Phanikumar_TransportPhenomenaNotes_01Feb2010

MM2040: Introduction to Transport Phenomena

G. PhanikumarDepartment of Metallurgical and Materials Engineering

Indian Institute of Technology MadrasChennai 600036 India

e-mail: [email protected]: http://mme.iitm.ac.in/gphani

course website: http://mme.iitm.ac.in/moodle

February 1, 2010

Contents

1 Introduction to Tensors 7

1.1 Subscript notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Kronecker Delta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.3 Levi-Civita Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.5 Co-ordinate transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.6 Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.6.1 Scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.2 Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.3 Bisor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.4 Trisor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.5 Tetror . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

1.6.6 Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.7 Operations on tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.8 Types of tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.8.1 Symmetric Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.8.2 Anti-symmetric Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.8.3 Isotropic tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1.9 Tensors that represent physical properties . . . . . . . . . . . . . . . . . . . . . 17

1.10 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 Navier-Stokes Equations 20

2.1 Types of specification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.2 Continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.3 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2.4 Equation of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2

CONTENTS

2.5 Stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.6 Strain rate tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.7 Relation between stress and strain-rate . . . . . . . . . . . . . . . . . . . . . . . 25

2.8 Navier-Stokes equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.9 Significance of the linear relation and Viscosity . . . . . . . . . . . . . . . . . . . 28

2.10 N-S equations in cylindrical co-ordinate system . . . . . . . . . . . . . . . . . . 29

3 Specific cases of fluid flow 31

3.1 Boundary Conditions and Problem Definitions . . . . . . . . . . . . . . . . . . . 31

3.2 Trivial case reproducing Newton’s Law . . . . . . . . . . . . . . . . . . . . . . . 32

3.3 Film flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.4 Flow between two plates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.6 Flow through a pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.8 Creeping flow over a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.10 Creeping flow through a porous medium . . . . . . . . . . . . . . . . . . . . . . 46

3.11 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4 Correlations for turbulent regime 49

4.1 Friction factor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.2 Flow through tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.3 Flow across a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

4.4 Flow through a porous medium . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.5 Flow through a packed bed of spheres . . . . . . . . . . . . . . . . . . . . . . . . 52

4.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5 Energy Transport 55

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

5.2 Fourier’s first law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5.3 Fourier’s second law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5.4 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

6 Heat transfer in solids 60

6.1 Steady state 1D heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

6.1.1 Across a rectangular slab . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

Transport Phenomena Notes 3 G. Phanikumar

CONTENTS

6.1.2 Across a cylindrical wall . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

6.1.3 Across a spherical shell . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

6.1.4 Point effect of diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

6.1.5 Across a planar composite wall . . . . . . . . . . . . . . . . . . . . . . . 63

6.1.6 Across a cylindrical composite wall . . . . . . . . . . . . . . . . . . . . . 64

6.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6.3 Transient 1D heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6.3.2 Interface dominated . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

6.3.3 Conduction dominated . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

6.5 Moving boundary condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

6.5.1 Solidification: mould and solid conductivity controlled . . . . . . . . . . 69

6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

7 Heat tranfer with advection term 72

7.1 Steady state heat transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

7.1.1 Heat transfer normal to plug flow . . . . . . . . . . . . . . . . . . . . . . 72

7.1.2 Heat transfer along plug flow . . . . . . . . . . . . . . . . . . . . . . . . 73

7.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

7.3 Heat transfer in a smooth pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

7.3.1 Bulk temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

7.4 Definitions of some non-dimensional numbers . . . . . . . . . . . . . . . . . . . . 77

7.5 Forced convection correlations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

8 Mass Transfer 81

8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

8.2 Governing equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

8.2.1 Diffusivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

8.3 Solid state diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

8.3.1 Fixed boundary compositions . . . . . . . . . . . . . . . . . . . . . . . . 84

8.3.2 Fixed total solute content . . . . . . . . . . . . . . . . . . . . . . . . . . 85

8.3.3 Flux and concentration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

8.4 Mass transfer with advection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

8.4.1 Stagnant layer approach . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

8.4.2 Mass transfer coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

G. Phanikumar 4 Transport Phenomena Notes

CONTENTS

8.4.3 Sherwood number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

8.4.4 Chilton-Colburn Analogy . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

8.5 Reaction mass transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

A Derivations 94

A.1 The quotient rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

A.2 Symmetric tensors are diagonalisable . . . . . . . . . . . . . . . . . . . . . . . . 94

A.3 Levi-Civita tensor is isotropic . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

A.4 General form of isotropic tensor of order four . . . . . . . . . . . . . . . . . . . . 96

A.5 Simplification of tensor properties for crystals . . . . . . . . . . . . . . . . . . . 99

A.6 Change of variable with multiple integrals . . . . . . . . . . . . . . . . . . . . . 100

A.7 Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

A.8 Reynold’s transport theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

A.9 RTT and Continuity Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

A.10 Cauchy’s stress principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

A.11 Stress is a tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

A.12 Stress tensor is symmetric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

A.13 Meaning of terms in strain rate tensor . . . . . . . . . . . . . . . . . . . . . . . 105

A.14 Velocity gradient is a tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107


List of Figures

1.1 Co-ordinate Transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.1 Newton’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.2 Velocity and Stress distribution in the Newton’s Law problem . . . . . . . . . . 33

3.3 Schematic of a film flow problem . . . . . . . . . . . . . . . . . . . . . . . . . . 35

3.4 Velocity and Stress distribution in a film flow problem . . . . . . . . . . . . . . . 36

3.5 Flow in a channel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.6 Flow in a channel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.7 Mixing film . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.8 Squeegee device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.9 Pipe Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.10 Solution to Pipe Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.11 Axial film flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.12 Leaking Tank . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.13 Cubic network of pipes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.14 Channel flow between porous walls . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.15 Creeping flow around a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

3.16 Porous medium and its approximation as a bundle of tubes . . . . . . . . . . . . 46

6.1 Heat flow across a slab . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

6.2 Heat flow across a cylindrical wall . . . . . . . . . . . . . . . . . . . . . . . . . . 61

6.3 Point effect of diffusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

6.4 Temperature profile during solidification . . . . . . . . . . . . . . . . . . . . . . 68

6.5 Temperature profile during solidification . . . . . . . . . . . . . . . . . . . . . . 70

8.1 Fixed Compositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

8.2 Stagnant layer approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

8.3 Surface Renewal Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

A.1 Analysis of deformation of a fluid element . . . . . . . . . . . . . . . . . . . . . 106

6

�

Chapter 1

Introduction to Tensors

Most of the discussion in the notes will assume a cartesian co-ordinate system unless otherwise mentioned.

1.1 Subscript notation

Introduced by G. Ricci and popularised by Einstein. Each subscript (or index) runs from 1 to the dimension of the space in consideration. Since most of the time a 3D space is referred to, the subscripts run from 1 to 3. Following rules apply to the notation:

1. Cartesian Summation Convention: Subscripts that are repeated are called dummy sub-scripts and should be summed over the range that the subscript can take.

2. Subscripts that are not repeated are called free subscripts.

3. Subscripts can come anywhere in an expression.

4. Subscripts after a comma indicate differentiation.

5. Free subscripts on either sides of the ’=’ sign should match.

Subscript notation is useful to represent three and higher dimensional entities. It simplifies expressions.

Examples

• Vector ui = (u1 u2 u3) = �u = u1x1 + u2x2 + u3x3

• Matrices and Tensors

a11 a12 a13

aij = a21 a22 a23

a31 a32 a33

• Nabla or Del or Grad:

∂ ∂ ∂∇ = x1 + x2 + x3 = ∇i∂x1 ∂x2 ∂x3

7

�

� �

CHAPTER 1. INTRODUCTION TO TENSORS

• Gradient: ∂φ ∂φ ∂φ ∂φ∇φ = x1 + x2 + x3 = = ∇iφ = φ,i ∂x1 ∂x2 ∂x3 ∂xi

• Divergence: ∂u1 ∂u2 ∂u3∇ · u = + + ∂x1 ∂x2 ∂x3

∂uiDiv(u) = = ui,i

∂xi

• Inner product is an operation where there is a contraction of the number of subscripts. For two vectors, it is also called as dot product.

aibi = a1b1 + a2b2 + a3b3

2 2 2 aiai = a1 + a2 + a3

cj = aijbi c1 = a11b1 + a21b2 + a31b3

. . .

• Outer product is an operation where there is a an expansion of the number of subscripts.

a1b1 a1b2 a1b3 aibj = a2b1 a2b2 a2b3

a3b1 a3b2 a3b3

1.2 Kronecker Delta

Definition: 1 if i = j

δij = 0 if i = j

1 0 0 δij = 0 1 0

0 0 1

1.2.1 Examples

• Trace of a matrix aTr(a) = aijδij = aii = a11 + a22 + a33

Tr(a) = aii

Trace of δij is δii = 3.

• δij is used to simplify expressions involving dummy indices. Consider bik = aijδjk Expand the expression for a given i, k and convince yourself that

aikδij = ajk

piδij = pj


� � �

� � �

1.3. LEVI-CIVITA SYMBOL

1.3 Levi-Civita Symbol

Also called as permutation matrix.

Definition: 1 if i, j, k appear cyclic

ǫijk = −1 if i, j, k donot appear cyclic 0 if i = j or j = k or k = i

ǫ123 = ǫ231 = ǫ312 = 1 ǫ132 = ǫ213 = ǫ321 = −1 ǫ111 = ǫ112 = ǫ211 = ǫ121 = ǫ113 = ǫ311 = ǫ131 = 0 ǫ222 = ǫ221 = ǫ122 = ǫ212 = ǫ223 = ǫ322 = ǫ232 = 0 ǫ333 = ǫ331 = ǫ133 = ǫ313 = ǫ332 = ǫ233 = ǫ323 = 0

Permutation matrix will help in writing expressions in a simplified manner.

One can write ǫijk also in terms of a triple product as follows:

ǫijk = [xi, xj , xk]

1.3.1 Examples

• Curl :p = ∇× u

∂uk pi = ǫijk∇juk = ǫijkuk,j = ǫijk

∂xj

∂ukCurl(u) = ǫijk

∂xj

• Cross Product : p = u× vpi = ǫijkujvk

• Condition for coplanarity of three vectors ai, bj and ck is:

ǫijkaibjck = 0

• Determinant of a matrix a : Det(a) = IaijI

Det(a) = ǫijka1ia2ja3k

• Relation between δij and ǫijk :

ǫijkǫklm = δilδjm − δimδjl

The values of RHS are +1 if i = l and j = m and i = j−1 if i = m and j = l and i = j (1.1)

0 for any other combination


� �

� �

� �

� �

� � �

� � � � � � �

� � �

� � �


For the first case, it turns out that ǫijk = ǫlmk = ǫklm and the indices are non-repeating. Thus, whether they are cyclic or not, LHS is ǫ2 and is +1. For the second case, klm

ǫijk = ǫmlk = −ǫklm and the indices are non-repeating. Thus, LHS is −ǫklm 2 and is −1.

ǫijkǫijm = 2δkm

ǫijkǫijk = 6

Based on the relations given above, the following vector identities can be derived using subscript notation.

∇ · (∇φ) = ∇2φ (1.2)

∂ ∂ ∂2φ∇ · (∇φ) = φ = = ∇2φ∂xi ∂xi ∂x2

i

∇× (∇φ) = 0 (1.3)

∂ ∂φ ∂2φpi = ∇× (∇φ) = ǫijk = ǫijk

∂xj ∂xk ∂xj∂xk

For each term with the index i, there are two non zero terms on the RHS to be summed up. While the order of differentiation is immaterial, ǫijk is asymmetric about the indices j, k. Hence the RHS will vanish.

Take for example,

∂2φ ∂2φ ∂2φ ∂2φp1 = ǫ1jk = ǫ123 + ǫ132 = (ǫ123 + ǫ132) = 0

∂xj∂xk ∂x2∂x3 ∂x3∂x2 ∂x2∂x3

Similarly the other terms will also vanish.

1.4 Exercises

1. Prove the following vector identities using subscript notation.

∇ · (∇× f) = 0 (1.4)

∇(∇ · f) = ∇× (∇× f) + ∇2f (1.5)

∇(φψ) = φ∇ψ + ψ∇φ (1.6)

∇ · (φf) = φ(∇ · f) + f · ∇φ (1.7)


� � �

� � � � � �

� � � � � � � � � �

� � � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � � � � �

� � � � � �

� � � �

1.5. CO-ORDINATE TRANSFORMATIONS

∇× (φf) = ∇φ× f + φ(∇× f) (1.8)

∇ · (f × g) = g · (∇× f) − f · (∇× g) (1.9)

∇× (f × g) = f(∇ · g) − g(∇ · f) + (g · ∇)f − (f · ∇)g (1.10)

∇(f · g) = f × (∇× g) + g × (∇× f) + f · (∇g) + g · (∇f) (1.11)

1 ∇(f · f) = f ×∇× f + (f · ∇)f (1.12) 2

(f × g) × h = (f · h)g − (g · h)f (1.13)

f × (g × h) = (f · h)g − (f · g)h (1.14)

(f × g) · (f × g) = |f |2|g|2 − (f · g)2 (1.15)

2. (Exercise 2.23.1 of [Ari62]) Show how to find the vector which lies in the intersection of

the plane of a and b with the plane of c and d.

1.5 Co-ordinate transformations

Cartesian co-ordinate system xi is used. Consider the following co-ordinate transformation where the xy plane is rotated about z-axis clockwise by an angle θ.

The new axes (starred) are given in terms of the old axes (unstarred) by the following relations:

x∗ 1 = x1cosθ + x2sinθx∗ 2 = −x1sinθ + x2cosθ (1.16) x∗ 3 = x3

or ∗ x1 = T11x1 + T12x2 + T13x3

x∗ 2 = T21x1 + T22x2 + T23x3 (1.17)∗ x3 = T31x1 + T32x2 + T33x3

x∗ 1 cosθ sinθ 0 x1

x∗ 2 = −sinθ cosθ 0 x2 (1.18)

x∗ 0 0 1 x33

or

x∗ = (1.19) p Tpixi



Figure 1.1: Co-ordinate Transformation

The transformation matrix Tpi used in the above expression is by the convention that first index represents the row and the second index, the column. In one of the reference texts, viz., Aris’ book [Ari62], equation (2.11.1) shows that this convention is reversed. As a result the indices appear swapped in the definitions. This use of different convention is clear also from equation (A.6.4) in the same book.

When we express xi ∗ in terms of xi, the transformation matrix can be written as:

∗

1 ∗

1 ∗

1∂x ∂x ∂x

∂x1 ∂x2 ∂x3

∗

∂x

∗

2

∗

2 ∗

2∂x ∂x ∂x ∂x1 ∂x2 ∂x3

pTpi =

(1.20) =∂xi

∗

3 ∗

3 ∗

3∂x ∂x ∂x

∂x1 ∂x2 ∂x3

We have chosen cartesian co-ordinate systems which have the following properties of orthogo-nality:

∗ ∗ ∗ ∗ ∗ ∗ x · x = 1 x · x3 = 0 x · x2 = 01 1 2 3 ∗ ∗ ∗ ∗ ∗ ∗ x · x = 1 x · x1 = 0 x · x3 = 0 (1.21) 2 2 3 1 ∗ ∗ ∗ ∗ ∗ ∗ x · x = 1 x · x 0 x · x 03 3 1 2 = 3 1 =

Expressing x∗ in terms of x and using equations 1.17 and 1.21, we get:


1.6. TENSORS

T11T12 + T21T22 + T31T32 = 0T12T13 + T22T23 + T32T33 = 0T13T11 + T23T21 + T33T31 = 0

(1.22) T11T11 + T12T12 + T13T13 = 1 T21T21 + T22T22 + T23T23 = 1T31T31 + T32T32 + T33T33 = 1

or

TijTik = δjk (1.23)

Rotating the co-ordinate system back...

Similarly, we can express x in terms of x∗ . Inverse transformation relation is given using the transpose of the transformation matrix as follows. Transpose of a matrix is nothing but the same matrix with the indices swapped around.

∗ xi = Dipxp

The co-ordinate transformation matrix D for new to old system is related to the matrix for old to new the following manner.

T−1 T TDip = ip = ip = Tpi

xi = x∗ = x∗Dipˆp Tpiˆp

xi = Tpix∗ p (1.24)

Note: Compare equations 1.19 and 1.24 and notice which of the indices is being repeated in the RHS.

Exploiting the orthogonality of the old co-ordinate system:

x1 · x1 = 1 x2 · x3 = 0 x3 · x2 = 0 x2 · x2 = 1 x3 · x1 = 0 x1 · x3 = 0 (1.25) x3 · x3 = 1 x1 · x2 = 0 x2 · x3 = 0

TjiTki = δjk (1.26)

1.6 Tensors

Many entities, such as those that have a physical meaning, are independent of the co-ordinate system we choose to represent them. If n is the dimension of the space we are concerned, the entities can be classifed - according to the number of components (ni) they would have - as tensors of order i.


�


1.6.1 Scalar

Scalar is a tensor of order zero. Only one value would suffice to describe it at any location.Scalars are invariant under co-ordinate transformations.

Definition: Scalar is an entity that remains invariant under all co-ordinate transformations.

Eg: Temperature T , Composition of - say - A species CA, Pressure in a fluid p, Density ρ, etc.,

1.6.2 Vector

Vector is a tensor of order one. It requires as many values / components as the dimension of the space (which is three for most of us) ie., a set of three components in a 3D space.

Definition: Vector is an entity that transforms the following way under a co-ordinate trans-formation:

a∗ p = Tpiai (1.27)

Eg: Velocity of fluid u = ui, Temperature gradient ∂T , Composition gradient ∂CA , Displacement ∂xi ∂xi

ui, Electric Potential difference P , Electric Current Ji etc.,

1.6.3 Bisor

Bisor is a tensor of order two. All tensors of order 2 or more are called as just tensors. Specific names for tensors of order 2 and above are often considered old fashioned. A tensor of order 2 requires 9 components to describe it in general in 3D.

Definition: Tensor of order two is an entity that transforms the following way under a co-ordinate transformation:

∗ apq = TpiTqjaij (1.28)

Eg: Stress σij , Strain eij, Electrical Conductivity σij , Electrical Resistivity ρij , Diffusivity Dij , Thermal Conductivity kij, Magnetic Permeability µij , Dielectric Permittivity ǫij , Thermal Expansion Coefficient αij, Gyration Tensor gij

1.6.4 Trisor

Trisor is a tensor of order three. 27 components!

Definition: Tensor of order three is an entity that transforms the following way under aco-ordinate transformation:

a∗ pqr = TpiTqjTrkaijk (1.29)

Eg: Piezoelectric coefficient eijk,

1.6.5 Tetror

Tetror is a tensor of order four. 81 components!

a∗ pqrs = TpiTqjTrkTslaijkl (1.30)

Eg: Elastic Modulus cijkl


1.7. OPERATIONS ON TENSORS

1.6.6 Tensor

In general, a tensor is an entity that follows the following generalised rule for any co-ordinate transformation. The number of indices of a indicate the order of the tensor which is the same as the number of times T occurs in the expression.

a∗ = (1.31) pqrst... TpiTqjTrkTslTtm...aijklm...

1.7 Operations on tensors

• Sum and difference of two tensors (of same order) is also a tensor. In the following example, if bij and cij are tensors, then aij and dij are also tensors.

aij = bij + cij

dij = bij − cij

• An outer product of a tensor of order m with a tensor of order n will give a tensor of order m+ n.

aijk = bijck

• An inner product of a tensor of order m with a tensor of order n will give a tensor of order |m− n|.

ai = bijkcjk

• Contraction theorem : If aijkl.. is a tensor of order m, then the entity obtained by repeating any two subscripts is a tensor of order m − 2. Eg., if dijkl is a tensor of order 4, then by repeating two of its indices (say, third and fourth), the entity obtained aij is a tensor of order 4 − 2 = 2.

aij = dijkk

A corollary to the above theorem is obtained when we take a tensor of order two and repeat the indices. i.e., Trace of a second order tensor aii is tensor of order zero or scalar or invariant across co-ordinate transformations.

• Quotient law of tensors: If there is an entity representable by (subscript notation as) aij relative to any cartesian co-ordinate system and if aijbi is a vector where bi is any

arbitrary vector then aij is a tensor of order two. Proof is given in section A.1.

Combining with the theorem on outer product, quotient law can be extended to tensors of higher order.

1.8 Types of tensors

1.8.1 Symmetric Tensor

a is a symmetric tensor if aij = aji



Theorem: For every second order symmetric tensor, there exists a co-ordinate system relative to which, the matrix of the components of the tensor is diagonal 1 . Proof is given in section A.2.

1.8.2 Anti-symmetric Tensor

a is a skew-symmetric or anti-symmetric tensor if aij = −aji

The vector formed by the components of an anti-symmetric tensor:

ω1

ωk = ω2

ω3

0 ω3 −ω2

Ωij = −ω3 0 −ω1

ω2 −ω1 0

1 ωk = ǫkijΩij

2

Also,

Ωij = ǫijkωk

The two tensors ωk and Ωij are called dual tensors.

Theorem: Every second order tensor is expressible as a sum of a symmetric tensor and an anti-symmetric tensor.

1 1 aij = (aij + aji) + (aij − aji) (1.32)

2 2

1.8.3 Isotropic tensors

Definition: A tensor is isotropic if its elements donot change under any co-ordinate transfor-mation2 .

• Kronecker delta is an isotropic tensor of order two. The general form of an isotropic tensor of order two is

aij = aδij

• Levi-Civita density is an isotropic tensor of order three. Proof is given in section A.3 It is also referred to as alternating tensor or permutation tensor.

Relation between ǫijk and δij and the proof are given in equation 1.1.

• δijδkl is an isotropic tensor of order four.

Most general form of an isotropic tensor of order four, where µ1, µ2 and µ3 are constants:

1This theorem has important implications when applied to the stress tensor in the mechanical behaviour of materials. The diagonal terms are called principal stresses.

2involving only rotations and reflections but not expansions or contractions ie., Det(Tij) is unity


1.9. TENSORS THAT REPRESENT PHYSICAL PROPERTIES

µijkl = µ1δijδkl + µ2δikδjl + µ3δilδjk (1.33)

Proof is given in section A.4.

1.9 Tensors that represent physical properties

• Constitutive relations are those that describe a physical phenomenon by connecting the cause with the effect through a material property.

In a constitutive relation, the entity that connects a tensor of order m with a tensor of order n is of the order m+ n by the quotient law of tensors.

• Neumann’s principle : The symmetry group of any physical property of a crystal comprises the point symmetry group of the crystal.

A tensor that represents a physical property of a material should have at least the sym-metry of the material.

• Pierre Curie symmetry principle (1894): Effect is at least as symmetric as the cause. The symmetry of a crystal of known symmetry in the presence of external fields is given by the intersection of the symmetries of the crystal and the fields.

Examples of common physical phenomena, adapted from [PT83].

Property

Pyroelectric effect

Thermal conduction

Electrical conduction

Thermal expansion

Direct piezoelectric effect

Inverse piezoelectric effect

Linear electrooptical effect

Elasticity

Equation

ΔPi = γiΔT

dT qi = −kij dxj

Ji = σijEj

eij = αijΔT

Pi = dijkσjk

ejk = lijkEi

Δηij = rijkEk

eij = cijklσkl

Quantities

electric polarisation, pyroelectric coeffi-cient and change in temperature

heat flux, thermal conductivity and temperature gradient

electric current, electrical conductivity and electrical field

thermal strain, thermal expansion coeffi-cient and change in temperature

electric polarization, piezoelectric coeffi-cient and stress tensor

strain, inverse piezoelectric coefficient and electric field

polarization constants, tensor of linear electrooptical effect and electric field

strain, elastic compliance and stress

Using the Neumann’s principle, the tensor representing a property of a material should have the same symmetry as that of the material itself.


�

� �

� �


Liquids and glasses are isotropic ie., they have infinite symmetry 3 . The stuff remains the same in all the directions. So the physical property should also remain the same in all directions. So, the tensor representing the physical property of this stuff should be isotropic. That is, any arbitrary transformation of the co-ordinate system should leave the matrix containing the values of physical property unchanged. It is possible only if the property is representable by an isotropic tensor of corresponding order. Thus, for second order tensors such as thermal conductivity, diffusivity etc., of liquids only one value is necessary.

aij = aδij

Polycrystalline materials have crystals in all possible directions and as a bulk they behave as if the material is isotropic. Most of the engineering materials of interest are polycrystalline. Exceptions are highly textured materials (eg. after rolling) and single crystals (for turbine blades or semiconductor industry).

Take, for example, the second order tensor representing the thermal conductivity of a crystal. It should in general have 6 components (9 components become 6 thanks to the tensor being symmetric). Using the Curie principle, it can be shown as in section A.5 that for cubic crystals one needs to specify only one value for thermal conductivity.

1.10 Exercises

1. Prove that gradient of a continuous scalar function is a vector.

∂φai = ∇iφ = (1.34)

∂xi

2. Prove that the magnitude of a vector is invariant under co-ordinate transformations.

3. Prove that δij is (an isotropic) tensor of order two.

4. Prove that if ui and vj are vectors, then the dyad aij = uivj is a tensor of order two.

5. (Exercise 2.42.1 of [Ari62]) Prove that for any vector a, ǫijkak are the components of a second order tensor.

6. (Exercise 2.42.3 and 2.44.2 of [Ari62]) If r2 = xkxk and f(r) is any twice differentiable function, show that the following nine derivatives are components of a tensor.

f ′(r) xixj f ′(r)′′ (r) −f2

+ δij r r r

. Show also that the trace of that tensor is the following.

1 d df2r2r dr dr

7. Prove that strain rate or velocity gradient is a tensor of order two.

∂ui eij =

∂xj

3Only on a long range. They do have a short range order allegedly close to icosahedral in case of liquid metals


1.10. EXERCISES

8. If σij is a tensor, prove that σii, the trace of the matrix of σij is invariant under co-ordinate transformations. 4

9. Prove that ǫijk is a tensor of order three. Clue: Use the subscript notation for determinant of the transformation matrix. When two rows of a matrix are same, the determinant vanishes.

10. Prove that the dyad δijδkl is (an isotropic) tensor of order four. Considering that δij is an isotropic and hence a symmetric tensors, how many combinations of the four indices can you get so that you can arrive at possible isotropic tensors of order four.

11. If aij and bkl are tensors of order two, prove that the dyad aijbkl is a tensor of order four. Prove that aijbjk is a tensor of order two. Also, that a : b = aijbji is a scalar.

12. Use the Neumann’s principle to prove that the number of components necessary to de-scribe the thermal conductivity of a tetragonal crystal is two. You can assume that the thermal conductivity is a symmetric tensor to start with, thanks to Onsager’s the-ory [Ons31a, Ons31b].

4This quantity has a special meaning in mechanical behaviour of materials.


� ��

Chapter 2

Navier-Stokes Equations

2.1 Types of specification

Consider a flow field and trace a particle that moves with the flow. The flow field can be specified in two ways. In the Eulerian type of specification, the velocities are specified in terms of position (xi) and time (t). It gives the spacial distribution of velocity ui(xi, t) similar to density, temperature and pressure at any given instant.

In Lagrangian type of specification, the dynamical history of a specific piece of material (fluid element) is given. The flow field of the element is given in terms of the position (of the center of mass) of the element and time as vi(ai, t).

Eulerian specification gives spacial gradients of velocities directly. Lagrangian specification helps trace the path of an element directly. Unless otherwise specified, one usually uses to the Eulerian frame of reference.

In the Lagrangian specification, since the velocity vi(ai, t) is specified at the centre of gravity of the fluid element, acceleration of a fluid element is given by

dvi ∂vi = (2.1)

dt ∂t

In the Eulerian specification, since the velocity (ui(xi, t)) is specified in terms of the absolute position where as the location of the fluid element changes as dictated by the flow (see section A.6), acceleration of the fluid element is given as below:

dui ∂ui ∂uj ∂ui = + ui = + uj∇jui (2.2)

dt ∂t ∂xj ∂t

or

du ∂u= + (u · ∇)u (2.3)

dt ∂t

Proof of the same is given in section A.7.

We now define the material derivative which is a time derivative following the motion of the fluid as

20

� �

�

�

� �

� �

� �

� � � �

� �

� �

� �

� �

� �

� �

2.2. CONTINUITY EQUATION

D ∂= + (u · ∇)

Dt ∂t(2.4)

2.2 Continuity equation

Consider a fluid element of surface area S and volume V though which a flow u is taking place. Conservation of mass requires that the increase in amount of fluid in the element is equal to the amount brought in by the fluid flow (in the absence of sources and sinks that are singular). By convention, the surface normal n points outwards and hence the conservation can be written as:

∂ρdV = − ρu · ndS (2.5)

∂tV S

Using the Gauss theorem1 to convert the surface integral to volume integral,

∂ρdV = − ∇ · (ρu)dV (2.6)

∂tV V

For this relation to be valid at all locations in the fluid,

∂ρ+ ∇ · (ρu) = 0 (2.7)

∂t

Use the identity 1.7 to expand the above equation as

∂ρ+ (u · ∇)ρ+ ρ∇ · u = 0 (2.8)

∂t

Identifying the definition of material derivative D and dividing by ρ,Dt

1Dρ+ ∇ · u = 0 (2.9)

ρ Dt

∇ · u is called as rate of dilation or rate of expansion and is indicated by a symbol Δ.

Δ = ∇ · u = ∇iui = ui,i (2.10)

Equations 2.7 and 2.9 are two alternate forms of the continuity equation.

A fluid is called as incompressible if the density does not change due to changes in pressure (or) if the rate of change of density following the flow is zero. For incompressible fluids, the continuity equation reduces to:

∇ · u = 0 (2.11)

Sections A.6 and A.7 state the same thing more rigorously.

1Actually Gauss-Ostrogradsky theorem


�

� �

� �

CHAPTER 2. NAVIER-STOKES EQUATIONS

2.3 Forces

Long range forces decrease slowly with increase in distance between the interacting elements. These are called body forces or volume forces. Examples such as gravity (due to density ρgradients), electromagnetic (in metals carrying electric currents) and fictitious (centrifugal or coriolis) act on the whole of the fluid element and are usually proportional to the size of the volume element (δV ). They can be represented as:

Fi(xi, t)ρδV

Eg. gravity pointing vertically downwards :

Fi = gx2

Short range forces decrease rapidly with increase in distance between the interacting elements and are of molecular origin. Examples such as forces applied on surfaces (normal and shear), Marangoni forces (due to surface tension gradients) act on a thin layer of the fluid and are called surface forces. Local short range forces exerted by the fluid on different surface elements (δA) can be represented as:

σ(xi, nj, t)δA

where, nj is the unit normal to the surface element δA. σ is called the local stress.

2.4 Equation of motion

Consider a small fluid element of volume dV and area dS on a surface with normal ni. Using the material derivative D introduced earlier, the rate of change of momentum of a small fluid

Dt

element of volume dv is given by:

DρuidV

Dt V

Using the Reynold’s transport theorem [A.8] and the continuity equation, one can bring the material derivative inside the integral as illustrated in section [A.9].

D

Dt

�

V

ρuidV =

�

V

ρDui

DtdV

The total force that acts on dV is the sum of body forces and surface forces:

FiρdV + σijnjdSV S

Using Gauss theorem,

∂σij σijnjdS = dv (2.12)

V S ∂xj


Azif

Comment on Text

typo

� � �

2.5. STRESS TENSOR

From Newton’s second law, the rate of change of momentum is equal to the total force acting on the element. Hence,

Dui ∂σij ρdV = FiρdV + dV (2.13)

Dt ∂xjV V V

Since the three integrands are being summed over the same volume element, it will be applicable at any location in the fluid if the equation applies to the integrands themselves:

Dui ∂σij ρ = Fiρ+ (2.14)

Dt ∂xj

We now need to express σij in a way that can minimise the number of unknown parameters in the above equation of motion.

2.5 Stress tensor

Section A.10 stating the Cauchy’s stress principle and section A.11 prove that stress is a tensor. We can also use conservation of angular momentum to show that the stress tensor is symmetric as described in section A.12. Hence it can be expressed as

σ11 σ12 σ13

σij = σ12 σ22 σ23 (2.15)

σ13 σ23 σ33

Since it is always possible to find a co-ordinate system such that the matrix containing the elements of a symmetric tensor is diagonal, we can write σij as the follows.

σ11 0 0 σij = 0 σ22 0 (2.16)

0 0 σ33

or

1

1

3σkk 0 0 σ11 −

3σkk 0 0

σij = 0 31σkk 0 + 0 σ22 − 1

3σkk 0 (2.17)

0 0 31σkk 0 0 σ33 −

31σkk

where

σkk = σ11 + σ22 + σ33 (2.18)

1 σij = σkkδij + dij (2.19)

3

We define static pressure of the fluid with the convention that positive pressure is that which acts to compress a fluid element,


� �

� �

� � �


3

1 p = − σkk (2.20)

so that

σij = −pδij + dij (2.21)

dij is called the deviatoric part of the tensor and leads only to volume conserving deformation of the fluid elements ie., flow of the fluid. Static pressure, on the other hand, leads to a shape-conserving change in the volume element. It is easy to note that

dkk = 0 (2.22)

We have separated the stress into two terms:

• pressure term (−pδij) that tends to change the volume of a fluid element

• deviatoric stress term (dij) that tends to change the shape of a fluid element while keeping the volume constant

By analysing all modes of change of shape of a fluid element, we can relate the deviatoric stress with the terms that quantify the shape changes of a fluid element. For non-zero velocity field, because the fluid element translates as time progresses, we will notice that rate of change of shape is more appropriate to analyse a fluid element.

2.6 Strain rate tensor

The meaning of velocity gradient or a strain rate term ∂ui is given in section A.13. Proof that ∂xj

velocity gradient is a tensor of order two is given in section A.14. Strain rate or velocity gradient is represented as below and can be shown to be a second order tensor and thus expressible as a sum of symmetric and anti-symmetric tensors.

∂ui

∂xj = eij + Ωij (2.23)

eij = 1

2

∂ui

∂xj + ∂uj

∂xi (2.24)

Ωij = 1

2

∂ui

∂xj − ∂uj

∂xi (2.25)

Define ω as

ω = ∇× u (2.26)

or

ωi = ǫijkuj,k = ǫijk ∂uj

∂xk (2.27)


Azif

Highlight

trace = 0

Azif

Sticky Note

2.7. RELATION BETWEEN STRESS AND STRAIN-RATE

so that,

1 Ωij = − ǫijkωk (2.28)

2

2.7 Relation between stress and strain-rate

A relation between the deviatoric stress and strain-rate (velocity gradient) is necessary to proceed further to be able to use the equation of motion to solve for the velocity ui. We have to adopt a phenomenological approach.

By definition, dij is the deviatoric stress which implies that it is zero for a stationary fluid.

The velocities and thus the velocity gradients ∂ui are zero for a stationary fluid. ∂xj

The deviatoric stress represents the frictional interaction between different layers of the fluid and is assumed to be dependent only on the instantaneous and local distribution of the velocities.

From the above observations, we may assume that the deviatoric stress and the strain-rate are directly and linearly related to each other. Since both the quantities are tensors of order two, the entity that connects them both must be a tensor of order four.

∂ukdij = Aijkl (2.29)

∂xl

Since the above equation connects an effect with a cause, it is a constitutive relation and the entity Aijkl is a physical parameter. The material in concern is a fluid in which a directionality can safely be assumed to be absent ie.,

fluids are isotropic

Hence, Aijkl must possess the same properties as that of an isotropic tensor of order four. From the properties of tensors, we know that the general form of an isotropic tensor of order four [A.4] is

Aijkl = µ1δijδkl + µ2δilδkj + µ3δikδjl (2.30)

Since deviatoric stress dij is symmetric, interchanging the subscripts i and j should keep the quantity identical. In the above equation, this applies also to the R.H.S. ie.,

Aijkl = Ajikl ⇒ µ2 = µ3 (2.31)

Aijkl = µ1δijδkl + µ2 (δilδkj + δikδjl) (2.32)

Aijkl is now symmetrical in k and l also.


� �


Expanding the above equation,

dij = Aijkl (ekl + Ωkl) (2.33)

Since Aijkl is now symmetrical in k and l, when it is multiplied by an entity that is anti-symmetric about k and l and the terms are summed over k and l they vanish. Thus, the Ωkl

term drops out.

dij = (µ1δijδkl + µ2 (δilδkj + δikδjl)) ekl (2.34)

dij = µ1δijδklekl + µ2 (δilδkjekl + δikδjlekl) (2.35)

We have already (2.10) defined compressibility or rate of dilation or rate of expansion as

∂u1 ∂u2 ∂u3 ekk = + + = Δ (2.36)

∂x1 ∂x2 ∂x3

For the first term, we use the contraction theorem [section 1.7] to get

dij = µ1δijΔ + 2µ2eij (2.37)

Recalling that dii = 0,

dii = µ13Δ + 2µ2Δ = (3µ1 + 2µ2)Δ = 0 (2.38)

We would like the above equation to be true also for incompressible fluids ie., also when Δ = 0. It can be true only when the term in parantheses vanishes.

Stokes’ Assumption: For monoatomic fluids since there is no conversion of translational energy into vibrationary / rotationary energies, bulk viscosity can be assumed to be zero:

(3µ1 + 2µ2) = 0

or

2 µ1 = − µ2 (2.39)

3

Watch out for fluids for which Stokes’ assumption is not valid.

Calling the one constant parameter as µ, the equation becomes

2 dij = − µδijΔ + 2µeij (2.40)

3

1 dij = 2µ eij − Δδij (2.41)

3

Now that we have an expression for dij in terms of velocity gradients, we can substitute the same in the equation of motion.


� � ��

� � � � � �

� � � � � �

� � � �

� �

� �

� �

2.8. NAVIER-STOKES EQUATIONS

2.8 Navier-Stokes equations

Combining equation of motion (2.14) and the expressions for stress tensor (2.21 and 2.41), we get,

Dui ∂ 1 ρ = ρFi + −pδij + 2µ eij − Δδij (2.42) Dt ∂xj 3

Expanding the term eij and using the kronecker delta to contract subscripts, we get:

Dui ∂p ∂ ∂ui ∂ ∂uj ∂ 2 ρ = ρFi − + µ + µ + − µΔ (2.43) Dt ∂xi ∂xj ∂xj ∂xj ∂xi ∂xi 3

Since the order of differentiation should not matter and if µ is not a function of location,

Dui ∂p ∂ ∂ui ∂ ∂uj ∂ 2 ρ = ρFi − + µ + µ + − µΔ (2.44) Dt ∂xi ∂xj ∂xj ∂xi ∂xj ∂xi

Dui ∂p ∂ ∂ui ∂ ∂ 2 ρ = ρFi − + µ + (µΔ) + − µΔ (2.45) Dt ∂xi ∂xj ∂xj ∂xi ∂xi

Dui ∂p ∂ ∂ui 1 ∂ρ = ρFi − + µ + (µΔ) (2.46) Dt ∂xi ∂xj ∂xj ∂xi

The above set of three equations (for i = 1, 2, 3) corresponding to the three components of the velocity of the fluid (u1, u2 and u3) is called Navier-Stokes equations.

If the fluid being considered is incompressible, we can set the last term to zero and obtain the N-S equation with variable property.

A further simplification can be done starting from equation 2.46 as follows. Most of the fluids under normal flow conditions are incompressible ie., Δ = 0. The Navier-Stokes equations for incompressible fluid flow with constant viscosity are obtained by taking µ out of the derivative and setting Δ = 0 in the equation 2.46.

Dui ∂p ∂2ui ∂2ujρ = ρFi − + µ + µDt ∂xi ∂x2

j ∂xj∂xi

∂2

ρ = ρFi − + µ + µDui ∂p ui ∂ ∂uj

Dt ∂xi ∂x2 j ∂xi ∂xj

∂2

ρ = ρFi − + µ + µ (Δ) Dui ∂p ui ∂

Dt ∂xi ∂x2 j ∂xi

Dui ∂p ∂2uiρ = ρFi − + µ (2.47) Dt ∂xi ∂xj

2

Expanding the material derivative D and writing in vector notation: Dt

3

3

3


Azif

Highlight

� �

� �

� �

� �

� �

� �


ρ

ρ

∂u1 1 ∂p µ∇2+ (u · ∇)u1 = F1 − + u1∂t ∂x1 ρ∂u2 1 ∂p µ∇2+ (u · ∇)u2 = F2 − + u2∂t ∂x2 ρ∂u3 1 ∂p µ∇2+ (u · ∇)u3 = F3 − + u3 (2.48) ∂t ρ ∂x3 ρ

These can be expanded to give the N-S equations for incompressible fluids of constant property using the definition of kinematic viscosity as follows:

µν =

ρ

∂u1 ∂u1 ∂u1 ∂u1 1 ∂p ∂2u1 ∂2u1 ∂2u1 + u1 + u2 + u3 = F1 − + ν + + (2.49)

∂t ∂x1 ∂x2 ∂x3 ρ ∂x1 ∂x21 ∂x2

2 ∂x23

∂2 ∂2 ∂2∂u2 ∂u2 ∂u2 ∂u2 1 ∂p u2 u2 u2 + u1 + u2 + u3 = F2 − + ν + + (2.50)

∂t ∂x1 ∂x2 ∂x3 ρ ∂x2 ∂x2 ∂x2 ∂x2 1 2 3

∂2 ∂2 ∂2∂u3 ∂u3 ∂u3 ∂u3 1 ∂p u3 u3 u3 + u1 + u2 + u3 = F3 − + ν + + (2.51)

∂t ∂x1 ∂x2 ∂x3 ρ ∂x3 ∂x2 ∂x2 ∂x2 1 2 3

The continuity equation 2.9 and the three N-S equations 2.46 are solved together to obtain fluid flow. In these four equations, we have four variables (p, u1, u2, u3) - thus the problem is well defined.

