shell momentum balances and velocity distributions in laminar flow

8/11/2019 Shell Momentum Balances and Velocity Distributions in Laminar Flow

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Chapter 2

Shell Momentum Balances

and Velocity Distributionsin Laminar Flow


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1.2. MOLECULAR TRANSPORT

Viscous force works if there is velocity gradientin the fluid. Viscous forces per unit area (viscous stress) may be

expressed as x, yor z.

Each viscous stress has 3 components, e.g. x has components

xx ,xy andxz Pressureand viscous stressboth are combined to form

molecular stress

ij

=pij

+ ij

where i and j may be x, y or z

ijisKronecker delta, 1 if i=j and 0 if ij, because pressureworks on the change of velocity at streamwise direction.


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ii=p+ ii = normal stress, i.e. stress in i directiondue to

momentum transfer (as a result of addition of pressure) in

i direction

ij=ij = shear stress. i.e. stress inj directiondue to

momentum transfer (velocity gradient) in i direction =

shear stress

Shear stress: Force per unit area that is exerted parallel to

the surface on which it acts. In this figure ij = yx = - dvx/dy (Newton law)

Momentum transferis due to velocity gradient from

higher to lower velocitiesDirection of momentum transfer

(through a plane normal to ydirection) fromhigher to lower velocities

generating a force (CAUSE)

direction of shearstress (force/area).

Area may be normal

to the force in case of

normal stress or

parallel to the force in

case of shear stress

Vectorof

momentum flux is

determined by the

2ndsubscript.A

momentum

balance contains

terms with thesame 2ndsubscript


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Normal stress: Force/unit area that is exerted normal to the

surface on which it acts. Because of this force, the velocity

changes in the direction of the force. Pressure is a normal stress

xx=p+ xx,yy=p+ yy, zz=p+ zzare normal stresses, while

xy =yx,xz =zx,yz =zyareshear stresses. These quantities

which have 2 subscripts are called tensors. A tensorrequires information of magnitude, direction of

momentum transferand direction of the force to specify it,

while a vectorrequires magnitude and direction

These stresses cause molecular (diffusional) transport.

direction of

momentum transfer

direction of shear

stress


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1.7. CONVECTIVE TRANSPORT

Momentum can be transportedby fluid flow bulk. This

process is called convective transport.

Volume ratethrough a unit area x direction is vx. This

volume flows with mass flux v. Momentum fluxcarrying

this mass flux through this area is vxv. With the same fashion, momentum fluxes carrying the same

mass flux through area yandzdirections are vyvdan

vzv.

Those vector vxv, vyv, vzvrespectively have components

in the directions ofx,yandz, e.g. vxvhas components

vxvx , vxvy andvxvz , but vbelongs to a mass flux, so

their components must be present in a mass balance

(inseparable)


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vxvy is the convective momentum flux carried by a

momentum through a plane normal to x componentand

carrying the y-componentof amass flux

Because vv has 2 subscripts to denote a plane whichnormal to mass flux direction and direction of convection

velocity, it is called tensor.

direction of

momentum transfer

(through a plane

normal to x-

direction) carrying

mass(CAUSE)

mass flux in y direction (part of inseparablex-, y-,

z-direction mass flux). Inseparablemeans all

these directions belong to amass flux and are

always present in one mass balance

Vectorof momentum flux is determined by

the second subscript.A momentum

balance contains terms with the same 2nd

subscript


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Only the second

component of tensor

vzis used becausemass is flowing only

inzdirection


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MOLECULAR AND CONVECTIVE

TRANSPORT

Table 1.7-1 Summary of components of convective

momentum flux

Dimension of vectors: force/area or

momentum flux (= force/area)

Direction ofmomentum

transfer as

CAUSE

Direction of mass-

carrying velocity

(momentum transfer)as CAUSE


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= Diffusive momentum-

flux tensor

F t f ibl fl id d i l i 2 3 di i


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For a system of compressible fluid, and involving 2 or 3 dimensions,

modified Newtons law of viscosity in Cartesian, cylindrical and

spherical cordinates are as follows


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2.1. SHELL MOMENTUM BALANCES

