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Resultant force at elbow from water flow

  1. Apr 12, 2016 #1
    1. The problem statement, all variables and given/known data
    water flows in an elbow was aligned horizontally , of angle 135 degree . If the water volume is section 1 and 2 is 0.2m^3 , the elbow is 12kg , flow rate is 0.4(m^3)/s , find the resultant force

    2. Relevant equations


    3. The attempt at a solution
    i found x and y component of the force
    for x-component , i have P1A1 - P2A2costheta -Fx = ρQ(v1-v2costheta)
    (150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-45) -Fx = (1000)(0.4)( 3.18- 12.7cos-45)
    Fx= 19170N to the left
    for y component ,
    P2A2sintheta +Fy -12(9.81) = ρQ(v2 sin theta)
    (90x10^3)(pi)[(200x10^-3 / 2 )^2 ](sin-45) +Fy -12(9.81) = (1000)(0.4)(12.7sin45)
    Fy = 1711 N upwards
    FR = 19246N
     

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  3. Apr 12, 2016 #2

    BvU

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    Are there relevant equations ?

    Is there a question ?

    Strange you use ## -F_x## but ##+F_y##. I think like that you let the resultant force point the wrong way

    Advice: with a bit of ##\TeX## this would look a lot more agreeable to the eye ... o0)
     
  4. Apr 12, 2016 #3
    it's okay to use use different sign for Fx and Fy , right ? when the ans that we get has different sign with the initial Fx and Fy , it means it's opposite direction to the initial direction of Fx and Fy that we assume , right ?
     
  5. Apr 12, 2016 #4
    other than the sign of Fx and Fy , is my equation okay ?
     
  6. Apr 14, 2016 #5
    it's okay to use use different sign for Fx and Fy , right ?
     
  7. Apr 14, 2016 #6

    BvU

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    Do you get the correct direction, then ?
     
  8. Apr 14, 2016 #7
    Is my idea correct?? I don't have the ans, I hope experts like you can check for me.....
     
  9. Apr 14, 2016 #8

    BvU

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    Thank you for your confidence, but I was just trying to read what you wrote and I noticed a strange asymmetry...that has disappeared since I ##TeX##ed your working -- it appears you simply took negatives on both sides for the y equation ?
    (do you see how much easier the typeset equations can be read ?)

    In combination with the picture I translate/guess that they want us to calculate the force this elbow exerts on the pipe segment that comes in from the left ?

    1. The problem statement, all variables and given/known data
    see above; establish a coordinate system, e.g. x is to the right, z is upwards (as in your drawing).
    ##P_1, P_2##
    ##\vec A_1, \vec A_2##
    ##\vec v_1, \vec v_2##
    ##\theta = -135^\circ ## (why -45##^\circ##?)
    ##F_x##
    etc

    2. Relevant equations
    I still miss relationships here: continuity, weight = mg, momentum balance ?

    3. The attempt at a solution

    for x-component , $$P_1A_1 - P_2A_2\cos\theta - F_x = ρQ(v_1-v_2\cos\theta)$$
    (150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-45) -Fx = (1000)(0.4)( 3.18- 12.7cos-45)
    Fx= 19170N to the left
    for y component ,
    $$-P_2A_2\sin\theta -F_y +12(9.81) = ρQ(-v_2 \sin \theta)$$
    (90x10^3)(pi)[(200x10^-3 / 2 )^2 ](sin-45) +Fy -12(9.81) = (1000)(0.4)(12.7sin45)
    Fy = 1711 N upwards
    FR = 19246N

    I notice you do charge for the elbow weight, but not for the water in the elbow (0.2 m3 respectively -- meaning 0.2 and 0.2 I suppose, so some 400 kg !).

    The outcome still surprises me -- I expect the water to push to the right and upwards, not to the left and upwards, but maybe the mass term dominates and it's to the right and downwards...

    My aim is not to do the exercise for you, but to ask and challenge (and tease) until you have put together an answer you yourself are convinced must be correct :smile:
     
  10. Apr 14, 2016 #9
    i mean measure from another direction , so i get 45 degree , is it wrong ?
     

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  11. Apr 14, 2016 #10

    BvU

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    What would you fill in for a 45 degree elbow ? Doesn't the direction of the water flow change from 0 to -135 degrees ?

