Analysis of a Differential Fluid Element in Laminar Flow

T

he analysis of afluid-flow situation may follow two different paths. One type of analysis has been discussed at length in Chapters 4–6 in which the region of interest has been a definite volume, the macroscopic control volume. In analyzing a problem from the standpoint of a macroscopic control volume, one is concerned only with gross quantities of mass, momentum, and energy crossing the control surface and the total change in these quantities exhibited by the material under consideration. Changes occurring within the control volume by each differential element offluid cannot be obtained from this type of overall analysis.

In this chapter, we shall direct our attention to elements offluid as they approach differential size. Our goal is the estimation and description offluid behavior from a differential point of view; the resulting expressions from such analyses will be differential equations. The solution to these differential equations will giveflow information of a different nature than that achieved from a macroscopic examination. Such information may be of less interest to the engineer needing overall design information, but it can give much greater insight into the mechanisms of mass, momentum, and energy transfer.

It is possible to change from one form of analysis to the other—that is, from a differential analysis to an integral analysis by integration and vice versa, rather easily.¹

A complete solution to the differential equations offluidflow is possible only if the flow is laminar; for this reason only laminar-flow situations will be examined in this chapter. A more general differential approach will be discussed in Chapter 9.

▶ 8.1

FULLY DEVELOPED LAMINAR FLOW IN A CIRCULAR CONDUIT OF CONSTANT CROSS SECTION

Engineers are often confronted withflow offluids inside circular conduits or pipes. We shall now analyze this situation for the case of incompressible laminar flow. In Figure 8.1, we have a section of pipe in which theflow is laminar and fully developed—that is, it is not influenced by entrance effects and represents a steady-flow situation.Fully developedflowis defined as that for which the velocity profile does not vary along the axis offlow.

1This tranformation may be accomplished by a variety of method, among which are the method of vector calculus. We shall use a limiting process in this text.

99

We now consider the cylindrical control volume offluid having an inside radius,r, thicknessDr, and lengthDx. Applying Newton’s second law to this control volume, we may evaluate the appropriate force and momentum terms for thexdirection. Starting with the control-volume expression for linear momentum in thexdirection,

SFxˆ Z Z

c:s:rvx…v

^n†dA‡_@^@_t^{Z Z Z}_c:v:^{rvxdV (5-5a)}

and evaluating each term as it applies to the control volume shown, we have SFxˆP…2prDr†jx P…2prDr†jx‡Dx‡t^rx…2prDx†jr‡Dr t^rx…2prDx†jr

Z Z

c:s:vxr…v

^{n†dAˆ …rvx†…2prDrvx†jx}‡Dx …rvx†…2prDrvx†jx

and

@

@t Z Z Z

c:v:vxrdV ˆ0 in steady flow.

The convective momentumflux

…rvx†…2prDrvx†jx‡Dx …rvx†…2pDrvx†jx

is equal to zero as, by the original stipulation that flow is fully developed, all terms are independent ofx. Substitution of the remaining terms into equation (5-5a) gives

‰P…2prDr†jx‡Dx P…2prDr†jxŠ ‡trx…2prDx†jr‡Dr trx…2prDx†jr ˆ0 Canceling terms where possible and rearranging, wefind that this expression reduces to the form

rPjx‡Dx Pjx

Dx ‡…rtrx†jr‡Dr …rtrx†jr

Dr ˆ0

Evaluating this expression in the limit as the control volume approaches differential size— that is, asDxandDrapproach zero, we have

rdP dx‡ d

dr…rtrx† ˆ0 (8-1)

Note that the pressure and shear stress are functions only ofxandr, respectively, and thus the derivatives formed are total rather than partial derivatives. In a region of fully developed flow, the pressure gradient,dP/dx, is constant.

The variables in equation (8-1) may be separated and integrated to give trxˆ dP

dx r 2‡C1

r x

Figure 8.1 Control volume for flow in a circular conduit.

The constant of integrationC₁may be evaluated by knowing a value oftrxat somer. Such a condition is known at the center of the conduit,rˆ0, where for anyfinite value ofC₁, the shear stress,trx, will be infinite. As this is physically impossible, the only realistic value for C₁is zero. Thus, the shear-stress distribution for the conditions and geometry specified is

trxˆ dP dx

r

2 (8-2)

We observe that the shear stress varies linearly across the conduit from a value of 0 atrˆ0, to a maximum atrˆR, the inside surface of the conduit.

