Differential Equations of Fluid Flow

▶

C H A P T E R

9

Differential Equations

which states that

net rate of mass flux out of control volume 8<

:

9=

;‡

rate of accumulation of mass within control volume 8<

:

9=

;ˆ0

The massfluxr(vn) at each face of the control volume is illustrated in Figure 9.1. The mass within the control volume isrDxDyDz, and thus the time rate of change of mass within the control volume is

@

@t…rDxDyDz†

The student is reminded that the density in general can vary from point to point—that is, rˆr(x, y, z, t).

The net massflux out of the control volume in the xdirection is …rvxjx‡Dx rvxjx†DyDz

in theydirection

…rvyjy‡Dy rvyjy†DxDz and in thezdirection

…rvzjz‡Dz rvzjz†DxDy

The total net massflux is the sum of the above three terms. Substituting into equation (4-1) yields

…rvxj_x‡Dx rvxj_x†DyDz‡ …rvyj_y‡Dy rvyj_y†DxDz

‡ …rvzjz‡Dz rvzjz†DxDy‡@

@t…rDxDyDz† ˆ0 The volume does not change with time, so we may divide the above equation byDxDyDz. In the limit asDx,Dy, andDzapproach zero, we obtain

@

@x…rvx† ‡ @

@y…rvy† ‡ @

@z…rvz† ‡@r

@t ˆ0 (9-1)

Thefirst three terms comprise the divergence of the vectorrv. The divergence of a vector is the dot product withÑ:

divAÑ

^A

The student may verify that thefirst three terms in equation (9-1) may be written asÑrv and thus a more compact statement of the continuity equation becomes

Ñ

^rv‡@r_@t ^{ˆ0 (9-2)}

The continuity equation above applies to unsteady, three-dimensionalflow. It is apparent that whenflow is incompressible, this equation reduces to

Ñ

^{vˆ0 (9-3)}

whether theflow is unsteady or not.

Equation (9-2) may be arranged in a slightly different form to illustrate the use of the substantial derivative. Carrying out the differentiation indicated in (9-1), we have

@r

@t‡vx@r

@x‡vy@r

@y‡vz@r

@z‡r @vx

@x ‡@vy

@y‡@vz

@z

ˆ0

Thefirst four terms of the above equation comprise the substantial derivative of the density, symbolized asDr/Dt, where

D Dtˆ @

@t‡vx @

@x‡vy @

@y‡vz @

@z (9-4)

in Cartesian coordinates. The continuity equation may, thus be written as Dr

Dt‡rÑ

^{vˆ0 (9-5)}

When considering the total differential of a quantity, three different approaches may be taken. If, for instance, we wish to evaluate the change in atmospheric pressure,P, the total differential written in rectangular coordinates is

dPˆ@P

@tdt‡@P

@xdx‡@P

@ydy‡@P

@zdz

wheredx,dy, anddzare arbitrary displacements in thex,y, andzdirections. The rate of pressure change is obtained by dividing through by dt, giving

dP dt ˆ@P

@t ‡dx dt

@P

@x‡dy dt

@P

@y‡dz dt

@P

@z (9-6)

As afirst approach, the instrument to measure pressure is located in a weather station, which is, of course,fixed on Earth’s surface. Thus, the coefficientsdx/dt, dy/dt, dz/dtare all zero, and for afixed point of observation the total derivative,dP/dt,is equal to the local derivative with respect to time@P/@t.

A second approach involves the pressure-measuring instrument housed in an aircraft, which, at the pilot’s discretion, can be made to climb or descend, orfly in any chosenx, y, z direction. In this case, the coefficientsdx/dt, dy/dt, dz/dtare thex, y,andzvelocities of the aircraft, and they are arbitrarily chosen, bearing only coincidental relationship to the air currents.

The third situation is one in which the pressure indicator is in a balloon that rises, falls, and drifts as influenced by the flow of air in which it is suspended. Here the coefficients dx/dt, dy/dt, dz/dtare those of theflowand they may be designatedvx,vy, andvz, respectively.

This latter situation corresponds to the substantial derivative, and the terms may be grouped as designated below:

dP dt ˆDP

Dt ˆ @P

@t

|{z}

local rate of change of

pressure

‡vx@P

@t ‡vy@P

@y‡vz@P

@z

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

rate of change of pressure due to motion

(9-7)

The derivativeD/Dtmay be interpreted as the time rate of change of afluid orflow variable along the path of afluid element. The substantial derivative will be applied to both scalar and vector variables in subsequent sections.

