PROBLEMSFORCHAPTER22-1Byconsiderationofthecylindricalelementalcontrolvolumeasshownbelow,usethe conservation of mass to derive the continuity equation in cylindricalcoordinates.

ContinuityequationThenetfluxofmassenteringtheelementequaltotherateofchangeofthemassoftheelement.

𝑚KL − 𝑚NOP =𝜕𝜕𝑡𝑚TUTVTLP

𝑚 = massflowrate = 𝜌𝐴𝑉

𝑉 = velocityoffluid

(a) Massflowrate,directionof𝑣]:

𝑚KL − 𝑚NOP = 𝜌 𝑟𝑑𝜃𝑑𝑧 𝑣] − 𝜌𝑣] +𝜕𝜕𝑟

𝜌𝑣] 𝑑𝑟 𝑟 + 𝑑𝑟 𝑑𝜃𝑑𝑧

= 𝜌 𝑟𝑑𝜃𝑑𝑧 𝑣] − 𝜌𝑣]𝑟 + 𝜌𝑣]𝑑𝑟 +𝜕𝜕𝑟

𝜌𝑣] 𝑟𝑑𝑟 +𝜕𝜕𝑟

𝜌𝑣] 𝑑𝑟𝑑𝑟 𝑑𝜃𝑑𝑧

𝑑𝑟𝑑𝑟 = 0,toosmall

= 𝜌𝑣]𝑟𝑑𝜃𝑑𝑧 − 𝜌𝑣]𝑟 + 𝜌𝑣]𝑑𝑟 +𝜕𝜕𝑟

𝜌 𝑟𝑑𝑟 𝑑𝜃𝑑𝑧

= 𝜌𝑣]𝑟𝑑𝜃𝑑𝑧 − 𝜌𝑣]𝑟 +𝜕𝜕𝑟

𝜌𝑣]𝑟 𝑑𝑟 𝑑𝜃𝑑𝑧

= −𝜕𝜕𝑟

𝜌𝑣]𝑟 𝑑𝑟𝑑𝜃𝑑𝑧

(b) Massflowrate,directionof𝑣e:

𝑚KL − 𝑚NOP = 𝜌𝑣e𝑟𝑑𝜃𝑑𝑟 − 𝜌𝑣e +𝜕𝜕𝑧

𝜌𝑣e 𝑑𝑧 𝑟𝑑𝜃𝑑𝑟

= −𝜕𝜕𝑧

𝜌𝑣e 𝑟𝑑𝜃𝑑𝑟𝑑𝑧

(c) Massflowrate,directionof𝑣f:

𝑚KL − 𝑚NOP = 𝜌𝑣f𝑑𝑧𝑑𝑟 − 𝜌𝑣f +𝜕𝜕𝜃

𝜌𝑣f 𝑑𝜃 𝑑𝑟𝑑𝑧

= −𝜕𝜕𝜃

𝜌𝑣f 𝑑𝜃𝑑𝑟𝑑𝑧

(d)

