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Bernoulli equation exercise from Fanning and Moody

  1. Aug 17, 2016 #1
    1. The problem statement, all variables and given/known data
    I would really appreciate it if you could give me a hand with this exercise, not sure on what I've done.
    Data:
    kden.JPG
    Moody:
    kzen.JPG

    L=55*10-3m
    D=10-1m
    k=0.0002m

    2. Relevant equations
    ##Re=\frac{D*u*ρ} {μ}##
    ##Re=\frac{4*m} {pi*D*μ}##
    Relative roughness ##ξ=\frac k D## where k=rougness and D=diameter
    f=fanning friction factor from Moddy chart as ξ ∩ Re
    K=loss coefficient =##\frac{f*4L} {D}+∑Ki##
    where Ki=other losses
    ##Δhloss=\frac{K*u^2} {2g}##
    Bernoulli:##\frac {P1} {ρg} ##+h1 +##\frac {u1^2} {2g} ## = ##\frac {P2} {ρg} ##+h2 +##\frac {u2^2} {2g} +Δhloss##

    3. The attempt at a solution
    i)Re=?
    Q=0.05m^3/s
    μ=1.2*10-6m2/s

    ##Re=\frac{4*m} {pi*D*μ}##
    where m=ρ*Q
    m=103*0.05
    m=50 Kg/s
    =>##Re=\frac{4*50} {pi*10^-1*1.2*10^-6}##
    =>Re=530516477 turbulent flow

    ii)f=?fanning friction
    ##ξ=\frac k D## =##ξ=\frac {0.0002} {10^-1}##
    ξ=2*10-3
    => f=0.006 from ξ∩Re on Moody Chart

    iii)Assumptions
    A: hA=7m PA=? uA=0
    B hB=0m(datum) PB=0(atomospheric) uB=0;

    :##\frac {P1} {ρg} ##+h1 +##\frac {u1^2} {2g} ## = ##\frac {P2} {ρg} ##+h2 +##\frac {u2^2} {2g} +Δhloss##

    Pluggin in the assumptions this means:
    ##Δhloss=\frac {PA} {ρg} +hA##
    =>##PA=(Δhloss-hA)*ρg##
    Δhloss=K*u2/2*g
    K=##\frac{f*4L} {D}+∑Ki##=##\frac{0.006*4*55*10^-3} {10^-1}+0.5+3*0.8+2*0.25+0.95##

    ##K=4.36##

    Q=0.05 m^3/s=>##u=\frac {Q} {A}## ##u=\frac {0.05} {pi*(10^-1)^2/4}##
    u=6.36m/s

    ##Δhloss=\frac {K*u^2} {2g}##=##Δhloss=\frac {4.36*6.36^2} {2*9.81}##

    Δhloss=8.98m

    ##PA=(Δhloss-hA)*ρg##

    ##PA=(8.98-7)*10^3*9.81##

    =>PA=19423.8 Pa
     
    Last edited by a moderator: Aug 17, 2016
  2. jcsd
  3. Aug 18, 2016 #2
    The Reynolds number is calculated incorrectly. You are given the kinematic viscosity, not the dynamic viscosity.
     
  4. Aug 18, 2016 #3
    ##Re=\frac{D*u} {∂}##
    where D=diameter u=velocity ∂=kinematic viscosity
    u=Q/A=6.36m/s

    ##Re=\frac{10^{-1}*6.36} {1.2*10^{-6}}##

    Re=530000=>Re=5.3*105

    ##ξ=\frac{k} {D}=2*10^{-3}##=0.002

    ξ∩Re=>f=0.006 from Moody Chart
     
  5. Aug 18, 2016 #4
    This is better. I haven't checked the rest of the analysis.
     
  6. Aug 18, 2016 #5
    Thank you for your reply.
    I would really appreciate it , if you can and have time , if you could check the rest of the exercise, as I am not sure if it is correct.
    In both cases the f=0.006
     
  7. Aug 18, 2016 #6
    It looks OK to me.
     
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