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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)
π·ππ βπππππ π = π΄ β π β 0.025 =π4 (0.06)
!ππ = 8.84 π/π
π π =ππ·π =
8.84 β 0.061.005 β 10!! = 527877
SinceReynoldsnumberisgreaterthan2000,flowisturbulent.π π > 10!πΏπ· = 120πΏ = 7.2 π
SinceLengthofpipeis50m,whichisgreaterthan7.2m,assumptionofdevelopedflowisacceptable.
2. 7.94(samenumberinFourthEdition)a) π π = !"
!= !.!"#β!.!"
!"!!= 1000lessthan2000soflowislaminar
RelativeRoughness:!!= !.!"
!"= 0.00625
Frictionfactor:π = !"!!"
= !"β!"!!
!.!"#β!.!"= 0.064
b) π π = !"!= !.!"β!.!"
!"!!= 10000TurbulentflowsouseMoodyDiagram
RelativeRoughness:!!= !.!"
!"= 0.00625
Moodydiagram:π = 0.04
3. 7.108(samenumberinFourthEdition)
a) π· = 0.66[π!.!" !!!
!!!
!.!"+ ππ!.!( !
!!!)!.!]!.!"
β! =βππΎ =
2000009810 = 20.38 π
π· = 0.66[0!.!"πΏπ!
πβ!
!.!"
+ 10!! β 0.002!.!100
9.81 β 20.38
!.!
]!.!" = 0.032π
Notethattheotherpartshavesamemethodbutdifferingspecificweightsforeachtypeofmedium.
4. 7.113(samenumberinFourthEdition)
β! =βππΎ =
1009810 = 0.01 π
HydraulicDiameter:π· = 4 !"#$!"#$%"&"#
= 4 (!.!"β!.!)!(!.!"!!.!
= 0.057πFlowRate:
π = β0.965[π !!!!!]!.! ln !
!.!!+ !.!"!!!
!!!!!
!.!= β0.965[9.81 !.!"#
!β!.!"!
]!.! ln !!.!!
+
!.!"(!"!!)!β!!.!"β!.!"#!!.!"
!.!= 7.31 β 10!! !
!
!
5. 7.117(samenumberinFourthEdition)
π! =ππ΄!
= 63.66ππ
Fromcontinuity:π! = π!!!!
!!!= 15.92 π/π
Idealgaslawtocalculatedensityofair:π = !!"#!!!"!#!"
= !"!.!!!"!.!"#β!"#
= 1.799 !"!!
Headlossduetosuddenenlargement:β! =!!!!!
!!
πΎ! = (1βπ!!
π!!)! = 0.5625
β! = 116.18Bernoullienergyequation:!!
!"+ !!!
!!+ π! = !!
!"+ !!!
!!+ π! + β!
π! = 51.366 πππ6. 7.121(samenumberinFourthEdition)
Manometer:π! + πΎ!π» = πΎ!!"#π» + π!βπ = 123606π»
π =ππ΄ = 4.77 π/π
UsingenergyequationandrearrangingtosolveforK:πΎ !!
!!= β!
!
IfH=4cm:πΎ !.!!!
!β!.!"= !"#$%$!
!"#$
πΎ = 0.435πΎ = 0.869
7. 7.124(samenumberinFourthEdition)
EnergyEquation:π» = πΎ!" + 2πΎ!"#$% + πΎ!"!!
!!+ π !
!!!
!!
Assumefrictionfactorf=0.020NominallosscoefficientsK:πΎ!"#$% = 1.0
πΎ!" = 0.8πΎ!" = 1.0
2 = 0.8+ 2(1.0)+ 1π!
2 β 9.81+ 0.02200.04
π!
2 β 9.81π = 1.69 π/π
CheckAssumption:CalculateReynoldsπ π = !"!= !.!"β!.!"
!.!"β!"!!= 59298
RelativeRoughness:AssumesmoothpipeMoodydiagram:π = 0.019sinceitisclosetotheassumptionOK
π = π΄π =π4 β 0.04
! β 1.69 = 0.0021 π!/π 8. 7.127(samenumberinFourthEdition)
EnergyEquation:Sincepoint1and3areopentoatmosphereandvelocityatpoint1iszero,theenergyequationreducesto:
π» = πΎ!" + 2πΎ!"#$% + πΎ!"π!
2π + ππΏπ·π!
2π
NominallosscoefficientsK:πΎ!"#$% = 1.39πΎ!" = 0.8πΎ!" = 1.0
Assumef=0.20
6 = 0.8+ 2(1.39)+ 1.0π!
2 β 9.81+ 0.026.8+ 2π»0.03
π!
2 β 9.81
ApplyBernoulliatpoint2andexit:!!"#$!"
+ !!"#$!
!!+ π§!"#$ =
!!!"+ !!!
!!+ π§! +
πΎ!"#$%!!
!!+ π !
!!!
!!
0 + 0 + 0 =1.13 πππ
(1000)(9.81)+
π!!
2 β 9.81+ 6 + π» + (1.39)
π!
2π+ 0.02
6 + π»0.03
π!
2 β 9.81
Solvingsystemofequations:π» = 0.64 ππ = 3.348 π/π
π π =ππ·π =
3.348 β 0.031.31 β 10!! = 7.7 β 10!
πππππ¦ π·ππππππ: π = 0.018Recalculateusingnewfrictionfactor:π» = 0.78 π
π = 3.477 π/π
π π =ππ·π =
3.477 β 0.031.31 β 10!! = 7.9 β 10!
πππππ¦ π·ππππππ π = 0.018Frictionfactorhasconvergedsoπ» = 0.78 π
9. ProblemDuetolengthofsolution,pleasefollowlinkbelowforcompletesolution:
http://web.eng.fiu.edu/arleon/courses/Hydraulic_engineering/Quizzes/Quiz_3_4_sol.pdf10. 11.41(samenumberinFourthedition)
UsingEPANET: