Lecture Notes of MSE 250 (Spring 2014)
INTRODUCTION TO MATERIALS KINETICS AND TRANSPORT PHENOMENA
YANFEI GAO, [email protected]
Department of Materials Science and Engineering University of Tennessee, Knoxville
MSE 250 – Spring 2014 2
CONTENT
1. MATHEMATICAL REVIEW 4
1.1 Motivation 4 1.2 Vectors and Tensors 4 1.3 Differential Operator and Divergence Theorem 6 1.4 Curvilinear Coordinates 9 1.5 Differential Equations 10
2. MOMENTUM TRANSFER 12
2.1 Fluid Flow Characteristics 12 2.2 Newton’s Law of Viscosity 13 2.3 Mass Balance and Equation of Continuity 15 2.4 Momentum Balance Equation 19 2.4.1 Special case: pipe flow 21 2.5 Problem Formulation and Dimensionless Parameters 25 2.6 Momentum Boundary Layer and Transfer Coefficient 28 2.7 Summary 32
3. HEAT TRANSFER 33
3.1 Fourier’s Law of Heat Conduction 33 3.2 Energy Balance Equation 33 3.2.1 Dimensionless parameters 36 3.3 Thermal Boundary Layer and Energy Transfer Coefficient 37 3.3.1 Heat transfer correlations 39 3.4 Example Problems 40 3.5 Thermal Buoyancy Convection 47 3.6 Summary 49
4. MASS TRANSFER 51
4.1 Ficks’s Law of Diffusion 51 4.2 Species Mass-Balance Equation 52 4.2.1 Dimensionless parameters 54 4.2.2 Example problems 55 4.3 Transport Phenomena in a Rising Film 59 4.4 Diffusion Boundary Layer, Mass Transfer Flux and Coefficient 62 4.5 Solutal Buoyancy Convection 65 4.6 Kirkendall Effect and Inter-Diffusion 66 4.7 Summary 67
5. TRANSPORT AND MATERIALS PROCESSING 69
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5.1 Similarities of Three Transport Phenomena and Chilton-Colburn Analogy 69 5.2 Solidification and Phase Transformation 72 5.3 Thermocapillary Convection 75 5.4 Surface Treatment 76
6. CHEMICAL REACTION KINETICS 79
6.1 Thermodynamics and Kinetics 79 6.2 Chemical Equilibrium 80 6.3 The Rate of Reaction 87 6.4 Catalysis 92 6.5 Summary 97
APPENDICES
A. Numerical Methods (provided separately) A.1 MATLAB basics and exercise A.2 Solving differential equations A.3 Using commercial software B. Designing a Cooling Tube 98 C. Homework Solutions (provided separately) D. Sample Midterm Exam (provided separately) E. Sample Final Exam (provided separately)
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1. MATHEMATICS REVIEW Mathematical descriptions of transport phenomena involve vectors, tensors, differential equations, and curvilinear coordinates such as cylindrical and spherical ones. A brief review here summarizes the Appendix A in the textbook by Kou.
1.1 Motivation From university physics course, we know the difference between speed and velocity. The former is a scalar, and the latter is a vector. In a flow field, the velocity depends on the position and time, so that it is a velocity field. We have learned from MSE 201 that there are tensile stress and shear stress. Therefore, we use a tensor (or a matrix) to describe all the stress components.
In many material phenomena, we often encounter a number of physical quantities that can be classified into scalars, vectors, and tensors. If we want to determine some quantities from the knowledge of other quantities, we must use some physics principles (such as kinematics, kinetics, material behavior), and also appropriately pose a solvable problem from these principles and the boundary/initial conditions. The central theme of this course is to explain this procedure. What is translated into is usually a set of ordinary or partial differential equations (ODE or PDE), a knowledge of which is a fundamental technique for engineering students. It should be noted that this course is more on physics principles and problem formulation; solving differential equations is not our concern – there are a lot of tools and software available.
1.2 Vectors and Tensors Usually, we are concerned with a 2nd order tensor, which can be represented by a matrix. For example, stress tensor, etc. In textbooks and papers, vectors and tensors are usually given in bold face. In handwriting, we use tilde underneath a symbol to denote vectors and tensors.
order/rank name example 0 scalar physical quantities of magnitude only, e.g., temperature,
energy, density, etc. 1 vector magnitude and direction, e.g., displacement, velocity,
force, etc. 2,3,… tensor multiple components and multiple directions
Vector is commonly represented by a directed line segment whose length represents the magnitude and whose orientation in space represents the direction. In the three-dimensional space, the Cartesian (rectangular) coordinates can be constructed by the basis vectors, xe , ye ,
and ze , which have the following properties:
(1) magnitude 1x y z e e e
(2) directions along x, y, and z axes, respectively
(3) orthonormal to each other
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An arbitrary vector v can be represented by its projections on these basis vectors:
x x y y z zv v v v e e e ,
where xv , yv , and zv are the components. For convenience, a vector can be written as
, ,x y zv v vT
v , where “T” means transpose.
Scalar product (or called dot product)
The dot product of two vectors is defined by
cos , v w v w v w ,
where denotes the length of a vector, and ,v w is the angle formed by v and w . This
operation has the commutative property (i.e., v w w v ), and the distributive property (i.e.,
u v w u v u w ). Note that associative property for arithmetic operation,
x y z x y z , does not make sense for dot product. In the expanded form, since 1x x e e ,
0x y e e , etc., we get
, ,
x x y y z z x x y y z z
x
x x y y z z x y z y
z
v v v w w w
w
v w v w v w v v v w
w
v w e e e e e e
Vector product (or called cross product)
The cross product of two vectors is defined by
sin , v w v w v w n ,
where n is a unit vector. The three vectors, v , w , and n form a right-handed screw. In the expanded form, since x y z e e e , y x z e e e , 0x x e e , etc., we get
x
y
z
ez
ey
ex
v
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det
x x y y z z x x y y z z
y z z y x z x x z y x y y x z
x y z x y z
x y z x y z
x y z x y z
v v v w w w
v w v w v w v w v w v w
v v v v v v
w w w w w w
v w e e e e e e
e e e
e e e e e e
where det is the determinant of a matrix. Expansion of det can be obviously seen from
the above equation.
In the three-dimensional case, a second-rank tensor has 9 components, which can be arranged into a 3x3 matrix:
xx xy xz
yx yy yz ij
zx zy zz
τ ,
where i denotes the row index, and j denotes the column index.
The index notation is easier to manipulate and memorize. The dot product of a tensor and a vector is given by
3
1
xx xy xz x xx x xy y xz z
yx yy yz y yx x yy y yz z ij jj
zx zy zz z zx x zy y zz z
v v v v
v v v v v
v v v v
τ v
Note that for arbitrary v , τ v v τ , unless Tτ τ .
Tensor product
The tensor product of two vectors is defined as
x x x y x z
y x y y y z i j
z x z y z z
v w v w v w
v w v w v w v w
v w v w v w
vw
It is clear that Twv vw .
1.3 Differential Operator and Divergence Theorem
Differentiation of a function is defined by
, ,0
, ,lim |y z fixedx
f x y z f
x x
. Then we can
arrange the partial derivatives into a vector, denoted by f :
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x y z
f
xf f f f
fy x y z
f
z
e e e ,
which is called the gradient of the scalar function f, since it is the slope vector. We can define the differential operator ( , del, nabla, gradient) by
x y z
x
y x y z
z
e e e
Treating as a vector leads to the following representations:
divergence
( )x y z x x y y z z
yx z
v v vx y z
vv v
x y z
v e e e e e e
,
curl ( ) ( ) ( )
x y z
y yx xz zx y z
x y z
v vv vv v
x y z y z z y x y
v v v
e e e
v e e e ,
Laplacian 2 2 2
22 2 2
f
xf f f f
f fx y z y x y z
f
z
,
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yx z
j yx z
i
yx z
vv v
x x xv vv v
x y y y
vv v
z z z
v
3
1
yxxx zx
ij xy yy zy
i i
yzxz zz
x y z
x x y z
x y z
τ
The divergence theorem (or called Gauss theorem) in the three-dimensional space is represented by
A
d dA
v v n ,
where n is the outward surface normal of the control volume . If we view v as the velocity field, then the left hand side (LHS) represents the volume integral of the divergence of the velocity, and the right hand side (RHS) represents the surface integral of the normal velocity. The physical meaning of the RHS is the net flow out of the control volume through the surface A . Since the mass is conserved, the mass exchange on the surface (i.e., RHS) is balanced by the chance in the volume (i.e., LHS), thus leading to the name of “divergence” for v .
The general Gauss theorem is given by replacing by n and reducing the integration by one dimension. For example,
v
n
Ω
dA
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A
sd sdA
n ,
A
d dA
τ n τ .
Equivalently, we can replace n in a surface integral by , and change the integral to volume.
The Gauss theorem in 1D corresponds to the Netwon-Lebniz relationship, namely,
b
a
dfdx f b f a
dx ,
since “surface normal” at boundary x b is +1, and that at x a is -1.
1.4 Curvilinear Coordinates Besides the Cartesian coordinates, other systems can be chosen, clearly for convenience purpose. The most commonly used ones are cylindrical and spherical coordinates. The radial displacement vector,
x y zx y z r e e e ,
will be represented by:
Cylindrical coordinates:
cos
sin
x r
y r
z z
Spherical coordinates:
sin cos
sin sin
cos
x r
y r
z r
In cylindrical coordinates, note that r and r have different meanings.
x
y
z
ez
ey
ex
r er
eθ ez
θ r z
y
z
er
eθ
eφ θ
r
cylindrical coordinates spherical coordinates
φ
φ: azimuth angle θ: zenith angle φ: longitude π/2-θ: latitude
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It should be noted that, in cartesian coordinates, each point in the space has the same set
of basis vectors. This means that ... 0x x
x y
e e
. However, in cylindrical and spherical
coordinates, the basis vectors depend on the position. This can be illustrated in the following
picture, leading to r
e
e and r
e
e .
In this course, whenever curvilinear coordinates are used, always look up the expanded forms in the textbook. For instance, using Table A.5.1 ~ A.5.3 in textbook Pages 617-618, we get
1 1r zrv v v
r r r z
v .
1.5 Differential Equations Differential equations are equations that relate functions and their derivatives. We restrict our attention to linear algebraic equations:
(a) linearity: 2
0 1 2 3 4 2...... 0
f f f fa f a a a a
x y z x
(b) algebraic:
0
0
0
yxxx zx
xy yy zy
yzxz zz
x y z
x y z
x y z
Example 1 (A.6-1, textbook page 618)
2
20
d T
dz , with 1T T at 1z z and 2T T at 2z z
x
y
r
er eθ
θ
polar coordinates (2D)
er+der
eθ+deθ
dθ
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The general solution of this differential equation is 1 2T c z c , where 1c and 2c are constants to
be determined by the boundary conditions. Clearly two boundary conditions are needed. The final solution is
1 2 1 2 2 1
1 2 2 1
T T T z T zT z
z z z z
.
Example 2
2
2
T T
t x
, 0 iT x t T x , initial condition
1 1
2 2
,
,
T x x t T t
T x x t T t
, boundary condition
The solution method for differential equation is not required. The general solution will be provided to you in homework and exams. However, you should know the following:
Any differential equation has infinite number of solutions, and the general solution has undetermined constants.
Our goal is to find a particular solution that satisfies the prescribed initial/boundary conditions.
The initial/boundary conditions are given so that the solution will be unique. For instance, in the above example, if we remove one initial or boundary condition, there will be infinite number of solutions. Also if we add one more boundary condition, it is redundant.
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2. MOMENTUM TRANSFER (FLUID FLOW)
2.1 Fluid Flow Characteristics Let’s first take a look at the Reynolds experiment (circa 1880), which illustrates the characteristics of momentum transfer in fluid flow. Two types of fluid flow can be observed:
laminar flow: two adjacent fluid elements will flow together in a layer-by-layer fashion;
turbulent flow: inertia effect is significant, and the motion of these fluid elements is random.
Without going into mathematical details, we can tell the following:
(1) fluid velocity, V increasing V more likely to get turbulent flow
(2) fluid density, increasing larger inertia, so more likely to have turbulent flow
(3) viscosity, increasing more “sticky”, meaning stronger constraints on a fluid
element by its surroundings, so less likely to get turbulent flow.
(4) length scale, D This is related to the so-called boundary effect. The larger the solid body, the more likely to get turbulent flow at the location faraway from the head.
Consequently, the characteristics of fluid flow can be determined by the Reynolds number
ReD
D V
,
in which subscript “D” corresponds to the length D. This number is dimensionless. Once we have characterized the relationship between flow behavior and Reynolds number, which is usually represented as engineering diagrams, we can use it for design purpose. For the cylindrical tube flow, the transition from laminar to turbulent flow occurs at ,Re 2100D crt ; for flow over a
sphere, the transition occurs at ,Re 1 10D crt .
dye lines laminar
turbulent
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2.2 Newton’s Law of Viscosity The Reynolds number basically describes the competition between the inertia force and the force acting between fluid elements. Without the latter, there will be no “constraints” to the random motion. The interaction between neighboring fluid elements is the viscous force.
Referring to the above schematic, the shear force per unit area, acting on the upper layer (B) by the lower layer (A), is given by
zdv
dy .
Since 0zdv dy , the force acting on B due to A is thus positive (i.e., in the z direction, as we
would intuitively expect). You will soon see that this is stress component yz .
The above relationship between viscous stress and velocity gradient gives the 1D form of the Newton’s law of viscosity, which is phenomenological (i.e., based on extensive measurements). The Newtonian fluid is the fluid that obeys this law. Since the unit of stress is
2N m , and the unit of velocity gradient is 1zdv dy s , the unit of viscosity, , is
2
Ns Pa s
m .
A fluid with 0 is called inviscid fluid.
In 3D, we define the stress tensor:
xx xy xz
yx yy yz ij
zx zy zz
τ ,
where the first index, i , denotes the surface that has norml ie , and the second index, j , denotes
that the force is in the direction of je .
moving plate
y
z
A
B
vz
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It is very important to notice that the stress component, ij , is the force per unit area
acting on the surrounding by this differential element. That is, the differential volume
element exerts a force T, ,yx yy yz dA on the surrounding. Therefore, the Newton’s law of
viscosity in this textbook has the negative sign. In some other books, the convention is different.
