Unsteady Laminar Flow in a Tube

8/12/2019 Unsteady Laminar Flow in a Tube

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Laminar flow that starts from a state of rest

in a straight circular tube

R. Shankar Subramanian

Department of Chemical and Biomolecular EngineeringClarkson University

Here, we consider unsteady laminar flow in a straight circular tube. We shall use a cylindrical

polar coordinate system ( ), ,r z in which z represents the direction of flow, and r and stand for the radial and angular directions, respectively.

We make the following simplifying assumptions.

1. Incompressible flow: Continuity reduces to 0 =v

2. Newtonian flow with constant viscosity

3. Neglect end effects:z

=

0

v

(fully developed flow)

4. Symmetry in (known as axial symmetry): ; 0v

= =

0

v

The incompressible version of the equation of continuity, in conjunction with assumptions 3 and

4, yields

( ) 0rrvr

=

(1)

Therefore, rrv must be constant across the cross-section of the tube. Because it is zero at

0r= , it must be zero everywhere. This implies that 0r

v = . Therefore, the flow is

unidirectional andz

v is the only non-zero velocity component.

Rr

zP0 PL


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The Navier-Stokes equation, subject to assumptions 1 and 2, can be written in component form

in cylindrical polar coordinates. The components in the rand -directions yield the result thatthe dynamic pressure P is uniform in those directions. Therefore, the dynamic pressure can, atbest, depend only on tand z . The component in the z -direction yields the following governing

equation for the velocity ( ),zv t r in the tube.

z zv vP rt z r r r

= +

(2)

By rearranging this equation as

z zv vP

rz r r r t

=

(3)

and recognizing that the left side can only depend on t and z , while the right side can onlydepend on tand r, we conclude that each can, at best, depend only on t. It can be shown that

the pressure gradient becomes steady rapidly in such a system, so that we can replace Pz

by its

steady representation0LP P P

L L

= where we have defined 0 LP P P = .

Therefore, Equation (2) can be rewritten as

z zv vP

rt L r r r

= +

(4)

The initial and boundary conditions are written as follows.

Initial Condition:

( )0, 0zv r = (5)Boundary Conditions:

( ),0 is finitezv t (6)

(No Slip) ( ), 0zv t R = (7)

Now, we proceed to scale this problem by defining the following dimensionless variables.

( )( )2 2

; ; ,/ 4

zvt r

T y V T yR R PR L

= = =

In the above, is the kinematic viscosity ( )/ . You may wonder about the choice of thescale for the velocity, and also about the factor 4 that is introduced in that scale. The velocity


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scale is obtained by considering the steady problem. In that problem, the characteristic velocity

is of the order ( )2 /PR L . The factor 4 is used to make the steady solution appearespecially simple, as you will see a bit later.

Non-dimensionalization leads to

14

V Vy

T y y y

= +

(8)

The scaled velocity field satisfies the following initial and boundary conditions.

Initial Condition:

(0, ) 0V y = (9)

Boundary Conditions:

( ),0 is finiteV T (10)

No Slip:

( ,1) 0V T = (11)

From the discussion in class, we know that directly applying separation of variables to Equations

(8) to (11) will fail because of the appearance of an inhomogeneity in the governing equation.Therefore, we first write the solution as the sum of a steady state solution and a transient

solution.

( ) ( ) ( ), ,s tV T y V y V T y= + (12)

The steady solution must satisfy Equation (8) without the time derivative.

14 0s

dVdy

y dy dy

+ =

(13)

along with the boundary conditions

( )0 is finitesV (14)and

(1) 0s

V = (15)


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We first substitute the solution in Equation (12) into Equations (8) to (11), and make use of the

fact that the steady solution must satisfy Equation (13) and the boundary conditions in Equations

(14) and (15). This yields the following governing equation and initial and boundary conditionsfor the transient part of the solution.

1t tV VyT y y y

= (16)

( )(0, )t sV y V y= (17)

( ),0 is finitetV T (18)

( ,1) 0t

V T = (19)

You can see that the consequence of separating the solution into a steady and a transient part is toyield a homogeneous differential equation for the transient contribution. The boundary

conditions in the y -coordinate are unaffected, and amenable to writing a product class solution.The initial condition is different from that on the complete velocity field. In fact, it is the

inhomogeneity in this initial condition that produces a non-trivial solution for the transient

problem. We can obtain the steady field by integrating Equation (13) and applying theboundary conditions. This leads to

( ) 21sV y y= (20)

The solution of Equations (16) through (19) is accomplished by separation of variables. If we

substitute a trial solution of the form ( ) ( ) ( ),tV T y G T y= into Equation (16) we find that

the functions ( )G T and ( )y must satisfy the following differential equations.

