hw 6 solutions -...
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FloridaInternationalUniversityDepartmentofCivilandEnvironmentalEngineering
CWR3201FluidMechanics,Fall2018Instructor:ArturoS.Leon,Ph.D.,P.E.,D.WRE
TA:ThaoDo,CEEUndergraduate
HomeworkAssignment6SolutionsMechanicsofFluids(Fifthedition),byM.C.Potter,D.C.WiggertandB.H.Ramadan.
1. 7.20(samenumberinFourthEdition)
𝐷𝑖𝑠ℎ𝑐𝑎𝑟𝑔𝑒 𝑄 = 𝐴 ∗ 𝑉 → 0.025 =𝜋4 (0.06)
!𝑉𝑉 = 8.84 𝑚/𝑠
𝑅𝑒 =𝑉𝐷𝜈 =
8.84 ∗ 0.061.005 ∗ 10!! = 527877
SinceReynoldsnumberisgreaterthan2000,flowisturbulent.𝑅𝑒 > 10!𝐿𝐷 = 120𝐿 = 7.2 𝑚
SinceLengthofpipeis50m,whichisgreaterthan7.2m,assumptionofdevelopedflowisacceptable.
2. 7.94(samenumberinFourthEdition)a) 𝑅𝑒 = !"
!= !.!"#∗!.!"
!"!!= 1000lessthan2000soflowislaminar
RelativeRoughness:!!= !.!"
!"= 0.00625
Frictionfactor:𝑓 = !"!!"
= !"∗!"!!
!.!"#∗!.!"= 0.064
b) 𝑅𝑒 = !"!= !.!"∗!.!"
!"!!= 10000TurbulentflowsouseMoodyDiagram
RelativeRoughness:!!= !.!"
!"= 0.00625
Moodydiagram:𝑓 = 0.04
3. 7.108(samenumberinFourthEdition)
a) 𝐷 = 0.66[𝑒!.!" !!!
!!!
!.!"+ 𝜈𝑄!.!( !
!!!)!.!]!.!"
ℎ! =∆𝑝𝛾 =
2000009810 = 20.38 𝑚
𝐷 = 0.66[0!.!"𝐿𝑄!
𝑔ℎ!
!.!"
+ 10!! ∗ 0.002!.!100
9.81 ∗ 20.38
!.!
]!.!" = 0.032𝑚
Notethattheotherpartshavesamemethodbutdifferingspecificweightsforeachtypeofmedium.
4. 7.113(samenumberinFourthEdition)
ℎ! =∆𝑝𝛾 =
1009810 = 0.01 𝑚
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HydraulicDiameter:𝐷 = 4 !"#$!"#$%"&"#
= 4 (!.!"∗!.!)!(!.!"!!.!
= 0.057𝑚FlowRate:
𝑄 = −0.965[𝑔 !!!!!]!.! ln !
!.!!+ !.!"!!!
!!!!!
!.!= −0.965[9.81 !.!"#
!∗!.!"!
]!.! ln !!.!!
+
!.!"(!"!!)!∗!!.!"∗!.!"#!!.!"
!.!= 7.31 ∗ 10!! !
!
!
5. 7.117(samenumberinFourthEdition)
𝑉! =𝑄𝐴!
= 63.66𝑚𝑠
Fromcontinuity:𝑉! = 𝑉!!!!
!!!= 15.92 𝑚/𝑠
Idealgaslawtocalculatedensityofair:𝜌 = !!"#!!!"!#!"
= !"!.!!!"!.!"#∗!"#
= 1.799 !"!!
Headlossduetosuddenenlargement:ℎ! =!!!!!
!!
𝐾! = (1−𝑟!!
𝑟!!)! = 0.5625
ℎ! = 116.18Bernoullienergyequation:!!
!"+ !!!
!!+ 𝑍! = !!
!"+ !!!
!!+ 𝑍! + ℎ!