2.9 Significance of the linear relation and Viscosity

The linear relation between the deviatoric stress tensor and the strain rate tensor (velocity gradient) is physically meaningful for a large number of fluids. Consider the two dimensional case of pure shear stress applied on a layer of liquid. Newton has observed that the shear stress is directly proportional to the velocity gradient.

∂u1d12 = d21 = µ (2.52)

∂x2

The proportionality constant µ is defined as the viscosity of the liquid. Fluids that obey this linear relation are called Newtonian fluids. Examples are water, air and most gases (in most of the situations except under shock wave) and liquid metals.

Based on their viscosity and the way they flow, materials can be classified as follows:

Newtonian fluid : Strain rate or velocity gradient is linearly dependent on the shear stress as σ12 = µu1,2. Eg: air and most of the gases, water, oils of low molecular weight and liquid metals. Often σ12 is referred to as τxy.


�

�

�

� �

� �

2.10. N-S EQUATIONS IN CYLINDRICAL CO-ORDINATE SYSTEM

Bingham Plastic : Flow starts above a critical shear stress which is then linear with the strain rate. σ12 = σ0 + µu1,2. Eg: drilling muds, peat slurries, margarine, chocolate mixtures, greases, soap, grain-water suspensions, toothpaste, paper pulp, and sewage sludge.

Pseudoplastic : Shear thinning material - viscosity decreases at higher strain rate.

Dilatant : Shear thickening material - viscosity increases at higher strain rate.

Thixotropic : Viscosity decrease with time.

Rheopectic : Viscosity increases with time.

Viscoelastic : The material returns back to its original shape after the stress is removed.

Viscosity is a strong function of temperature. In liquid metals viscosity follows an Arrhenius

relation. In situations where the temperature is roughly constant through out the flow, viscosity can be taken as constant. For fluids such as air and water, viscosity is negligible for most of the situations. Such fluids are called inviscid.

2.10 N-S equations in cylindrical co-ordinate system

We should choose a co-ordinate system with an orientation that best captures the symmetry of the problem and simplifies the final form of the solution.

Equations 2.48 are written for constant properties (ρ and µ) and using the operators ∇ and ∇2 . These operators convey a meaning independent of co-ordinate system and enable us to assign a meaning to each term in the equation.

One can (with some endurance) derive the equations by expressing x1, x2, x3 in terms of r, θ, zfor cylindrical co-ordinate system or r, θ, φ for spherical co-ordinate system and derive the necessary relations. The expansion of these operators and the N-S equations for cylindrical co-ordinate system can be borrowed from Appendix A of [BSL02]. Following are the expressions that will be of use to us later.

ˆu = urr + uθθ + uzz (2.53)

∂ 1 ∂ ∂ˆ∇ = r + θ + z (2.54) ∂r r ∂θ ∂z

∂ uθ ∂ ∂u · ∇ = ur + + uz (2.55)

∂r r ∂θ ∂z

1 ∂ ∂ 1 ∂2 ∂2

∇2 = r + + (2.56) r ∂r ∂r r2 ∂θ2 ∂z2

Most of the time we are concerned about incompressible fluids of approximately constant prop-erties. Hence we can borrow the N-S equations in cylindrical co-ordinate system for momentum transfer with these assumptions as reproduced below:


� � � �

� � � �

� � � �


∂ur ∂ur uθ ∂ur ∂ur uθ 2 1 ∂p

+ ur + + uz − = Fr − ∂t ∂r r ∂θ ∂z r ρ ∂r

∂ 1 ∂ {rur} 1 ∂2ur 2 ∂uθ ∂2ur+ν + − + (2.57)

∂r r ∂r r2 ∂θ2 r2 ∂θ ∂z2

∂uθ ∂uθ uθ ∂uθ uruθ ∂uθ 1 ∂p+ ur + + + uz = Fθ −

∂t ∂r r ∂θ r ∂z ρr ∂θ

∂ 1 ∂ {ruθ} 1 ∂2uθ 2 ∂ur ∂2uθ +ν + + + (2.58)

∂r r ∂r r2 ∂θ2 r2 ∂θ ∂z2

∂uz ∂uz uθ ∂uz ∂uz 1 ∂p+ ur + + uz = Fz −

∂t ∂r r ∂θ ∂z ρ ∂z

1 ∂ ∂uz 1 ∂2uz ∂2uz+ν r + + (2.59)

r ∂r ∂r r2 ∂θ2 ∂z2


� �

� �

� �

�

�

�

Chapter 3

Specific cases of fluid flow

3.1 Boundary Conditions and Problem Definitions

Fluid-Solid: Due to van der Waals attractions, wetting and any other atomistic phenomena, liquid tends to stick to solid and a relative motion is not possible1 . In most of the situation this phenomena of a solid preventing a liquid in contact with it from having a motion relative to it is called no slip condition. Often the solid wall is stationary making the velocity of liquid at the wall zero.

u| = u|liquid, interface solid

Fluid-Liquid: Using the arguments similar to above, there is no relative motion between two layers of liquids in contact with each other2 . Additionally, since most liquids wet each other, the shear stress at the interface of two liquids has a unique value.

τ | = τ |liquid1, interface liquid2, interface

For Newtonian fluids, if we take the interface to be at zero,

∂ui � ∂ui �

µ1 = µ2∂xj ∂xjxj→−0 xj→+0

Fluid-Gas: Because the density of gas is usually much smaller than that of liquid, it cannot sustain any shear stress at the top of the liquid layer and will lead to surface deformation. Thus, the shear stress at a free surface is zero.

τ |liquid, free surface = 0

For Newtonian fluids, if we take the interface to be at zero,

∂ui �

= 0 ∂xj xj→0

Steady state: Time derivative of the velocity is to be taken zero.

∂ui = 0

∂t

1except under situations where surface tension plays a major role 2except in situations where the two liquids in contact with each other are immiscible and interfacial tension

plays a major role

31

000000000000000000111111111111111111

000000000000000000000000000000000000000000000000000000

000000000000000000000000000000000000000000000000000000

CHAPTER 3. SPECIFIC CASES OF FLUID FLOW

Unidirectional flow: Velocity has only one component and the other components are to be taken zero.

u2 = u3 = 0 = u1

Fully developed flow: The velocity has no variation along the direction of the flow.

∂ui = 0

∂xi

Validity of solution: The analytical solutions given in this chapter are applicable for laminar

regime of fluid flow where the flow can be visualised as layers of liquid moving with respect to each other and the effects of wall penetrate far into the liquid. When the intertial forces acting on the fluid are far greater than the viscous forces, such an assumption is not valid and the flow is said to be turbulent. A transition from laminar to turbulent regimes is governed by the non-dimensional number indicating the ratio of intertial to viscous forces and is named after Reynolds.

ρDu0 Du0Re = =

µ ν

D is the characteristic length scale (diameter for a tube flow, width of channel for flow between two parallel plates etc.,), u0 is the characteristic velocity (typically the average velocity or far field velocity) and ν is the kinematic viscosity.

Range of Re for validity of a laminar solution for a problem of certain geometry is obtained from careful experiments.

Equivalent diameter: For tubes of non-circular cross sections or for other geometries, the equiv-alent diameter can be defined as

4 × cross sectional area De =

wetted perimeter

Thus, for the case of flow between two parallel plates separated by a distance of 2δ that is much smaller than the width of the plates W , De = 4δ.

3.2 Trivial case reproducing Newton’s Law

111111111111111111111111111111111111111111111111111111

111111111111111111111111111111111111111111111111111111

x1

x2

u1,max

Figure 3.1: Newton’s Law

Figure 3.1 shows the problem definition.

Assumptions:


� �

3.2. TRIVIAL CASE REPRODUCING NEWTON’S LAW

• Flow is unidirectional. Only u1 is to be known, u2 and u3 are zero.

∂u1• Flow is steady state. = 0. ∂t

• Flow is fully developed. ∂u1 = 0. ∂x1

• Flow is entirely due to the top surface being moved at u1,max and no body force or pressure gradients exist.

Boundary Conditions:

• No slip condition at bottom layer: at y = 0, u1 = 0.

• No slip condition at top layer: at y = δ, u1 = u1,max.

Solution: Use N-S equation for u1 and eliminate terms as per the assumptions above.

δ

∂u1

∂t+ u1

∂u1

∂x1

+ u2 ∂u1

∂x2

+ u3 ∂u1

∂x3

= F1 − 1 ∂p

ρ ∂x1

+ ν∂2u1

∂x2 1

+ ∂2u1

∂x2 2

+ ∂2u1

∂x2 3

(3.1)

ν∂2u1

∂x2 2

= 0 (3.2)

or u1 = Ax2 + B (3.3)

Using the boundary conditions,

A = u1,max

δ

B = 0

or y

u1 = u1,max δ

Using Newton’s law ∂u1

σ21 = τyx = µ∂x2

τyx = µu1,max

The solution is plotted schematically in figure 3.2.

x1

δ u1,max

u τyx

x2 x2 x2

Figure 3.2: Velocity and Stress distribution in the Newton’s Law problem


�


Convention: σ12 or τ12 is on plane 1 and in the direction 2. As for the sign, there are two different convections adapted in the literature.

Convention +:

In the above case, we have used the convention that the shear stress is positive and it is exerted by the top layer on the fluid beneath it so that Newton’s law is written as τ21 = µ∂u1 .

∂x2

Shear stress τyx is stress exerted on plane y in the positive direction x by the layer at greater yon the layer at lesser y.

This is the convention adapted in this handout.

Convention -:

This is favoured by [BSL02] for the reasons quoted below.

Shear stress τxy is stress exerted on plane x in direction y by the layer at lesser y on the layer at greater y.

Quoting from [BSL02] section 1.2, page 19:

Note on the Sign Convention for the Stress Tensor We have emphasised in connection with Eq. 1.1-2 (and in the generalization in this section) that τyx

is the force in the positive x direction on a unit area perpendicular to the ydirection, this being the force per unit area exerted by the fluid in the region of the lesser y on the fluid of greater y. In most fluid dynamics and elasticity books, the words ”lesser” and ”greater” are interchanged and Eq 1.1-2 is written as τyx = +µ(dvx/dy). The advantages of the sign convention used in this book are: (a) the sign convention used in Newton’s law of viscosity is consistent with that used in Fourier’s law of heat conduction and Fick’s law of diffusion; (b) the sign convention used for τij is the same as that for convective momentum flux ρvv (see section 1.7 and Table 19.2-2); (c) in Eq 1.2-2, the terms pδij and τij have the same sign affixed, and the terms pand τii are both positive in compression (in accordance with common usage in thermodynamics); (d) all terms in the entropy production in Eq. 24.1-5 have the same sign. Clearly the sign convention in Eqs. 1.1-2 and 1.2-6 is arbitrary, and either sign convention can be used, provided the physical meaning of the sign convention is clearly understood.

Note: Figure 2.8 of [Gas92] shows a symmetric plot of velocity profile as well as shear stress profile for a channel flow. There is an error in the shear stress profile. Watch out!

3.3 Film flow

Consider the case of a film of liquid falling on an inclined plane as shown in figure 3.3.

Assumptions:

• Flow is unidirectional. Only u1 needs to be known, u2 = u3 = 0.

∂u1• Flow is steady state: ∂t

= 0.

∂u1• Flow is fully developed i.e., it varies only along x2 but not along x1 or x3. ∂x1 = 0 for all

x1.

• The only driving force for the film to fall is gravity: F = g cos θx1 + g sin θx2.


� �

� �

3.3. FILM FLOW

Figure 3.3: Schematic of a film flow problem

Boundary conditions:

• Thickness of the film is δ.

• No slip condition at the bottom of the plane: at x2 = δ, u1 = 0.

∂u1• At x2 = 0, there is a free surface on which the shear stresses are zero. µ∂x2

= 0 at x2 = 0.

Solution: Use N-S equation for u and eliminate terms as per the assumptions above.

∂u1 ∂u1 ∂u1 ∂u1 1 ∂p ∂2u1 ∂2u1 ∂2u1 + u1 + u2 + u3 = F1 − + ν + + (3.4)

∂t ∂x1 ∂x2 ∂x3 ρ ∂x1 ∂x12 ∂x2

2 ∂x23

∂2u10 = g cos θ + ν (3.5)

∂x22

∂2u1 ρg cos θ= −

∂x22 µ

ρg cos θx22 u1 = − + C1x2 + C2

2µ

Using the boundary conditions, C1 = 0.

ρg cos θδ2

C2 = 2µ

Hence, ρδ2g cos θ (x2

)2

u1 = 1 −2µ δ

ρδ2g cos θu1|max =

2µ


σ21 = τyx = µ∂x2


0000000000000000011111111111111111

0000000000000000011111111111111111

000000000000000000000000000000000000000000000000000

000000000000000000000000000000000000000000000000000


τyx = −ρg cos θx2

� δ u1dx2 ρδ2g cos θ 2

u1 = 0 = = u1|max δ 3µ 3

If W is the width of the plane, mass flow rate M is:

ρ2δ3Wg cos θM = ρWδu1 =

3µ

The solutions are plotted schematically in figure 3.4.

x2x2 x2

δ

u1,max u

x1 τyx

Figure 3.4: Velocity and Stress distribution in a film flow problem

Validity: 4δρu1

Re = ≤ 25 µ

3.4 Flow between two plates

pH

x2

111111111111111111111111111111111111111111111111111

111111111111111111111111111111111111111111111111111

L

x1 2δ

pL

Figure 3.5: Flow in a channel

Figure 3.5 shows the problem definition. We choose the axes to make the maximum out of thesymmetry of the problem.

Assumptions:

• Flow is unidirectional. Only u1 is to be known, u2 and u3 are zero.

∂u1• Flow is steady state. = 0. ∂t


� �

3.4. FLOW BETWEEN TWO PLATES

• Flow is fully developed. ∂u1 = 0. ∂x1

∂p pH−pL• Pressure gradient is constant − = ∂x1 L


• No slip condition at bottom layer: at y = −δ, u1 = 0.

• No slip condition at top layer: at y = δ, u1 = 0.

Use N-S equation for u1 and eliminate terms as per the assumptions above.

∂u1 ∂u1 ∂u1 ∂u1 1 ∂p ∂2u1 ∂2u1 ∂2u1 + u1 + u2 + u3 = F1 − + ν + + (3.6)

∂t ∂x1 ∂x2 ∂x3 ρ ∂x1 ∂x2 ∂x2 ∂x2 1 2 3

∂2u1 1 ∂p pL − pHν = = (3.7) ∂x2

2 ρ ∂x1 ρLor

∂2u1 = −2A (3.8)

∂x22

pH − pLA =

2Lµ

u1 = −Ax12 + Bx1 + C

Using the boundary conditions,

pH − pL δ2 − x2

2

u1 = L 2µ


σ21 = τ21 = µ∂x2

pH − pLτ21 = (−2x2)

2L

Maximum velocity:

δ2pH − pL u1|max =

L 2µ

Average velocity:

δ22 pH − pL u1 = u1|max =

3 L 3µ

The solutions are plotted schematically in figure 3.6.

Interpretation of τ21. At x2 = 0, τ21 = +Aµδ. Recollecting convention + we are using for Newton’s law, the shear stress is exerted by the layer at greater x2 on the layer at lesser x2 ie., by the liquid on the wall. At x2 = δ, τ21 = −Aµδ. The shear stress is exerted by the layer at greater x2 on the layer at lesser x2 ie., by the wall on the liquid (thus the negative sign).


00

00

11

11

000

000

111

111

0000000000000000000000000000000000

000000000000000000000000000000000000000000000000000


x2 x2

δ

u1,max

u

x1

x2

τyx

Figure 3.6: Flow in a channel

3.5 Exercises

1. Mixing film: Problem 2B.3 and 2B.4 of [BSL02]. An incompressible Newtonian fluid is contained between two long plates of width W = 1m (along z) and a distance B = 1 mm apart as shown in figure 3.7 below. The plate on the right is moved upwards at a velocity v0. Gravity g = 9.81m s−2 acts along y axis downwards, density of fluid ρ = 1000 kg m−3 and viscosity of fluid 0.01Pa s. Assume the flow is uniaxial, steady state and fully developed. (a) Calculate v0 such that the net volume flow rate across a y plane is zero. (b) What is the ratio of the width of the fluid layer that flows downwards to the width of the fluid layer that flows upwards. (c) Sketch a schematic of the flow field.

1111111111111111111111111111111111

111111111111111111111111111111111111111111111111111

y

B

g

v0

x

Figure 3.7: Mixing film

2. Slag film: Problem 2.5 of [Gas92]. A layer of molten slag of density 2700kg m−3 and viscosity 0.3 Pa s is being transferred from one reverberatory furnace to another by flow down a plane between the two furnaces inclined at 10o to the horizontal. The plane is 5m in width and 5m in length and the mass flow rate of the slag is 7.5 kg s−1 . Neglecting the end effects, calculate (a) thickness of the slag layer (b) average linear flow velocity of the slag and (c) mean residence time of slag on the plane. (d) Fraction of slag that remains on the plane for times equal to or greater than the mean residence time. Answer: (a) 4.77 × 10−3 kg m−1 s−1 (b) 0.116m s−1 (c) 43 s (d) 0.423

3. Squeegee device: A continuous sheet (1.5m wide) of metal is cold-rolled by passing vertically between rolls at a constant speed of 0.3m s−1 . Before entering the rolls, the sheet passes through a tank of lubricating oil equipped with a squeegee device that coats both sides of the sheet uniformly as it exits. The amount of oil that is carried through can be controlled by the squeegee device. Determine the mass rate of oil as a function


000000000000000000111111111111111111

000000000000000000000000000000000000000000000000000000

000000000000000000000000000000000000000000000000000000

3.6. FLOW THROUGH A PIPE

of thickness of oil film that usually range between 0 mm and 0.6 mm. Properties of the lubricating oil: ρ = 962 kg m−3 , µ = 4.1 × 10−3 Pa s.

Squeegee

d

Figure 3.8: Squeegee device

3.6 Flow through a pipe

pH

Fz

pL

111111111111111111111111111111111111111111111111111111

111111111111111111111111111111111111111111111111111111

L

R

z

r

Figure 3.9: Pipe Flow

Figure 3.9 shows the problem definition.

Assumptions:

• Flow is unidirectional: Only uz is to be known, ur and uθ are zero

∂uz• Flow is steady state: = 0∂t

• Flow is fully developed: ∂uz = 0∂z

• Pressure gradient is constant: ∂p = Δp ∂z L



� � � �

� �

� �

� �

� �

� �

� �

� �

� �


• No slip condition at pipe wall: at r = R, uz = 0.

• Finite velocity at center: at r = 0, uz = ∞.

Use N-S equation in cylindrical co-ordinate system for uz and eliminate terms as per the as-sumptions above.

∂uz ∂uz uθ ∂uz ∂uz 1 ∂p+ ur + + uz = Fz −

∂t ∂r r ∂θ ∂z ρ ∂z

1 ∂ ∂uz 1 uz uz∂2 ∂2

+ν r + + (3.9) r ∂r ∂r r2 ∂θ2 ∂z2

1 Δp µ 1 ∂ ∂uz0 = Fz − + r

ρ L ρ r ∂r ∂r

ρFz 1 Δpuz = − − r2 + C1 ln(r) + C2

4µ 4µ L

Using boundary conditions, C1 = 0

ρFz 1 ΔpR2C2 = −

4µ 4µ L

ρFz 1 Δpuz = − (R2 − r2)

4µ 4µ L

Using Newton’s law ∂uz

σ21 = τzr = µ∂r

ρFz 1 Δpτzr = − (−2r)

4µ 4µ L

The solution is plotted schematically in figure 3.16.

