unidirectional transport

8/10/2019 Unidirectional Transport

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Chapter 3

Unidirectional transport

In this chapter, we consider transport in which there is a variation in themass, momentum and temperature elds in only one dimension. The anal-ysis is considerably simplied in this case, since there is only one spatialcoordinate to be considered. However, the examples solved here illustratethe basic principles of the solution of more complex problems in multipledimensions, which involve shell balances to derive differential equations forthe concentration, velocity and temperature elds, and then an integrationprocedure for determining the variations in the concentration, velocity andtemperature.

3.1 Solutions of the diffusion equation

3.1.1 Unsteady transport into an innite uidMass transfer

Consider a at surface in the x y plane located at z = 0 through which thesolvent diffuses into the uid. The plane surface is assumed to be of inniteextent in the x y plane, and the height of the uid supported by the planeis also considered to be of innite extent. Initially, the concentration of the solute in the solvent at the surface is c. At time t = 0, the soluteconcentration at the surface is instanteneously increased to c0, and thereis diffusion of the solute into the solvent. We would like to determine thevariation of the concentration of the solute with time.

As the diffusion proceeds, there is transport of solute from the surface

1


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2 CHAPTER 3. UNIDIRECTIONAL TRANSPORT

to the solvent, resulting in an increase in the concentration near the surface.However, the concentration far from the surface z remains unchangedat c = c. The conditions for the concentration eld at the spatial bound-aries and at the initial time are

c = c as z at all tc = c0 at z = 0 at all t > 0c = c at t = 0 for all z > 0 (3.1)

The problem is simplied by dening a non-dimensional concentration

c = ccc0 c

(3.2)

The conditions for c are

c = 0 as z at all tc = 1 at z = 0 at all t > 0c = 0 at t = 0 for all z > 0 (3.3)

Heat transfer

The conguration consists of a at surface in the xy plane of innite extent,and a uid of conductivity K and specic heat C p in the half space z > 0.Initially, the temperature of the uid and the surface is T . At time t = 0,the temperature of the surface is instanteneously increased to T 0. Determinethe temperature as a function of time.

As conduction proceeds, there is a transport of energy from the surface tothe uid, resulting in an increase in the temperature of the uid. However,the temperature at a large distance from the surface, z , remains at T =T . Therefore, the conditions for the temperature at the spatial boundariesof the uid and at initial time are

T = T as z at all tT = T 0 at z = 0 at all t > 0

T = T at t = 0 for all z > 0 (3.4)

The problem is simplied by dening a non-dimensional temperature

T = T

T

T 0 T (3.5)


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3.1. SOLUTIONS OF THE DIFFUSION EQUATION 3

The conditions for T are

T = 0 as z at all tT = 1 at z = 0 at all t > 0T = 0 at t = 0 for all z > 0 (3.6)

Momentum transfer

The conguration consists of an innite uid in the z > 0 half space boundedby an innite at surface in the x z plane. The uid and the surface areinitially at rest. At time t = 0, the plane is instanteneously moved witha constant velocity U in the x direction. Determine the uid velocity as afunction of time.

As time proceeds, the momentum that is transported from the surfacediffuses through the uid, resulting in uid motion. However, the uid ata large distance from the surface z

remains at rest. If u

x is the uid

velocity in the x direction, it is convenient to dene a non-dimensional uidvelocity ux = ( ux /U ). The conditions for the non-dimensional uid velocityat the spatial boundaries of the ow and at initial time are

ux = 0 as z at all tux = 1 at z = 0 at all t > 0ux = 0 at t = 0 for all z > 0 (3.7)

Shell balance

In all three cases, the boundary and initial conditions 3.3, 3.6 and 3.7 areidentical in form. Since the transport in all three cases involves diffusion inthe direction perpendicular to the surface, and no variation along the surface,it is anticipated that the differential equation governing the transport willalso be identical in form. The differential equation for the concentration eldis rst derived using a shell balance, and the analogous equations for heatand momentum transfer are provided.

Consider a shell of thickness z in the z coordinate as shown in gure 3.1,and of thickness x and y in the x y plane. There is a transport of massacross the surfaces of the shell due to diffusion, which results in a change inthe concentration in the shell. We consider the variation in the concentra-tion within this control volume over a time interval t. Mass conservation


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requires that

Accumulation of massin the shell =

Input of mass into shell

Output of mass from shell

(3.8)

The accumulation of mass in a time t is given by Accumulation of mass

in the shell = ( c(x ,y,z , t + t)c(x,y,x,t )) x y z (3.9)The mass ux at the surface at z is given by

jz = Dcz z

(3.10)

and the mass entering the shell through the surface at z in a time interval t is given by the product of the mass ux, the area of transfer and the timeinterval t,

Input of mass into shell = t x y D

cz z

(3.11)

In a similar manner, the mass leaving the surface at z + z is given by

Output of mass from shell = t x y D

cz z+ z

(3.12)

Substituting equations 3.9, 3.11 and 3.12 into equation 3.8, and dividing by x y z t, we obtain

c(x ,y,z , t + t) c(x,y.z,t ) t = 1 z D cz z+ z D cz z (3.13)The limit t 0 and z 0 is taken, to obtain a partial differentialequaiton for the concentration eld.

ct

= z

Dcz

(3.14)

If the diffusion coefficient is independent of the spatial position, the differen-tial equation 3.14 reduces to

ct = D 2

cz 2 (3.15)


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c*=1T*=1u*=1x

x

zz

z+ z

z=0

c*=0T*=0u*=0x

z> n n ty

Figure 3.1: Conguration for similarity solution for unidirectional transport.