9.1 The Differential Continuity Equation ◀ 109

▶ 9.2

NAVIER–STOKES EQUATIONS

The Navier–Stokes equations are the differential form of Newton’s second law of motion.

Consider the differential control volume illustrated in Figure 9.1.

The basic tool we shall use in developing the Navier–Stokes equations is Newton’s second law of motion for an arbitrary control volume as given in Chapter 5

SFˆ Z Z

rv v

^n†dA‡_@^@_t^{Z Z Z rvdV (5-4)}

which states that

sum of the external forces acting on the

c:v: 8<

:

9=

;ˆ net rate of linear momentum efflux

‡

time rate of change of linear momentum

within the c:v: 8<

:

9=

; As the mathematical expression for each of the above terms is rather lengthy, each will be evaluated separately and then substituted into equation (5-4).

The development may be further simplified by recalling that we have, in the prior case, divided by the volume of the control volume and taken the limit as the dimensions approach zero. Equation (5-4) can also be written

Dx;Dlimy;Dz®0

∑F

DxDyDzˆ lim

Dx;Dy;Dz®0

RRrv…v

^n†dA

DxDyDz ‡ lim

Dx;Dy;Dz®0

@=@tRRR rvdV DxDyDz

°1 °2 °3

(9-8) 1. Sum of the external forces. The forces acting on the control volume are those due to the normal stress and to the shear stress, and body forces such as that due to gravity.

Figure 9.2 illustrates the various forces acting on the control volume. Summing the forces in thexdirection, we obtain

∑F_xˆ …sxxj_x‡Dx sxxj_x†DyDz‡ …tyxj_y‡Dy tyxjy†DxDz

‡…tzxjz‡Dz tzxjz†DxDy‡g_xrDxDyDz

whereg_xis the component of the gravitational acceleration in thexdirection. In the limit as the dimensions of the element approach zero, this becomes

Dx;Dlimy;Dz®0

∑F_x

DxDyDzˆ@sxx

@x ‡@tyx

@y ‡@tzx

@z ‡rg_x (9-9) Similar expressions are obtained for the force summations in theyandzdirections:

Dx;Dlimy;Dz®0

∑Fy

DxDyDzˆ@txy

@x ‡@syy

@y ‡@tzy

@z ‡rg_y (9-10)

Dx;Dlimy;Dz®0

∑Fz

DxDyDzˆ@txz

@x ‡@tyz

@y ‡@szz

@z ‡rg_z (9-11) 2. Net momentumflux through the control volume. The net momentumflux through

the control volume illustrated in Figure 9.3 is

Dx;Dlimy;Dz®0

RRrv…v

^n†dA

DxDyDz ˆ lim

Dx;Dy;Dz®0

…rvvxjx‡Dx rvvxjx†DyDz DxDyDz

‡…rvvyjy‡Dy rvvyjy†DxDz DxDyDz

‡…rvvzj_z‡Dz rvvzj_z†DxDy DxDyDz

ˆ @

@x…rvvx† ‡ @

@y…rvvy† ‡ @

@z…rvvz†

(9-12) y

x z

y x

z Figure 9.2 Forces acting on a differential control volume.

y

z

x

Figure 9.3 Momentumflux through a differential control volume.

9.2 Navier–Stokes Equations ◀ 111

Performing the indicated differentiation of the right-hand side of equation (9-12) yields

Dx;Dlimy;Dz®0

RRrv…v

^n†dA

DxDyDz ˆv @

@x…rvx† ‡ @

@y…rvy†@

@z…rvz†

‡r vx@v

@x‡vy@v

@y‡vz @v

@z

The above term may be simplified with the aid of the continuity equation:

@r

@t ‡ @

@x…rvx† ‡ @

@y…rvy† ‡@

@z…rvz† ˆ0 (9-1) which, upon substitution, yields

Dx;Dlimy;Dz®0

RRrv…v

^n†dA

DxDyDz ˆ v@r

@t‡r vx@v

@x‡vy@v

@y‡vz@v

@z

(9-13)

3. Time rate of change of momentum within the control volume. The time rate of change of momentum within the control volume may be evaluated directly:

Dx;Dlimy;Dz®0

@=@tRRR vrdV

DxDyDz ˆ…@=@t†rvDxDyDz DxDyDz ˆ @

@trvˆr@v

@t ‡v@r

@t (9-14) We have now evaluated all terms in equation (9-8):

°1

Dx;Dylim;Dz®0

∑F DxDyDzˆ

@sxx

@x ‡@tyx

@y ‡@tzx

@z ‡rg_x

ex

@txy

@x ‡@syy

@y ‡@tzy

@z ‡rg_y

ey

@txz

@x ‡@tyz

@y ‡@szz

@z ‡rg_z

e_z 8>

>>

>>

>>

><

>>

>>

>>

>>

:

9>

>>

>>

>>

>=

>>

>>

>>

>>

;

…9-9† …9-10† …9-11†

°2 lim

Dx;Dy;Dz®0

RRrv…v

^n†dA

DxDyDz ˆ v@r

@t ‡r vx @v

@t‡vy @v

@y‡vz @v

@z

(9-13)

°3 lim

Dx;Dy;Dz®0

@=@tRRR rvdV DxDyDz ˆr@v

@t‡v@r

@t (9-14)

It can be seen that the forces are expressed in components, whereas the rate-of-change-of- momentum terms are expressed as vectors. When the momentum terms are expressed as components, we obtain three differential equations that are the statements of Newton’s second law in thex,y,andzdirections:

r @vx

@t ‡vx@vx

@x ‡vy@vx

@y‡vz@vx

@z

ˆrg_x‡@sxx

@x ‡@tyx

@y ‡@tzx

@z (9-15a)

r @vy

@t ‡vx@vy

@x ‡vy @vy

@x‡vz@vy

@z

ˆrg_y‡@txy

@x ‡@syy

@y ‡@tzy

@z (9-15b)

r @vz

@t ‡vx@vz

@x ‡vy@vz

@x‡vz@vz

@z

ˆrg_z‡@txz

@x ‡@tyz

@y ‡@szz

@z (9-15c)

It will be noted that in equations (9-15) above, the terms on the left-hand side represent the time rate of change of momentum, and the terms on the right-hand side represent the forces. Focusing our attention on the left-hand terms in equation (9-15a), we see that

@vx

@t

|fflfflfflfflfflffl{zfflfflfflfflfflffl}

local of change

ofvx

‡ vx@vx

@x ‡vy@vx

@y ‡vz@vx

@x

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

rate of change in vxdue to motion

ˆ @

@t‡vx @

@x‡vy @

@y‡vz @

@z

vx

Thefirst term,@vx/@t, involves the time rate of change of v_xat a point and is called the local acceleration.The remaining terms involve the velocity change from point to point— that is, the convective acceleration. The sum of these two bracketed terms is the total acceleration. The reader may verify that the terms on the left-hand side of equations (9-15) are all of the form

@

@t‡vx @

@x‡vy @

@y‡vz @

@z

vi

where viˆvx,vy, orvz. The above term is the substantial derivative ofvi. When the substantial derivative notation is used, equations (9-15) become

rDvx

Dt ˆrg_x‡@sxx

@x ‡@tyx

@y ‡@tzx

@z (9-16a)

rDvy

Dt ˆrg_y‡@txy

@x ‡@syy

@y ‡@tzy

@z (9-16b)

and

rDvz

Dt ˆrg_z‡@txz

@x ‡@tyz

@y ‡@szz

@z (9-16c)

Equations (9-16) are valid for any type offluid, regardless of the nature of the stress rate-of-strain relation. If Stokes’s viscosity relations, equations (7-13) and (7-14) are used for the stress components, equations (9-16) become

rDvx

Dt ˆrg_x @P

@x

@

@x 2 3mÑ

^v

‡Ñ

^m@v_@x^‡Ñ

^{mÑvx† (9-17a)}

rDvy

Dt ˆrg_y @P

@x

@

@y 2 3mÑ

^v

‡Ñ

^m@_@^v_y^‡Ñ

^{mÑvy† (9-17b)}

and

rDvz

Dt ˆrg_z @P

@z

@

@z 2 3mÑ

^v

‡Ñ

^m@_@^v_z^‡Ñ

^{mÑvz† (9-17c)}

9.2 Navier–Stokes Equations ◀ 113

The above equations are called the Navier–Stokes¹ equations and are the differential expressions of Newton’s second law of motion for a Newtonianfluid. As no assumptions relating to the compressibility of thefluid have been made, these equations are valid for both compressible and incompressibleflows. In our study of momentum transfer we shall restrict our attention to incompressible flow with constant viscosity. In an incompressible flow, Ñ

