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Calculating the pressure drop in a contraction?

  1. Dec 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Water flows at room temperature in a horizontal pipe with mean velocity of 1 m/s. Then a sudden contraction of the pipe changes the diameter to half the upstreams diameter (##\frac{D_1}{D_2} = 2##). What is the pressure drop (##p_1 - p_2##) from a point (1) upstream of the contraction to a point downstream of the contraction (2)?

    2. Relevant equations
    Mass conservation
    Energy equation

    3. The attempt at a solution
    This is my attempt:

    Unfortunately, the correct answer is 10500 Pa and I have no idea what I did wrong.
  2. jcsd
  3. Dec 11, 2013 #2

    rude man

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    Hmm. I got 8500 Pa. In your expression for the correction factor I think you need to use v2, not v1. That adds an extra 1000 Pa to your 7500 Pa. Can't imagine where the other 2000 Pa would come from ...

    See attachment, page 20.

    Attached Files:

  4. Dec 11, 2013 #3
    Yeah, I accidentally used ##V_1## instead of ##V_2##.

    I now have a pressure drop of 11500 Pa:

    It's still not 10500 Pa however.
  5. Dec 11, 2013 #4


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    I don't think you can use K = 0.5 for the sudden contraction within a closed pipe. The factor K = 0.5 is based on having a free surface present in the fluid reservoir before the fluid enters a pipe. Because there is a closed pipe on both sides of the sudden contraction, the flow will be different. The flow streamlines start to contract before the abrupt change in diameters occurs and continue to contract for a short distance after the change has been passed, leading to the creation of a vena contracta. The head loss factor K for such sudden contractions depends not only on the ratio of the diameters of the pipes ahead of and after the contraction, but also on the flow velocity as well.

    The attachment discusses this phenomenon in more detail and presents a table of K-factors for sudden pipe contractions on p. 17, Table 10.3. I believe the correct K-factor for the diameter ratio and flow velocity for this case is closer to 0.37 than 0.50, which I calculate gives you close to the pressure drop of 10500 Pa.

    Attached Files:

    • ch10.pdf
      File size:
      559.2 KB
  6. Dec 11, 2013 #5
    Yeah, that definitely works.
    My only problem is however that I would like to use information given in the book for my lecture only. So I'm trying to figure out where in my book I can come up with this 0.37 value.

    This is the closest thing that I found in my textbook:

    There is the graph below which shows the contraction and expansion coefficient in terms of the area ratio.
    In my case, my area ratio is 0.25 however I'm not sure which of the two graphs I'm supposed to look at? The text does not describe anything and the only thing that it indicates is two arrows pointing at opposite directions.
    It seems clear that it is the graph for which arrow points to the left that gives a coefficient of 0.37 for AR=0.25

    But what is that other graph that points to the right and which one do I use when I'm sitting at my exam?
    I'd just like to understand this for the future lol. The text in the book does not mention anything about these graphs.
  7. Dec 11, 2013 #6


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    You enter the attached figure in Post #5 with the area ratio. If you have a sudden contraction, you read the K-factor using the lower curve and take the value of K from the left vertical axis. If you have a sudden expansion of area, you read the value of the K factor from the right vertical axis using the upper curve. The arrows are there to indicate which vertical axis to use.
  8. Dec 11, 2013 #7
    Ohh, now I get it. So the arrows are just pointing at which axis to read.
    Exams are really lowering my brain capacity. Thanks! :P
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