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Minor loss due to expansion

  1. May 2, 2016 #1
    1. The problem statement, all variables and given/known data
    from the previous thread , https://www.physicsforums.com/threads/minor-loss-in-pipe.869148/
    , i know that formula of loss due to contraction can also be expressed as k[( mean velocity )^ 2 ] / 2g
    how about the formula of minor loss due to expansion , can we expressed as k[( mean velocity )^ 2 ] / 2g ??
    2. Relevant equations


    3. The attempt at a solution
     

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    Last edited: May 2, 2016
  2. jcsd
  3. May 2, 2016 #2
    is it feasible ?
     
  4. May 2, 2016 #3

    BvU

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    What mean velocity did you have in mind ? ##V_1+ V_2\over 2\ \ ## ?

    Or ##k_1 ## in combination with ##V_1## ?

    The old thread is a 'special case' with one of the V = 0
     
  5. May 2, 2016 #4
    Really, the old thread has one of the v =0 ? At which region?
    For the loss due to expansion, I mean (V1 +V2 ) /2
     
    Last edited by a moderator: May 2, 2016
  6. May 2, 2016 #5
    in the old thread , how can one of the v = 0 ? it is [(Vc -V2)^2]/ 2g
     
  7. May 2, 2016 #6
    What do you mean by combination of k and v1 ???
     
  8. May 3, 2016 #7

    BvU

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    in the reservoir
    well, the link gives the expressions. You can try to wiggle ##\ \
    \left (V_1+ V_2\over 2\right )^2 \ \ ## into this but i doubt that succeeds...
     
  9. May 3, 2016 #8

    BvU

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    I meant the one in post #1
     
  10. May 3, 2016 #9
    So , its not possible to change the to change the original formula of head loss due to expansion into
    k[( mean velocity )^ 2 ] / 2g??
     
  11. May 3, 2016 #10
    it doesnt state v = 0 , it just stated mean velocity in 165
     
  12. May 3, 2016 #11

    BvU

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    Ah ! I was mistaken in post #3. In 165 there is no counterpart of the ##V_1## in 166 -- so I wanted to consider ##V_1## as zero in picture 165. But it plays no role in either case: the ##v## involved are ## V_c## and ##V_2## in both cases, so 165 and 166 say the same (as haru indicated and you Ok'd)

    (we are now talking about pictures in another thread - never mind :smile:) .

    In 166 you see a ## k_L = \left ( {\displaystyle A_2\over \displaystyle A_c} - 1\right )^2##.

    In the current thread the idea is that there is no contraction at CD and that means ##V_1## plays the role of ##V_c## (and ##A_1## is ##A_c##).

    ##v_1## and ##v_2## are linked through continuity (##V_1 A_1 = V_2 A_2##) so -- analogous to 166 -- you get a ##k_L## for ##V_2## but you can also get a (different) ##k_L## for ##V_1##.

    'Mean velocity in the pipe' may have caused confusion: what is meant is volume flow divided by area (i.e. the velocity averaged over the cross section -- so not lengthwise). In other words: there is no place for (V1 +V2 ) /2
     
  13. May 3, 2016 #12
    ok , i understand that the velocity is average velocity across the area , but can still i say that the head loss due to contraction expansion is k[( mean velocity^2] / 2g ???
     
    Last edited: May 3, 2016
  14. May 3, 2016 #13
    sorry , pls ignore my several posts before this . do you mean the mean velocity is not suitable and should be replaced by v1(mean velocity at region 1 ) or v2( mean velocity at region 2 ) to avoid confusion ?
     
    Last edited: May 3, 2016
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