Further information may be obtained if we substitute the Newtonian viscosity relationship—that is, assuming thefluid to be Newtonian and recalling that the flow is laminar:

trxˆmdvx

dr (8-3)

Substituting this relation into equation (8-2) gives mdvx

dr ˆ dP dx

r 2 which becomes, upon integration,

vxˆ dP dx

r² 4m‡C2

The second constant of integration,C2, may be evaluated, using the boundary condition that the velocity,vx, is zero at the conduit surface (the no-slip condition),rˆR. Thus,

C₂ˆ dP dx

R² 4m and the velocity distribution becomes

vxˆ dP dx

1

4m…R² r²† (8-4)

or

vxˆ dP dx

R²

4m 1 r

R

2

(8-5) Equations (8-4) and (8-5) indicate that the velocity profile is parabolic and that the maximum velocity occurs at the center of the circular conduit whererˆ0. Thus,

vmaxˆ dP dx

R²

4m (8-6)

and equation (8-5) may be written in the form vxˆvmax 1 r

R

2

(8-7) Note that the velocity profile written in the form of equation (8-7) is identical to that used in Chapter 4, example 2. We may, therefore, use the result obtained in Chapter 4, example 2

vavgˆvmax

2 ˆ dP

dx R²

8m (8-8)

8.1 Fully Developed Laminar Flow in a Circular Conduit of Constant Cross Section ◀ 101

Equation (8-8) may be rearranged to express the pressure gradient, dP/dx, in terms ofvavg: dP

dxˆ8mvavg

R² ˆ32mvavg

D² (8-9)

Equation (8-9) is known as the Hagen–Poiseuille equation, in honor of the two men credited with its original derivation. This expression may be integrated over a given length of conduit tofind the pressure drop and associated drag force on the conduit resulting from theflow of a viscousfluid.

The conditions for which the preceding equations were derived and apply should be remembered and understood. They are as follows:

1. Thefluid a. is Newtonian

b. behaves as a continuum 2. Theflow is

a. laminar b. steady

c. fully developed d. incompressible

▶ 8.2

LAMINAR FLOW OF A NEWTONIAN FLUID DOWN AN INCLINED-PLANE SURFACE

The approach used in Section 8.1 will now be applied to a slightly different situation—that of a Newtonianfluid in laminarflow down an inclined-plane surface. This configuration and associated nomenclature are depicted in Figure 8.2. We will examine the two-dimensional case—that is, we consider no significant variation in thezdirection.

The analysis again involves the application of the control-volume expression for linear momentum in thexdirection, which is

SFxˆ Z Z

c:s:vxr…v

^n†dA‡_@^@_t^{Z Z Z}_c:v:^{rvxdV (5-5a)}

L y

x

g

Figure 8.2 Laminarflow down an inclined-plane surface.

Evaluating each term in this expression for the fluid element of volume (Dx)(Dy)(1) as shown in thefigure, we have

SF_xˆPDyj_x PDyj_x‡Dx‡tyxDxj_y‡Dy tyxDxj_y‡rgDxDysinq Z Z

c:s:rvx…v

^{n†dAˆrv2xDyjx‡Dx rv2xDyjx}

and

@

@t Z Z Z

c:v:rvxdVˆ0

Noting that the convective-momentum terms cancel for fully developedflow and that the pressure–force terms also cancel because of the presence of a free liquid surface, we see that the equation resulting from the substitution of these terms into equation (5-5a) becomes

tyxDxjy‡Dy tyxDxjy‡rgDxDysinqˆ0 Dividing by (Dx)(Dy) (1), the volume of the element considered, gives

t^yxjy‡Dy t^yxjy

Dy ‡rgsinqˆ0 In the limit asDy®0, we get the applicable differential equation

d

dytyx‡rgsinqˆ0 (8-10)

Separating the variables in this simple equation and integrating, we obtain for the shear stress t^yxˆ rgsinqy‡C1

The integration constant,C₁, may be evaluated by using the boundary condition that the shear stress, tyx, is zero at the free surface, yˆL. Thus the shear-stress variation becomes

t^yxˆrgLsinq 1 y L

h i

(8-11) The consideration of a Newtonianfluid in laminarflow enables the substitution ofm(dvx/dy) to be made fortyx, yielding

dvx

dy ˆrgLsinq

m 1 y

L

h i

which, upon separation of variables and integration, becomes vxˆrgLsinq

m y y² 2L

‡C²

Using the no-slip boundary condition—that is,vxˆ0 atyˆ0—the constant of integra- tion, C₂, is seen to be zero. Thefinal expression for the velocity profile may now be written as

vxˆrgL²sinq m

y L

1 2

y L

2

(8-12) 8.2 Laminar Flow of a Newtonian Fluid Down an Inclined-Plane Surface ◀ 103

The form of this solution indicates the velocity variation to be parabolic, reaching the maximum value

vmaxˆrgL²sinq

2m (8-13)

at the free surface, yˆL.