The concentration equation 3.15 can also be expressed in terms of the scaledconcentration variable, c , dened in 3.2,

ct

= D 2c

z 2 (3.16)

A similar shell balance procedure can be carried out for momentum trans-fer, and the equations for the temperature and velocity elds are

T t

= z

DT T z

(3.17)

Though the nal result for the momentum transfer process is exactlyanalogous to equations 3.16 and 3.17, the procedure is slightly different, andso we provide a brief outline of the calculation. First, note that there are


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now two directions in the problem. Since momemtum is a vector, there is adirection associated with the momentum itself, which is the x direction. Thesecond is the direction of diffusion, which is the z direction. The fundamentalbalance relation, equivalent of equation 3.8, is

Rate of change of x momentumin the shell

= Sum of forcesin x direction (3.18)

The rate of change of momentum in the differential volume of thickness z about z in a time interval t is given by,

Rate of change of x momentumin the shell

= ux x y z

t (3.19)

The forces acting on the two surfaces at z and z + z are the products of theshear stress xz and the surface area ( x y). It is important to keep trackof the directions of the forces in this case, since the force is a vector. Theshear stress xz is dened as the force per unit area in the x direction actingat a surface whose outward unit normal is in the positive z direction. For thesurface at z + z , the outward unit normal is in the + z direction, as shown ingure 3.1, and therefore the force per unit area at this surface is + xz |z+ z .For the surface at z , the outward unit normal is in the z direction, andtherefore the force per unit area at this surface is xz |z . Finally, the rateof change of momentum, which is the change in momentum per unit time, isgiven by ( ux )(A z )/ t. Therefore, the momentum balance equation is,

(A z ) ux

t = A( xz |z+ z xz |z) (3.20)

Dividing throughout by A z , we obtain,

ux t

= xz |z+ z xz |z

z (3.21)

Taking the limit t 0 and z 0, we obtain the partial differentialequation,u xt =

xzz (3.22)


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The shear stress is given by the product of the viscosity and the gradient of the velocity,

xz = lim z0

ux z

= u xz

(3.23)

With this, the governing equation for the velocity eld becomes,

u xt

= z

u xz

(3.24)

Note that the differential equation derived above has the same form as theconcentration and energy diffusion equations 3.16 and 3.17, though it wasderived from a force balance. This indicates that the diffusion process is thesame for mass, momentum and energy. However, it should be noted thatmomentum could be transmitted by pressure forces in addition to viscousforces, and there is no analogue of pressure in mass and energy transport.

When the velocity ux is scaled by U , the velocity of the bottom surface,the momentum equation becomes,

u xt

= z

u xz

(3.25)

where = ( /rho ) are the thermal and momentum diffusivities respectively.

Solution procedure

Since the conservation equations 3.16, 3.17 and 3.25 are identical in form, andthe boundary conditions 3.9, 3.11 and 3.12 are also identical, the same solu-

tion procedure can be used for all these, and the solutions for the concentra-tion, velocity and temperature les, expressed in terms of the dimensionlessvariables c , T and ux , turn out to be identical.

In order to solve the concentration equation 3.16 with the boundary con-dition 3.9, it is rst important to realise that there no intrinsic length scalein the problem, because the uid and the at plate are of innite extent.Since the concentration c is dimensionless, there are only three dimensionalvariables z , t and D in the problem. These contain two dimensions, Land T ,and it is possible to construct only one dimensionless number, = ( z/ Dt ).Therefore, just from dimensional analysis, it can be concluded that the con-centration eld does not vary independently with z and t, but only on thecombination = ( z/ Dt ). If this inference is correct, it should be possible to


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express the conservation equation 3.16 in terms of the variable alone. Whenz and t are expressed in terms of , the concentration equation becomes

z

2D 1/ 2t3/ 2c

= DDt

2c 2

(3.26)

After multiplying throughout by t, the equation for the concentration eldreduces to

2

c

+ 2c 2

= 0 (3.27)

Equation 3.27 validates the inference that the non-dimensionalised concen-tration eld is only a function of . It is also necessary to transform theboundary conditions 3.9 into conditions for the coordinate. The trans-formed boundary conditions are

c = 0 as z

at all t

as

c = 1 at z = 0 at all t > 0 at = 0c = 0 at t = 0 for all z > 0 as (3.28)

It is useful to note that the original conservation equation, 3.16, is a secondorder differential equation in z and a rst order differential equation in t, andso this requires two boundary conditions in the z coordinate and one initialcondition. The conservation equation expressed in terms of is a secondorder differential equation, which requires just two boundary conditions for . Equation 3.28 shows that one of the boundary conditions for z andthe initial condition t = 0 turn out to be identical conditions for

,

and therefore the transformation form ( z, t) produces no inconsistency in theboundary and initial conditions.

Equation 3.27 can be easily solved to obtain

c ( ) = C 1 + C 2 d exp 2

2 (3.29)

The constants C 1 and C 2 are determined from the conditions c = 1 at = 0,and c = 0 for , to obtain

c ( ) = 2

(z/ Dt ) d exp

2

2 (3.30)


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From the solution 3.30, it can be inferred that there is no intrinsic lengthscale in the system, but the length scale for the z coordinate is a function of time, and is proportional to Dt at time t. Thus, the length scale over whichdiffusion has taken place increases proportional to t1/ 2. Thus, equation 3.30will be a solution for the concentration diffusion in a conguration boundedby two plates, so long as the distance between the two plates L is largecompared to Dt .

3.1.2 Steady diffusion into a falling lm

This problem is a simplication of the actual diffusion in a falling lm, whichinvolves a combination of convection and diffusion. A thin lm of uid owsdown a vertical surface. One side of the lm is in contact with a gas whichis soluble in the liquid, and as the liquid ows down, the gas is dissolvedin the liquid. The concentration of gas in the liquid at the entrance is c,

while the concentration of gas at the liquid-gas interface is c0. The differ-ence in concentration between the initial concentration in the liquid and theconcentration at the interface acts as a driving force for diffusion. The z coordinate is perpendicular to the gas-liquid interface, which is located atz = 0. We also assume that the penetration depth for the gas into the liquid(to be determined a little later) is small compared to the thickness of theliquid lm, so that the boundary conditions in the z coordinate given by 3.9are applicable. In addition, the boundary condition in the x coordinate is

c = c at x = 0 (3.31)

3.1.3 Diffusion in bounded channels

Next we consider the problem of diffusion in a channel bounded by two wallsseparated by a distance L, as shown in gure 3.2. Initially, the concentrationof the uid in the channel is equal to c. At initial time, the concentrationof the solute on the wall at z = 0 is instanteneously increased to c0, whilethe concentration on the surface at z = L remains equal to c. The problemis to nd the concentration eld as a function of time. Similar problems canbe formulated for heat and momentum transfer.