𝑝! = 51.366 𝑘𝑃𝑎6. 7.121(samenumberinFourthEdition)
Manometer:𝑝! + 𝛾!𝐻 = 𝛾!!"#𝐻 + 𝑝!∆𝑝 = 123606𝐻
𝑉 =𝑄𝐴 = 4.77 𝑚/𝑠
UsingenergyequationandrearrangingtosolveforK:𝐾 !!
!!= ∆!
!
IfH=4cm:𝐾 !.!!!
!∗!.!"= !"#$%$!
!"#$
𝐾 = 0.435𝐾 = 0.869
7. 7.124(samenumberinFourthEdition)
EnergyEquation:𝐻 = 𝐾!" + 2𝐾!"#$% + 𝐾!"!!
!!+ 𝑓 !
!!!
!!
Assumefrictionfactorf=0.020NominallosscoefficientsK:𝐾!"#$% = 1.0
𝐾!" = 0.8𝐾!" = 1.0
2 = 0.8+ 2(1.0)+ 1𝑉!
2 ∗ 9.81+ 0.02200.04
𝑉!
2 ∗ 9.81𝑉 = 1.69 𝑚/𝑠
CheckAssumption:CalculateReynolds𝑅𝑒 = !"!= !.!"∗!.!"
!.!"∗!"!!= 59298
RelativeRoughness:AssumesmoothpipeMoodydiagram:𝑓 = 0.019sinceitisclosetotheassumptionOK
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𝑄 = 𝐴𝑉 =𝜋4 ∗ 0.04
! ∗ 1.69 = 0.0021 𝑚!/𝑠8. 7.127(samenumberinFourthEdition)
EnergyEquation:Sincepoint1and3areopentoatmosphereandvelocityatpoint1iszero,theenergyequationreducesto:
𝐻 = 𝐾!" + 2𝐾!"#$% + 𝐾!"𝑉!
2𝑔 + 𝑓𝐿𝐷𝑉!
2𝑔
NominallosscoefficientsK:𝐾!"#$% = 1.39𝐾!" = 0.8𝐾!" = 1.0
Assumef=0.20
6 = 0.8+ 2(1.39)+ 1.0𝑉!
2 ∗ 9.81+ 0.026.8+ 2𝐻0.03
𝑉!
2 ∗ 9.81
ApplyBernoulliatpoint2andexit:!!"#$!"
+ !!"#$!
!!+ 𝑧!"#$ =
!!!"+ !!!
!!+ 𝑧! +
𝐾!"#$%!!
!!+ 𝑓 !
!!!
!!
0 + 0 + 0 =1.13 𝑘𝑃𝑎
(1000)(9.81)+
𝑉!!
2 ∗ 9.81+ 6 + 𝐻 + (1.39)
𝑉!
2𝑔+ 0.02
6 + 𝐻0.03
𝑉!
2 ∗ 9.81
Solvingsystemofequations:𝐻 = 0.64 𝑚𝑉 = 3.348 𝑚/𝑠
𝑅𝑒 =𝑉𝐷𝜈 =
3.348 ∗ 0.031.31 ∗ 10!! = 7.7 ∗ 10!
𝑀𝑜𝑜𝑑𝑦 𝐷𝑖𝑎𝑔𝑟𝑎𝑚: 𝑓 = 0.018Recalculateusingnewfrictionfactor:𝐻 = 0.78 𝑚
𝑉 = 3.477 𝑚/𝑠
𝑅𝑒 =𝑉𝐷𝜈 =
3.477 ∗ 0.031.31 ∗ 10!! = 7.9 ∗ 10!
𝑀𝑜𝑜𝑑𝑦 𝐷𝑖𝑎𝑔𝑟𝑎𝑚 𝑓 = 0.018Frictionfactorhasconvergedso𝐻 = 0.78 𝑚
9. ProblemDuetolengthofsolution,pleasefollowlinkbelowforcompletesolution:
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http://web.eng.fiu.edu/arleon/courses/Hydraulic_engineering/Quizzes/Quiz_3_4_sol.pdf10. 11.41(samenumberinFourthedition)
UsingEPANET:
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