𝑚𝑎𝑠𝑠 = 𝜌∀

= 𝜌 𝑟𝑑𝜃𝑑𝑟𝑑𝑧

𝜕𝜕𝑡𝑚 =

𝜕𝜕𝑡

𝜌𝑟𝑑𝜃𝑑𝑟𝑑𝑧

𝑚KL − 𝑚NOP =𝜕𝜕𝑡𝑚TUTVTLP

−𝜕𝜕𝑟

𝜌𝑣]𝑟 𝑑𝑟𝑑𝜃𝑑𝑧 −𝜕𝜕𝑧

𝜌𝑣e 𝑟𝑑𝜃𝑑𝑟𝑑𝑧 −𝜕𝜕𝜃

𝜌𝑣f 𝑑𝜃𝑑𝑟𝑑𝑧 =𝜕𝜕𝑡

𝜌𝑟𝑑𝜃𝑑𝑟𝑑𝑧

dividewith𝑑𝜃𝑑𝑟𝑑𝑧

−𝜕𝜕𝑟

𝜌𝑣]𝑟 −𝜕𝜕𝑧

𝜌𝑣e 𝑟 −𝜕𝜕𝜃

𝜌𝑣f =𝜕𝜕𝑡

𝜌𝑟

= 𝜌𝜕𝑟𝜕𝑡+ 𝑟

𝜕𝜌𝜕𝑡

𝜕𝑟𝜕𝑡= 0,Nochangesof𝑟regardingtothetime

0 = 𝑟𝜕𝜌𝜕𝑡+𝜕𝜕𝑟

𝜌𝑣]𝑟 +𝜕𝜕𝑧

𝜌𝑣e 𝑟 +𝜕𝜕𝜃

𝜌𝑣f

atau

0 =𝜕𝜌𝜕𝑡+1𝑟𝜕𝜕𝑟

𝜌𝑣]𝑟 +1𝑟𝜕𝜕𝑧

𝜌𝑣e 𝑟 +1𝑟𝜕𝜕𝜃

𝜌𝑣f

0 =𝜕𝜌𝜕𝑡+1𝑟𝜕𝜕𝑟

𝜌𝑣]𝑟 +𝜕𝜕𝑧

𝜌𝑣e +1𝑟𝜕𝜕𝜃

𝜌𝑣f

Thisisthecompressibleequationofcontinuityincylindricalpolarcoordinates.

2-2Simplifytheequationofcontinuityincylindricalcoordinates 𝑟, 𝜃, 𝑧 tothecaseofsteadycompressibleflowinpolarcoordinates l

le= 0 andderiveastream

functionforthiscase.Fromquestion(2-1):

0 =𝜕𝜌𝜕𝑡+1𝑟𝜕𝜕𝑟

𝜌𝑣]𝑟 +𝜕𝜕𝑧

𝜌𝑣e +1𝑟𝜕𝜕𝜃

𝜌𝑣f

Forpolarcoordinate, 𝜕𝜕𝑧

= 0

𝑣e = 0

𝜕𝜕𝑡= 0

Continuityequationbecomes:

0 =1𝑟𝜕𝜕𝑟

𝜌𝑣]𝑟 +1𝑟𝜕𝜕𝜃

𝜌𝑣f

=𝜕𝜕𝑟

𝜌𝑣]𝑟 +𝜕𝜕𝜃

𝜌𝑣f

𝑣] =𝑑𝜓𝑟𝑑𝜃

; 𝑣f = −𝑑𝜓𝑑𝑟

𝜓 = massflowrate = 𝜌𝐴𝑉

𝑉 =𝜓𝜌𝐴

𝑣] =1𝜌𝜕𝜓𝑟𝜕𝜃

=1𝜌𝑟𝜕𝜓𝜕𝜃

𝑣f = −1𝜌𝜕𝜓𝜕𝑟

lawanarahjam

2-3Simplifytheequationofcontinuityincylindricalcoordinatestothecaseofsteadycompressible flow in axisymmetric coordinates l

lf= 0 and derive a stream

functionforthiscase.Continuityequation:

0 =𝜕𝜌𝜕𝑡+1𝑟𝜕𝜕𝑟

𝜌𝑣]𝑟 +𝜕𝜕𝑧

𝜌𝑣e +1𝑟𝜕𝜕𝜃

𝜌𝑣f

Foraxisymmetricflow, 𝜕𝜕𝜃

= 0

𝑣f = 0

Steadyflow, 𝜕𝜕𝑡= 0

Continuityequationbecomes:

0 =1𝑟𝜕𝜕𝑟

𝜌𝑣]𝑟 +𝜕𝜕𝑧

𝜌𝑣e

=1𝑟𝜕𝜕𝑟

𝜌𝑣]𝑟 +1𝑟𝜕𝜕𝑧

𝜌𝑣e𝑟

Assumethat,𝑟 = constant

𝜌𝑣]𝑟 =𝜕𝜓𝜕𝑧

𝑣] =1𝜌𝑟𝜕𝜓𝜕𝑧

𝜌𝑣e𝑟 =𝜕𝜓𝜕𝑟

𝑣e =1𝜌𝑟𝜕𝜓𝜕𝑟

𝑣e = −1𝜌𝑟𝜕𝜓𝜕𝑟

2-4For steady incompressible flowwith negligible viscosity, show that theNavier-Stokesrelation(Eq.2-30)reducestotheconditionthats

t+ u

v

w+ 𝑔ℎisconstant

along a streamline of the flow,where h denotes the height of the fluid particleabove a horizontal datum. The is the weaker form of the so-called Bernoullirelation.Navier-Stokesequationcanbewrittenas:

𝜌𝑑𝑉𝑑𝑡

= 𝜌𝑔 − ∇𝑝 + 𝜇∇w𝑉Forsteady,incompressibleflowwithzeroviscosity;𝜇 = 0Navier-Stokesequationbecomes;

𝜌𝑑𝑉𝑑𝑡

= 𝜌𝑔 − ∇𝑝When you have an inviscid flow (when viscosity is zero and there is no heatconduction),thentheNavier-StokesequationreducestotheEulerequation.

Eulerequationcanbewrittenas:

−∇𝑝 − 𝜌𝑔 = 𝜌𝑑𝑉𝑑𝑡

dividebydensity

−∇𝑝𝜌− 𝑔 =

𝑑𝑉𝑑𝑡

weputadotproductwithdisplacement𝑑𝑠alongthestreamline.Where𝑑𝑠 = 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧Eulerequationbecomes:(Gravitybecomesnegativebecauseitactsreversefromthepositivedirectionofz-axis)

−∇𝑝𝜌∙ 𝑑𝑠 − 𝑔 ∙ 𝑑𝑠 =

𝑑𝑉𝑑𝑡

∙ 𝑑𝑠

Solvethefirstterm:

−∇𝑝𝜌∙ 𝑑𝑠 = −

1𝜌∇𝑝 ∙ 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧

= −1𝜌∂p𝜕𝑥

+∂p𝜕𝑦

+∂p𝜕𝑧

∙ 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧

= −1𝜌∂p𝜕𝑥𝑑𝑥 +

∂p𝜕𝑦𝑑𝑦 +

∂p𝜕𝑧𝑑𝑧

= −1𝜌𝑑𝑝

solvethesecondterm:

−𝑔 ∙ 𝑑𝑠 = −𝑔 ∙ 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 = −𝑔𝑑𝑧Gravityonlyworkonz-axis(verticaldirection)

Solvethethirdterm:

𝑑𝑉𝑑𝑡

∙ 𝑑𝑠 = 𝑢𝜕𝑉𝜕𝑥

+ 𝑣𝜕𝑉𝜕𝑦

+ 𝑤𝜕𝑉𝜕𝑧

∙ 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧

= 𝑉 ∙ ∇ 𝑉 ∙ 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧

=12∇ 𝑉w ∙ 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧

=12𝜕𝑉w

𝜕𝑥+𝜕𝑉w

𝜕𝑦+𝜕𝑉w

𝜕𝑧∙ 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧

=12𝜕𝑉w

𝜕𝑥𝑑𝑥 +

𝜕𝑉w

𝜕𝑦𝑑𝑦 +

𝜕𝑉w

𝜕𝑧𝑑𝑧

=12𝑑 𝑉w

SubstitutingallthreetermsinEulerequation:

−∇𝑝𝜌∙ 𝑑𝑠 − 𝑔 ∙ 𝑑𝑠 =

𝑑𝑉𝑑𝑡

∙ 𝑑𝑠

∇𝑝𝜌∙ 𝑑𝑠 +

𝑑𝑉𝑑𝑡

∙ 𝑑𝑠 + 𝑔 ∙ 𝑑𝑠 = 0

1𝜌𝑑𝑝 +

12𝑑 𝑉w + 𝑔𝑑𝑧 = 0

integratingthisequation,weobtain:

1𝜌𝑑𝑝 +

12𝑑 𝑉w + 𝑔𝑑𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑝𝜌+𝑉w

2+ 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Bernoulliequation:𝑝𝜌𝑔

+𝑉w

2𝑔+ 𝑔 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

2-11Thedifferentialequationforirrotationalplanecompressiblegasflowis:

𝜕w𝜙𝜕𝑡w

+𝜕𝜕𝑡

𝑢w + 𝑣w + 𝑢w − 𝑎w𝜕w𝜙𝜕𝑥w

+ 𝑣w − 𝑎w𝜕w𝜙𝜕𝑦w

+ 2𝑢𝑣𝜕w𝜙𝜕𝑥𝜕𝑦

= 0

where𝜙isthevelocitypotentialand𝑎the(variable)speedofsoundinthegas.In the spirit of Sec.2-9-2, nondimensionalize this equation and define anyparameterswhichappear.Nondimensionalvariablesare:

𝜙∗ =𝜙𝑢𝐿 𝑢∗, 𝑣∗, 𝑎∗ =

𝑢, 𝑣, 𝑎𝑢

𝑥∗, 𝑦∗ =𝑥, 𝑦𝐿 𝑡∗ =

𝑢𝑡𝐿

Itwillproduces:

𝜕w𝜙∗

𝜕𝑡∗w+𝜕𝜕𝑡

𝑢∗w + 𝑣∗w + 𝑢∗w − 𝑎∗w𝜕w𝜙∗

𝜕𝑥∗w+ 𝑣∗w − 𝑎∗w

𝜕w𝜙∗

𝜕𝑦∗w+ 2𝑢∗𝑣∗

𝜕w𝜙∗

𝜕𝑥∗𝑦∗w= 0

Nodimensionlessparametersappear.Actually,correlating𝑎∗withtemperatureandvelocitywouldinfactleadtoMachnumberandspecificheatratio.

𝑎 = speedofsound

2-13The equation of motion for free convection near a hot vertical plate forincompressibleflowwithconstantpropertiesare;

𝜕𝑢𝜕𝑥

+𝜕𝑣𝜕𝑦

= 0

𝑢𝜕𝑢𝜕𝑥

+ 𝑣𝜕𝑢𝜕𝑦

= 𝑔𝛽 𝑇 − 𝑇 + 𝑣𝜕w𝑢𝜕𝑥w

+𝜕w𝑢𝜕𝑦w

𝜌𝑐 𝑢𝜕𝑇𝜕𝑥

+ 𝑣𝜕𝑇𝜕𝑦

= 𝑘𝜕w𝑇𝜕𝑥w

+𝜕w𝑇𝜕𝑦w

Introducethedimensionlessvariables

𝑢∗ =𝑢𝐿𝜐 𝑣∗ =

𝑣𝐿𝜐 𝑥∗ =

𝑥𝐿 𝑦∗ =

𝑦𝐿 𝑇∗ =

𝑇 − 𝑇𝑇 − 𝑇

where𝐿is the lengthof theplate,𝜐iskinematicviscosity.Use thesevariable tonondimensionalize the free convection equations and define any parameterswhicharise.

Equation#1

𝜕𝑢𝜕𝑥

+𝜕𝑣𝜕𝑦 = 0

0 =𝜕 𝑢

∗𝜐𝐿

𝜕 𝑥∗𝐿+𝜕 𝑣

∗𝜐𝐿

𝜕 𝑦∗𝐿=𝜐𝐿w𝜕𝑢∗

𝜕𝑥∗+𝜐𝐿w𝜕𝑣∗

𝜕𝑦∗

0 =𝜕𝑢∗

𝜕𝑥∗+𝜕𝑣∗

𝜕𝑦∗

withnoparameterappearing.