Additional calculations may be performed to determine the average velocity, as was indicated in Section 8.1. Note that there will be no counterpart in this case to the Hagen– Poiseuille relation, equation (8-9), for the pressure gradient. The reason for this is the presence of a free liquid surface along which the pressure is constant. Thus, for our present case, flow is not the result of a pressure gradient, but rather the manifestation of the gravitational acceleration upon a fluid.

Example 1

Your lab partner picked an out capillary tube without recording the diameter from the literature in the package. It is very difficult to accurately measure the diameter of long capillary tubes, but it can be calculated. The capillary in question is 12 cm long, which is long enough to achieve fully developedflow. Water isflowing through the capillary in steady, continuous laminarflow at a temperature of 313 K and a velocity of 0.05 cm/s. The pressure drop across the capillary is measured to be 6 Pa. You can assume that the density of thefluid is unchanging for this analysis, and that the no-slip boundary condition applies. The diameter of the capillary can be calculated as follows:

The assumptions of the problem allow us to use the Hagen–Poiseuille equation:

dP

dz ˆ8mvavg

R² ˆ32mvavg

D² (8-9)

We can takedzto be the length of the capillary tube anddPto be the change in pressure orDP, thus, DP

L ˆ32mvavg

D² Solving for the diameter,

Dˆ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 32mvavgL

DP r

ˆ

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 32…6:5810 ⁴Pa_?s†…0:5 cm=s†…12 cm†

6 Pa s

ˆ0:145 cmˆ1:45 mm

▶ 8.3

CLOSURE

The method of analysis employed in this chapter, that of applying the basic relation for linear momentum to a small control volume and allowing the control volume to shrink to differential size, enables one to find information of a sort different from that obtained previously. Velocity and shear-stress profiles are examples of this type of information. The behavior of afluid element of differential size can give considerable insight into a given transfer process and provide an understanding available in no other type of analysis.

This method has direct counterparts in heat and mass transfer, where the element may be subjected to an energy or a mass balance.

In Chapter 9, the methods introduced in this chapter will be used to derive differential equations offluidflow for a general control volume.

▶

PROBLEMS

8.1 Express equation (8-9) in terms of theflow rate and the pipe diameter. If the pipe diameter is doubled at constant pressure drop, what percentage change will occur in the flow rate?

8.2 A 40-km-long pipeline delivers petroleum at a rate of 4000 barrels per day. The resulting pressure drop is 3.4510⁶Pa. If a parallel line of the same size is laid along the last 18 km of the line, what will be the new capacity of this network? Flow in both cases is laminar and the pressure drop remains 3.4510⁶Pa.

8.3 A 0.635-cm hydraulic line suddenly ruptures 8 m from a reservoir with a gage pressure of 207 kPa. Compare the laminar and inviscidflow rates from the ruptured line in cubic meters per second.

8.4 A common type of viscosimeter for liquids consists of a relatively large reservoir with a very slender outlet tube, the rate of outflow being determined by timing the fall in the surface level. If oil of constant density flows out of the viscosimeter shown at the rate of 0.273 cm³/s, what is the kinematic viscosity of the fluid? The tube diameter is 0.18 cm.

8 cm

55 cm Oil

8.5 Derive the expressions for the velocity distribution and for the pressure drop for a Newtonian fluid in fully developed laminar flow in the annular space between two horizontal, concentric pipes. Apply the momentum theorem to an annular fluid shell of thicknessDrand show that the analysis of such a control volume leads to

d

dr…rt† ˆrDP L

The desired expressions may then be obtained by the substitution of Newton’s viscosity law and two integrations.

8.6 A thin rod of diameter d is pulled at constant velocity through a pipe of diameterD. If the wire is at the center of the

pipe,find the drag per unit length of wire. Thefluidfilling the space between the rod and the inner pipe wall has densityrand viscositym.