The concentration eld is rst expressed in terms of the scaled concen-tration eld c by equation 3.2. The diffusion equation, obtained by a shellbalance as before, is given by equation 3.16. However, there is a modication


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c*=1T*=1u*=1x

x

zzz+ z

z=0

c*=0T*=0u*=0x

z=

Figure 3.2: Conguration for similarity solution for unidirectional transport.

in the boundary conditions,

c = 0 at z = L at all tc = 1 at z = 0 at all t > 0c = 0 at t = 0 for all z > 0 (3.32)

In this case, it is not possible to effect a reduction to a similarity form,because there is an additional length scale L in the problem, and so the z

coordinate can be scaled by L. A scaled z coordinate is dened as z = ( z/L ),and the diffusion equation in terms of this coordinate is

ct

= DL2

2cz 2

(3.33)

The above equation suggests that it is appropriate to dene a scaled timecoordinate t = ( Dt/L 2), and the conservation equation in terms of thisscaled time coordinate is

ct

= 2cz 2

(3.34)

The boundary conditions, in terms of the scaled coordinates z and t , are

c = 0 at z = 1 at all t


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c = 1 at z = 0 at all t > 0c = 0 at t = 0 for all z > 0 (3.35)

It is expected that in the long time limit, t , the concentration eldwill attain a steady state value cs which is independent of time. This steadystate concentration eld is obtained by solving 3.16 with the time derivativeset equal to zero, and the solution for the steady concentration eld is alinear concentration prole,

c = C 1z + C 2 (3.36)

The constants C 1 and C 2 are obtained from the boundary conditions at z = 0and z = 1,

cs = (1 z ) (3.37)The variation of concentration with time can be determined by dividing

the concentration into a steady and an unsteady part,c = cs + cu , (3.38)

where cu is the difference between the actual concentration and the concen-tration at steady state. The reason for this decomposition will become cleara little later. The conservation equation for the unsteady concentration eldis identical to that for the original concentration eld, because ( cs /t ) = 0and ( 2cs /z 2) = 0,

cut

= 2cuz 2

(3.39)

However, the boundary condition for cu is different from that for the c , andis obtained by subtracting cs from c at the boundaries,

cu = 0 at z = 1 at all tcu = 0 at z = 0 at all t > 0

cu = cs at t = 0 for all z > 0 (3.40)Equation 3.16 can be solved by the method of separation of variables,

where the unsteady concentration eld is expressed as two functions, one of which is only a function of t , while the other is only a function of z .

cu = ( t )Z (z ) (3.41)


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This is inserted into the conservation equation 3.16, and the equation isdivided by the production Z , to obtain

1

ddt

= 1Z

d2Z dz 2

(3.42)

In equation 3.42, the left side is only a function of t , while the right sideis only a function of z . From this, it can be inferred, as follows, that thesetwo functions have to be constants independent of z and t . To infer this,assume that these two functions are not constants, and that the left sidevaries as t is varied, and the right side varies as z is varied. In this case,if we keep z a constant and vary t , then the left side of 3.42 varies, whilethe right side remains a constant, and so the equality is destroyed. The onlyway for the equality to hold, if the left side is only a function of t and theright side is only a function of z , is if the two sides are constants.

The solution for Z is rst obtained by solving

1Z

d2Z dz 2

= 2 (3.43)where is a positive constant. The reason for choosing the right side of 3.43 to be negative will become apparent a little later. The solution for thisequation is

Z = C 1 sin (z ) + C 2 cos (z ) (3.44)

where C 1 and C 2 are constants to be determined from the boundary con-ditions. The boundary condition cu = 0 (Z = 0) at z = 0 is satised forC 2 = 0. The boundary condition cau = 0 at z = 1 is satised if = ( n ).Therefore, the solution for Z which satised the boundary conditions in thez coordinate is

Z = C 1 sin (nz ) (3.45)

where n is an integer.The solution for can now be obtained from the equation

1

ddt

= 2 = n22 (3.46)This equation is solved to obtain

= C 3 exp (n22t ) (3.47)


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The nal solution for cu = ( t )Z (z ) is

cu = C exp (n22t )sin(nz ) (3.48)The solution 3.48 contains the integer n which is as yet unspecied, and it isveried that the solution 3.48 satises the equation 3.46 for any value of n.The most general solution is one which contains a linear combination of thesolution 3.48 for different values of n,

cu =

n =1C n exp(n22t )sin(nz ) (3.49)

The values of the coefficients C n have to be determined from the boundarycondition that has not been used so far,

cu = (1 z ) for t = 0

n =1C n sin(nz ) =

(1

za) (3.50)

The coefficients can be determined because the functions sin ( nz ) satisfyorthogonality conditions,

1

0dz sin (nz )sin(mz ) = 0 for n = m

= (1 / 2) for n = m (3.51)

To use this condition, the left and right sides of 3.50 are multiplied bysin(mz ), and integrated over the interval 0 z 1, to obtain

C m = 2 1

0 dz sin (mz )(1 z )=

1m

(3.52)

The reason for choosing the right side of 3.46 to be negative is now ap-parent. If we had chosen the right side of 3.46 to be positive, the solution forZ consists of an exponentially growing and an exponentially decaying func-tion. In this case, it can easily be veried that it is not possible to satisfythe boundary condtions c = 0 at z = 0 and at z = 1 simultaneously. Inaddition, the solution for in equation 3.47 would have been a function thatis exponentially increasing in time, and therefore, there is no steady solutionin this case.


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3.1.4 Unsteady diffusion in a cylinder

A cylindrical container of radius R containing uid with temperature T isdipped into a bath of uid containing solute with temperature T 0. Assumethat the bath is large enough that the conduction of heat into the uid does

not appreciably change the temperature of the bath, and that there is noresistance to heat transfer in the walls of the container. The temperature inthe cylinder as a function of time and position is to be determined.