The Newton’s law of viscosity in 3D is given by
0 02
0 03
0 0
.
yx x x x z
y y yz
z z
vv v v v v
x x y x z x
v v vv
y y y z
v vsymm
z z
v
τ v
v
When 0 v , the above equation can also be written as T τ v v .
We also not the following definitions: dynamic viscosity , Pa s ; and kinematic
viscosity
, 2m s .
Application of Newton’s law of viscoity
If we know the velocity field, we can determine the stress and force acting on the surroundings, by using the Newton’s law of viscosity. In the internal pipe flow,
Given: 22
2[1 ( ) ]z
Q rv
R R , where Q is the volume flow rate
Question: calculate the friction force on the tube
x
τyy
τyx
τyz
y
z
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Solution method: use the Newton’s law of viscosity
Step (1): choose appropriate coordinates, which is obviously cylindrical
Step (2): Refer to Table 1.1, and decide which equation is relevant
Eq. [F] in Table 1.1-2, textbook page 13: [ ]z rrz
v v
r z
,
Step (3): Substitute the velocity field into the Newton’s law of viscosity
2
2 2 4
2 4(1 )z
rz
v Q d r Qr
r R dr R R
,
Step (4): Consider a fluid element near the wall, and the shear stress acting on the wall is
3
4rz r R
Q
R
,
so that the friction force (positive since it’s in z direction) is
2
82z rz r R
LQF RL
R
.
2.3 Mass Balance and Equation of Continuity Mass balance is based on the conservation of mass, i.e., the mass added and removed from a “control volume” should be equal to the net change of the total mass. In the following schematic, the fluid flow is visualized by a set of streamlines, which are a family of curves showing the pattern of fluid flow. The slope of the curves is in the direction of the velocity vector at every point.
The surface area can be divided into
wallA , where mass exchange does not occur 0 v n
inA , where mass flows into 0 v n
outA , where mass flows out of 0 v n
r
z
τrz(r)
vz(r)
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Considering a differential element dA with outward unit normal vector n , we get
Outward mass flow rate through dA : dA v n
Inward mass flow rate through dA : dA v n
where v v n n has no contribution to the mass exchange at the surface.
The principle of mass conservation gives
Rate ofRate of Rate of
mass = - sink sourcemass in mass out
accumulation
which can be written as
in outA A
d dA dAt
v n v n .
Clearly, we do not need to distinguish inA and outA , thus leading to
A
d dAt
v n integral form
We can also present the principle of mass conservation in the overall form:
( ) ( )av in av out
dMV A V A
dt overall form
where M is the total mass in the control volume, and avV is the average flow rate (not signed).
Using the Gauss theorem, we can convert a surface integral into a volume integral,
v
n
Ω
dA
wall
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( )A A
dA dA d
v n n v v ,
so that the integral form of the mass conservation can be written as
0dt
v .
Because the integral equation is valid for arbitrary , the integrand should be zero at everywhere, i.e.,
0t t
v v v differential form
which is also called equation of continuity.
It can be shown that the incompressible fluid (i.e., .const ) is divergence free (i.e., 0 v ). The reverse is also true.
Example: “Creeping” flow around a sphere ( Re 1R )
Given:
3
3
3 11 cos
2 2
3 11 sin
4 4
0
r
R Rv v
r r
R Rv v
r r
v
y
z
θ
v∞
φ
x
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Questions:
(1) validate that this velocity field satisfies the equation of continuity (assuming const )
(2) calculate stress tensor using the Newton’s law of viscosity
Solution method: use equation of continuity and Newton’s law of viscosity
Answer:
(1) Spherical coordinates are used, and the equation of continuity is given in textbook Page 32. We need to prove
22
1 1 1( ) ( sin ) ( ) 0
sin sinrr v v vr r r r
v ,
in which
3 3
2 22 2 2 2
cos1 1 3 1 3( ) 1 cos 2
2 2 2 2r
vR R Rr v v r R R
r r r r r r r r
3
3
sin1 1 3 11 sin sin
sin sin 4 4
3 11 2sin cos
sin 4 4
v R Rv
r r r r
v R R
r r r
The above two terms cancel each other, so that we confirm that 0 v .
(2) Using Newton’s law of viscosity in Page 14, and only considering one component give
1 r
r
V Vr
r r r
,
in which
3 3
2 4 2 4
1 3 1 1 3sin sin
4 4 2
v R R R Rr rv v
r r r r r r r r r
3 3
3 2 4
1 3 1 cos 1 31 sin
2 2 2rv R R R R
v vr r r r r r r
Consequently, the shear stress component r is
4
3sin
2r
v R
R r
.
This positive quantity denotes that the shear stress is along the longitude from south to north.
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2.4 Momentum Balance Equation The derivation of momentum balance equation almost entirely follows the procedure in Section 2.3. Considering the schematic shown below, we can write down the description of the momentum balance:
Rate ofRate of Rate of
momentum sum of forces= momentum - momentum
accumulation acting oninto out of
in
It should be noted that such a representation of Newton’s law of motion follows the Euler view,
while Lagrangian view follows an object. The left-hand-side is represented by dt
v .
Considering a differential element dA with outward unit normal vector n , we get
v
n
Ω
dA
wall
Outward volume flow rate through dA : dAv n
Outward mass flow rate through dA : dA v n
Outward momentum flow rate through dA : dA v v n
Inward volume flow rate through dA : dA v n
Inward mass flow rate through dA : dA v n
Inward momentum flow rate through dA : dA v v n
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Consequently, the rate of momentum exchange at the control surface can be written as
in outA A A
dA dA dA v v n v v n v v n .
Next, we consider the forces acting on the control volume, including pressure force, viscous force, body force, and/or others.
The pressure force acting on the control volume by the surrounding is
p
A
p dA F n ,
where negative sign means the pressure force is in the opposite direction of n .
The viscous force is calculated from viscous stress and area. Referring to the following schematic, we know that the viscous force acting on the surrounding at the surface with yn e is
0
1
0
xx xy xz xy
yx yy yz yy
zx zy zz zy
dxdz dxdz
In general, the viscous force acting on the surrounding at a differential surface element with outward normal n is dAn τ -- called Cauchy rule. We will not prove this in this course. Therefore, the viscous force exerted on by the surrounding is (note τ is symmetric)
v
A
dA F τ n .
A body force is the force acting on the body of the control volume, such as gravity, electromagnetic force, etc. This can be written as
x
τyy
τyx
τyz
y
z
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b bd
F f ,
where bf is the body force density. For example, b f g for gravity force, and L f J B for
Lorentz force where J is the current density vector and B the magnetic flux vector.
In summary, the momentum balance equation is
b
A A A
d dA p dA dA dt
v v v n n τ n f integral form
Using divergence theorem, this equation can be converted into
bd d pd d dt
v vv τ f ,
0bpt
v vv τ f . differential form
Furthermore, for constant and , using Newton’s law of viscosity and the equation of continuity (i.e.,), the momentum balance equation can be written as
2bp
t
v
v v v f . Navier-Stokes equation
The expanded forms are given in Table 1-5, textbook pages 53-55.
In the derivation of the Navier-Stokes equation, we have used the property, vv v v v v , which can be proved from its component form. The conversion
from 2 τ v is less obvious, but can be proved as follows. From T τ v v ,
we get
3 3
1 1
2 23 3
2 21 1
ki k ii
k kk k i k
k i i
k ki k k k
v v
x x x x
v v v
x x x x
τ
where the last step uses 3
1
0k
k k
v
x
v .
2.4.1 Special Case: Pipe Flow
For the pipe flow, we make the following assumptions: (1) flow is perpendicular to the cross sections at inlet and outlet (i.e., //v n ), and (2) is uniform. Taking the entire pipe volume as the control volume, the momentum balance equation is given by
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, , , ,
in out
in out p v b
A A
in av in av in out av out av out p v b
dv dA v dA
dt
a v A a v A
P
v v F F F
v v F F F
,
where P is the total momentum, in
in Av v n ,
outout A
v v n (which are therefore unsigned), and
the correction factor a is defined according to
1
1 1
1 1
A
A AA A A
av av
v dAv dA A vdA dA
vdA dA
Aav
vv v
v
v
namely,
1
av
A
v vdAA
,
A v dAa
vdA dA
v
v.
For a circular pipe flow, we get
2
20
2
20
12
12
R
R
v r rdrR
a
v r rdrR
,
so that
laminar flow, 2
1r
v rR
, 4
3a
turbulent flow, 1 7
1r
v rR
, 1a
uniform flow, v r const , 1a
This simply means that mean square is not equal to square mean. For example, consider three
numbers 1, 2, and 6, so that we will get
2 2 21 2 63
21 2 63
41
27a
.
Example 1: Thrust on a pipe bend (textbook, page 41)
Consider a uniform flow through a pipe bend of uniform cross-section A, with constant flow rate Q, Determine the force acting on the bend by the fluid flow at steady state.
The control volume is taken as the dashed area, . At steady state, 0d dt , so that
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2 2 20 cos sinx x yAv Av Av e e e F .
The force acting on the bend (i.e., surrounding) by the fluid is
2 21 cos sinx yAv Av F F e e
Example 2: Impinging jet flow (textbook, page 38)
Horizontal turbulent liquid jet is directed normal to a vertical plate. The subsequent flow is parallel to the plate. Determine the pressure force acting on the plate due to the fluid flow at steady sate. (Note that if the problem does not specify gravity, don’t consider it.)
Using the momentum balance in the z-direction gives
20 0 zAv F ,
so that the force acting on the plate is 2
2z z
QF F v A
A , where Q Av .
Example 3: Problem 1.4-6 (textbook, page 105)
Choosing the control volume as shown in Fig. P1.4-6 (Page 107) and using the mass balance equation (Page 28 and 29, textbook), we get the following:
v
Ω
z
vin
vout
θ
Ω x
y
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The mass flow rate of the incoming turbulent liquid jet is 214 jD v
The mass flow rate of the jet leaving the orifice is 224 jD v
The mass flow rate of the deflected flow along the cone surface is: 2 21 24 4j jD v D v
.
The velocity of the deflected flow can be decomposed into a radial component sinjv ,
and z -component cosjv . Therefore, the corresponding momentum flow out of is
2 21 2( )( cos sin )
4 4j j j z j rD v D v v v e e
The momentum-balance equation (1.4-8, textbook page 38) in the ze direction gives
2 2 2 21 2 1 20 ( ) ( ) ( ) cos 0 0
4 4 4 4j j j j j j j pzD v v D v v D v D v v F
where we neglect viscous force and gravity force. Consequently, the z-component of the force acting on the cone by the liquid is
2 2 21 2( )(1 cos )
4z pz jF F v D D .
Example 4: Problem 1.4-9, textbook page 107
This problem is the same as 1.4-2 (page 103) except that the dish moves at a constant velocity dv in the z-direction. The Hint in textbook is “Let the control volume move with the
dish”. This is equivalent to choosing a new coordinate z* (fixed on the dish):
*dz z v t .
Since dv is a constant, *z is also an inertial coordinate system.
In *z coordinate, the incoming flow velocity is j dv v . Similar to Example 3, the mass
flow rate of the deflected liquid is
2 21 2( ) ( )
4 4j d j dD v v D v v .
The momentum-balance equation for the moving in the *z direction gives
2 21 2
2 21 2
04 4
cos4 4
j d j d j d j d
j d j d j d pz
D v v v v D v v v v
D v v D v v v v F
MSE 250 – Spring 2014 25
so that the z-component of the force acting on the dish by the liquid is
2 2 21 2( ) ( )(1 cos )
4z pz j dF F v v D D
2.5 Problem Formulation and Dimensionless Parameters Here we summarize Sections 2.2-2.4, giving three sets of governing equations:
Newton’s law of viscosity
2
23
xxx
v
x
v , T τ v v ,
yxxy yx
vv
y x
Mass balance (equation of continuity)
0A
d dAt
v n , 0
t
v
Momentum balance
b
A A A
d dA p dA dA dt
v v v n n τ n f
2bp
t
v
v v v f
Denote the characteristic velocity by V and the characteristic length by L , so that the characteristic time is L V . In the Navier-Stokes equation:
Inertia force: 2v V
t L
Viscous force: 22
Vv
L
Gravity force: bf g g
Consequently, we can define the following dimensionless parameters:
Reynolds number: 2
2
inertia forceRe
viscous force
V L VL LV
V L
Froude number: 2 2inertia force
Fr =gravity force
V L V
g gL
MSE 250 – Spring 2014 26
The physical meaning of these numbers is obvious. For example, a large Reynolds number means that the viscous force can be neglected when compared to the inertia force, so that the fluid flow is more likely to be turbulent.
The three sets of governing equations, plus appropriate boundary/initial conditions, will lead to a solvable problem. We use the following example to illustrate this solution procedure.
Laminar flow through a vertical tube (textbook, page 64)
Given: steady state
laminar flow
incompressible Netwonian fluid
long tube of radius R and length L
Question: determine the velocity distribution
Procedure (you can also refer to the diagram in textbook, page 61):
Step (i): Choose cylindrical coordinates
Step (ii): steady state 0t ; ignore end effects; and are constants
Step (iii): Equation of Continuity
0t
v Table 1.3-1, textbook page 32
1 1
( ) ( ) ( ) 0r zrv v vt r r r z
Since =const and 0rv v (from symmetry argument), we get
0zv
z
z zv v r
If zv depends on z, the fluid will not be able to maintain a constant density.
r
z
τrz(r)
vz(r)
L
MSE 250 – Spring 2014 27
Step (iv): Equation of motion + Newton’s law of viscosity = Navier-Stokes Equation
Using Table 1.5-2, in the cylindrical coordinates, the z-component of the N-S equation is
2 2
2 2 2
1 1z z z z z z zr z z
vv v v v v v vpv v r g
t r r z z r r r r z
0( )1 Lz p p gLdvd dpr g
r dr dr dZ L
Step (v): The governing equation is an ordinary differential equation. The commonly used boundary conditions are given in §1.5.3, textbook page 57, including: (1) no-slip boundary condition at the wall, (2) symmetry boundary condition, and (3) free surface condition.