^{v ˆ}0. Equations (9-17) thus become

rDvx

Dt ˆrg_x @P

@x‡m @²vx

@x² ‡@²vx

@y² ‡@²vx

@z²

(9-18a)

rDvy

Dt ˆrg_y @P

@y‡m @²vy

@x² ‡@²vy

@y² ‡@²vy

@z²

(9-18b)

rDvz

Dt ˆrg_z @P

@z‡m @²vz

@x² ‡@²vz

@y² ‡@²vz

@z²

(9-18c) These equations may be expressed in a more compact form in the single vector equation:

rDv

Dt ˆrg ÑP‡mѲv (9-19)

The above equation is the Navier–Stokes equation for an incompressibleflow. The Navier–Stokes equations are written in Cartesian, cylindrical, and spherical coordinate forms in Appendix E. As the development has been lengthy, let us review the assumptions and, therefore, the limitations of equation (9-19). The assumptions are

1. incompressibleflow 2. constant viscosity 3. laminarflow²

All of the above assumptions are associated with the use of the Stokes viscosity relation. If the flow is inviscid (mˆ0), the Navier–Stokes equation becomes

rDv

Dt ˆrg ÑP (9-20)

which is known as Euler’s equation. Euler’s equation has only one limitation, that being inviscidflow.

Example 1

Equation (9-19) may be applied to numerous flow systems to provide information regarding velocity variation, pressure gradients, and other information of the type achieved in Chapter 8. Many situations are of sufficient complexity to make the solution extremely difficult and are beyond the scope of this text. A situation for which a solution can be obtained is illustrated in Figure 9.4.

1L. M. H. Navier, Mémoire sur les Lois du Mouvements des Fluides,Mem. de l’Acad. d. Sci.,6, 398 (1822); C. G.

Stokes, On the Theories of the Internal Friction of Fluids in Motion,Trans. Cambridge Phys. Soc.,8(1845).

2Strictly speaking, equation (9-19) is valid for turbulentflow, as the turbulent stress is included in the momentum flux term. This will be illustrated in Chapter 12.

Figure 9.4 shows the situation of an incompressiblefluid confined between two parallel, vertical surfaces. One surface, shown to the left, is stationary, whereas the other is moving upward at a constant velocityv0. If we consider thefluid Newtonian and theflow laminar, the governing equation of motion is the Navier–Stokes equation in the form given by equation (9-19).

The reduction of each term in the vector equation into its applicable form is shown below:

rDv Dt ˆr

(@v

@t‡vx@v

@t‡vy@v

@t‡vz@v

@t )

ˆ0 rgˆ rgey

ÑPˆdP dye_y wheredP/dyis constant, and

mѲvˆmd²vy

dx²ey

The resulting equation to be solved is

0ˆ rg dP

dy‡md²vy

dx² This differential equation is separable. Thefirst integration yields

dvy

dx ‡x

m rg dP dy

ˆC1

Integrating once more, we obtain

vy‡x²

2m rg dP

dy

ˆC1x‡C2

The integration constants may be evaluated, using the boundary conditions thatvyˆ0 atxˆ0, andvyˆv0atxˆL. The constants thus become

C1ˆv0

L ‡ L

2m rg dP

dy

and C2ˆ0 The velocity profile may now be expressed as

vyˆ 1 2m

(

rg dP dy

)

fLx x²g

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

°1

‡v0

x

|{z}L

°2

(9-21) y

x L Fluid

Figure 9.4 Fluid between two vertical plates with the one on the left

stationary and the other on the right moving vertically upward with velocityu0. 9.2 Navier–Stokes Equations ◀ 115

It is interesting to note, in equation (9-21), the effect of the terms labeled°1 and°2 , which are added. Thefirst term is the equation for a symmetric parabola, the second for a straight line. Equation (9-21) is valid whetherv0is upward, downward, or zero. In each case, the terms may be added to yield the complete velocity profile. These results are indicated in Figure 9.5. The resulting velocity profile obtained by superposing the two parts is shown in each case.

Euler’s equation may also be solved to determine velocity profiles, as will be shown in Chapter 10. The vector properties of Euler’s equation are illustrated by the example below, in which the form of the velocity profile is given.

