HW 6 Solutions -...

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

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

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:

http://web.eng.fiu.edu/arleon/courses/Hydraulic_engineering/Quizzes/Quiz_3_4_sol.pdf10. 11.41(samenumberinFourthedition)

UsingEPANET: