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Head loss due to sudden expansion

  1. Aug 28, 2007 #1
    Hi all,

    I'm sure a lot of you know about the head loss due to sudden expansion:

    Hl = (1/2g)*(v1-v2)^2

    This equation can be derived from Bernoulli, continuity and momentum balans equations. The underlying assumption in deriving this equation is that the pressure just after the expansion (say, p0) is equal to the pressure before the expansion (p1).

    Now, I was wondering how accurate is this assumption? And what is the underlying physics behind this assumption?

    best regards,
  2. jcsd
  3. Aug 28, 2007 #2


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    I don't see where you get the g from? Also, the pressure just after expansion isn't assumed to be the same as the pressure before. In fact, if a and b are just before and after the expansion,

    [tex] \frac{p_a - p_b}{\rho} = \frac{a_b v_b^2 - a_a v_a^2}{2} + h_e[/tex]

    If you disregard the correction factors and use the momentum and continuity equation, you'll end up with

    [tex] h_e = \frac{(v_a-v_b)^2}{2}[/tex]
  4. Aug 28, 2007 #3
    Well, the g should be there, otherwise the dimensions of your equation does not match:

    m (n.e.) m^2/s^2

    (how do you post an equation in a thread by the way?)

    About the pressure: yes the (static) pressure is not the same between a and b. What I mean is the (static) pressure at border between small area and big area (so at position before b but just after a) is assumed to be equal to the (static) pressure at a. This is the underlying assumption in the momentum balans equation to get to the head loss equation I post earlier (it is far easier to explain if I can post a picture, but I do not know how).

    best regards,
  5. Aug 28, 2007 #4


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    Well, they do match. Both the RHS and LHS of my equation has the dimension [tex](m/s)^2[/tex]

    Click here for the tutorial thread. You can also click on the images to see the code.

    I don't quite understand your question. To post a picture, look for the "manage attachments" button below the box where you type your text.
  6. Aug 28, 2007 #5


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    The expression for a sudden expansion that I am used to seeing is

    [tex]h_L=K_L \left(\frac{V_1^2}{2g}\right)[/tex]


    [tex]K_L = \left(1-\frac{A_1}{A_2}\right)^2[/tex]

    According to my references, the sudden expansion is the only case in which the simple analysis actually matches actual measurements.
  7. Aug 28, 2007 #6


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    Yeah, it's the same thing, with a factor of g throughout.

    In fact, I did this very experiment in my momentum transfer lab this week, and the experimental value of k_l matched well with the analysis.
  8. Aug 29, 2007 #7


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    Well, there we have it. I never did that particular experiment in my school days so it's nice to have real experimental back up to the books.
  9. Aug 29, 2007 #8
    Bernoulli equation with pressure loss:

    [tex] p_{1}+0.5\rho v_{1}^{2}=p_{2}+0.5\rho v_{2}^{2}+p_{L}[/tex]

    With [tex]p_{L}[/tex] the pressure loss. The head loss, [tex] h_{L} [/tex] can be computed by dividing the equation with [tex] \rho g [/tex], and after re-arranging:

    [tex] h_{L} =\frac{p_{1}-p_{2}}{\rho g }+\frac{v_{1}^{2}-v_{2}^{2}}{2g }[/tex]

    Applying the momentum balans on the Control Volume between position 1 and 2 (see attachment):

    [tex] p_{1}A_{1}+p_{0}\left(A_{2}-A_{1}\right)-p_{2}A_{2}=\rho \left(A_{2}v_{2}^{2}-A_{1}v_{1}^{2}\right)[/tex]

    And here when you make an assumption that the pressure [tex]p_{0}[/tex] equals the pressure [tex]p_{1}[/tex]. And after using continuity equation the equation becomes:

    [tex] \frac{p_{2}-p_{1}}{\rho g }=\frac{v_{2}\left(v_{1}-v_{2}\right)}{g}[/tex]

    Fill this in to the expression of the head loss (second equation), voila, you get the head loss due to sudden expansion:

    [tex] h_{L} =\frac{\left(v_{1}-v_{2}\right)^{2}}{2g }[/tex]

    Now, this expression of the head loss is derived after we make an above mentioned assumption regarding the pressures. How accurate is this assumption then? I mean if the two pressures are not the same, the equation will have different form.

    best regards,

    Attached Files:

  10. Jul 16, 2010 #9


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    actually, the "head" has dimensions of length ...
  11. Mar 15, 2011 #10
    hi all..
    i wanted to know the value of "k" for practical uses. i was calculating head loss in expansion after main inlet valve in hydro plant.
    Going by penstock manual from Central water commission, New Delhi, the value of k =(1-a1/a2)sin(theta)
    where ;
    theta=one half of flare angle, a taper of 1:10 is recommended.
    Some colleague say the value of k is to be taken from practical values or practical charts.
    Last edited: Mar 15, 2011
  12. Mar 19, 2011 #11
    can any one derive for me the darcy weisbach formular for head loss
  13. May 21, 2013 #12
    Long time passed since your question, Dear previah
    This is for the people who is concerned to head loss in sudden expansioning.

    I am sure you're misunderstanding the underlying pressure equality assumption.

    This assumption says "the pressure at the annular area, A2-A1,is equal to the pressure in A1" for the calculation of pressure difference perpendicular to control volume. This annular area contacts turbulent eddies.

    So, as you say, " the pressure just after the expansion (say, p0) is equal to the pressure before the expansion (p1)" has completly different meaning.
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