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CHAPTER 1
BASIC CONCEPTS
1.1 Convection Heat Transfer
Examine thermal interaction between a surface and an adjacent moving fluid.
1.2 Important Factors in Convection Heat Transfer
Three factors play major roles in convection heat transfer:
(1) Fluid motion
(2) Fluid nature
(3) Surface geometry.
1.3 Focal Point in Convection Heat Transfer
The determination of temperature distribution in a moving fluid can be expressed as
T
T ( x, y, z, t )
(1.1)
1.4 The Continuum and Thermodynamic Equilibrium Concepts
Continuum model:
Material is composed of continuous matter.
Behavior of individual molecules is ignored.
Valid when molecular man free path
is small relative to the characteristic
dimension of the system (small Knudsen number, Kn 10 1 ).
Knudson number: defined as
Kn
De
(1.2)
Thermodynamic equilibrium:
Depends molecular interaction with an adjacent surface.
At thermodynamic equilibrium:
(1) Fluid and adjacent surface have the same velocity (no velocity slip).
(2) Fluid and adjacent surface have the same temperature (no temperature jump).
Valid for
Kn 10
3
1.5 Fourier’s Law of Conduction
Gives a relationship between heat transfer rate or heat flux and temperature:
(1.3b)
2
qx
k
T
,
x
qy
k
T
,
y
qz
k
T
z
(1.8)
Used in the formulation of the governing equation for temperature distribution.
1.6 Newton's Law of Cooling
This law defines the heat transfer coefficient h in terms of heat flux and surface
temperature
qs
h (Ts
T )
(1.9)
Used in the formulation of boundary conditions.
Used to determine the rate of transfer or heat flux when temperature distribution is
difficult to obtain analytically.
1.7 The Heat Transfer Coefficient h
Defined in (1.9).
h is not a property. It depends on:
(1.10)
h f (geometry, fluid motion, fluid properties, T )
It is determined analytically using (1.8) and (1.9). This requires the determination of
temperature distribution.
h
k
T ( x,0, z )
y
Ts T
(1.12)
It is determined experimentally using Newton’s law of cooling (1.9).
Table 1.1 gives typical values for h. Use this table as a guide only.
1.8 Radiation: Stefan-Boltzmann Law
Emissivity : radiation surface property
Absorptivity : surface property defined as the fraction of radiation energy incident
on a surface which is absorbed by the surface
q12 is the net heat exchanged by radiation between two bodies.
Special case: a small surface which is completely enclosed by a much larger surface,
q12 is given by Stefan-Boltzmann radiation law
q12
1
A1 (T14
T24 )
(1.14)
3
Subscript 1 in (1.14) refers to the small surface. Temperature is in absolute
degrees.
1.9 Differential Formulation of Basic Laws
Various formulation procedures:
Differential
Integral
Variational
Finite difference
Differential formulation:
Based on the assumption of continuum.
The basic laws are applied to an infinitesimal element.
The result is a partial differential equation which is valid at every point.
1.10 Mathematical Background

(a) Velocity Vector V

V
(b) Velocity Derivative
ui v j

V
x
u
i
x
(1.15a)
wk
v
j
x
w
k
x
(1.15b)
k
(1.16)
(c) The Operator
Cartesian:
x
i
y
j
z
Cylindrical:
r
ir
1
r
1
r
i
i
z
(1.17)
iz
Spherical:
r
(d) Divergence of a Vector
ir
1
r sin
i
(1.18)
4

div.V

V
u
x
v
y
w
z
(1.19)
u
x
v
y
w
z
(1.20)
(e) Derivative of the Divergence

V
x
x
or

V
x

V
x
(1.21)
(f) Gradient of Scalar. The gradient of a scalar, such as temperature T, is a vector given by
Grad T
T
i
x
T
T
j
y
T
k
z
(1.22)
(g) Total Differential and Total Derivative
Consider the variable f:
f
Total differential of f:
and t:
f ( x, y, z, t )
This is the total change in f resulting from changes in x, y, z
f
dx
x
df
f
dy
y
f
dz
z
f
dt
t
Total derivative
df
dt
Df
Dt
u
f
x
v
f
y
f
z
f
y
w
f
z
convective derivative
w
f
t
(1.23)
where
u
f
x
v
f
t
Example: f
local derivative
u in (1.23) gives
du
dt
Du
Dt
u
u
x
v
u
y
w
u
z
u
t
(1.24)
Cylindrical coordinates:
dv r
dt
Dv r
Dt
vr
vr
r
v
r
vr
v2
r
vz
vr
z
vr
t
(1.25a)
5
dv
dt
Dv
Dt
dv z
dt
vr
Dv z
Dt
v
r
vr
v
r
v
vz
r
v
r
T
x
v
vrv
r
vz
vz
vz
vz
z
v
z
v
t
vz
t
(1.25b)
(1.25c)
Example: f = T
dT
dt
DT
Dt
u
T
y
w
T
z
T
t
(1.26)
1.11 Units
Basic SI units:
Length (L): meter (m)
Time (t): second (s)
Mass (m): kilogram (kg)
Temperature (T): kelvin (K)
Derived units:
Force (newton): N = kg-m /s2
Energy (joules): J = N m = kg-m2 /s2
Power (watts): W = J/s = N m/s = kg-m2 /s3
1.12 Problem Solving Format
Convection problems lend themselves to a systematic solution procedure. Use the following
format when solving problems. Read details of each step and take advantage of illustrative
examples and posted solutions to homework problems to develop skills in using this
problem solving methodology. The following is an outline of this method:
(1) Observations
(2) Problem Definition
(3) Solution Plan
(4) Plan Execution
(i) Assumptions
(ii) Analysis
(iii) Computations
(iv) Checking
(5) Comments
CHAPTER 2
DIFFERENTIAL FORMULATION
OF THE BASIC LAWS
2.1 Introduction
Differential formulation of basic laws:
Conservation of mass
Conservation of momentum
Conservation of energy
2.2 Flow Generation
(i) Forced convection. Motion is driven by mechanical means.
(ii) Free (natural) convection. Motion is driven by natural forces.
2.3 Laminar vs. Turbulent Flow
Laminar flow: no fluctuations in velocity, pressure, temperature, …
Turbulent flow: random fluctuations in velocity, pressure, temperature, …
Transition from laminar to turbulent flow: Determined by the Reynolds number:
Flow over a flat plate: Ret
V xt /ν
Flow through tubes: Ret = u D
v
500,000
2300
u
u
turbulent
laminar
t
t
Fig. 2.1
2.4 Conservation of Mass: The Continuity Equation
2.4.1 Cartesian Coordinates
2
The principle of conservation of mass is applied to an element dxdydz
Rate of mass added to element - Rate of mass removed from element =
Rate of mass change within element
my
y
dy
dx
mx
( my )
y
mx
x
dy
( mx )
dy
y
my
(a )
(b )
Fig. 2.2
(2.1)
Expressing each term in terms of velocity components gives continuity equation
t
x
u
v
y
w
z
(2.2a)
0
This equation can be expressed in different forms:
t
u
x
v
y
w
u
x
z
v
y
w
z
0
(2.2b)
or
D
Dt
or
t

V
0
(2.2c)

V
0
(2.2d)
For constant density (incompressible fluid):

V
2.4.2 Cylindrical Coordinates
0
(2.3)
3
t
1
r r
1
r
r vr
v
vz
z
(2.4)
0
2.4.3 Spherical Coordinates
1
t
r
2
r
1
r sin
r 2 vr
1
r sin
v sin
v
0
(2 .5)
2.5 Conservation of Momentum: The Navier-Stokes Equations of Motion
2.6 2.5.1 Cartesian Coordinates
Application of Newton’s law of motion to the
element shown in Fig. 2.5, gives

F

( m)a
dz
y
dy
(a)
dx
Application of (a) in the x-direction, gives
x
Fx
(b)
( m)a x
z
Fig. 2.5
Each term in (b) is expressed in terms of flow
field variables: density, pressure, and velocity components:
Mass of the element:
m
(c)
dxdydz
Acceleration of the element a x :
ax
du
dt
Du
Dt
Substituting (c) and (d) into (b)
Du
dxdydz
Dt
Fx
(e)
Forces:
(i) Body force
Fx
body
(ii) Surface force
g x dxdydz
(g)
u
u
x
v
u
y
w
u
z
u
t
(d)
4
Summing up all the x-component forces shown in Fig. 2.6 gives
xx
δFx surface
x
yx
zx
y
z
dxdydz
(h)
Combining the above equations
Du
Dt
gx
y-direction:
Dv
Dt
gy
z-direction:
Dw
Dt
gz
x-direction:
yx
zx
x
y
z
xy
yy
zy
xx
(2.6a)
By analogy:
x
y
xz
yz
x
y
(2.6b)
z
zz
(2.6c)
z
IMPORTANT
THE NORMAL AND SHEARING STRESSES ARE EXPRESSED IN TERMS
OF VELOSICTY AND PRESSURE. THIS IS VALID FOR NEWTOINAN
FLUIDS. (See equations 2.7a-2.7f).
THE RESULTING EQUATIONS ARE KNOWN AS THE NAVIER-STOKES
EQUAITONS OF MOTION
SPECIAL SIMPLIFIED CASES:
(i) Constant viscosity

DV
Dt

g

p

V
1
3

V
2
(2.9)
(2.9) is valid for: (1) continuum, (2) Newtonian fluid, and (3) constant viscosity
(ii) Constant viscosity and density

DV
Dt

g

V

p
2
(2.10)
(2.10) is valid for: (1) continuum, (2) Newtonian fluid, (3) constant viscosity and
(4) constant density.
The three component of (2.10) are
x:
u
t
u
u
x
v
u
y
w
u
z
gx
p
x
2
u
x
2
2
u
y
2
2
u
z
2
(2.10x)
5
v
t
y-
z-
w
t
u
u
v
x
v
v
y
w
v
z
gy
p
y
w
x
v
w
y
w
w
z
gz
p
z
2
2
v
2
v
v
x2
y2
z2
2
2
2
w
x2
w
y2
(2.10y)
w
z2
(2.10z)
2.5.2 Cylindrical Coordinates
The three equations corresponding to (2.10) in cylindrical coordinates are (2.11r), (2.11 ),
and (2.11z).
2.5.3 Spherical Coordinates
The three equations corresponding to (2.10) in spherical coordinates are (2.11r), (2.11 ),
and (2.11 ).
2.6 Conservation of Energy: The Energy Equation
dz
y
2.6.1 Formulation
dy
dx
The principle of conservation of energy is
applied to an element dxdydz
x
A
B
Rate of change of
internal and kinetic
energy of element
z
Net rate of internal and kinetic
Fig. 2.7
energy transport by convection
C
Net rate of heat
addition by conduction
The variables u, v, w, p, T, and
(2.14)
_
D
Net rate of work done by
element on surroundin gs
are used to express each term in (2.14).
Assumptions: (1) continuum, (2) Newtonian fluid, and (3) negligible nuclear,
electromagnetic and radiation energy transfer.
Detailed formulation of the terms A, B, C and D is given in Appendix A
The following is the resulting equation
DT
Dp
k T
T
Dt
Dt
(2.15) is referred to as the energy equation
cp
is the coefficient of thermal expansion, defined as
(2.15)
6
1
(2.16)
T p
The dissipation function
is associated with energy dissipation due to friction. It is
important in high speed flow and for very viscous fluids. In Cartesian coordinates
is given by
2
2
3
u 2
x
u
x
v
y
v
y
w
z
2
w 2
z
u
y
2
v
x
v
z
w
y
2
w
x
u 2
z
(2.17)
2
2.6.2 Simplified Form of the Energy Equation
Cartesian Coordinates
(i) Incompressible fluid. Equation (2.15) becomes
cp
DT
Dt
(2.18)
k T
(ii) Incompressible constant conductivity fluid. Equation (2.18) is simplified further if
the conductivity k is assumed constant
cp
DT
Dt
T
y
w
k
2
(2.19a)
T
or
cp
T
t
u
T
x
v
T
z
2
T
x
2
k
2
T
2
y
2
z2
T
(2.19b)
(iii) Ideal gas. (2.15) becomes
cp
DT
Dt
k T
Dp
Dt
or
cv
DT
Dt
k T
p
(2.22)

V
(2.23)
Cylindrical Coordinates. The corresponding energy equation in cylindrical
coordinate is given in (2.24)
Spherical Coordinates. The corresponding energy equation in cylindrical
coordinate is given in (2.26)
7
2.7 Solutions to the Temperature Distribution
The flow field (velocity distribution) is needed for the determination of the
temperature distribution.
IMPORTANT:
Table 2.1 shows that for constant density and viscosity, continuity and
momentum (four equations) give the solution to u, v, w, and p. Thus for this
condition the flow field and temperature fields are uncoupled (smallest
rectangle).
For compressible fluid the density is an added variable. Energy equation and
the equation of state provide the fifth and sixth required equations. For this
case the velocity and temperature fields are coupled and thus must be solved
simultaneously (largest rectangle in Table 2.1).
TABLE 2.1
Basic law
No. of
Equations
Unknowns
p
u
v
w
1
u
v
w
Momentum
3
u
v
w
Equation of State
1
T
p
Viscosity relation
( p, T )
1
T
p
1
T
p
Energy
1
Continuity
Conductivity relation
k
k ( p, T )
TT
k
p
k
2.8 The Boussinesq Approximation
Fluid motion in free convection is driven by buoyancy forces.
Gravity and density change due to temperature change give rise to buoyancy.
According to Table 2.1, continuity, momentum, energy and equation of state must be
solved simultaneously for the 6 unknowns: u. v, w, p, T and
Starting with the definition of coefficient of thermal expansion
1
T p
, defined as
(2.16)
8
or
1
T
(f)
T
This result gives
(2.28)
(T T )
Based on the above approximation, the momentum equation becomes



DV
1
gT T
p p
v 2V
Dt
(2.29)
2.9 Boundary Conditions
(1) No-slip condition. At the wall, y
0

V ( x,0, z, t )
(2.30a)
0
or
u( x,0, z, t )
v( x,0, z, t )
w( x,0, z, t )
(2) Free stream condition. Far away from an object ( y
0
(2.30b)
)
(2.31)
u( x, , z, t ) V
Similarly, uniform temperature far away from an object is expressed as
(2.32)
T ( x, , z, t ) T
(3) Surface thermal conditions. Two common surface thermal conditions are used in the
analysis of convection problems. They are:
(i) Specified temperature. At the wall:
T ( x,0, z , t )
Ts
(2.33)
(ii) Specified heat flux. Heated or cooled surface:
k
T ( x,0, z, t )
y
qo
(2.34)
2.10 Non-dimensional Form of the Governing Equations: Dynamic and Thermal
Similarity Parameters
Express the governing equations in dimensionless form to:
(1) identify the governing parameters
(2) plan experiments
(3) guide in the presentation of experimental results and theoretical solutions
Dimensional form:
Independent variables: x, y, z and t
9
Unknown variables are: u, v , w, p and T . These variables depend on the four
independent variables. In addition various quantities affect the solutions. They
are
p , T , V , Ts , L, g , p , and
Fluid properties c p , k, , , and
Geometry
2.10.1 Dimensionless Variables
Dependent and independent variables are made dimensionless as follows:

V*

V
,
V
p
*
(p
p )
V
2
,
T
*
(T T )
,
(Ts T )

g*

g
,
g
(2.35)
V
x
y
z
, y*
,
,
t*
t
x*
z*
L
L
L
L
Using (2.35) the governing equations are rewritten in dimensionless form.
2.10.2 Dimensionless Form of Continuity
D
Dt
*

V*
0
(2.37)
No parameters appear in (2.37)
2.10.3 Dimensionless Form of the Navier-Stokes Equations of Motion