Example 2

A rotating shaft, as illustrated in Figure 9.6, causes thefluid to move in circular streamlines with a velocity that is inversely proportional to the distance from the shaft. Find the shape of the free surface if thefluid can be considered inviscid.

As the pressure along the free surface will be constant, we may observe that the free surface is perpendicular to the pressure gradient. Determination of the pressure gradient, therefore, will enable us to evaluate the slope of the free surface.

+

= = =

+ +

Figure 9.5 Velocity profiles for one surface moving upward, downward, or stationary.

Streamlines

Figure 9.6 Rotating shaft in afluid.

Rearranging equation (9-20), we have

ÑPˆrg rDv

Dt (9-20)

The velocityvˆAe_q/r, whereAis a constant, when using the coordinate system shown in Figure 9.7. Assuming that there is no slip between thefluid and the shaft at the surface of the shaft, we have

v… † ˆR wRˆA R and thusAˆwR²and

vˆwR² r e_q

The substantial derivativeDv/Dtmay be evaluated by taking the total derivative dv

dt ˆ wR²

r² e_qr_‡wR² r

de_q dt wherede_q=dtˆ qe_ r:The total derivative becomes

dv

dtˆ wR²

r² re_ _q wR² r q_er

Now thefluid velocity in therdirection is zero, andq_ for thefluid isv/r, so dv

dt _fluidˆDv

Dtˆ wR²

r² verˆ w²R⁴ r³ er

This result could have been obtained in a more direct manner by observing thatDv/Dtis the localfluid acceleration, which for this case is v²e_r/r. The pressure gradient becomes

ÑPˆ rgez‡rw²R⁴er

r³

From Figure 9.8, it can be seen that the free surface makes an anglebwith theraxis so that tanbˆrw²R⁴

r³rg

ˆw²R⁴ gr³ ez

z

r

er

Figure 9.7 Cylindrical coordinate system for rotating shaft andfluid.

9.2 Navier–Stokes Equations ◀ 117

▶ 9.3

BERNOULLI’S EQUATION

Euler’s equation may be integrated directly for a particular case,flow along a streamline. In integrating Euler’s equation, the use of streamline coordinates is extremely helpful.

Streamline coordinates s and n are illustrated in Figure 9.9. The s direction is parallel to the streamline and thendirection is perpendicular to the streamline, directed away from the instantaneous center of curvature. Theflow andfluid properties are functions of position and time. Thus,vˆv(s, n, t),andPˆP(s, n, t).The substantial derivatives of the velocity and pressure gradients in equation (9-20) must be expressed in terms of streamline coordinates so that equation (9-20) may be integrated.

Following the form used in equations (9-6) to obtain the substantial derivative, we have dv

dt ˆdv dt‡s_@v

@s‡n_@v

@n

As the velocity of thefluid element has componentss_ˆv;n^? ˆ0;the substantial derivative of the velocity in streamline coordinates is

Dv Dtˆ@v

@t ‡v@v

@s (9-22)

z

r Free surface

er

r³

Figure 9.8 Free-surface slope.

y

z

x

en v es

Streamlines

Figure 9.9 Streamline coordinates.

The pressure gradient in streamline coordinates may be written as ÑPˆ@P

@ses‡@P

@nen (9-23)

Taking the dot product of equation (9-20) withe_sds, and using equations (9-22) and (9-23), we obtain

r @v

@t

^esds‡v@v_@s

^esds

ˆrg

^{esds @}_@^P_s^es‡@_@^P_n^en

^esds

or, as@v=@s

^esˆ@=@s…v

^{es† ˆ@v=@s;we have}

r @v

@t

^esds‡_@^@_s ^v₂^{2 ds}

ˆrg

^{esds @}_@^P_n^{ds (9-24)}

Selectinggto act in the ydirection, we havegesdsˆ g dy. Forsteady incompressible flow, equation (9-24) may be integrated to yield

v²

2 ‡gy‡P

rˆconstant (9-25)

which is known as Bernoulli’s equation. The limitations are 1. inviscidflow

2. steadyflow

3. incompressibleflow

4. the equation applies along a streamline

Limitation 4 will be relaxed for certain conditions to be investigated in Chapter 10.

Bernoulli’s equation was also developed in Chapter 6 from energy considerations for steady incompressibleflow with constant internal energy. It is interesting to note that the constant internal energy assumption and the inviscidflow assumption must be equivalent, as the other assumptions were the same. We may note, therefore, that the viscosity in some way will effect a change in internal energy.

