EKC314: TRANSPORT PHENOMENA Core Course …chemical.eng.usm.my/notes/HEKARL/notes/EKC314_Transport Phenomena.pdfEKC314: TRANSPORT PHENOMENA Core Course for B.Eng.(Chemical Engineering)

EKC314: TRANSPORT PHENOMENACore Course for

B.Eng.(Chemical Engineering)Semester II (2008/2009)

Dr. Mohamad Hekarl [email protected]

School of Chemical Engineering

Engineering Campus, Universiti Sains Malaysia

Seri Ampangan, 14300 Nibong Tebal

Seberang Perai Selatan

Penang. EKC314-SCE – p. 1/54

Syllabus

1. Introduction and Concepts

2. Momentum Transport

Viscosity and and mechanism of momentumtransport

Newtonian and non-newtonian fluidsFlux and gradients

Velocity distribution in in laminar and turbulent flows

Boundary layer

Velocity distribution with more than oneindependent variables

EKC314-SCE – p. 2/54

Syllabus

3. Equation of changes in Isothermal Systems;

Interphase transport in isothermal system.

Macroscopic balances for isothermal systems.

EKC314-SCE – p. 3/54

Introduction and Concepts

What are Transport Phenomena?1. Fluid dynamics2. Heat transfer3. Mass transfer

They should be studied together since:1. they occur simultaneously2. basic equations that described the 3 transport

phenomena are closely related3. mathematical tools required are very similar4. molecular mechanisms underlying various transport

phenomena are very closely related

EKC314-SCE – p. 4/54


Macroscopic level

Microscopic level

Molecular level

Write down balance equation; macroscopic balances.

Equations should describe; mass, momentum, energy and angular momentum within the system.

No need to understand the details of the system.

Examination of the fluid mixture in a small region within the equipment.

Can write down the equation of change that describe mass, momentum, energy and angular momentum change within the small region

Aim is to get information about velocity, temperature, pressure and concentration profile within the system.

To seek a fundamental understanding of the mechanism of mass, momentum, energy and angular momentum in terms of molecular structure and intermolecular forces

Involve some theoretical physics and physical chemistry work EKC314-SCE – p. 5/54


Mixture of gases

Macroscopic

Microscopic

Molecular

Heat added to the system

EKC314-SCE – p. 6/54


The flow of fluid are studied in 3 different parts whichconsist of:

flow of pure fluids at constant temperature-withemphasis on viscous and convective momentumtransport

flow of pure fluids with varying temperature-withemphasis on conductive, convective and radiativeenergy transport

flow of fluid mixtures with varying composition-withemphasis on diffusive and convective mass transport

EKC314-SCE – p. 7/54


Conservation Laws (mass):

Consider two colliding diatomic molecules (N2 and O2).

The conservation of mass can be written as;

mN + mO = m′

N + m′

O

for a system with no chemical reaction;

mN = m′

N

andmO = m

′

O

EKC314-SCE – p. 8/54


Conservation Laws (momentum):

According to the law of conservation of momentum, thesum of the momenta of all atoms before collision mustequal that after the collision.

The conservation of momentum can be written as;

mN1rN1

+ mN2rN2

+ mO1rO1

+ mO2rO2

= m′

N1r′N1

+ m′

N2r′N2

+ m′

O1r′O1

+ m′

O2r′O2

EKC314-SCE – p. 9/54



rN1is the position vector for atom 1 of molecule N and

rN1is its velocity.

It can be written as;

rN1= rN + RN1

rN1is the sum of the position vector for the centre of

mass and the position vector of the atom w.r.t thecentre of mass.

EKC314-SCE – p. 10/54



Also RN2= −RN1

Then the conservation equation can be simplified as;

mN rN + mOrO = mN r′N + mOr′O

EKC314-SCE – p. 11/54


How to study Transport Phenomena?

Read the text with pencil and paper in hand-work through details of the

mathematical developments

Refer back to any maths textbook just to brush up on culculus, differential

equations, vectors etc.

Make a point to give a physical interpretation of key results-get into habit of relating physical ideas to the

equations

Ask whether results seem reasonable. If they do not agree with intuition, it is important to find out which is correct

Make a habit to check the dimensions of all results. This is a very good way

to locate errors in derivations

EKC314-SCE – p. 12/54


Recap: Vector Calculus

If given a vector of a form;

a = [a1, a2, a3] = a1i + a2j + a3k

consists of the real vector space R3 with vector

addition defined as;

[a1, a2, a3] + [b1, b2, b3] = [a1 + b1, a2 + b2, a3 + b3]

and scalar multiplication defined by;

c[a1, a2, a3] = [ca1, ca2, ca3]

EKC314-SCE – p. 13/54


The dot product of two vectors is defined by;

a · b = |a||b|cosγ = a1b1 + a2b2 + a3b3

where γ is the angle between a and b. This gives thenorm or length |a| of a with a formula;

|a| =√

a · a =√

a21 + a2

2 + a33

If a · b = 0 therefore a and b orthogonal.