For the fluid flow in the circular tube, we get
0zv , at r R
0zdv
dr , at 0r .
For a general description of boundary conditions, please refer to textbook.
Step (vi): Solving the ODE,
0 2 2
4L
z
p p gLv R r
L
which is a parabolic distribution.
Step (vii): The shear-stress distribution can be calculated from the velocity field, using the Newton’s law of viscosity. The only nontrivial term is
0( )
2Lz
rz
p p gLdvr
dr L
.
Step (viii): The average velocity is
20( )
8L
av
p p gLv R
L
,
and the volume flow rate is
40[( ) ]
8Lp p gL
Q RL
,
which is called Hagen-Poiseuille law. Since 4Q R , if your blood vessel is blocked, so that, say
2R R , in order to keep the same Q, your blood pressure needs to be 16 times larger. Of course, this is an overestimate, since the blood vessel is elastic.
MSE 250 – Spring 2014 28
2.6 Momentum Boundary Layer and Transfer Coefficient
External flow: Consider a fluid flow with uniform velocity v approaching a stationary
flat plate in the flow direction. Because of viscosity, the velocity near the plate reduces from v
to 0. The transition region is called the momentum boundary layer (Prandtl, 1904). Its thickness is taken as the distance where 99%zv v .
With increasing distance from the leading edge, the effect of this boundary is more significant. In other words, a larger region of fluid will be disturbed. If the length is sufficiently long, the flow characteristics in the boundary layer changes from laminar to turbulent. The transitioned length
corresponds to a critical Reynolds number, 5Re 2 10z
v z .
Suppose we can solve the Prandtl’s boundary layer problem, giving
3
3 1
2 2zv y y
v
,
4.64z
v
.
The local friction coefficient is
0
2
| 0.6461 Re2
yz yfz
z
CV
, laminar flow, 5Re 2 10z ,
where Rez z v . Experimental measurements for turbulent flow give
1 50.00592 Refz zC , turbulent flow, 5 75 10 Re 10z
v∞
transition
turbulent boundary layer
laminar boundary layer
y z
MSE 250 – Spring 2014 29
Internal flow (textbook page 23, Fig. 1.1-16): Consider a fluid flow into a circular pipe. A laminar boundary larger develops at the entrance. For the fully developed flow, we have learned that the velocity profile is parabolic. Typically the length of the developing flow is negligible.
Internal pipe flow
Friction factor is defined by
0 02
( ) ( )
2L L
av
p gh p gh Df
v L
,
We have solved this problem for laminar flow, giving
20( )
8L
av
p p gLv R
L
,
so that
64 64
Reav D
fD v
.
For turbulent flow, ~ ReDf has to be determined experimentally. Such a relationship is
called the Moody’s diagram, which also includes the effect of surface roughness.
1 4 4
1 5 4
64 Re , Re 2100
0.316Re , 2100 Re 2 10
0.184Re , Re 2 10
D D
D D
D D
f
The last two lines are simply curve fitting – the power exponents do not have particular meaning.
MSE 250 – Spring 2014 30
Flow past a sphere
Similar to the circular pipe flow, one can solve the problem of laminar flow around a sphere. The derivation is much more lengthy, and it does not contain more information than our circular pipe-flow example. To this end, let’s directly jump to the final result.
The pressure force acting on the sphere by the fluid is
342
3pzF R g RV ,
the viscous force is
4vzF RV .
The total force is
346
3zF R g RV ,
where the first term is the Archimedes buoyancy force (since it is related to gravity), and the second term is called kinetic force (Stokes’ law).
The so-called drag coefficient is given by
MSE 250 – Spring 2014 31
2 2 2
6 241 1 Re2 2
DD
D
F RVC
A V R V
,
where ReD
D V
. This is only valid for laminar flow.
Similar to the Moody’s diagram, we can get a chart in textbook page 85, giving
3 5
5
24 Re , Re 1
18.5Re , 2 Re 500
0.44, 500 Re 2 10
D D
D D D
D
C
.
Again the last two lines are simply curve fitting. The relationship of Re ,D DC roughness is very
useful in designing golf balls.
Now let’s consider some applications. In the micro- and nano-fluidics, we get
3
2 1 1
1 1 1Re 1 2100
10
L V m g cm m s
gcm s
laminar flow
2 2
52
(1 / )10 1
9.8 1r
V m sF
gL m s m
inertia force >> gravity force
In addition, using the Stokes’ law, for micro-particles,
3
6 18F1
4 Re3
D r
s
F Rv
F R g
Example: 1.8-1, textbook page 881
Given: copper cooling tube, L, D, h
water, ,
applied pressure, in outp p
Question: water flow rate?
friction factor?
Solution method: using Moody’s diagram.
The relationship between ~ ReDf can be written as Re ~ ReD D f , where
1 We will present a thorough review of this problem in our Appendix B.
MSE 250 – Spring 2014 32
0 0
2Re [( ) ( )]D L L
D Df p gh p gh
L
.
In this problem, ReD f can be calculated from knowns, so that the Reynolds number can be
determined. Thus the water flow rate v can be calculated. Then we can use Moody’s diagram to get ReD and f .
2.7 Summary What are required?
(1) Be able to explain how to formulate a solvable problem (e.g., circular pipe flow).
(2) Understand the physical meanings of each term in the governing equations, as well as the Reynolds and Froude numbers.
(3) Understand the correlation between friction/drag coefficients and Re number.
What are not required?
(1) You don’t need to memorize these equations. But be able to find them in book.
(2) Solving differential equation is not required.
(3) It’s optional to you to read examples in the textbook that are not covered in my note. The exams will be on the level of homework, and will be limited to the lecture note.
Additional Note:
From engineering point of view, solving the governing equations in lecture note page 25 is rarely done analytically because of complex geometry. One can use engineering software to deal with these problems. But such numerical methods have difficulties in dealing with compressible fluid, turbulent flow, fluttering, etc. As we have seen from Section 2.6, we rely on engineering diagrams such as Moody’s work for typical problems such as pipe flow.
MSE 250 – Spring 2014 33
3. HEAT TRANSFER Discussion in this chapter is similar to that in Chapter 2, so that we will not spend too much time here. We have learned before that there are three types of heat transfer:
Conduction: heat transfer due to the temperature gradient, regardless of fluid motion;
Convection: heat transfer across a moving fluid in which a temperature gradient exists;
Radiation: heat transfer between two surfaces at different temperatures by means of electromagnetic waves in the medium in between. Since it rarely occurs in materials processing, radiation will not be discussed in our course.
3.1 Fourier’s Law of Heat Conduction Consider a one-dimensional bar with a temperature gradient, and define the heat flux, q , as the amount of heat transferred per unit area per unit time,
22
Jq W m
m s .
Experimentally, one can find that the larger the temperature gradient, the larger the heat flux. Given a temperature gradient, steel conducts “much faster” than wood. Consequently, it is intuitive to derive a phenomenological law:
y
dTq k
dy ,
where k is the thermal conductivity (a material property), and k J m s K . This is the
Fourier’s law of heat conduction, i.e., the counterpart of the Newton’s law of viscosity. In 3D,
k T q ,
where , ,x y zq q q q . Refer to Table 2.1-2 (textbook page 120) for expansion in cylindrical and
spherical coordinates.
We can also define the thermal diffusivity, , by
V
k
C
,
where is the material density and VC is the specific heat (under constant volume).
3.2 Energy Balance Equation The derivation of energy balance equation almost entirely follows the procedure for the derivation of mass balance and momentum balance equations. Considering the schematic shown below, the 1st law of thermodynamics states that:
MSE 250 – Spring 2014 34
Rate of Rate of Rate ofRate of Rate of
energy energy energy= energy carried energy carried
accumulation conducted conductedby flow into by flow out of
in into out of
Rate ofRate of
work doneheat generation
on thein
surrounding
The energy density is 21
2t Ve C T v , where is the potential energy per unit mass.
Consequently, the total energy is t tE e d
, and the left hand side is te dt
.
Considering a differential element dA with outward unit normal vector n , we get
v
n
Ω
dA
wall
Outward volume flow rate through dA : dAv n
Outward mass flow rate through dA : dA v n
Outward momentum flow rate through dA : dA v v n
Outward energy flow rate through dA : te dA v n
MSE 250 – Spring 2014 35
Consequently, the rate of energy exchange at the control surface by convection can be written as
in out
t t t
A A A
e dA e dA e dA v n v n v n .
For the contribution due to heat conduction, if 0 q n , thermal energy is conducted out, and the heat conduction rate is dAq n . If 0 q n , the rate of heat conduction into is
dA q n . Consequently, the total rate of energy exchange by conduction is
* *in out AA A
dA dA dA q n q n q n .
The condition of 0 q n corresponds to the adiabatic/insulated surface.
The pressure is in the opposite direction of n . Therefore if 0 v n , p dAv n is the work (per unit time) done by the fluid on the surrounding. If 0 v n , the surrounding does the work (per unit time) of p dA v n on the fluid. The total rate of pressure work on the surrounding is
out inA A A
p dA p dA p dA v n v n v n ,
which will decrease the energy in .
If the fluid flows over a turbine, the shaft work done by the fluid on the turbine is sW .
The viscous force acting on the surrounding due to the fluid is dAτ n , so that the rate of viscous work done by the fluid on the surrounding is
A A
dA dA τ n v τ v n .
This equality is due to the symmetric property of τ .
The rate of heat generation is written as S sd
.
Consequently, the energy balance equation is
Inward volume flow rate through dA : dA v n
Inward mass flow rate through dA : dA v n
Inward momentum flow rate through dA : dA v v n
Inward energy flow rate through dA : te dA v n
MSE 250 – Spring 2014 36
t t
A A
s
A A
e d e dA dAt
p dA dA W sd
v n q n
v n τ v n integral form
In most materials processing applications, the kinetic energy and potential energy can be ignored, since the flow velocity is very low. And also in these applications, pressure, viscous, and shaft work can be ignored, since we are not dealing with a hydraulic problem. To this end, the integral energy-balance equation is simplified as
V V
A A
C Td C T dA dA sdt
v n q n ,
which can converted into
V VC T C T st
v q .
Assuming constant VC and k in the Fourier’s law of conduction leads to
0 V V
V V V V
V V
C T C T st
TC C T C T C T s
t tT
C C T k T st
v q
v v q
v
so that
2V
TC T k T s
t
v differential form
or written as
2
V
T sT T
t C
v .
Refer to Table 2.3.1 (textbook page 142) for representations in curvilinear coordinates.
3.2.1 Dimensionless Parameters
From the governing equations,
V V
A A
C Td C T dA dA sdt
v n q n ,
MSE 250 – Spring 2014 37
2V
TC T k T s
t
v ,
we can easily determine:
convection: VC T dA v n , or VC T v with magnitude of VC V T
L
conduction: dAq n , or 2k T with magnitude of 2
Tk
L
Consequently, we can define the thermal Peclet number by
T 2V VC V T L C LVconvective heat transport LV
Peconductive heat transport k T L k
,
and Prandtl number by
PrV
v viscous diffusivity
k C thermal diffusivity
.
Clearly Re PrTPe . Note that the Prandtl number only combines material properties, so that it
does bear similar physical meanings as Re and TPe numbers.
3.3 Thermal Boundary Layer and Energy Transfer Coefficient A uniform fluid flow approaching a semi-infinite plate will lead to the momentum boundary layer. Fluid flow carried thermal energy. Suppose that the faraway temperature is T
and the plate temperature is sT . At the solid/liquid interface, the fluid temperature reaches sT ,
and there is a temperature gradient from sT to T . This region is called the thermal boundary
layer. The thickness T is typically taken as the distance where s
s
T T
T T
=99%. If we solve a
boundary value problem, we will prescribe boundary conditions:
T T and 0dT
dy at Ty .
With increasing distance from the leading edge of the plate, the thickness of the thermal boundary layer increases.
MSE 250 – Spring 2014 38
The solution of this problem under laminar flow condition is quite complicated. We directly write down the solution:
1 3PrT
, Pr
, 4.64Rez
z , Rez
z V
Physically, it means that T is a result of convective heat transfer in competition to the
conductive heat transfer.
Recall the local momentum transfer coefficient, defined by
0
0
yz yf
z y
Cv v
, where 0
0
zyz y
y
v
y
.
Similarly, the heat transfer coefficient is defined by
0y y
s
qh
T T
, where 0
0
y yy
Tq k
y
.
Or we can rewrite these equations into:
0y sy
q h T T , for flow over a plate,
r s avr Rq h T T
, for pipe flow.
The above representation is also called the Newton’s law of cooling.
v∞
momentum boundary layer y
z
T∞
thermal boundary layer y
z TS
δT
MSE 250 – Spring 2014 39
3.3.1 Heat Transfer Correlations
In general, convective heat transfer can be determined by solving the governing equations for fluid flow and heat transfer. These theoretical predictions, however, are valid only in limited conditions – correlations are better verified and determined experimentally. These correlations are useful for studying convective heat transfer.
Forced Convection over a Flat Plate
For the forced convection over a flat plate, the heat transfer coefficient under laminar flow condition can be calculated, given by
3
3 1
2 2s
s T T
T T y y
T T
, 0 3
2
Ty y
s T
k kh
T T
.
Using the relationship between T and , we will get the so-called Nusselt number,
1 3 1 20.323Pr Rez
hzNu
k .
We can also define an average heat transfer coefficient by
0
1 L
L zh h dzL
.
After some numerical manipulations, we get the average Nusselt number,
1 2 1 30.664Re PrLL L
h LNu
k .
The above relations are valid for 0.6 Pr 50 and 5Re 2 10z .
For liquid metals and semiconductors, which are of our interests, we have Pr 1 , so that the above relationships are not valid. An alternative solution is
1 2 1 20.565 Re Pr 0.565z z zNu Pe ,
which is valid for Pr 0.05 and 100Pe .
Forced Convection Normal to a Cylinder
For the flow of air normal to a cylinder of diameter D , one experimental fit gives
1 3Re PrbD DNu a ,
which is valid for 50.1 Re 3 10D and Pr 0.7 . The fitting constants a and b can be found in
a table in textbook (page 169).