A cylindrical coordinate system is used, as shown in gure 3.3, where z is the axial coordinate and r is the radial coordinate. There is no variationof temperature in the z coordinate, since the cylinder is considered to be of innite extent. In order to obtain a differential equation for the temperaturevariation in the cylinder, a shell of thickness r and height z at radius ris considered. The terms in the balance equation 3.8 for the energy in thecylindrical shell are as follows.

Accumulation of energyin the shell = C p(T (x ,y,z , t + t)T (x,y,x,t ))2r r z (3.53)

The total energy entering the shell at r is the product of the heat ux, thesurface area, and the time interval t,

Input of energy into shell = K

T r

(2r z t)r

(3.54)

Similarly, the total energy leaving the shell at r + r is

Output of

energy from shell=

K

T

r(2r z t)

r + r(3.55)

When these are inserted into the conservation equation 3.8, and divided by r t, the net energy balance for the shell is

C pr(T (x ,y,z , t + t) T (x,y,x,t ))

t =

1 r

KrT r r + r Kr

T r r

(3.56)Taking the limit r 0 and t 0, the partial differential equation forthe temperature eld is

T t =

DT r

r r

T r (3.57)


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r

z

z

R

r r T*=0

Figure 3.3: Heat diffusion into a cylinder.

It is important to note that there is a variation in the surface area of the

shell as r varies, and there is a variation in the total energy transportedthrough the shell due to the variation in this surface area. This leads to amore complicated form for the diffusion term in 3.57, in comparison to thesecond derivative encountered in the diffusion from a at plane, 3.17.

The conservation equation 3.57 can be scaled as follows. The scaledtemperature eld is dened by equation 3.5, while the scaled distance andtime coordinates are dened as r = ( r/R ) and t = ( tR 2/D T ), since Ris the length scale in the radial direction. The conservation equation 3.57,expressed in these coordinates, is

T t =

1r

r r

T r (3.58)


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The boundary conditions are

T = 0 at r = 1 for all tT = 1 at t = 0 for all r

T r = 0 at r = 0 for all t (3.59)

Equation 3.58 is solved using the method of separation of variables. Thenal steady state solution is one in which the temperature is uniform, T = 0,throughout the cylinder. The unsteady solution is solved using the substitu-tion

T = R(r )( t ) (3.60)

where R is a function of the radial coordinate, and is a function of time.The form 3.60 is inserted into the temperature equation, 3.58, and the re-sulting equation is divided by R(r )( t ) to obtain

1

t

= 1R

1r

r

rRr

(3.61)

The left side of the above equation is only a function of time, while the rightside is only a function of r . Therefore, the equality can only be satised if both sides are equal to constants. The right side of the above equation isrst solved,

1R

d2Rdr 2

+ 1r

dRdr

= 2 (3.62)

A special function solution for equation 3.62 can be obtained by recastingthe equation as

r 2d2Rdr 2

+ rdRdr

+ 2r R = 0 (3.63)

The solution of the above equation is in the form of Bessel functions,

R = C 1J 0(r ) + C 2Y 0(r ) (3.64)

The values of the constants C 1 and C 2 are determined from the boundaryconditions at r = 0 and r = 1. At r = 0, the radial derivative of J 0(r ) isidentically zero, whereas the radial derivative of Y 0(r ) approaches innityfor r 0. Therefore the boundary condition at r = 0 can be satised only


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if C 2 = 0. The second boundary condition, T = 0 at r = 1, is used todetermine the value of in equation 3.62,

J 0() = 0 (3.65)

There are multiple solutions for this equation, the rst few of which are givenby. The equation for ,1

ddt

= 2n (3.66)can be solved to obtain

= exp ( 2n t ) (3.67)With this, the solution for the temperature eld is

T =

n =1C n J 0(n r )exp( 2n t ) (3.68)

The coefficients C n should satisfy the initial condition at t = 0,

n =1

C n J 0(n r ) = 1 (3.69)

The values of the coefficients can be determined using orthogonality condi-tions for the Bessel functions, which in this case are

1

0r dr J 0(n r )J 0(m r ) =

12

(J 1(n ))2 for m = n

= 0 for m = n (3.70)In order to determine the coefficients, the right and left sides of 3.70 aremultiplied by r J 0(m r ), and integrated from r = 0 to r = 1, to obtain

12

(J 1(n ))2C n = 1

0r dr J 0(n r ) (3.71)

3.1.5 Oscillatory ow

This example is used to illustrate the use of complex variables in problemswhere the forcing on the uid is oscillatory in time. Consider the at platebounding a uid in the half space z > 0, shown in gure 3.4, with the


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modication that the plate has an oscillatory velocity U = U 0 cos (t). Thedifferential equation for the velocity eld is given by equation 3.25,

u xt

= 2uxz 2

(3.72)

The boundary conditions in this case, analogous to 3.7, are

ux = U cos (t) at z = 0ux = 0 at z = L (3.73)

The solution procedure is simplied if the boundary condition is recast asfollows. The complex velocity eld ux is dened as a velocity eld thatsatises the same differential equation as 3.72,

ux

t =

2ux

z 2 (3.74)

but which satises the boundary conditions

ux = U exp(t) at z = 0ux = 0 at z = L (3.75)

where is the square root of 1. It is apparent that the solution to thedifferential equation 3.74, with the boundary condition 3.75, is the real partof the solution to the differential equation 3.72 with the boundary condition3.73. Since dealing with exponential functions is easier than dealing withsines and cosines, it is more convenient to solve equation 3.74 with boundaryconditions 3.75 for ux , and then take the real part of the solution to obtainthe solution ux of equation 3.72.