EKC314-SCE – p. 14/54


Example of a dot product is given by;

W = p · d

which is work done by a force p in a displacement d.

EKC314-SCE – p. 15/54


The cross product v = a × b is a vector of length;

|a × b| = |a||b|sinγ

and perpendicular to both a and b such that a, b, v

form a right-handed triple.

This can also be written in the form of;

a × b =

i j k

a1 a2 a3

b1 b2 b3

The cross product is anticommutative a× b = −b× a

and not associative.EKC314-SCE – p. 16/54


For a vector function given by

v(t) = [v1(t), v2(t), v3(t)] = v1(t)i + v2(t)j + v3(t)k

Then the derivative is;

v′ =dv

dt= lim

∆t→0

v(t + ∆t) − v(t)

∆t

therefore;

v′ = [v′

1, v′

2, v′

3] = v′

1i + v′

2j + v′

3k

EKC314-SCE – p. 17/54


Vector function r(t) can be used to represent a curve C

in space.

Then r(t) associates with each t = t0 in some intervala < t < b the point of C with position vector r(t0).

The derivative r′(t) is a tangent vector of C.

If a vector in a Cartesian coordinate is given by;

v(x, y, z) = [v1(x, y, z), v2(x, y, z), v3(x, y, z)]

= v1(x, y, z)i + v2(x, y, z)j + v3(x, y, z)k

EKC314-SCE – p. 18/54


Therefore, the partial derivative of v can be obtainedby;

∂v

∂x=

(

∂v1

∂x,∂v2

∂x,∂v3

∂x

)

=∂v1

∂xi +

∂v2

∂xj +

∂v3

∂xk

EKC314-SCE – p. 19/54


Gradient of a function f can be written as;

grad f = ∇f =

[

∂f

∂x,∂f

∂y,∂f

∂z

]

Therefore, using the operation, the directionalderivative of f in a direction of a unit vector b can beobtained by;

Dbf =df

ds= b · ∇f

EKC314-SCE – p. 20/54


Divergence of a vector function v can be written as;

div v = ∇ · v =∂v1

∂x+

∂v2

∂y+

∂v3

∂z

And the curl of v is given as;

curl v = ∇× v =

i j k∂∂x

∂∂y

∂∂z

v1 v2 v3

EKC314-SCE – p. 21/54


Some basic formula for grad, div and curl;

Grad:∇(fg) = f∇g + g∇f

∇(f/g) = (1/g2)(g∇f − f∇g)

Div:div(fv) = fdiv v + v · ∇f

div(f∇g) = f∇2g + ∇f · ∇g

EKC314-SCE – p. 22/54


Div and Grad (Laplacian):

∇2f = div(∇f)

∇2(fg) = g∇2f + 2∇f · ∇g + f∇2g

Div and Curl:

curl(fv) = ∇f × v + fcurl v

div(u × v) = v · curl u − u · curl v

EKC314-SCE – p. 23/54


Extra:curl(∇f) = 0

div(curl v) = 0

EKC314-SCE – p. 24/54


Example 1:Let a particle A of mass M be fixed at point P0 and let aparticle B of mass m be free to take up various positions Pin space. Then A attracts B. According to Newton’s Lawof Gravitation, the corresponding gravitational force p isdirected from P to P0 and its magnitude is proportional to1

r2 , where r is the distance between P and P0. This can bewritten as;

|p| =c

r2

EKC314-SCE – p. 25/54


Hence p defines a vector field in space. Using Cartesiancoordinates, such that P0(x0, y0, z0) and P (x, y, z), thereforethe distance r can be determined as;

r =√

(x − x0)2 + (y − y0)2 + (z − z0)2

For r > 0 in vector form;

r = [x − x0, y − y0, z − z0] = (x − x0)i + (y − y0)j + (z − z0)k

EKC314-SCE – p. 26/54


Therefore;|r| = r

and (−1/r)r is a unit vector in the direction of p. The minussign indicates that p is directed from P to P0. Thus,

p = |p|(

−1

rr

)