Other examples can be found in Section 2.6 in the textbook.
MSE 250 – Spring 2014 40
3.4 Example Problems
For pipe flow, we often use the flow speed at the entrance and exit, so that inv v n
and outv v n . Following the above procedure of deriving energy balance equation, we get
* *
viscous work shaft work ...
tav V av av V avin out
n nin out
av avin out
dEv AC T v AC T
dtq A q A
pv A pv A
or, for simplicity,
tav V av av V avin out
dEv AC T v AC T Q S
dt , overall form
where Q is the rate of heat flow due to conduction, and S is the rate of heat generation.
One hindsight is the definition of the average temperature. Recall that the definition of
average velocity is the ratio of total volume flow rate to the area, 1
avv vdAA
. Similarly, we
get the following:
V
av
V
C TvdAT
C vdA
.
and for constant and VC , av
TvdAT
vdA
. Obviously, in general, 1
avT TdAA
.
Example 1: Heat transfer in a fluid flow through a pipe (textbook, page 133)
In our example in textbook page 88, using Moody diagram, we have found out a relation between applied pressure & the flow rate, given geometry and material properties. The cooling tube is intended to cool down the chamber wall. Here we investigate how this is achieved. Consider a fluid flow through a circular pip wall radius R and length L. the pipe wall temperature
vin vout Ω
MSE 250 – Spring 2014 41
is .T The heat transfer coefficient is U. The mass flow rate is m. The average inlet temperature
is 0T .
Questions: (1) the average outlet temperature LT =?
(2) the rate of heat exchange between the inside and outside at steady state.
Solution:
Choosing 2R dz as the control volume gives the overall energy balance equation:
( ) ( )Tav V av in av V av out
dEv AC T v AC T Q S
dt ,
where Q and S are gains. From the given conditions, the mass flow rate is avm Av , steady
state means 0TdE dt , and there is no heat source, 0S . Consequently,
0 ( )V av V av avmC T mC T dT dQ .
According to the definition of the heat transfer coefficient, r s avr Rq h T T
, so that
2r s avr RdQ A q Rdz U T T
.
Note that we have neglected the heat conduction at the inlet and outlet. Then the governing equation becomes
0 ( ) 2 ( )V av V av av avmC T mC T dT RdzU T T .
The resulting boundary value problem is
2 ( )avV av
dTmC RU T T
dz
0avT T at 0z ,
z
Ω
z+dz
z
Tav
m, T0 m, TL
T∞
MSE 250 – Spring 2014 42
and the solution is
0 0
21 expL
V
RULT T T T
mC
.
The total heat conduction rate between pipe and fluid is determined by
00 V V L totalmC T mC T Q ,
so that 0total V LQ mC T T . The relationship between totalQ and m is shown by the following
picture, plotted in normalized form, i.e., 02totalQ
RUL T T vs
2VmC
RUL. The following picture
has the form of 1 exp 1 1x x .
0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
mCv/2RUL
Qto
tal/2 R
UL
(T
-T0)
Example 2: Counter-flow heat exchanger (textbook, page 135)
Consider a double-pipe heat exchanger of length L. Hot fluid flows through the inner pipe, which is cooled by the cold liquid flowing through the outer pipe. As shown by the schematic, we are provided with 1hT , 2hT , 1cT , 2cT , L and R. Furthermore, we assume uniform
flow, heat transfer coefficient U between the two pipes, and ignore heat conduction between the outer pipe and the surrounding.
Question: find the steady-state heat exchange rate eQ as a function of the inlet and outlet
temperatures and U. (Be careful that the mass flow rates, hm , cm , are not provided.)
MSE 250 – Spring 2014 43
Choose the entire inner pipe as control volume, and we get
1 20 ( )h V h h V h em C T m C T Q .
Choose the entire outer pipe as control volume, and we get
2 10 c V c c V c em C T m C T Q .
Consequently,
2 1 2 1
1 1 1h h c c
h V c V e
T T T Tm C m C Q
.
Now choosing the differential volume as control volume gives rise to
0 ( ) ( 2 )( )
0 ( ) ( 2 )( )h V h h V h h h c
c V c c c V c h c
m C T m C T dT U Rdz T T
m C T dT m C T U Rdz T T
so that
1 1
2h c
h c h V c V
d T TRU dz
T T m C m C
.
Combining the above results gives
2 2 1 1
2 2 1 1
2ln
h c h ce
h c h c
T T T TQ RLU
T T T T
.
z
Ωin
z+dz
z
Tav
mh Th1
mc Tc2
T∞
Ωout
Ωout
mh Th2
mc Tc1
Th1 Tc1
Th2 Tc2
MSE 250 – Spring 2014 44
Example 3: Heat transfer with laminar flow in a tube (textbook, page 160)
Assume that the fluid flow is not affected by heat transfer. The laminar velocity field of an incompressible, Newtonian fluid has been determined before,
2
2
21V
z
Q rv
R R
,
with volume flow rate of VQ .
Step (i): cylindrical coordinates
Step (ii): energy-balance equation (refer to page 142 in textbook)
2 2
2 2 2
1 1v r z
vT T T T T T TC v v k r
t r r z r r r r z
.
Step (iii): 0rv v , zv r , ,T r z .
Step (iv): Since the heat transfer in the z direction is dominated by convection, we can assume
2
2 V z
T Tk C v
z z
,
therefore,
1
z
T Tv r
z r r r
.
Step (v): Boundary conditions (refer to page 144 in textbook) are
0T
r
, at r=0
RT T , at r=R.
For a general description of boundary conditions, please refer to the textbook.
Referring to Sections 2.1.3 and 2.1.4 (textbook), we learned that: in the case of a constant heat flux r r R
q
, the heat transfer coefficient h is constant in the thermally fully developed region,
and the gradient, T z , is independent of r. Consequently, we solve the above differential equation with respect to r, and the solution is
4 2 2 44
3 48
VR
Q TT T R R r r
R z
.
Now find out the representation of avT and r R
T
r
in terms of T
z
:
MSE 250 – Spring 2014 45
2
V
r R
QT T
r R z
,
11
48V
av R
Q TT T
z
.
Consequently, the heat transfer coefficient is
2 48
11 48 2 11
Tr VR r R
R av R av V
k kQ Rq kh
T T T T Q R
,
and the Nusselt number (normalized heat transfer coefficient) is
4.36D
hDNu
k ,
where 2D R . This relation only applies for laminar, thermally fully developed flow with uniform surface heat flux.
The validity of the above equation is restricted to a certain range of Reynolds number and Prandtl number. In general, we get
Re, PrNu Nu
which should be determined experimentally.
Example 4: Heat conduction into a semi-infinite solid (textbook, page 152)
Consider a semi-finite solid, [0, )x , with initial temperature iT . At t=0, the surface at
x=0 is suddenly raised to sT and will be kept at this value for t>0. Find out the temperature
evolution , ?T x t .
This is a solid, so convective heat transfer does not happen. The governing equation can be simplified into
2
2v
T TC k
t x
.
This equation can also be derived by considering the energy balance in a differential element, dx , given by
2
2
V x x dx
x x dx
TC dx q q
tT T
k kx x
Tk dx
x
MSE 250 – Spring 2014 46
The last step uses the Taylor expansion.
For this partial differential equation, we need two boundary conditions and one initial condition:
0,
,
, 0
s
i
i
T x t T
T x t T
T x t T
The solution is not required. It should be noted that this is a standard hyperbolic PDE, and there are many ways to solve it. The solution is
4
s
i s
T T xerf
T T t
,
where the error function is
2
0
2 terf e dt
.
Example 5: Bernoulli equation (textbook, page 128)
This is only for your information, and will not be required in the exams. From the integral form of the energy balance equation,
T
Ts
Ti
increasing time
x
x x+dx
q q+dq
MSE 250 – Spring 2014 47
t t
A A
s
A A
e d e dA dAt
p dA dA W sd
v n q n
v n τ v n
if q, viscous work, shaft work, and s can be neglected, under steady state condition, we get
0t
A
pe dA
v n .
When 1 2T T and 1 2
vA vA , we obtain
2 21 1 1 2 2 2
1 1
2 2v p v p ,
21
2v p const .
3.5 Thermal Buoyancy Convection Refer to textbook, Section 4.2 in page 288. Buoyancy convection is due to the coupling between thermal transfer and momentum transfer. The variation of temperature causes density change, thus driving fluid flow. For example, the increase of temperature leads to the density increase, so that the fluid will rise up (due to gravity), leading to the so-called natural convection.
Define the thermal expansion coefficient as
,
1T
p cT
under constant pressure and composition. The first order Taylor expansion gives
0 0 0( )T T T .
Consequently, the Navier-Stokes equation with buoyancy force is given by
20 0 0( )Tp T T
t
v
v v v g g ,
which, together with
2V
TC T k T s
t
v ,
form a set of coupled equations for the velocity and temperature fields.
MSE 250 – Spring 2014 48
Note that the thermal buoyancy force scales as 0( )T g T T , and the viscous force as 2V L . However, there is no applied velocity field, so that it’s better to use L to replace V .
Thus we can define the thermal Grashof number by
3
0 03 2
T TT
g T T g T T Lthermal buoyancy forceGr
viscous force L
,
and the thermal Rayleigh number by
PrT TRa Gr .
Without explaining this in details, we simply write down the relationship:
, PrTNu Nu Gr .
For a limited parameter regime, analytical solution can be found. Practically, we curve fit to extensive experimental measurements. Refer to Section 4.2.6 in textbook for a compilation of common cases.
Example: Natural convection over a vertical plate
Consider a vertical plate at 360K, 0.3m long by 0.1m wide (i.e., vertical length 0.3L m and 20.03A m ), in the air at 290K. This is a natural convection problem, in which the heated air will rise up because of the reduction of density. The distribution of velocity and temperature in the momentum and thermal boundary layers are given in the following schematic.
This is a very interesting problem that combines both momentum and thermal transfer phenomena, which are also coupled. For the natural convection, ,Nu Nu Pr Gr , meaning that
the heat transfer is determined as long as the Prandtl and Grashof numbers are given.
Particularly for this example, an empirical relationship gives
1 2
1 44
0.902
4 0.861L
L
Nu Pr
Gr Pr
,
in the range of 0.00835 1000Pr and 4 1010 10GrPr . For this problem, we are given 0.703Pr , 2 7 1 39.85 10Tg K m , and 3 1 128.1 10k Wm K . Please compute the heat
transfer coefficient h and the total heat flux Q . (Hint: Compute h from LNu , and use the
definition of h to compute q and Q qA . Only consider one side of the plate.)
MSE 250 – Spring 2014 49
Solution: The given Pr and Gr are in the range of validity of equation
1 2
1 44
0.902
4 0.861L
L
Nu Pr
Gr Pr
. Therefore, we get 55.8LNu . The heat transfer coefficient is
1 15.23L
kh Nu Wm K
L ,
and the total heat flux is
2 2 10.09 5.23 360 290 11.0Q Ah T m Wm K K W .
3.6 Summary Some basic requirements include
Be able to understand the homework problems and sample exam problems.
Understand the physical meaning of each term in the governing equations, as well as the meaning of dimensionless numbers.
Understand the relationship Ref , Re, PrNu , Gr, PrNu .
Ts=360K
T∞=290K
L=0.3m
δ δT
vx
T-T∞
y
x
MSE 250 – Spring 2014 50
Momentum Transfer
Heat Transfer
Newton’s law of viscosity: (1) definition of τ
(2) T τ v v
(3)
Fourier’s law of conduction: (1) definition of q
(2) k T q
(3) Vk C
Equation of Continuity:
A
d dAt
v n
0t
v
n/a
Momentum-Balance Equation
A
b
A A
d dAt
p dA dA d
v v v n
n τ n f
2pt
v
v v v g
Energy-Balance Equation
t t
A A
S
A A
e d e dA p dAt
dA dA W S
v n v n
q n τ v n
2V
TC T k T s
t
v
Average Value 1
av
A
A
v vdAdA
Average Value 1
av V
AV
A
T vC TdAvC dA
Dimensionless Numbers
ReLV
2
FrV
gL
Dimensionless Numbers
TPeLV
Pr
Momentum Boundary Layer
Thermal Boundary Layer
Momentum Transfer Coefficient
0 0
0 0
z
yz y yf
z zy y
vy
Cv v v v
Example: circular pipe (Moody diagram)
Heat Transfer Coefficient
0 0y y y
s s
Tk
q yh
T T T T
Example: circular pipe
MSE 250 – Spring 2014 51
4. MASS (SPECIES) TRANSFER The equation of continuity in Chapter 2 only deals with the total density of the fluid. Here we are interested in the individual species. In analogy to the heat transfer, we get
Diffusion: mass/species transfer in a stationary solid or fluid with a concentration gradient
Convection: mass/species transfer due to fluid flow
For example, it will take a long time for sugar crystals in coffee to diffuse uniformly to the entire cup. Stirring the liquid promotes the convective transfer of sugar to everywhere.
4.1 Fick’s Law of Diffusion
Similar to the Newton’s law of viscosity and the Fourier’s law of conduction, we can postulate a relationship between the “diffusion rate” and the concentration gradient. Define the following physical quantities:
Aw : mass (or weight) fraction of the species A,
A Aw : density of species A
Ayj : diffusion flux in the y direction,
where
amount of material diffused
diffusion flux =unit area × unit time
,
so that 2A
kgj
m s .
The Fick’s law of diffusion (also called Fick’s first law) gives
AAy A
dwj D
dy ,
y
z
wA jAy
MSE 250 – Spring 2014 52
where AD is the diffusion coefficient of species A, and 2AD m s . The vector form in three
dimension is given by
A A AD w j .
You may postulate the other relationship, A
Ay A
d wj D
dy
. Usually, the change of
is negligible, so that A Ad w dy dw dy . In addition, we should regard the Fick’s law as a
phenomenological one. Either form is okay, as long as it fits to the experiments.
The Fick’s law of diffusion in molar representation is
A A AcD x j ,
where
Aj : molar diffusion flux
Ax : mole fraction of species A
c : moles of all species per unit volume
A Ac cx : moles of species A per unit volume
Whether the diffusion flux is represented by Aj or by Aj can be clearly determined from the
context. Therefore, in this course, we do not distinguish them for the sake of simplicity.
Diffusion Mechanism
This has been discussed in MSE 201, including vacancy diffusion and interstitial diffusion. Direct exchange of atoms is almost impossible because of high activation energy.
It should be pointed out that the true “driving force” for diffusion is actually the gradient of the chemical potential. For ideal solution, it degenerates to the concentration gradient.
4.2 Species Mass-Balance Equation The mass balance of species A can be represented by the following equation:
Rate of Rate of Rate of Rate of
species A species A species A species A=
accumulation exchange by exchange by generation
in convection diffusion in
The left hand side is given by Adt
.
MSE 250 – Spring 2014 53
Considering a differential element dA with outward unit normal vector n , we get
Consequently, the exchange rate of species A on the control surface by convection is
A
A
dA v n .
For the contribution due to diffusion, if 0A j n , the species A is diffused out, and the
species transfer rate is A dAj n . If 0A j n , the rate of species A conducted into is A dA j n .
Consequently, the exchange rate of species A on the control surface by diffusion is
v
n
Ω
dA
wall
Inward volume flow rate through dA : dA v n
Inward mass flow rate through dA : dA v n
Inward momentum flow rate through dA : dA v v n
Inward energy flow rate through dA : te dA v n
Inward species flow rate through dA : A dA v n
Outward volume flow rate through dA : dAv n
Outward mass flow rate through dA : dA v n
Outward momentum flow rate through dA : dA v v n
Outward energy flow rate through dA : te dA v n
Outward species flow rate through dA : A dA v n
MSE 250 – Spring 2014 54
A
A
dA j n .
In general, the rate of exchange of
momentum
energy
species
by
viscous force
conduction
diffusion
is
A
A
A
A
dA
dA
dA
τ n
q n
j n
.
The rate of species generation is represented by A AR r d
.
Consequently, the species balance equation is
A A A A
A A
d dA dA r dt
v n j n integral form
which can be converted into
AA A A A Ar r
t
v j n ,
where An is the mass flux.
Assuming constant and AD and using the Fick’s first law and equation of continuity
give rise to
2AA A A AD r
t
v differential form
which is also called Equation of Species Continuity.
The species overall mass-balance equation is
, ,
, ,
Aav A av av A av A Ain out
A av A av A Ain out
dMv Aw v Aw J R
dt
mw mw J R
overall form
4.2.1 Dimensionless Parameters
Here we summarize the key equations involved in the three transport phenomena:
Continuity 0 v (constant )
Momentum Balance 2pt
v
v v v g (constant and )
MSE 250 – Spring 2014 55
Energy Balance 2V
TT T s C
t
v (constant VC and k )
Species Mass-Balance 2AA A A A
cc D c r
t
v (constant and AD )
where
: dynamic viscosity
: kinematic viscosity (viscous diffusivity)
k : thermal conductivity
V
k
C
: thermal diffusivity
AD : species diffusivity
Materials Properties
Prandtl number: viscous diffusivity
Prthermal diffusivity
Schmidt number: viscous diffusivity
species diffusivity A
ScD
"convection" "conduction"
Reynolds number: 2
2
inertia forceRe
viscous forceL
V L LV
V L
Thermal Peclet number:
2
convective heat transferRe Pr
diffusive heat transferV
T
C V T L LVPe
k T L
Solutal Peclet number:
2
convective species transferReSc
diffusive species transferA
sA A A
V L LVPe
D L D
4.2.2 Example Problems
Example 1 (textbook, page 216): Fluid composition in a mixing tank
MSE 250 – Spring 2014 56
We use this simple example to illustrate the use of species overall mass-balance equation. Given: 1m , 1Aw , 2m , 2Aw , 3m , 0M , 0Aw , assuming uniform concentration Aw inside the tank,
and ignoring diffusion at the inlets and outlet, we want to determine M(t)=? and wA(t)=?
Using mass balance equation gives
1 2 3
dMm m m
dt ,
so that 1 2 3 0 0M m m m t M mt M .
Using species mass-balance equation gives
1 1 2 2 3A A A A
dMw m w m w m w
dt ,
so that
0 1 1 2 2 3A
A A A A
dwmt M mw m w m w m w
dt .
Together with the initial condition,
0A Aw w at 0t ,
we can determine Aw t .
Example 2 (textbook, page 234): Self-diffusion of a radioactive isotope transfer
Two solids of pure A are sandwiched together, and in between there is a thin film (with negligible thickness) of radioactive isotope A*. The total quantity of A* per unit area (in the y-z plane) is M . Diffusion takes place at 0t . What is the evolution of concentration profile, i.e.,
* ,Ac x t ?
This is a solid, so convective species transfer does not happen. The governing equation can be simplified into
m1, wA1
m2, wA2
m3, wA
M, wA
MSE 250 – Spring 2014 57
2
* ** 2
A AA
c cD
t x
.
This equation can also be derived by considering the species balance in a differential element, dx , given by
**, *,
2* * *
* * * 2
AA x A x dx
A A AA A A
x x dx
c dxj j
t
c c cD D D dx
x x x
The last step uses the Taylor expansion.
For the above diffusion equation, the associated boundary condition is
* , 0Ac x t ,
and the initial condition is
* , 0Ac x t M x ,
where x is the Dirac delta function,
0, 0
, 0
xx
x
, and 1x dx
.
This is a classic problem, and the solution is
2
***
exp42
AAA
M xc
D tD t
,
which suggests a method to measure *AD . Taking the log function of both sides gives
cA*
t1
t2
t1<t2
x
A A A*
MSE 250 – Spring 2014 58
2
**
log4A
A
xc
D t ,
so that the measurement of concentration profiles (of this radioactive isotope), against 2x t , can be used to determine the material diffusivity. Refer to textbook page 237 for an example.
Example 3 (textbook, page 238): Interstitial diffusion couple with constant AD
Two interstitial alloys of different initial compositions 1Ac and 2Ac are bonded together.
Diffusion starts at 0t . What is the evolution of concentration profile, i.e., ,Ac x t ?
The governing equation is
2
2A A
A
c cD
t x
.
The initial condition is
1 2 1, 0A A A Ac x t c c c H x ,
where H x is the Heaviside function
1, 0
0, 0
xH x
x
The boundary conditions are
1 20,2
A AA As
c cc x t c
,
2,A Ac x t c , 1,A Ac x t c .
The solution is given by
2 1 4
A As
A A A
c c xerf
c c D t
,
where the error function is
2
0
2 terf e dt
.
It should be noted that this problem is equivalent to the problem in page 45 (this lecture note).
MSE 250 – Spring 2014 59
4.3 Transport Phenomena in a Rising Film Now let’s discuss a problem that essentially covers all the three sets of transport phenomena. This was one of the final exam problems in Spring 2006.
Flow of a Rising Film (see textbook page 68)
When being pulled out of a liquid reservoir, a solid will tract a liquid film on its surface. Clearly the viscous force keeps the film from running off the solid (due to gravity force). We assume a uniform film thickness, and neglect the end effects. Find out the steady-state velocity distribution in the film.
Step (i): Choose rectangular coordinates. From symmetry argument, we get
0x yv v , ,z zv v y z .
Step (ii): Using equation of continuity and assuming const give
0zv
z
,
so that z zv v y .
Step (iii): Using equation of motion (Navier-Stokes equation) gives
2 2 2
2 2 2z z z z z z z
x y z z
v v v v v v vpv v v g
t x y z z x y z
,
which will be simplified into
2
2zd v
gdy
.
Together with boundary conditions:
0zdv
dy , at 0y traction-free boundary condition
zv V , at y L no-slip boundary condition
we find out the velocity distribution:
2 2
2z
gv V y L
.
For a general description of boundary conditions, please refer to the textbook.
Heat Loss from a Rising Film (see textbook page 155)
MSE 250 – Spring 2014 60
In addition to the rising film problem, we consider the heat transfer. The liquid temperature is initially iT . When exposed to the gas with temperature fT , the liquid near the free
surface will be cooled. Clearly, the temperature profile is ,T y z .
Using the equation of energy gives
2 2 2
2 2 2V x y z
T T T T T T TC v v v k
t x y z x y z
,
which can simplified into
2 2
2 2z
T T Tv
z y z
.
It should be noted again that the equation of motion and the equation of energy are one-way coupled. That is, the velocity field is solely determined from the equation of motion, and then is used as input for the equation of energy.
Comparing z
Tv
z
and 2
2
T
z
gives the thermal Peclet number, 2 2
zT
v T z LVPe
T z
.
If 1TPe , 2
2
T
z
can be neglected, indicating a more significant convective heat transfer than
y=0
y
z
vz(y)
y=L
V
cover
Gas
Liquid
y=0
y
z
vz(y)
y=L
V
cover
Gas at Tf
Liquid at Ti
T(y,z)
T(y,z)
cA(y,z) cA0
y=0
y
z
vz(y)
y=L
V
cover
Gas A at cA0
Liquid B with cA=0
cA(y,z) cA0
MSE 250 – Spring 2014 61
the conductive heat transfer. This also implies that TPe should be a “vector”, since, in this case,
1TPe does not mean that 2
2z
T Tv
z y
.
According to the textbook page 156, the boundary conditions are
, 0 iT y z T ,
0,
0, f
T y zk h T y z T
y
,
, iT y L z T .
Refer to textbook page 144. The above boundary condition at y=0 is the convection exchange condition. A much better boundary condition involves the solution in the gas, and continuity conditions such as
I II ,T T I II
T Tk k
y y
.
Diffusion into a Rising Film (see textbook page 250)
As the solid is moving upwards, the liquid film is exposed to gas A of concentration 0Ac .
Find out the steady-state concentration distribution of A in the liquid film.
Clearly, the distribution is ,Ac y z . Using the species continuity equation gives
2 2 2
2 2 2A A A A A A A
x y z A
c c c c c c cv v v D
t x y z x y z
,
which can simplified into
2 2
2 2A A A
z A
c c cv D
z y z
.
Similarly, the comparison between Az
cv
z
and 2
2A
A
cD
z
gives the solutal Peclet number,
2 2z A
sA A A
v c z LVPe
D c z D
. If 1sPe ,
2
2A
TD
z
can be neglected, indicating a more significant
convective species transfer than the diffusive species heat transfer.
According to the textbook page 251, the boundary conditions are
, 0 0Ac y z ,
MSE 250 – Spring 2014 62
00,A Ac y z c ,
, 0Ac y L z .
Or we can assume that L is much larger than the regime where Ac varies rapidly from 0Ac to
zero, so that we can use the boundary condition , 0Ac y z .
Refer to textbook page 231. We can use the convection exchange condition, given by
0
0
AA m A A
y
cD k c c
y
.
4.4 Diffusion Boundary Layer, Mass Transfer Flux and Coefficient
Diffusion (or Concentration) Boundary Layer
This is similar to the momentum and thermal boundary layers. Consider a fluid approaching a semi-infinite plate in the flow direction. The plate is at concentration ASw , while it
is Aw at faraway. The coupling of diffusion and convection leads to the concentration profile as
shown above. The layer becomes thicker away from the leading edge. Practically, we define the boundary layer thickness by
99%A AS
A AS
w w
w w
, 0Aw
y
, at cy .
Mass Transfer Flux
We will proceed to make the analogy to momentum and thermal transfer phenomena. That is, we will relate the “degree” of mass transfer to some dimensionless parameters. However, it is quite tricky here. Consider a binary solution. Let Av and Bv be the velocities of species A
and B with respect to the fixed (Lagrangian) coordinates. They are due to both convective flow and diffusion. Therefore, the mass flux of species A is defined by
wA∞
diffusion (or concentration) boundary layer
y z
wAS
δc
MSE 250 – Spring 2014 63
A A A A A n v v j ,
where A A B BA A B B
A B
w w
v v
v v v . This is one big difference between heat and species
transfer phenomena.
Mass Transfer Coefficient
Since Ay A Aj D w y , we want to determine Aj from concentration difference
instead of concentration gradient. So that we define the mass transfer coefficient, mk , by
0Ay m As A m As Ay
n k k w w .
If 0yv , then 0 0Ay Ayy y
n j , and
0Ay y
mAs A
jk
w w
.
For internal pipe flow, ,
Ar r Rm
As A av
jk
w w
, where ,
A
A av
w dAw
dA
v n
v n
Refer to the schematic in page 62 (this lecture note). The solution procedure has been documented in textbook page 222 and page 252, which are
3
3 1
2 2zv y y
v
, 4.64
Rezz
,
33 1
2 2A As
A As c c
w w y y
w w
, 1 3c Sc
,
where ASc D . Therefore the diffusion boundary layer thickness is
1 34.64c
zSc
v
,
and the resulting mass transfer coefficient is
0 3
2
AA
y Am
As A c
wD
y Dk
w w
.
Define the local Sherwood number:
MSE 250 – Spring 2014 64
mz
A
k zSh
D .
For the forced convection over a flat plate, under laminar condition, we get
1 3 1 230.332
2m
z zA c
k z zSh Sc Re
D ,
which is valid for 0.6 50Sc and 5Re 2 10z . For turbulent flow, extensive experiments
give the following empirical relation:
4 5 1 30.0288Rez zSh Sc ,
which is valid for 0.6 50Sc and 5Re 2 10z .
Momentum Transfer Coefficient (textbook §1.8)
External flow over a flat plate:
0
2120
yz yfzc
v
Drag coefficient for flow over a sphere:
21
2
DD
f
Fc
A v
Friction factor for pipe flow:
0 0
212
L L
av
p gh p gh Df
v L
,
which relates to the Reynolds number, Ref , as shown by the Moody’s diagram.
Heat Transfer Coefficient (textbook §2.6)
Definition of heat transfer coefficient is given by
0y y
s
qh
T T
,
which is also called the Newton’s law of cooling. The normalized form is the Nusselt number:
z
hzNu
k .
MSE 250 – Spring 2014 65
4.5 Solutal Buoyancy Convection The Navier-Stokes equation becomes
20 0T s A Ap T T c c
t
v
v v v g g ,
where the thermal expansion coefficient is
,
1T
P cT
,
and the solutal expansion coefficient is
,
1s
A P Tc
.