AND BOUNDARY CONDITIONS

Momentum balance at steady-statecondition:

(rate of momentum in)

(rate of momentum out) +

(sum of forces acting on system) =0 (2.1-1)

Forces included in the balance are pressure force (acting

on the surface) and gravity force (acting on the entire

fluid volume)

In momentum balance, the flow sometimes is driven by

shear, by pressure or by gravity (shear-driven flow,

pressure-driven flow or gravity-driven flow)


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Procedure to solve the problem of viscous flow

Write Eq. (2.1-1) forshellforfinite thickness as arepresentative part of a system in which the velocityvaries.

Change the thickness approaching zero (0).

Make differential equation describing distribution ofmomentum flux

Insert Newton equation to obtain differential equation forvelocity distribution

Information from integrationscan be used to calculatevarious quantities, such as average velocity, maximumvelocity, volumetric rate, pressure drop and forces on

fluid boundaries


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Some constants arising in the integrations are evaluatedusingboundary conditions.

Boundary conditions generally used are

At interface solid-fluid, fluid velocity = surface (solid)velocity on which the fluid attaches

At a liquid-gasinterfacial plane of constant x, the shearstress is zero. For example xy or xz= zero, provided thatthe gas-side velocity gradient is not too large. This isreasonable, since the viscosities of gases are much less than

those of liquids At a liquid-liquidinterfacial plane of constant x, thetangential velocity components vyand vzare continuousthrough the interface (the "no-slip condition") as are alsothe molecular stress-tensor components xy andxz .


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2.2. FLOW OF A FALLING FILM (the case

of gravity-driven flow)

Observe the fluid with lengthLat position far away from

the endsof the wall so that disturbance effects at inlet and

outlet points are negligible.

Momentum balance in z direction on the system withthickness zis developed between planes atz=0 toz=L.

Fluid flow has width =Wat y direction.


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For first instance, assume that

there are momentum transfers

across each surface due to

mass fluxand due to velocity

gradient

In this case, convection

(mass flux) is only in z-

direction. The causesof

momentum transfer could be

in x, y and z directions.

If a component ofthe

causes are non existence,

then, ignore the related term

Gravity-driven, not pressure-driven flow


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Force/area Force/area Force/volume

All terms in

momentum

balance have

the same 2nd

subscript


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Gravity-driven, not pressure-

driven flow


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Therefore, velocity distribution is parabolic (see Fig. 2.2-

3) Maximum velocity, vz,maxoccurs at x=0, is

. (2.2-19)

Average velocity, vzalong the film is

. (2.2-20)

2

z,max

g cosv

2

W

z

0 0

z zW

0

0 0

1 22 2

0

v dx dy1

v v dxdx dy

g cos x x g cos1 d

2 3


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Volumetric rate Q obtained from integration of velocity

distribution

. (2.2-21)

Film thickness as a function of vz, Q and massa rate

(W) (=vz) are

. (2.2-22)

W 2 3

z z0 0

g W cosw v dx dy W v

3

z3 3

2

3 v 3 w

gcos gWcos


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z-component of force on the surface (solid) is obtained by

integrating momentum flux on the interfaceof solid-

liquid.

. (2.2-23)

Fz= the weight of the entire fluid along L of the film.

L W L W

zz xxz x

0 0 0 0

dvF dy dz dy dzdx

g cos(LW)( ) g LW cos


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Examp le 2.2-1. Calcu lat ion o f Film

Velocity

Oil has kinematic viscosity ()=2x10-4 m2.sec-1and density0.8x103kg/m3. What is the film mass rate, w, verticallyflowing on the wall such that =2.5mm?

Answer:

According to Eq. (2.2-20) with cos =1and =/,

.

