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Minor loss due to Sudden contraction (help needed)

  1. Mar 25, 2008 #1
    Hi, I am an undergraduate, doing an experiment regarding of minor losses due to sudden contraction. However, I found that my theoretical loss is higher than the experimental loss.

    The flow rate = 3.28*10^-4 (m^3/s)
    (Inner diameter) D1= 3.652*10^-2 m
    (Inner diameter) D2= 1.539*10^-2 m

    by using the energy equation
    [tex]\frac{P_1}{\gamma}+\frac{V_1^2}{2g}=\frac{P_2}{\gamma}+\frac{V_2^2}{2g}+h_L[/tex]
    rearranged
    [tex]\frac{P_1}{\gamma}-\frac{P_2}{\gamma}=\frac{V_2^2}{2g}-\frac{V_1^2}{2g}+h_L[/tex]

    I assumed that major head loss/friction head loss negligible since the length between to the two points is only 50mm.

    SO [tex]h_L[/tex]= minor loss

    [tex]h_L=K_L\frac{V^2}{2g}[/tex]

    Based on the charts given in the book, Fluid Mechanics by McGrall Hill. I found that [tex]K_L[/tex] roughly equals to 0.45 and V=the velocity of the smaller pipe.

    By using all these, I calculated the pressure difference = 0.22m

    But my experimental result = 0.159m

    Could anyone help me with this, I expected my experimental results would yield bigger pressure difference, yet it gave me less.

    I built a simple piezometer to measure the pressure difference between the two points.
     
    Last edited: Mar 25, 2008
  2. jcsd
  3. Mar 25, 2008 #2

    stewartcs

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    Probably due to the resistance coefficient (K). The K value will depend on the angle of the contraction. The book value you used is probably for a different angle.

    CS
     
  4. Mar 25, 2008 #3
    My set up is build to study sudden contraction not gradual contraction, the chart in the book is also for sudden contraction.

    I have include an attachment of the chart I used.
     

    Attached Files:

  5. Mar 25, 2008 #4

    stewartcs

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    Sorry, didn't see the sudden part above. However, the angle is still used in the calculation of K. If it is sudden, then the angle is just 180 degrees.

    Your discrepancy may be due to you reading the value off of the chart and not actually calculating it.

    Just calculate it yourself and don't use the chart.

    CS
     
  6. Mar 25, 2008 #5
    Sorry for sounding a bit ignorant but I am not very familiar with the part about calculating the K for sudden contraction.
     
  7. Mar 25, 2008 #6

    FredGarvin

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    According to Crane's, for a 180° contraction (actually for angles 45° to 180°), the loss coefficient is:

    [tex]k=\left[\frac{0.5(1-\beta^2)\sqrt{sin\left(\frac{\theta}{2}\right)}}{\beta^4}\right][/tex]

    Where

    [tex]\beta[/tex] is the diameter ratio
    [tex]\theta[/tex] is the angle of contraction
     
    Last edited: Mar 25, 2008
  8. Mar 25, 2008 #7

    stewartcs

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    Ah, Fred beat me to it.

    BTW the K is with respect to the larger diameter pipe.

    CS
     
    Last edited: Mar 25, 2008
  9. Mar 25, 2008 #8
    So what you mean is that the K I found through Crane's, the V, I should use is with respect to the velocity of the larger diameter pipe?

    [tex]h_L=K_L\frac{V^2}{2g}[/tex]

    In that case value I found for [tex]h_L[/tex] is 0.065m, Some difference between the one I calculated using the chart, which is 0.071m.

    By the way, I am not that familiar with Crane's, would like to read more about it.
     
  10. Mar 25, 2008 #9

    FredGarvin

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  11. Mar 25, 2008 #10
    Thanks, no wonder I've never seen it. Anyway, still it didn't explain the difference between theoretical values and experimental results. I would have expected increment in pressure loss in experimental results but it came out as the other way round.
     
  12. Mar 25, 2008 #11

    stewartcs

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    You may have some instrumentation errors too. It might be worth trying some different configurations to see if you obtain comparable results.

    Also, remember that mathematical models of natural phenomenon don't always turn out like one would think.

    Hope that helps.

    CS
     
  13. Mar 25, 2008 #12
    My experiment is based on water by the way. It was to investigate how eccentricity affects minor loss due to sudden contraction. I built a concentric setup as control.

    Well, I have been thinking of using a simple U-shape manometer but I don't seem to be able to find a suitable mano fluid. Therefore, I tuned out building a simple piezometer instead.

    Anyway, I've made a simple sketch of my setup, please comment if I've made any mistakes. The assumed the "h" in the sketch as pressure difference between two points.
     

    Attached Files:

  14. Mar 26, 2008 #13

    FredGarvin

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    What has been the problem with the manometer fluid?

    The only thing with your set up may be, depending on the lengths of the sections, is that you are dealing with a entrance region in stead of a sudden contraction. In other words, have you tried measuring the pressure a bit farther down stream of the contraction to ensure that you have re established fully developed? That may be a big source of error. Try moving your pressure measurements farther upstream and downstream from the contraction and see if that helps your results.
     
  15. Mar 26, 2008 #14
    I had... i made 2 more points further down with an increment of 1 cm of from each point.

    Are you suggesting it is due to vena contractra? But won't it be that the region at vena contractra contribute more pressure drop than after the flow fully developped?

    Are you suggesting that I should move the point to where the flow fully developed after the contraction? If so, then with turbulent flow, the lenght of the pipe would be very long.
     
    Last edited: Mar 26, 2008
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