The non-dimensional velocity ux is dened as ux = ( ux /U ), while thescaled z and time coordinates are dened as z = ( z/L ) and t = ( t/L 2).The momentum conservation equation and boundary conditions in terms of this scaled velocity are

uxt

= 2uxz 2

(3.76)

ux = exp ( t ) at z = 0ux = 0 at z = L (3.77)


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u*=0x~

x

zz

z+ z

z=0u*=1x~

z>Infinity

u =U cos( t)x

Figure 3.4: Oscillatory ow at a at surface.

where the scaled frequency is given by = ( L2/ ).The differential equation 3.76 for the velocity eld is a linear differential

equation, since all terms in the equation contain only the rst power of ux .This rst order differential equation is driven by a wall which is oscillatorywall velocity with scaled frequency . When a linear system is driven bywall motion of frequency , the response of the system also has the samefrequency . (This is not true if the system is non-linear, since forcing of acertain frequency will generate response at different harmonics of this basefrequency). Therefore, the time dependence of the velocity eld in the uidcan be considered to be of the form

ux = ux(z )exp( t ) (3.78)

When this form is inserted into the differential equation 3.76, and divided byexp( t ), the resulting equation is an ordinary differential equation for ux .

ux = 2uxz 2

(3.79)

The boundary conditions for ux (3.77) become

ux = 1 at z = 0


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ux = 0 at z = L (3.80)

This equation is easily solved to obtain

ux(z ) = C 1 exp ( z ) + C 2 exp( z ) (3.81)where C 1 and C 2 are determined from the boundary conditions at z = 0and z = 1,

ux (z ) = exp( z ) exp( (2 z ))

1 exp(2 )(3.82)

The physical velocity eld, which is the real part of the product of ux andexp( t ), is

ux (z ) = Realexp( z ) exp( (2 z ))

1 exp (2 ) exp( t ) (3.83)

Before solving the above equation, it is useful to examine some limitingcases for the frequency of oscillation. In the limit 0, the solution forthe uid velocity, 3.83, is

ux (z ) = (1 z )exp( t ) (3.84)The physical reason for this result is as follows. The scaled frequency of oscillation is obtained by dividing the frequency, , by the inverst of thetime required for the momentum to diffuse across the length of the channel,(L2/ ). If the scaled frequency is small, then the time period of variationof the velocity of the bottom plate is long compared to the time required formomentum to diffuse across the channel. In this case, it is expected, that

the velocity prole at any time instant along the oscillation cycle is identicalto the velocity for a ow driven by a steady plate velocity which is equal tothe instanteneous value of the plate velocity at that instant in the cycle. Inthe limit 1, the uid velocity eld is given by

ux(z ) = exp ( z ) (3.85)In this case, the velocity eld decreases over a distance z (1/ ) fromthe surface. This is because the frequency of oscillation is large comparedto the time required for diffusion of momentum across the channel, and themomentum diffuses only to a distance comparable to ( L/ ). Beyond thisdistance, the momentum generated during the positive and negative parts of a cycle cancel out, and the uid velocity approaches zero.


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3.2. EFFECT OF BULK FLOW AND REACTION IN MASS TRANSFER 21

Waterx

z

z=L

z=0

Dry air

Figure 3.5: Diffusion with bulk ow.

3.2 Effect of bulk ow and reaction in masstransfer

In this section, the special effects of bulk ow and reactions on the solutionsfor unidirection mass transfer problems are examined.

3.2.1 Diffusion in a stagnant lm

Water evaporates from a container through a stagnant air lm through aglass tube into dry air owing at the top of the tube, as shown in gure 3.5.If the mole fraction at the surface of the liquid surface is the saturation molefraction xW s , and the dry air owing past the tube does not contain anywater, what is the concentration prole of water in the glass tube?

Though the air in the tube is stationary, the mean velocity across anyhorizontal surface in the tube is not zero, because of the ow of water vapour


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across the surface. The ux of water across a surface, j W z , contains a com-ponent due to the bulk ow, as well as a component due to the diffusion of water across the surface.

jW z =

cDW A

dxW

dz + xW ( jW z + jAz ) (3.86)

The last term on the right side of equation 3.86 is the ux of water due to thebulk ow, where the total molar ow rate is the sum of the uxes of water( jW z ) and air ( jAz ). In this particular case, the ux of air is identically zero,and so the ux of water vapour across a surface is given by

jW z = c

1 xW dxW dz

(3.87)

The mass balance equation is obtained by writing a ux balance across asection of thickness z of the tube, which at steady state provides,

jW z |z+ z j W z |z = 0 (3.88)If the above equation is divided by z the differential equation for the uxin the limit z 0 is

dj W zdz

= ddz

11 xW

dxW dz

= 0 (3.89)

This equation is solved to obtain

log (1

xW ) = A1z + A2 (3.90)

The constants A1 and A2 are determined from the boundary conditions xW =xW s at z = 0 and x = 0 at z = l,

(1 xW )(1 xW s )

=(1 xW f )(1 xW s )

z/l

(3.91)

3.2.2 Diffusion with homogeneous reaction

A gaseous reactant A dissolves in a liquid B, and undergoes a rst orderreaction A+ B

AB, in a tank of height L, as shown in gure 3.6. The mass

balance equation, 3.14 has to be modied in this case due to the presence of


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3.3. EFFECT OF PRESSURE ON MOMENTUM TRANSPORT 23

a consumption term due to chemical reaction. The mass balance equationat steady state takes the form

jAz |z j Az |z+ z kcAS z = 0 (3.92)

This can be reduced to a differential equation by dividing throughout byS z , and taking the limit z 0,dj Azdz kC A = 0 (3.93)

If the concentration of A is small, the ux of A is given by

jAz = DABdcAdz

(3.94)

Inserting this into the concentration equation 3.93, we get

DAB d2

cAdz 2 + kC A = 0 (3.95)

The boundary conditions are

cA = cA0 at z = 0 jAz = 0 at z = L (3.96)

The solution that satised both these conditions is

C AC A0

= cosh ((kL2/D AB )1/ 2(1 (z/L )))

cosh (kL2/D AB )1/ 2 (3.97)

3.3 Effect of pressure on momentum trans-port

The momentum balance condition states that the rate of change of momen-tum is equal to the applied force.

Rate of momentum in

Rate of momentum out +

Sum of forcesacting on the system = 0

(3.98)


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z

z=0

z=L

zz+ z

c=c at z=

j =0 at z=Lz

Figure 3.6: Diffison with homogeneous chemical reaction.