= − c

r3r = −c

x − x0

r3i − c

y − y0

r3j − c

z − z0

r3k

EKC314-SCE – p. 27/54


Example 2:From the former example, by the Newton’s Law ofGravitation, the force of attraction between 2 particles isgiven as;

p = − c

r3r = −c

(

x − x0

r3i +

y − y0

r3j +

z − z0

r3k

)

where r is a distance between two particles P0 and P ofthe given coordinates. Thus,

r =√

(x − x0)2 + (y − y0)2 + (z − z0)2

EKC314-SCE – p. 28/54


The important observation now is

∂

∂x

(

1

r

)

=−2(x − x0)

2[(x − x0)2 + (y − y0)2 + (z − z0)2]3/2= −x − x0

r3

Similarly;∂

∂y

(

1

r

)

= −y − y0

r3

and∂

∂z

(

1

r

)

= −z − z0

r3

EKC314-SCE – p. 29/54


From here, it is observed that p is the gradient of the scalarfunction;

f(x, y, z) =c

r

and f is a potential of that gravitational field. ApplyingLaplace Equation of the form;

∂2f

∂x2+

∂2f

∂y2+

∂2f

∂z2= 0

EKC314-SCE – p. 30/54


Applying this to the unit vector, gives for the individualcomponent;

∂2

∂x2

(

1

r

)

= − 1

r3+

3(x − x0)2

r5

∂2

∂y2

(

1

r

)

= − 1

r3+

3(y − y0)2

r5

∂2

∂z2

(

1

r

)

= − 1

r3+

3(z − z0)2

r5

EKC314-SCE – p. 31/54


Combining the 3 gives;(

∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)

c

r= 0

or it can be written in the form of;

∇2f = 0

which is normally termed as the Laplacian operator of afunction.

EKC314-SCE – p. 32/54


Example 3:From Example 2, it is known that the gravitational force p isthe gradient of the scalar function f(x, y, z) = c

rwhich

satisfies Laplace’s equation ∇2f = 0. Therefore,

div p = 0

for r > 0.

EKC314-SCE – p. 33/54


Example 4:Consider a motion of a fluid in a region R having nosources and sinks in R (no points at which the fluid isproduced or disappeared). Assuming that the fluid iscompressible and it flows through a small rectangular boxW of dimension ∆x, ∆y and ∆z with edges parallel to thecoordinate axes. W has a volume of ∆V = ∆x∆y∆z.Let v is the velocity vector of the form given by;

v = [v1, v2, v3] = v1i + v2j + v3k

EKC314-SCE – p. 34/54


Setting

u = ρv = [u1, u2, u3] = u1i + u2j + u3k

where ρ is the density of the fluid. Consider the flow of fluidout of the box W is through the left face whose area is∆x∆z and the components v1 and v3 are parallel to thatface and contribute nothing to that flow. Thus, the mass offluid entering through that face during a short time interval∆t is approx. given by;

(ρv2)y∆x∆z∆t = (u2)y∆x∆z∆t

EKC314-SCE – p. 35/54


And the mass of fluid leaving the opposite face of the boxW, during the same time interval is approx. given by;

(u2)y+∆y∆x∆z∆t

Therefore, the difference is in the form of;

∆u2∆x∆z∆t =∆u2

∆y∆V ∆t

where∆u2 = (u2)y+∆y − (u2)y

EKC314-SCE – p. 36/54


Two other pairs in x and z directions are taken andcombined in the form given by;

(

∆u1

∆x+

∆u2

∆y+

∆u3

∆z

)

∆V ∆t

The loss of mass in W is caused by the rate of change ofthe density and therefore equals to;

−∂ρ

∂t∆V ∆t

EKC314-SCE – p. 37/54


Equating both equations gives;(

∆u1

∆x+

∆u2

∆y+

∆u3

∆z

)

∆V ∆t = −∂ρ

∂t∆V ∆t

Diving the above equation by ∆V ∆t and let ∆x, ∆y, ∆zand ∆t approaching 0 leads to;

(

∂u1

∂x+

∂u2

∂y+

∂u3

∂z

)

= −∂ρ

∂t

EKC314-SCE – p. 38/54


Or in its simplified form;

div u = −∂ρ

∂t

or

div(ρv) = −∂ρ

∂t

which consequently becomes;

∂ρ

∂t+ div(ρv) = 0

which is the condition for the conservation of mass orthe continuity equation of a compressible fluid flow!

