FundamentalsofFlowinClosed Conduits
2.1.Fromthegivendata:D1=0.1m,D2=0.15m,V1=2m/s.Usingthesedata,thefollowing preliminarycalculationsareuseful:
Volumetricflowrate,Q,isgivenby Q=A1V1=(0.007854)(2)=0.0157m3/s
Accordingtothecontinuityequation,
= 00157 0.01767=0889m/s
At20◦C,thedensityofwater,ρ,is998kg/m3,andthemassflowrate,˙m,isgivenby
m=ρQ=(998)(0.0157)=15.7kg/s
2.xxFromthegivendata:D1=0.2m,D2=0.3m,andV1=0.75m/s.Usingthesedata,the followingpreliminarycalculationsareuseful:
(a)Accordingtothecontinuityequation,
(0.03142)(0.75) 007069 =0333m/s
(b)Thevolumeflowrate,Q,isgivenby
=(0.03142)(0.75)=0.02357m3/s=23.6L/s
2.2.Fromthegivendata:D1=200mm,D2=100mm,V1=1m/s,and
Theflowrate,Q1,inthe200-mmpipeisgivenby Q1=A1V1=(00314)(1)=00314m3/s
andhencetheflowrate,Q2,inthe100-mmpipeis
Theaveragevelocity,V2,inthe100-mmpipeis
2.3.Thevelocitydistributioninthepipeis
andtheaveragevelocity,V,isdefinedas
CombiningEquations1to3yields
2.4.
Theflowrate,Q,isthereforegivenby
3/s
2.5.D=0.2m,Q=0.06m3/s,L=100m,p1=500kPa,p2=400kPa,γ=9.79kN/m3 R= D 4 = 02 4 =005m ∆h= p1 γ p2 γ = 500400 979 =102m τ0= γR∆h L = (979×103)(005)(102) 100 =49.9N/m2 A= πD2 4 = π(0.2)2 4 =0.0314m2 V= Q A = 0.06 00314 =1.91m/s f= 8τ0 ρV2 = 8(49.9) (998)(191)2=0.11
2.6.T=20◦C,V=2m/s,D=0.25m,horizontalpipe,ductileiron.Forductileironpipe,ks= 0.26mm,and ks D = 026 250 =000104
Re=ρVD µ = (9982)(2)(025) (1.002×103)=4981×105
FromtheMoodydiagram: f=00202(flowisnotfullyturbulent)
UsingtheColebrookequation,
Substitutingforks/DandRegives
Bytrialanderrorleadsto f=0.0204
UsingtheSwamee-Jainequation,
whichleadsto
=0.0205
Theheadloss,hf,over100mofpipelineisgivenby
Thereforethepressuredrop,∆p,isgivenby
∆p=γhf=(9.79)(1.66)=16.3kPa
Ifthepipeis1mloweratthedownstreamend,fwouldnotchange,butthepressuredrop, ∆p,wouldthenbegivenby ∆p=γ(hf1.0)=9.79(1.661)=6.46kPa
2.7.Fromthegivendata:D=25mm,ks=0.1mm,θ=10◦ ,p1=550kPa,andL=100m.At 20◦C,ν=1.00×106m2/s,γ=9.79kN/m3,and ks D = 01 25 =0004 A= π 4 D2 = π 4(0.025)2=4.909×
Theenergyequationappliedover100mofpipeis
whichsimplifiesto p2=p1γ(z2z1)γhf p2=5509.79(100sin10◦)9.79(8.46×108fQ2) p2=380.08.28×109fQ2
(a)ForQ=2L/min=3333×105m3/s, V= Q A = 3333×105 4.909×104=0.06790m/s Re= VD ν = (006790)(0025) 1×106 =1698
SinceRe<2000,theflowislaminarwhenQ=2L/min.Hence, f= 64 Re = 64 1698 =0.03770 p2=3800828×109(003770)(3333×105)2=380kPa
Therefore,whentheflowis2L/min,thepressureatthedownstreamsectionis380kPa ForQ=20L/min=3333×104m3/s, V= Q A = 3.333×104 4909×104=06790m/s