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Parallel flow exercise

  1. Aug 14, 2016 #1
    1. The problem statement, all variables and given/known data
    I would really appreciate if you could give me some hints regarding what exactly to iii)
    0wODAs9.jpg

    2. Relevant equations
    we know they are in parallel
    so Q=Q1+Q2+.....+Qn
    delta hloss=delta hloss1+deltahloss2+......
    delta P=delta P1=delta P2=......


    3. The attempt at a solution
    We know flow is turbulent hence
    delta h loss=32*f*L*Q^2/pi^2*g*D^5
    where f=fanning friction factor
    L=length
    Q=vol flowrate
    g=gravit accel
    D=diameter

    We have 3 pipes in parallel
    so Q1=QA+QB+QC

    I need to show that QB~0.3Q1
     
  2. jcsd
  3. Aug 14, 2016 #2
    What are the limiting value of the fanning friction factor at high Reynolds numbers for surface roughnesses of 0.001, 0.007, and 0.035? Using these values, what is the ##\Delta P## for each pipe as a function of Q for that pipe?
     
  4. Aug 14, 2016 #3
    Well, we know that Q=pi*R^4/8Mew * deltaP/L but I guess that was for laminar flow , so it won't work here.
     
  5. Aug 14, 2016 #4
    That's only for laminar flow.
     
  6. Aug 14, 2016 #5
    for Turbulent we know delta h loss=32*f*L*Q^2/pi^2*g*D^5
    delta h loss=K*Q^2
     
  7. Aug 14, 2016 #6
    We will be working with the first formula. So, please answer my questions in post #2.
     
  8. Aug 14, 2016 #7
    I will be using the moody chart to get the fanning friction factors:
    for relative rougnhess of 0.001 we have f=0.005
    for relative roughness of 0.007 we have f=0.009
    for relative roughness of 0.035 we have f=0.015

    Now delta P as a function of Q for each pipe
    Because it is in parallel delta PA=delta PB=delta PC

    delta P=32*ro*f*L*Q^2/pi^2*D^5
     
  9. Aug 14, 2016 #8
    OK, now, for the friction factors of the 3 pipes and the given dimensions of the three pipes, express ##\Delta P/\rho## in terms of Q for each of the three pipes.
     
  10. Aug 14, 2016 #9
    delta P/ro=(32*f*L/pi^2*D^5)*Q^2

    and deltaPA=deltaPB=deltaPC because in parallel

    for pipe A
    delta PA/ro=(32*0.005*L/pi^2*D)*Q^2
     
  11. Aug 14, 2016 #10
    I need you to plug in the numbers.
     
  12. Aug 14, 2016 #11
    for pipe A
    delta PA/ro=(32*0.005*L/3.14^2*D)*Q^2=(0.16*L/9.86*D^2)*Q^2

    for pipe B
    delta PB/ro=(32*0.009*L/3.14^2*D)*Q^2=(0.288*L/9.86*D^2)*Q^2

    for pipe C
    delta PC/ro=(32*0.015*L/3.14^2*D)*Q^2=(0.48*L/9.86*D^2)*Q^2
     
  13. Aug 14, 2016 #12
    The figure gives a value of L and D for each pipe. Plug those in.
     
  14. Aug 14, 2016 #13
    for pipe A
    delta PA/ro=(32*0.005*L/3.14^2*D)*Q^2=(0.16*L/9.86*D^2)*Q^2
    delta PA/ro=(0.16*10/9.86*(4*10^-2)^2)*Q^2=101.32*QA^2

    =101.32*QA^2

    for pipe B
    delta PB/ro=(32*0.009*L/3.14^2*D)*Q^2=(0.288*L/9.86*D^2)*Q^2
    delta PB/ro=(0.288*4/9.86*(6*10^-2)^2)*Q^2=32.42*QB^2

    =32.42*QB^2

    for pipe C
    delta PC/ro=(0.48*6/9.86*(10*10^-2)^2)*Q^2=29.18*QC^2

    =29.18*QC^2
     
  15. Aug 14, 2016 #14
    OK. So
    $$h_A=h=K_AQ_A^2$$
    $$h_B=h=K_BQ_B^2$$
    $$h_C=h=K_CQ_C^2$$where $$K_A=101.32$$$$K_B=32.42$$$$K_C=29.18$$
    Now you just repeat what you did in the previous part of the problem.
     
  16. Aug 14, 2016 #15
    Thank you, what exactly do you mean by this?
     
  17. Aug 14, 2016 #16
    $$Q_A=\frac{\sqrt{h}}{\sqrt{K_A}}$$$$Q_B=\frac{\sqrt{h}}{\sqrt{K_B}}\tag{1}$$$$Q_C=\frac{\sqrt{h}}{\sqrt{K_C}}$$
    $$Q=Q_A+Q_B+Q_C=Q=\sqrt{h}\left[\frac{1}{\sqrt{K_A}}+\frac{1}{\sqrt{K_B}}+\frac{1}{\sqrt{K_C}}\right]$$
    $$\sqrt{h}=\frac{Q}{\left[\frac{1}{\sqrt{K_A}}+\frac{1}{\sqrt{K_B}}+\frac{1}{\sqrt{K_C}}\right]}\tag{2}$$
     
    Last edited: Aug 15, 2016
  18. Aug 14, 2016 #17
    Thank you
     
  19. Aug 15, 2016 #18
    sqrt(h)=sqrt(delta h loss)
    what exactly it is sqrt(h)?.I do not have value for it
    Have everything except Q and sqrt(h)
    Problem told me to show that QB~0.3Qtotal
     
  20. Aug 15, 2016 #19
    ##\sqrt{h}## is exactly what you said, the square root of the head loss. What do you get if you eliminate ##\sqrt{h}## between Eqns. 1 and 2 in post #16. C'mon man, it's just algebra. How were you able to solve part ii if you didn't do this?
     
  21. Aug 15, 2016 #20
    Yes, I understood this.However I do not understand how I am supposed to show that QB=0.3Q1 as I don't have numerical values for delta h.
     
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