DV *
Gr *  *
1 *2 
* *
T
g
P
V*
Re
Dt *
Re 2
(2.38)
Constant (characteristic) quantities combine into two governing parameters:
V L
Re
Gr
V L
Reynolds number (viscous effect)
(2.39)
, Grashof number (free convection effect)
(2.40)
v
g Ts
T L3
v2
,
2.10.4 Dimensionless Form of the Energy Equation
Consider two cases:
(i) Incompressible, constant conductivity
10
DT *
Dt
1
RePr
*
*2
T*
Εc
Re
*
(2.41a)
Constant (characteristic) quantities combine into two additional governing
parameters:
Pr
Εc
cp
/
k
V
c p (Ts
k / cp
v,
Prandtl number (heat transfer effect)
(2.42)
2
T )
,
Eckert number (dissipation effect – high speed, large viscosity)
(2.43)
(ii) Ideal gas, constant conductivity and viscosity
DT *
1
RePr
Dt
*2
T*
Εc
Dp *
Dt
*
Εc
Re
*
(2.41b)
No new parameters appear.
2.10.5 Significance of the Governing Parameters
Dimensionless temperature solution:
T*
f ( x * , y * , z * , t * ; Re, Pr , Gr , Ec )
(2.45)
NOTE:
Six quantities: p , T , Ts , V , L, g and five properties c p , k, , , and
replaced by four dimensionless parameters: Re, Pr, Gr and Ec.
, are
Special case: negligible free convection and dissipation: Two governing
parameters:
T*
f ( x * , y * , z * , t * ; Re, Pr )
(2.46)
Geometrically similar bodies have the same solution when the parameters are the
same.
Experiments and correlation of data are expressed in terms of parameters rather
than dimensional quantities.
Numerical solutions are expressed in terms of parameters rather than dimensional
quantities.
2.10.6 Heat Transfer Coefficient: The Nusselt Number
Local Nusselt number:
Nu x = f ( x* ; Re, Pr , Gr , Ec )
Special case: negligible buoyancy and dissipation:
(2.51)
11
Nu x = f ( x* ; Re, Pr )
(2.52)
Free convection, negligible dissipation
Nu x = f ( x* ; Gr , Pr )
(2. 53)
For the average Nusselt number, x * is eliminated in the above.
2.9
Scale Analysis
A procedure used to obtain order of magnitude estimates without solving governing
equations.
CHAPTER 3
EXACT ONE-DIMENSIONAL SOLUTIONS
3.1 Introduction
Exact solutions for simple cases are presented.
Objective is to:
Understand the physical significance of each term in the equations of continuity,
Navier-Stokes, and energy.
Identify the conditions under which certain terms can be neglected.
General procedure: first determine the flow field and then the temperature field.
3.2 Simplification of the Governing Equations
Simplifying Assumptions
(1) Laminar flow
(2) Constant properties
(3) Parallel streamlines (fully developed flow):
v
(3.1)
0
Continuity (two-dimensional, constant density):
u
x
0
(3.2)
0
(3.3)
It follows that
2
u
x2
(3.1)-(3.3) result in significant simplifications.
(4) Negligible axial variation of temperature. For axial flow, this condition leads to
T
x
0
(3.4)
(3.4) is exact for certain channel flows and a reasonable approximation for others.
The following are conditions that may lead to the validity of (3.4):
(i) Parallel streamlines.
(ii) Far away from the entrance region of a channel (infinitely long channels).
2
(iii) Uniform surface conditions.
If (3.4) is valid everywhere:
2
T
x2
0
Rotating flows, Fig. 3.2: The streamlines are concentric circles
T
0
(3.6)
0
(3.7)
and
2
T
2
3.3 Exact Solutions
Applications of simplifications of Section 3.2.
3.3.1 Couette Flow
This is shear driven flow
Fluid is set in motion by moving channel surface
Streamlines are parallel
No axial variation
Many terms in the Navier-Stokes equations and energy equation drop out
Example 3.1: Couette Flow with Dissipation
Upper plate moves with velocity U o
Moving plate at temperature To
Taking into consideration dissipation
Determine temperature distribution
and the rate of heat transfer at the
moving plate
Assume laminar flow
Solution
Review all assumptions
Analysis
Starting with the energy equation
(3.5)
3
T
t
cp
u
T
x
T
y
v
2
T
z
w
k
2
T
x2
2
T
y2
T
(2.19b)
z2
Dissipation:
2
2
3
u 2
x
u
x
2
v
y
v
y
w
z
w 2
z
u
y
v
x
2
v
z
w
y
2
u 2
z
w
x
(2.17)
2
Simplification of the above equations:
(1) Incompressible fluid:
t
x
y
(a)
0
z
(2) Infinite plates:
x
z
(b)
w 0
(3) Continuity and no-slip condition:
v
(f)
0
Navier-Stokes equation (2.10x):
u
t
u
u
x
v
u
y
w
u
z
p
x
gx
2
u
x
2
2
u
2
y
2
z2
u
(2.10x)
Introduce the above simplifications into (2.10x):
d 2u
dy 2
(j)
0
Solution to (j) is
u
C1 y C 2
(k)
Use boundary conditions on u to determine the two constants. Solution becomes
u
Uo
y
H
(3.8)
Dissipation function simplifies to:
u
y
2
(n)
4
Use (3.8)
U o2
(o)
H2
Energy equation (2.19b) simplifies to
k
d 2T
U o2
dy 2
H2
0
(p)
Solution is
T
To
U o2
k
1
1
2
y2
H2
(3.9)
Heat flux at the moving plate:
q (H )
k
dT ( H )
dy
Use (3.9)
q (H )
U o2
H
Example 3.2: Flow in a Tube at Uniform Surface Temperature
Study this example with attention to physical conditions and how they lead to
simplifications of the governing equations.
Follow the procedure of Example 3.1 above
3.3.3 Rotating Flow
Note that all angular variation of velocity, pressure and temperature vanish in
Concentric rotating flows.
Example 3.3: Lubrication Oil Temperature in Rotating Shaft
Study this example with attention to physical conditions and how they lead to
simplifications of the governing equations.
Follow the procedure of Example 3.1
(3.10)
CHAPTER 4
BOUNDARY LAYER FLOW:
APPLICATION TO EXTERNAL FLOW
4.1 Introduction
Navier-Stokes equations and the energy equation are simplified using the boundary layer
concept.
Under special conditions certain terms in the equations can be neglected.
Two key questions:
(1) What are the conditions under which terms in the governing equations can be
dropped?
(2) What terms can be dropped?
4.2 The Boundary Layer Concept: Simplification of the Governing Equations
4.2.1 Qualitative Description
Consider convection over a semiinfinite plate (Fig. 4.1).
Under certain conditions the effect of
viscosity is confined to a thin region
near the surface called the velocity or
viscous boundary layer, .
Under certain conditions the effect of thermal interaction is confined to a thin region
near the surface called the thermal boundary layer t .
Conditions for the formation of the two boundary layers:
Velocity boundary layer conditions:
(1) Slender body
(2) High Reynolds number (Re > 100)
Thermal boundary layer conditions:
(1) Slender body
(2) High product of Reynolds and Prandtl numbers (Re Pr > 100)
Peclet Number
Pe
RePr
V L cp
k
c pV L
k
(4.1)
2
NOTE:
Fluid velocity at the surface vanishes.
Large changes in velocity across .
Large changes in temperature across
t.
Viscosity plays no role outside .
4.2.2 The Governing Equations
Assumptions: (1) Steady, (2) two-dimensional, (3) laminar, (4) uniform properties, (5) no
dissipation, and (6) no gravity.
Governing equations:
u
x
u
u
v
y
u
y
1 p
x
ν
v
v
y
1 p
y
ν
cp u
T
x
T
y
k
u
x
v
v
x
v
(2.3)
0
2
u
2
x
2
y2
2
2
v
x2
2
T
x2
u
v
y2
2
T
y2
(2.10x)
(2.10y)
(2.19)
4.2.3 Mathematical Simplification
Simplify the above equations based on boundary layer approximations.
4.2.4 Simplification of the Momentum Equations
(i) Intuitive Argument
Follow the intuitive argument leading to:
2
u
2
x
2
y2
u
(4.2)
and
p
y
0
(4.3)
Therefore
p
p(x)
It follows that
p
x
dp
dx
dp
dx
(2.10x) simplifies to the following boundary layer x-momentum equation
(4.4)
3
u
u
x
1 dp
x
u
y
v
2
ν
u
y2
(4.5)
(ii) Scale Analysis
We start by assuming that
(4.6)
1
L
Follow scale analysis leading to equations (4.2)-(4.4)
Follow scale analysis leading to:
1
L
(4.14b)
Re L
(4.14b) shows that (4.6) is valid when Re L
1.
(4.14b) is generalized as
1
x
(4.16)
Re x
4.2.5 Simplification of the Energy Equation
The energy equation for two-dimensional constant properties flow is
c
u
T
x
v
2
T
y
k
2
T
x2
T
y2
(2.19)
(2.19) is simplified for boundary layer flow using two arguments:
(i) Intuitive Argument
Follow the intuitive argument leading to:
2
2
T
x2
T
y2
(4.17)
(2.19) simplifies to the following boundary layer energy
T
u
x
2
T
v
y
T
y2
(4.18)
(ii) Scale Analysis
We start by assuming that
t
L
1
Follow scale analysis leading to equations (4.17) and (4.18)
Follow scale analysis for the validity of (4.19). Two cases are considered:
(4.19)
4
Case (1):
t
. Follow the argument leading to:
1
t
L
(4.24)
PrRe L
Thus
t
The criterion for
t
PrRe L
1 when
L
1
(4.25)
: Taking the ratio of (4.24) to (4.14b) gives
1
t
(4.27)
Pr
Thus
Case (2):
t
Pr
when
t
1
(4.28)
. Follow the argument leading to:
1
t
L
Pr
1/3
(4.31)
Re L
Thus
t
1 when Pr 1/3 Re L
L
The criterion for
t
1
(4.32)
: Taking the ratio of (4.31) to (4.14b)
1
Pr 1/3
t
(4.33)
Thus
when Pr 1/3
t
(4.34)
1
4.3 Summary of Boundary Layer Equations for Steady Laminar Flow
Review all assumptions leading to the following boundary layer equations:
Continuity:
u
x
v
y
(2.3)
0
x-Momentum:
u
u
x
1 dp
dx
u
y
v
ν
2
u
y2
(4.13)
Energy:
u
T
x
Note the following:
The continuity is not simplified.
v
T
y
2
T
y2
(4.18)
5
Solution to inviscid flow outside the boundary layer gives pressure gradient needed
in (4.13).
To include buoyancy effect, add [
g (T T ) ] to (4.13).
4.4 Solutions: External Flow
For constant properties, velocity distribution is independent of temperature.
First obtain the flow field solution and then use it to determine temperature
distribution.
4.4.1 Laminar Boundary Layer Flow over Semi-infinite Flat Plate: Uniform Surface
Temperature
The basic problem is shown in Fig. 4.5.
The plate is at uniform temperature T s .
Upstream temperature is T .
Apply all the assumptions summarized in
Section 4.3.
Governing equations: (continuity, momentum, and energy) are given in (2.3), (4.13),
and (4.18).
(i) Velocity Distribution.
Determine:
Velocity and pressure distribution.
Boundary layer thickness (x) .
Wall shear
o (x ).
(a) Governing equations and boundary conditions
u
x
u
u
x
v
u
y
v
y
(2.3)
0
1 dp
dx
ν
2
u
y2
(4.13)
The velocity boundary conditions are:
u(x,0)
0
(4.35a)
v(x,0)
0
(4.35b)
u( x, ) V
(4.35c)
u (0, y ) V
(4.35d)
(b) Scale analysis: boundary layer thickness, wall shear and friction coefficient.
We showed that
6
1
x
(4.16)
Re x
Define Darcy friction coefficient C f
o
Cf
(4.37a)
(1 / 2) V 2
Follow scale analysis leading to
1
Cf
(4.37b)
Re x
(c) Blasius solution: similarity method
Equations (2.3) and (4.13) are solved analytically by Blasius.
For inviscid flow over flat plate
u = V , v = 0, p = p = constant
(4.38)
Thus the pressure gradient is
dp
dx
(4.39)
0
(4.39) into (4.13)
u
u
x
u
y
v
ν
2
u
(4.40)
y2
(2.3) and (4.40) are solved by the method of similarity transformation.
The basic approach is combining the two independent variables x and y into a single
variable (x, y) and postulate that u/V depends on only.
For this problem the correct form of the transformation variable is
( x, y)
V
y
(4.41)
νx
Assume
u
V
df
d
(4.42)
Using (4.41) and (4.42), integration of the continuity gives
1 ν
2 V x
v
V
df
d
Transform all derivatives in terms of f and
2
d3 f
d 3
f( )
f
(4.43)
, substitute (4.42), (4.43)
d2 f
d 2
Boundary conditions (4.35a-4.35d) transform to
0
(4.44)
7
df (0)
0
d
f (0) 0
(4.45a)
(4.45b)
df ( )
1
d
df ( )
1
d
(4.45c)
(4.45d)
NOTE:
The momentum is transformed into
an ordinary differential equation.
Table 4.1
Blasius solution [1]
Boundary conditions (4.35c) and
(4.35d) coalesce into a single
condition.
(4.44) is solved by power series.
The solution is presented in Table
4.1.
From Table 4.1 we obtain
x
5.2
Re x
(4.46)
Scaling gives
1
x
(4.16)
Re x
From Table 4.1 we obtain
Cf
0.664
Re x
(4.48)
Scaling gives
Cf
1
Re x
y
0.0
0.4
0.8
1.2
1.6
2.0
2.4
2.8
3.2
3.6
4.0
4.4
4.8
5.0
5.2
5.4
5.6
6.0
7.0
8.0
(4.37b)
(ii) Temperature Distribution.
Determine:
Temperature distribution.
Thermal boundary layer thickness
t.
Heat transfer coefficient h(x).
Nusselt number Nu(x).
(a) Governing equation and boundary conditions
V
f
x
0.0
0.02656
0.10611
0.23795
0.42032
0.65003
0.92230
1.23099
1.56911
1.92954
2.30576
2.69238
3.08534
3.28329
3.48189
3.68094
3.88031
4.27964
5.27926
6.27923
df
d
u
V
0.0
0.13277
0.26471
0.39378
0.51676
0.62977
0.72899
0.81152
0.87609
0.92333
0.95552
0.97587
0.98779
0.99155
0.99425
0.99616
0.99748
0.99898
0.99992
1.00000
d2 f
d
2
0.33206
0.33147
0.32739
0.31659
0.29667
0.26675
0.22809
0.18401
0.13913
0.09809
0.06424
0.03897
0.02187
0.01591
0.01134
0.00793
0.00543
0.00240
0.00022
0.00001
8
u
T
x
v
2
T
y
T
y2
(4.18)
The boundary conditions are:
T ( x,0)
Ts
(4.49a)
T ( x, ) T
(4.49b)
T (0, y) T
(4.49c)
(b) Scale analysis: Thermal boundary layer thickness, heat transfer coefficient and
Nusselt number
Return to the results of Section 4.2.5:
Case (1):
t
(Pr <<1)
1
t
x
Case (2):
t
PrRe x
(4.50)
(Pr >>1)
1
t
x
Pr
1/3
(4.51)
Re x
Scale analysis for h. Begin with
h
T ( x,0)
y
k
Ts T
(1.10)
Using the scales, the above gives
h
k
(4.52)
t
Case (1):
t
(Pr <<1). Substituting (4.50) into (4.52)
h
k
PrRe x ,
x
for Pr <<1
(4.53)
Defining the local Nusselt number Nu x as
Nu x
hx
k
(4.54)
Substituting (4.53) into (4.54)
Nu x
Case (2):
t
Pr 1/2 Rex ,
for Pr <<1
(4.55)
(Pr >>1). Substituting (4.51) into (4.52)
h
k 1/3
Pr
Re x , for Pr >>1
x
The corresponding Nusselt number is
(4.56)
9
Nu x
Pr
1/3
Re x , for Pr >>1
(4.57)
(c) Pohlhausen’s solution: Temperature distribution, thermal boundary layer
thickness, heat transfer coefficient, and Nusselt number
Equation (4.18) is solved analytically by Pohlhausen using similarity transformation.
Define
T
T
Ts
Ts
(4.58)
(4.58) into (4.18)
2
u
x
v
(4.59)
y2
y
Boundary conditions (4.49) become
0
(4.60a)
(x, ) 1
(4.60b)
(0, y) 1
(4.60c)
(x,0)
Combine x and y into a single variable (x, y) given by
( x, y)
V
y
(4.41)
νx
Assume
( x, y)
( )
Velocity components u and v in (4.59) are given by Blasius solution
u
V
v
V
df
d
1 ν
2 V x
(4.42)
df
d
f
(4.43)
0
(4.61)
(4.41)-(4.43) into (4.59)
d2
d 2
Pr
d
f( )
2
d
Using (4.41), the three boundary conditions (4.60a-4.60c) transform to
0
(4.62a)
( ) 1
(4.62b)
( ) 1
(4.62c)
(0)
Integration details of (4.61) are found in Appendix B. The temperature solution is
10
d2 f )
d 2
( ) 1
d f
d 2
0
Surface temperature gradient is
0.332
d2 f
d 2
d
d
1 .0
Pr
Pr
(4.63)
Pr
2
d (0)
d
Pr
100 10
1 0.7(air)
0.8
(4.64)
0.1
0.6
d
T Ts
T Ts 0.4
The integrals in (4.63) and (4.64)
are evaluated numerically.
Boundary layer thickness t is
determined from Fig. 4.6. The
edge of the thermal layer is
defined as the distance y where
T T . This corresponds to
T
T
Pr
0.2
0
2
4
0.01
6
8
y V
10
12
Fig. 4.6 Pohlhausen ' s solution for temper ature
distribution for laminar flow over a
semi - infinte isothermal flat plate
Ts
Ts
1 , at y
t
(4.65)
The heat transfer coefficient h is determined using equation (1.10)
T ( x,0)
y
k
Ts T
h
(1.10)
Using (4.41) and (4.58) into the above
h( x ) k
V
d (0)
νx d
(4.66)
Average heat transfer coefficient
L
1
L
h
h( x )dx
(2.50)
0
Substituting (4.66) into (2.50) and integrating
h
14
x
2
k
d (0)
Re L
L
d
The local Nusselt number is obtained by substituting (4.66) into (4.54)
(4.67)
11
d (0)
Re x
d
Nu x
(4.68)
The corresponding average Nusselt number is
d (0)
Re L
d
Table 4.2 gives d (0) / d for various values of Pr.
Nu L
(4.69)
2
d (0)
d
d (0)
d
0.564 Pr 1/ 2 ,
0.332 Pr 1/ 3 ,
d (0)
d
Pr < 0.05
(4.71a)
0.6 < Pr < 10
(4.71b)
