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Finding speed of water leaving a pressurized vessel

  1. Sep 6, 2014 #1
    1. The problem statement, all variables and given/known data

    A large tank of water has a hose connected to it, as shown in the figure. The tank is sealed at the top and has compressed air between the water surface and the top. When the water height h has the value 3.50m , the absolute pressure p of the compressed air is 4.20 ×105Pa. Assume that the air above the water expands at constant temperature, and take the atmospheric pressure to be 1.00 ×105Pa.

    You can't see the figure obviously but it looks like a large cylindrical tank 4 meters high with pressurized air at the top. At the bottom is an outlet pipe that bends upward like an S and is open to the atmosphere 1 meter above the bottom of the tank.

    A) What is the speed with which water flows out of the hose when h = 3.5 m?

    B) As water flows out of the tank, h decreases. Calculate the speed of flow for h = 3.1m.

    C) Calculate the speed of flow for h = 2 m.

    D) At what value of h does the flow stop?




    2. Relevant equations

    Bernouli's -> P1 + (1/2)ρ(v1)^2 + ρgh1 = P2 + (1/2)ρ(v2)^2 + ρg(h2)

    pv =nRT at some point?


    3. The attempt at a solution

    I am so stuck on this that I wasn't able to do part A even, but I feel that once I understand how to do part A I will understand how to do the other parts. I went back to material from last semester to try and refresh on the pressure and flow rate concepts but it didn't spark any new ideas.

    I tried the following:

    P1 = 4.20*10^5 Pa = 4.145 atm
    ρ= 1 for water in this case
    h1= 3.5m
    v1= 0 (my professor told us that the water movement at the surface inside the tank would be negligible, so I'm thinking that means that this v1 would be 0)

    P2 = 1*10^5 Pa = .9869 atm
    h2 = 1m (for outlet of pipe leaving tank)
    v2 = ??? (attempting to figure out, as speed of water at outlet would be speed of water entering pipe)

    4.145 atm + (1/2)(1 g/cm^3)(0^2) + (1 g/cm^3)(9.8 m/s^2)(3.5m) = .9869 atm + (1/2)(1g/cm^3)(v2^2)+(1g/cm^3)(9.8m/s^2)(1m)

    Did that math and I got v2 = 7.44 m/s, but the answer is 26.2 m/s.

    Any and all help would be greatly appreciated.
     
  2. jcsd
  3. Sep 7, 2014 #2
    Never mind, I think I got it. I just needed to double check all the proper units to use, and I mistakenly converted Pa to atm.
     
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