Taking the characteristic velocity as L , we can define the solutal Grashof number by
3
0 02 2
solutal buoyancy force
viscous forces A A s A A
s
g c c g c c LGr
V L
,
and define the solutal Rayleigh number by
s sRa Gr Sc .
Forced Convection Nu = Nu Pr, Re Sh = Sh Sc, Re
Natural Convection TNu = Nu Pr,Gr
example in textbook page 303
sSh = Sh Sc,Gr
example in textbook page 315
From engineering point of view, we want to find out A
q
j from
A
T
. First, we need
to determine Nu
Sh
from the above table, and calculate m A
kNuh z
k DSh
z
. Second, the
heat/diffusion flux can be calculated by A m A
q h T
j k
.
MSE 250 – Spring 2014 66
4.6 Kirkendall Effect and Inter-Diffusion We have discussed two diffusion mechanisms: vacancy diffusion and interstitial diffusion. Surprisingly, the diffusion mechanisms were only confirmed in 1950s. Before that, the commonly accepted mechanism is the direct exchange of atoms. The milestone work that leads to the development of vacancy diffusion mechanism is the experiment by Kirkendall and his student in 1947.
It is obviously reasonable to assume a constant AD for diffusion of interstitial atom A in
a lattice of larger atoms. However, for diffusion in substitutional solid solution, the inter-diffusion of all species must be considered. In Kirkendall experiment, a diffusion couple of brass (70% Cu + 30% Zn in atomic percent) and pure copper are joined together with molybdenum wires sandwiched in between. Consequently, there will be diffusion of zinc atoms from left to right. If the diffusion mechanism is direct exchange of atoms, there will be equal amount of copper atoms moving from right to left, and therefore the Mo markers will be stationary. However, it is found that Mo markers move to the left. The reason is qualitatively explained as follows. The diffusion mechanism is by vacancies. Zn diffuses more quickly in Cu than Cu does in Zn, i.e., Zn CuD D . Consequently, vacancies are created on the side of the more rapidly
diffusing components, while on the other side the material “swells”. The excess atoms at the right (or the excess vacancies at the left) cause a convection of materials to the left, which carries the Mo markers.
Let’s consider the concentration field of species A,
A Ac cx ,
where c is moles of atoms (A+B) per unit volume, and Ax is mole fraction of species A. The
convection-conduction equations are (remember that A A Ac n v j ):
brass (70% Cu+ 30% Zn) copper
molybdenum markers
MSE 250 – Spring 2014 67
AA A A
BB B B
cc cD x
tc
c cD xt
v
v
which can be rewritten as
A B A B A A B Bc c c c cD x cD xt
v .
Assuming A B A Bc c c c x x const leads to
A A B B A B AD x D x D D x v .
Since Zn CuD D , we get a nontrivial convective flow. For the schematic in the previous page, it
gives 0xv since 0Znx dx .
Combining the above equations give the Darken’s equation:
AA A B A A A
ccx D D x cD x
t
,
AA
ccD x
t
,
where the inter-diffusion coefficient is defined as
B A A BD x D x D .
4.7 Summary Some basic requirements include
Be able to understand the homework problems and sample exam problems.
Understand the physical meaning of each term in the governing equations, as well as the meaning of dimensionless numbers.
Understand the relationship Sh Re,Sc , Sh Gr,Sc .
MSE 250 – Spring 2014 68
MSE 250 – Spring 2014 69
5. TRANSPORT AND MATERIALS PROCESSING
5.1 Similarities of Three Transport Phenomena and Chilton-Colburn Analogy
During our discussion in Chapters 2-4, we have repeatedly highlighted the similarities of these three transport phenomena. We have noticed
zyz
y
AAy A
dv
dy
dTq k
dy
dwj D
dy
which can be described by
Flux ofproportional transport
transport = ×constant property
property
.
The transport-property balance equation is described as follows:
rate of net rate ofrate of rate of
= inflow by + other +accumulation generation
convection net inflow
Example 1: Transport through two layers of materials (textbook, page 282)
The following three problems are similar to each other:
Two immiscible liquids (I and II) of different viscosities are held between two horizontal plates, (a) the lower plate moving at a constant velocity V .
Two plates of different thermal conductivities are in perfect contact, the top surface kept at a constant temperature 1T and the bottom at 2T .
Two plates containing species A and of different diffusivities are in perfect contact, the top surface kept at a constant concentration 1Aw and the bottom at 2Aw .
As shown in the table in textbook page 283, the governing equations are Laplacian:
2
20zd v
dy ,
2
20
d T
dy ,
2
20Ad w
dy ,
with “no-slip” boundary conditions:
MSE 250 – Spring 2014 70
0z topv , z bottom
v V ,
1topT T , 2bottom
T T ,
1A Atopw w , 2A Abottom
w w ,
and “continuity” boundary conditions:
z zI IIv v , z z
I II
dv dv
dy dy ,
I II
T T , I II
dT dTk k
dy dy ,
A AI IIw w , A A
A A
I II
dw dwD D
dy dy .
Another set of commonly used boundary conditions are
0 τ n (free surface)
0 q n (adiabatic (insulated) surface)
0A j n (sealed surface)
Similar expressions can be found for convection exchange conditions.
Chilton-Colburn Analogy
The coefficients of transfer are defined by
coefficient flux at the interface
=of transfer difference in transport property
,
for example,
0
0
yz yfc
v
, and 12
ff
cC
v
0
0
y yq
hT T
,
0
0
Ay ym
A A
jk
w w
.
MSE 250 – Spring 2014 71
Consequently, we write
j ,
0 0
0 0
y yj y
k
.
As we have discussed before, for external laminar flow over a flat plate, at the distance z from the leading edge,
1 20.323Re2
fzz
C , ( Rez
z v
)
1 3 1 20.323Pr Rez
hz
k , ( Pr
)
1 3 1 20.323 Remz
A
k zSc
D . (
A
ScD
)
Consequently,
2 3 2 3 1 21 1Pr 0.323Re
2 Pr Re Refz m
zz A z
C k zhzSc
k D Sc ,
and
2 3 2 3Pr2
fz m
V
C khSc
v C v
.
The boxed equation, known as the Chilton-Colburn analogy, is often written as
2
fzH D
Cj j ,
with Hj : j -factor for heat transfer, and Dj : j -factor for mass transfer.
The Chilton-Colburn analogy for momentum, heat, and mass transfer has been derived here on the basis of laminar flow over a flat plate. However, it has been observed to be a reasonable approximation in laminar and turbulent flow in systems of other geometries, provided no form drag is present. Form drag, which has no counterpart in heat and mass transfer, makes
2fC greater than Hj and Dj , for example, in flow around (normal to) cylinders.
The Chilton-Colburn analogy is useful in that it allows one unknown transfer coefficient to be evaluated from another transfer coefficient which is known or measured in the same geometry.
MSE 250 – Spring 2014 72
5.2 Solidification and Phase Transformation One central line in MSE is the capability to tune the materials microstructure by processing methods. In the textbook, you will see solidification processes in Chapter 6 such as Bridgman crystal growth, floating zone technique, etc. A question that arises naturally is what the condition at the solid/liquid interface is. Answering this question will help us to deal with:
the temperature field that is coupled with evolving microstructure
the effects of temperature gradients on microstructural evolution
the coupling of thermal and momentum transport to enable desirable microstructural development
...
For instance, in the directional solidification, the temperature gradient is forced to be almost along a certain direction, so that the two-phase alloy will develop into lamellae or fiber microstructure.
MSE 250 – Spring 2014 73
Example (1): Solidification of a pure species (textbook page 321)
As schematically shown above, the boundary conditions for the Solid/Liquid interface are
mT T
S fL S
T Tk H k
n n
n R ,
where mT is the melting point, fH is the heat of fusion ( 0fH ), and T n T n . The
second condition, often called Stefan condition, states that the heat conducted from the melt
(liquid) to the interface, L
Tk
n
, and the heat of fusion released at the interface, S fH n R ,
must be conducted away from the interface into the solid, S
Tk
n
, in order to satisfy the
conservation of energy at the interface.
Solid Liquid
n
R T
x
Tm
MSE 250 – Spring 2014 74
Example 2: Solidification of a solution (textbook page 342)
Now consider a binary solution. A the Solid/Liquid interface, the temperature is *T , the solute concentration of the melt *
,A liquidw , and the solute concentration of the solid * *
, 0 ,A solid A liquidw k w , where 0k is the equilibrium segregation coefficient.
Similar to the Stephen condition, we get
* *, 0 ,
A AS A liquid A liquid A A
L S
w ww k w D D
n n
n R ,
which means that the solute released at the interface, * *, 0 ,S A liquid A liquidw k w n R , will be
diffused away into the liquid, AA
L
wD
n
, and into the solid, AA
S
wD
n
. This will help us
draw the concentration profile qualitatively.
Solid Liquid
n
R Aw
x
*0 ,A liquidk w
*,A liquidw
00 ,A liquidk w
0,A liquidw
Aw
T
*0 ,A liquidk w *
,A liquidw 00 ,A liquidk w
0,A liquidw
*T
MSE 250 – Spring 2014 75
It should be noted that temperature *T should be determined from a heat transfer analysis, as shown in the previous example. Thus a length scale is associated with the
temperature gradient. Another length scale is associated with A
TD
T
,
5.3 Thermocapillary Convection Along a liquid/gas interface, significant surface tension gradients are often present due to the presence of temperature and/or composition gradients. These surface tension gradients can induce significant shear stresses and thus fluid flow along the interface – a phenomenon called thermocapillary or Marangoni convection. More details can be found in textbook, Section 5.2.
Consider a rectangular container with left wall at a higher temperature leftT and right wall
at a lower temperature rightT . The bottom wall is insulated. The boundary condition at the
liquid/gas interface is
x Ayx
A
v wT
y x T x w x
,
which can be seen from a free body diagram of the differential element dx . In this example, 0T x (from the boundary condition) and 0T (as for most materials), we will thus
get a positive shear stress, 0yx , which induces a thermocapillary convection as shown below.
Now consider the thermocapillary convection in a vertical liquid bridge. The lower rod is at low temperature, and the upper part at high temperature. The liquid/gas interface has the following boundary condition:
s Ans
A
v wT
n s T s w s
.
For a pure material, 0ns since 0T and 0T s . Thus the liquid is pulled downward
along the free surface. The resulting flow pattern is as depicted schematically.
Tleft hot
x
Tright cold
y
τyx γ+dγ γ
MSE 250 – Spring 2014 76
5.4 Surface Treatment Example 1: Surface carburization/doping or heating
Doping of impurity atoms into the surface is a very useful engineering practice, for example, for semiconductors and surface carburization. When we discuss these problems before, we typically assume a one-dimensional situation. How about the diffusion flux (or equivalent, the heat flux) is localized on the surface?
Using the example in textbook page 461, we consider a semi-infinite solid subjected to a heat source q of radius a on the surface. The governing equation is
2 2 2
2 2 2
T T T T
t x y z
.
Let’s consider an auxiliary problem. That is, there is a point heat source located at point , ,0x y . At time 0t , this point heat source releases instantaneously an amount of heat Q . Therefore, the initial and boundary conditions are
, , ,0 iT x y z T , except at , ,0x y ,
, , , iT t T ,
and the energy conservation requirement is
V iC T T dxdydz Q
.
cold
hot
s n
MSE 250 – Spring 2014 77
The solution is
2 2 2
3 2
, , ,
24 exp
4
i
V
T T Q f x x y y z t
x x y y zQ t
C t
.
If the heat is given continuously at a constant rate starting from time 0t , then linear superposition gives
0
, , ,time
iT T Q f x x y y z t dt ,
where Q is the amount of heat given by the point source per unit time. For a distributed heat source, the solution is
0
, , , ,current time y x
i t y xT T q x y f x x y y z t dx dy dt
.
Example 2: Convection in a weld pool (textbook, page 382)
Consider the convection in the weld pool in arc welding without a filler wire. The first driving force is the buoyancy force. The melt at the center is hotter and hence lighter than the melt at the boundary of the weld zone.
The second driving force is the Lorentz force. The density of the electric current is much higher near the center of the pool surface than near the pool boundary. As such, a Lorentz force is created by the nonuniform electric current field and the magnetic field it induces. The force pushes the melt near the center flowing inward and downward.
The third driving force is the surface tension force. The thermocapillary convection leads to the flow of melt near the center surface towards the surrounding.
MSE 250 – Spring 2014 78
Example 3: Materials processing in space
Obviously the buoyancy convection is completely gone. The handout picture shows that the specimen welded on Earth is that redisual vortices left behind after the electron beam passed were frozen into the grain structures. This typically occurs because the high melting point and thermal conductivity of tantalum result in rapid cooling of the molten region. In the same experiment conducted on Skylab (NASA), the residual vortices were erased in the grain structure of the specimen welded in space.
MSE 250 – Spring 2014 79
6. CHEMICAL REACTION KINETICS As we have discussed in the beginning of this course, MSE 250 replaces ChE 200, Chemical Engineering Fundamentals, and ChE 240, Fluid Flow and Heat Transfer. There are a number of novel materials processing techniques which involve chemical reaction, e.g., chemical vapor deposition. Also modern chemical reactions are typically occurring at surfaces. Surface physics and chemistry is an integral part of MSE, particularly because we are moving towards smaller and smaller objects.
Based on the textbook, “Fundamentals of Chemical Reaction Engineering”, by M.E. Davis and R.J. Davis, McGraw-Hill, 2003, this chapter will offer some basic principles and glance through some examples in chemical reaction.
6.1 Thermodynamics and Kinetics Two fundamental questions include:
What changes are expected to occur in a chemically reacting system?
How fast will these changes occur?
We will use the ammonia reaction to illustrate these two questions. Ammonia is critical to produce nitrogen-rich fertilizers. However, N2 and H2 cannot react “easily”. The iron catalyst is introduced, which is a substance that increases the rate of reaction without being consumed. The atomistic mechanism for
2 2 33 2N H NH ,
is shown below.