Pr >10
(4.71c)
0.339 Pr 1/ 3 ,
Table 4.2
d (0)
Pr
d
0.001
0.01
0.1
0.5
0.7
1.0
7.0
10.0
15.0
50
100
1000
0.0173
0.0516
0.140
0.259
0.292
0.332
0.645
0.730
0.835
1.247
1.572
3.387
4.4.2 Applications: Blasius Solution, Pohlhausen’s Solution, and Scaling
Review Examples 4.1, 4.2, and 4.3. They illustrate the application of Blasius
solution, Pohlhausen’s solution, and scaling to the solution of convection problems.
4.4.3 Laminar Boundary Layer Flow over Semi-infinite Flat Plate: Variable Surface
Temperature
Surface temperature varies with axial
distance x according to
Ts ( x) T
Cx n
(4.74)
Assumptions: see Section 4.3.
(i) Velocity Distribution. Blasius flow field solution is applicable to this case:
u
V
v
V
where the similarity variable
1 ν
2 V x
is defined as
df
d
(4.42)
df
d
f
(4.43)
12
( x, y)
V
y
(4.41)
νx
(ii) Governing Equations for Temperature Distribution. Based on the assumptions
listed in Section 4.3, temperature is governed by energy equation (4.18)
u
T
x
2
T
y
v
T
(4.18)
y2
The boundary conditions for this problem are:
T ( x,0)
(iii) Solution. Define
Ts
Cx n
T
(a)
T ( x, ) T
(b)
T (0, y) T
(c)
as
T
T
Ts
Ts
(4.58)
Assume
( x, y)
(4.75)
( )
Using (4.41)-(4.43), (4.58), (4.74) and (4.75), energy equation (4.18) transforms to (see
Appendix C for details)
d2
d 2
nPr
df
(1
d
Pr
d
f( )
2
d
)
0
(4.76)
Boundary conditions (a)-(c) become
0
(4.77a)
( ) 1
(4.77b)
( ) 1
(4.77c)
(0)
Local heat transfer coefficient and Nusselt number are determined using (1.10)
h
T ( x,0)
y
k
Ts T
(1.10)
Using (4.41), (4.58) and (4.72) into the above
T ( x,0)
y
Cx n
V
d (0)
νx d
Substituting into (1.10)
h( x )
k
V
ν
d (0)
x d
(4.78)
The average heat transfer coefficient for a plate of length L is defined in equation (2.50)
13
L
1
L
h
h( x)dx
(2.50)
0
(4.78) into (2.50)
h
k
d (0)
Re L
L
d
2
(4.79)
(4.78) into (4.54) gives the local Nusselt number
d (0)
Re x
d
Nu x
(4.80)
The corresponding average Nusselt number is
d (0)
Re L
d
Key factor: surface temperature gradient d (0) / d .
Nu L
(4. 81)
2
2.0
(iv) Results. (4.76) is solved
numerically subject to boundary
conditions (4.77). Fig. 4.8 gives
d (0) / d for three Prandtl numbers.
30
d (0)
d
10
1.0
Pr
0
Fig.4.8
0.5
n
1.0
0.7
1.5
d (0)
for plate with varying surface temperature,
d
Ts T
Cx n [4]
4.4.4 Laminar Boundary Layer Flow over a Wedge: Uniform Surface Temperature
Assumptions: listed in Section 4.3.
x-momentum equation:
u
u
x
v
u
y
1 dp
dx
ν
2
u
y2
(4.13)
Inviscid flow solution for V (x) is
V ( x)
Cx m
C is a constant and m is defined as
(4.82)
14
m
(4.83)
2
Application of (4.13) to the inviscid flow outside the viscous boundary layer, gives
1 dp
dx
V
x
V
Substituting into (4.13)
u
u
x
v
2
u
V
y
V
x
u(x,0)
0
(4.85a)
v (x,0)
0
(4.85b)
ν
u
(4.84)
y2
The boundary conditions are
u( x, ) V ( x)
Cx m
(4.85c)
Solution to the velocity distribution is obtained by the method of similarity. Define
( x, y)
V ( x)
νx
y
y
C
ν
x (m
as
1) / 2
(4.86)
Assume
u
V ( x)
dF
d
(4.87)
Continuity equation (2.3), (4.86), and (4.87) give the vertical velocity component v
v
V ( x)
ν
m 1
1 m
F
xV ( x) 2
1 m
dF
d
(4.88)
0
(4.89)
(4.82) and (4.86)-(4.88) into (4.84)
d 3F
d 3
m 1 d 2F
F
2
d 2
dF
m
d
2
m
This is the transformed momentum equation. Boundary conditions (4.85) transform to
dF (0)
d
F (0)
dF ( )
d
0
0
1
(4.90a)
(4.90b)
(4.90c)
Solution. (4.89) is integrated numerically. The solution gives F ( ) and dF / d .
These in turn give the velocity components u and v.
Temperature distribution. Start with the energy
15
2
u
x
v
(4.59)
y2
y
Boundary conditions
0
(4.60a)
(x, ) 1
(4.60b)
(0, y) 1
(4.60c)
(x,0)
Where
is defined as
T
T
Ts
Ts
(4.58)
( )
(4.75)
Assume
is defined in (4.86). (4.86)-(4.88) and (4.75) into (4.59) and (4.60)
d2
d 2
Pr
d
(m 1) F ( )
2
d
(0) 0
0
(4.91)
(4.92a)
( ) 1
(4.92b)
( ) 1
(4.92c)
Solution. Separating variables in (4.91), integrating twice and applying boundary
conditions (4.92), gives the temperature solution as
exp
( )
1
exp
0
(m 1) Pr
2
(m 1) Pr
2
F ( )d
d
0
(4.93)
F ( )d
d
0
(4.93) gives the temperature gradient at the surface
d (0)
d
exp
0
(m 1)Pr
2
1
F ( )d
d
0
The integrals in (4.93) and (4.94) are evaluated numerically.
Results for d (0) / d are given in Table 4.3
(4.94)
16
Table 4.3 Surface temperature gradient
d (0)
and surface velocity
d
gradient F (0) for flow over an isothermal wedge
m
0
0.111
0.333
1.0
wedge angle
d (0) / d
F (0)
0
0.3206
0.5120
0.7575
1.2326
o
/ 5 (36 )
/ 2 (90o)
(180o)
at five values of Pr
0.7
0.8
1.0
5.0
10.0
0.292
0.331
0.384
0.496
0.307
0.348
0.403
0.523
0.332
0.378
0.440
0.570
0.585
0.669
0.792
1.043
0.730
0.851
1.013
1.344
Heat transfer coefficient h and Nusselt number Nu. Equation (1.10) gives h
T ( x,0)
y
k
Ts T
h
(1.10)
Using (4.58), (4.75) and (4.86) into (1.10) gives
h( x )
k
V ( x) d (0)
νx d
(4.95)
(4.95) into (4.54) gives the Nusselt number
Nu x
d (0)
Re x
d
(4.96)
where Re x is the local Reynolds number defined as
Re x
xV (x)
ν
(4.97)
CHAPTER 5
APPROXIMATE SOLUTIONS:
THE INTEGRAL METHOD
5.1 Introduction
Seek approximate solution when:
Exact solution is unavailable.
Form of exact solution is not suitable or convenient.
Solution requires numerical integration.
The integral method gives approximate solutions.
5.2 Differential vs. Integral Formulation
Example: boundary layer flow, Fig. 5.1.
Differential formulation, Fig. 5.1a: the basic laws are formulated for a differential
element dx dy.
Solutions satisfy the basic laws exactly (at every point).
Integral formulation, Fig. 5.1b: the basic laws are formulated for the element
.
dx
Solutions satisfy the basic laws in an average sense (for section
).
5.3 Integral Method Approximation: Mathematical Simplification
Reduction in the number of independent variables.
Reduction of the order of the governing differential equation may result
5.4 Procedure
Integral solutions are obtained for the velocity and temperature fields.
The following procedure is used in obtaining integral solutions:
2
(1) Integral formulation of the basic laws:
Conservation of mass, momentum, and energy.
(2) Assumed velocity and temperature profiles:
Several options. Polynomials are used in Cartesian coordinates.
Assumed velocity and temperature profiles should satisfy known boundary
conditions
Assumed profile contains an unknown parameter or variable.
(3) Determination of the unknown parameter or variable:
Integral form of the basic law gives the unknown parameter or variable.
5.5 Accuracy of the Integral Method
Different assumed profiles give different solutions and accuracy.
Errors are acceptable in many engineering applications.
Accuracy is not very sensitive to the form of an assumed profile.
No procedure is available for identifying assumed profiles that will result in the most
accurate solutions.
5.6 Integral Formulation of the Basic Laws
5.6.1 Conservation of Mass
Boundary layer flow over porous plate of
porosity P with mass injection.
Conservation of mass for element
dx ,
shown in Fig. 5.2 and enlarged in Fig. 5.3,
gives
dm x
(a)
dme
dx dmo
dx
dme
( x)
dme
d
dx
u dy dx
(5.1)
v o Pdx
mx
dmx
dx
dx
mx
0
dme is the external mass flow rate into
element.
dmo
Fig. 5.3
5.6.2 Conservation of Momentum
Application of the momentum theorem in the x-direction to the element
Fx
M x (out ) M x (in)
Axial velocity u varies with x and y.
dx
(a)
3
Pressure p varies with x only (boundary layer approximation).
dp
)d
2
(p
V ( x) me
p
p
d
( p )dx
dx
Mx
Mx
dx
o (1
dM x
dx
dx
dx
P)dx
(a) forces
(b) x momentum
Fig. 5.4
M x = x-momentum, given by
( x)
u 2 dy
Mx
(c)
0
o
is wall shear, given by
u x,0
y
o
(d)
Equation (a) gives
( x)
dp
dx
u x,0
1 P
y
d
dx
( x)
u 2 dy V
d
x
dx
0
u dy V
x
Pv o
(5.2)
0
NOTE:
(1)
(2)
(3)
(4)
Equation (5.2) is the integral formulation of conservation of momentum.
Equation (5.2) applies to laminar as well as turbulent flow.
Although u is a function of x and y.
Evaluating the integrals in (5.2) results in a first order ordinary differential equation with
x as the independent variable.
Special Cases:
(i) Case 1: Incompressible fluid
Boundary layer approximation gives
dp
dx
dp
dx
The x-momentum equation for boundary layer flow is
(4.12)
4
u
x
u
v
1 dp
x
u
y
ν
2
u
(4.5)
y2
Applying equation (4.5) at y
dp
dx
dp
dx
V ( x)
Substituting (5.3) into (5.2) and noting that
dV
dx
(5.3)
is constant
( x)
dV
V ( x)
dx
ν
u x,0
y
1 P
d
dx
( x)
u 2 dy V
x
d
dx
u dy V
0
x Pv o
(5.4)
0
(ii) Case 2: Incompressible fluid and impermeable flat plate
For boundary layer flow over a flat plate
dV
dx
dp
dx
dp
dx
0,
P
(e)
0
For an impermeable plate
vo
(f)
0
(e) and (f) into (5.4)
x
v
u x,0
y
d
V
dx
x
udy
0
d
dx
u 2 dy
(5.5)
0
5.6.3 Conservation of Energy
Application of conservation of energy to the
element t dx , neglecting changes
in kinetic and potential energy, axial
conduction, and dissipation:
dE x
dx dEe
dx
t
Ex
Ex
Based on these assumptions,
conservation of energy for the
element gives
dEc
dEe
dx
(a)
dEo
dEx
dx
dx
dEc
dEe = energy added by external mass
dEo = energy added by injected mass
dEo
Fig.5.6
E x = energy convected with boundary layer flow
Formulating each term in (a)
dEc
k (1 P)
T x,0
dx
y
(b)
5
dEe
t ( x)
d
c pT
dx
u dy dx c p T
(c)
vo Pdx
0
dEo
(d)
c pTo vo Pdx
t ( x)
Ex
(e)
c p uT dy
0
Substituting (b)-(e) into (a)
T x,0
y
k1 P
t
d
dx
( x)
c p uTdy
t
d
c pT
dx
0
( x)
udy
c p v o P To
T
(5.6)
0
NOTE:
Equation (5.6) is integral formulation of conservation of mass and energy.
Although u and T are functions of x and y, evaluation of the integrals gives a first
order ordinary differential equation with x as the independent variable.
Special Case: Constant properties and impermeable flat plate
Setting P 1 in (5.6)
T x,0
y
t ( x)
d
dx
u (T
T )dy
(5.7)
0
5.7 Integral Solutions
Flow field solution.
Temperature field solution.
5.7.1 Flow Field Solution: Uniform Flow over a
Semi-Infinite Plate
Integral form of governing equation:
x
v
u x,0
y
V
d
dx
udy
x
d
dx
0
u 2 dy (5.5)
0
Assumed velocity profile
u( x, y)
Boundary conditions
a0 ( x) a1 ( x) y a2 ( x) y 2
a3 ( x) y 3
(a)
6
(1) u(x,0)
0
(2) u( x, ) V
(3)
u ( x, )
y
2
(4)
u ( x,0)
y2
0
0
NOTE: The fourth boundary condition is obtained by setting y = 0 in (2.10x)
Boundary conditions give the four coefficients. Thus
u
V
3 y
2
3
1 y
2
(5.9)
Note that the assumed velocity is in terms of the unknown variable
(x).
Boundary layer thickness. Use (5.5) to determine (x). Substituting (5.9) into (5.5)
3
vV 1
2
39 2 d
V
280
dx
Separating variables, integrating and noting that
(0)
140 v
13 V
d
0
(b)
0
x
dx
0
Evaluating the integrals and rearranging
280 / 13
4.64
Re x
Re x
x
(5.10)
Friction coefficient. (5.10) into (5.9) gives u as a function of x and y. With the
velocity distribution determined, friction coefficient C f is obtained using (4.36) and
(4.37a)
u
x,0
y
3v
o
Cf
2
2
V
x
V /2
V /2
Using (5.10) to eliminate
(x)
Cf
Compare with Blasius solution:
0.646
Re x
(5.11)
7
5.2
x
Re x
,
Blasius solution
(4.46)
,
Blasius solution
(4.48)
and
Cf
0.664
Re x
Note the small error in prediction C f .
5.7.2 Temperature Solution and Nusselt
Number: Flow over a Semi-Infinite Plate
(i) Temperature Distribution
A leading section of the plate of length x o
is insulated and the remaining part is at
uniform temperature Ts .
Assume laminar, steady, two-dimensional,
constant properties boundary layer flow and
neglect axial conduction and dissipation.
Determine t , h(x), and Nu(x).
Must determine flow field u( x, y) and temperature T(x,y).
Flow field solution of Section 5.7.1 applies to this case, equation (5.9).
Equation (5.7) gives the integral formulation of conservation of energy for this
problem
t ( x)
T x ,0
y
d
dx
u (T
T )dy
(5.7
0
Assumed temperature profile
T ( x, y )
Boundary conditions
(1) T ( x,0)
(2) T ( x,
(3)
T ( x,
y
2
(4)
Ts
t)
t)
T ( x,0)
y2
T
0
0
b0 ( x) b1 ( x) y b2 ( x) y 2
b3 ( x) y 3
(a)
8
NOTE: The fourth condition is obtained by setting y
0 in the energy equation (2.19).
Boundary conditions give the four coefficients. Thus
T ( x, y )
Ts
(T
Ts )
1 y3
2 t3
3 y
2 t
(5.13)
(5.9) and (5.13) into (5.7), evaluating the integral, gives
3
2
T
Ts
t
d
(T
dx
2
3
20
Ts ) V
t
3
280
4
t
(5.14)
Simplification of (5.14). Note that
t
1 , for Pr
(5.15)
1
It follows that
3
280
2
3
20
t
t
(5.14) simplifies to
2
d
V
dx
10
t
t
(b)
where
280 x
13 V
(c)
Boundary condition
t ( xo )
0
(h)
Solution to (b)
t
4.528
x
Pr 1/3 Re x 1/2
1
xo
x
3/ 4
1/ 3
(5.17b)
(ii) Nusselt Number. Local Nusselt number is defined as
Nu x
hx
k
(j)
h is the local heat transfer coefficient given by
T ( x,0)
y
Ts T
k
h
(k)
Using (5.13) into (k)
h( x )
3 k
2 t
(5.19)
9
Eliminating
t
by using (5.17b)
k
h( x) 0.331 1
x
xo
x
1/ 3
3/ 4
Pr 1/3 Re x1/2
(5.20)
Substituting into (j)
Nu x
0.331 1
xo
x
3/ 4
1/ 3
Pr 1/3 Re x1/2
(5.21)
(iii) Special Case: Plate with no Insulated Section
Set x o
0 in (5.17b), (5.20) and (5.21)
4.528
t
x
Pr
1/3
(5.23)
Re x 1/2
k
h( x) 0.331 Pr 1/3 Re x1/2
x
(5.24)
Nu x
0.331Pr 1/3 Re x
1/2
(5.25)
Examine the accuracy of the local Nusselt number. For Pr 10 equation (4.72c)
gives Pohlhausen’s solution
Nu x
0.339 Pr 1/ 3 Re x , for Pr 10
(4.72c)
Comparing this result with integral solution (5.25) shows that the error is 2.4%.
Example 5.1: Laminar Boundary Layer Flow over a Flat Plate:
Uniform Surface Temperature
This is a repeat of the Section 5.7.1 and 5.72, using assume linear velocity and
temperature profiles.
A linear profile gives less accurate flow and heat transfer results.
The procedure of the previous sections is repeated in this example.
The following is a summary of the results.
Assumed velocity
u
Boundary conditions
(1) u(x,0) 0
(2) u( x, ) V
Velocity solution
a0
a1 y
(b)
10
u
V
y
(c)
Integral solution to
x
12
Re x
(5.26)
b1 y
(f)
Assumed temperature
T
b0
Boundary conditions
(1) T ( x,0) Ts
(2) T ( x, t ) T
Temperature solution
T
Ts
(T
Ts )
y
(g)
t
Integral solution to
t
12ν
1
1/3
V
Pr
t
x 1 ( xo / x) 3 / 4
1/ 3
(o)
Solution to local Nusselt number
Nu x
0.289 Pr 1/3 Re x 1 ( xo / x) 3 / 4
Special Case: no insulated section, set xo
Nu x
1/ 3
(5.27)
0 in (5.27) gives
0.289 Pr 1/3 Re x
(5.28)
/ x and Nu x / Pr 1 / 3 Rex1/2 with integral
Comments. Table 5.1 compares exact solutions for
results for the case of a plate with no insulated section based on assumed linear and
polynomial profiles
Note that the integral method gives more accurate prediction of Nusselt number than
of the boundary layer thickness .
Table 5.1
Solution
Exact (Blasius/ Pohlhausen)
3rd degree polynomial
Linear
x
Re x
5.2
4.64
3.46
Nu x
Pr 1 / 3 Re1 / 2
0.332
0.339
0.289
11
5.7.3 Uniform Surface Flux
Plate with an insulted leading section of
length x o .
Plate is heated with uniform flux q s
along its surface x xo .
Steady, two-dimensional, laminar flow.
Determine surface temperature and the local Nusselt number.
Surface temperature is unknown.
Solution
Newton’s law of cooling gives
h( x )
qs
Ts ( x ) T
Introducing the definition of the Nusselt number
Nu x
qs x
k Ts ( x ) T
(b)
Need surface temperature Ts (x). Use the integral form of the energy equation to
determine Ts (x)
T x,0
y
t ( x)
d
dx
u (T
T )dy
(5.7)
0