ρFz 1 ΔpR2 uz|max = −

4µ 4µ L

Average flow velocity uz: � R

uz2πrdr 10 uz = �

= uz|max R 22πrdr0

Volume flow rate V : ρFz 1 Δp

V = πr2uz = − πR4

8µ 8µ L

Mass flow rate M (Hagen-Poiseuille Equation):

M = ρV = ρ2Fz − ρ Δp

πR4

8µ 8µ L


� �

3.7. EXERCISES

uz,max z

r

R

u

τzr

Figure 3.10: Solution to Pipe Flow

3.7 Exercises

1. Axial film flow: Problem 2B.7 of [BSL02]. A cylindrical rod of radius kR moves axially with velocity v = v0 along the axis of a cylindrical cavity of radius R as shown in figure below. Assuming that the pressure at both ends of the cavity is same and fluid flows through the annular region only because of the rod motion. (a) Find the velocity distribution in the narrow annular region. (b) Find the mass flow rate through the annular region. (c) Obtain the viscous force acting on the rod over a length L. Answer: (a)

uz ln (r/R) =

u0 ln k

(b)

M = πR2u0ρ 1 − k2

− 2k2

2 ln (1/k)

(c) 2πLµu0

Fz = ln (1/k)

2R 2kR

v0

Figure 3.11: Axial film flow

2. Leaking tank: Water of viscosity µ = 0.01Pa s and density ρ = 1000 kg m−3 from a large tank is to be transported using a rigid smooth tube of 1 cm dia connected at the bottom across a distance of L = 100m as shown in figure 3.12. Assuming that at steady state the tank is large enough that the height of water in it (h) does not change much in time, what should be h such that the flow rate in the tube is one litre per minute.



h

2R

L

Figure 3.12: Leaking Tank

3. Cubic network of pipes: Problem 2B.12 of [BSL02]. A fluid is flowing in laminar flow from A to B through a network of tubes, as depicted in figure 3.13. Obtain an expression for the mass flow rate w of the fluid entering at A (or leaving at B) as a function of the modified pressure drop pA − pB. Neglect the disturbances at the various tube junctions. Neglect any body forces that may act on some of the segments due to their orientation. Answer:

˙ 3π (pA − pB)R4ρM =

20µL

Figure 3.13: Cubic network of pipes

4. Blasius equation: Section 4.1 of [Gas92]. For steady viscous flow through a circular tube, the axial velocity profile is given approximately by the Blasius’ equation below.

( )mru = u0 1 −

R

For turbulent flow, m = 1/7. Assuming the density is constant, what is the average velocity? Is it closer to the maximum than in the case of Poiseuille flow? Comment. Answer: 0.817u0

5. Channel flow between porous walls: Problem 3B.16 of [BSL02]. Newtonian fluid flows through a rectangular channel with porous walls with a height h and infinite ex-tensions along x1 and x3 directions as shown in figure 3.14. The fluid leaks through the walls at a constant velocity of vw. The plane flow is steady, density ρ and viscosity µ are constant and there are no body forces. The flow takes place due to a constant pressure


3.7. EXERCISES

gradient ΔLp . Velocity is fully developed along x1. (a) Use the continuity equation to

obtain u2(x2). (b) Write the Navier-Stokes equation for u1 after simplification. (c) As-suming no slip boundary conditions applicable for u1, show that the velocity distribution can be given by the expression below. (d) Show that in the limit vw → 0, we retrieve the velocity profile of a plane channel flow. Note the location of axes different from the ones used in the class.

Δp h x2 1 − evwx2

ν

u1(x2) = − vwhL ρvw h ν1 − e

vw

vw

x2

x1

Figure 3.14: Channel flow between porous walls

6. Euler equation: Euler equation governs the flow of inviscid fluids i.e., fluids with nearly zero viscosity. (a) Use the information given at the end of the question paper and write down Euler equation in one dimension case for flow in a vertical tube due to both gravity as well as pressure gradient. (b) Show that in steady state, it reduces to the following expression popularly known as Bernouli equation.

2p v+ gz + z = constant

ρ 2

7. Couette flow: The Navier-Stokes equation for the θ component of velocity in cylindrical coordinate system is given to you. Annular region between two coaxial cylindrical surfaces of radii ri and ro (for inner and outer radii, respectively) is filled with a fluid. The outer cylinder is rotating with an angular velocity of Ωo and the inner cylinder is rotating with an angular velocity of Ωi, both clockwise. Assume that the fluid flow is uniform, steady state and fully developed and is only due to the rotation of outer cylinder. (a) Determine the velocity distribution in the liquid in the annular region. (b) Show that the result will approximate to Newton’s Law when the difference of radii is very small compared to either radii.

8. Channel flow of two stratified immiscible fluids: Section 2.5 of [BSL02]. Two im-miscible incompressible fluids of viscosities µ1 (bottom) and µ2 (top) are flowing between two stationary parallel plates under a pressure gradient Δ

LP . Assume µ1 > µ2. The thick-

ness of the two layers δ is same. (a) Derive an expression for the shear stress as a function


� �

� � � �

� �


of distance between the two plates. (b) Determine the location of zero shear stress from the bottom. (c) Draw schematically the shear stress profile.

3.8 Creeping flow over a sphere

When Re < 0.1, the flow around a smooth sphere is such that the viscous effects are present all around the sphere and no flow separation takes place. For such a flow regime called creeping

flow. around a sphere, we are interested not in the flow distribution around it but the viscous drag. The velocity of a sphere falling in a liquid column (terminal velocity) or the far field velocity of a fluid flowing around a sphere is u∞ as shown in the figure 3.15.

z

Normal pressure Tangential shear stressCreeping flow

θ v∞

Figure 3.15: Creeping flow around a sphere

We borrow the expressions for pressure and stress for creeping flow around a sphere from section 2.6 of [BSL02].

Pressure: 3 µu∞ R

2

p = p0 − ρgz − cos θ2 R r

Stress:

τrr = 3µv∞

R

− R

r

2

+ R

r

4

cos θ

3 µu∞ R4

τrθ = sin θ2 R r

The normal force along z due to the z component of pressure and normal stress τrr is given as:

� π � 2π

Fn = (p|r=R + τrr|r=R) cos θR2 sin θdθdφ0 0

Fn =4 πR3ρg + 2πµRu∞

3

The first term 43πR3ρg is called buoyancy force and the second term 2πµRu∞ is called form

drag.


3.9. EXERCISES

The tangential force along z due to z component of shear stress τrθ is given as:

� π � 2π

Ft = τrθ|r=R sin θR2 sin θdθdφ0 0

Ft = 4πµRu∞

This term 4πµRu∞ is called friction drag.

The weight of the falling sphere should equal the sum of form drag and friction drag:

4 πR3ρsg =

4 πR3ρg + 6πµRu∞

3 3

Stokes Law: 4 πR3 (ρs − ρ) g = 6πµRu∞

3

3.9 Exercises

1. Example 3.2 of [Gas92]. When a hollow sphere of diameter 0.005m and composite density 1500 kg m−3 is dropped into a column of oil of density 888 kg m−3, it attains a terminal velocity of 0.01m s−1 . (a) Calculate the viscosity of the oil. (b) Comment if the estimate is valid. Answer: µ = 0.834Pa s.

2. Example 3.3 of [Gas92]. The product of the deoxidation of liquid steel by the addition of aluminum is solid alumina. It forms throughout the liquid and floats to the top. This process is performed for a fixed duration of time after which the floating mass is skikked off. Those particles that take longer to float will remain in the solidified steel as inclusions and are detrimental to the mechanical properties. Assuming that the solid alumina forms as small spheres, (a) calculate the size of the smallest sphere that can float to the surface of a 1.5m deep quiescent liquid steel in 20min. (b) Comment if the estimate is valid. Answer: R = 33.2µm.

3. Problem 3.1 of [Gas92]. Viscosities of experimental glasses are being determined by mea-suring the terminal velocity of a platinum sphere falling through a column of molten glass. When dropped into a column of standard glass of viscosity 10Pa s and density 2500 kg m−3, the measured terminal velocity is 0.0258 m s−1 . When dropped into an ex-perimental glass of density 3000 kg m−3, the measured velocity is 0.0168 m s−1 . Calculate the viscosity of the experimental molten glass. Density of platinum is 21 450 kg m−3 . Answer: µ = 14.96Pa s.

4. Problem 3.4 of [Gas92]. Small glass spheres of density 2620 kg m−3 are allowed to fall through CCl4 of density 1590 kg m−3 and viscosity 9.58 × 10−4 Pa s. Calculate (a) the maximum diameter of sphere for which the flow obeys Stoke’s law (b) the terminal velocity that a sphere of this diameter attains. Answer: (a) 46.8µm (b) 1.28mm s−1


Azif

Highlight


= dp

Figure 3.16: Porous medium and its approximation as a bundle of tubes

3.10 Creeping flow through a porous medium

A porous medium is characterised by the fraction of voids or porosity ǫ.

volume of voids ǫ =

volume of porous body

For a given porosity, the distribution of porosity could be such that the surface area is a second variable. Thus a porous medium is also characterised by surface area per unit volume of solid S0.

wettable surface area S0 =

volume of solid

If a porous medium can be imagined to act like a bundle of thin tubes, then the equivalent diameter could be defined as hydraulic radius:

volume of voids Rh =

wettable surface area

Using the definition of ǫ, volume of voids is V ǫ. Using the definition of S0, the wettable surface area is S0V (1 − ǫ).

ǫRh = (3.10)

S0(1 − ǫ)

If A is the cross sectional area of the porous body, then Aǫ is the cross sectional area of the voids through which the liquid can flow. If u is the actual velocity of the fluid through the void, we can define the average (superficial) velocity us through the entire porous body such that the volume flow rate V is same:

˙usA = ¯ VuAǫ =

or us = ¯ (3.11) uǫ

We take clue from the Poiseuille flow the connects the pressure gradient with the flow through a tube:

Δp R2 1 Δp R2

u = = (3.12) L 8µ K1 L µ


3.10. CREEPING FLOW THROUGH A POROUS MEDIUM

Darcy’s Law: Considering a porous medium as a bundle of tubes, the volume flow rate is proportional to the cross sectional area and the pressure gradient. The proportionality constant is called as permeability coefficient.

ΔpV = kDA (3.13)

L

Considering that the nature of flow through a porous medium is a lot tortuous than through a tube, the constant K1 in equation 3.12 is taken not as 8 but 4.2 as has also been validated through experiments.

Now substituting equations 3.11, 3.10 into 3.12 with 4.2 as the constant, we get the Blake-

Kozeny equation:

1 Δp ǫ3 us = (3.14)

4.2 L µS02 (1 − ǫ)2

Validity:

Evaluating Re for porous medium taking the equivalent diameter as 2Rh,

uDρ¯ ρusRe = = 2

µ µS0(1 − ǫ)

We define the Reynolds number for porous medium as:

Rec = ρus

µS0(1 − ǫ) (3.15)

The Blake-Kozeny equation is valid for Rec < 2

Packed bed of spheres: If the porous medium is made of spherical particles of diameter dp, then S0, the surface area per unit volume of solid can be estimated directly since total wettable area is the sum of surface area of all spheres and total volume of solid is the sum of volume of all spheres.

4πR2 3 6 S0 = 4 = =

3πR3 R dp

Substituting this in Blake-Kozeny equation, we get the following equation:

Δp 150µus (1 − ǫ)2

= (3.16) L dp

2ǫ3

Evaluating Re for porous medium made of spherical particles (packed bed of spheres) and taking the equivalent diameter as dp,

¯ ρusdpuDρRe = =

µ µ3(1 − ǫ)

We define the Reynolds number for packed bed of spheres as:



ρusdpReE = (3.17)

µ(1 − ǫ)

3.11 Exercises

1. Problem 3.12 of [GP94]. Preliminary experimental studies have shown that the porosity in a newly developed packed bed reactor is ǫ < 0.6. The pellets have a diameter of 30 mm and the reducing gas flows through the bed at a rate of 0.025 kg s−1 . The reactor has 3 m × 3 m square cross section and is 15 m in height. A constant pressure difference of 690Pa is maintained between the inlet and outlet nozzles, and it may be assumed that the temperature is uniform throughout the reactor. You are required to evaluate the bed porosity. The properties of the gas are µ = 2.07 × 10−5 Pa s and ρ = 1.2 kg m−3 . Comment if the solution is valid. Answer: ǫ = 0.054.


Chapter 4

Correlations for turbulent regime

At large Reynolds numbers, the flow distribution cannot be obtained through a simple analytical expression as it could be time dependent. In such cases, we are interested in the force associated with the flow and its effectiveness in imparting kinetic energy to the fluid. For this, we define a quantity called friction factor and express it as a function of non-dimensional quantities such as Re and relative roughness ζ . For laminar flow, the expression for friction factor must result back to the analytical expressions derived for the situation.

4.1 Friction factor

Friction factor f is defined in the following expression:

1 Fk = fA ρu2 (4.1)

2

1ρu2 is the kinetic energy per unit volume of the fluid with average velocity u.2

For internal flow, A is the wetted area.

For external flow or flow over submerged objects, A is the area projected on a plane perpen-dicular to the velocity u.

Fk is the force associated with the fluid flow. For flow through a tube, it would be the pressure times the cross-sectional area over which it acts. For a sphere falling through a liquid, it would be the buoyancy force.

4.2 Flow through tube

For internal flow through a tube of diameter D, length L due to pressure gradient ΔLp ,

D2 1 2Δp · π = f · πDL · ρu4 2

or

1 Δp Df = (4.2) 1 24 L ρu

2

49

� �

CHAPTER 4. CORRELATIONS FOR TURBULENT REGIME

For laminar regime, we know that

R2Δpu =

L 8µ

Substitute the following in equation 4.1:

A = 2πRL

Fk = ΔpπR2

to get

2Δp ρu= f

L R(4.3)

Eliminating ΔLp using the expression for Poiseuille flow, we get

8µ µ 16 f = = 16 =

ρRu ρDu Re

Friction factors for pipe flow:

Laminar flow through smooth pipe (Re < 2100):

16 f =

Re

Turbulent flow through smooth pipe (3000 < Re < 105):

f = 0.0791Re−0.25

Turbulent flow through rough pipe (4 × 104 < Re < 108):

f−0.5 = −3.6 log10

ζ1.11

+ 6.9

3.7 Re

Here, ζ is relative roughness

4.3 Flow across a sphere

For laminar regime, we know that the force associated with the flow is given by Stoke’s law.

Substitute the following in equation 4.1:

A is projected area for flow over submerged objects:

A = πR2

u = u∞

Fk = 6πµRu∞


�

4.4. FLOW THROUGH A POROUS MEDIUM

to get

1 6πµRu∞ = f · πR2 · ρu2

2 ∞

Simplifying, we get

24 f =

Re

Friction factors for flow over sphere:

F = f πR2 1 ρu∞

2 (4.4) 2

Friction Factor Re range Remarks f = 24/Re Re < 0.1 Laminar f = 18.5Re−0.6 2 < Re < 500 Turbulent

( )2

f = 24/Re+ 0.5407 Re < 6000 Turbulent

f = 0.44 500 < Re < 2 × 105 Newtons Law

4.4 Flow through a porous medium

For creeping flow through a porous medium, we can use the Darcy’s law to approximate the porous medium to be a bundle of tubes and obtain the following relation from equation 3.14.

Use the definition of friction factor applied for tube flow and expressions as used in the derivation of 3.14:

D2

Δp · π h = f · πDhL · 1 ρu2

4 2

Δp ρu2 s = f ·

L ǫ2Rh

Thus,

Δp S0 (1 − ǫ) = f

ǫ3 ρu2

s (4.5) L

Substituting equation 3.14 in the above to eliminate pressure drop,

(1 − ǫ)S0µ 4.2 f = 4.2 =

ρus Rec

Friction factors for flow through porous medium:

For Rec < 2: 4.2

f = Rec



For 2 < Rec < 1000: 4.2

f = + 0.292 Rec

For 1000 < Rec < 105: f = 0.292

4.5 Flow through a packed bed of spheres

For creeping flow through a packed bed of sphere, we can use the porous medium equation 3.14 and substitute expression for S0.

6 S0 =

dp

Substitute the same in equation 4.5 to get

Δp

L= f

6 (1 − ǫ)

dpǫ3 ρu2

s (4.6)

We define 6f as fE so that:

Δp

L= fE

(1 − ǫ)

dpǫ3 ρu2

s (4.7)

Eliminating ΔLp , we get

(1 − ǫ)µ 150 fE = 150 =

ρusdp ReE

Friction factors for flow through packed bed of spheres:

For ReE < 10: 150

fE = ReE

For 10 < ReE < 1000: 150

fE = + 1.75 ReE

For 1000 < ReE < 105: fE = 1.75

4.6 Exercises

1. Flow past infinite cylinder: Problem 6B.9 of [BSL02]. The flow past a long cylinder is very different from the flow past a sphere. It is found that, when the fluid approaches a velocity u∞, the kinetic force acting on a length L of the cylinder is given by:

4πµu∞LFk =

ln (7.4/Re)


4.6. EXERCISES

The Reynolds number is defined here as Re = Du∞ρ/µ and the above equation is valid up to Re = 1. For this range, what is the formula for the friction factor as a function of the Reynolds number?

2. Examples 4.1, 4.3 of [Gas92]. (a) Calculate the pressure drop required to pass water at 300 K through a smooth pipe of inside diameter 0.05 m at the rate of 1.5 × 10−3 m3 s−1 . ρ = 997 kg m−3 , µ = 8.57 × 10−4 Pa s. (b) What would be the answer if the pipe is not smooth but has a relative roughness of 0.002? Answer: (a) 3.8 × 104 Pa (b) 4.64 × 104 Pa.

3. Examples 4.2, 4.4 of [Gas92]. (a) Calculate the flow rate of water at 300 K in a smooth pipe of 0.07 m inner diameter when the pressure drop per unit length is 125 Pa/m. (b) What would be the answer if the pipe has a relative roughness of 0.002. Answer: (a) 3.7 kg s−1 (b) 3.17 kg s−1 .

4. Problem 4.2 of [Gas92]. Water at 300 K is pumped at an average linear flow velocity of 2 m s−1 through a 30 m length of horizontal pipe of inside diameter 0.025 m and relative roughness of 0.004. (a) Calculate the pressure drop over the length of pipe. (b) The rough pipe is replaced by a smooth walled pipe of that diameter, which with the same pressure drop, gives the same average linear flow velocity. Calculate the required diameter of the smooth walled pipe. Answer: (a) 72.07 kPa (b) 0.0183m.

5. Problem 4.4 of [Gas92]. The average flow velocity of water at 300 K in a smooth pipe of internal diameter 0.07 m is 1m s−1 when the rate of pressure drop is 125Pa m−1 . Calculate the average flow velocity if the relative roughness were 0.002. Answer: 0.83m s−1 .

6. Example 4.7 of [Gas92]. Calculate the terminal velocity attained by a steel sphere of diameter 0.01 m when it is dropped in still air. The density of steel is 7500 kg m−3 , density of air is 1.177 kg m−3 and viscosity of air is 1.85 × 10−5 Pa s. Answer: 43.5m s−1 .

7. Problem 4.5 of [Gas92]. Density of lead is 11 340 kg m−3 . A lead shot of diameter 3 mm is fired upward into still air at 300 K over open water. Calculate (a) the drag force that the shot experiences as it leaves the gun barrel at a velocity of 150m s−1 (b) the terminal velocity that it attains when it falls through the still air and (c) the terminal velocity that it attains when it falls through the still water. Answer: (a) 0.0412 N (b) 29.3m s−1 (c) 0.9618m s−1 .

8. Settling ratio is defined as the ratio of sizes of two particles that settle at the same time. It is useful to know the settling ratio of two minerals in a given medium so that they could later be mechanically separated using a sieve. Assuming smooth spherical shapes of particles, estimate the settling ratio of mineral A to mineral B in water for (a) laminar flow regime and (b) turbulent flow regime. The properties of water are ρ = 997 kg m−3 , µ = 8.57 × 10−5 Pa s. Density of minerals are ρA = 2700 kg m−3 and ρB = 3400 kg m−3 . Answer: (a) 1.19 (b) 1.41.

9. Problem 4.8 of [Gas92]. Air at 300 K and an average pressure of 101.2 kPa is flowing through a packed bed of spheres of 1 cm diameter. The bed is cylindrical in geometry with 0.1 m in dia and 0.2 m in height and a porosity of 0.35. Calculate the pressure drop across the bed required to give a mass flow rate of air of 0.05 kg s−1 . Answer: 21 kPa.



10. Problem 3.13 of [GP94]. Molten aluminum is passed through a horizontal filter bed of Al2O3 spheres in order to remove drossy oxides from the aluminum. The filter bed comprises of two different packings arranged in series. The first packing encountered by the flow captures large drossy particles and the second packing captures the smaller drossy particles. Given the lengths as L1 = 0.7L2, ǫ1 = ǫ2, dp,1 = 2dp,2, compute the ratio of the pressure drop across the first and second beds for the cases of (a) creeping flow through the bed (b) fully turbulent flow through the bed. Answer: (a) 0.175 (b) 0.35


Chapter 5

Energy Transport

5.1 Introduction

Energy transport or heat transfer is basically through two means: conduction and radiation. The third mode of heat transfer by convection could be imagined as a combination of conduc-tion and advection. Conduction is a mode of energy transport by vibrational, rotational and translational degrees of freedom of atoms. The first two are predominant modes of energy ex-change between atoms in solid and liquid states while the third is predominant in gaseous state. Energy transport by radiation involves emission and absorption of electromagnetic radiation as photons typically in the infrared wavelength range. Taking cue from Lambert’s law that the attenuation of electromagnetic radiation in a medium is exponential (qz = q0 exp (−mz)), the absorption coefficient m being higher for medium of higher density, one can consider that radiative heat transfer is mostly a surface phenomena for solids and liquids. Thus, it is often taken in to account in the boundary conditions.

