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Bernoulli's Equation on water hose

  1. Nov 3, 2008 #1
    Water flows at 0.67 m/s through a 3.0 cm diameter hose that terminates in a 0.35 cm diameter nozzle. Assume laminar non-viscous steady-state flow.
    (a) At what speed does the water pass through the nozzle?
    49.22 m/s

    (b) If the pump at one end of the hose and the nozzle at the other end are at the same height, and if the pressure at the nozzle is 1 atm, what is the pressure at the pump?

    I have found the answer for part A.
    I have set up an equation for part B:
    I have to find P1
    P2 = 1 atm
    v1 = .67 m/s
    v2 = 49.22 m/s
    let D = density of water = 1000kg/m^3

    P1 + 1/2 D * V1^2 = P2 + 1/2 D * V2^2
    P1 = P2 + 1/2 D * V2^2 - 1/2 D * V1^2
    P1 = 1atm + 1/2(1000kg/m^3) * (2422.6084 m/s) - 1/2(1000kg/m^3) * (.4489m/s)

    How do I convert this or get answer for P1 in terms of atm?
  2. jcsd
  3. Nov 4, 2008 #2
    Most of your figures are in SI units. So I'd recommend working in SI, then converting at the end if you need the answer in terms of atm.

    Your table of information / equation table should give you 1 atm in SI units. (If you don't have one, find the table of information / equation table you'll have in your exam, print it out, and refer to it constantly, so that you get used to where everything is!)
  4. Nov 5, 2008 #3
    1 atm = 101.325 kPa
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