Nitrogen molecules are dissociated on the iron surface. One nitrogen atom and three hydrogen atoms form a NH3 molecule and leave the surface.
Chemical equilibrium is the state in which a chemical reaction proceeds at the same rate as the reverse reaction. The rates of the forward and reverse reactions are the same, and the amounts of reactants and products do not change. However, reactions never stop; they are in dynamic equilibrium. The above ammonia reaction can be represented by
312 2 32 2N H NH ,
which is an elementary step (reversible). The rate of reaction can be significantly increased with the introduction of catalysis, but the equilibrium conditions will not be changed.
N2 H2 NH3
MSE 250 – Spring 2014 80
The comparison of thermodynamics and kinetics can also be illustrated by the following two examples. First, in classic mechanics, the equilibrium occurs at the minima on the energy landscape. However, the pathway leading us to the minimum (global or local) is governed by kinetics. We may be trapped in a metastable state (local minimum). Or the rate to reach the equilibrium is so low that practically we are at unstable state. Second, in MSE 201, we have learned the phase diagram and the temperature-time-transformation (TTT) diagram. The equilibrium phases, their compositions, and their relative fractions can be read from the phase diagram, which, however, does not tell us how we reach the equilibrium. Using the Fe-C phase diagram as an example, cooling a γ phase to room temperature, the equilibrium condition is fully-separated α and Fe3C phases. But we can develop a lot of interesting microstructures, such as pearlite and martensite, which are actually not on the phase diagram. In other words, they are not in equilibrium.
6.2 Chemical Equilibrium Consider a closed system with the following quantities:
E: internal energy of the system
P: pressure
V: volume
H: enthalpy, H E PV ,
T: absolute temperature
G: Gibbs free energy, G H TS
dE TdS PdV
dH TdS VdP
dG SdT VdP
For a solution of multiple species, we can write down:
...A A B BdG dn dn SdT VdP ,
where in is the moles of species i, and i is the partial molar free energy of species i, or called
chemical potential. The chemical equilibrium is attained if the total change of Gibbs free energy at fixed P and T is zero.
Thermodynamics of Chemical Equilibrium
Let’s consider a chemical reaction:
312 2 32 2N H NH ,
which can generally be written as
MSE 250 – Spring 2014 81
1
0NCOMP
i ii
A
,
iA : the i -th component
i : stoichiometric coefficient
NCOMP: number of components.
During the reaction, suppose that *n mol of NH3 is produced, so that *12 n mol of N2 and *3
2 n
mol of H2 are used. The change of free energy is thus
3 2 2
* * *312 2NH N HG n n n .
At equilibrium, 3 2 2
312 2 0NH N H . Consequently, the general equilibrium condition is
1
0NCOMP
i ii
G
.
Thus the outstanding question is how to calculate the chemical potential? Generally,
0 lni i iRT a ,
where 0i is the standard chemical potential in a reference state, and ia is the activity of species.
In our course, we limit our interest to i ia X , so that
1
0 0
1 1 1
0 0
1 1
ln ln
ln ln
0
i
NCOMP
i ii
NCOMP NCOMP NCOMP
i i i i i ii i i
NCOMPNCOMP
i i ii i
G
RT a RT X
G RT a RT X
Consequently, under given pressure and temperature,
chemical equilibrium 1
i
NCOMP
X ii
K X const
,
so that xK is called equilibrium constant. For endothermic reactions (i.e., which require thermal
energy), xK increases with the increase of pressure or temperature.
Example 1: Consider the ammonia reaction at 300atm and 723K. The equilibrium constant 2.64XK . The initial concentrations are
20.25NX and
20.75HX . Find out the final
equilibrium concentration.
MSE 250 – Spring 2014 82
Suppose that we start with 1 mol of reactants, and mole of 2N has been reacted at
equilibrium. Thus,
2
2
3
0.25 0.25
0.75 0.75 3
0 2
____________________
1 1 2
N
H
NH
total
and the equilibrium constant is
3
2 2
1 2 3 21 2 3 2
2 1 22.64
0.25 0.75 31 2 1 2
NHX
N H
XK
X X
.
Solving the above equation leads to =0.1312. Therefore, at equilibrium,
2
16%eqNX ,
248%eq
HX , 3
36%eqNHX .
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
time
X
H2
NH3
N2
In practice, how do we increase the reaction efficiency? The only way is to increase XK .
For the example of ammonia reaction, this can be achieved by increasing pressure. It is obvious from the definition of XK , if the equilibrium lies towards the right hand side (products), XK will
be large, and the reaction is called irreversible. A small XK means that the equilibrium lies at the
left hand side, and the reaction almost does not proceed.
MSE 250 – Spring 2014 83
Example 2 (page 10 in the textbook)
where the circle-equal sign means “equilibrated”. CO reacts with the stoichiometric amount of air. At 1atm, 3
1 8.26 10XK and 2 0.7XK . Find out the equilibrium concentration
2 1
12 2 12
2
2 2
1
122
___________________________________
1.88 1.88
0.5 0.5
1 1
0
0 2
___________________________________
3.38 3.38
Species Initial Equilibrium
N
O
CO
CO
NO
total
According to the given condition, 1 mol of CO reacts with some amount of air that has 0.5 mol of O2. The molar fractions of N2 and O2 in air are approximately 79% and 21%, respectively. From the above table, at equilibrium
2 2
2
1
21
1 22
1 1 11 2 2 1 22 2 2
3
2 3.38
1.88 3.38 0.5 3.38
8.26 10
NOX
N O
XK
X X
2
2
2 1 2
12 22
1 21 1 1
2 2 2 1 22 2 2
3.38
1 3.38 0.5 3.38
0.7
COX
CO O
XK
X X
which lead to 1 0.037 and 2 0.19 .
Reaction Network and Intermediates
N2 + O2 2NO
CO + ½ O2 CO2
MSE 250 – Spring 2014 84
As a chemical reaction proceeds, a number of elementary reactions (or called elementary steps) occur, which represent the irreducible molecular events. This is called reaction network. For example, consider the oxidation of methane near 700K and at 1atm. The reaction
4 2 2 22 2CH O CO H O
is not elementary. If we are going to apply the equilibrium constant and to analyze the rate of reaction, we refer to page 6 of the textbook:
4 2 2 2
2 2 2 2
12 22
12 2 2 22
CH O CH O H O
CH O O CO H O
CO O CO
H O H O O
In this reaction network, we have three intermediates:
2CH O : formaldehyde
2 2H O : hydrogen peroxide
CO : carbon monoxide
Page 6 of the textbook also shows that the elementary steps may involve CHO and 2HO .
However, they have short lifetime, so that they are called reactive intermediates. We will not discuss these lines in our class.
Extent of Reaction
Recall the reaction
312 2 32 2N H NH ,
and we can find that
3 32 2 2 2 2 12 1 2 1
312 2 1
NH NHN N H H n t n tn t n t n t n t
.
Motivated by this example, we can define the extend of reaction by
0i i
i
n t nt
.
If there are NRXN reactions and NCOMP components in the system, then
,1
0NCOMP
i j ii
A
, 1, 2,...j NRXN ,
MSE 250 – Spring 2014 85
0,
1
NRXN
i i i j jj
n n
,
where j t is the extend of reaction for the j-th reaction. Clearly, we can use d dt as a
measure of the rate of reaction. It’s also obvious that is extensive. If we double the amount of the mass, will be doubled.
Fractional Conversion
For the ammonia reaction, if there are 1 mole of N2 and 3 moles of H2 at t=0, the maximum allowable amount of NH3 will be 2 moles, i.e., max 2 . If there are 1.1 mol of N2
and 3 mol of H2 at t=0, the maximum allowable amount of NH3 will be 2 mol, i.e., max 2 .
In general, reactants are not initially prepared in stoichiometric amounts, and the reactant in the least amount determines the maximum value of the extent of reaction,
0max0 l ln ,
where l is the limiting reactant.
The fractional conversion is defined as
0
0 0 0max
1l l ll ll
l l l l
t n t n n tf t
n n n
At equilibrium, the extent of reaction, eq is less than max , or equal to max when the reaction
is completely towards the right hand side.
max
max . ., 1
eq
eqX
reversible reaction
irreversible reaction i e K
Alternatively,
0 1eql lf t f .
Molar Expansion Factor
Consider a reaction:
... ...aA bB sS wW
Without loss of generality, we assume that species A is the limiting reactant.
MSE 250 – Spring 2014 86
0 0
0 0
0 0
0 0
0
0 0
0
...
...
______________________
... ...
1
A A A A
bB B A Aa
sS S A Aa
wW W A Aa
I I
total total A A
total A A
n n n f
n n n f
n n n f
n n n f
n n inert components
s w a bn n n f
a
n f
where the molar expansion factor A is defined by
0
00
1 1AA i A i
i itotal A
nX
n a
Example 3 (textbook page 15)
Consider the reaction
12 5 2 4 22N O N O O ,
in a closed chamber (fixed volume) with 2 5
0.5N OX and 2
0.5NX . If the reaction completes,
what will be the pressure increase?
Assuming ideal gas, the pressure is proportional to the total moles.
2 5 2 50
0
1totalN O N O
total
nPf
P n
Since 2 5
1N Of (complete reaction) and
2 5
10.5 1 0.5 1 0.25
1N O
,
the pressure will increase by 25%.
MSE 250 – Spring 2014 87
6.3 The Rate of Reaction
The reaction rate can be described by d dt , which, however, is an extensive quantity. Therefore, it is preferable to define the volumic rate of reaction by
1 1
1
1
i
i
i
i
i i
i i
dn tdr
V dt V dt
d cV
V dt
dc c dV
dt V dt
and
moler
volume time
.
General Properties of the Rate Function for a Single Reaction
Understanding and controlling the rate of reaction is the central problem of the chemical reaction applications. Here we will introduce some empirical relations (see §1.4 of the textbook).
Rule 1 As reaction proceeds, increases and r decreases monotonically.
Rule 2 For an irreversible (one-way) reaction, ,ir k T F c T , where k T is called
rate constant.
Rule 3 The rate constant generally depends on the absolute temperature, T, following the Arrhenius law (1889),
exp gk A E R T ,
where A is a pre-exponential factor, and E is the activation energy. Refer to textbook page 22 for a comparison to experiments.
Rule 4 Frequently, F is only a function of ic , and
Φ
t
r
t
MSE 250 – Spring 2014 88
ii i
i
F c c ,
where the exponent i is called the order of reaction. In general, i i and 2i . The Law
of Mass Action, proposed by Guldberg and Waage (1867), give
i
i ii
F c c (for reactants only).
Rule 5 When a reaction is two-way (i.e., reversible), its rate can generally be expressed bas the difference between the rates in the forward direction and reverse direction,
r r r .
For example, the reaction aA bB sS wW has the following rate:
a b s wA B S Wr k c c k c c .
Unimolecular Reaction
Considering the unimolecular reaction,
A products,
and using the Guldberg-Waage law give
1 1i i
Ai
dn dnr kc
V dt V dt ,
so that
0
, variable
1 , since 1
, const
, const and
AA
A AA A
A
AA
AA A A g
dnkn V
dtdf n
k f fdt n
dckc V
dtdP
kP V c P R Tdt
This is called the first-order reaction system.
Bimolecular Reaction
Using the Guldberg-Waage law for the following reaction,
A B products,
MSE 250 – Spring 2014 89
give A Br kc c , so that
, variable
, const
, const and
AA B
AA B
AA B A A g
g
dnV kn n V
dtdc
kc c Vdt
dP kP P V c P R T
dt R T
Trimolecular Reaction
For the following reaction,
A B C products,
we get
1 A
A B C
dnr kc c c
V dt
2
2
, variable
, const
, const and
AA B C
AA B C
AA B C A A g
g
dnV kn n n V
dtdc
kc c c Vdt
dP kP P P V c P R T
dt R T
Trimolecular reactions are very rare, since the probability of three objects colliding with sufficient energy and in the correct configuration for reaction to occur is low.
Example: Validation of Arrhenius law (textbook page 27 and page 54)
Consider the decomposition reaction:
2 5 2 4 2
1
2N O N O O ,
which occurs in a closed system with constant T and V. By measuring the pressure evolution, we can obtain the reaction constant. The results are given in the following table. Please show the principle of this method, and using these measurements to verify the Arrhenius law.
MSE 250 – Spring 2014 90
1
5
5
4
4
3
___________________
288 1.04 10
298 3.38 10
313 2.47 10
323 7.59 10
338 4.87 10
T K k s
At 0t , suppose that 0n mole of N2O5 is admitted into the closed system. This is a first-
order reacting system, so that
2 5
2 5
N ON O
dnkn
dt ,
or, since 2 5
0
1 N Onf
n ,
1df
k fdt
.
The initial condition is 0 0f t . Therefore,
2 5
1 expN Of kt .
The molecular expansion factor is
2 5
2 5
2 5
0 10 2
00
1 11 1
1 2N O
N O iitotal N O
n n
n n
,
so that
2 5 2 5 2 50
11 1
2total
N O N O N Ototal
nf f
n .
The system is an ideal gas with constant T and V, so that
2 50
0
11 1.5 0.5exp
2total
N Ototal
nPf kt
P n .
Thus the rate constant can be obtained by the pressure measurement. The results, as shown in the above table, can be nicely fit to the Arrhenius law, as shown in the following picture. The curve fitting adopts the method of least squares (i.e., the linear regression method).
MSE 250 – Spring 2014 91
2.8 2.9 3 3.1 3.2 3.3 3.4 3.5 3.6
x 10-3
10-6
10-5
10-4
10-3
10-2
10-1
1/T
k
Exothermic and Endothermic Reactions
Consider a reversible reaction:
1
2
k
kA B S W ,
at equilibrium,
1 1 2 1
2 2
expS WC
A B g
c c k A E EK
c c k A R T
.
If 2 1 0E E , the reaction is exothermic. In order to increase the yield (i.e., a larger CK ), a
low T is preferred. In order to increase the reaction rate, a large T, however, is preferred. So that there is always a trade-off between the equilibrium yield and the reaction rate.
If 2 1 0E E , the reaction is endothermic. Both reaction rate and yield increase with
temperature increase.