u( x, y) for a third degree polynomial is given by (5.9)
u
V
3 y
2
1 y
2
3
(5.9)
Assume temperature T ( x, y)
T
Boundary conditions
(1)
k
(2) T x,
T x,0
y
t
T
qs
b0
b1 y b2 y 2
b3 y 3
(c)
12
(3)
T x,
y
2
(4)
t
0
T x,0
y2
0
Application of boundary conditions give the coefficients in (c)
2
3
T ( x, y ) T
Surface temperature. Set y
1 y 3 qs
3 t2 k
y
t
(5.29)
0 in the above
Ts ( x )
T ( x,0)
2 qs
3 k
T
(5.30)
t
(5.30) into (b)
3
2
Nu x
Must determine
x
t
(5.31)
x
t.
(5.9) and (5.29) into (5.7), evaluate the integral
d
dx
V
2
t
1
10
Simplify for Prandtl numbers larger than unity,
10
t
t
/
(e)
1
3
d
dx
V
3
1
140
t
t
Thermal boundary layer thickness. Integrating and use boundary condition
t ( x o ) 0 , gives
1/ 3
t
10
V
(x
xo )
(j)
Use (5.10) to eliminate in (j), rearrange
t
3.594
x
Pr 1/3 Re 1/2
x
xo
x
1
1/ 3
(5.32)
Surface temperature. (5.32) into (5.30) gives
Ts ( x ) T
q
2.396 s 1
k
xo
x
Local Nusselt number. (5.32) into (5.31) gives
1/ 3
x
Pr
1/3
Re 1/2
x
(5.33)
13
Nu x
0.417 1
xo
x
1/ 3
Pr 1/3 Re 1/2
x
Special Case: Plate with no insulated section, set xo 0 in (5.33) and (5.34)
q
x
Ts ( x ) T
2.396 s
1/3
k Pr Re1/2
x
Nu x
0.417 Pr 1/3 Re1/2
x
(5.34)
(5.35)
(5.36)
Compare with differential formulation solution:
Nu x
0.453 Pr 1/3 Re1/2
x
(5.37)
CHAPTER 6
HEAT TRANSFER IN CHANNEL FLOW
6.1 Introduction
(1) Laminar vs. turbulent flow
Flow through tubes, transition Reynolds number Re D t is
Re Dt
uD
2300
(6.1)
(2) Entrance vs. fully developed region
Classification based on velocity and temperature profiles:
(i) Entrance region
(ii) Fully developed region
(3) Surface boundary conditions
Two common boundary conditions::
(i) Uniform surface temperature
(ii) Uniform surface heat flux
(4) Objective.
Objective depends on surface thermal boundary condition:
(i) Uniform surface temperature. Determine axial variation of:
(1) Mean fluid temperature
(2) Heat transfer coefficient
(3) Surface heat flux
(ii) Uniform surface flux. Determine axial variation of:
(1) Mean fluid temperature
(2) Heat transfer coefficient
(3) Surface temperature
6.2 Hydrodynamic and Thermal Regions: General Features
Fluid enters with uniform velocity and temperature.
(1) Entrance region. Extends from the inlet to the section where the boundary layer
thickness reaches the center of channel.
(2) Fully developed region. This zone follows the entrance region.
2
6.2.1 Flow Field
(1) Entrance Region (Developing Flow, 0
x
Lh ).
Name: hydrodynamic entrance region.
Length: L h (hydrodynamic entrance length).
Streamlines are not parallel.
Core velocity u c increases with distance
Pressure decreases with distance ( dp / dx 0 ).
D/ 2.
r
uc
Vi
x
u
u
Lh
(2) Fully Developed Flow Region. x
fully developed
Lh
Fig. 6.1
Streamlines are parallel ( vr 0).
u / x 0 for two-dimensional incompressible fluid.
6.2.2 Temperature Field
(1) Entrance Region (Developing Temperature, 0
x
Lt )
Name: Thermal entrance region.
Length: Lt (thermal entrance length).
Core temperature Tc is uniform, Tc Ti .
D/2
t
(2) Fully Developed Temperature Region. x
r
Tc
Ts
Ts
T
Vi
Ti
Lt
x
Ts
t
Temperature varies radially and axially,
T / x 0.
fully developed
Lt
Fig. 6.2
6.3 Hydrodynamic and Thermal Entrance Lengths
6.3.1 Scale Analysis
(1) Hydrodynamic Entrance Length Lh .
Starting with external flow result (4.16)
1
x
Applying (4.16) to a tube at x
(4.16)
Re x
Lh :
D and Re Lh
Substituting (b) into (4.16) and rearranging
Re D
Lh
D
(b)
3
1/ 2
Lh / D
Re D
(6.2)
~1
(2) Thermal Entrance Length Lt .
Starting with external flow result (4.24)
t
Applying (4.24) at x
~ L Re L1 / 2 Pr
1/ 2
(4.24)
Lt :
t
D and Re Lt
Re D
Lt
D
(b)
Substituting (b) into (4.24) and rearranging
Lt / D
Re D Pr
1/ 2
(6.3)
~1
(6.2) and (6.3) give
Lt
~ Pr
Lh
(6.4)
6.3.2 Analytic and Numerical Solutions: Laminar Flow
(1) Hydrodynamic Entrance Length L h .
Results for L h :
Lh
De
Table 6.1
Entrance length coefficients C h and C t [1]
(6.5)
C h Re D e
Ct
Table 6.1 gives C h
Compare with scaling:
Lh / D
Re D
geometry
uniform
surface flux
uniform
surface
temperature
0.056
0.043
0.033
0.09
0.066
0.041
0.085
0.057
0.049
0.075
0.042
0.054
0.011
0.012
0.008
1/ 2
(6.2)
~1
a
a
b
Rewrite (6.5)
a/b =1
a
1/ 2
Lh / De
Re De
Ch
Ch
1/ 2
(a)
Example: Rectangular channel,
aspect ratio 2, Table 6.1 gives
C h 0.085. Substituting this value
into (a), gives
a/b = 2
b
a
b
a/b = 4
4
1/ 2
Lh / De
Re D e
0.085 1 / 2
(b)
0.29
Scaling replaces 0.29 by unity.
(2) Thermal Entrance Length Lt .
Lt depends on surface boundary conditions: Two cases: (i) Uniform surface
temperature. (ii) Uniform surface flux.
Solution
Lt
De
(6.6)
C t PrRe D
Table 6.1 gives C t for both cases.
Compare with scaling. Rewrite (6.6)
Lt / De
PrRe D
1/ 2
Ct
1/ 2
(c)
Scaling gives
1/ 2
Lt / D
Re D Pr
(6.3)
~1
Example: Rectangular channel, aspect ratio 2, Table 6.1 gives Ct
this value into (c), gives
0.049. Substituting
1/ 2
Lt / De
PrRe De
0.049 1 / 2
(d)
0.22
Scaling replaces 0.22 be unity.
Turbulent flow: L
Lh
Lt
L
D
(6.7)
10
6.4 Channels with Uniform Surface Heat Flux q s
Inlet mean temperature: Tmi
Tm (0) .
L
Determine:
(1) Total heat transfer rate q s .
(2) Mean temperature variation Tm (x).
(3) Surface temperature variation Ts (x ).
Total heat transfer rate
Tmi
0
x
qs
Fig. 6.3
Tm (x)
5
qs
q s As
(6.8)
qs P x
As = surface area
P = perimeter
Mean temperature Tm (x) . Conservation of energy between inlet and section x:
qs
qs P x
mcp [Tm ( x) Tmi ]
or
Tm ( x)
qs P
x
mc p
Tmi
(6.9)
Surface temperature Ts (x) . Newton’s law of cooling gives
qs
h( x) Ts ( x) Tm ( x)
or
Ts ( x)
qs
h( x )
Tm ( x)
Using (6.9)
Ts ( x)
Tm i
qs
Px
mc p
1
h( x )
(6.10)
NOTE:
Determining Ts (x) requires knowing h(x).
To determine h(x): Must know if:
Flow is Laminar or turbulent.
Entrance or fully developed region
6.5 Channels with Uniform Surface Temperature
Inlet mean temperature: Tmi
Tm (0) .
Ts
Determine:
(1) Mean temperature variation Tm (x).
(2) Total heat transfer rate q s between x
location x.
(3) Surface heat flux variation q s (x).
Tmi
0
dx
dqs
Tm
Tm
Conservation of energy to element
m c p dTm
m
0 and
Mean temperature variation Tm (x).
dqs
Tm (x)
x
dx
(a)
Fig. 6.4
dTm
dx
dx
6
Newton's law:
dqs
h( x) Ts
Tm ( x) Pdx
(b)
P
h( x)dx
m cp
(c)
Combine (a) and (b)
dTm
Ts Tm ( x)
Integrating (c)
x
T ( x) Ts
ln m
Tmi Ts
P
m cp
Definite h
(6.11)
h( x)dx
0
x
h
1
x
h( x)dx
(6.12)
0
(6.12) into (6.11), solve Tm (x)
Tm ( x)
(Tmi Ts ) exp[
Ts
Ph
x]
mcp
(6.13)
NOTE:
Determining Tm (x) requires knowing h(x).
To determine h(x): Must know if:
Flow is laminar or turbulent.
Entrance or fully developed region
Heat transfer rate. Conservation of energy:
qs
(6.14)
m c p [Tm ( x) Tmi ]
Surface heat flux.: Newton’s law:
q s ( x)
h( x)[Ts
Tm ( x)]
(6.15)
6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number Nu D
6.6.1 Scale Analysis
Estimate h(x) and Nu D .
Tube: radius ro , surface temperature T s , mean temperature Tm .
Fourier’s law and Newton’s law:
h
Scaling (6.16)
T (ro , x)
k
r
Tm Ts
r
qs
ro
0
Tm
(6.16)
Fig. 6.5
Ts
7
Tm
k
Ts
t
h~
Tm
Ts
or
k
h~
(6.17)
t
The Nusselt number
hD
k
Nu D
Use (6.17)
D
Nu D ~
(6.18)
t
Fully developed region:
t (x ) ~
D, equation (6.18) gives
Nu D ~ 1 (fully developed)
Entrance region: Need to scale
t
(
t (x ) <
(6.19)
D).
For external flow
t
~ x Pr
1/ 2
Re x 1/ 2
(4.24)
(4.24) into (6.18)
Nu D ~
D 1 / 2 1/2
Pr Re x
x
(c)
Expressing Re x in terms of Re D
Re x
ux
ν
uD x
ν D
Re D
x
D
(d)
Substitute (d) into (c)
D
Nu D ~
x
1/2
Pr 1 / 2Re1/2
D
(6.20a)
Rewrite
Nu D
PrRe D
x/D
1/ 2
~1
Scaling estimates (6.19) and (6.20) will be compared with exact solutions.
(6.20b)
8
6.6.2 Basic Considerations for the Analytical Determination of Heat Flux, Heat
Transfer Coefficient and Nusselt Number
r
qs
Need to determine velocity and temperature distribution.
Assume: fully developed velocity
Neglect axial conduction
Section outline:
ro
0
Tm
Definitions
Governing equations for determining:
Ts
Fig. 6.5
(i) Surface heat flux
(ii) Heat transfer coefficient
(iii) Nusselt number
(1) Fourier’s law and Newton’s law.
Surface heat flux. Fourier’s law gives surface heat flux q s
qs
k
T x, ro
r
(a)
Define dimensionless variables
T Ts
,
Ti Ts
x/D
,
Re D Pr
r
,
ro
R
vx
v
k
Ts
ro
Ti
vx
,
u
vr
vr
,
u
Re D
uD
ν
(6.21)
Substitute into (a)
qs ( )
0( ,1)
R
(6.22)
Heat transfer coefficient. Define h
h
q" s
Tm Ts
(6.23)
Combine (6.22) and (6.23)
h( )
where
m
k (Ts Ti )
ro (Tm Ts )
( ,1)
R
k
ro
1
m( )
( ,1)
R
(6.24)
is defined as
m
Tm Ts
Ti Ts
(6.25)
Nusselt number. Define:
Nu ( )
h( ) D
k
h( )2ro
k
(6.26)
( ,1)
R
(6.27)
(6.24) into (6.26)
Nu ( )
2
m( )
9
(2) The Energy Equation. Review assumptions on energy equation (2.24).
T
r
c p vr
T
z
vz
k
2
1
T
r
r r
r
T
(2.24)
z2
Replace z by x, use dimensionless variables:
vx
2 Re D Pr vr
4
R
R R
R
R
1
( Re D Pr) 2
2
2
(6.28)
where
Pe
Re D Pr , Peclet number
(6.29)
Pe
(6.30)
Neglect conduction for
PrRe D
100
Thus, under such conditions, (6.28) becomes
vx
2 Re D Pr vr
4
R
R R
R
R
(6.31)
(3) Mean (Bulk) Temperature Tm . Define:
ro
mc p Tm
c p v x T 2 rdr
(a)
v x 2 rdr
(b)
0
Mass flow rate m is given by
ro
m
0
(b) into (a), assume constant properties
ro
v x Trdr
Tm
0
(6.32a)
ro
v x rdr
0
Dimensionless form:
1
v x R dR
m
Tm Ts
Ti Ts
0
1
v x R dR
0
6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region
6.7.1 Definition of Fully Developed Temperature Profile
(6.32b)
10
Far away from the entrance ( x / d
fully developed.
0.05Re D Pr ), temperature profile becomes
To define fully developed temperature, introduce the dimensionless temperature
Ts ( x ) T ( r , x )
Ts ( x) Tm ( x)
(6.33)
Fully developed temperature is defined as a profile in which
is independent of x:
(r )
(6.34)
0
(6.35)
(6.34) gives
x
(6.33) and (6.35) give
Ts ( x) T (r , x)
x Ts ( x) Tm ( x)
x
Expand and use the definition of
(6.36a)
0
in (6.33)
dTs
dx
T
x
(r )
dTs
dx
dTm
dx
0
(6.36b)
6.7.2 Heat Transfer Coefficient and Nusselt Number
Examine h and Nu in the fully developed region.
Fourier’s and Newton’s law:
T (ro , x)
r
Tm Ts
k
h
(6.16)
Use (6.33) to eliminate T (ro , x) / r . (6.16) gives
h
k
d (ro )
= constant
dr
(6.37)
IMPORTANT CONCLUSOIN:
THE HEAT TRANSFER COEFFICIENT IN THE FULLY DEVELOPED REGION
IS CONSTANT INDEPENDET OF LOCATION.
Nusselt number
Nu D
hD
k
D
d (ro )
dr
(6.38)
Scaling estimate based on limiting case of entrance region:
Nu D ~ 1 (fully developed)
Scale estimate based on fully developed region:
(6.19)
11
Scale T (ro , x) / r as
T (ro , x) Ts Tm
~
r
D
Substitute into (6.16)
h~
k
D
(6.39)
Substitute (6.39) into (6.38)
(fully developed)
Nu D ~ 1
(6.40)
6.7.3 Fully Developed Region for Tubes at Uniform Surface flux
r
Determine:
qs
T
(i) Surface temperature Ts (x).
(ii) Heat transfer coefficient.
u
0
D
x
qs
Fig. 6.6
Newton’s law
qs
h Ts ( x) Tm ( x)
(a)
Since q s and h are constant it follows that
Ts ( x) Tm ( x) constant
(b)
Differentiate
dTs
dx
dTm
.
dx
(c)
dTs
dx
(d)
(c) into (6.36b)
T
x
(c) and (d)
T
x
dTs
dx
dTm
(for constant q s )
dx
(6.41)
Unknowns: T (r, x), Tm (x) and Ts (x)
Conservation of energy:
q s Pdx mc pTm
or
mc p Tm
m
dTm
dx
dx
Tm
Tm
dx
qs
Fig. 6.7
dTm
dx
dx
12
dTm
dx
qs P
= constant
mc p
(6.42)
dTs
dx
dTm q s P
=
= constant
dx mc p
(6.43)
Substitute (6.42) into (6.41)
T
x
Integrate(6.43)
qs P
x C1
mc p
Tm ( x)
(e)
Use inlet condition
Tm (0)
Tmi
(f)
Solution (e) becomes
Tm ( x)
qs P
x
mc p
Tmi
(6.44)
Need to determine T (r, x) and Ts (x). This requires solving the differential form of
the energy equation.
Set vr
0 in energy equation (2.24)
T
x
c p vx
k
T
r
r r
r
(6.45)
Fully developed flow axial velocity
vx
r2
2u 1
ro2
(6.46)
(6.43) and (6.46) into (6.45)
c p 2u 1
However, m
ro2 u and P
r2
2
ro
qs P
m c p
k
T
r
r r
r
(g)
2 ro , equation (g) becomes
4q s
r2
1 2
ro
ro
k
T
r
r r
r
(6.47)
Boundary conditions are:
T (0, x)
0
r
T (ro , x)
k
qs
r
Integrate (6.47)
(6.48a)
(6.48b)
13
4
r2
qs
ro
2
r4
T
r
kr
2
4ro
Boundary condition (6.48a) gives f (x)
(h)
f x
0. Equation (h) becomes
4q s
kro
T
r
r3
r
2
4ro2
Integrate again
r2
4
4q s
kro
T (r , x)
r4
(6.49)
g ( x)
16ro2
The integration “constant” is g (x) . Use Tm (x) to determine g (x). Substitute (6.46) and
(6.49) into (6.32a)
7 ro q s
(6.50)
Tm ( x)
g ( x)
24 k
Equate (6.44) and (6.50) gives g (x)
g ( x)
Tmi
7 ro q s
24 k
Pq s
x
mc p
r4
16ro
7 ro q s
24 k
11 ro q s
24 k
Pq s
x
mc p
(6.51)
(6.51) into (6.49)
T (r , x) Tmi
Set r
4q s
kro
r2
4
2
Pq s
x
mc p
(6.52)
ro in (6.52) to obtain Ts (x)
Ts ( x)
Tmi
(6.53)
(6.44), (6.52) and (6.53) into (6.33) gives (r )
24 1 2
(r ) 1
r
11 ro2
r4
24 Pq s
7
x
x
11 mc p
11
2
4ro
(6.54)
Differentiate (6.54) and substitute into (6.38) gives
Nu D
48
11
4.364
(6.55)
NOTE:
(6.55) applies to laminar fully developed velocity and temperature in tubes with
uniform surface heat flux.
The Nusselt number is independent of Reynolds and Prandtl numbers.
Scaling gives Nusselt as
Nu D ~ 1
(6.40)
14
This compares favorable with (6.55).
6.7.4 Fully Developed Region for Tubes at Uniform Surface Temperature
Determine: Nusselt number
Solve the energy equation for the fully developed region
Neglect axial conduction and dissipation.
Energy equation: set vr 0 in (2.24)
c p vx
T
x
k
T
r
r r
r
(6.45)
Boundary conditions
T (0, x)
0
r
T (ro , x) Ts
(6.56a)
(6.56b)
Axial velocity for fully developed flow is
vx
2u 1
r2
(6.46)
ro2
Use (6.36a) to Eliminate T / x in (6.45)
Ts ( x) T (r , x)
x Ts ( x) Tm ( x)
x
For uniform Ts ( x)
0
(6.36a)
Ts , above gives
T
x
Ts T (r , x) dTm
Ts Tm ( x) dx
(6.57)
(6.46) and (6.57) into (6.45)
2 cpu 1
r 2 Ts T (r , x) dTm
ro2 Ts Tm ( x) dx
k
T
r
r r
r
(6.58)
Solution: (6.58) was solved using an infinite power series. Solution gives the Nusselt
number as
Nu D
3.657
(6.59)
6.7.5 Nusselt Number for Laminar Fully Developed Velocity and Temperature in
Channels of Various Cross-Sections
Table 6.2 lists Nusselt numbers for channels of various cross-sections.
Two cases: (1) uniform surface heat flux and (2) uniform surface temperature.
Nusselt number of Non-circular channels is based on the equivalent diameter.
Scaling estimate:
Table 6.2
Nu D ~ 1 (fully developed)
(6.40)
Nusselt number for laminar fully developed
conditions in channels of various cross-sections [3]
Nusselt number Nu D
Table 6.2: Nusselt number ranges from 2.46
to 8.235.
a
b
Channel geometry
Uniform
surface
flux
Uniform
surface
temperature
4.364
3.657
1
3.608
2.976
2
4.123
3.391
4
5.331
4.439
8
6.49
5.597
8.235
7.541
3.102
2.46
6.8 Thermal Entrance Region: Laminar Flow
through Tubes
a
6.8.1 Uniform Surface Temperature: Graetz
Solution
b
a
Laminar flow.
b
Fully developed inlet velocity.
b
a
Neglect axial conduction (Pe > 100).
a
Uniform surface temperature Ts .
b
Fully developed flow:
vr
0
15
(3.1)
Axial velocity
vz
1 dp 2 2
(r ro )
4 dz
(3.12)
r
Ts
(3.12) expressed in dimensionless form
vx
vx
u
2
2(1 R )
T
Ti
(6.61)
u
0
x
t
(3.1) and (6.61) into energy equation (6.31)
1
1 R2
2
1
R
R R
R
Fig. 6.8
(6.62)
Boundary conditions
( ,0)
R
0
(6.63a)
( ,1)
0
(6.63b)
(0, R) 1
(6.63c)
Analytic and numerical solutions to this problem have been obtained.
Review analytic solution leading to:
16
(i) Mean temperature,
m(
)
Gn
8
m
2
2
exp( 2
(6.66)
)
n
n
n 0
