feb 18, 2014 #3
    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?
     
  5. Feb 18, 2014 #4
    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.
     
  6. Feb 18, 2014 #5
    Theoretical flow rate is already given 0.012 m^3s^-1. I need to calculate the actual flow rate
     
  7. Feb 18, 2014 #6
    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
     
  8. Feb 18, 2014 #7
    I am really confused now, my professor said that was the theoretical flow rate
     
  9. Feb 18, 2014 #8
    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
     
  10. Feb 18, 2014 #9
    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.
     
  11. Feb 18, 2014 #10
    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
     
  12. Feb 19, 2014 #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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