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Venturi Meter (Air speed)

  1. Sep 22, 2014 #1
    1. The problem statement, all variables and given/known data

    Air is blown through a horizontal Venturi tube. The diameters of the narrow and wider sections of the tube are 1.0 cm and 3.0 cm respectively, and the height h of the mercury column is measured to be 1.00 mm. What are the air speeds in the wider and the narrower sections of the tube? The density of mercury is 13,600 kg/m3. (Assume the density of air is 1.29 kg/m3.)

    Air is blowing through the wider section to the smaller section.
    2. Relevant equations
    p=rho
    P1 + (1/2)(p)(v1)^2 + pgy1 = P2 + (1/2)(p)(v2)^2 + pgy2
    A1v1=A2v2
    P1 = Po + pgh
    A(circle)=pi(r)^2

    3. The attempt at a solution
    Formulas:
    y1=y2 so the pgy1 and pgy2 can cancel and v2=A1v1/A2
    We can rearrange the equation to give:
    P1-P2 = (1/2)(p)(v1)^2((A1/A2)^2-1)
    v1=((P1-P2)/((1/2)(p)((A1/A2)^2-1))^(1/2)

    P1-P2
    Po +pgh1 - Po - pgh2
    Change in pressure: pg(h1-h2)

    v2=A1v1/A2

    What we have:
    r1= 3.0cm/2/100= 0.015m
    r2= 1.0cm/2/100= 0.005m
    (P1-P2)= change in pressure
    change in height = 1.00mm/1000= 0.001m
    p(mercury)= 13600kg/m^3
    p(air)= 1.29kg/m^3

    ------------------------------------------------------------------

    Change in pressure:
    pg(h1-h2)

    (13600kg/m^3)(9.8m/s^2)(0.001m)= 133.28pa

    A1= (0.015m)^2(pi)= 7.068583471x10^-4 m^2
    A2= (0.005m)^2(pi)= 7.853981634x10^-5 m^2

    v1=((P1-P2)/((1/2)(p)((A1/A2)^2-1))^(1/2)

    (133.28pa/(1/2)(1.29kg/m^3)((7.068583471x10^-4 m^2/7.853981634x10^-5 m^2)^2-1))^(1/2)
    v1= 1.607154546 m/s

    v2=A1v1/A2
    (7.068583471x10^-4 m^2)(1.607154546 m/s)/(7.853981634x10^-5 m^2)
    v2=14.46439092 m/s

    I am not sure if I am even doing this right. :(
    Sorry if formatting is hard to follow/ugly, I am still getting used to the new changes, don't know where everything is yet.
     
  2. jcsd
  3. Sep 22, 2014 #2
    I think this is the correct answer, I was plugging in everything at once on my calculator and I might have been doing something wrong so I was getting a lot of different answers. But I'd still like this to be checked. :)

    Edit: If the change in height was 2.66mm, would v1 and v2 be:

    v1: 2.621065424 m/s
    v2: 23.58958882 m/s

    ?
    (Used same equations as above)
     
  4. Sep 22, 2014 #3
    Nvm, I am right, I don't think I've made any mistakes.
     
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