(ii) Local Nusselt number, Nu ( )
G n exp( 2
2
n
)
n 0
Nu
(66.7)
Gn
2
2
n 0
exp( 2
2
n
)
n
(iii) Average Nusselt number, Nu ( )
h ( )D
k
(f)
and G n for 0
n 10. Table 6.4 gives Nu ( ) and
Nu ( )
RESULTS
Table 6.3 lists values of
n
Nu ( ) at selected values of the axial distance .
Fig. 6.9 gives the variation of Nu ( ) and Nu ( ) along a tube.
Table 6.4
Table 6.3
Uniform surface temperature [4]
n
0
1
2
3
4
5
6
7
8
9
10
n
2.70436
6.67903
10.67338
14.67108
18.66987
22.66914
26.66866
30.66832
34.66807
38.66788
42.66773
Gn
0.74877
0.54383
0.46286
0.41542
0.38292
0.35869
0.33962
0.32406
0.31101
0.29984
0.29012
Local and average Nusselt
number for tube at uniform
surface temperature [5]
=
x/D
Re D Pr
0
0.0005
0.002
0.005
0.02
0.04
0.05
0.1
Nu ( )
Nu ( )
12.8
8.03
6.00
4.17
3.77
3.71
3.66
3.66
19.29
12.09
8.92
5.81
4.86
4.64
4.15
3.66
17
Average Nu
Nusselt number
Local Nu
Fig. 6.9
x/D
ReD Pr
Local and average Nusselt number for
tube at uniform surface temepratu re [4]
NOTE:
(1) The average Nusselt number is greater than the local Nusselt number.
(2) Asymptotic value of Nusselt number of 3.657 is reached at
0.05 . Thus
(6.69)
Nu( ) 3.657
(3) Evaluate fluid properties at the mean temperatures Tm , defined as
Tm
Tmi
Tmo
6.8.2 Uniform Surface Heat Flux
Repeat Graetz entrance problem
replacing the uniform surface
temperature with uniform heat flux.
(6.70)
2
r
qs
T
Ti
u
0
D
x
t
qs
Inlet velocity is fully developed.
Fig. 6.10
Energy equation is
1
1 R2
2
1
R
R R
R
(6.62)
18
Boundary conditions
( ,0)
)
R
(6.71a)
0
q s ro
k (Ti Ts )
( ,1)
R
(6.71b)
(6.71c)
(0, R) 1
Solution.
Local Nusselt number:
1
Nu ( )
hx
k
11
48
1
2
An exp( 2
2
Table 6.5
(6.72]
)
n
n 1
n
The average Nusselt number is given by
Nu ( )
hx
k
11
48
1
2
The eigenvalues
in Table 6.5
Limiting case:
An
n 1
2
n
1 exp( 2
2
2
n
1
)
2
n
(6.73]
and the constant An are listed
(fully developed)
Nu ( )
Uniform surface flux [4]
48
4.364
11
1
2
3
4
5
6
7
8
9
10
2
n
25.6796
83.8618
174.1667
296.5363
450.9472
637.3874
855.8495
1106.3290
1388.8226
1703.3279
(6.74)
Average Nu
Nusselt number
Local Nu
x/D
ReD Pr
Fig. 6.11 Local and average Nusselt number for
tube at uniform surface heat flux [4]
An
0.198722
0.069257
0.036521
0.023014
0.016030
0.011906
0.009249
0.007427
0.006117
0.005141
CHAPTER 7
FREE CONVECTION
7.1 Introduction
7.2 Features and Parameters of Free Convection
(1) Driving Force.
Requirements
(i) Gravitational field
(ii) Density change with temperature
(2) Governing Parameters. Two parameters:
(i) Grashof number
T ) L3
g (Ts
Grashof number = GrL
2
(7.1)
(ii) Prandtl number
is the Coefficient of thermal expansion, also known as compressibility factor.
For ideal gases it is given by
1 , for ideal gas
T
(2.21)
Rayleigh number
Ra L
GrL Pr
g (Ts T ) L3
ν2
Pr =
g (Ts T ) L3
ν
(7.2)
(3) Boundary Layer.
Flow: Laminar, turbulent, or mixed.
Boundary layer approximations are valid for Ra x
10 4.
(4) Transition from Laminar to Turbulent Flow.
For vertical plates: transition Rayleigh number, Rax t , is
Rax t
109
(7.3)
(5) External vs. Enclosure Free Convection.
(i) External free convection: surface is immersed in infinite medium.
(ii) Enclosure free convection. Free convection takes place inside closed volumetric regions.
(6) Analytic Solutions.
Velocity and temperature fields are coupled.
Momentum and energy equation must be solved simultaneously.
2
7.3 Governing Equations
Approximations:
(1) Constant density, except in evaluating gravity forces.
(2) The Boussinesq approximation (relates density change to temperature change).
(3) No dissipation.
Continuity, momentum, and energy equations are obtained from equations (2.2),
(2.29) and (2.19), respectively
u
x
u
x
u
u
y
v
u
v
x
v
y
u
x
1
y
T
x
v
(p
(7.4)
0
1
g (T T )
v
v
y
(p
p )
v(
2
T
y
v(
p )
T
x2
2
v
x2
2
u
2
x
2
y2
2
v
y2
u
)
)
T
y2
(7.7)
Continuity equation (7.4) is unchanged
x-component of the Navier-Stokes equations simplifies to
u
x
v
u
y
v
βg T T
2
u
y2
(7.8)
Energy equation (7.7)
u
T
x
v
T
y
2
α
T
y2
(7.9)
(7.4), (7.8), and (7.9) contain three unknowns: u, v, and T.
Momentum and energy are coupled.
7.4 Laminar Free Convection over a Vertical Plate: Uniform Surface Temperature
Uniform temperature T s (Fig. 7.1).
Infinite fluid at temperature T .
Determine: velocity and temperature distribution.
7.4.1 Assumptions. Note all assumptions listed in this section.
7.4.2 Governing Equations
(7.6)
2
7.3.1 Boundary Layer Equations
u
(7.5)
3
u
x
u
u
x
v
v
y
u
y
(7.4)
0
2
v
βg T T
u
y2
(7.8)
2
u
where
v
x
α
y
(7.10)
y2
is defined as
T T
Ts T
(7.11)
7.4.3 Boundary Conditions.
Velocity:
(1)
(2)
(3)
(4)
u(x,0) 0
v(x,0) 0
u(x, ) 0
u(0, y) 0
Temperature:
(5) (x,0) 1
(6) (x, ) 0
(7) (0, y) 0
7.4.4 Similarity Transformation. Introduce the similarity variable
Grx
4
1/ 4
y
x
(7.14)
where
g (Ts T ) x 3
Grx
(7.15)
ν2
Let
( )
(7.16)
Grx d
x d
(7.20)
( x, y)
u
2v
v
(Grx )1/ 4
x
Continuity gives
v
1/ 4
(4)
d
d
(7.21)
3
(7.20) and (7.21) into (7.8) and (7.10) and using (7.11) and (7.16), gives
d3
d
3
3
d2
d
2
d
2
d
2
0
(7.22)
4
d2
d
2
3Pr
d
d
0
(7.23)
Transformation of boundary conditions:
Velocity:
d (0)
0
d
(2) (0) 0
d ( )
(3)
0
d
d ( )
(4)
0
d
(1)
Temperature:
(1) (0) 1
(2) ( ) 0
(3) ( ) 0
The problem is characterized by a
single parameter which is the Prandtl
number.
7.4.5 Solution.
(7.22) and (7.23) and their
five boundary conditions are
solved numerically.
The solution is presented
graphically in Figs. 7.2 and
7.3. Fig. 7.2 gives the
5
7.4.6 Heat Transfer Coefficient and Nusselt Number.
Fourier’s law and Newton’s law:
T ( x,0)
y
Ts T
k
h
Express in terms of
(7.24)
and
h
k dT d (0)
T d
d
y
Ts
Use(7.11) and (7.14)
h
1/ 4
k Grx
x
4
d (0)
d
(7.25)
Local Nusselt number
Nu x
Grx
4
hx
k
1/ 4
d (0)
d
(7.26)
Average h
1
L
h
Table 7.1 [1,2]
L
h( x)dx
(2.50)
0
(7.25) into (2.50), and performing the integration
h
4 k GrL
3L 4
1/ 4
d (0)
d
(7.27)
Average Nusselt number is
Nu L
hL
k
4 GrL
3 4
1/ 4
d (0)
d
(7.28)
Solution depends on a single parameter which is the
Prandtl number.
d (0)
Numerical solution gives
, listed in Table 7.1.
d
Special Cases
Very small and very large Prandtl numbers:
Nu x
Nu x
0.600 ( PrRa x )1/4 ,
0.503( PrGrx )1/4 ,
Pr
0
Pr
(7.29a)
(7.29b)
7.5 Laminar Free Convection over a Vertical Plate: Uniform
Surface Heat Flux
Assumptions: Same as constant temperature plate.
Pr
_ d (0)
0.01
0.03
0.09
0.5
0.72
0.733
1.0
1.5
2.0
3.5
5.0
7.0
10
100
1000
0.0806
0.136
0.219
0.442
0.5045
0.508
0.5671
0.6515
0.7165
0.8558
0.954
1.0542
1.1649
2.191
3.9660
d
d 2 ( 0)
d 2
0.9862
0.676
0.6741
0.6421
0.5713
0.4192
0.2517
0.1450
6
Surface boundary conditions
k
T ( x,0)
y
(7.30)
qs
Surface flux is specified.
Determine: Surface temperature Ts (x) and local Nusselt number Nu x .
Solution by similarity transformation.
Solution:
Surface temperature
Ts ( x) T
5
ν 2 (q s ) 4
gk 4
1/ 5
x
( 0)
(7.31)
Local Nusselt number
g qs
Nu x
5ν 2 k
1/ 5
x
4
1
(0)
(7.32)
Table 7.2 [4]
(0) is a dimensionless parameter which depends on the
Prandtl number and is given in Table 7.2 [4].
Correlation equation for (0)
(0)
4 9 Pr
1/ 2
5Pr
10Pr
10
1/ 5
2
, 0.001 Pr 1000
(7.33)
Properties at the film temperature T f
Tf
T
Pr
0.1
1.0
7.34)
Ts (L / 2) / 2
7.6 Inclined Plates
Vertical plate solutions of Sections 7.4 and 7.5 apply
to inclined plates, with g replaced by g cos .
This approach is recommended for
60o .
7.7 Integral Method
7.7.1 Integral Formulation of Conservation of Momentum
Assume:
t
(a)
100
(0)
- 2.7507
- 1.3574
- 0.76746
- 0.46566
7
Application of the momentum theorem in the x-direction to the element
7.6
Fx
dx , Fig.
(b)
M x (out ) M x (in)
dx is enlarged in Fig. 7.7
dM x
dx
dx
Mx
p
d
( p ) dx
dx
dy
o dx
dx
gdxdy
M x dW
( p dp / 2)d
p
Fig. 7.7
p
p
dp
d
2
d
p dx
dx
p
o dx
Mx
dM x
dx
dx
Mx
(c)
Simplify
dp
o dx
dM x
dx
dx
dW
(d)
Wall shearing stress
u x,0
y
o
(e)
Weight of element
dW
dx
g dy
(f)
0
The x-momentum of the fluid entering element
( x)
u 2 dy
Mx
(g)
0
(e), (f) and (g) into (d)
u x,0
y
dp
dx
d
dx
gdy
0
u 2 dy
(h)
0
Combine pressure and gravity terms
dp
dx
dp
dx
g
(i)
8
Multiply by
and rewrite as integral
dp
dx
g
(j)
gdy
0
(j) into (h)
u x ,0
y
g
(
d
dx
)dy
0
u 2 dy
(k)
0
Express density difference in terms of temperature change
(T
(2.28)
T )
(2.28) into (k)
ν
u ( x,0)
y
g (T T ) dy
0
d
dx
u 2 dy
(7.35)
0
(7.35) applies to laminar as well as turbulent flow.
7.7.2 Integral Formulation of Conservation of Energy
Assume:
(1)
(2)
(3)
(4)
No changes in kinetic and potential energy
Negligible axial conduction
Negligible dissipation
Properties are constant
Forced convection formulation of conservation of energy, (5.7), is applicable to free
convection
T x,0
y
( x)
d
dx
u (T
T )dy
(7.36)
0
7.7.3 Integral Solution
Vertical plate, Fig. 7.6.
Uniform surface temperature T s .
We assumed
t . Thus we have two equations, (7.35) and (7.36) for the
determination of a single unknown .
Since both (7.35) and (7.36) must be satisfied, we introduce another unknown as
follows:
Assumed Velocity Profile:
u x, y
a0 ( x) a1 ( x) y a2 ( x) y 2
Boundary conditions on the velocity
(1) u(x,0)
0
(2) u(x, ) 0
a3 ( x) y 3
(a)
9
(3)
u( x, )
y
2
(4)
u ( x,0)
y
2
0
g
(Ts
T )
Applying the four boundary conditions gives a n . Equation (a) becomes
g (Ts T )
4ν
u
2
y
2
y
1
(b)
Let
g (Ts T )
4ν
u o ( x)
2
(c)
(b) becomes
u
y
uo ( x)
1
y
2
(7.37)
Treat uo (x) as the second unknown function, independent of .
Assumed Temperature Profile:
T ( x, y)
b0 ( x) b1 ( x) y b2 ( x) y 2
(d)
The boundary conditions are
(1) T ( x,0) Ts
(2) T ( x, ) T
T ( x, )
(3)
0
y
Application of the above boundary conditions gives
T ( x, y) T
(Ts T ) 1
y
2
(7.38)
Heat Transfer Coefficient and Nusselt Number
T ( x,0)
y
Ts T
k
h
(7.24)
(7.38) into (7.24)
h
2k
( x)
(7.39)
Thus the local Nusselt number is
Nu x
Must find and uo (x) and
(x).
hx
k
2
x
( x)
(7.40)
10
Solution
(7.37) and (7.38) into (7.35)
ν
uo
g (Ts T )
2
y
1
d u o2
dx 2
dy
0
y
y2 1
4
dy
(e)
0
Evaluate the integrals
1 d 2
uo δ
105 dx
1
βg Ts T
3
δ
ν
uo
δ
(7.41)
(7.37) and (7.38) into (7.36)
( x)
2 (Ts T )
1
d uo
T )
dx
(Ts
y
y 1
4
dy
(f)
0
Evaluate the integrals
1 d
uo
60 dx
1
(7.42)
(7.41) and (7.42) are two equation for (x) and uo (x).
Assume a solution of the form
u o ( x)
Ax m
(7.43)
( x)
Bx n
(7.44)
A, B, m and n are constants.
substitute (7.43) and (7.44) into (7.41) and (7.42)
2m n 2 2m
A Bx
105
n 1
m n
ABx m
210
1
g To T Bx n
3
1
x
B
n 1
A m
vx
B
n
n
(7.45)
(7.46)
Exponents in each equation must be identical. Thus
2m n 1 n
m n 1
m n
(g)
n
(h)
1
4
(i)
Solve (g) and (h) for m and n gives
m
1
, n
2
(i) into (7.45) and (7.46) gives A and B
A 5.17v Pr
and
20
21
1/ 2
g (Ts T )
v2
1/ 2
(l)
11
20
21
B 3.93 Pr -1/2 Pr
1/ 4
g (Ts T )
1/ 4
(m)
v2
(i) and (m) into (7.44)
20 1
3.93
1
21 Pr
x
1/ 4
( Ra x )
1/ 4
(7.47)
(7.47) into (7.40)
1/ 4
20 1
0.508
1
21 Pr
Nu x
( Ra x )1/ 4
(7.48)
7.7.4 Comparison with Exact Solution for Nusselt Number
(7.26) is the exact solution to the local Nusselt number
1/ 4
Grx
4
Nu x
d (0)
d
(7.26)
Rewrite (7.26) as
Grx
4
1/ 4
d (0)
d
Nu x
(7.49)
Rewrite (7.48)
Grx
4
1/ 4
Nu x
20 1
0.508
1
21 Pr
The right hand side of (7.49) and (7.50) are
compared in Table 7.3.
The exact solution for Pr 0
Nu x
exact
0.600 ( PrRa x )1/4 , Pr
integral
0.514( PrRa x )1 / 4 , Pr
0
Exact and integral solutions for Pr
Nu x
exact
Nu x
integral
0.503( Ra x )1/4 ,
0
(7.51a)
are
Pr
(7.29b)
0.508( Ra x )1 / 4 , Pr
(7.51b)
NOTE: The error ranges from 1% for Pr
14% for Pr 0 .
(4 Pr )1/ 4
(7.50)
Table 7.3
Pr
_ d (0)
0.01
0.03
0.09
0.5
0.72
0.733
1.0
1.5
2.0
3.5
5.0
7.0
10
100
1000
0.0806
0.136
0.219
0.442
0.5045
0.508
0.5671
0.6515
0.7165
0.8558
0.954
1.0542
1.1649
2.191
3.9660
d
0.508
20 1
1
21 Pr
(7.29a)
0
Applying integral solution (7.47) to Pr
Nu x
1/ 4
to
0.0725
0.1250
0.2133
0.213
0.4627
0.5361
0.5399
0.6078
0.7031
0.7751
0.9253
1.0285
1.1319
1.2488
2.2665
4.0390
1/ 4
(4 Pr )1/4
Chapter 8: Convection in External Turbulent Flow
8.1.
Introduction
Turbulent flow is disordered, with random and unsteady velocity fluctuations; hence,
exact predictions cannot be determined.
Turbulence affects local velocity distribution, drag force, and heat transfer in both
natural and industrial processes.
Understanding of turbulent flow leads to the ability to make design improvements to
either reduce or enhance turbulent effects
Our understanding still relies on empirical data and rudimentary conceptual drawings,
and, more recently, computer simulations.
Exact solutions are not possible.
Chapter Focus: Wall-bounded shear flows
8.1.1. Examples of Turbulent Flows
(a) Mixing Processes
Combustion processes – proper mix of fuel and air is one of the requirements for
combustion efficiency
Chemical processing, such as the production of polymers
Laminar mixing occurs when liquids are viscous and/or slowly mixed, and can be
problematic
(b) Free Shear Flow
Jet Flows (refer to fig. 8.1a): jet energy is dissipated to the surrounding fluid
Turbulent wake (refer to fig. 8.1b): transfers energy between an object and the
ambient flow, contributes to an object’s drag
Smoke stack exhaust is dispersed by turbulence (refer to figure 8.c)
(c) Wall-Bounded Flows
Flow of air over a flat plate or airfoil
Flow of fluid in a pipe
Irregular or random motions cause the shape of the velocity profile and the boundary
layer edge location to change with time
Instantaneous velocity profiles are time-averaged for simplicity: u
Turbulent velocity can be decomposed into steady (mean, u ) and unsteady
(fluctuating, u ) components
Note: Refer to fig. 8.2 for details on the velocity profiles and unsteady components
The mixing of velocity fluctuations in turbulent flow creates a steeper profile than that of
a laminar flow, with a larger boundary layer and higher wall shear stress.
Turbulent fluctuations enhance momentum transfer between the surface and the flowing
fluid, resulting in higher skin friction; this suggests that surface fluctuations similarly
enhance heat transfer.
Note: See figure 8.3 for a comparison of laminar and turbulent velocity profiles.
By understanding turbulence, we can alter designs to take advantage of turbulent effects,
such as increased heat transfer and reduced drag.
Note: See the golf ball example in fig. 8.4 for an example of drag reduction via
turbulence.
8.1.2. The Reynolds Number and the Onset of Turbulence
The Reynolds number is the ratio of inertial to viscous forces, and indicates the onset of
turbulence.
Reynolds equation:
uD
Re D
(8.1)
The onset of turbulence for flow through tubes is approximately Ret
uD /
2300 .
For uniform flow over a semi-infinite flat plate, the onset of turbulence is approximately
Ret V xt /
500,000 .
Viscous forces dominate at low flow velocity.
At high velocity, inertial forces acting on individual particles dominate; the flow
amplifies these disturbances, creating a more chaotic flow.
Turbulence initiates near the wall in wall-bounded flows.
Note: See fig. 8.5 for the development of turbulent flow over a semi-infinite flat plate.
8.1.3. Eddies and Vorticity
Eddies are regions of intermittent, swirling patches of fluid
An eddy is a particle of vorticity,