The governing equation for heat transfer by conduction can be derived similar to the way the governing equations for momentum transfer (Navier-Stokes equations) were derived. In that process we had followed the following sequence of steps:

• Balance equation or conservation principle

• Gauss theorem to convert surface integral to volume integral

• Linear constitutive relation

• Curie principle to simplify property tensor

We will follow the same steps to derive the governing equation for energy transport.

The balance equation for heat transfer is the first law of thermodynamics or the conservation of energy. If dU is the change in the internal energy, dq is the heat flow into the system, −pdV is the work done by the system and dWother is the work done by the system other than reduction in volume,

dU = dq − pdV + dWother

dWother is usually zero for the kind of situations we are interested in.

Using definition of enthalpy H as H = U + PV ,

55

� � � �

� � �

� � �

CHAPTER 5. ENERGY TRANSPORT

dH = dU + pdV + V dp = dq − pdV + pdV + V dp = dq + V dp

Writing H as a function of T and p,

∂H ∂HH = dT + dp

∂T p ∂p T

Since pressure is usually left constant at atmospheric pressure, H is usally Hp. Thus,

Hp = dq = ρCpdT = ρCp(T − Tref) (5.1)

If the volumetric heat generation term is g and Ji is the surface heat flux through a surface of area dS and surface normal ni, then we can write the energy balance for a control volume of volume dV as

dHdV = gdV − JinidS

dt V V S

Using equation 5.1 and making use of Reynold’s transport theorem (section A.8) to take the material derivative inside the integral:

dρCp(T − Tref)dV = gdV − JinidS (5.2)

dtV V S

Using the Gauss theorem to convert the surface integral to volume integral,

�

V

d

dtρCp(T − Tref)dV =

�

V

gdV − �

V

∂Ji

∂xi

dV (5.3)

Since the integration is over the same control volume dV , we can equate the integrands.

d

dtρCpT = g − ∂Ji

∂xi (5.4)

To obtain a relation between the surface heat flux Ji and its effects, we seek a linear constitutive equation as given in the following section.

5.2 Fourier’s first law

Since surface heat flux Ji tends to increase temperatures locally leading to temperature differ-ences in the body, it can be related to temperature gradients. The most general way of such a relation could be as given below:

∂TJj = −kij

∂xi

The thermal conductivity (kij) is a tensor of order 2. Usually, the materials of interest are polycrystalline metallic materials and liquids, both of which can be considered as isotropic in


� �

�

� �

�

� �

�

� �

5.2. FOURIER’S FIRST LAW

most of the situations1 . Using Curie principle, thermal conductivity being a material property, it must exhibit at least as much symmetry as the material itself. Thermal conductivity for an isotropic material will be an isotropic tensor and can be written as kij = kδij .

For single crystalline materials that are anisotropic, one can use symmetry arguments to reduce the number of independent values necessary to write the thermal conductivity tensor. Lars

Onsager’s reciprocal relations come of great use here in stating that for crystals of rotational symmetry such as 3, 4 and 6 fold, the property tensor can be represented by a symmetric matrix. Using the theorem that all symmetric tensors can be diagonalised (section A.2), and applying the four fold symmetry on the tensor to reduce the number of independent values to one as shown in section A.5, we can write the thermal conductivity tensor as following for polycrystalline materials, liquids and single crystalline materials with four fold symmetry :

kij = kδij

However, for single crystalline materials that donot possess cubic symmetry such as graphite, thermal conductivity should be written as a tensor (with two or more components).

Thus, for the kind of materials that we interested in (i.e., isotropic), the flux can be expressed in terms of temperature gradient as :

∂TJj = −kδij

∂xi

or

First law of Fourier heat conduction:

∂TJi = −k (5.5)

∂xi

The Fourier’s equation can be written in a co-ordinate system independent general form as:

q = −k∇T

such that it can be written for different co-ordinate systems as follows:

Cartesian (x1, x2, x3):

∂T ∂T ∂Tq = −k x1 + x2 + x3

∂x1 ∂x2 ∂x3

Cylindrical (r, θ, z):

∂T 1 ∂T ∂Tˆq = −k r + θ + z∂r r ∂θ ∂z

Spherical (r, θ, φ):

∂T 1 ∂T 1 ∂Tˆ ˆq = −k r + θ + φ∂r r ∂θ r sin θ ∂φ

1Highly textured polycrystalline materials and liquid crystals are exceptions


� �

� �

� �

� �

� � � �

� � � � � �

� � �

� �

� � � �

� � � � � �

CHAPTER 5. ENERGY TRANSPORT

5.3 Fourier’s second law

Substituting equation 5.5 in to 5.4:

d ∂ ∂TρCpT = g − −k

dt ∂xi ∂xi

dρCpT ∂ ∂T= k + g (5.6)

dt ∂xi ∂xi

kFor constant properties, using thermal diffusivity α = ρCp

∂T g+ u · ∇T = α∇2T + (5.7)

∂t ρCp

Expanding the ∇ operator for different co-ordinate systems:

Cartesian:

∂T ∂T ∂T ∂T ∂2T ∂2T ∂2T g+ u1 + u2 + u3 = α + + + (5.8)

∂t ∂x1 ∂x2 ∂x3 ∂x2 ∂x2 ∂x2 ρCp1 2 3

Cylindrical:

∂T ∂T uθ ∂T ∂T 1 ∂ ∂T 1 ∂2T ∂2T g+ ur + + uz = α r + + + (5.9)

∂t ∂r r ∂θ ∂z r ∂r ∂r r2 ∂θ2 ∂z2 ρCp

Spherical:

∂T ∂T uθ ∂T uφ ∂T+ ur + + =

∂t ∂r r ∂θ r sin θ ∂φ

1 ∂ ∂T 1 ∂ ∂T 1 ∂2T gα

2 r2 + sin θ + + (5.10)

r ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin2 θ ∂φ2 ρCp

Recognising that for solids, u = 0, leading to d = ∂ + u · ∇ = ∂ , we can use co-ordinate dt ∂t ∂t

system independent notation to write the Fourier’s second law as:

∂T g= α∇2T + (5.11)

∂t ρCp

The expanded forms for the three co-ordinate systems of interest are:

Cartesian (x1, x2, x3): ∂T ∂2T ∂2T ∂2T g

= α + + + (5.12) ∂t ∂x2 ∂x2 ∂x2 ρCp1 2 3

Cylindrical (r, θ, z):

∂T 1 ∂ ∂T 1 ∂2T ∂2T g= α r + + + (5.13)

∂t r ∂r ∂r r2 ∂θ2 ∂z2 ρCp

Spherical (r, θ, φ):

∂T 1 ∂ 2∂T 1 ∂ ∂T 1 ∂2T g= α r + sin θ + + (5.14)

∂t r2 ∂r ∂r r2 sin θ ∂θ ∂θ r2 sin2 θ ∂φ2 ρCp


�

�

�

�

�

�

�

�

�

�

5.4. BOUNDARY CONDITIONS

5.4 Boundary conditions

Neumann boundary condition: Heat flux is specified at the boundary.

Dirichlet boundary condition: Temperature is specified at the boundaries.

Newton’s Law of cooling: Heat flux as a function of boundary temperature is specified at theboundary.

The rate of heat transfer from a surface of a solid to the fluid it is in contact with is proportionalto the difference in temperatures of the fluid and the solid.

q = h (Ts − T∞)

The heat transfer coefficient (h) is a property dependent on several factors including velocity of the fluid, geometry of the surface and thermophysical properties of the two materials.

A heat flux balance at the surface will connect the Fourier’s equation with the Newton’s law of cooling as follows:

∂Tq|x=0 = h (Ts − T∞) = −k |x=0

∂x

Since h is always defined as a positive quantity irrespective of the direction of heat flow,

−k ∂T |x=0 ∂xh = Ts − T∞

Interface resistance: When smooth surfaces of two solids are in perfect contact, the interface can be said to be at one temperature. However, if the surfaces are not smooth and the contact is not perfect, there could be a jump in the interface temperatures2 and the resistance to heat flow across the interface is characterized by a parameter h, interfacial heat transfer coefficient. The heat flux across the two interfaces can then be written similar to Newton’s law as

q|interface = h (Ts1 − Ts2)

Radiative heat flux: For gases at high temperatures or surfaces of solids or liquids at high temperatures, radiation can be a significant mode of energy transport. When expressed as a boundary heat flux, it can written as:

q = eσSBT4

σSB is the Stefan-Boltzmann constant and e is the emissivity. Radiative heat flux from surface 1 to surface 2 (of areas A1 and A2) that have view factors F12 and F21 is

q12 = A1F12(T14 − T2

4) = A2F21(T14 − T2

4)

2at a macro-scale. The temperature will be continuous without a jump when measured at micro-scale. Kapitza resistance is an exception


� �

Chapter 6

Heat transfer in solids

6.1 Steady state 1D heat transfer

6.1.1 Across a rectangular slab

T

T1

T2

Δx

x1 x2

x

Figure 6.1: Heat flow across a slab

At steady state, if q is the rate of heat transfer (Js−1) across a slab area A:

q ∂T= −k

A ∂x

If the surface temperatures of the slab are T1 and T2 at x1 and x2, respectively, then:

T2 x2−qdT = dx

AkT1 x1

q qT2 − T1 = (x1 − x2) = − Δx

Ak Ak

T1 − T2 =

Δx(6.1)

q Ak

Taking the analogy of electricity, T1 − T2 is the driving force similar to voltage, q is the flux similar to current and Δx is the resistance. Now the resistance to heat flow is known, they can

Ak

combined in serial and parallel similar to the circuits in electricity.

60

6.1. STEADY STATE 1D HEAT TRANSFER

As can be seen from the equation for flux:

q T1 − T2 = k

A Δx

doubling the slab thickness Δx will halve the heat flux.

6.1.2 Across a cylindrical wall

T

T1

T2

ΔR

R1 R2

x

Figure 6.2: Heat flow across a cylindrical wall

At steady state, if q is the rate of heat transfer (Js−1) across a cylindrical wall of (variable) area A = 2πrL:

q q ∂T= = −k

A 2πrL ∂r

Heat flow across a hollow cylindrical wall of inner radius R1 and outer R2 at temperatures T1

and T2, respectively, is then given by:

� T2 −q � R2 drdT =

T1 2πLk R1

r

−q R2T2 − T1 = ln

2πLk R1

R2T1 − T2 =

ln R1 (6.2)

q 2πLk

Drawing the analogy to electricity as above, the resistance to heat flow across a cylindrical wall R2ln

is R1 .2πLk


q q T1 − T2 T1 − T2|r=R1 = = k = kR2 ΔRA 2πLR1 R1 ln R1 ln (1 + )R1 R1


� �

� �

�

�

�

� ��

�

�

�

� �

CHAPTER 6. HEAT TRANSFER IN SOLIDS

Doubling the slab thickness ΔR will not necessarily halve the heat flux. It can be noted that in the limit of large curvature, R1 → ∞, ΔR is a small quantity and ln (1 + ΔR) can be

R1 R1

approximated to ΔR leading to the above expression as: R1

q T1 − T2 T1 − T2|r=R1→∞ = kΔR

= kA R1 ΔR

R1

which is the limit where a cylindrical wall of thickness ΔR can be approximated to a rectangular slab of same thickness.

6.1.3 Across a spherical shell

At steady state, if q is the rate of heat transfer (Js−1) across a spherical shell of (variable) area A = 4πr2:

q q ∂T= = −k

A 4πr2 ∂r

Heat flow across a hollow sphere of inner radius R1 and outer R2 at temperatures T1 and T2, respectively, is then given by:

T2 R2−q drdT =

4πk r2 T1 R1

−q 1 1 T2 − T1 = −

4πk R1 R2

T1 − T2 1 1 1 = − (6.3)

q 4πk R1 R2

Drawing the analogy to electricity as above, the resistance to heat flow across a cylindrical wall ( )

is 1 4πk

1 R2

− 1 R1

.

q (T1 − T2)R2 T1 − T2 ΔR T1 − T2 = k = k 1 + = k

A r=R1→∞ ΔRR1 ΔR R1 R1→∞ ΔR

which is the limit where a spherical shell of thickness ΔR can be approximated to a rectangular slab of same thickness.

6.1.4 Point effect of diffusion

When the temperature differences and the properties are kept constant, heat flux (heat per unit time per unit area) depends on the geometry - with the following geometries in the decreasing order of effectiveness: point, line, plane, edge, corner. Such a sequence of effectiveness of thermal diffusion arises from the amount of space available for exchange of energy. Point effect of diffusion comes of use when interpreting defects in casting.



�

�

�

� �

6.1. STEADY STATE 1D HEAT TRANSFER

Doubling the slab thickness ΔR will not necessarily halve the heat flux. It can be noted that ΔRin the limit of large curvature, R1 → ∞, R1

is a small quantity and goes to zero leading to the above expression as:

q= k

T1 − T2

A r=R1→∞ ΔR

which is the limit where a spherical wall of thickness ΔR can be approximated to a slab of same thickness.

The rationale behind this is the what we call as point effect of diffusion. Taking the wall thickness to be same, we plot the three geometries below:

Point effect of diffusion

0

1

2

3

4

5

6

0 20 100

Den

omin

ator

in

fl ux

term

Δx = 5 ΔR = 5, cylindrical

ΔR = 5, spherical

40 60 80

radius of curvature R

Figure 6.3: Point effect of diffusion

6.1.5 Across a planar composite wall

Using the electrical analogy of the previous section, flux boundary condition Aq = h (Ts − T∞)

1can be written such that the resistance to heat flow is .Ah

At steady state, if q is the rate of heat transfer (Js−1) across a planar composite wall of constant area A:

q T2 − T1 T3 − T2 T4 − T3 = hb (Tb − T1) = −k12 = −k23 = −k34 = ha (T4 − Ta)

A Δx12 Δx23 Δx34

or

q 1 Δx12 Δx23 Δx34 1 Ta − Tb = + + + +

A ha k12 k23 k34 hb

Analogy with electricity: h 1 and Δ

kx act as resistances. Temperature is analogous to voltage

difference (driving force for the current). Aq is analogous to current.


� �


6.1.6 Across a cylindrical composite wall

Using the electrical analogue of the previous section, the heat flow across a composite hollow cylindrical wall is given by:

r1 r2 r3 q 1 ln ln ln 1r2 r3 r4Ta − Tb = + + + + 2πL r1h1 k12 k23 k34 r4h4

6.2 Exercises

1. Heat is flowing at steady state through an annular wall of a cylinder of inner radius R1

and outer radius R2. Thermal conductivity of the wall material is approximated to be linear function of temperature with a value of k1 at the inner wall temperature T1 and k2

at the outer wall temperature T2. (a) Derive an expression for the heat flow across the cylindrical shell. (b) Simplify the expression for the case when the wall thickness is very small compared to the radius of the cylinder.

2. Example 6.3 of [Gas92]. Hot water flows through a glass tube of inner radius 3 cm and outer radius 5 cm. The temperatures of the inner and outer surfaces of the tube are, respectively, 90 ◦C and 85 ◦C and the mean thermal conductivity of the glass is 0.84W m−1 K−1 . (a) Calculate the rate of heat loss from the tube per unit length. (b)By how much is the rate of heat loss decreased if the wall thickness of the tube is doubled keeping the inner radius same. Answer: (a) 51.7W m−1 (b) 31.1W m−1

3. An iron slab of thickness 2 cm carries an electric current that generates heat at a rate of 10 × 106 W m−3 . If the right side face of the slab is at 20 ◦C, what should be the temperature of the left side face such that all the heat flux in the slab is from the left face to the right face at steady state? Properties of steel are k = 45W m−1 K−1 , ρ = 7210 kg m−3 and Cp = 750 J kg−1 K−1 .

4. Example 6.4 of [Gas92]. A furnace wall consists of 15 cm thick silica brick (k = 1.1W m−1 K−1), a 5 cm glass fibre (k = 0.035W m−1 K−1 and a 1 cm steel (k = 45W m−1 K−1). Inner temperature of the furnace is 500 ◦C and heat transfer coefficient on the inner wall is 15W m−2 K−1 . Ambient temperature is 20 ◦C and heat transfer coefficient on the outer wall is 20W m−2 K−1 . (a) Calculate the power loss per unit area of furnace wall at steady state. (b) Temperatures at each junction. Answer: (a) 285.4W m−2 (b) Ti = 481 ◦C, T1 = 442 ◦C, T2 ≈ T3 ≈ 34 ◦C.

5. A 50 cm long tubular furnace is made of a 1 cm thick silica tube of 10 cm inner radius kept in a 5 mm thick steel tube of 15 cm outer dia with glass wool in between. Knowing that the glass wool would melt at 800 ◦C and that furnace converts all its electrical power to heat energy, (a) what is the highest temperature that can be set in the furnace and (b) what is the power consumption at that temperature setting? Ignore end effects and assume steady state. The heat transfer coefficients for inside and outside of the furnace can be taken to be 20 W m−2 K−1 . Assume room temperature to be 300K .

6. Example 6.7 of [Gas92]. A copper wire of diameter 1mm has an insulating plastic sheath of 0.5 mm around it. The heat transfer coefficient on the surface exposed to ambient air at 30 ◦C is 8W m−2 K−1 . If the insulating plastic softens above 100 ◦C, calculate the maximum current that can be passed through the wire. Properties of copper are


6.3. TRANSIENT 1D HEAT TRANSFER

k = 380W m−1 K−1 , σ = 1.96 × 10−8 Ω m. Properties of the insulating plastic are k =0.35W m−1 K−1 and σ ≈ 0Ω m.Answer: 14.3 A

6.3 Transient 1D heat transfer

6.3.1 Introduction

Within the domain of unidirectional heat transfer where the transient evolution of temperatures is of interest, one can think of two extreme cases. The first where the heat transfer from the solid is limited by thermal diffusion through its bulk. This is when the removal of heat from the interface is not constrained. An example could be a hot ceramic piece kept in front of a fan.

The second case is when bulk diffusion of heat within the solid is not constrained at all but the heat transfer at the interface is limited by the surrounding medium. An example could be a hot copper block kept in still air.

In the second case, one can assume that no thermal gradients would be present within the solid and use what is known as lumped heat capacitance method to obtain the variation of average temperature of the solid with time. This forms the interface dominated heat transfer.

In the first case, one needs to solve the equation of thermal diffusion in solid to determine the thermal gradients that drive the heat flux across the interface of the solid with the ambient medium. This forms the conduction dominated heat transfer.

6.3.2 Interface dominated

The change in the temperature of an object immersed in a fluid at different temperature can be estimated easily if the thermal conductivity of the object is high enough that no thermal gradients develop within the object.

If the characteristic length scale of the object is L, we require that the inner temperature of the object Ti is as close to the surface temperature Ts as possible. A balance of flux at the surface gives:

Ti − Tsk = h (Ts − T∞)

Lor

Ti − Ts hL=

Ts − T∞ k

We define Biot Number as hL

Bi = k

If Bi ≤ 0.1, Ti is close enough to Ts that thermal gradients within the object can be neglected so that Newtonian cooling is applicable. In such as case, we can write,

dThA (T − T∞) = −ρCpV

dt


� �

�


� T tdT −hA= dt

T0 T − T∞ ρCpV 0

or T − T∞ −hAt

ln = T0 − T∞ ρCpV

or T − T∞ hAt

= exp − T0 − T∞ ρCpV

Defining V/A as L, the characteristic length scale,

ht hL αt= = Bi.Fo

ρCpL k L2

where, we define Fourier number as:

αt ktFo = =

L2 ρCpL2

Thus, when Bi ≤ 0.1,

T − T∞ = exp (−Bi.Fo)

T0 − T∞

6.3.3 Conduction dominated

In problems where heat transfer is dominated by conduction, temperature as a function of time and spatial coordinates can be obtained by solving the Fourier heat conduction equation.

For the problem of semi-infinite domain where temperature at one end is fixed and we are interested in its evolution as a function of distance and time, the governing equation can be written for a one dimensional case as follows.

∂T ∂2T= α

∂t ∂x2

While variable separation is one way of solving this equation, it leads to summation of a series. We use a co-ordinate transformation to make the equation simpler.

