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Perforated Pipe Flow, closed end.

  1. Nov 16, 2008 #1
    1.
    A hydroponic garden uses the 10 m long perforated pipe system to deliver water at 20ºC. The pipe is 5cm in diameter and contains a circular hole every 20 cm. A pump delivers water at 75 kPa (gage) at the entrance, while the other end of the pipe is closed. pressure near the closed end of the perforated “manifold” is surprisingly high and there will be too much flow trough the holes near that end. One remedy is to vary the hole size along the pipe axis. Make a design analysis to pick the optimum hole-size distribution that will make the discharge flow rate as uniform as possible along the pipe axis. You are constrained to pick hole sizes that correspond only to commercial metric drill-bit sizes available to the typical machine shop.



    2. Relevant equations

    1) bernoulli's modified energy equation for head loss:
    P1/gamma + V1^2/2*g + z1 = P2/gamma + V2^2/2*g + z2 + hf(headloss)

    Where P=gage pressure kpa, V= average velocity in pipe, Gamma=rho*g, z= vertical position in meters, g=gravity. rho=density, 1 and 2 denote different places of interest in center of pipe.

    2) Head loss equation with respect to friction factor:
    hf= f*L*V^2/(2*g*D)

    where f= friction factor, L = length of pipe where flow is being analyzed, D is diameter of pipe, g= gravity,

    3) V= 4Q/(pi*D^2)
    where Q= flow rate in m^3/s

    4) Reynolds #, (Re)= rho*V*D/mu
    mu= viscosity

    turbulent flow Re>2300
    Laminar flow Re<2300

    5) for Laminar flow, f=64/Re
    for Turbulent flow, f= determined from moody chart

    Assume smooth pipe curve for moody chart.

    3. The attempt at a solution

    First of all I don't think that the pump has much to do with the problem, its just there to tell you what the pressure is at the start of the 10m section. Secondly I know that if all the exit holes are the same diameter, the pressure is a function of x from x=0 to x=10. I also know that the x direction velocity in the pipe will be a function of x, with Vmax at x=0, and V=0 at x=10. So I know in order to even out the flow through each exit hole, I need to vary the size from biggest to smallest from x=0 to x=10.

    I have several theories on how to solve it, I just don't know how to go about it. first of all, my equation for velocity as function of x is V(x)=Vmax*(10-x)/10
    I thought I might be able to plug this into eq. 1) where V2 at end of pipe=0, and z1=z2. Then substite eq 2) into eq 1). From this stuff make a function of flow Q(x) as function of x, and f. Then pick f value, Re value off moody chart, and iterate to find Q(x), then some how use That equation to adjust the diameter of each hole.....

    I was also thinking, that If I have Q(x) and I want all the flow rates to be the same, could I take the derivative of Q, dQ/dx, and set it equal to zero to solve for something, because then flow rate won't change? or something like that?

    I don't know for sure, any input would be greatly appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

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