▶ 9.4

SPHERICAL COORDINATE FORMS OF THE NAVIER–STOKES EQUATIONS³

Examples 1 and 2 gave examples of the use of the Navier–Stokes equations in rectangular and cylindrical coordinates, respectively. The use of spherical coordinates is significantly more complex, but is extremely useful.

Consider a solid sphere of radiusRrotating in a large body of stagnantfluid as shown in Figure 9.10. This is an example of“creepingflow”and will be revisited in Chapter 12. The sphere rotates steadily about its vertical axis with an angular velocity W(rad/sec) in an infinite Newtonian liquid of viscosity m. Our goal is to calculate the velocity in the f-direction. To begin the problem, we need to use the Navier–Stokes equation in spherical coordinates:

3R. B. Bird, W. E. Stewart, and E. N. Lightfoot,Transport Phenomena, Wiley, New York, 2007; W. Deen, Analysis of Transport Phenomena, Oxford University Press, 2012.

9.4 Spherical Coordinate Forms of the Navier–Stokes Equations ◀ 119

The rand qcomponents of the Navier–Stokes equations are both zero, so we only require thef-direction equation (from Appendix E):

r @vf

@t ‡vr@vf

@r ‡vq

r

@vf

@q ‡ vf

rsin…q†

@vf

@f ‡vfvr

r ‡vfvq

r cot…q†

ˆ 1

rsin…q†

@P

@frg_f

‡m 1 r²

@

@r r²@vf

@r

‡ 1 r²sinq

@

@q sinq@vf

@q

‡ 1 r²sin²q

@²vf

@f²

v_f

r²sin²…q†‡ 2 r²sin…q†

@vr

@f‡ 2 cos…q†

r²sin²…q†

@v_q

@f

In this analysis we will neglect gravity and pressure, and assume steady, fully developed one-dimensional flow and that there is no velocity in the rand q directions.

We can also assume symmetry about thezaxis (see Appendix B, Figure B.3), which allows simplification by removing all the _@f^@ terms indicating that there is no dependence on the anglef. The resulting equation is

0ˆm 1 r²

@

@r r²@vf

@r

‡ 1 r²sinq

@

@q sinq@vf

@q

vf

r²sin²…q†

This equation is a partial differential equation forvf, wherevfis a function ofrandq and can be solved using an equation solver or handbook⁴by realizing that it can be rewritten with

vfˆf…r†sinq as an ordinary differential equation in the form

0ˆ d dr r²df

dr

2f

This is annth-order equidimensional equation (also called a Cauchy-type equation) with a solution of the formf ˆrⁿwithnˆ1 and 2 resulting in a general solution of the form:

f…r† ˆC₁r‡C₂ r²

4M. Abramowitz and I. A. Stegun,Handbook of Mathematical Functions, Dover Publications, 1972, p. 17.

r y

x

z

R

Figure 9.10 A sphere rotating in a stagnantfluid.

Where C1 and C2 are integration constants. The resulting equation forvf is vfˆC1rsinq‡C2

r₂sinq

To solve this equation we need two boundary conditions. From Figure 9.10 we obtain B:C: 1:vf ˆRWsinq at rˆR

B:C: 2:vf ˆ0 at rˆ

∞

Thefirst boundary condition states that as the sphere rotates at an angular velocity ofW, and as a result of the no-slip boundary condition, the fluid at the surface is at the same velocity as the surface itself, which is equal toRWsinq. The second boundary condition states that far away from the sphere, thefluid is at rest and is not affected by the rotation of the sphere.

Applying B.C. 2 results inC1ˆ0, and applying B.C. 1, gives RWˆ0‡C₂

R² C2 ˆR³W This results in the desired velocity equation:

vfˆR³W r² sinq

▶ 9.5

CLOSURE

We have developed the differential equations for the conservation of mass and Newton’s second law of motion. These equations may be subdivided into two special groups:

@r

@t‡Ñ

^{rvˆ0 (9-26)}

…continuity equation† Inviscidflow

rDv

Dt ˆrg ÑP (9-27)

…Euler’s equation† Incompressible, viscousflow

Ñ

^{vˆ0 (9-28)}

…continuity equation†

rDv

Dt ˆrg ÑP‡mѲv (9-29)

…Navier−Stokes equation for incompressible flow† In addition, the student should note the physical meaning of the substantial derivative and appreciate the compactness of the vector representation. In component form, for example, equation (9-29) comprises some 27 terms in Cartesian coordinates.