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This balance equation is written for a control volume of uid which is in theform of a thin shell. Momentum enters or leaves the control volume due touid ow into or out of the control volume, or due to the stresses acting onthe surface due to viscosity. The forces acting on the system are usually thegravitational force or the centrifugal force. Momentum balances are usuallyeasy to apply only if the streamlines are straight. Applying momentumbalances to systems with curved streamlines is more difficult, as we shall seein the last example of this section.

The procedure for solving problems with momentum balance is to writethe momentum balance equation for a shell of nite thickness, and thenlet the thickness go to zero. In this limit, the difference equations for thevelocity eld across a nite shell reduces to differential equations for thevelocity elds. These can then be solved, subject to appropriate boundaryconditions, in order to determine the velocity elds.

The boundary conditions generally involve specifying the velocity or stress

elds at the boundaries of the ow. These conditions depend on the surfaceadjoining the liquid at its boundaries.

1. If there is a solid surface adjacent to the uid, the appropriate boundarycondition is the no slip condition which states that the velocity of theuid at the surface is equal to the velocity of the surface itself.

2. At a liquid - gas interface, the momentum ux, and consequently thevelocity gradient, in the liquid side is assumed to be zero, because theviscosity of the gas is small compared to that of the liquid.

3. At a liquid - liquid interface, the momentum ux and the velocity arecontinuous across the interface.

Falling lm

We rst consider the ow of a falling lm along an inclined plane, as shownin gure 3.7. This type of ow is often encountered in cooling towers, evap-oration chambers and gas absorption equipment. The plane is inclined atan angle to the vertical, and we consider a section L of the lm which issufficiently far removed from the entrance and exit that the velocity proleis independent of variation in the z direction along the ow. The only non -zero component of the velocity, uz , is a function of the coordinate x alone.


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g

v (x)z

z+ z

x

z

z=0

z=L

z

Figure 3.7: Flow down an inclined plane.


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A momentum balance is set up for a thin shell of uid of thickness xbounded by the planes z = 0 and z = L. The various components of themomentum balance in the z direction are

Rate of z

momentum inacross surfaceat x

(LW ) xz |x

Rate of zmomentum outacross surfaceat x + x

(LW ) xz |x+ x

Rate of zmomentum inacross surface

at z = 0

(W xvz ) vz|z=0

Rate of zmomentum outacross surfaceat z = L

(W xvz ) vz|z= L

Gravity forceacting onthe uid

(WL x)(g cos( ) (3.99)

When these terms are substituted into the momentum balance equation, weget

LW xz |x + LW xz |x+ x + W xv2z z=0 W x v2z z= L + LWxWg cos( ) = 0(3.100)

Dividing throughout by LW x and taking the limit as x 0, we getlim

x0 xz |x+ x xz |x

x+ g cos( ) = 0 (3.101)

The rst term on the left of the above equation is the derivative of the shearstress, and so this can be written as

d xzdx + g cos( ) = 0 (3.102)


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This can be integrated to give

xz = g cos( )x + C 1 (3.103)where C 1 is a constant of integration. The constant of integration can be de-termined by making use of the boundary condition at the liquid gas interface

xz = 0 at x = 0 (3.104)

With this, the equation for the shear stress becomes

xz = g cos( )x (3.105)If the uid is Newtonian, the momentum ux is related to the velocity

gradient by

xz = dvzdx

(3.106)

This is easily integrated to give an expression for the velocity

vz = g cos( )

2x2 + C 2 (3.107)

The constant of integration can be determined using the condition that atx = h, vz = 0, since the surface at x = h is a solid surface and a noslip boundary condition is applicable here. With this, the equation for thevelocity prole is

vz = gh2 cos( )

21

xh

2 (3.108)

Hence the velocity prole is a parabolic prole.Various parameters can be determined once this velocity prole is known.

1. the maximum velocity, vzm , is clearly at x = 0

vzm = gh2 cos( )

2 (3.109)

2. The total ow rate is determined from

Q = h

0dx

W

0dyvz

= gWh3 cos( )

3 (3.110)


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3. The mean velocity can be calculated from

vz = Q

h

= gWh2 cos( )

3 (3.111)

4. The lm thickness can be expressed in terms of the ow rate as

h = 3Q

gW cos ( )

1/ 3

(3.112)

5. The total force on the inclined surface in the z direction is given by

F = L

0dz

W

0dy xz |x= h (3.113)

= ghLW cos ( ) (3.114)

This is just equal to the weight of the uid in the z direction understeady ow conditions.

The momentum balance equation in the x direction can also be writtenin an analogous fashion.

Rate of xmomentum inacross surface

at x

(LW ) xx |x

Rate of xmomentum outacross surfaceat x + x

(LW ) xx |x+ x

Gravity forceacting onthe uid

(W L x)(g sin( ) (3.115)

The momentum balance equation obtained from the above equation is

LW xx |x + LW xx |x+ x + LWxWg sin( ) = 0 (3.116)


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r

x

x

rrR

Figure 3.8: Flow in a circular tube.

Since xx = p for a Newtonian uid, the above equation reduces to

dpdx = g sin( ) (3.117)

The variation in the pressure in the lm is then given by

p = gx sin( ) (3.118)

The above analytical results are valid only if the ow is laminar andthe streamlines are smooth, so that the ow can be considered steady. Theseconditions are satised for the slow viscous ow of a thin lm. As the velocityincreases or the lm thickness increases, it has been found that there is a

transition from a laminar ow ith straight streamlines to a laminar ow withrippling and then to a turbulent ow. The conditions under which thesetransitions occur is determined by the Reynolds number, Re = 4h vz / .A laminar ow without rippling is observed for Re < 10, while there isrippling for 10 < Re < 1000. The ow becomes turbulent when the Reynoldsnumber increases beyond about 1000.Example: Flow through a circular tube.

The ow through a circular tube is often encountered in engineering ap-plications, and the laminar ow can be analysed using shell momentum bal-ances. The only new feature here is the cylindrical coordinate system that isused for the analysis, as shown in gure 3.8.