EKC314-SCE – p. 39/54


If the flow is steady, or it is independent of time, thus;

∂ρ

∂t= 0

and therefore the continuity equation reduces into

div(ρv) = 0

If the density ρ is constant, which means the fluid isincompressible, therefore the continuity equation becomes;

div v = 0

EKC314-SCE – p. 40/54

Viscosity and the Mechanisms ofMomentum Transport

Consider a parallel plates with area A separated bydistance Y with a type of fluid;

The system is initially at rest

At time t = 0, the lower plate is set in motion in thepositive x direction at a constant velocity V .

As time proceeds, the fluid gains momentum and thelinear steady-state velocity is established.

At steady motion, a constant force, F is required tomaintain the motion of the lower plate and this can beexpressed as;

F

A= µ

V

YEKC314-SCE – p. 41/54


Y

V

x

y vx(y)

vx(y,t)

t < 0

t = 0

Small t

Large t

Fluid initially at rest

Lower plate set in motion

Velocity build up in unsteady flow

Final velocity distribution in steady flow

V

V

EKC314-SCE – p. 42/54


Consider a parallel plates with area A separated bydistance Y with a type of fluid;

Force in the direction of x perpendicular to the ydirection given by F/A is replaced by τyx.

The term V/Y is replaced by −dvx/dy and thereforethe equation becomes;

τyx = −µdvx

dy

This means that the shearing force per unit area isproportional to the negative of the velocity gradient i.e.Newton’s Law of Viscosity.

EKC314-SCE – p. 43/54


Similarly, let the angle between the fixed plate and themoving boundary δγ and the distance from the origin isδx.

Therefore,

tan δγ =δx

Y

for a very small angle,

δx

Y= δγ

EKC314-SCE – p. 44/54


Butδx = V δt

thus,

δγ =V δt

Y

Taking limit on on the above terms;

limδt→0

δγ

δt=

V

Y=

dv

dy

EKC314-SCE – p. 45/54


Hence;δγ

δt∼ dv

dy

Butγ ∼ τ

Thus,

τ ∼ dv

dy

EKC314-SCE – p. 46/54


Which then becomes

τ = −µdv

dy

with µ as the proportionality constant.

The minus sign represents the force exerted by thefluid of the lesser Y on the fluid of greater Y.

EKC314-SCE – p. 47/54


Shear thinning

Newtonian

Shear thickening

τ

dγ/dt

EKC314-SCE – p. 48/54


Generalisation of Newton’s Law of Viscosity.

Consider a fluid moving in 3-dimensional space withrespect to time t.

Therefore, the velocity components are given asfollows;

vx = vx(x, y, z, t) vy = vy(x, y, z, t) vz = vz(x, y, z, t)

In the given situation, there will be 9 stresscomponents, τij.

EKC314-SCE – p. 49/54


Based on the given diagram:

The pressure force-always perpendicular to theexposed surface.

Thus, the force per unit area on the shaded surface willbe a vector pδx (pressure multiplied by the unit vectorδx in the x direction)

Same goes to the y and z directions.

EKC314-SCE – p. 50/54


The velocity gradients within the fluid are neitherperpendicular to the surface element nor parallel to it,but rather at some angle to the surface.

The force per unit area, τx exerted on the shaded areawith components (τxx, τxy and τxz).

This can be conveniently represented by standardsymbols which include both types of stresses(pressure and viscous stresses);

πij = pδij + τij

where i and j may be x, y or z.

EKC314-SCE – p. 51/54


Here, δij is the Kronecker delta, which is 1 if i = j andzero if i 6= j.

The term πij can be defined as;force in the j direction on a unit area perpendicularto the i direction, where it is understood that thefluid in the region of lesser xi is exerting the forceon the fluid of greater xi.flux of j-momentum in the positive i direction – thatis, from the region of lesser xi to that of greater xi.

EKC314-SCE – p. 52/54


Summary of the components of the molecular stresstensor.

Dir. Vector forceComponents of forces

x-component y-component z-component

x πx = pδx + τx πxx = p + τxx πxy = τxy πxz = τxz

y πy = pδy + τy πyx = τyx πyy = p + τyy πyz = τyz

z πz = pδz + τz πzx = τzx πzy = τzy πzz = p + τzz

EKC314-SCE – p. 53/54

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EKC314: TRANSPORT PHENOMENA Core Course …chemical.eng.usm.my/notes/HEKARL/notes/EKC314_Transport Phenomena.pdfEKC314: TRANSPORT PHENOMENA Core Course for B.Eng.(Chemical Engineering)