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Head loss in pipe

  1. Apr 14, 2016 #1
    1. The problem statement, all variables and given/known data
    estimate the energy head lost along a short length of pipe suddenly enlarging a diameter of 350mm to 700mm which discharges 0.7(m^-3) of water per second .
    the solution given is Q1= Q2 = 0.7(m^3)/s
    0.7 = pi ((350x10^-3)^2 ) V1 / 4 , V1= 7.28m/s
    0.7 = pi ((700x10^-3)^2 ) V2 / 4 , v2 = 1.82m/s
    [(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss
    head loss = 2.53m

    2. Relevant equations


    3. The attempt at a solution
    the bernoulli's equation is P1/ ρg + (V1^2) /2g + z1 = P2/ ρg + (V2^2) /2g + z2, by writing [(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss , the author assume P1 = P2 , how could it be possible ? since the inlet is smaller than outlet , the pressure at inlet should be lower , right ? the velocity at inlet is smaller
     
  2. jcsd
  3. Apr 14, 2016 #2
    The head loss is ##\frac{(P_1-P_2)}{\rho g}##
     
  4. Apr 14, 2016 #3
    but i dont have P1 and P2.... how to solve this ?
     
  5. Apr 14, 2016 #4
    You're solving algebraically for the entire quantity ##\frac{(P_1-P_2)}{\rho g}## (which is the head loss), without having to know either P1 or P2 individually. You are aware that the pipe is assumed to be horizontal so that z1 = z2, correct? And you are aware of the definition of head, correct?
     
  6. Apr 14, 2016 #5
    yes , [(v1)^2 ] / 2g = [(v2)^2 ] / 2g +head loss is it correct ??
     
  7. Apr 14, 2016 #6
    No. There should be a minus sign in front of head loss.
     
  8. Apr 14, 2016 #7
    the energy flow from point 1 to 2 , so we should add the head loss at point 2 , right ?
     
  9. Apr 14, 2016 #8
    The head loss is defined as the decrease in head from the upstream point to the downstream point.
     
  10. Apr 14, 2016 #9
    ya , point 1 is upstream , point 2 is downstream , right ?
     
  11. Apr 14, 2016 #10
    why the head loss is ##\frac{(P_1-P_2)}{\rho g}##?
    shouldn't it be P1/ ρg + (V1^2) /2g + z1 = P2/ ρg + (V2^2) /2g + z2 + head loss ???? we cancel our z1 and z2 , since they are the same
     
  12. Apr 14, 2016 #11
    What is your understanding of the term head, and what is your understanding of the term head loss?
     
  13. Apr 14, 2016 #12
    A portion of energy is lost to the resistance to flow....
     
  14. Apr 14, 2016 #13
    That is not the general definition of head loss and, in this problem, the resistance to flow is implicitly assumed to be zero. More generally, head loss is the decrease in ##P/(\rho g)+z## between inlet and outlet.
     
  15. Apr 14, 2016 #14
    can you help how to solve this question ?
     
  16. Apr 14, 2016 #15
    sorry , i missed out part of thge question . the full question given is
    estimate the energy head lost along a short length of pipe suddenly enlarging a diameter of 350mm to 700mm which discharges 0.7(m^-3) of water per second .. If the pressure at entrance is 105N/ (m^2) . Find the pressure at exit
     
  17. Apr 14, 2016 #16
    so the equation should be (105x10^3)/ (1000x9.81) + (7.28^2) / (2x9.81) +z1 = P2 / (1000x9.81) + (1.82^2) / (2x9.81) + head loss , am i right ?
    cancel off z1 and z2 , we still have unknown P2 , how to find the head loss?
     
  18. Apr 14, 2016 #17
    No. It's not correct.

    $$\frac{P_1}{\rho g}+z_1+\frac{(7.28)^2}{(2)(9.81)}=\frac{P_2}{\rho g}+z_2+\frac{(1.82)^2}{(2)(9.81)}$$
    $$\frac{P_1}{\rho g}-\frac{P_2}{\rho g}=head\ loss=\frac{(1.82)^2}{(2)(9.81)}-\frac{(7.28)^2}{(2)(9.81)}=-2.53m$$So,
    $$\frac{P_2}{\rho g}=\frac{P_1}{\rho g}+2.53m$$
     
  19. Apr 14, 2016 #18
    pls refer to the link diagram 6.5 , the formula fo hf is : CjUGeji.png
    http://www.efm.leeds.ac.uk/CIVE/CIVE1400/Examples/eg6_ans.htm
     
  20. Apr 14, 2016 #19
    upload_2016-4-15_12-31-7.png
     
  21. Apr 14, 2016 #20
    View attachment 99132
     
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