Equation#2

𝑢𝜕𝑢𝜕𝑥

+ 𝑣𝜕𝑢𝜕𝑦

= 𝑔𝛽 𝑇 − 𝑇 + 𝑣𝜕w𝑢𝜕𝑥w

+𝜕w𝑢𝜕𝑦w

LHS:

𝑢 =𝑢∗𝜐𝐿 𝑣 =

𝑣∗𝜐𝐿

𝜕𝑢𝜕𝑥 =

𝜕 𝑢∗𝜐𝐿

𝜕 𝑥∗𝐿=𝜐𝐿w𝜕𝑢∗

𝜕𝑥∗

𝜕𝑢𝜕𝑦 =

𝜐𝐿w𝜕𝑢∗

𝜕𝑦∗

𝜐𝐿𝑢∗

𝜐𝐿w𝜕𝑢∗

𝜕𝑥∗+𝜐𝐿𝑣∗

𝜐𝐿w𝜕𝑢∗

𝜕𝑦∗=𝜐w

𝐿𝑢∗𝜕𝑢∗

𝜕𝑥∗+ 𝑣∗

𝜕𝑢∗

𝜕𝑦∗

RHS:

𝑔𝛽 𝑇 − 𝑇 + 𝜐𝜕w𝑢𝜕𝑥w

+𝜕w𝑢𝜕𝑦w

𝑇∗ =𝑇 − 𝑇𝑇 − 𝑇

𝜕w𝑢𝜕𝑥w

=𝜕𝜕𝑥

𝜐𝐿w𝜕𝑢∗

𝜕𝑥∗ Note:

=𝜕𝜕𝑥∗

𝜕𝑥∗

𝜕𝑥𝜐𝐿w𝜕𝑢∗

𝜕𝑥∗ 𝑥∗ =

𝑥𝐿

=𝜕𝜕𝑥∗

𝜕𝑢∗

𝜕𝑥∗𝜕𝑥∗

𝜕𝑥𝜐𝐿w

𝜕𝜕𝑥

𝑥∗ =𝜕𝜕𝑥

𝑥𝐿

=𝜕𝑥∗

𝜕𝑥𝜐𝐿w

𝜕𝜕𝑥∗

𝜕𝑢∗

𝜕𝑥∗ =

1𝐿𝜕𝑥𝜕𝑥

=1𝐿𝜐𝐿w𝜕w𝑢∗

𝜕𝑥∗w =

1𝐿

=𝜐𝐿𝜕w𝑢∗

𝜕𝑥∗w

𝜕w𝑢𝜕𝑦w

=𝜐𝐿𝜕w𝑢∗

𝜕𝑦∗w

𝑔𝛽 𝑇 − 𝑇 + 𝜐𝜕w𝑢𝜕𝑥w

+𝜕w𝑢𝜕𝑦w

= 𝑔𝛽𝑇∗ 𝑇 − 𝑇 + 𝜐𝜐𝐿𝜕w𝑢∗

𝜕𝑥∗w+𝜐𝐿𝜕w𝑢∗

𝜕𝑦∗w

= 𝑔𝛽𝑇∗ 𝑇 − 𝑇 +𝜐w

𝐿𝜕w𝑢∗

𝜕𝑥∗w+𝜕w𝑢∗

𝜕𝑦∗w

Finalexpression:

𝜐w

𝐿𝑢∗𝜕𝑢∗

𝜕𝑥∗+ 𝑣∗

𝜕𝑢∗

𝜕𝑦∗ = 𝑔𝛽𝑇∗ 𝑇 − 𝑇 +

𝜐w

𝐿𝜕w𝑢∗

𝜕𝑥∗w+𝜕w𝑢∗

𝜕𝑦∗w

Dividealltermswithv

𝑢∗𝜕𝑢∗

𝜕𝑥∗+ 𝑣∗

𝜕𝑢∗

𝜕𝑦∗ =

𝐿

𝜐w𝑔𝛽 𝑇 − 𝑇 𝑇∗ +

𝐿

𝜐w𝜐w

𝐿𝜕w𝑢∗

𝜕𝑥∗w+𝜕w𝑢∗

𝜕𝑦∗w

𝑢∗𝜕𝑢∗

𝜕𝑥∗+ 𝑣∗

𝜕𝑢∗

𝜕𝑦∗ =

𝐿

𝜐w𝑔𝛽 𝑇 − 𝑇 𝑇∗ +

𝜕w𝑢∗

𝜕𝑥∗w+𝜕w𝑢∗

𝜕𝑦∗w

Simplificationcanbemadeas:

𝐿

𝜐w𝑔𝛽 𝑇 − 𝑇 𝑇∗ = 𝐺𝑟𝐿𝑇∗

which;

𝐺𝑟 =𝐿3

𝜐2𝑔𝛽 𝑇0 − 𝑇1

𝐺𝑟isknownasGrashofnumber.

Equation#3

𝜌𝑐 𝑢𝜕𝑇𝜕𝑥

+ 𝑣𝜕𝑇𝜕𝑦

= 𝑘𝜕w𝑇𝜕𝑥w

+𝜕w𝑇𝜕𝑦w

(1)

𝑢∗ =𝑢𝐿𝜐→ 𝑢 =

𝑢∗𝜐𝐿 𝑥∗ =

𝑥𝐿→ 𝑥 = 𝑥∗𝐿

𝑣∗ =𝑣𝐿𝜐→ 𝑣 =

𝑣∗𝜐𝐿 𝑦∗ =

𝑦𝐿→ 𝑦 = 𝑦∗𝐿

𝑇∗ =𝑇 − 𝑇𝑇 − 𝑇

→ 𝑇 = 𝑇 + 𝑇∗ 𝑇 − 𝑇

𝜕𝑇𝜕𝑥 =

𝜕 𝑇 + 𝑇∗ 𝑇 − 𝑇𝜕 𝑥∗𝐿

=1𝐿𝜕𝑇𝜕𝑥∗

+𝑇 − 𝑇𝐿

𝜕𝑇∗

𝜕𝑥∗

𝜕𝑇𝜕𝑥 =

𝑇 − 𝑇𝐿

𝜕𝑇∗

𝜕𝑥∗

𝜕𝑇𝜕𝑦 =

𝑇 − 𝑇𝐿

𝜕𝑇∗

𝜕𝑦∗

𝜕w𝑇𝜕𝑥w

=𝜕𝜕𝑥

𝜕𝑇𝜕𝑥

=𝜕𝜕𝑥

𝑇 − 𝑇𝐿

𝜕𝑇∗

𝜕𝑥∗

=𝑇 − 𝑇𝐿

𝜕𝜕𝑥∗

𝜕𝑥∗

𝜕𝑥𝜕𝑇∗

𝜕𝑥∗

=𝑇 − 𝑇𝐿

𝜕𝑥∗

𝜕𝑥𝜕𝜕𝑥∗

𝜕𝑇∗

𝜕𝑥∗

=𝑇 − 𝑇𝐿

1𝐿𝜕w𝑇∗

𝜕𝑥∗w

𝜕w𝑇𝜕𝑥w

=𝑇 − 𝑇𝐿w

𝜕w𝑇∗

𝜕𝑥∗w

𝜕w𝑇𝜕𝑦w

=𝑇 − 𝑇𝐿w

𝜕w𝑇∗

𝜕𝑦∗w

SubstituteinEq.(1):