Consider a cylindrical shell of thickness r and length x. The balance


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equation for the x component of the momentum can be written as,

Rate of changeof x momentum = Sum of forces l (3.119)

The rate of change of momentum in a time interval t within the differential

volume under consideration can be written as,Rate of changeof x momentum =

( ux ) t

(2r r z (3.120)

where ( ux ) is the change in the momentum per unit volume in the timeinterval t.

There are four bounding surfaces for the differential volume under con-sideration, two or which are at perpendicular to the x axis and located atx and x + x, and two of which are perpendicular to the radial co-ordinateand are located at r and r + r . The forces acting on the surfaces at x andx + x can be separated into two parts, the rst due to the pressure acting

on the surfaces, and the second due to the ux of momentum due to uidmotion. The forces due to uid pressure can be written as,

Force due to pressureon surface at x = p(r,,x )(2r r )

Force due to pressureon surface at x + x = p(r,,x + x)(2r r ) (3.121)

Note that there is a negative sign for the force at x+ x, because the pressurealways acts along the inward unit normal at the surface, and the inward unitnormal at x + x is in the negative x direction. There is an additional forcedue to the ow of momentum into the differential volume through the surfaceat x and the ow of momentum out of the differential volume through thesurface at x + x. This force is given by the product of the momentum ux(momentum transported per unit area per unit time) and the surface area.The momentum ux is the product of the momentum density ( ux ) (per unitvolume) and the normal velocity to the surface ux . This is analogous to themass ux, which is the product of the concentration (mass density) c andthe normal velocity. Therefore, the force due to the ow of momentum is,

Force due to momentum owon surface at x = (( ux)ux (2r r )

Force due to momentum owon surface at x + x = p(r,,x + x)(2r r )(3.122)


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The volumetric ow rate is determined by integrating the velocity proleover the radius of the tube.

Q = (P 0 P L )R4

8L (3.130)

This shell balance analysis is also valid only for steady ows, where thestreamlines are straight. This occurs in a fully developed ow, away fromthe entrance or exit of the pipe, and also in a the Reynolds number is lessthan 2100, where the Reynolds number is dened as Re = vmax R/ , wherevmax is the maximum velocity and R is the radius of the pipe.

Oscillatory ow in a pipe

An oscillatory pressure gradient ( p/L ) = k cos(t) is applied across a pipeof length L. Determine the velocity prole in the pipe.

The differential equation for the velocity prole is, The momentum equa-tion in the ow x direction is

u xt

1Re

2uxx 2

+ 2ux

r 2=

px

(3.131)

Setting ur = 0 for a unidirectional ow, and neglecting variations in thestreamwise direction, the equation for the velocity prole is

u xt

1Re

1r

ddr

rduxdr

= cos(t) (3.132)

where Re = ( R2/ ). Use the solution for an oscillatory ow,

ux = ux exp(t) (3.133)

The equation for ux becomes

ux 1Re

1r

ddr

rduxdr

= 1 (3.134)

The solution for ux is divided into a homogeneous and a particular solution.The homogeneous solution is given by

uxh = CJ 0( Rer ) (3.135)


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x

z

Figure 3.9: Flow of immiscible uids.

where J 0 is the Bessel function. The particular solution is given by

uxp = (3.136)The complete solution is

ux = J 0( r )J 0( )

1 (3.137)

Adjacent ow of two immiscible liquids.

This is an example which illustrates the application of boundary conditionsbetween two immiscible liquids. Two immiscible liquids ow through a chan-nel of length L and width W under the inuence of a pressure gradient, asshown in gure 3.9. The ow rates are adjusted such that the channel is half lled with uid I (more dense phase) and half lled with uid II (less densephase). It is necessary to nd the distribution or velocity in this case.

The momentum balance is similar to that for the ow in a pressure gra-dient shown in the last example. The pressure balance reduces to

d xzdx

+ p0 pL

L = 0 (3.138)

This equation is valid in either phase I or phase II. Integration gives tworelations in the two phases.

(I )xz = p0 pL

Lx + C (I )

(II )xz = p0 pLL x + C (II ) (3.139)


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We can make use of one of the boundary conditions, that the stress is equalat the interface, to relate C (I ) and C (II ) .

Atx = 0 (I )xz = (II )xz

C (I ) = C (II ) (3.140)

Using Newtons law of viscosity to relate the stress to the strain rate, we get

(I )dv(I )zdx

= p0 pL

Lx + C (I )

(II )dv(II )z

dx =

p0 pLL

x + C (II ) (3.141)

This can be integrted to give

v(I )z = p0 pL2(I )L

x2 + C (I ) x

(I ) + C (I )2

v(II )z =

p0 pL2(II )L

x2 + C (I ) x

(I ) + C (II )2

(3.142)

There are three constants in the above equations, which are determined usingthe three available boundary conditions.

Atx = 0 v(I )z = v(II )z

Atx = b v(I )z = 0Atx = b v(II )z = 0 (3.143)

These boundary conditions provide three relationships between the constantsC (I ) , C (I )2 and C

(II )2 .

C (I )2 = C (II )2

0 = p0 pL2(I )L

b2 C (I )b

(I ) + C (I )2

0 = p0 pL2(II )L

b2 + C (I )b

(I ) + C (I )2 (3.144)

These can be solved to obtain

C 1 = ( p0 pL )b

2L(I ) (II )(I ) + (II )

C (I )2 =

( p0

pL )b2

2(I )L 2 (I )

(I ) + (II ) (3.145)


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Therefore, the velocity proles in each layer are

v(I )z = ( p0 pL )b2

2(I )L 2(I )

(I ) + (II )+

(I ) (II )(I ) + (II )

xb

x2

b2

v(I )z =

( p0

pL )b2

2(II )L 2(II )

(I ) + (II ) +(I )

(II )

(I ) + (II )xb

x2

b2 (3.146)

Exercises

1. Consider a long and narrow channel two - dimensional of length L andheight H , where H L. The ends of the channel are closed so thatno uid can enter or leave the channel. The bottom and side walls of the channel are stationary, while the top wall moves with a velocityV (t). Since the length of the channel is large compared to the height,the ow near the center can be considered as one dimensional. Nearthe ends, there will be some circulation due to the presence of the side

walls, but this can be neglected far from the sides. For the ow farfrom the walls of the channel,

(a) Write the equations for the unidirectional ow. What are theboundary conditions? What restriction is placed on the velocityprole due to the fact that the ends are closed and uid cannotenter or leave the channel?