To get mass flowrate, value of W(width of the wall) mustbe inserted. This applies to laminar flow

3 3 33

z z 4

1 1

(2.5x10 ) 0.8x10 9.8WgWw v A v W

3 3 2x10

0.204Wkg m sec


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To check laminarity, calculate Re

.

z

3 4

4 v 4 0.2044 w / WRe 5.1

0.8x10 2x10

For conduits other than pipe,Re= 4 x hydraulic radius x x vz/

Hydraulic radius = cross-sectional area/wetted perimeter = (.W)/W=


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2.3. FLOW THROUGH A CIRCULAR

TUBE (the case of combined gravity-driven

and pressure-driven flow)

We considersteady statelaminar flow for fluid with

constant density in the very long tube of lengthLand

radiusR. Very long means no end effect.

We consider on this system the cylindricalshellwiththickness rand lengthL(see Fig. 2.3-1).


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We can take the shell the

whole length of cylinder,

L, because it is assumed

that the velocity profile

remains the same from

the top to the bottom.

Convection is only in z

direction.

Datum line


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Momentum rate in through cylinder surface at r. .

Momentum rate out through cylinder surface at r+r.

.

Momentum rateinto the surface atz=0

Momentum rate outto the surface atz=L

All terms have the same 2ndsubscript.

2 rz rrL

2

rz r r

rL

02 zz zr r

2 zz z Lr r


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By dividing Eq. (2.3-6) by 2Lrand taking limit r

0.

(2.3-7)

By changing to differential form, then Eq. (2.3-7)

becomes

. (2.3-8)

0

0

lim

rz r r rz r zz zzz z L

r

r rg r

r L

0zz zzz z Lrz( r ) g r

t L


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.

.

. (i) because vr= 0, we can drop the term vrvzin Eq. 2.3-

9a;

(ii) because vz=vz(r), the term vzvzwill be the same at

both ends of the tube (iii) because vz=vz(r), the term -2dvz/dzwill be the same

at both ends of the tube =0.

Hence Eq. 2.3-8 simplifies to

vz= vz(r), vr= 0, v= 0, andp=p(z).


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wherep0pL+ gL= (p0- g0) - (pL- gL) = P0- PL

Eq 2.3-10 may be integrated to become

.

Constant C1is evaluated by using the boundary conditionsAt r=0, rz= 0, then C1= 0 and Eq. (2.3-11) becomes

.

p gz P


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Velocity distribution is shown in Fig. 2.3-2.

Newton viscosity law for this situation is

.

Substitution this relationship to Eq. (2.3-12) gives

.

Integrations results in

.

zrz

dv

dr


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This equation is called Hagen-Poiseuille law. Two forces

acting on the system: pressure force + gravity force

z-component of fluid force acting on the wetted wall,Fz=

integration of momentum over the wetted area: .


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Validity of Eq. (2.3-22)

Laminar flowRe < 2100

Density is constant (incompressible flow)

The flow is steady-state

The fluid is Newtonian ,

End-effect is negligible

The fluid behaves continuum

No slip on the wall

zrz

dv

dr


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Examp le 2.3-1. Determ ination of

Viscos i ty from Capi l lary Flow Data

Glycerine (CH2OH.CHOH.CH2OH) at 26.5oC is flowing

through a horizontal tube 1 ft long and with 0.1 in. inside

diameter. For P= 40 psi, the volume flow rate w/is

0.00398 ft3/min. The density of glycerine at 26.5oC is 1.261

g/cm3. From the flow data, find theviscosity of glycerine

in centipoises and in Pa.

SOLUTION

From the Hagen-Poiseuille equation (Eq. 2.3-21), we find


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.

To check whether the flow is laminar, we calculate the

Reynolds number


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.

Hence the flow is indeed laminar. Furthermore, the

entrance length is

Hence, entrance effects are not important, and the

viscosity value given above has been calculated properly


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Examp le 2.3-2. Compress ible Flow in a

Horizon tal Circu lar Tube

Obtain an expression for the mass rate of flow wfor an

ideal gas in laminar flow in a long circular tube. The flow

is presumed to be isothermal.

Assume that the pressure change through the tube is notvery large, so that the viscosity can be regarded a constant

throughout


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This problem can be solved approximately by assuming

that the Hagen-Poiseuille equation (Eq. 2.3-21) can be

applied over a small length dzof the tube as follows:

.

To eliminate in favor ofp, we use the ideal gas law in

the formp/=p0/0, or = 0p/p0, wherep0and 0are

the pressure and density atz= 0. This gives

.