𝜌𝑐 𝑢𝜕𝑇𝜕𝑥

+ 𝑣𝜕𝑇𝜕𝑦

= 𝑘𝜕w𝑇𝜕𝑥w

+𝜕w𝑇𝜕𝑦w

LHS = 𝜌𝑐𝑢∗𝜐𝐿

𝑇 − 𝑇𝐿

𝜕𝑇∗

𝜕𝑥∗+𝑣∗𝜐𝐿

𝑇 − 𝑇𝐿

𝜕𝑇∗

𝜕𝑦∗

= 𝜌𝑐𝜐 𝑇 − 𝑇

𝐿w𝑢∗𝜕𝑇∗

𝜕𝑥∗+ 𝑣∗

𝜕𝑇∗

𝜕𝑦∗

RHS = 𝑘𝑇 − 𝑇𝐿w

𝜕w𝑇∗

𝜕𝑥∗w+𝜕w𝑇∗

𝜕𝑦∗w

=𝑇 − 𝑇𝐿w

𝑘𝜕w𝑇∗

𝜕𝑥∗w+𝜕w𝑇∗

𝜕𝑦∗w

LHS=RHS

𝜌𝑐𝜐 𝑇 − 𝑇

𝐿w𝑢∗𝜕𝑇∗

𝜕𝑥∗+ 𝑣∗

𝜕𝑇∗

𝜕𝑦∗ =

𝑇 − 𝑇𝐿w

𝑘𝜕w𝑇∗

𝜕𝑥∗w+𝜕w𝑇∗

𝜕𝑦∗w

𝑢∗𝜕𝑇∗

𝜕𝑥∗+ 𝑣∗

𝜕𝑇∗

𝜕𝑦∗ =

𝑘𝜌𝑐𝜐

𝜕w𝑇∗

𝜕𝑥∗w+𝜕w𝑇∗

𝜕𝑦∗w

𝑘𝜌𝑐𝜐

=1𝑃𝑟

(Prandtlnumber)

2-14Laminar flow in the entrance to a pipe, as shown below, the entrance flow isuniform,𝑢 = 𝑈andtheflowdownstreamisparabolicinprofile,

𝑢 𝑟 = 𝐶 𝑟w − 𝑟w Usingtheintegralrelation,sec.2-13,showthattheviscousdragexertedonthepipewallsbetween0andxisgivenby:

𝐷𝑟𝑎𝑔 = 𝜋𝑟w 𝑝 − 𝑝£ −13𝜌𝑈w

UseReynoldstransporttheorem:

𝑢 𝑟 = 𝐶 𝑟Nw − 𝑟w

NeedtodeterminethevalueofconstantC

massflowratein=massflowrateout

𝜌N𝐴N𝑢N = 𝜌𝑢𝑑𝐴 Circulararea:

= 𝜌N 𝐶 𝑟𝑜2 − 𝑟2 2𝜋𝑟𝑑𝑟 𝐴 = 𝜋𝑟w

Assumeitisincompressible, 𝜌N = constant 𝑑𝐴𝑑𝑟

= 2𝜋𝑟

𝐴N𝑢N = 𝐶 𝑟𝑜2 − 𝑟2 2𝜋𝑟𝑑𝑟 𝑑𝐴 = 2𝜋𝑟𝑑𝑟

= 2𝜋𝐶 𝑟𝑜2𝑟 − 𝑟3 𝑑𝑟¤

= 2𝜋𝐶 𝑅w𝑅w

2−𝑅¦

4

= 2𝜋𝐶𝑅¦

2−𝑅¦

4

= 2𝜋𝐶𝑅¦

4

𝜋𝑅w 𝑢N = 2𝜋𝐶𝑅¦

4

= 𝜋𝐶𝑅¦

2

𝐶 =𝜋𝑅w𝑢N𝜋

2𝑅¦

𝐶 =2𝑢N𝑅w

𝑢 𝑟 = 𝐶 𝑟Nw − 𝑟w

𝑢 𝑟 =2𝑢N𝑅w

𝑟Nw − 𝑟w

FD

n

𝑝£𝐴£𝑝N𝐴N

Viscousdragordragforce,FD:

𝐹 =𝜕𝜕𝑡

𝑚𝑣 =𝑑𝑑𝑡

𝑣𝜌𝑑∀ + 𝑣𝜌𝑣𝑑𝐴©ª©«

0 𝑥

𝐹 =𝑑𝑑𝑡

𝑣𝜌𝑑∀©«

+ 𝑣𝜌𝑣L𝑑𝐴©ª

steadyflow=0

𝑢 𝑟 =2𝑢N𝑅w

𝑟Nw − 𝑟w =2𝑢𝑅w

𝑅w − 𝑟w

nu(r)u0

𝐹 = 𝑣𝜌𝑣𝑑𝐴©ª

−𝐹¬ + 𝑝N𝐴N − 𝑝£𝐴£ = 𝑢𝜌 −𝑢 𝐴N + 𝑢 𝑟 𝜌𝑢 𝑟 𝑑𝐴¤

−𝐹¬ + 𝜋𝑅w 𝑝N − 𝑝£ = −𝑢w𝜌𝜋𝑅w + 𝑢 𝑟w𝜌2𝜋𝑟𝑑𝑟

¤

𝑢 𝑟 w𝜌2𝜋𝑟𝑑𝑟¤

= 𝜌

𝜕𝑢𝑅w

𝑅w − 𝑟ww

2𝜋𝑟𝑑𝑟

= 𝜌4𝑢w

𝑅¦𝑅¦ + 𝑟¦ − 2𝑅w𝑟w 2𝜋𝑟𝑑𝑟

=4𝑢w

𝑅¦𝜌2𝜋 𝑅¦𝑟 + 𝑟 − 2𝑅w𝑟 𝑑𝑟

𝑢 𝑟 w𝜌2𝜋𝑟𝑑𝑟¤

=

4𝑢w

𝑅¦𝜌2𝜋

𝑅¦𝑟w

2+𝑟®

6− 2𝑅w

𝑟¦

4

¤

=4𝑢w

𝑅¦𝜌2𝜋

𝑅®

2+𝑅®

6− 2

𝑅®

4

=4𝑢w

𝑅¦𝜌2𝜋

𝑅®

6

𝑢 𝑟 w𝜌2𝜋𝑟𝑑𝑟¤

= 𝜌𝜋𝑢w

43𝑅w

−𝐹¬ + 𝜋𝑅w 𝑝N − 𝑝£ = −𝑢w𝜌𝜋𝑅w + 𝑢 𝑟w𝜌2𝜋𝑟𝑑𝑟

¤

= −𝜌𝜋𝑢w𝑅w + 𝜌𝜋𝑢w43𝑅w

=13𝜌𝜋𝑢w𝑅w

𝐹¬ = 𝜋𝑅w 𝑝N − 𝑝£ −13𝜌𝜋𝑢w𝑅w

𝐹¬ = 𝜋𝑅w 𝑝N − 𝑝£ −13𝜌𝑢w

2-18Flowthroughawell-designedcontractionornozzleisnearlyfrictionless.Supposethatwaterat20°Cflowsthroughahorizontalnozzleataweightflowof50N/s.Ifentranceandexitdiametersare8cmand3cm,respectively,andtheexitpressureis1atm,estimatetheentrancepressurefromBernoulli’sequation.

𝜌µ¶PT] = 998 kg m

50𝑁𝑠ialahweightflowrate

𝑁𝑠= 𝑚𝑔

50 = 𝑚𝑔 = 𝜌𝐴𝑉𝑔 = 998

𝜋4

0.08 w 𝑣 9.81

𝑣 = 1.02m/s

𝐴𝑉 = 𝐴w𝑉w

𝑉w =𝐴𝐴w𝑉 =

𝜋4 0.08

w

𝜋4 0.03

w1.02

𝑉w = 7.25m/s

𝑝𝜌𝑔

+𝑉12

2𝑔+ 𝑧 =

𝑝w𝜌𝑔

+𝑉22

2𝑔+ 𝑧w

𝑝𝜌𝑔 =

101350𝜌𝑔

+7.25 22𝑔

−1.02 22𝑔

= 12.978

𝑝 = 12.978 𝜌𝑔 = 127.06kPa