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Fluids: calculate pressure loss due to turns in pipe flow

  1. Apr 6, 2005 #1
    Hi,

    I am trying to estimate the flow rate in a long pipe of rectangular cross section. The pipe is 100 m long and has twenty 180 degree turns. I have been able to calculate the pressure drip for a 100m straight pipe, but I have not been able to estimate the losses due to the turns. Is there a way to estimate these? I would appreciate any help.
     
  2. jcsd
  3. Apr 6, 2005 #2

    Clausius2

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    Pressure drops are generally calculated by means of experimental tables. Search for one correspondent to your geometry.
     
  4. Apr 6, 2005 #3

    FredGarvin

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    The following is based on loss coeficients for a 90° bend but can calculate any bend angle:

    [tex]K_B = (n-1)(.25*\pi f_T \frac{r}{d} + .5 K) + K[/tex]
    Where:
    [tex]K_B[/tex] = Resistance coefficient for overall pipe bend.
    [tex] n[/tex] = # of 90° bends (for a single 180° bend, n=2).
    [tex]\pi[/tex] = well......pi
    [tex]f_T[/tex] = Friction factor in turbulent zone.
    [tex]r[/tex]= radius of bend (in same units as d).
    [tex]d[/tex]= inside diameter of pipe (same unit as r).
    [tex]K[/tex]= Loss coefficient for a 90° bend based on table below.

    90° Bend Loss Coefficients:
    r/d = 1, K = 20[tex]f_T[/tex]
    r/d = 1.5, K = 14[tex]f_T[/tex]
    r/d = 2, K = 12[tex]f_T[/tex]
    r/d = 3, K = 12[tex]f_T[/tex]
    r/d = 4, K = 14[tex]f_T[/tex]
    r/d = 6, K = 17[tex]f_T[/tex]
    r/d = 8, K = 24[tex]f_T[/tex]
    r/d = 10, K = 30[tex]f_T[/tex]
    r/d = 12, K = 34[tex]f_T[/tex]
    r/d = 14, K = 38[tex]f_T[/tex]
    r/d = 16, K = 42[tex]f_T[/tex]
    r/d = 20, K = 50[tex]f_T[/tex]

    Ref: K factor Table, Crane Flow of Fluids Through Valves, Fittings and Pipes.
     
  5. Feb 26, 2009 #4
    That's great its just what I was lloking for, but what is the friction factor Ft?

    I'm assuming you use the resistance coefficient Kb as in loss = Kb*(V^2)/(2*g)

    Thanks
     
  6. Feb 26, 2009 #5
    I think this refers to the darcy weisbach friction factor. It can be found using the colebrook equations or the swaimee-jain approximation for turbulent flows or just 64/(Reynolds number) for laminar flows
     
  7. Apr 30, 2009 #6
    Don't forget to adjust your calculated pipe area for the non circular cross section. Calculate the hydraulic diameter of an equivelant round pipe and use that in your calculations.

    Hydraulic diameter for a non circular cross section is given by Dh = (4A) / U, where A is the cross section area and U is the wetted perimeter.

    For a rectangular cross section this becomes Dh = (2LW) / (L+W)

    For a square cross section where L = W, this reduces to Dh = L.
     
  8. Nov 11, 2011 #7
    Hi, great help from the above posts....

    I'd like to go a step further in this direction of discussion....

    As seen in the attached picture, i basically have a small reservoir of fluid, whose properties are known...and it is made to flow through the pipe and enter a microchannel, whose resistance to flow is known, say R.

    Now, as seen, there will be head losses due to the sudden contraction of fluid path from reservoir to pipe, due to the bend, and due to the sudden expansion into the microchannel, and also viscous resistance, ofcourse.

    The head loss due to all the above can be calculated in terms of the inlet and outlet velocities through each of them separately.

    However, my goal is to make a simple equation, which takes into account all these losses and relates the height/elevation of the reservoir from the micro-channel and the final flow rate through the microchannel achieved, without having to see the velocities at each step...

    can anyone give me a clue how to do this?
     

    Attached Files:

  9. Nov 15, 2011 #8
    OK, so someone could atleast tell me if the following is correct:

    Let's assume datum is at outlet of micro channel, height of fluid surface of reservoir is H, and reservoir area far bigger than micro channel area, this allow us to neglect head loss from reservoir.

    First write Bernoulli's equation between the free surface reservoir(s) and entrance of elbow bend(1).
    p1-ps/ρg = (zs-z1) - V1^2/2g
    p1-ps/ρg = H - V1^2/2g

    Next write energy equation between elbow bend (1) and outlet of micro channel(2).
    p1-p2/ρg = (z2-z1) + f(L/D) (V^2/2g) + Kmc(V^2/2g) + K1(V^2/2g) + ...Kn(V^2/2g), z1 = z2
    p1-p2/ρg = f(L/D) (V^2/2g) + Kmc(V^2/2g) + K1(V^2/2g) + ...Kn(V^2/2g)
    where,
    Kmc = loss coefficient for micro channel
    K1 = loss coefficient of elbow
    K2, K3, ..Kn = other coefficient losses
    Note: p2 = ps = Patm
    Combining the two equations gives,
    (fL/D + Kmc+K1+...Kn + 1) V^2/2g = H
    or
    V^2 = (2gH) / (fL/D + Kmc+K1+...Kn + 1)
    V = [(2gH) / (fL/D + Kmc+K1+...Kn + 1)]^0.5
    Flow rate
    Q = V*A_mc = [(2gH) / (fL/D + Kmc+K1+...Kn + 1)]^0.5 * A_mc
    A_mc = area of micro channel outlet

    If you just want looking for total losses, then.
    HL = (f(L/2gD+ Kmc+K1+...Kn)*Q^2
    HL = Kt*Q^2
     
  10. Nov 17, 2011 #9
    He wanted to estimate the losses. For estimation purposes is is much cheaper to just use equivalent length tables, and then use your friction factor to find the losses in an equivalent length of straight pipe.
     
  11. Nov 17, 2011 #10
    what does the equivalent length mean? Does it apply to all the types of losses?..i thought friction factors only for viscous losses...please clarify.
    However, please note that I'm trying to account for all the losses and have the only variables as my Q(flow rate) and h...height of the reservoir above the microchannnel
     
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