9.5 Closure ◀ 121

▶

PROBLEMS

9.1 Apply the law of conservation of mass to an element in a polar coordinate system and obtain the continuity equation for a steady, two-dimensional, incompressible flow.

9.2 In Cartesian coordinates, show that v @

@x‡vy @

@y‡vz@

@z

may be written (v

Ñ). What is the physical meaning of the term (v

^Ñ)?

9.3 In an incompressible flow, the volume of the fluid is constant. Using the continuity equation,Ñ

^vˆ 0, show that thefluid volume change is zero.

9.4 FindDv/Dtin polar coordinates by taking the derivative of the velocity. (Hint:vˆvr(r,q, t)er‡vq(r, q,t)eq. Remember that the unit vectors have derivatives.)

9.5 Forflow at very low speeds and with large viscosity (the so-called creepingflows) such as occur in lubrication, it is possible to delete the inertia terms,Dv/Dtfrom the Navier– Stokes equation. Forflows at high velocity and small viscos- ity, it is not proper to delete the viscous term vѲv. Explain this.

9.6 Using the Navier–Stokes equations and the continuity equation, obtain an expression for the velocity profile between twoflat, parallel plates.

9.7 Does the velocity distribution in Example 2 satisfy continuity?

9.8 The atmospheric density may be approximated by the relation r ˆ r0 exp( y/b), whereb ˆ 22,000 ft. Determine the rate at which the density changes with respect to body falling atvfps. Ifv ˆ 20,000 fps at 100,000 ft, evaluate the rate of density change.

9.9 In a velocityfield wherevˆ400[(y/L)²e^x‡(x/L)²e^y] fps, determine the pressure gradient at the point (L,2L). Theyaxis is vertical, the density is 64.4 lb_m/ft³and theflow may be consid- ered inviscid.

9.10 Write equations (9-17) in component form for Cartesian coordinates.

9.11 Derive equation (2-3) from equation (9-27).

9.12 In polar coordinates, the continuity equation is 1

r

@

@r…rvr† ‡1 r

@vq

@q ˆ0

Show that

a. ifvqˆ0, thenvrˆF(q)/r b. ifvrˆ0, thenvqˆf(r)

9.13 Using the laws for the addition of vectors and equation (9- 19), show that in the absence of gravity,

a. Thefluid acceleration, pressure force, and viscous force all lie in the same plane

b. In the absence of viscous forces thefluid accelerates in the direction of decreasing pressure

c. A staticfluid will always start to move in the direction of decreasing pressure

9.14 Obtain the equations for a one-dimensional steady, vis- cous, compressible flow in the x direction from the Navier– Stokes equations. (These equations, together with an equation of state and the energy equation, may be solved for the case of weak shock waves.)

9.15 Obtain the equations for one-dimensional inviscid, unsteady, compressibleflow.

9.16 Using the Navier–Stokes equations as given in Appen- dix E, work Problems 8.17 and 8.18.

9.17 Using the Navier–Stokes equations,find the differential equation for a radialflow in whichv_zˆvqˆ0, andvrˆf(r).

Using continuity, show that the solution to the equation does not involve viscosity.

9.18 Using the Navier–Stokes equations in Appendix E, solve Problem 8.13.

9.19 For the flow described in Problem 8.13, obtain the differential equation of motion ifvqˆf(r, t).

9.20 Determine the velocity profile in afluid situated between two coaxial rotating cylinders. Let the inner cylinder have radius R1,and angular velocityW1; let the outer cylinder have radiusR2

and angular velocityW2.

9.21 Beginning with the appropriate form of the Navier–

Stokes equations, develop an equation in the appropriate coor- dinate system to describe the velocity of afluid that isflowing in the annular space as shown in thefigure. Thefluid is Newtonian, and is flowing in steady, incompressible, fully developed, laminar flow through an infinitely long vertical round pipe annulus of inner radius RI and outer radius RO. The inner cylinder (shown in thefigure as a gray solid) is solid, and the fluidflows between the inner and outer walls as shown in the figure. The center cylinder moves downward in the same direction as the fluid with a velocity v0. The outside wall of the annulus is stationary. In developing your equation,