    And: do you see how simple (simple) ##TeX## is ?
     
  12. Apr 14, 2016 #11
    why not the angle = positive 135 degree ? why gt negative sign ??? and do you mean i also missed out the water weight in my calculation for Fy , right ? i have put the -12(9.81) means the elbow weight
     
  13. Apr 14, 2016 #12
    sorry , i noticed that cos (-135) = cos 135
     
  14. Apr 14, 2016 #13
    after making correction , my ans is
    my (150x10^3)(pi)[(400x10^-3 / 2 )^2 ] -(90x10^3)(pi)[(200x10^-3 /2 )^2 ](cos-135) -Fx = (1000)(0.4)( 3.18- 12.7cos-135)
    so , 20848N - Fx = 4864N
    so, Fx = 15984N act to right , is it correct ?
    for vertical forces , i have
    -12(9.81) +(90x10^3)(pi)( [(200x10^-3)/ 2 ]^2 )sin45 -Fy -1000(9.81)(0.2) = 1000(0.4)(12.7sin45)
    -81 - Fy = 3592N
    Fy = -3673N , so Fy = 3673N upwards
    anything wrong with this ?
     
    Last edited: Apr 14, 2016
  15. Apr 15, 2016 #14
    is there anything wrong with the working above ?
     
  16. Apr 16, 2016 #15

    haruspex

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    The left hand side is the net force on the water in the elbow, with positive being to the right. With that sign convention, what is the expression for the rate of increase in momentum?

    I'm not sure how to interpret the information regarding the volume of the elbow. I suspect it has gained in translation, and should say that the volume between cross sections 1 and 2 is 0.2m3. As it stands, it sounds like there are two sections (i.e. lengths of pipe) each containing 0.2m3.
     
  17. Apr 16, 2016 #16

    BvU

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    Don't have my Bird Stewart & Lightfoot at hand here, so I'm still wondering what relationship is being worked out :smile: . Well, re-read post #8.
    I can decipher $$
    P_1A_1 - P_2A_2 \cos(-{3\over 4}\pi) -F_x = \rho Q(v_1-v_2\cos(-{3\over 4}\pi))
    $$with
    ##P_1 = 1.5 \times 10^5## N/m2, ## \quad P_2## = 1.5 \times 10^5## N/m2,
    ##\vec A_{1, x} = 0.04\pi ## m2, ##\quad \vec A_{2,x} = 0.01\pi\cos(-{3\over 4}\pi)##
    ##\rho = 1000 ## kg/m3,
    ## Q = 0,4 ## m3/s,
    ##\vec v_{1, x}, \vec v_{2,x}## ? Do you want me to check those numbers ?
    ##- F_x## meaning: if ## F_x## if the left pipe has to pull on the elbow, ## F_x## comes out positive ?
    etc

    For the z direction I would indeed stick to the 0.2 m3 between A1 and A2 .

    (It is seldom a good idea to rename given coordinate systems...)
     
  18. Apr 16, 2016 #17
    since flow rate = 0.4(m^3) /s , Q= A1V1 0.4= 0.04\pi ## m2 (V1) , V1 = 3.18m/s
    Q= A2V2 0.4= 0.01(pi) V2 , V2 = 12.7m/s ,
     
  19. Apr 16, 2016 #18
    in my previous calculation, my Fx = 16182N to the left , anything wrong with it ?
     
  20. Apr 16, 2016 #19

    BvU

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    Depends what you mean with Fx
     
  21. Apr 16, 2016 #20
    It looks like there have been significant difficulties with getting the trigonometry correct in this problem. Whenever that happens to me, I opt for expressing everything in terms of unit vectors in the x and y directions. This simplifies things considerably. foo9008, what is the unit vector perpendicular to the pipe at cross section 2 in terms of the unit vectors in the x and z directions? The force balance can then be expressed vectorially in terms of the unit vectors.

    I also noticed that the Bernoulli equation has not been applied to this problem yet.
     
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