Put x

η = √ 2 αt

leading to ∂η 1

= √ ∂x 2 αt

∂η −η=

∂t 2t

∂T ∂T 1 = √

∂x ∂x 2 αt

∂2T ∂2T 1 =

∂x2 ∂x2 4αt


�

6.4. EXERCISES

Equation 6.3.3 becomes ∂2T ∂T

= −2η∂η2 ∂η

or ¨ T

= −2ηT

Integrating once,

log T = −η2 + constant

or

T = C1e−η2

Integrating once more,

T = η=η

C1e−η2

dη + C2 η=0

The integral cannot be simplified and is often given as a tabulated function with the following definition and properties

� η2 erf(η) = √ e−η2

dηπ 0

∂ 2 −η2

erf(η) = √ e∂η π

erf(0) = 0

erf(∞) = 1

erf(−η) = −erf(η)

Thus, the solution of 6.3.3 in a semi-infinite domain can be written as :

xT = Aerf( √ ) +B

2 αt

The constants A and B can be determined using the boundary conditions.

6.4 Exercises

1. A long copper wire of 2 mm diameter is exposed to air stream at a temperature of 400 K. After a minute, the average temperature of the wire increased from 280 K to 350 K. (a) Estimate the average heat transfer coefficient on the surface (b) Using the Biot number, comment on the validity of your solution.


000000000000000000000000000000000000000000000000000000

��

�

�

�


2. Small droplets of a molten glass maintain their amorphicity if they cool at a rate of atleast 10K s−1 measured at 1070K. For a spherical droplet with 0.1mm diameter, what is the required heat transfer coefficient to achieve the minimum cooling rate? The quench environment is maintained at 293K. Properties of glass are ρ = 3000 kg m−3 , Cp = 840 J kg−1 K−1 , k= 17W m−1 K−1 . Verify if your answer is valid for lumped capacitance method to work.

3. A typical human body generates about 100W due to metabolism. In order to keep the body temperature at 37 ◦C, the heat must be dissipated through various mechanisms such as convective heat transfer, radiation, evaporation and conduction that are dynamically balanced depending on the outside temperature. If the heat loss is significant and the core temperature drops below 21 ◦C, death occurs. Assuming the surface area of a human body to be about 2m2, estimate how long a human can stay alive in (a) quiescent water at 10 ◦C where h = 230W m−2 K−1 and (b) in flowing water at 10 ◦C moving at 0.25m s−1

where h = 580W m−2 K−1 . Assume Cp of human body to be about 3470 J kg−1 K−1 .

6.5 Moving boundary condition

Problem: Liquid is poured at TM in a thick mould kept at T0. Assuming that all the latent heat is extracted through mould, arrive at the rate of solidification.

111111111111111111111111111111111111111111111111111111

T

−x x

TM

(0, T0)(−∞, T0)

Figure 6.4: Temperature profile during solidification

Solution:

1Solidification is usually antiparallel to the direction of heat flux

Heat flux at x = 0 is ∂T

J = −km x∂x x=0

Latent heat release per unit volume of solid formed is ρsΔHf

Let the heat transfer be across a constant mould-solid surface area of A

Balancing the two,

∂VAJ = −ρsΔHf

∂t

1Solidification in undercooled melts is an exception


�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

6.5. MOVING BOUNDARY CONDITION

since a positive J implies heat flux in +x direction leading to a negative rate of solidification.

∂T ∂VAkm = ρsΔHf

∂x ∂tx=0

Writing the error function solution to the temperature in the mould (properties of mould indicated by subscript m),

TM − T −x= erf( √ )

TM − T0 2 αmt

Substituting,

2 1 ∂VAkm(TM − T0)√ √ = ρsΔHf

π 2 αmt ∂t

or

1 ∂V (TM − T0)� 1 = √ kmρmCpm √

A ∂t ρsΔHf π t

Define heat diffusivity HD as

HD = kmρmCpm

Integrating from V = 0 at t = 0 to V = V at t = t,

√ = √ t

V 2 (TM − T0)HDm

A π ρsΔHf

Chvorinov’s Rule: √V ∝ tA

6.5.1 Solidification: mould and solid conductivity controlled

Problem: Liquid at TM is poured in to a mould at T0. Heat transfer is controlled by conduction through solid as well as mould. If the mould-solid interface temperature is TS after a thickness of M of solid has formed, derive expressions to estimate the two quantities.

Boundary conditions:

BC1: T = T0 at x = −∞ BC2: T = TS at x = 0 and t = τ

BC3: Flux balance ∂T ∂T−km = −ks∂x ∂xx→−0 x→+0

BC4: T = TM at x = M and t = τ

BC5: Flux balance ∂T ∂V

Aks = ρsΔHf∂x ∂tx=M


000000000000000000000000000000000000000000000000000000000000000000000000

� �

� �


111111111111111111111111111111111111111111111111111111111111111111111111

T

−x x

(M, TM )

(0, TS)

(−∞, T0)

Figure 6.5: Temperature profile during solidification

Solution:

Using BC1 and BC2, write the error function solution for heat conduction in the mould as:

TS − T −x= erf( √ )

TS − T0 2 αmt

Similarly the error function solution for heat conduction in the solid is:

xT = Aerf √ + B

2 αst

Using BC2 and BC4 and naming M 2 √

αsτ as φ, write the solution for temperature in the solid as

TM − TS xT = TS + erf √

erf(φ) 2 αst

BC3 provides an estimate of TS for a given φ:

TS − T0 TM − TS 1 km √ = ks √

παmτ παsτ erf(φ)

Using the definition of heat diffusivity,

HDsHDm(TS − T0) = (TM − TS)erf(φ)

Call

HDs p = HDmerf(φ)

pTM + T0TS =

p+ 1

BC5 provides an estimate for φ as a function of physical parameters.


6.6. EXERCISES

Aks 2 √ πe−φ2 TM − TS

erf(φ)2 √ αst

= ρsΔHf ∂V

∂t

Recognising that V A

= M ,

ks 2 √ e−φ2 (TM − TS)√ = ρsΔHf

∂M

π 2 αsterf(φ) ∂t

Integrating the equation from M = 0 at t = 0 to M = M at t = τ ,

√ 2 −φ2 (TM − TS) τ

ks √ e √ = ρsΔHfMπ αserf(φ)

Simplifying,

φ2 Cps(TM − TS)φerf(φ)e = √

ΔHf π

Using a table / plot of φerf(φ)eφ2 as a function of φ, one can look up for the RHS of above √

equation for the corresponding φ which gives the relation M = 2φ αsτ .

6.6 Exercises

1. A two inch slab of aluminium is cast in a mold made of silica sand on one side and an unknown material on the other side. The cast slab is sectioned to look at the plane of last solidification that can be identified by porosity. If it is known that the plane is located 30mm away from the silica side of the mould, (a) estimate the heat diffusivity of the unknown mould material. From the properties of mould materials given below, (b) guess which one comes closest. Neglect superheat.

Typical values of physical properties: Material k/W m−1 K−1 ρ/kg m−3 Cp/J kg−1 K−1

Silica 0.6 1500 1160 Mullite 0.37 1600 770 Zircon 1.0 2720 840

2. A liquid metal (L) at temperatute Tp is poured into a mould (m) kept at temperature T0. (a) If the mould-metal surface reaches thermal equilibrium instantly, find its temperature Ts when the freezing is yet to start. (b) A component is formed from an alpha brass alloy poured into a mould kept at 25 ◦C. If the freezing takes place over a temperature range from 1055 ◦C to 1045 ◦C, fine the minimum pouring temperature to prevent instantaneous freezing. Properties are as follows. km = 1.6 W m−1 K−1 , kL = 109W m−1 K−1 , ρm = 3.2 × 103 kg m−3 , ρL = 8.52 × 103 kg m−3 , Cp,m = 1 × 103 J kg−1 K−1 , Cp,L = 385 J kg−1 K−1 .


� �

�

�

�

�

Chapter 7

Heat tranfer with advection term

7.1 Steady state heat transfer

7.1.1 Heat transfer normal to plug flow

Plug flow is defined as spatially constant fluid flow. One usually approximates flow regimes in a chemical reactor to be plug flow. Also, flow across a porous medium under constant pressure drop leads to constant flow.

The equation for heat transfer is given as:

∂T ∂T ∂T ∂T ∂2T ∂2T ∂2T g+ u1 + u2 + u3 = α + + +

∂t ∂x1 ∂x1 ∂x1 ∂x2 ∂x2 ∂x2 ρCp1 2 3

Let us assume that the constant flow is along x1 and the steady state unidirectional heat transfer is along x2. Setting these values, we can see that the governing equation reduces to:

∂2Tk + g = 0 ∂x2

2

Thus, plug flow normal to steady state unidirectional heat transfer has no effect on the tem-perature profile. Since there are no velocity gradients in a plug flow, viscous dissipiation is also not considered. Thus, often, g = 0.

In the case of a velocity profile such as channel flow normal to steady state unidirectional heat transfer, the equation remains the same except that the source term g can be given by the viscous dissipation. Assuming that the fluid is Newtonian and the flow to be channel flow, we can write:

∂2T u02

k + µ = 0 ∂x2

2 δ2

We can solve the above equation subjec to the following boundary conditions:

BC1: T | = T0x2=0

BC2: ∂T

= 0 ∂y x2=δ

72

� �

�

7.2. EXERCISES

( )2k x2 1 x2(T − T0) = −

µu02 δ 2 δ

7.1.2 Heat transfer along plug flow

Let us assume that the constant flow and the steady state unidirectional heat transfer are both along x1. We can write the governing equation as:

∂T ∂2T gu1 = α + ∂x1 ∂x2

1 ρCp

If we ignore heat generation term due to viscous dissipation, then the solution can be written subject to the following boundary conditions.

BC1: T |x1=0 = T1

BC2: T |x1=δ = T2

T − T1 exp u1x1 − 1 =

(

(

α )

)

T2 − T1 exp uα 1δ − 1

One can see that as u1 → 0, the solution approaches the linear profile as in steady state conduction for unidirectional heat transfer.

7.2 Exercises

1. A journal bearing of outer diameter 20 cm has a separation between the rotating shaft and the stationary journal of 1 mm filled with a lubricant liquid. The angular velocity of the shaft is 1000 rpm. Assuming that the shaft is kept at 20 ◦C and the journal at 30 ◦C, determine the temperature distribution along the radial direction in the lubricant layer taking viscous dissipation into account. The properties of the lubricant are: ρ = 900 kg m−3 , k = 0.15W m−1 K−1 , µ = 0.8 Pa s. Calculate the maximum temperature in the liquid lubricant layer. Assume the liquid to be flowing with usual limitations as that of Couette flow.

7.3 Heat transfer in a smooth pipe

Energy transport in fluids involves the complete form of equation 5.7.

∂T g+ u · ∇T = α∇2T + (7.1)

∂t ρCp

We will limit this chapter to the cases with the following assumptions:

• u is known analytically


�

� � � �

� �

CHAPTER 7. HEAT TRANFER WITH ADVECTION TERM

• Flow is unidirectional, steady state and fully developed

• Heat transfer is steady state

Problem: For a fluid flowing unidirectionally through a pipe at steady state and with fully developed velocity profile and a given entry temperature, what is T (r, z) for a given surface heat flux. In addition to the assumptions above, we make the following ones too.

Assumptions:

( )2 • u = 1 − r zumax R

k ∂T • Surface heat flux is constant. q0 = ∂r

= constant

• Cylindrical symmetry. Nothing happens along θ direction.

We can write the governing equation as follows:

( )2r ∂T 1 ∂ ∂T1 − = α rumax

R ∂z r ∂r ∂r

subject to the following boundary conditions:

BC1: Finite temperature at the center of pipe

T (r = 0, z) = finite

BC2: Constant wall heat flux ∂T

q0 = −k∂r

BC3: Entry temperature T (r, z = 0) = T0

Before attempting to solve the above differential equation, it is a good idea to non-dimensionalize the variables that could possibly simplify the derivation as well as the final expressions. In this process, we recognise that q0

kR has units of temperature, R is the characteristic length scale,

τ0 = R2 is the characteristic time scale and that along z the fluid moves a characteristic distance

α

of umaxτ0 that can be used to scale z.

Scaling:

T ∗ = T − T0

q0R k

∗ rr =

Rzα

z∗ = R2umax

leading to

∂T ∂T ∗ kRumax =

∂z ∂z∗ αq0


� �

�

�

�

�

� �

� �

� �

� �

� �

7.3. HEAT TRANSFER IN A SMOOTH PIPE

∂T ∂T∗ k=

∂r ∂r∗ q0R2

Thus the governing equation can be non-dimensionalised to:

∂T ∗ 1 ∂ ∂T ∗

(1 − r∗2) = r∗

r∗ ∂z∗ ∂r∗ ∂r∗

The above equation needs to be solved subject to the following boundary conditions:

BC1:

T ∗(r∗ = 0, z∗) = finite

BC2: ∂T ∗

= 1 ∂r∗ r ∗=1

BC3:

T ∗(r∗ , z∗ = 0) = 0

As we would notice later, BC3 should actually be stated in integral form to state that the starting temperature is not given as a function of r∗ but as a flow averaged (bulk) value.

We propose the following solution:

T ∗ = C0z∗ + φ(r∗)

Substituting the solution into the governing equation,

1 ∂ ∂φ(1 − r∗2)C0 = r∗

r∗ ∂r∗ ∂r∗

∗Integrating once w.r.t r ,

r∗ ∂φ

∂r∗ = C0

r∗2

2 − r

∗4

4 + C1

Integrating once more,

φ = C0 r∗2

4 − r

∗4

16 + C1 ln r

∗ + C2

Thus, the solution is:

T ∗ = C0z∗ + C0

r∗2

4 − r

∗4

16 + C1 ln r

∗ + C2 (7.2)

BC1 implies C1 = 0

BC2 implies C0 = 4

T ∗ = 4z∗ + r∗2

1 − r

∗4

4 + C2 (7.3)


� �

� �

� �

� �

� �


BC3 is not useful in arriving at C2. We need one more boundary condition. We use the overall balance of energy for the same.

BC4: The overall energy entering the pipe at a length z is given by

� 2π � R

Qz = ρCpT (r, z)u(r)dr rdθ0 0

The overall energy entering the pipe at a length z = 0 is given by

� 2π � R

Q0 = ρCpT0u(r)dr rdθ0 0

The overall balance Qz −Q0 = 2πRzq0 can now be written as:

� 2π � R ( )2rρCp (T − T0)umax 1 − dr rdθ = 2πRzq0

R0 0

or

� R ( )2r Rzq0 Rq0 zα(T − T0) 1 − rdr = =

0 R umaxρCp k umax

This can be expressed in the non-dimensional form as:

� 1

T ∗(1 − r∗2)r∗dr∗ = z∗ (7.4) 0

Substituting 7.3 in to 7.4,

� 1 1 � �

∗2 ∗4 ∗24z∗ + r − r + C2 1 − r r∗dr∗ = z∗ 40

or

� 1 1 � �

∗2 ∗4 ∗2r − r + C2 1 − r r∗dr∗ = 0 40

or

� 1C2 ∗2 1 �

∗2�

+ r − r∗4 1 − r r∗dr∗ = 0 4 40

7We use the above integral BC4 to arrive at C2 = 24

Thus, the final solution for temperature distribution in a fluid flowing through a tube with constant heat flux is

1 7∗ ∗ ∗2 ∗4T = 4z + r − r − 4 24


� �

� �

7.4. DEFINITIONS OF SOME NON-DIMENSIONAL NUMBERS

7.3.1 Bulk temperature

We now define bulk temperature or flow averaged temperature Tb as

< Tuz >2π R

T (r, z)u(r)dr rdθ= = 0 0Tb � 2π � R< uz > u(r)dr rdθ

0 0

For our case of unidirectional, fully developed flow,

� R T (r, z)u(r)rdr

0Tb = � R

u(r)rdr0

Numerator is

R22πRzq02π R 2πRzq0

+ T0 u(r)dr rdθ = + 2πρCpumax T0ρCp 0 0 ρCp 4

Denominator is R2

2πumax 4

4q0zTb = T0 +

ρCpRumax

Surface temperature is given by setting r = R.

4q0z q0R 11 q0D 11 Ts = T0 + + = Tb +

ρCpRumax k 24 k 48

Rearranging,

(Ts − Tb) k 11 =

q0D 48

or

q0D 48 Nu = =

(Ts − Tb) k 11

We define the heat transfer coefficient for tube flow as:

q0h =

(Ts − Tb)

7.4 Definitions of some non-dimensional numbers

Biot Number: hL

Bi = ks


� �

Fourier Number:


αtFo =

L2

Grashof Number: gβ(Ts − T∞)L3

GrL = ν2

Rayleigh Number: gβ(Ts − T∞)x3

Rax = να

Nusselt Number: hL

NuL = kf

Peclet Number: V L

PeL = α

Prandtl Number: ν

Pr = α

Stanton Number: h

St = ρV Cp

7.5 Forced convection correlations

External flow over flat plate at uniform surface temperature:

• Laminar external flow for Pr ≥ 0.6 :

hxxNux = = 0.332Re1x

/2Pr1/3

k

• Laminar external flow for 0.6 < Pr < 50 :

hxxNux = = 0.664Re1/2Pr1/3

k x

• Laminar Flow of liquid metal for Pr ≤ 0.05, Pex ≥ 102 where Pex = RexPr :

Nux = 0.565Pe1x/2

• Laminar flow of fluids of all Prandtl numbers, Pex ≥ 102 :

0.3387Rex 1/2Pr1/3

Nux = ( )2/3 1/4

0.0468 1 +Pr

• Turbulent flow for 0.6 < Pr < 60, Re < 108:

Nux = StRexPr = 0.0296Re4x/5Pr1/3


� �

� �

� �

7.5. FORCED CONVECTION CORRELATIONS

• Laminar + Turbulent flow where the representative Rex,c for transition is taken as 5×105 , 0.6 < Pr < 60, 5 × 105 < ReL < 108 :

( )

¯ 4/5NuL = 0.037ReL − 871 Pr1/3

Flat plate with uniform surface flux:

• Laminar flow, Pr ≥ 0.6 :

Nux = 0.453Re1x/2Pr1/3

• Turbulent flow, 0.6 < Pr < 60 :

Nux = 0.0308Re4x/5Pr1/3

External flow over circular cylinder:

• Correlation for wide range of parameters:

¯NuD =

hD= CRemPr1/3

Dk

ReD C m0.4 − 4 0.989 0.33 4 − 40 0.911 0.385 40 − 4000 0.683 0.466 4 × 103 − 4 × 104 0.193 0.618 4 × 104 − 4 × 105 0.027 0.805

When properties are not constant over the temperature range involved, for 0.7 < Pr < 500 and 1 < ReD < 106 ,

14Pr

NuD = CRemPrn D Prs

If Pr ≤ 10, n = 0.37 and if Pr ≥ 10, n = 0.36

ReD C m 1- 40 0.75 0.4 40 - 103 0.51 0.5 2 × 105 - 106 0.076 0.7

Comprehensive equation for all ReD and Pr > 0.2

0.62Re1/2Pr1/3 ReD¯NuD = D

455

8

0.3 + 1 +14( )2/30.41 +

Pr

282000


� �

� �

� �


External flow over a sphere :

• For 0.71 < Pr < 380, 3.5 < ReD < 7.6 × 104, 1 < µ/µs < 3.2 :

¯ 1/2 2/3NuD = 2 + 0.4ReD + 0.06ReD Pr0.4 µ

µs

• For freely falling droplets of liquids:

NuD = 2 + 0.6Re1/2Pr1/3

D

Internal flow through circular tube:

• Laminar, fully developed flow, uniform heat flux, Pr ≥ 0.6:

NuD = 4.36

14

• Laminar, fully developed flow, uniform surface temperature, Pr ≥ 0.6:

NuD = 3.66

• Turbulent, fully developed flow, 0.7 ≤ Pr ≤ 16700, ReD ≥ 104 , L/D ≥ 10 :

0.14 µ

µs

• Liquid metals, Turbulent, fully developed, uniform heat flux, 3.6×103 < ReD < 9.05×105 , 102 < PeD < 104:

0.827 NuD = 4.82 + 0.0185 (ReDPr)

• Liquid metals, Turbulent, fully developed, uniform surface temperature, PeD > 102:

NuD = 5.0 + 0.025 (ReDPr)0.8

7.6 Exercises

1. A solar water heater is made of a parabolic mirror that concentrates a total heat flux of 2 kW s−2 on to a thin copper tube that carries the water. If the inlet water is at 20 ◦C and at a flow rate of 0.02 kg s−1 and the outlet shall have 60 ◦C, what should be the length

48 of the tube? Assuming fully developed flow of water through the tube with Nu = 11

applicable, estimate the surface temperature at the outlet. Properties of water are: Cp = 4200 J kg−1 K−1 , k = 0.67W m−1 K−1 , µ = 0.35 × 10−3 Pa s, Pr = 2.2.