8.7 The viscosity of heavy liquids, such as oils, is frequently measured with a device that consists of a rotating cylinder inside a large cylinder. The annular region between these cylinders is filled with liquid and the torque required to rotate the inner cylinder at constant speed is computed, a linear velocity profile being assumed. For what ratio of cylinder diameters is the assumption of a linear profile accurate within 1% of the true profile?

8.8 Two immiscible fluids of different density and viscosity are flowing between two parallel plates. Express the boundary conditions at the interface between the two fluids.

8.9 Determine the velocity profile forfluidflowing between two parallel plates separated by a distance2h. The pressure drop is constant.

8.10 Fluidflows between two parallel plates, a distancehapart.

The upper plate moves at velocity, v0; the lower plate is stationary. For what value of pressure gradient will the shear stress at the lower wall be zero?

8.11 Derive the equation of motion for a one-dimensional, inviscid, unsteady compressibleflow in a pipe of constant cross- sectional area neglect gravity.

8.12 A continuous belt passes upward through a chemical bath at velocityv0and picks up afilm of liquid of thicknessh, density,r, and viscositym. Gravity tends to make the liquid drain down, but the movement of the belt keeps thefluid from running off completely. Assume that the flow is a well- developed laminar flow with zero pressure gradient, and that the atmosphere produces no shear at the outer surface of thefilm.

a. State clearly the boundary conditions atyˆ0 andyˆhto be satisfied by the velocity.

b. Calculate the velocity profile.

c. Determine the rate at whichfluid is being dragged up with the belt in terms ofm, r, h,v0.

8.13 The device in the schematic diagram on the next page is a viscosity pump. It consists of a rotating drum inside of a stationary case. The case and the drum are concentric. Fluid enters at A,flows through the annulus between the case and the drum, and leaves at B. The pressure at B is higher than that at A, the difference beingDP. The length of the annulus isL. The width of the annulushis very small compared to the diameter of the drum, so that theflow in the annulus is equivalent to theflow between twoflat plates.

Problems ◀ 105

Assume theflow to be laminar. Find the pressure rise and efficiency as a function of theflow rate per unit depth.

A B

h R

Drum

Case

8.14 Oil is supplied at the center of two long plates. The volumetricflow rate per unit length isQand the plates remain a constant distance,b, apart. Determine the vertical force per unit length as a function of theQ,m,L, andb.

8.15 A viscous film drains uniformly down the side of a vertical rod of radiusR. At some distance down the rod, the film approaches a terminal or fully developedflow such that the film thickness,h, is constant andvzˆf(r). Neglecting the shear resistance due to the atmosphere, determine the velocity distri- bution in thefilm.

8.16 Determine the maximumfilm velocity in Problem 8-15.

8.17 Benzene flows steadily and continuously at 100°F through a 3000-ft horizontal pipe with a constant diameter of 4 in. The pressure drop across the pipe under these conditions is 300 lbf/ft². Assuming fully developed, laminar, incompressible flow, calculate the volumetricflow rate and average velocity of thefluid in the pipe.

8.18 A Newtonianfluid in continuous, incompressible laminar flow is moving steadily through a very long 700-m, horizontal pipe. The inside radius is 0.25 m for the entire length, and the pressure drop across the pipe is 1000 Pa. The average velocity of thefluid is 0.5 m/s. What is the viscosity of thisfluid?

8.19 Benzene, which is an incompressible Newtonianfluid, flows steadily and continuously at 150°F through a 3000-ft pipe with a constant diameter of 4 in. with a volumetricflow rate of 3.5 ft³/s. Assuming fully developed laminarflow, and that the no-slip boundary condition applies, calculate the change in pressure across this pipe system.

8.20 You have been asked to calculate the density of an incompressible Newtonianfluid in steadyflow that isflowing continuously at 250°F along a 2500-ft pipe with a constant diameter of 4 in. and a volumetricflow rate of 2.5 ft³/s. The only fluid properties known are the kinematic viscosity, 7.14 10 ⁶ft²/sec, and surface tension 0.0435 N/m. Assuming that theflow is laminar and fully developed with a pressure drop of 256 lbf/ft², calculate the density of this fluid if the no-slip boundary condition applies.

▶

C H A P T E R

9

Differential Equations