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What rate does water flow through the pipes?

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data

    A horizontal pipe 10.9 cm in diameter has a smooth reduction to a pipe 4.50 cm in diameter. If the pressure of the water (density = 1000 kg/m3) in the larger pipe is 7.50×104 Pa and the pressure in the smaller pipe is 5.70×104 Pa, at what rate (as mass per unit time) does water flow through the pipes?

    2. Relevant equations

    Bernoulli's equation. Po+1/2(rho)v^2+(rho)gh= pfinal +1/2(rho)vfinal^2+(rho)(g)(h final)

    3. The attempt at a solution
    if i sub in the necessary parts we know that h will be the same so that becomes a constant. Po is 7.50×104 Pa and Pfinal is 5.70×104 Pa so if we simploify the above equation we get P1-P2 = ½[tex]\rho[/tex]v22-½[tex]\rho[/tex]v12.so i solved for [tex]\Delta[/tex]v by [tex]\sqrt{\frac{P1-P2}{0.5*\rho}}[/tex] i got an answer of 6 but am unsure of the units to put in the answer..if this is actually the right working..pls. help
     
  2. jcsd
  3. Jun 10, 2009 #2

    rl.bhat

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    The rate of flow of water can be written as
    m = Avρ
    m^2 = A^2*v^2*ρ^2
    1/2*m^2/A^2*ρ = 1/2*ρ*v^2
    Substitute this value in the equation to find m.
     
  4. Jun 10, 2009 #3
    i'm a little confused what value am i substituting in the equation to find m??and if i use the equation 1/2*m^2/A^2*ρ = 1/2*ρ*v^2

    (m2/A2)*(1000)= (1000)v2

    would my A= sum of both pipes diameter ....and what would v equal???l
     
  5. Jun 10, 2009 #4

    rl.bhat

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    P1 - P2 =1/2*ρ*v2^2 - 1/2*ρ*v1^2 = 1/2*m^2/ρ*(1/A2^2 - 1/A1^2)
     
  6. Jun 10, 2009 #5
    P1 - P2 = 1/2*m^2/ρ*(1/A2^2 - 1/A1^2)
    delta p =18000
    A1= 9.331x10^-3
    A2= 1.5904 x10^-3

    i sub it in and get a value of 137030.15 ..does that look right??
     
  7. Jun 10, 2009 #6

    rl.bhat

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    I am getting 68644 m^3/s
     
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