NuD = 0.027ReD 4/5Pr

13


Chapter 8

Mass Transfer

8.1 Introduction

Mass transfer topics fall into the following four major categories.

• Solid state diffusion: discussed mainly in the context of physical metallurgy. We can look at the governing equation and couple of sample solutions.

• Convective mass transfer: when flux is more important than the acutal composition distributions.

• Reaction with a generation term: when solute is generated or consumed at a location because of a reaction. The rate of formation of a certain product is of interest here. We can use balance of fluxes taking into account the fact that more than one species are involved in the transport.

• Solute redistribution: as it occurs during solidification. We will look at a sample solution.

Unlike the parameters used for Momentum transfer (Velocity) and heat transfer (Temperature), the parameters used for mass transfer are varied in their physical meaning as well as in their units.

• Composition (mass per unit volume): default variable for us

• Atom/Mole fraction or percentage or ppm: popular in physical metallurgy

• Weight fraction or percentage or ppm: a practicable quantity used in process metallurgy

• Partial pressure: useful unit when dealing with gases

In our discussion, unless otherwise specified, we deal with the composition as our variable and convert the rest of the quantities to composition when necessary.

In addition, we must also specify for which of the species are we writing down the balance equation. Meaning, instead of C as our variable, really we have Ci as our set of variables where i goes over the range of number of species participating in the mass transfer. Unless specified, we will be dealing with a situation where the balance is written down only for one species.

81

�

� �

� � � �

�

�

� � �

CHAPTER 8. MASS TRANSFER

8.2 Governing equation

First we write down the balance of mass for a control volume and then write the same in a geometry free form using integrals.

Rate of increase in mass = influx of mass + rate of generation

dCdV = − j · ndS +ˆ gdV

dt V S V

Take the material derivative into the integral using the Reynold’s Transport Theorem. Convert the suface integral into volume integral using Divergence theorem. Recognize the balance of integrands as every integral is over the same arbitrary control volume.

dC= −∇ · j + g

dt

Expanding the material / complete derivative

( )∂C+ u · ∇ C = −∇ · j + g (8.1)

∂t

We now seek a linear constitutive relation to determine j. Refer to the section 43.14 of Feynman Lecture Series for more discussion at this point. Flux of a species is usually expressed as a product of composition of that species and the velocity of that species.

j = vC

If velocity of a species is the effect, then the cause is a gradient in chemical potential - meaning that atoms move to reduce the free energy of the system. The quantity that connects a cause and effect is a material property and one can use relevant theories to deduce the minimum number of entities necessary to represent that property.

∂µv = −M

∂x

M is called the mobility tensor. Using Onsager’s reciprocal relations [Ons31a, Ons31b], one can say that M is a symemtric tensor. It is isotropic for isotropic media. For our discussion, we are usually limited to liquids, gases, polycrystalline materials that can be approximated to istotropic media.

∂µj = −MC

∂x

Using an ideal solution approximation,

µ = µ0 + RT lnXγ = µ0 + RT (lnX + ln γ)

Where, X is mole fraction and is proportional to C and γ is the activity of the species in the solution.


� �

� � ��

�

8.2. GOVERNING EQUATION

∂µ ∂µ ∂C ∂µ ∂C−MC = −MC = −M∂x ∂C ∂x ∂ lnC ∂x

Since composition can be related as

C = XMwρ

∂µ ∂µ=

∂ lnC ∂ lnX

∂µ ∂ ln γ−M = −MRT 1 + = −D∂ lnX ∂ lnX

Where, D is the diffusivity of the solute in the solution. Thus, the Fick’s first law of solute diffusion goes as

∂Cj = −D (8.2)

∂x

Substituting the same in the balance equation,

( )∂C ∂C+ u · ∇ C = ∇ · D + g (8.3)

∂t ∂x

Assuming the D is constant over the range of variation of C, absence of any generation term and that the domain is solid, we get the Fick’s second law of solute diffusion

∂C= D∇2C (8.4)

∂t

8.2.1 Diffusivity

Unlike the relevant properties for momentum and heat transfer, the kind of diffusivity to be used depends on the situation of mass transfer problem. Following are some examples.

• Diffusivity in gas: Using kinetic theory of gases,

1DAA = λu

3

Where, λ is mean free path and u is the average velocity of the atoms in the gas.

kBTλ = √

2πσ2 ρA

8kBNTu =

πMA

Thus, the diffusivity in gas goes as T 3/2 .



• Diffusivity in gas entrapped in a pore: In case the gas is entrapped in a pore, instead of the mean free path from the kinetic theory of gases, we are supposed to use the pore diameter. This means the temperature dependence of Diffusivity goes as T 1/2 .

• Liquid Diffusivity: Using the Einstein equation,

kBT = 6πrAµBDAB

• Solid Diffusivity: Arrhenius variation

−Q RT D = D0e

The activation energy for Diffusion in solid Q depends on the way diffusion takes place and following are some situations where they are different:

– Diffusion aided by non-equilibrium defect concentration eg., due to quenched-in va-cancies

– Pipe diffusion aided by dislocations

– Grain boundary aided diffusion

– Surface diffusion

– Stress induced diffusion

– Bulk diffusion

8.3 Solid state diffusion

We consider two simple cases of solid state diffusion.

8.3.1 Fixed boundary compositions

In the domain, composition is fixed with the following boundary conditions: C = Cs at x = 0 and t > 0 C = C∞ at x = ∞ C = Cs at t = ∞ and x > 0

The governing equation is

∂C ∂2C= D

∂t ∂x2

By using a variable substition η = √x , one can integrate the equation subject to the above 2 Dt

boundary conditions as:

C − C∞ = 1 − erf

x√ = erfc x√

Cs − C∞ 2 Dt 2 Dt


� �

8.3. SOLID STATE DIFFUSION

C

Cs

C∞

x 0

Figure 8.1: Fixed Compositions

8.3.2 Fixed total solute content

In the domain, the total solute is fixed with the following boundary condition:

� ∞

Cdx = M−∞

The governing equation to be solved is same as Fick’s second law of solute diffusion. In order to figure out the form of C that will satisfy the governing equation as well as the boundary condition, we use the property of erf(x).

� ∞ 22 x 1 erf(∞) = 1 = √ exp(− ) √ dx

π 0 4Dt 2 Dt

Using algebraic manipulations,

∞ 2 ∞M x√ exp(− )dx = M = Cdx−∞ 2 πDt 4Dt −∞

One can verify that the following form of solution for C will not only satisfy the boundary condition as given above but also the governing equation.

2M xC = √ exp(− )

2 πDt 4Dt

8.3.3 Flux and concentration

Most of the time, one is interested in determining flux of a species as a function of the concen-tration of that species.



2

Sievert’s Law For absorption of gases in metals, one can use the following expression to obtain the concentration.

1 H2 → [H ]

Equilibrium constant is given by

[H ]K = √

pH2

If the concentration of hydrogen is given in ppm and the partial pressure in atm, then K would √ be in ppm/ atm. One must watch the units of K to know the convention followed. This is because pressure could be in units such as bar (atm), torr (mmHg), mbar or Pa.

Conversion of partial pressure of a gas into concentration is done by approximating the gas to be an ideal gas.

pAV = nART

nA pACA = =

V RT

Here, CA is given in mol/vol and may need to be converted to mass/vol using molar weight and density.

8.4 Mass transfer with advection

In a situation where the domain consists of species A and B and we are interested in the flux of species A, there are following possible situations.

• Species B is not present / species A is very dilute in B and thus B can be said to not take part in the diffusion process. In this situation, the flux is given by Fick’s first law.

• Species B is present and takes part in the diffusion process. Additionally, an assumption can be made that net flux of B is zero and species B moves only to replace species A. This approach is called stagnant layer approach.

• Species B is present and the domain is not stagnant. In such a situation, the influence of bulk velocity on the flux of both species be determined separately.

The first case is simple as discussed in the previous section. For the second case, we can proceed by assuming that the flux is one dimensional and the velocity of the species A is v which is same (but opposite in direction) for species B.

We can first derive the expression of flux to accommodate effect of advection into Fick’s

first law.

Consider the mass balance equation in cartesian co-ordinate system:

∂CA ∂CA ∂CA ∂CA ∂ ∂CA ∂ ∂CA ∂ ∂CA + u + v + w = DAB + DAB + DAB + g

∂t ∂x ∂y ∂z ∂x ∂x ∂y ∂y ∂z ∂z


� � � � � �

� � � � �

� � �

�

8.4. MASS TRANSFER WITH ADVECTION

Assuming the fluid to be incompressible, the continuity equation can be written as:

∂u ∂v ∂w+ + = 0

∂x ∂y ∂z

Multiplying the above equation with CA and additing to the balance equation,

∂CA ∂CA ∂u ∂CA ∂v ∂CA ∂w ∂ ∂CA ∂ ∂CA ∂ ∂CA +u +CA +v +CA +w +CA = DAB + DAB + DAB

∂t ∂x ∂x ∂y ∂y ∂z ∂z ∂x ∂x ∂y ∂y ∂z ∂z

∂CA ∂uCA ∂vCA ∂wCA ∂ ∂CA ∂ ∂CA ∂ ∂CA + + + = DAB + DAB + DAB

∂t ∂x ∂y ∂z ∂x ∂x ∂y ∂y ∂z ∂z

∂CA ∂ ∂CA ∂ ∂CA ∂ ∂CA = DAB − uCA + DAB − vCA + DAB − wCA

∂t ∂x ∂x ∂y ∂y ∂z ∂z

( ) ( )∂CA = ∇ · DAB∇CA − uCA = ∇ · −jA

∂t

Thus, the modified Fick’s first law that takes advection into account can be given as:

jA = −DAB∇CA + uCA

Where, jA is the net flux of species A taking the advection into account.

8.4.1 Stagnant layer approach

Consider a situation such as in the figure 8.2. At the bottom layer, species A is being generated and we are interested in the net flux of A and its distribution in the layer above. It can be taken that in the stagnant layer, there is no net flux of B. The net flux equations are:

∂CAjA = −DAB + uCA

∂x

∂CBjB = −DAB + uCB = 0

∂xor

DAB ∂CB u =

CB ∂x

If only species A and B are present,

CA + CB = 1

such that ∂CB ∂CA

= − ∂x ∂x

Substituting in the equation of net flux of A,


0000000000011111111111

0000000000000000011111111111111111

000000

111111

00000000000

� �

� �


B

B B

0000000000000000000000000000000000000000000000000000000

111111111111111111111111111111111111111111111111111111111111111111

BA

A

B

Figure 8.2: Stagnant layer approach

∂CA DAB ∂CAjA = −DAB − CA

∂x (1 − CA) ∂x

jA = −DAB 1

(1 − CA)

∂CA

∂x

At steady state,

∇ · jA = 0

or

∂ 1 ∂CA = 0

∂x (1 − CA) ∂x

Integrating twice with the boundary conditions:

• Considering that species B is being flushed at the top, concentration of species A at x = L, the height of the layer can be taken as CAL, usually zero.

• At x = 0, CA = CA0

we obtain the solution for distribution of CA in the stagnant layer as:

1 − CA x 1 − CAL ln = ln

1 − CA0 L 1 − CA0



Using the above expression,

∂CA (1 − CA) 1 − CAL = − ln

∂x L 1 − CA0

Substiting the same to obtain jA,

DAB 1 − CAL jA = ln

L 1 − CA0

In the limit of CA0 = 0 and the value of CAL being small, the above equation reduces to CAL jA = −DAB L

, which is the same as Fick’s first law valid for the case of no advection. Thus, the stagnant layer approach with very dilute species can be approximated to the regime where Fick’s law is applicable even though the other species is also taking part in the diffusion process.

It is easy to see that instead of concentrations, if one were to use partial pressures of the respective species, one would have the following equations valid:

pA + pB = p

pA = RTCA

p− pA x p− pAL ln = ln p− pA0 L p− pA0

pDAB p− pAL jA = ln

RTL p− pA0

jA when expressed in rate of loss of mass can be accuarately measured and thus provides an excellent means to obtain DAB by weight loss method using a setup similar to that of figure 8.2.

8.4.2 Mass transfer coefficient

In many situations, it may be favourable to define a mass transfer coefficient akin to heat transfer coefficient such that one can experimentally or otherwise determine it. The definition is as given below:

jA = kA (CA − CA0)

If the units of CA, as used commonly, are mass/vol then the units of kA are same as those of velocity. Once the mass transfer coefficient is known, flux can be determined using the above equation. In situations such as solid state diffusion, kA can be determined by evaluating the slope of the species distribution (say, error function) itself.

Eg., for transient diffusion in solid / quiescent liquid whose bulk composition is CA0 and is in contact with atmosphere with zero concentration of species A as illustrated in the figure, we can write the following flux balance.

Writing the composition distribution for species A in the domain,


� �

�

�

�

�

�

�

�

�

�

�

�

�


CA0

CA = 0 jA = kA(CA0 − 0)

CA

x

Figure 8.3: Surface Renewal Approach

CA x= erf √

CA0 2 DABt

Evaluating the flux at the interface at time t,

„ «2

√ x∂CA 2 − 1 2 DABtkA,t (CA0 − 0) = jA = | − DAB | = DABCA0 √ e √

∂x πx→0 DABtx→0

DAB kA,t = 2

πt

This is the instantaneous mass transfer coefficient. The time averaged coefficient will be ob-tained by averaging the coefficient over the time interval t = 0 to t = t.

DAB kA =

πt

In case of a steadily falling (along z axis) film of liquid, the quantity z plays the role of uz,max

time. The z averaged mass transfer coefficient will then be given as

DABuz,max kA =

πz

8.4.3 Sherwood number

Like the heat transfer coefficient is obtained in non-dimensionalized form by encapsulating it with the thermal conductivity and characteristic length scale as Nusselt number, mass transfer


�

�

�

�

�

�

�

�

�

�

�

�


coefficient is also obtained in non-dimensionalized form by encapsulating it with the solute diffusivity and characteristic length scale as Sherwood number.

kALcSh =

DAB

Sherwood number is determined experimentally as a function of Reynold’s number and Schmidt number to obtain correlations for advection aided mass transfer.

νSc =

DAB

Examples:

Mass transfer from a sphere:

13ShD = 2 + CRem

DSc

The constants C and m depend on the range of Re and Sc.

Mass transfer from a plate of length L (for Re < 2 × 105):

ShL = 0.664ReL 0.5Sc

8.4.4 Chilton-Colburn Analogy

13

In case the transport properties are known for one quantity (say, heat) and we are interested in the transport of another quantity (say, solute), we can make use of the similarity expressions thanks to the fact that under laminar conditions and when fully developed profiles are present, similarity relations are valid.

The problem is to connect friction factor, heat transfer coefficient and mass transfer coefficient at an interface.

If the velocity, temperature and solute profiles are fully developed, then the following three expressions will all be the same functions (eg., parabolas for a tube):

v T − Ts CA − CA0 = =

v∞ T∞ − Ts CA,∞ − CA0

Equating the slopes at the interface since the profiles are the same:

∂ v ∂ T − Ts ∂ CA − CA0 = =

∂x v∞ x→0 ∂x T∞ − Ts x→0 ∂x CA,∞ − CA0 x→0

Take the case of a hypothetical fluid that has Pr = Sc = 1 ie.,

µCp µ= = 1

k ρDAB

or


�

�

�

�

�

�

�

�

�

�

�

�


kµ = = ρDAB

Cp

Multiply these quantities with the respective slopes of the profiles as:

∂ v k ∂ T − Ts ∂ CA − CA0 µ = = ρDAB ∂x v∞ Cp ∂x T∞ − Ts ∂x CA,∞ − CA0 x→0x→0 x→0

τ q ρjA = =

v∞ Cp(T∞ − Ts) (CA∞ − CA0)

Dividing each quantity with ρv∞,

1 τ q jA = = 1ρv2 ρCp(T∞ − Ts)v∞ (CA∞ − CA0)v∞2 ∞2

Recognize the following quantities:

Skin friction coefficient: τ

= f1ρv2 2 ∞

Heat transfer coefficient: q

= h(T∞ − Ts)

Mass transfer coefficient: jA

= kA(CA∞ − CA0)

f h kA = =

2 ρCpv∞ v∞

We define Stanton number as:

hSt =

ρCpv∞

The following analogy is called as Reynold’s analogy:

f kA = St =

2 v∞

Chilton-Colburn realized that the above expression can be made to be applicable even whenPr and Sc are not unity but for a range of 0.6 < Pr < 100 and 0.6 < Sc < 2500 provided a

23

23correction factor of Pr for heat transfer and Sc for mass tranfer are used:

23

23

f kA = StPr = Sc

2 v∞

Each of the above three quantities are called as j-factors.


8.5. REACTION MASS TRANSFER

h= PrρCpv∞

23

23jH = StPr

kAjD = Sc

v∞

23

In situations where the analogy is applicable, one can obtain the mass transfer coefficient from the heat transfer coefficient or the skin friction factor by using the above analogy and the appropriate property values.

8.5 Reaction mass transfer

In a reaction, we are interested in the rate of formation of a particular product, usually a condensed one. It can usually be obtained by knowing the flux of that species by balacing with the flux of a diffusing species and applying appropriate boundary conditions. The compositions of the participating species are related by the equilibrium constant of the reaction. Consider the following example.

A(g) → nB(g) + C(s)

[B]n pBn

K = = [A] pA

pA + pB = p

One can solve for pB by solving the above two equations.

One has to know the value of K at a given temperature and should watch out the units of Kto know the convention for the units of [A] and [B].

Once the value of K is known, it is possible to obtain the flux of one of the species, say A, using either the Fick’s law or the stagnant layer approach depending on how dilute the system is. The fluxes of the diffusing species A and B are related to as:

jB = njA = njC

If B is a dilute species in a stagnant layer of A of height L and a total pressure p, one can use the following expression for jB

pDAB p− pBL jB = ln

RTL p− pB0

The quantity jC can then be converted to appropriate rate of formation as mass or volume or moles or thickness as desired using the properties of C.


Appendix A

Derivations

A.1 The quotient rule

Adapted from section 2.62 of [Ari62].

Let bi be any vector and aij be a matrix of nine numbers (i, j = 1, 2, 3). If a relation could be found such that aijbi = cj is vector, then aij is a tensor of order 2.

Consider a co-ordinate transformation with the tranformation matrix Tij . Since b and c are vectors,

b∗ p = Tpibi or bi = Tpibp ∗ (A.1)

c∗ q = Tqjcj

In the new co-ordinate system:

∗ b∗ ∗ apq p = cq = Tqjcj = Tqjaijbi = TqjaijTpibp ∗

(a∗ = 0pq − TpiTqjaij)b∗ p

Since b∗ p is only the arbitrary vector bi in new co-ordinate system which is independent of aij

the only way the above equation can hold is if

∗ a = TpiTqjaij pq

which is the definition for the entity aij to be a tensor of order 2.s

A.2 Symmetric tensors are diagonalisable


If aij is a tensor and bi is an arbitrary vector then by quotient rule, cj = aijbi is a vector. For some bi, the vector ci could be in the same orientation such that if λ is the ratio of their magnitudes, aijbi = λbj = λδijbi .

(aij − λδij) bi = 0

94

A.2. SYMMETRIC TENSORS ARE DIAGONALISABLE

This is a set of three unknowns (a1, a2 and a3) and for a solution to exist, λ must satisfy the following equation:

det (aij − λδij) = 0

λ3 −Xλ2 + Y λ− Z = 0

X = aii = Trace(aij)

Y = a22a33 − a23a32 + a11a33 − a13a31 + a22a11 − a12a21

Z = det(aij)

X, Y and Z are three invariants of aij under rotation of co-ordinate system.

The above is called the characteristic equation of the tensor aij and the three values of λ its characteristic values / latent roots / eigen values.

For the three eigen values λ(p), p = 1 : 3, the corresponding characteristic vectors (bi(p), p = 1 : 3 be denoted as Tj1, Tj2 and Tj3.

b1(1) b1(2) b1(3) T11 T12 T13

= T = b2(1) b2(2) b2(3) T21 T22 T23 (A.2) b3(1) b3(2) b3(3) T31 T32 T33

aijTj1 = λ1Ti1 (A.3) aijTj2 = λ2Ti2

Multiply the first equation with Ti2 and the second equation with Ti1.

aijTj1Ti2 = λ1Ti1Ti2 (A.4) aijTj2Ti1 = λ2Ti2Ti1

Transpose the second equation (swap the indices i and j in second equation):

aijTj1Ti2 = λ1Ti1Ti2 (A.5) ajiTi2Tj1 = λ2Tj2Tj1

If aij is symmetric, then

aijTj1Ti2 = λ1Ti1Ti2 (A.6) aijTi2Tj1 = λ2Tj2Tj1

ie.,

aijTj1Ti2 = λ1Ti1Ti2 = λ2Tj2Tj1

Since λ1 and λ2 are distinct, the above equation can be true only if


�

�

�

�

�

�

�

�

�

�

�

�

APPENDIX A. DERIVATIONS

Ti1Ti2 = δij (A.7)

If we choose a co-ordinate transformation with the transformation matrix to contain the ele-ments of Tij made out of the three eigen vectors,

∗ apq = TipTjqaij = λpTipTiq = λpδpq

ie., apq is diagonal.

