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Moving water using pressure/vacuum

  1. Jan 25, 2010 #1
    I am trying to move water using a vacuum pump. I have a vacuum pump hooked up to a tank that has one inlet, one exit with a length of hose and check valves on both. What I am doing is creating a vacuum in the tank, causing water to fill the tank. Then using the pump to generate pressure in the tank to discharge the water. I know the pressures/vacuum required to lift/push the water for my suction and discharge heights. As well as the flow rate of the pump. What I need to do is calculate the flow rate of the water into/out of the tank. My pump supplier has said to use Boyle's Law (P1V1=P2V2). The time this yeilds is close to real world results, however, in my mind it doesn't add up. It would make sense if I was moving air, but I am using the air to move water. The flow rate of the water has to be the same at both ends of the hose, as water will not compress or expand. I have since started trying to use Bernoulli's Equation (V^2/2+P/d+gZ=const), this makes more sense as it relates velocity (which leads to flow rate) to pressure. Since I am using pressure differential to move the water, I feel this is the equation I should be using. When I do all the calculations the results from Bernoulli's is way faster then the real world expermental data I have. If anyone could give any insight or a push in the right direction that would be great.
     
  2. jcsd
  3. Jan 25, 2010 #2
    Show us your calculations and maybe we can say something about them. Otherwise there's nothing to say, anything would be just guessing about what you've done.
     
  4. Jan 25, 2010 #3
    Here is more data and explanation of my problem.

    GIVEN INFORMATION:
    Suction height = 2m
    Discharge height = 3m
    Flow rate (air) on suction side of pump = 600 m^3/hr
    Volume of tank = .95 m^3
    Atmospheric Pressure = 101325 Pa
    Density of water = 1025 kg/m^3
    Acceleration of Gravity = 9.81 m/s^2
    Area of Hoses = .051 m^2

    FORMULAS:
    Boyle’s Law: P1V1 = P2V2
    Bernoulli’s Equation: V1^2/2 + P1/d + gZ1 = V2^2/2 + P2/d + gZ2
    Force: F = m * a = P * a

    CALCULATIONS:
    Pressure Required To Start Moving the Water Columns:
    By using the sum of forces acting on the water column I get (Note: the area of the hose cancels across the equations):

    P(suction) = P(atm) – d * h(suction) * g = 101325 Pa – (1025 kg/m^3 * 2 m * 9.81 m/s^2) = 81214 Pa

    P(discharge) = P(atm) + d * h(discharge) * g = 101325 Pa + (1025 kg/m^3 * 3 m * 9.81 m/s^2) = 131491 Pa

    Boyle’s Law Method:

    V1 = (P2/P1)V2 = (81214 Pa / 101325 Pa) 600 m^3/hr = 481 m^3/hr

    Time to fill the tank = .95 m^3 / 481 m^3/hr *3600 s/hr = 7.1 sec this time ignores the time needed to create the vacuum in the tank.

    Here’s where it gets a little tricky, as I actually have two tanks. One is filling while the other is empting. Because I am only using one pump P1 for the discharge volume has to be based on the vacuum required for the suction height.

    V2 = (P1/P2)V1 = (81214 Pa / 131491 Pa) 600 m^3/hr = 371 m^3/hr

    Time to empty the tank = .95 m^3 / 371 m^3/hr *3600 s/hr = 9.2 sec this time ignores the time needed to create the pressure in the tank.

    The times needed to build the vacuum and pressure required are easily calculated using The Ideal Gas Law (PV=nRT) and can be added to the above times.

    Bernoulli’s Equation Method:

    V1^2/2 + P1/d + gZ1 = V2^2/2 + P2/d + gZ2

    I assumed that velocity of the surface of the pond I’m sucking out of is very small compared to that at the end of the hose going into the tank and I set my datum as being that surface. Then the Equation reduces down to:

    V2 = SQRT(2*(P1/d – P2/d – gZ2)) = SQRT(2*(101325/1025 – 81214/1025– (9.81)(2))) = .031 m/s

    Time to fill the tank = .95 m^3 / (.031 m/s * .051 m^2) = 601 sec this time ignores the time needed to create the vacuum in the tank.

    I assumed that velocity of the surface of the tank is very small compared to that at the end of the hose going out of the tank and I set my datum as being the surface of the water in the tank. Then the Equation reduces down to:

    V2 = SQRT(2*(P1/d – P2/d – gZ2)) = SQRT(2*(131491/1025 – 101325/1025– (9.81)(3))) = .022 m/s

    Time to fill the tank = .95 m^3 / (.022 m/s * .051 m^2) = 847 sec this time ignores the time needed to create the vacuum in the tank.

    Now these times seem really long. However, the pump does not stop generating vacuum/pressure when we get to the required vacuum/pressure needed to move the water column. As the pump generates greater vacuum/pressure the velocity will increase, thus reducing the time. I need to generate a curve based on the flow into the tank, empty volume left in the tank and the vacuum/pressure in the tank at that instant given the vacuum/pressure curves for the pump. When I did this the times worked out to be rather fast compared with the Boyle’s Law method and the experimental data.

    I am really just looking for some insight into this problem. Both methods yield data but which one is correct. Boyle’s Law is usually used with gases but yields closer results, while Bernoulli’s deals with liquids, relates pressure to velocity but seems to generate times that are too fast.

    Any help with this would be greatly appreciated.
     
  5. Jan 25, 2010 #4
    maybe I'm just dense today, but I don't see what you're trying to do here. Your original post had a tank and a vacuum pump, now you have a pond, and two tanks and pumps... can you make a sketch of the configuration.

    Your first two equations show a pressure difference between suction and discharge of 5 meters, yet you say the elevation of one is 2 meters and the other is 3 meters... What are these 2 and 3 meter elevations with respect to?? Is there a water level in the tank? Sorry, but I just don't have a picture in my mind of your set up...
     
  6. Jan 25, 2010 #5
    Here is a picture of what I have.

    system  2.JPG
     
  7. Jan 25, 2010 #6
    I started looking at the bernoulli calc you did. The difference between the tank pressure you calculated (81214 Pa) and the pressure at the suction end of the hose (atmospheric, 101325 Pa) is just equal to the elevation head (the 2 meters) - so you have just enough vacuum to raise the column to the height of the tank, with essentially nothing left over for velocity. So the flow should be zero.

    I think you need to re-think what the tank pressure is. It begins at atmospheric, right? And then as the vacuum pump draws on the tank, the tank pressure drops and the water column starts to rise. When the tank pressure is "minus two meters" the column has reached the tank. How low will the tank pressure go? Will it evacuate the tank vapor faster than the liquid enters?

    Sorry - I'm at work and shouldn't be looking at this right now. Have to get back to this later...
     
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