2 0dG

GdT

+ = (21)

21 0d d

yy dy dy

+ =

(22)

The solution of Equation (21) is2T

G Ce = (23)


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where C is an arbitrary constant of integration. The general solution of Equation (22) can beobtained by using Frobenius series. For our purposes, we simply use the final result without

going through the solution process.

( ) ( )1 0 2 0C J y C Y y = + (24)

where 0J and 0Y are known as the Bessel functions of the first and second kinds, respectively,

of order zero. Information about these functions can be obtained from Abramowitz and Stegun

(1965). In general, the Bessel functions ( )pJ y and ( )pY y of integer order p are thetwo linearly independent solutions of Bessels equation

( )2 21

0d d

y py dy dy

+ =

(25)

Here are sample graphs showing the behavior of the functions 0 ( )J x and 1( )J x , followed by

( )0Y x and ( )1Y x .


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For our purposes, it is important only to note that the functions ( )pY x are singular at 0x= ,that is, they approach minus infinity as x approaches zero. This is not compatible with the

boundary condition stated in Equation (18) that tV remain finite at 0y= . Therefore, the

constant 2C in Equation (24) must be set equal to zero. This leaves us with a product class

solution

( ) ( )2

0, T

tV T y Be J y

= (26)

where the product of the two arbitrary constants Cand 1C is written as a new constant B . We

know that this solution satisfies the governing differential equation for ( ),tV T y and thecondition that the solution must remain finite at 0y= . Now, we need to apply the remainingconditions. The no-slip boundary condition at the wall that leads to Equation (19) can be

satisfied by requiring

( )0 0J = (27)

As you can see from the sketch showing the behavior of 0J , this equation has more than one

root. In fact, it has an infinite number of positive roots that come paired with negative roots of

the same magnitude. It is known that ( ) ( )0 0J x J x = so that we can simply use the infinite

set of positive roots, labeled ( )1,2,3...n n = . The first 20 roots of Equation (27), labeled

0,sj may be found in Table 9.5, page 409, in Abramowitz and Stegun (1965). Youll see that


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0,20 0,19j j . Therefore, additional roots can be generated by approximating the intervalbetween each by .

Corresponding to each root of Equation (27), the product form appearing in Equation (26)with a

multiplicative constantnB

will satisfy the governing equation and the two boundary conditions

in the y -coordinate. We still need to satisfy the initial condition given in Equation (17). This

can be done by using a sum of all these solutions (recall that in a linear problem, we can use

superposition), which leads us to write the solution in the form

( ) ( )2

0

1

, n Tt n n

n

V T y B e J y

=

= (28)

All that remains is the determination of the constantsn

B . We can obtain them by applying the

initial condition.

( ) ( ) ( )01

0,t s n n

n

V y V y B J y

=

= = (29)

To evaluate the constants, we need to use the following results from Hildebrand (1976). More

general results can be found in Abramowitz and Stegun (1965).

( ) ( )1

0 0

0

0,n my J y J y dy m n = (30)

( ) ( )

( )

21

12

0 0

0

provided 02

n

n n

Jy J y dy J

= = (31)

( ) ( )1p p

p p

dy J y y J y

dy = (32)

( ) ( ) ( )1 12

p p p

pJ x J x J x

x ++ = (33)

The procedure is as follows. Begin with Equation (29) in the form

( ) ( )01

s n n

n

V y B J y

=

= (34)


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Multiply both sides by ( )0 myJ y where m is some integer, and integrate with respect to y from 0y= to 1. Use Equations (30) and (31) to obtain

( ) ( ) ( )

1

02

01

2m s m

mB y V y J y dyJ =

(35)

Evaluate the integral in Equation (35) by splitting it into the sum of two integrals. One of these

can be found immediately by using Equation (32). To find the other, youll need to useintegration by parts in conjunction with Equation (32). The final result, after simplification using

Equation (33), is

( )3 1

8m

m m

BJ

= (36)

Substitution of this result after replacing the index m with n permits us to write the completesolution from Equation (12) as follows.

( ) ( )

( )

202

31 1

, 1 8 nn T

n n n

J yV T y y e

J

=

= (37)

The velocity field, calculated from Equation (37) at a few selected values of T, is displayedbelow.


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References

F.B. Hildebrand,Advanced Calculus for Applications, Prentice-Hall, Englewood Cliffs, 1976.

M. Abramowitz and I.A. Stegun, Handbook of Mathematical Functions, Dover, New York,

1965.

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Unsteady Laminar Flow in a Tube