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Venturi meter pressure difference

  1. Jan 4, 2017 #1
    1. The problem statement, all variables and given/known data
    mt2YJJE.png

    2. Relevant equations
    p=ρgh
    Q=A1U1=A2U2
    Bernoulli:
    p1+½ρU12+ρgz1=p2+½ρU22+ρgz2

    3. The attempt at a solution

    So I got that the pressure difference is gh(ρm-ρ). So I rearranged Bernoulli for p1-p2 to get (assuming the potential is the same I eliminated it),

    p1-p2=½ρU22-½ρU12
    I carried on rearranging to get:
    √(2gh(ρm-ρ))/ρ = Q/A2-Q/A1

    I am unsure how to eliminate A1 so I can proceed with the question
     
  2. jcsd
  3. Jan 4, 2017 #2

    TSny

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    Gold Member

    Looks like you made an error in the rearranging. I'm not sure, but it appears that you assumed that ##\sqrt{U_2^2 - U_1^2} = U_2 - U_1##

    How can you express A1 in terms of A2 and the diameters D1 and D2?
     
  4. Jan 5, 2017 #3
    Oh I see, if I use A1U1=A2U2 And rearrange for U2, and use d2/D2 for the area ratio?
     
  5. Jan 5, 2017 #4

    TSny

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    Yes. But you might want to rearrange for U1. Depends on how you are doing the algebra to get to the result.
     
  6. Jan 5, 2017 #5
    Thank you! I got it now! :)
     
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