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I Effective Resistance (Fluid Mechanics)

  1. Oct 31, 2018 #1
    Imagine you have a vertical pressure head 2m tall with water flowing down to height 0m and emptying. Through this length we have 3 identical tubes of resistance R and length L. Assuming I know the volumetric flow rate(Q) as a function of height, how can I calculate the resistance of each tube? The flow rate decreases linearly with height as the fluid empties and pressure head drops. Should I use the average flow rate on the height interval [0m to 2m] and the max pressure head in my calculation? Or does the resistance change as a function of water level? Intuitively, the tube resistance should be an internal property such that it shouldn't vary. Any guidance would be much appreciated.

    Relevant Equations

    R=P/Q where P=pgh
  2. jcsd
  3. Oct 31, 2018 #2
    Can you please provide a diagram?
  4. Oct 31, 2018 #3
    I assume this should be treated like a circuit in series.

    image1 (3).jpeg
  5. Oct 31, 2018 #4
    Well, your approach is definitely correct. It will help to derive some model equations for this system. This will help to analyze the data. Would you like to begin by writing down some equations? I will be pleased to assist with this.

  6. Oct 31, 2018 #5
    Thank you very kindly. This is what I have so far.

    Rtotal = ∑Ri = 3R ("resistors in series")

    Q without any resistance is 9.5 x 10^-6 m^3/s

    Height (cm) = [13.5 27 40.5 54 67.5 81 94.5 108 121.5 135 148.5 162 175.5]
    Flow Rate given 3R (mL/s) = [0.007 0.008 0.009 0.012 0.018 0.025 0.033 0.042 0.052 0.065 0.079 0.0928 0.116]
    Flow Rate given 2R (mL/s)= [0.0132 0.024 0.04 0.050 0.066 0.079 0.093 0.104 0.122 0.129 0.134 0.140 0.146]

    Note how Q is proportional to height of the water level as it empties.

    Now we need to use R = P/Q = pgh/Q. How would you proceed?
  7. Oct 31, 2018 #6
    I would do something like this: The relationship between the rate of change of height and the flow rate is $$A\frac{dh}{dt}=-Q$$where A is the cross sectional area of the vertical pipe. The pressure at the bottom of the vertical pipe is $$p=\rho g h$$The relationship between the pressure and the flow rate is the one you have given: $$Q=\frac{p}{R}$$So, if we eliminate p and Q from these equations, we obtain:$$A\frac{dh}{dt}=-\frac{\rho gh}{R}$$The solution to this differential equation is $$h=h_0e^{-\frac{\rho gt}{AR}}$$ So, of up plot h vs t on a semi-log plot (i.e., using a semilog scale for h), you should obtain a straight line with a slope of ##\frac{\rho g}{AR}##. From this slope, you can determine R. Following this approach would allow you to include all the data at once.
  8. Nov 1, 2018 #7
    Thank you. This is helpful. The other alternative is to assume that (1/Rtot) is slope "m" if you plot H vs Q with H along the x axis.
  9. Nov 1, 2018 #8
    The down side of this latter approach is that you have to differentiate H numerically to get Q.
  10. Nov 7, 2018 #9
    Maybe one small remark: the resistance in pipe flow is not linear, the head loss ΔH is proportional to the square of the discharge, inserting ΔH=fQ|Q| in the equations will solve the problem correctly, f is the resistance factor depending on pipe length, diameter, cross sectional area and wall surface roughness.
    So you obtain the following differential equation:
    Which leads to the following solution:

    with ΔH(0) the head at t=t0
    There is one import restriction to this approach: the surface area A should be much larger than the cross-sectional area of the pipes (it is implicitly assumed that the flow velocity in the reservoir is much smaller that the flow velocity in the pipes. When you want to determine the resistance factor of the three individual pipes you will need to do additional measurements with respect to the head loss over each individual pipe.). The above approach only results in the overall resistance for the combination of them. (I just hope the equations are readable, I have some trouble to get them readable in this post).

    Attached Files:

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