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The mass rate of flow wis the same for allz (=constant).Hence Eq. 2.3-27 can be integrated fromz = 0 toz =Ltogive

.

Sincep02pL

2= (p0+pL)(p0-pL), we get finally

.

whereavg= (0+L)is the average density calculatedat the average pressurepavg= (p0+pL).

2 4 FLOW THROUGH AN ANNULUS (the


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2.4. FLOW THROUGH AN ANNULUS (the

case of combined pressure and gravity-driven

flow) We now solve another viscous flow problem in

cylindrical coordinates, namely the steady-state axial flowof an incompressible liquid in an annular region betweentwo coaxial cylinders of radii RandRas shown in Fig.

2.4-1. The fluid is flowing upward in the tube that is, inthe direction opposed to gravity.

We make the same postulates as in 2.3: vz= vz(r), vr= 0,v= 0, andp=p(z) and assume that velocity profile

remains unchanged alongz. Then when we make amomentum balance over a thin cylindrical shell of liquid,we arrive at the following differential equation:


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This differs from Eq. 2.3-10 only in that P=p+ gzhere,

since the coordinatez is in the direction opposed to

gravity. Integration of Eq. 2.4-1 gives

The constant C1cannot be determined immediately, since

we have no information about the momentum flux at the

fixed surfaces r= R and r=R.

P0PL = p0+ g0 - pL- gL


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The only difference between this equation and Eq. 2.4-2 isthat the constant of integration C1has been eliminated infavour of a different constant . The advantage of this is thatwe know the geometrical significance of .

We now substitute Newton's law of viscosity, rz= -(dvz/dr), into Eq. 2.4-4 to obtain a differential equation forvz.

.


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Integration of this first-order separable differential

equation then gives

.

We now evaluate the two constants of integration, and

C2by using the no-slip condition on each solid boundary:

.


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Substitution of these boundary conditions into Eq. 2.4-6then gives two simultaneous equations:

.

From these the two integration constants and C2arefound to be

.

These expressions can be inserted into Eqs. 2.4-4 and 2.4-6to give the momentum-flux distribution and the velocitydistribution as follows:


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.

Note that when the annulus becomes very thin (i.e.,

only slightly less than unity and 1- 0), these resultssimplify to those for a plane slit (see Problem 2B.5).

It is always a good idea to check "limiting cases" such as

these whenever the opportunity presents itself


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The lower limit of 0 is not so simple, because theratio ln(R/r)/ln(l/) will always be important in a regionclose to the inner boundary. Hence Eq. 2.4-14 does notsimplify to the parabolic distribution.

Once we have the momentum-flux and velocitydistributions, it is straightforward to get other results ofinterest:


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.

The equations derived above are valid only for laminarflow. The laminar-turbulent transition occurs in theneighborhood of Re = 2000, with the Reynolds number

defined as Re = 2R(1 - )vz/.

Hydraulic radius =4(R2- (R)2)/(2R + 2R) =2R(1 - )

2 5 FLOW OF TWO ADJACENT


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2.5. FLOW OF TWO ADJACENT

IMMISCIBLE FLUIDS

Two immiscible, incompressible liquids are flowing inthez direction in a horizontal thin slit of lengthL andwidth Wunder the influence of a horizontal pressuregradient (po-pL)/L.

The fluid flow rates are adjusted so that the slit is halffilled with fluid I (the more dense phase) and half filledwith fluid II (the less dense phase).

The fluids are flowing sufficiently slowly that noinstabilities occur-that is, that the interface remains

exactly planar. It is desired to find the momentum-flux and velocity

distributions.


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A differential momentum balance leads to the following

differential equation for the momentum flux (assume that

no change of velocity profile at z direction):

.

This equation is obtained forboth phase I and phase II.

Integration of Eq. 2.5-1 for the two regions gives

.


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momentum flux xzis continuous through the fluid-fluid

interface

BC. 1, at x = 0, xzI= xz

II (2.5-4)

Therefore, C

I

= CII


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shell momentum balances and velocity distributions in laminar flow

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