(b) If the wall is given a steady velocity V which is independent of time, solve the equations (neglecting the time derivative term).Calculate the gradient of the pressure.

(c) If the wall is given an oscillating velocity V cos(t), obtain anordinary differential equation to obtain the velocity prole. Getan analytical solution for this which involves the constants of ite-gration. Use the boundary conditions to determine all unknownconstants.

2. Wire-coating of dies Consider a cylinder in a thin annular region,as shown in gure 3.10. The cylinder is pulled with a constant velocityV. The pressure is equal on both sides of the cylinder. Determine theuid velocity, and the ow rate.

3. A uid is contained in the annular region between two concentric cylin-ders of radius R1 and R2 moving with angular velocities 1 and 2.


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Outer radius R

Inner radius k R

V

Pressure pPressure p

Figure 3.10: Wire coating of dies.

The gravitational eld acts along the axis of the cylinders as shown ingure. The vessel is tall enough that the ow can be considered unidi-rectional when the distance from the bottom is large compared to thegap width ( R1R2). In this case, choose a coordinate system and writedown the mass and momentum conservation equations. Solve these forthe pressure and velocity elds. Can you nd the equation for the freesurface?

4. A resistance heating appratus for a uid consists of a thin wire im-mersed in a uid. In order to design the appratus, it is necessary todetermine the temperature in the uid as a function of he heat uxfrom the wire. For the purposes of the calculation, the wire can beconsidered of innite length so that the heat conduction problem is

effectively a two dimensional problem. In addition, the thickness of the wire is considered small compared to any other length scales in theproblem, so that the wire is a line source of heat. The wire and theuid are initially at a temperature T 0. At time t = 0, the current isswitched on so that the wire acts as a source of heat, and the heattransmitted per unit length of the wire is Q. The heat conduction inthe uid is determined by the unsteady state heat conduction equation

t T = K 2T (3.147)

and the heat ux (heat conducted per unit area) is

K T (3.148)


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where K is the thermal conductivity of the uid.

(a) Choose an appropriate coordinate system, and write down theunsteady heat conduction equation.

(b) What are the boundary conditions? Give special attention o theheat ux condition at the wire, and note that the wire is consideredto be of innitesimal radius.

(c) Solve the heat conduction equation using the simplest method,and determine the temperature eld in the uid.

(d) Use the boundary conditions to determine the constants in theexpression for the temperature eld.

5. Consider a two dimensional incompressible ow in a diverging channelwith subtended angle bounded by solid walls as shown in the gure

below. The ux of uid through the channel is Q. The two dimensional(r, ) polar coordinate system used for the analysis is also shown in thegure. Assume a one dimensional velocity eld ur = 0 and u = 0. Forthis case,(a) From the mass conservation equation, determine the form of the

velocity in the radial direction.

(b) Write down the momentum conservation equations in the r and directions and simplify assuming that the Reynolds number issmall, so that inertial terms can be neglected. Simplify the equa-tions. What are the boundary conditions?

(c) Eliminate the pressure from the momentum equations to obtainan equation for the velocity eld.

(d) Solve the equation to obtain an expression for the velocity. Howmany constants are there, and how are they determined? Do not determine the constants.

(e) If we assume that the Reynolds number is high, so that poten-tial ow conditions apply, what are the governing equations andboundary conditions?

(f) What are the velocity and pressure elds in this case?


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r

Wall

Wall

6. An ideal vortex is a ow with circular streamlines where the particlemotion is incompressible and irrotational. The velocity prole obeysthe equation in cylindrical coordinates:

v = 2r

(3.149)

with vr = vz = 0. At the origin, the above equation indicates that thevelocity becomes innite. But this is prohibited because viscous forcesbecome important and the ow is rotational in a small region near thecore.

Consider an ideal vortex in which the velocity is given by the aboveequation for t < 0, and the core velocity is constrained to be zero att = 0. Find the velocity prole for t > 0. Assume that v is the onlynon - zero velocity component.

7. A cubic solid of side a is initially held at a temperature T 0. At timest 0, its lateral faces are held at temperatures T A , T B , T 0 and T 0 as il-lustrated in gure 3.11. The top and bottom faces are insulated so thatno heat is transferred through them. The cube has heat conductivityC , density and thermal conductivity K .

(a) Solve for the steady state temperature in the cube.

(b) Show how the transient problem may be set up in a form to whichseparation of variables can be applied.


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q=0

q=0

T0 T0T

T

A

B

Figure 3.11: Conduction from a cube.

8. A rotating cylinder geometry consists of a cylinder of radius R andheight H , lled with uid, with two end caps. The cylinder rotates withan angular velocity , while the end caps are stationary. Determinethe uid velocity eld using separation of variables as follows.

(a) Choose a coordinate system for the problem. Clearly, the onlynon-zero component of the velocity is u . Determine the boundaryconditions for this component of the velocity.

(b) Write down the mass balance condition for an incompressible uid.For a uni-directional ow in which the density is a constant, whatdoes this reduce to?

(c) Use a shell balance to determine the conservation equation for thevelocity. Can you eliminate the pressure term using a result fromthe mass balance condition?

(d) Solve the conservation equation at steady state using the methodof separation of variables. Frame the orthogonality conditionswhich would be required to solve the problem.

Data:


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(a) Bessel equation:

x2d2ydx2

+ xdydx

+ ( x2 n2)y = 0

Solution: y = A1J n (x) + A2Y n (x)

where J n (x) is bounded for x 0, and Y n (x) is bounded forx . tem Modied Bessel equation:x2

d2ydx2

+ xdydx (x

2 + n2)y = 0

Solution:y = A1I n (x) + A2K n (x)

where I n (x) is bounded for x

0, and K n (x) is bounded for

x .

unidirectional transport

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