Although the Arrhenius law is empirical, it can be supported by the transition-state theory. This is out the cope of our class. Essentially, the transition state during reaction sets up an activation barrier on the energy landscape.
MSE 250 – Spring 2014 92
6.4 Catalysis Let’s first use several examples to illustrate reactions and elementary steps.
Example 1: In this reaction, O is a reactive intermediate. It is an open sequence, in which the reactive intermediate is not produced in other steps.
3 2
3 2 2
3 2
______________
2 3
O O O
O O O O
O O
Example 2: Here O and N are reactive intermediates. It is a chain sequence, since reactive intermediates are reproduced so that a cycle forms.
2
2
2 2
_______________
2
O N NO N
N O NO O
N O NO
Example 3: The atomic site on a catalyst surface, denoted by *, and *O are reactive intermediates. It is a chain sequence, and also a catalytic sequence, since * remains after the reaction.
2 2
2
* *
* *
H O H O
O CO CO
Chain and catalytic reaction cycles provide energetically favorable pathways for the reaction to proceed. As shown in the schematic (next page), the catalytic reaction only requires very low activation energies.
energy
reaction coordinate
endothermic
A+B
S+W
energy
reaction coordinate
exothermic
A+B
S+W
MSE 250 – Spring 2014 93
Ozone Decomposition (textbook page 103)
Now consider the decomposition of ozone,
3 22O O O .
Using the notation in textbook, we get the rate of the direct reaction
3d dr k O O ,
where .. is the number density (molecule/cm3) and dr is in units of molecule/cm3/s. The rate
constant is known in the units of cm3/s/molecule:
11 23001.9 10 expk
T
,
where T is in Kelvin. Obviously, the decomposition of ozone at atmospheric conditions is quite low.
However, the decomposition of ozone dramatically changes in the presence of chlorine atoms (catalyst):
1
2
3 2
2
3 2
________________
2
k
k
Cl O O ClO
ClO O O Cl
O O O
where
111
1405 10 expk
T
cm3/s/molecule
102
2201.1 10 expk
T
cm3/s/molecule
energy
reaction coordinate
O + O3
2 O2 ClO
MSE 250 – Spring 2014 94
The rate of the catalyzed reaction can be analyzed in the following way.
Using the Guldberg-Waage law gives
31 3
1 3 2
21 3 2
d Ok Cl O
dtd ClO
k Cl O k ClO Odt
d Ok Cl O k ClO O
dt
In most cases, it is convenient to make the steady-state approximation for the reactive
intermediates, i.e.,
0d ClO
dt . The detailed derivation is given in textbook page 110, and we
won’t spend time on it. The final result is
1 2 3
1 3 2c
k k O O Cl ClOr
k O k O
.
Assuming 3O O , 33 10Cl ClO O , and 200T K , we get
190c
d
r
r .
Please read the vignette in textbook page 105 for some interesting information.
Heterogeneous Catalysis
Homogeneous catalysis denotes the reaction of species with the spatially uniform distribution. Heterogeneous catalysis usually occurs in the presence of many phases. Typically, it involves:
adsorption of reactants from a fluid phase onto a solid surface
surface reaction of adsorbed species
desorption of products into the fluid phase
The adsorption/desorption processes can be analyzed in a similar way to the chemical reaction.
Consider the isomerization of molecule A to B (textbook page 172),
* *
* *
* *
A A
A B
B B
Oftentimes, one or several elementary steps are the rate-determining step. Other steps are almost in equilibrium.
MSE 250 – Spring 2014 95
(1) If the rate of adsorption is rate determining, then we get
and the reactions are described by
1 1 *r r k A ,
2
*
*a
AK
B , (note a “mistake” here)
0* * *A .
Thus the reaction rate is given by
1 0
2
*
1 a
k Ar
K B
.
(2) If the surface reaction is rate determining, the following sequence is appropriate:
The rate expression for this case is
2 2 *r r k A ,
1
*
*
AK
A ,
3
*
*
BK
B ,
0* * * *A B ,
so that
2 0
1 3
*
1
k Ar
K A K B
.
(3) If the desorption of product is assumed to be rate determining, we get
1
2
3
* *
* *
* *
K
k
K
A A
A B
B B
1
2
* *
* *a
k
K
A A
A B
MSE 250 – Spring 2014 96
and the rate is
3 3 *r r k B ,
1
*
*a
BK
A ,
0* * *B ,
so that
3 1 0
1
*
1a
a
k K Ar
K A
.
The above three cases can be differentiated by the following plots.
Transport-Limited Reaction
Consider a surface reaction with a gradient of concentration near the surface. In ideal case, the reaction rate should be equal to the mass flux, thus,
,s As c A bulk Asr k c k c c ,
where the mass transfer coefficient, ck , can be
obtained from Sherwood number, ,Sh Re Sc .
( ck is the parameter mk in our original definition.)
For transport-limited phenomenon, we
1
2
* *
* *
aK
k
A B
B B
r
[A], when [B]0
case (1)
case (2) or (3)
r
[B]
case (3)
case (1) or (2)
cAs
cA,bulk
catalyst surface
diffusion boundary layer
MSE 250 – Spring 2014 97
have a small ck . Since ck is usually not very temperature dependent, such a limit usually occurs
at high temperatures. On the other limit, a small sk corresponds to the surface reaction limited
phenomenon.
6.5 Summary
Chemical Equilibrium
Equilibrium constant, XK (its derivation from Gibbs free energy is not required)
Elementary step and stoichiometric coefficients
Extent of reaction, 0
i i
i
n t n
and max0 eqt
Fractional Conversion,
0max
1 ll
l
t n tf
n
and 0 1eql lf t f
Molar expansion factor
Reaction Rate
Definition, 1 1 i
i
dndr
V dt V dt
Five rules, particularly Arrhenius law and Guldberg-Waage law
Example on lecture note page 90
Catalysis
Open, chain, and catalytic sequences
Energetics: schematic of energy ~ reaction coordinate
Kinetics (not required in our course)
MSE 250 – Spring 2014 98
Appendix B: Designing a Cooling Tube
Cooling water runs through a tube, which is exposed to a hot environment at high temperature.
Given conditions:
steady state
ignore heat conduction inside the water
material properties , , k
ReDf or equivalent relationship for turbulent flow in smooth pipe
Re, PrNu for turbulent flow in smooth pipe
Tunable parameters:
p , pressure drop
D, tube diameter (or R, radius)
L, tube length
Objective of design:
Our goal is to realize a large rate of heat exchange between the water inside the tube and the environment surrounding the tube.
Solution procedure:
(1) Find out the velocity or mass flow rate from a known pressure drop, material properties, and geometric parameters. (Please refer to lecture note page 31 and textbook page 88-89 for details.)
This is done by using the relationship between ReD and ReD f for turbulent flow in a
smooth pipe, namely,
log Re 0.380 1.115log ReD D f , (1)
where
2
ReD
D pDf
L
. (2)
Consequently, we get first calculate ReD f from geometry, material, and pressure drop
information, and then use the above relationship to find out ReD . Subsequently, the mass flow
rate is
2 Re4 4av Dm D v D . (3)
MSE 250 – Spring 2014 99
(2) Find out the heat transfer coefficient between the pipe flow and the surrounding. (Please refer to textbook page 170-171 for details.)
Here we use the empirical equation [2.6-26] in textbook page 171, giving
8
23
8
Re 1000 Pr
1 12.7 Pr 1
fD
Df
Nu
, (4)
which is applicable for 0.5 Pr 2000 and 62300 Re 5 10D . Consequently, using f and
ReD from the first step, we can compute the Nusselt number. Thus the heat transfer coefficient
can be obtained from
D
kh Nu
D . (5)
(3) Find out the rate of heat exchange between the pipe flow and the surrounding. (Please refer to lecture note page 40-42 or textbook page 133-134 for details.)
The total heat conduction rate between pipe and fluid is given by
0
21 exptotal V
V
RhLQ mC T T
mC
, (6)
where h is the heat transfer coefficient obtained from the previous step.
Properties of Eq. (6) can be shown in the figure in next page. The solid line plots
02totalQ RUL T T against 2VmC RUL . Clearly, with the increase of m (while other
parameters are fixed), the rate of heat exchange increases rapidly when 2VmC RUL is very
small, but slowly when 2VmC RUL is large (when compared to unity). The dashed line plots
0total VQ mC T T against 2VmC RUL . So if the tube surface area, 2 RL , increases, the rate
z
Ω
z+dz
z
Tav
m, T0 m, TL=?
T∞
MSE 250 – Spring 2014 100
of heat exchange will increase. However, a change of R or L will unavoidably affect m, as will be explained later.
0 2 4 6 8 100
0.2
0.4
0.6
0.8
1
mCv/2RUL
Qtotal
/2RUL(T-T
0)
Qtotal
/mCv(T
-T
0)
Example (1): Consider the following parameters:
Geometry: D=0.4cm, L=20m,
Material: =1×10-2gcm-1s-1, =1g/cm3, Pr=7,
k=6×10-3Wcm-1K-1, Cv=4.2Jg-1K-1,
Applied conditions: p =2.758×106gs-2cm-1, 0 100T T K .
Using Eq. (2) gives
3 6 2 1
2 1 1 3
2Re
0.4 1 2 2.758 10 0.4
10 1 2000
1328.6
D
D pDf
L
cm g cm g s cm cm
g cm s g cm cm
The Reynolds number is given by Eq. (1), namely,
MSE 250 – Spring 2014 101
0.380 1.115log ReRe 10 7288D f
D
,
so that the friction factor is
2 2Re 1328.6
0.0332Re 7288
D
D
ff
.
Consequently, the mass flow rate is, according to Eq. (3),
2 1 1
Re4
0.4 10 7288422.9
Dm D
cm g cm s
g s
The Nusselt number is calculated from Eq. (4),
23
8
23
8
0.03328
0.03328
Re 1000 Pr
1 12.7 Pr 1
7288 1000 7
1 12.7 7 1
57.5
fD
Df
Nu
and the heat transfer coefficient is
3 1 1
2 1
6 1057.5
0.4
0.863
D
k W cm Kh Nu
D cm
W cm K
Eventually, Eq. (6) gives
0
1 1 1
2 1
1 1 1
3
21 exp
22.9 4.2 100
2 0.2 0.863 20001 exp
22.9 4.2
9.62 10
total VV
RhLQ mC T T
mC
g s J g K K
cm W cm K cm
g s J g K
W
MSE 250 – Spring 2014 102
Example (2): Based on example (1), what if we double the diameter of the tube, D?
3 6 2 1
2 1 1 3
2Re
0.8 1 2 2.758 10 0.8
10 1 2000
3757.8
D
D pDf
L
cm g cm g s cm cm
g cm s g cm cm
0.380 1.115log ReRe 10 23230D f
D
2 2Re 3757.8
0.0262Re 23230
D
D
ff
2 1 1
Re4
0.8 10 232304
146
Dm D
cm g cm s
g s
23
8
23
8
0.02628
0.02628
Re 1000 Pr
1 12.7 Pr 1
23230 1000 7
1 12.7 7 1
173.8
fD
Df
Nu
3 1 1
2 1
6 10173.8
0.8
1.304
D
k W cm Kh Nu
D cm
W cm K
0
1 1 1
2 1
1 1 1
4
21 exp
146 4.2 100
2 0.4 1.304 20001 exp
146 4.2
6.13 10
total VV
RhLQ mC T T
mC
g s J g K K
cm W cm K cm
g s J g K
W
In this example, the increase of D leads to the increase of m and h, although it might be possible to have a decrease in h. In the representation of totalQ , the increase of m or R or h will lead to
increase of totalQ .
D increases Re increases
D increases m increases
D increases Nu increases more than D increases, so that h increases
MSE 250 – Spring 2014 103
Example (3): Based on example (1), what if we double the pressure drop, Δp?
3 6 2 1
2 1 1 3
2Re
0.4 1 2 5.516 10 0.4
10 1 2000
1879
D
D pDf
L
cm g cm g s cm cm
g cm s g cm cm
0.380 1.115log ReRe 10 10725D f
D
,
2 2Re 1879
0.0307Re 10725
D
D
ff
.
2 1 1
Re4
0.4 10 10725433.7
Dm D
cm g cm s
g s
23
8
23
8
0.03078
0.03078
Re 1000 Pr
1 12.7 Pr 1
10725 1000 7
1 12.7 7 1
84.5
fD
Df
Nu
3 1 1
2 1
6 1084.5
0.4
1.27
D
k W cm Kh Nu
D cm
W cm K
0
1 1 1
2 1
1 1 1
4
21 exp
33.7 4.2 100
2 0.2 1.27 20001 exp
33.7 4.2
1.42 10
total VV
RhLQ mC T T
mC
g s J g K K
cm W cm K cm
g s J g K
W
In this example, the increase of Δp leads to the increase of m and h. In the representation of
totalQ , the increase of m or h will lead to increase of totalQ .
Δp increases Re increases
Δp increases m increases
Δp increases h increases
MSE 250 – Spring 2014 104
Example (4): Based on example (1), what if we double the pipe length, L?
3 6 2 1
2 1 1 3
2Re
0.4 1 2 2.758 10 0.4
10 1 4000
939.5
D
D pDf
L
cm g cm g s cm cm
g cm s g cm cm
0.380 1.115log ReRe 10 4952D f
D
,
2 2Re 939.5
0.036Re 4952
D
D
ff
.
2 1 1
Re4
0.4 10 49524
15.6
Dm D
cm g cm s
g s
23
8
23
8
0.0368
0.0368
Re 1000 Pr
1 12.7 Pr 1
4952 1000 7
1 12.7 7 1
38.1
fD
Df
Nu
3 1 1
2 1
6 1038.1
0.4
0.572
D
k W cm Kh Nu
D cm
W cm K
0
1 1 1
2 1
1 1 1
3
21 exp
15.6 4.2 100
2 0.2 0.572 40001 exp
15.6 4.2
6.55 10
total VV
RhLQ mC T T
mC
g s J g K K
cm W cm K cm
g s J g K
W
In this example, the increase of L leads to the decrease of m and h. In the representation of totalQ ,
the decrease of m or h will lead to decrease of totalQ .
L increases Re decreases
L increases m decreases
L increases h decreases
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