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Water Propulsion Device - geometry configuratiton optimization problem

  1. Jun 17, 2014 #1
    Dear physicist,

    I am designing a device that will use firetruck/hydrant water pressure (~150 psi) and fire hose (5" diameter) to lift up a device in the air and into a burning room. Please advice me and help understand how water flow, pressure and areas influence thrust. My goal is to input pressure and area of source -> math model -> output payload, thrust, max height achieved. I have the control over design of the exit flows and their configuration.

    Here is the inspiration video:
    http://www.liveleak.com/view?i=66e_1397564023

    This is a 2D mathematical model that I have for payload calculations.
    m_total = mass of hose + mass of water
    m_total = x*cross sectional area*density + constant (hose weight per m)*(x^2+y^2)^1/2
    where y is height and x is the horizontal distance the device needs to travel (~ 4m - from hydrant to the building which is standard street distance away)
    Payload = g*m_total (750N or 170 lb for y = 12.2 m or 4 story)
    The pump does work of lifting the water vertically but the device needs to carry water weight when it moves horizontally away from the source. x is a horizontal projection.

    Now for the hard part:

    For simplicity and for ease of control, a four valve/nozzle configuration is my choice. I attached a link to a picture to better describe it. --> http://imgur.com/qkK8C3e

    Maybe its better to attach the source hose from the side instead of the bottom. It will make it difficult for manufacture and control so if its not very influential on thrust, then a bottom configuration would work better.

    I've been using "Carried by impulse: The physics of water jetpacks"
    written by Matthew Vonk and Peter Bohacek from Physics Today to estimate my thrust capabilities. I think I can use this formula to have 4 equal areas for exit. (thrust 215 lb). v = Q/A

    Possible acceleration of the drone = F_net (Thrust - payload)/m_total = 2.62 m/s^2
    (something is not right about this)

    Pressure loss from height:
    P source = density*g*hmax+P atm ---> hmax is where you lose all pressure. (95 m with 150 psi P source which doesnt seem right either) How does pressure influence the thrust?

    I am not sure about my calculations. Please advice, I would really appreciate it. I will be testing a smaller design with a house water hose. I will provide pictures and data to who ever will help out.
     

    Attached Files:

    Last edited: Jun 17, 2014
  2. jcsd
  3. Jun 17, 2014 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    Seems kind of wasteful and limited in usefulness. Why not just use a standard ladder truck?
     
  4. Jun 17, 2014 #3
    Berkeman, I would like to calculate the actual numbers and possibilities. If you like to help me just for the though exercise, I would appreciate it.
     
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