A.3 Levi-Civita tensor is isotropic


Since ǫijk is a tensor of order 3, for an arbitrary co-ordinate transformation Tij ,

T1p T1q T1r

TipTjqTkrǫijk = T2p T2q T2r (A.8) T3p T3q T3r

In the above determinant, if p = q or p = r or q = r, two columns being same will make the R.H.S. go to 0. If p, q and r are cyclic (e.g., 1, 2, 3) then the R.H.S. is the determinant of the transformation matrix itself which is 1. If p, q and r are anti-cyclic (e.g., 1, 3, 2), then the R.H.S. is the determinant of the transformation matrix (with two rows interchanged) which is -1.

These are the values defined for ǫpqr. Hence the R.H.S. can be equated to ǫ∗ . Thus, pqr

ǫ∗ pqr = TipTjqTkrǫijk

Since the above result is applicable for any Tij , ǫijk is isotropic.

A.4 General form of isotropic tensor of order four


An isotropic tensor aijkl of order 4 should be invariant under any co-ordinate transformationTij .

apqrs = TipTjqTkrTlsaijkl

We categorise the 81 components of aijkl : i, j, k, l = 1, 2, 3 in to classes to figure out the non-zero (independent) components.


A.4. GENERAL FORM OF ISOTROPIC TENSOR OF ORDER FOUR

Class component remark I II III(i) III(ii) III(iii) IV

T1111

T1112

T1122

T1221

T1212

T1123

All suffixes are same (nnnn) Three suffixes are same (nnnm) Two suffixes are same (nnmm) Two suffixes are same (nmmn) Two suffixes are same (nmnm) Only two suffixes are same (mmno)

Since class II doesnot include the elements of class I, introduce a tensor Xijkl defined as below:

if i = j = k = l Xijkl = 1 (A.9)

else Xijkl = 0

We now choose co-ordinate rotation operations so that Tij can take such values that will make a conclusion about the relation between components in each of the above mentioned classes is clear.

Operation Description form of Tij

A B C

Rotation about [111] axis by 120o

Rotation about [001] axis by 90o

Rotation to inverse the direction of [111]

T12 = T31 = T23 = 1 T12 = −T21 = T33 = 1 T13 = −T22 = T31 = −1



Class conclusion operation I T1111 = T2222

T1111 = T2222

T1111 = T3333

all equal Xijkl

A B C Overall conclusion Representation

II T1112 = T2223

T1112 = −T2221

T1112 = T3332

T2223 = T2221

all zero

A B C C Overall conclusion

III(i) T1122 = T2233

T1122 = T2211

T1122 = T3322

all equal δijδkl −Xijkl


III(ii) T1221 = T2332

T1221 = T2112

T1221 = T3223

all equal δilδjk −Xiljk


III(iii) T1212 = T2323

T1212 = T2112

T1212 = T3232

all equal δikδjl −Xikjl


IV T1123 = T2231

T1123 = −T2213

T1123 = T3321

T2231 = T2213

all zero

A B C C Overall conclusion

Since there is no obvious relation between the components of classes I, III(i), III(ii) and III(iii), we can represent aijkl by a linear combination of each class:

aijkl = µ1(δijδkl −Xijkl) + µ2(δikδjl −Xijkl) + µ3(δilδjk −Xijkl) + µ4Xijkl

or

aijkl = µ1δijδkl + µ2δikδjl + µ3δilδjk + (µ4 − µ1 − µ2 − µ3)Xijkl

Choose an operation D : arbitrary rotation by some angle. Given the form of Xijkl, the elements will not remain the same under such a transformation. Hence the last term in the above equation drops off giving the most general form of a fourth order isotropic tensor as:

aijkl = µ1δijδkl + µ2δikδjl + µ3δilδjk


A.5. SIMPLIFICATION OF TENSOR PROPERTIES FOR CRYSTALS

A.5 Simplification of tensor properties for crystals

Lars Onsager has proved ([Ons31a],[Ons31b]) using microscopic reversibility that for crystals with symmetries of order 3,4,6 etc., the thermal conductivity tensor is symmetric.

k11 k12 k13

kij = k12 k22 k23 (A.10)

k13 k23 k33

For cubic crystals, the form of the tensors simplifies even further. Take co-ordinate system rotations about x3 by 900 for which the transformation matrix is:

0 1 0 T = −1 0 0 (A.11)

0 0 1

By definition, if T is the transformation matrix of a co-ordinate axes rotation,

k∗ ij = TpiTqjkpq

Since the tensor in discussion is a property that should not change upon co-ordinate rotations that leave the crystal identical,

k∗ = k

Take k11 and expand the above definition of tensor:

k11 = Ti1Tj2kij = T21T21k22 = k22

or

k11 = k22

Similarly,

k21 = Ti2Tj1kij = T12T21k12 = −k12 = −k21

or

k21 = k12 = 0

Similarly rotating the co-ordinate system about the other two axes will show that all off diagonal terms of kij vanish and all diagonal terms are same.

k11 0 0 kij = 0 k11 0 = kδij (A.12)

0 0 k11

Thus, only one value of k is needed to completely specify the second order symmetric tensor property such as thermal conductivity of a cubic crystal.


�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�


A.6 Change of variable with multiple integrals

Adapated from section 3.16 of [Ari62]

In cartesian co-ordinates the element of volume dV is the volume of rectangular parallelepiped of sides dx1, dx2 and dx3:

dV = dx1dx2dx3

It is sometimes convenient to describe position by some other co-ordinates, say, ξ1, ξ2 and ξ3. The change of co-ordinate system must be given by fixing a point x such that

xi = xi(ξ1, ξ2, ξ3)

Small differences in the new co-ordinate system are given by

∂xidxi = dξj

∂ξj

If dx(j) are vectors with components (∂xi dξj) for j = 1, 2, 3 then the volume of the element is ∂ξj

dx(1) (dx(2) × dx(3))dV = ·

Using ǫijk to write the short form of determinant (triple product),

∂xi ∂xi ∂xidV = ǫijk dξ1 dξ2 dξ3

∂ξ1 ∂ξ2 ∂ξ3

dV = Jdξ1ξ2dξ3

where J is the Jacobian of the transformation of variables.

∂(x1, x2, x3) ∂xi ∂xj ∂xkJ = = ǫijk

∂(ξ1, ξ2, ξ3) ∂ξ1 ∂ξ2 ∂ξ3

In determinant form, it is

∂x1

∂ξ1 ∂x1

∂ξ2 ∂x1

∂ξ3

J = ∂x2

∂ξ1 ∂x2

∂ξ2 ∂x2

∂ξ3 (A.13)

∂x3 ∂x3 ∂x3

∂ξ1 ∂ξ2 ∂ξ3

Evidently, when two of the rows are same, the determinant is zero.

A.7 Dilation

Adapated from section 4.21 of [Ari62]


� � � �

��

��

� ��

� �

A.7. DILATION

or

dV = Jdξ1ξ2dξ3 = JdV0

dVJ =

dV0

The change in dilation as we follow the motion is given by the material derivative dJ .dt

d ∂xi ∂ dxi ∂ui = =

dt ∂ξj ∂ξj dt ∂ξj

Since ui is a function of x1, x2 and x3,

∂ui ∂ui ∂x1 ∂ui ∂x2 ∂ui ∂x3 ∂ui ∂xl = + + =

∂ξj ∂x1 ∂ξj ∂x2 ∂ξj ∂x3 ∂ξj ∂xl ∂ξj

Using differentiation by parts in

∂xi ∂xj ∂xkJ = ǫijk

∂ξ1 ∂ξ2 ∂ξ3

dJ ∂ui ∂xj ∂xk ∂xi ∂uj ∂xk ∂xi ∂xj ∂uk = ǫijk + +

dt ∂ξ1 ∂ξ2 ∂ξ3 ∂ξ1 ∂ξ2 ∂ξ3 ∂ξ1 ∂ξ2 ∂ξ3

dJ ∂ui ∂xl ∂xj ∂xk ∂xi ∂uj ∂xl ∂xk ∂xi ∂xj ∂uk ∂xl = ǫijk + +

dt ∂xl ∂ξ1 ∂ξ2 ∂ξ3 ∂ξ1 ∂xl ∂ξ2 ∂ξ3 ∂ξ1 ∂ξ2 ∂xl ∂ξk

Each of the terms in the above equation is expressible as a determinant shown in the previous section. The dummy index l can take values from 1 to 3. The first term is non zero only when l = i and the second when l = j and the thrid when l = k as the determinant goes to zero when two of its rows are same. Hence,

dJ ∂ui ∂vj ∂vk ∂xi ∂xj ∂xk = ǫijk + + + +

dt ∂xi ∂xj ∂xk ∂ξ1 ∂ξ2 ∂ξ3

or

dJ= J [∇iui]

dt

Divergence of a velocity field ∇iui can now be interpreted as the rate of dilation or rate of change of elemental volume following the flow path. Since incompressible fluids are defined as those with no dilatation during flow, ∇iui = 0 or ∇· u = 0 is the condition for incompressible fluid flow.


�

�

�

�

�

�

�

�

� �

� � ( )

� ( )

� �

� � �


A.8 Reynold’s transport theorem


Let f(x, t) be any function and V (t) be a closed volume moving with the fluid ie., consisting of the same fluid particles.

f(t) = f(x, t)dVV (t)

is a function of t that can be calculated. We are interested in the material derivative Df . Since Dt

the integral is varying over V (t), differentiation cannot be taken through the integral sign. If the differentiation is with respect to a volume in the material co-ordinate system (ξ1, ξ2, ξ3), it would be possible to interchange differentiation and integration since D is differentiation with

Dt

respect to time keeping ξ constant.

The transformation x = x(ξ, t) with V = JdV0 allows us to this, for V (t) has been defined as a moving material volume and so came from some fixed V0 at time t = 0.

D Df(x, t)dV = f(x(ξ, t), t)JdV0

Dt DtV (t) V0

Since the integral is over the same volume V0, we can take the differentiation operator inside the integral.

D df J + f dJ f(x, t)dV = dt

dV0Dt V0) V0 dt

� (

df )

= + f(∇ · u) JdV0 (A.14) V0 dt

= df + f(∇ · u) dVV0 dt

Using the expression for material derivative

D ∂= + (u · ∇)

Dt ∂t

D ∂ff(x, t)dV = + ∇ · (fu) dV

Dt ∂tV (t) V0

Apply Green’s theorem to the second term, with S(t) as the surface of the element following the fluid flow and n as the unit normal to S(t):

D ∂ff(x, t)dV = + fu · ndSˆ

Dt ∂tV (t) V0 S(t)

Rate of change of the integral of any function f within a moving element is the sum of integral of rate of change at a location and the net flow of f over the surface enclosing the element.


�

� �

�

� �

�

� ��

�

��

��

� � �

� � �

� � �

� � �

� �

�

�

A.9. RTT AND CONTINUITY EQUATION

A.9 RTT and Continuity Equation

Using f = ρ in the Reynold’s Transport Theorem given in section A.8, one recovers continuity equation.

Using f = ρG in RTT given in section A.8 where G is any dynamical property per unit mass (such as G = ui, momentum per unit mass) one can bring the material derivative inside the integration.

D ∂ρGρGdV = dV + ρGu · ndSˆ

Dt ∂tV V S

Using Gauss theorem to convert a surface integral to volume integral,

D ∂ρ ∂GρGdV = G + ρ dV + ∇ · (ρuiF ) dV

Dt ∂t ∂tV V V

Keeping ρui together and expanding the integrand of the second integral,

D ∂ρ ∂GρGdV = G + ρ dV + G∇ · (ρui) + (ρui) · ∇G dV

Dt ∂t ∂tV V V

Regrouping,

( )D ∂ρ ∂GρGdV = G + ∇ · (ρui) dV + ρ + ui · ∇ G dV

Dt ∂t ∂tV V V

Recognising that the term in square brackets in the first integral is zero because of continu-ity equation and the integrand of the second integral could be simplified using the material derivative,

D DGρGdV = ρ dV

Dt DtV V

A.10 Cauchy’s stress principle


Let n be the unit outward normal at a point of the surface S and σ(n) the force per unit area exerted there by the material outside S. Then Cauchy’s principle asserts that σ(n) is a function of the position x, the time t and the orientation n of the surface element. Thus the total internal force exerted on the volume V through its bounding surface S is

σ(n)dSS

If f is the external force per unit mass (e.g., f = −gx3), the total external force will be

ρfdVV


� �

� � �

�

� �

� �


The principle of conservation of linear momentum asserts that the sum of these two forces equals the rate of change of linear momentum of the volume.

dρvdV = ρfdV + σ(n)dS

dt V V S

If V is a volume of a given shape with a characteristic dimension d then V ∼ d3 and S ∼ d2 . As we let V shrink on a point but preserve the shape, the first two integrals decrease as d3

where as the last will as d2 . So,

1 lim σ(n)dS = 0 d→0 d2

S

ie., the stresses are locally in equilibrium.

A.11 Stress is a tensor

Adapted from section 5.12 of [Ari62] and section 1.3, pg.9 of [Bat67].

Stress σ is defined as force per unit area. Consider all the forces acting instantaneously on the fluid within an element of volume δV in the shape of a tetrahedron. The three orthogonal faces have areas δA1, δA2 and δA3 and unit outward normals as −a, −b and −c. The fourth inclined face has area δA and unit normal n. The resultant of surface forces is

σ(n)δA+ σ(−a)δA1 + σ(−b)δA2 + σ(c)δA3

In view of the orthogonality of three of the faces,

δA1 = a · nδAˆ ˆ

ˆδA2 = ˆb · nδAδA3 = c · nδAˆ ˆ

Thus the balance of forces along direction i can be written as

δA σi(n) − ajσi(x1) + bjσi(x2) + cjσi(x3) nj

As we shrink the volume (V ∼ d3), since the area shrinks only as S ∼ d2, the quantity in square brackets must go to zero for local equilibrium. ie.,

σi(n) = ajσi(x1) + bjσi(x2) + cjσi(x3) nj

If we represent the quantity in the flower brackets on R.H.S. as σij ,

σi(n) = σijnj

Since the vectors σi (surface force per unit area) and n (unit normal to the surface) donot depend on the choice of co-ordinate axes, the quantity connecting them σij must represent (i, j)-component of a axes-independent entity namely, a tensor of order 2. This is also true by the quotient rule of tensors.


A.12. STRESS TENSOR IS SYMMETRIC

A.12 Stress tensor is symmetric

Adapted from section 2.7 pg. 57 of [SAH89].

Consider stress at the center of a control volume of size δx1 × δx2 × δx3 as shown in the figure. The torque produced by the forces about an axis along x3 and through the center of gravity of the CV is

T = σ12δx1δx3δx2 − σ21δx2δx3δx1

or

T = (σ12 − σ21)δx1δx2δx3

The torque may be equated to the product of angular acceleration ( α3) and the moment of inertia taken about the previously mentioned axis (x3):

(σ12 − σ21)δx1δx2δx3 = ρδx1δx2δx3(δx1

2 + δx22)α3

12

or

(σ12 − σ21) = ρ

(δx12 + δx2

2)α312

As we shrink the CV to infinitesimal size, since the term in the bracket on the right goes to zero, if the L.H.S. were to remain finite α3 must blow up. It can be prevented only if L.H.S. is zero. ie.,

σ12 = σ21

or

The tensor σij is symmetric.

A.13 Meaning of terms in strain rate tensor

Figure A.1 shows how a fluid element deforms during its motion. The deformation can be analysed into different modes and the respective strains and strain rates can be expressed in terms of the velocity gradients as shown below. Initial Positions:

A x0 y0

B (x0 + Δx) y0

D x0 (y0 + Δy)

Velocities:

A u vB (u+ ∂u

∂xΔx) (v + ∂v

∂xΔx)

D (u+ ∂u ∂y

Δy) (v + ∂v ∂y

Δy)


� �

� �


A B

CD

Rotation DilationTranslation Shear

R

S

β Q

α P

Figure A.1: Analysis of deformation of a fluid element

Positions after dt:

P x0 + udt y0 + vdtQ (x0 + Δx) + (u+ ∂u

∂xΔx)dt y0 + (v + ∂v

∂xΔx)dt

S x0 + (u+ ∂u ∂y

Δy)dt (y0 + Δy) + (v + ∂v ∂y

Δy)dt

Dilational strain along x is PQx−AB . Shear strain along y is PQy .AB AB

PSy−AD Dilational strain along y is . Shear strain along x is PSx .AD AD

Dilational Strain rate: ∂u Δxdt 1 ∂u∂xe11 = =

Δx dt ∂x

∂v Δydt∂y 1 ∂v

e22 = Δy dt

= ∂y

Shear Strains:

α = ∂v ∂x

Δxdt= ∂vdt

Δx ∂x

β =

∂u ∂y

Δydt= ∂udt

Δy ∂y

Pure shear strain rate: 1 1 1 ∂v ∂u

e12 = 2(α+ β)

dt=

2 ∂x+ ∂y

Pure rotational rate: 1 1 1 ∂v ∂u

Ω12 = 2(α− β)

dt=

2 ∂x− ∂y


� � � �

A.14. VELOCITY GRADIENT IS A TENSOR

A.14 Velocity gradient is a tensor

Adapted from section 4.41 of [Ari62]

Consider two points P and Q at co-ordinates ξ and ξ + dξ. At time t they are to be found atx(ξ, t) and x(ξ + dξ, t).

Using only the first term of Taylor’s expansion,

∂xi xi(ξ + dξ, t) = xi(ξ, t) + dξj

∂ξj

So the displacement vector is

dx = x(ξ + dξ, t) − x(ξ, t)

where ∂xi

dxi = dξj∂ξj

Since dx and dξ are vectors, by quotient rule, the nine quantities ∂xi form the components of ∂ξj

a tensor of order 2.

Since we define velocity v as v = dx , the relative velocity of two fluid particles at ξ and ξ + dξdt

is

∂ui d ∂xidui = dξk = dξj

∂ξk dt ∂ξj

Using the inverse of the relation given above,

∂ui ∂ξk ∂uidui = dxj = dxj

∂ξk ∂xj ∂xj

Once again, by quotient rule, the components of ∂ui form a tensor or order 2. This is called ∂xj

velocity gradient tensor. Like any tensor of order two, velocity gradient tensor can be split into two tensors, one symmetric and one anti-symmetric.

∂ui 1 ∂ui ∂uj 1 ∂ui ∂uj= + + −

∂xj 2 ∂xj ∂xi 2 ∂xj ∂xi

We denote the symmetric part of the velocity gradient tensor as eij (strain rate tensor) and the anti-symmetric part as Ωij (vorticity tensor).

∂ui = eij + Ωij

∂xj



As shown in the section A.13, we can recognise that eij represents rate of strain (both dilational and shear) and Ωij represents rigid body rotation.


Bibliography

[Ari62] Rutherford Aris. Vectors, Tensors and the Basic Equations of Fluid Mechanics. Prentice Hall Inc, Eaglewood Cliffs, N.J., 1962.

[Bat67] G.K. Batchelor. An Introduction to Fluid Dynamics. Cambridge University Press, Eaglewood Cliffs, N.J., 1967.

[BSL02] R. Byron Bird, Warren E. Stewart, and Edwin N. Lightfoot. Transport Phenomena. John Wiley and Sons, 2002.

[Gas92] David R. Gaskell. An Introduction to Transport Phenomena in Materials Engineering. Macmillan Publishing Company, New York 10022, 1992.

[GP94] D.R. Geiger and G.H. Poirier. Transport Phenomena in Materials Processing. TMS, Warrendale, PA, USA, 1994.

[Ons31a] Lars Onsager. Reciprocal relations in irreversible processes - i. Physical Review, 37:405–426, 1931.

[Ons31b] Lars Onsager. Reciprocal relations in irreversible processes - ii. Physical Review, 38:2265–2279, 1931.

[PT83] N.V. Perelomova and M.M. Tagieva. Problems in Crystal Physics with solutions. Mir Publishers, Moscow, 1983.

[SAH89] Rolf H. Sabersky, Allan J. Acosta, and Edward G. Hauptmann. Fluid Flow : A first

course in Fluid Mechanics. Macmillan Publishing Company, New York, 3 edition, 1989.

109

Phanikumar_TransportPhenomenaNotes_01Feb2010

Documents

Transcript of Phanikumar_TransportPhenomenaNotes_01Feb2010