:

V
(8.2)
Eddies form in regions of velocity gradient.
Example: for 2D flow over a flat plate:
z
v
x
High shear stress results in high vorticity.
u
y
The formation and behavior of eddies within the boundary layer is not fully understood.
Vortex stretching increases the kinetic energy of the vortex, and is thought to be a major
mechanism for the main flow to transfer energy to the turbulence.
Note: Refer to fig. 8.6 for details of eddy formation, vortex formation and vortex
stretching in a wall-bounded flow.
Eddies provide bulk motion (advection) and mixing within the boundary layer.
Advection differentiates turbulent flow from laminar flow; laminar flow has no bulk
motion and relies solely on viscous diffusion to transfer momentum.
8.1.4. Scales of Turbulence
Viscous effects play a vital role in turbulence.
Most turbulence originates from shear flow.
Rotation of a fluid element occurs under the action of viscous shear.
Lewis Richardson proposed the concept of energy cascade in 1922:
Turbulence is comprised of a range of different eddy sizes.
Treating the eddy as a distinct fluid structure with:
Characteristic size: 
Characteristic velocity: u
Eddy turnover time: t
/u
Reynolds number for the eddy: Re
u /
Inertial forces in a turbulent flow cause large, unstable eddies with high kinetic
energy and high Reynolds numbers (on the scale of the main flow) to break up into
smaller and smaller eddies.
Energy is transferred as each eddy breaks into smaller ones.
The breaking up process continues until the Reynolds numbers of the eddies
approach unity; at this time, viscous forces dissipate the energy of the small eddies
into heat.
Note: Fig. 8.7 illustrates the cascade process.
Eddies decrease in size much faster than in velocity; therefore, the smaller eddies
experience very high velocity gradients and have high vorticity.
The largest eddies contain the bulk of the kinetic energy in a turbulent flow; the smallest
eddies contain the bulk of vorticity and therefore the mechanism of dissipation.
Andrey Kolmogorov proposed a model in 1942 based on the ideas above:
The largest eddies contain the bulk of the kinetic energy
The smallest scales reach a Reynolds number of unity prior to dissipating into heat
Kolmogorov’s relations:
/  ~ Re
3/ 4
(8.3a)
v / u ~ Re
1/ 4
(8.3b)
1/ 2
(8.3c)
/ t ~ Re
/  is the ratio of the length scales of the largest and smallest eddies
v / u is the ratio of their velocities
/ t is the ratio of their time scales
The Reynolds number is that of the largest eddy: Re
The variables
, v and
u /
are called the Kolmogorov microscales
The variables  , u and t are called the integral scales
Important points from Kolmogorov’s model:
There is a vast range of eddy sizes, velocities and time scales in turbulent flow,
making modeling difficult
The smallest eddies are not infinitely small, since they are dissipated into heat by
viscous forces.
The scale of the smallest eddies is determined by the scale of the largest eddies
through the Reynolds number.
Turbulent flow responds to an increase in velocity by producing smaller eddies,
hence, viscous dissipation is increased.
Faster turbulent flows have finer turbulent structure; this is observed in the real
world.
8.1.5. Characteristics of Turbulence
Turbulence is comprised of irregular, chaotic, three-dimensional fluid motion, but
containing coherent structures.
Turbulence occurs at high Reynolds numbers, where instabilities give way to chaotic
motion.
Turbulence is comprised of many scales of eddies, which dissipate energy and
momentum through a series of scale ranges.
The largest eddies contain the bulk of the kinetic energy and break up by inertial
forces.
The smallest eddies contain the bulk of the vorticity and dissipate into heat.
Turbulent flows are not only dissipative, but also dispersive through the advection
mechanism.
8.1.6. Analytical Approaches
The continuum hypothesis still applies to the governing equations of fluid mechanics for
turbulent flow.
This is based on the example in section 8.1.4, where the microscale for air flowing
over a 10 cm diameter cylinder at 10 m/s was approximately 20 microns.
The mean free path of air at standard conditions is 10-8 m; this is three orders of
magnitude smaller.
The computing power required to process a direct numerical simulation (DNS) using
Computational Fluid Dynamics (CFD) is enormous and exceeds practical limitations,
due to the vast range of scales of turbulence between the largest and smallest eddies.
An analytical approach is used to predict macroscale properties, such as velocity, drag
force and heat transfer.
Microscopic flow structure is ignored; turbulent fluctuations are analyzed statistically.
Important idealizations are used to simplify statistical analysis:
Homogeneous turbulence: turbulence whose microscale motion does not, on average,
change from location to location or from time to time
Isotropic turbulence: turbulence whose microscale motion does not, on average,
change as the coordinate axes are rotated
These idealizations are not necessarily realistic, but they can be approximated in the
laboratory, so experimental data can be compared to the simplified statistical analytic
flow models.
8.2.
Conservation Equations for Turbulent Flow
8.2.1. Reynolds Decomposition
Reynolds proposed that turbulent flow can be considered as the superposition of a timeaveraged and a fluctuating component.
Reynolds Decomposition Approach:
Each fluctuating property in the governing equations is decomposed into a time
averaged and a fluctuating component.
The entire equation is time averaged.
For an arbitrary property g:
g
g
g
(8.4)
Where g , the time-averaged component, is determined by:
g
1
g t dt
0
By definition, the time average of the fluctuating property, g , is 0:
(8.5)
1
g
g t dt
(8.6)
0
0
Useful averaging identities:
a
a
(a ) 2
ab
(a ) 2
ab
a b
a
t
ab
a b
0
(8.7a)
ab
ab
(8.7b)
(8.7c)
aa
0
(8.7d)
(8.7e)
a2
(a ) 2
(8.7g)
a
x
(8.7i)
a
t
(a ) 2
a
x
0
(8.7f)
(8.7h)
(8.7j)
An example proving identities (8.7a) and (8.7g) is given.
8.2.2. Conservation of Mass
Reynolds decomposition is applied to the Cartesian conservation of mass equation:
t
( v)
y
( u)
x
( w)
z
0
(2.2a)
Analysis is limited to incompressible, two-dimensional flow.
Substituting the Reynolds decomposed velocities, u u u , and v v v :
(u u )
x
(v v )
y
0
(a)
u
x
u
x
v
y
v
y
0
(b)
u
x
v
y
v
y
0
(c)
This is expanded:
The equation is time averaged:
u
x
Invoking identity (8.7h), (c) becomes:
u
x
u
x
v
y
v
y
0
(d)
Using identity (8.7a), u u and v v , and by (8.6), u v 0 , so (d)
reduces to the time averaged turbulent flow continuity equation, identical in form
to (2.2):
v
y
u
x
(8.8)
0
Subtracting (8.8) from (b) demonstrates that the divergence of the fluctuating
term is 0:
v
y
u
x
(8.9)
0
8.2.3. Momentum Equations
Reynolds decomposition is applied to the Cartesian momentum equations:
u
t
2
2
2
u
x
v
u
y
w
u
z
gx
p
x
u
x2
u
y2
u
z2
v
v
u
t
x
v
y
w
v
z
gy
p
y
v
x2
v
y2
2
v
u
2
2
v
z2
(2.10x)
(2.10y)
Assumptions
(1) Two-dimensional flow
(2) Incompressible flow
(3) Constant properties
(4) Body forces are neglected
(5) Flow is, on average, steady-state, so u and v are constant
Formulation
Equations (2.10x) and (2.10y) become:
u
t
u
u
x
v
u
y
p
x
v
t
u
v
x
v
v
y
p
y
2
u
2
x
2
y2
2
v
2
x
2
y2
u
v
(8.10x)
(8.10y)
From the product rule of differentiation, u( u / x) and v( u / y) in the xmomentum equation become:
u
u
x
u2
x
v
u
y
(u v)
y
u
x
u
u
(a)
v
y
Substituting (a) into the x-momentum equation (8.10x):
(b)
u2
x
u
t
(u v)
y
u
x
u
( A)
u
v
y
2
p
x
2
u
x2
u
(c)
y2
( B)
The terms (A) and (B) in (c) above are combined, and by conservation of mass:
v
y
u
x
u
(d)
0
The x-momentum equation reduces to:
(u v)
y
u2
x
u
t
2
p
x
2
u
x2
u
(8.11)
y2
The y-momentum equation reduces in a similar fashion
Reynolds decomposition is now performed on the simplified x- and y-momentum
equations, resulting in the turbulent x- and y-momentum equations for turbulent
flow:
2
u
u
x
v
u
y
p
x
u
x2
v
x
v
v
y
p
y
2
u
v
x2
2
u
y2
(u ) 2
x
v
uv
x
2
y2
uv
y
(8.12x)
(v ) 2
y
(8.12y)
Equations (8.12) are identical to equations (8.10) with two notable exceptions:
The transient terms u / t and v / t disappear.
More importantly, new terms indicating fluctuating velocity are introduced.
8.2.4. Energy Equation
Reynolds decomposition is applied to the Energy equation for incompressible flow:
cp
T
T
u
t
x
v
T
y
w
T
z
2
k
T
x2
2
T
y2
2
T
z2
(2.19b)
Assumptions
(1) Heat generation is neglected
(2) Thermal properties are considered constant
(3) Steady-on-average flow
(4) Two dimensional flow
(5) The dissipation function Φ is neglected; this is appropriate as long as the flow
is not highly viscous or compressible
Formulation
The Energy equation reduces to:
T
t
cp
u
T
x
v
2
T
y
k
2
T
x2
T
y2
(8.13)
After Reynolds decomposition and time averaging is applied (left as an exercise,
Problem 8.7), equation (8.13) becomes:
cp u
T
x
v
2
T
y
2
T
x2
k
T
y2
cp
uT
x
cp
vT
y
(8.14)
Equation (8.14) is almost identical to the steady-state Energy equation (8.10) with two
notable exceptions:
The transient term
c p T / t disappears
Two terms indicating fluctuating velocities and temperature are introduced, and
come out of the convective terms on the left side of the equation.
8.2.5. Summary of Governing Equations for Turbulent Flow
Continuity:
v
y
u
x
(8.8)
0
x-momentum:
2
u
u
x
v
u
y
p
x
u
x2
v
x
v
v
y
p
y
2
u
2
u
y2
(u ) 2
x
v
uv
x
uv
y
(8.12x)
(v ) 2
y
(8.12y)
vT
y
(8.14)
y-momentum:
v
x2
2
y2
Energy:
cp u
8.3.
T
x
v
T
y
2
k
T
x2
2
T
y2
cp
uT
x
cp
Analysis of External Turbulent Flow
Goal: Solve the governing equations for turbulent flow for forces of interest, such as
drag force and heat transfer.
Focus: Flow along a surface
Boundary layer concept is invoked
Similar development of boundary layer equations to those of laminar flow (Chapter 4),
with two important differences:
Turbulent flow governing equations contain time-averaged quantities (i.e. u , T ),
which are handled exactly like u and T previously
Turbulent equations contain fluctuating terms (i.e. u and T ) that require additional
consideration
8.3.1. Turbulent Boundary Layer Equations
(i) Turbulent Momentum Boundary Layer Equation
Turbulent flow over a flat plate is considered (refer to Fig. 8.8)
Assumptions
(1) Steady-state
(2) Incompressible flow
(3) Constant properties
(4) The boundary layer is thin,
L , so:
(8.15)
1
L
Formulation
Following the arguments used for the laminar boundary layer, the following scalar
arguments are made:
u ~V
(8.16a)
x~L
(8.16b)
y~
(8.16c)
Following an analysis similar to that in Section 4.2, the viscous dissipation terms in
(8.12x) compare as follows:
2
u
2
x2
y2
u
(8.17)
The pressure gradient in the y-direction is negligible:
p
y
(8.18)
0
The pressure gradient in the x-direction is expressed as:
p
x
The fluctuation terms
dp
dx
(u ) 2
and
x
dp
dx
uv
require additional scaling arguments.
y
(8.19)
There is no preferred direction to the fluctuations; this is equivalent to assuming the
turbulence is isotropic, therefore
u ~v
2
u
(8.20)
~uv
(8.21)
Comparing the fluctuating terms using scale analysis:
Since
(u ) 2 (u ) 2
~
x
L
(a)
u v u v (u ) 2
~
~
y
(b)
L , we can conclude that
(u ) 2
x
uv
y
(8.22)
Using the simplifications (8.17) and (8.22), the x-momentum equation for the turbulent
boundary layer reduces to:
u
x
u
v
u
y
dp
dx
2
u
y2
uv
y
(8.23)
(ii) Turbulent Energy Equation
The derivation for laminar boundary layer equations is again followed.
Formulation
Scaling arguments for the thermal boundary layer:
(8.16b)
x~L
y~
(8.24)
t
T ~ Ts T
(8.25)
The scale for the velocity sizes depend on
t
and
, as in Chapter 4.
The second derivative terms compare as follows
2
T
2
2
y2
x
The fluctuation terms
cp
x
uT
T
and
(8.26)
cp
y
vT
require additional scaling
arguments:
There is no preferred direction for the fluctuations, so:
u ~v
(8.20)
uT ~vT
(8.27)
The fluctuating terms compare as follows (it is left to the reader to derive this):
vT
y
uT
x
(8.28)
Applying simplifications (8.26) and (8.28), the energy equation for the turbulent
boundary layer reduces to:
cp u
T
x
v
T
y
2
T
y2
k
cp
vT
y
(8.29)
8.3.2. Reynolds Stress and Heat Flux
Equations (8.23) and (8.29), the x-momentum and energy equations for the turbulent
boundary layer, are written as follows to provide physical insight and a way to model the
time-averaged fluctuation terms, u v and v T :
u
u
x
cp u
u
y
v
T
x
v
dp
dx
T
y
y
y
k
T
y
u
y
uv
cp v T
(8.30)
(8.31)
The terms in parenthesis on the right side of equation (8.30) represent shear stress
The first term represents the molecular shear stress from the time-averaged velocity
Boussinesq suggested (1877) that the second term can be viewed as shear imposed
by the time-averaging velocity fluctuations.
Note: Refer to Fig. 8.9 and to the related discussion in the text for a visualization of
Boussinesq’s idea.
The time average fluctuation term u v is proportional to the velocity gradient, just
like for the viscous shear stress, suggesting u v behaves like shear in the flow.
uv
u
y
The apparent shear stress experienced by the flow is made up of two parts
Molecular shear imposed by the time-averaged velocity component
Turbulent shear stress
velocity fluctuations
u v , or Reynolds stress, imposed by the time-averaged
The apparent energy flux in the energy equation contains a turbulence-induced heat flux,
c p v T , called the turbulence heat flux or Reynolds heat flux.
8.3.3. The Closure Problem of Turbulence
Summary of turbulence boundary equations:
Continuity
v
y
u
x
(8.8)
0
x-momentum
u
u
x
v
u
y
dp
dx
y
u
y
uv
(8.30)
Energy
cp u
T
x
v
T
y
y
k
T
y
cp v T
(8.31)
The boundary conditions for these equations are:
u (x,0)
0
(8.31a)
v(x,0)
0
(8.31b)
u ( x, )
V
(8.31c)
u (0, y)
V
(8.31d)
T ( x,0)
Ts
(8.31e)
T ( x, ) T
(8.31f)
T (0, y )
(8.31g)
T
If we know the velocity field outside of the boundary layer, the pressure can be
determined:
The pressure gradient is expressed as
dp
dx
dp
dx
(8.32)
From inviscid flow theory, outside the boundary layer:
V
dV
dx
1 dp
dx
(8.33)
We are left with three equations, (8.8), (8.30) and (8.31), and five unknowns, u , v , T ,
u v , and v T ; this is the closure problem of turbulence.
The two terms u v and v T are nonlinear terms
There is no exact solution to the turbulence boundary layer equations
Modeling will facilitate numerical and approximate solutions
8.3.4. Eddy Diffusivity
Reynolds Stress, based on Boussinesq’s hypothesis, is modeled as
uv
u
y
M
(8.34)
is the momentum eddy diffusivity
M
M
is referred to as the eddy viscosity
The Reynolds heat flux is modeled as
cp v T
H
cp
H
T
y
(8.35)
is the thermal eddy diffusivity
cp
H
is referred to as the eddy conductivity
The boundary layer momentum and energy equations are written as:
u
u
x
cp u
u
y
v
T
x
v
dp
dx
T
y
y
These are simplified by dividing (8.36) through by
u
u
x
u
T
x
v
v
u
y
y
T
y
y
k
cp
H
and (8.37) by
u
y
T
y
(8.36)
(8.37)
cp :
u
y
M
H
y
M
T
y
(8.38)
(8.39)
The terms in brackets in (8.38) represent the apparent shear stress:
u
y
app
M
(8.40)
The terms in brackets in (8.39) represent the apparent heat flux:
qapp
cp
H
T
y
(8.41)
The negative sign in (8.41) assigns the correct direction to heat transfer.
and H are properties of the flow and are dependent on the velocity and temperature
fields, respectively
M
The number of unknowns has not been reduced; the velocity fluctuation terms have been
replaced with expressions containing different unknowns
Finding ways to evaluate
8.4.
M
and
H
is one of the main goals of turbulence research
Momentum Transfer in External Turbulent Flow
Section Focus: Finding ways to evaluate
M
and
H
Energy and momentum equations are decoupled
Energy equation solution requires knowledge of both velocity and temperature fields
Momentum equation requires knowledge of velocity field, so solving for momentum
transfer requires solutions of continuity and momentum equations only
This still leaves two equations and three unknowns, but the situation is
simplified; now only M requires modeling
8.4.1. Modeling Eddy Diffusivity: Prandtl’s Mixing Length Theory
Boussinesq postulated that
M
is constant
This does not allow u v to approach zero at the wall, which is unrealistic, since
turbulent fluctuations are expected to dampen out near the wall
Prandtl’s model reasons that fluid particle behavior is analogous to that of molecules in
the kinetic theory of gases.
Note: Refer to Fig. 8.10 for visual details, (i.e. directions) of variables discussed
below.
In a two-dimensional flow, Prandtl defines:
v as a velocity fluctuation forcing a particle towards the wall
 , the mixing length, as the distance the particle travels as a result of the velocity
fluctuation
The resulting velocity fluctuation u is approximated using Taylor series expansion:
u final
So, with u
uinitial
u
dy
y
(a)
u final uinitial,
u ~
u
y
(b)
Again assuming that the velocity fluctuations have no preferred direction, u ~ v , we
have
v ~
u
y
(c)
By the above scales, one argues that the scale of the turbulent stress term
u v ~ (u )( v ) ~ 
2
u
y
u v is:
2
(d)
Therefore, solving equation (8.34) for eddy viscosity gives:
-u v
u
~ 2
u/ y
y
M
(8.42)
The absolute value on the derivative in equation (8.42) ensures the eddy diffusion
remains positive.
Note that the mixing length itself must still be modeled, and is dependent on the type
of flow.
Prandtl’s model for mixing length for flow over a flat plat is:

where
(8.43)
y
is some constant
This implies that the mixing length approaches zero as y approaches zero.
Thus equation (8.42) becomes Prandtl’s mixing-length model:
2 2
M
y
u
y
However:
is unknown
No single value for
is effective throughout the boundary layer.
Before developing a suitable model for
understood. This is presented next.
, boundary layer behavior must be further
8.4.2. Universal Turbulent Velocity Flow Profile
In Chapter 5, an approximate solution for the integral form of the boundary layer
equations for laminar flow was found using a universal velocity profile u y ; this
approach is used here.
This approach also provides physical insight that can be applied to other solution
techniques.
Section Goal: Find a universal velocity profile function
(i) Large-Scale Velocity Distribution: “Velocity Defect Law”
Variables making up the velocity distribution are normalized:
y
becomes one axis
(8.44)
u / V becomes the other axis, where V is the velocity outside of the boundary
layer
Figure 8.11 shows velocity curves at different values of wall friction; if the plot were
truly universal, all curves would collapse into one.
Data is normalized by the wall friction factor, C f
o
(1 / 2) V 2
, where
0
is the
shear stress at the wall.
Further manipulations to the data are performed:
For convenience, we cast the data relative to the velocity outside the boundary
layer:
(8.45)
(u V )
A friction velocity is defined:
u*
o
/
(8.46)
Friction velocity can also be written as:
u*
V
Cf /2
(8.47)
Note: u * has the same dimension as velocity.
The velocity difference is normalized by u * to produce the velocity defect:
(u V )
u*
(8.48)
The velocity defect is plotted against y / in Figure 8.12, which also shows the
range of experimental data. Note that the curves from Fig. 8.11 collapse into a single
curve
Note: The defect plot does not show enough detail near the wall.
An improved plot will probably be logarithmic, and will require an entirely new set of
coordinates.
(ii) Wall Coordinates
The following coordinates will collapse the boundary layer velocity data into a single
curve reasonably well:
u
y
Similarly, v
v / u* , and x
u
u*
yu*
xu * / .
(8.49)
(8.50)
All of the above variables are dimensionless, and are called wall coordinates.
Note: Refer to Figure 8.13 to see a plot of data produced by Clauser using these
coordinates.
The boundary layer extends out to approximately
2000 to 5000
The plot includes data obtained from flow in a pipe as well as flow along a flat plate,
so the profile appears to be universal.
(iii) Near-Wall Profile: Couette Flow Assumption
The turbulent boundary layer momentum equation, (8.38), is invoked to develop a model
of the velocity profile near the wall.
Assumption: Flow is over a flat plate, so dp / dx
0.
The momentum equation is simplified by recognizing that the flow is nearly parallel to
the wall, so v ~ 0 .
Conservation of mass implies that u / x ~ 0 ; therefore the u component of velocity
does not change significantly along the wall.
This scaling argument suggests that the convective terms in the momentum equation,
u u / x and v u / y , are each approximately zero, therefore, in (8.38):
u
y
M
y
~ 0 near the wall.
The bracketed term is the apparent shear stress,
app /
(Eqn. 8.40)
The above relationship implies that the apparent stress is approximately constant (with
respect to y), resulting in the Couette Flow Assumption:
app
M
u
~ constant
y
(8.51)
Recall from Chapter 3: this result is similar to Couette Flow.
Since the local shear is constant in (8.51), we can replace
app
with its value at the wall,
o.
The presence of the eddy diffusivity makes (8.51) different from Couette flow, implying
that the shear stress is constant, but not necessarily linear.
The Couette Flow Assumption is used to develop a velocity profile at the wall:
Substituting the definitions of u and y into (8.51), it can be shown that:
1
M
u
y
After rearranging and integrating:
1
(8.52)
y
dy
u
1
M
(8.53)
/
0
This is a general expression for the universal velocity profile in wall coordinates
This integral can be evaluated more simply by dividing the boundary layer into two
near-wall regions:
(1) a region very close to the wall where viscous forces dominate
(2) a region where turbulent fluctuations dominate
(iv) Viscous Sublayer
The wall tends to damp out or prevent turbulent fluctuations, so viscous forces dominate
very close to the wall:
M .
The Couette Flow Assumption reduces to
u
1.
y
Integrating, with boundary condition u
u
0 at y
y , 0
y
0 yields:
7
(8.54)
This relation compares well to experimental data from y+ 0 to 7, which is called the
viscous sublayer.
Equation (8.54) is illustrated in Figure 8.13. Note that the curvature in the plot results
from the semi-logarithmic coordinates.
(v) Fully Turbulent Region: “Law of the Wall”
Further away from the wall, turbulent fluctuations (i.e., Reynolds stresses) dominate, so
.
M
The Couette Flow Assumption (8.52) therefore becomes:
M
u
(8.55)
1
y
As before, is constant; this is the same value as in the viscous sublayer (which is the
value at the wall), so discontinuity between regions is avoided.
Substituting wall coordinates into Prandtl’s mixing length theory (8.44) yields:
2
M
( y )2
Equation (8.56) is substituted into (8.55):
u
y
(8.56)
2
(y )
2
2
u
1
y
Solving the above for the velocity gradient:
u
y
1
y
(8.57)
Integrating the above results in the Law of the Wall:
u
1
ln y
(8.58)
B
is called von Kármán’s constant, and experiments show that
0.41 .
B is a constant of integration, and is found as follows:
The viscous sublayer and the Law of the Wall region appear to intersect at roughly
y u ~ 10.8 .
Using the above as a boundary condition, B
5.0 .
An approximation for the Law of the Wall region is:
u
2.44 ln y
5.0,
50
y
1500
(8.59)
(vi) Other Models
The velocity profile presented in the previous sections is a two-layer model, and is called
the Prandtl-Taylor model.
Three-layer models, like that of von Kármán, also exist.
van Driest’s continuous law of the wall is a single equation model that illustrates how
single equation models work:
The eddy diffusion must diminish as y approaches zero, so van Driest proposed a
mixing length model of this form:

y1 e
y/ A
(8.60)
The term in parentheses is damping factor that makes (8.60) approach zero at the
wall.
Using (8.60) with (8.42) in (8.51) yields:
app
2 2
y 1 e
y/ A 2
u
y
(8.61)
As y approaches zero, the eddy diffusivity approaches zero, leaving pure viscous
shear.
Transforming (8.61) into wall coordinates, and solving for u / y , yields:
u
y
2
1
2
1 4
y
2
1 e
y /A
(8.62)
2
For flow over a smooth, flat plate, van Driest used κ = 0.4 and A+ = 26.
When integrated numerically, Equation (8.62) produces the curve shown in Fig.
8.13.
D.B. Spalding’s model is commonly used for both flat plates and pipe flow:
y
where Spalding used
u
e
B
e
u
1
( u )2
2
u
( u )3
6
(8.63)
= 0.40 and B = 5.5
Note that that u+ is implicit in (8.63), so there is no closed-form solution and one
must numerically solve for u+
Reichardt developed a profile frequently applied to pipe flow:
u
where
1
ln 1
y
C1 e
y /X
y
e
X
0.33 y
(8.64)
= 0.40, C = 7.8, and X = 11.
(vii) Effect of Pressure Gradient
The velocity profiles modeled up to this point assume the pressure gradient is zero.
Figure 8.14 depicts how the velocity profile is affected by a pressure gradient.
The plot, for flow over a flat plate, shows that in the presence of an adverse pressure
gradient, the velocity profile beyond y 350 deviates from the Law of the Wall
model.
The deviation is referred to as a ―wake.‖
The region y
350 is commonly referred to as the wake region.
The region where the data continue to adhere to the Wall Law is called the overlap
region.
The wake increases with adverse pressure gradient, until separation, where the
velocity profile deviates even from the overlap region.
A slight wake exists for zero or even a strong favorable pressure gradient, although
the difference between the two sets of data is negligible.
Wake models are not addressed in this text.
The Law of the Wall-type models developed earlier model flat plate flow reasonably
well in the presence of zero pressure gradient.
A favorable pressure gradient is approximately what we encounter in pipe flow; this is
one reason why the models developed here apply as well to pipe flow.
8.4.3. Approximate Solution for Momentum Transfer: Momentum Integral Method
To obtain drag force on the surface of a body, the momentum integral equation is
invoked, as in Chapter 5.
In independent works, both Prandtl and von Kármán used this approach to estimate the
friction factor on a flat plate.
(i) Prandtl-von Kármán Model
Considering a flat, impermeable plate exposed to incompressible, zero-pressure-gradient
flow, the integral momentum equation reduces to equation (5.5):
( x)
u ( x,0)
d
V
y
dx
( x)
u dy
0
d
dx
u 2 dy
(5.5)
0
This equation applies to turbulent flows if the behavior of the flow on average is
considered, and the flow properties are interpreted as time-averaged values
Recall that the integral method requires an estimate for the velocity profile in the
boundary layer.
Prandtl and von Kármán used Blasius’ model for the shear at the wall of a circular pipe,
based on dimensional analysis and experimental data:
Cf
where C f
2
o
0.07910 Re D1 / 4 ,
4000 Re D
105
(a)
/ um2 , and um is the mean velocity over the pipe cross-section.
Based on the above, the velocity profile in the pipe could be modeled as
u
uCL
y
ro
1/ 7
(b)
where y is the distance from the wall, and u CL is the centerline velocity
This is the well-known 1/7th Law velocity profile, further discussed in Chapter 9.
In an external flow a mean velocity is not defined, nor is ro,, so the following
adjustments are made:
ro is approximated by the edge of the boundary layer
u CL is approximated as representing the free-stream velocity V
The velocity profile in the boundary layer is then modeled as:
u
V
y
1/ 7
(8.65)
The fundamental problem with using this model in the integral equation is that the
velocity profile gradient goes to infinity as y goes to zero.
Equation (8.65) cannot be used directly to estimate the wall shear in equation (5.5).
Prandtl and von Kármán modeled the wall shear differently:
Since the characteristics of the flow near the surface of the plate are similar to that of
pipe flow, the Blasius correlation was adapted to find an expression for the wall
shear on a flat plate:
Recasting (a) in terms of the wall shear and the tube radius:
1/ 4
0.03326
o
2
um
roum
It can be shown that for the 1/7th velocity profile, u m
0.8167uCL .
Recall, u CL is modeled as V
The expression for
0
is written as:
u ( x ,0 )
y
o
1/ 4
0.02333V
2
(8.66)
V
In terms the friction factor, this is:
Cf
o
2
V2
0.02333
V
1/ 4
(8.67)
This can now be used in the momentum equation.
An example (Example 8.2) is presented in which the expressions for the boundary layer
thickness and the friction factor for flow over a flat plate are developed using the 1/7th
power law for the velocity profile (8.65) and the expression for friction factor (8.67).
The following equations are found within the example:
7 d
72 dx
o
V2
4
5
5/ 4
72 V
0.02333
7
(8.68)
1/ 4
x
C
(8.69)
The expression for the boundary layer thickness is:
0.3816
x
Re1x / 5
The expression for the friction factor is:
(8.70)
Cf
0.02968
2
Re1x / 5
(8.71)
Note: Refer to Fig. 8.15 for additional considerations for the boundary layer over a
flat plate.
Note that, according to this model, the turbulent boundary layer
Re x
1/ 5
/x varies as
, as does the friction factor C f ; this is contrast to laminar flow, in which
/x
and C f vary as Re x 1 / 2 .
(ii) Newer Models
A limitation of the Prandtl-von Kármán model is that the approximation for the wall
shear, Eqn. (8.66), is based on limited experimental data, and is of limited applicability
even for pipe flow
White presents a method that makes use of the Law of the Wall velocity profile (8.59):
Because the wall coordinates u and y can be expressed as functions of
Cf /2 ,
an expression for the friction factor can be developed from the Law of the Wall (a
technique which is also seen in analysis of pipe flow – see Section 9.5).
Substituting the definitions of u and y , as well as u * , into the Law of the Wall
expression (8.59):
u
V
2
Cf
2.44 ln
Cf
yV
2
5.0
Any y value within the wall law layer would satisfy this expression, but a useful
value to choose is the edge of the boundary layer, where u ( y
) V , so:
1
Cf /2
where Re
V
2.44 ln Re
Cf
2
5.0
(8.72)
/
Equation (8.72) relates the skin friction to the boundary layer thickness, and can be used
in the integral momentum equation, though it’s cumbersome.
By curve-fitting values obtained from (8.72) over a range of values from
Re 104 to 107 , yielding the approximate relation:
Cf
0.02Re
1/ 6
(8.73)
The above is used to estimate wall shear using the integral method, along with the 1/7th
power law for the velocity profile, resulting in the following solutions to the integral
momentum equation:
0.16
x
(8.74)
Re1x / 7
Cf
0.0135
2
Re1x / 7
(8.75)
The above equations replace the less accurate Prandtl-von Kármán correlations; White
recommends these for general use.
Kestin and Persen used Spalding’s law of the wall for the velocity profile to develop a
more-accurate, but more cumbersome correlation; White simplified their model to:
Cf
0.455
(8.76)
2
ln (0.06Re x )
According to White, the above relation is accurate to Kestin and Persen’s model to
within 1%.
(iii) Total Drag
Total drag is found by integrating the wall shear along the entire plate.
Assumptions
Laminar flow exists along the initial portion of the plate.
The plate has width w.
Formulation
The total drag over the entire plate is:
xcrit
FD
L
o lam wdx
0
Dividing by
1
2
V 2A
1
2
o turb wdx
(8.77)
xcrit
V 2 wL , the drag coefficient C D is:
xcrit
CD
1
L
L
C f ,lamdx
0
C f ,turbdx
(8.78)
xcrit
Substituting Equation (4.48) for laminar flow and using White’s model (8.75) for
turbulent flow, we obtain with some manipulation:
CD
The above assumes xcrit
0.0315
Re1L/ 7
5 105 .
1477
Re L
(8.79)
8.4.4. Effect of Surface Roughness on Friction Factor
The previous models have all assumed flow over smooth walls.
The interaction between the turbulent flow and the complex, random geometric features
of a rough wall is the subject of advanced study and numerical modeling.
Crude modeling and experimental study give some physical insight.
k is defined as the average height of roughness elements on the wall
k is transformed into wall coordinates: k
ku* /
For small values of k , ( k
5 ), experiments show that the velocity profile and friction
factor are unaffected by roughness.
Roughness is contained within the viscous sublayer; disturbances are likely
dampened out by the viscosity-dominated flow.
For k
10 , the roughness extends beyond the viscous sublayer and the viscous
sublayer begins to disappear, likely due to enhanced mixing the roughness provides.
For k
70 , viscous effects are virtually eliminated and the flow is fully rough.
The shape of the velocity profile changes little after this roughness value, so it is
expected that increasing the roughness will not change the friction factor.
Refer to Figure 8.16 for an illustration of how the near-wall velocity profile is affected
by roughness.
Roughness tends to shift the Law of the Wall down and to the right
The shift in velocity profile means that the velocity gradient at the wall is greater,
and therefore the friction factor increases
The slope of the wall law curve is not affected by roughness
Correlations for friction factor on rough plates are highly dependent on roughness
geometry; there are few models.
For fully rough flow over sand-rough plates, White [14] suggests the following
correlation, based on a wall law velocity profile developed from experimental data
Cf
1.4 3.7 log10
x
k
2
,
x
k
Re x
1000
(8.80)
Note that this correlation does not include the Reynolds number
8.5.
Energy Transfer in External Turbulent Flow
From Chapter 2, the heat transfer for flow over a geometrically similar body like a flat
plate can be correlated through dimensionless analysis by
Nu x
f ( x * , Re, Pr )
where x * is the dimensionless location along the body
(2.52)
This neglects buoyancy and viscous dissipation
The momentum and thermal eddy diffusivities introduced in equations (8.38) and (8.39),
M and H , are used to create the turbulent Prandtl number, a new dimensionless
parameter:
M
Prt
(8.81)
H
There are several options to develop suitable models for turbulent heat transfer:
Find an analogy between heat and mass transfer
Develop a universal temperature profile and then attempt to obtain an approximate
solution for heat transfer using the integral method
The universal temperature profile may also lend itself to a simple algebraic method
for evaluating the heat transfer
More advanced methods, including numerical solutions to the boundary layer flow,
are not covered in this text
8.5.1. Momentum and Heat Transfer Analogies
Reynolds theorized that heat transfer and the frictional resistance in a pipe are
proportional, based on his study of turbulent flow in steam boilers
This implies that if the friction along a pipe wall is measured or predicted, heat transfer
can be determined by using a multiplying factor, allowing one to solve directly for heat
transfer.
(i) Reynolds Analogy
The Reynolds analogy for external flow is developed:
Assumptions
Flow is parallel and is over a flat plate
The pressure gradient dp/ dx is zero
Formulation
The boundary layer momentum and energy equations reduce to:
u
u
x
u
T
x
v
v
u
y
y
T
y
y
0,
T (y
M
u
y
(8.82a)
T
y
(8.82b)
0) Ts ,
(8.83a)
H
The boundary conditions are
u (y
0)
u(y
T (y
) V ,
(8.83b)
) T
The variables are normalized as follows:
v
,
V
u
,V
V
U
T
T
Ts
, X
Ts
x
, and Y
L
y
L
Equations (8.82) and boundary conditions (8.83) become:
U
U
U
X
x
U (Y
U
Y
V
V
0)
U (Y
1
M
V L Y
U
Y
(8.84a)
1
y
0,
) 1,
H
V L Y
(Y
(Y
(8.84b)
Y
(8.85a)
0) 0
(8.85b)
) 1
Note that normalizing the boundary conditions has made them identical.
Equations (8.84) can then be made identical if (
possible under two conditions:
M)
(
H),
which is
The kinematic viscosity and thermal diffusivity are equal:
(8.86)
This condition limits the analogy to fluids with Pr = 1
This also suggests that the velocity and thermal boundary layers are
approximately the same thickness,
t
The eddy diffusivities are equal
M
H
(8.87)
This is justified by arguing that the same turbulent mechanism—the motion
and interaction of fluid particles—is responsible for both momentum and heat
transfer; Reynolds essentially made this argument, so equation (8.87) is
sometimes referred to as the Reynolds Analogy.
This also means that the turbulent Prandtl number Prt is equal to 1
The analogy is now complete; the normalized velocity and temperature profiles,
U ( X , Y ) and ( X , Y ) are equal.
Refer to Figure 8.17 for an illustration.
Derivation of a relationship between the shear stress and heat flux at the wall:
The ratio of the apparent heat flux and shear stress (equations. 8.40 and 8.41) is
qapp / c p
(
(
app /
H)
T/ y
M) u/ y
(8.88)
By imposing conditions (8.86) and (8.87), the terms in parenthesis cancel and, after
substituting the dimensionless variables into (8.88):
qapp
c p (Ts T )
V
app
/ Y
U/ Y
(8.89)
Since the dimensionless velocity and temperature profiles are identical, their
derivates cancel
The ratio qapp / app is constant throughout the boundary layer, so this ratio can be
represented by the same ratio at the wall , and equation (8.89) becomes:
qo
c p (Ts T )
V
o
This can be recast into a more convenient form by substituting qo h(Ts T ) and
2
1
into the above, and rearranging:
o
2 Cf V
h
V cp
Cf
2
The terms on the left side can also be written in terms of the Reynolds, Nusselt, and
Prandtl numbers
Stx
Nu x
Rex Pr
Cf
(8.90)
2
where St x is called the Stanton number
Equation (8.90) is commonly referred to as the Reynolds Analogy; it can also be
derived for laminar flow over a flat plate for Pr =1.
The Reynolds Analogy is limited to Pr = 1 fluids; it is appropriate for gases, but not for
most liquids.
(ii) Prandtl-Taylor Analogy
The Reynolds analogy doesn’t account for the varying intensity of molecular and
turbulent diffusion in the boundary layer
Very close to the wall, molecular forces are expected to dominate:
M
, and
H
(8.91)
Further away from the wall, turbulent effects dominate:
M
, and
H
Neither condition above restricts us to Pr = 1 fluids.
(8.92)
Prandtl and Taylor independently divided the boundary layer into two regions:
A viscous sublayer where molecular effects (8.91) dominate
A turbulent outer layer, where (8.92) is assumed to hold
In order for an analogy to exist, the momentum and boundary layer equations, and their
boundary conditions, must be identical in both regions, so:
The viscous sublayer is defined as the portion of the boundary layer beneath y = y1,
where y1 is some threshold value, with boundary conditions:
u (0)
T (0) Ts ,
0,
u ( y1 )
u1 ,
T ( y1 )
T1
The following normalized variables make the boundary conditions and equation
(8.82) identical:
u
, V
u1
U
v
,
u1
T Ts
, X
T1 Ts
x
and Y
y1
y
.
y1
For the viscous sublayer, the ratio of the apparent heat flux and apparent shear stress
(Eqn. 8.86) leads to the following:
Ts T1
where qapp /
app
= qo /
o
qo
Pr u1
o cp
(8.93)
= constant
The outer layer closely resembles the Reynolds Analogy, with M
H (or Prt 1 ),
but this time we assume that the turbulent effects outweigh the molecular effects; this
region has boundary conditions:
u ( y1 )
u(y
u1 ,
T ( y1 ) T1 ,
) V ,
T (y
) T
The following normalizing variables make the analogy valid in this region:
U
u u1
, V
V u1
v - u1
,
V
u1
T
T
T1
, X
T1
x
, and Y
L
y
L
For the outer region, the ratio of the apparent heat flux and apparent shear stress
(equation 8.86) leads to:
T1 T
The ratio qapp /
app
qo
(V
c
o p
u1 )
(8.94)
is constant, so we have chosen the value at y=y1, which can be
represented by qo / o .
Adding (8.93) and (8.94) yields:
qo
V
o cp
Ts T
Substituting
o
1C
2 f
u1
Pr 1
V
1
V 2 into the above yields
Cf /2
qo
V c p (Ts T )
St
u1
( Pr 1) 1
V
The velocity at the edge of the viscous sublayer, u1 , is estimated using the universal
velocity profile (Fig. 8.13); a value of u y 5 is chosen to approximate the edge
of the viscous sublayer.
From the definition of u :
u
5
u1
V
5
u1
V
2
, or
Cf
Cf
(8.95)
2
Thus the Prandtl-Taylor analogy is:
Stx
Cf /2
Nu x
Rex Pr
5
Cf
2
(8.96)
( Pr 1) 1
(iii) von Kármán Analogy
Theodore von Kármán extended the Reynolds analogy even further to include a third
layer – a buffer layer – between the viscous sublayer and outer layer:
Stx
Cf /2
Nu x
Rex Pr
1 5
Cf
2
(8.97)
5Pr 1
( Pr 1) ln
6
Note: Refer to Appendix D for development.
(iv) Colburn Analogy
Colburn proposed a purely empirical modification to the Reynolds analogy that accounts
for fluids with varying Prandtl number, using an empirical fit of available experimental
data:
Stx Pr 2 / 3
The exponent (2/3) is entirely empirical
Cf
2
(8.98)
The Colburn Analogy yields acceptable results for Rex 107 (including the laminar flow
regime) and Prandtl number ranging from about 0.5 to 60.
An example, Example 8.3, is presented in which the average Nusselt number for heat
transfer along a flat plate of length L with constant surface temperature is determined,
using White’s model for turbulent friction factor and assuming the flow over the plate
has an initial laminar region.
Equations developed within the example:
The average heat transfer coefficient from Equation (2.50) is split into laminar and
turbulent regions:
xc
hL
1
L
L
hlam ( x)dx
0
(8.99)
hturb ( x)dx
xc
After several manipulations, substitutions and assumptions, the equation for the
average Nusselt number for heat transfer along a flat plate of length L with constant
surface temperature and initial laminar region is:
Nu L
0.0158ReL6 / 7
739 Pr1/ 3
(8.100)
If the laminar region had been neglected, the above would be:
Nu L
0.0158ReL6 / 7 Pr1/ 3
(8.101)
8.5.2. Validity of Analogies
Momentum-heat transfer analogies are frequently used to develop heat transfer models
for many types of flows and geometries.
Although derived for a flat plate, these analogies are considered generally valid for
slender bodies, where pressure gradient does not vary greatly from zero.
They are approximately valid for internal flows in circular pipes, although other
analogies have been developed specifically for internal flow (See Chapter 9).
Although they are derived assuming constant wall temperature, the above correlations
work reasonably well even for constant heat flux.
The large temperature variation near the wall means that the assumption of uniform
properties, particularly for Pr, becomes a weakness; this is overcome by evaluating
properties at the film temperature:
Tf
Ts T
2
(8.102)
The analogies were derived assuming that the turbulent Prandtl number is equal to unity.
Experimentally-measured values of Prt are as high as 3 very near the wall, though
outside the viscous sublayer the values range from around 1 to 0.7.
The turbulent Prandtl number seems to be affected slightly by pressure gradient,
though largely unaffected by surface roughness or the presence of boundary layer
suction or blowing.
A value of Prt 0.85 is considered reasonable for most flows, so the analogies
should be approximately valid for real flows.
Though only an empirical correlation, the Colburn analogy was shown to represent
experimental data well over a variety of fluids.
However, it has been demonstrated that the Colburn analogy under-predicts the Nusselt
number by 30-40% for fluids with Prandtl numbers greater than 7.
8.5.3. Universal Turbulent Temperature Profile
Development of a universal temperature profile in turbulent flow provides physical
insight.
An approximate temperature profile can be used in an integral approach to solve for the
heat transfer.
(i) Near Wall Profile
Beginning with the turbulent energy equation, assume that, near the wall, the velocity
component v ~ 0, as is the temperature gradient T / x .
The left-hand side of (8.39) approaches 0:
H
y
T
~ 0 , near wall
y
This implies that the apparent heat flux is approximately constant with respect to y:
qapp
H
cp
T
~ constant
y
(8.103)
Solve the above relation for the temperature profile:
Since qapp / c p is constant throughout this region, replace qapp with qo and
substitute wall coordinates into (8.103) and rearrange so both sides are
dimensionless:
T
c p u*
y
qo
(8.104)
H
(8.104) suggests a definition for a temperature wall coordinate:
T
(Ts T )
The above is cast into a simpler form:
c p u*
qo
(8.105)
T
(8.106)
y
H
Though the above can now be integrated, H is unknown, but substituting M into
(8.106) by using the definition of the turbulent Prandtl number (8.81) yields:
y
dy
T
(8.107)
H
0
This is a general expression for the temperature profile in wall coordinates.
The above is evaluated by dividing it into two regions, as was done with the
universal velocity profile.
(ii) Conduction Sublayer
Very close to the wall, molecular effects are expected to dominate the heat transfer, so
H .
By evoking this, (8.107) reduces to:
T
Pr dy
Pr y
C
The constant of integration, C, is found by applying the boundary condition that
T ( y 0) 0 , resulting in C 0 and a temperature profile in the conduction sublayer
of:
T
Pr y , ( y
y1 )
(8.108)
y1 is the dividing point between the conduction and outer layers
(iii) Fully Turbulent Region
Outside the conduction-dominated region close to the wall, turbulent effects dominate,
so H
The temperature profile (8.104) becomes:
y
T
T1
dy
(8.109)
H
y1
must be evaluated to evaluate (8.109); this is done by relating it to the momentum
eddy diffusivity:
H
H
Prt
M
Recall (8.44), the model for the eddy diffusivity from Prandtl’s mixing length theory:
u
y
2 2
y
M
(8.44)
This is written in terms of wall coordinates:
2
M
( y )2
u
(8.110)
y
The partial derivative u / y can be found from the Law of the Wall
Substituting the above and (8.58) into (8.109) yields:
y
Prt
T
(8.111)
dy
y
y1
Assume Prt and
are constants, so (8.111) becomes:
T
Prt
y
ln
y1
, y
y1
(8.112)
The temperature profile defined by Equations (8.108) and (8.112) depends on the fluid
(Pr), as well as the parameters Prt and .
Kays et al. assumed Prt = 0.85 and = 0.41, but found that the thickness of the
conduction sublayer ( y1 ) varies by fluid
White reports a correlation that can be used for any fluid with Pr 0.7:
T
Prt
13 Pr 2 / 3 7
ln y
(8.113)
In this model, Prt is assumed to be approximately 0.9 or 1.0
Figure 8.18 is a plot of this model, along with viscous sublayer, for various values of
Pr
Note that the temperature profile increases with increasing Prandtl number.
(iv) A 1/7th Law for Temperature
A simpler 1/7th power law relation is sometimes used for the temperature profile:
T
T
Ts
Ts
y
1/ 7
(8.114)
t
8.5.4. Algebraic Method for Heat Transfer Coefficient
The existence of a universal temperature and velocity profile makes for a fairly simple
method to estimate the heat transfer:
The definition of the Nusselt number, expressed using Newton’s law of cooling is:
hx
k
Nu x
qo x
(Ts T )k
(8.115)
The universal temperature profile T is to be invoked, so using the definition of T ,
equation (8.105), the free stream temperature is defined:
T
c p u*
(Ts T )
(Ts T )
qo
c pV
Cf /2
qo
(8.116)
where (8.47) was substituted for the friction velocity u *
Substituting (8.116) into (8.115) for (Ts T ) and rearranging yields:
c pV
Nu x
Cf /2 x
T k
Then, multiplying the numerator and denominator by
yields:
Re x Pr C f / 2
Nu x
(8.117)
T
The universal temperature profile, Equation (8.113),is used to evaluate T :
T
Prt
13Pr 2 / 3 7
ln y
(8.118)
A precise value for y is not easy to determine, but a clever substitution is made by
using the Law of the Wall velocity profile.
Equation (8.58) is evaluated in the free stream as:
u
1
ln y
(8.119)
B
Substituting (8.119) into (8.118) for ln y , the Nusselt number relation then becomes
Nu x
Re x Pr C f / 2
Prt (u
B ) 13Pr 2 / 3
7
The above is simplified using the definition of Stanton number, St Nu x /( Rex Pr ) ,
selecting B = 5.0 and Prt = 0.9, and noting that the definition of u leads to
u
2/Cf
Stx
0.9 13 Pr
Cf
2/3
/2
0.88 C f / 2
(8.120)
Note how similar this result is to the more advanced momentum-heat transfer analogies,
particularly those by Prandtl and Taylor (8.96) and von Kármán (8.97).
8.5.5. Integral Methods for Heat Transfer Coefficient
The universal temperature profile allows us to model heat transfer using the integral
energy equation.
A case of turbulent flow over a flat plate where a portion of the leading surface is
unheated is examined, illustrated in Figure 8.19.
The simplest solution is to assume the 1/7th power law for both the velocity and
temperature profiles
This is mathematically cumbersome, and is developed in Appendix E.
The result of the analysis is:
Stx
Nu x
Rex Pr
Cf
1
2
xo
x
9 / 10
1/ 9
(8.121)
where xo is the unheated starting length
Note that Equation (8.121) reduced to the Reynolds Analogy when xo = 0, (the Prandtl
number in that derivation was assumed to be 1).
The model has been used to approximate heat transfer for other fluids:
Equation (8.121) is expressed as:
Nu xo
Nu x
1
where Nu xo
0
xo / x
0
9 / 10 1 / 9
(8.122)
represents the heat transfer in the limit of zero insulated starting
length
In this form, other models for heat transfer, like von Kármán’s analogy, could be
used to approximate Nu xo 0 for Pr ≠ 1 fluids.
8.5.6. Effect of Surface Roughness on Heat Transfer
Surface roughness is important in such applications as turbomachinery, where rough
surfaces can be used to enhance heat transfer in the cooling of turbine blades.
Figure 8.16 shows that the viscous sublayer diminishes and disappears as roughness
increases, implying that the turbulent fluid elements exchange momentum with the
surface directly (i.e. profile or pressure drag) and the role of molecular diffusion (i.e.
skin friction) is diminished.
Heat transfer relies on molecular conduction at the surface, no matter how rough the
surface, or how turbulent the flow.
Fluid in the spaces between roughness elements is largely stagnant, and transfers heat
entirely by molecular conduction.
The conduction sublayer can be viewed as the average height of the roughness
elements.
The major resistance to heat transfer is formed by the stagnant regions between
roughness elements
Roughness cannot improve heat transfer as much as it increases friction.
This means that we cannot predict the heat transfer by using a friction factor for
rough plates along with one of the momentum-heat transfer analogies
In developing a model for heat transfer on a rough surface, the following are expected:
Roughness size has no influence until it extends beyond the viscous and conduction
sublayers.
The influence of roughness reaches a maximum beyond some roughness size (the
fully rough limit).
The Prandtl number should be present in any model, and fluids with higher Prandtl
number (lower conductivity) would be affected more by roughness
In these fluids, lower-conductivity fluid trapped between the roughness elements
will have a higher resistance to heat transfer.
The conduction sublayer is shorter for these fluids, so roughness elements
penetrate relatively further into the thermal boundary layer
A correlation for a rough plate was developed by Kays, et al. [29]:
St
Cf
2
Prt
C (ks )0.2 Pr 0.44 C f / 2
1
(8.123)
where k s k s u * / is based on the equivalent sand-grain roughness, k s , and C is a
constant that depends on roughness geometry
This model displays the expected behavior.
For high-Prandtl-number fluids the second term in the parentheses dominates
For sufficiently low-Pr fluids the second term diminishes, in spite of roughness size
Bogard, et al. showed a 50% increase in heat transfer on rough turbine blades
compared to heat transfer in smooth turbine blades, and that increasing roughness
beyond some value showed little increase in the heat transfer, which is consistent
with the above expectations.
Chapter 9: Convection in Turbulent Channel Flow
9.1
Introduction
Laminar channel flow was discussed in Chapter 6; many features of turbulent flow are
similar
Chapter begins with the criteria for fully developed velocity and temperature profiles
Chapter Focus: Analysis of fully developed flows
Analysis is limited to the following general boundary conditions:
(i) uniform surface temperature
(ii) uniform surface heat flux
9.2
Entry Length
Criteria for entry length was discussed in Chapter 6
As a rule of thumb, fully developed velocity and temperature profiles exist for
Lh
De
Lt
De
10
(6.7)
where D e is the hydraulic or equivalent diameter
De
4Af
P
where A f is the flow area and P is the wetted perimeter
Eq. (6.7) is recommended for Pr = 1 fluids
More elaborate correlations exist, especially for hydrodymanic entry length; the
following approximations are recommended:
From White
Lh
De
4.4 Re1D/e6
(9.1)
Lh
De
0.623Re1D/e4
(9.2)
From Latzko
Thermal entry length doesn’t lend itself to a simple, universally-applicable equation
since the flow is influenced by fluid properties and boundary conditions
Hydrodynamic entry length is much shorter for turbulent flow than for laminar, so much
so that sometimes it’s neglected from analysis
Thermal entry length is often important
Analysis of heat transfer in the thermal entry length is complicated and is not covered in
the text
9.3
Governing Equations
Figure 9.1 shows a circular pipe with the velocity in the x-direction is labeled as u
Assumptions:
Two-dimensional
Axisymmetric
Incompressible flow
9.3.1 Conservation Equations
After Reynolds-averaging, conservation of mass reduces to:
u
x
1
(rv r ) 0
r r
(9.3)
Using the same conditions, the Reynolds-averaged x-momentum equation reduces to:
u
u
x
vr
vr
r
T
x
vr
T
r
1 p
x
1
r(
r r
M
)
u
r
(9.4)
Conservation of energy becomes:
u
1
r(
r r
H
)
T
r
(9.5)
9.3.2 Apparent Shear Stress and Heat Flux
The apparent shear stress and heat flux are defined similarly to that of the flat plate
development:
app
q app
cp
(
M
(
u
r
)
H
)
(9.6)
T
r
9.3.3 Mean Velocity and Temperature
Mean velocity and bulk, or mean, temperature are used in correlations for predicting
friction and heat transfer in duct flow
Mean velocity is calculated by evaluating the mass flow rate in the duct
(9.7)
ro
m
um A
u (2 r )dr
0
Assuming constant density, this becomes
ro
1
um
ro
u (2 r )dr
ro2
0
2
ro2
u rdr
(9.8)
0
The mean temperature in the duct is evaluated by integrating the total energy of the flow:
ro
mc p Tm
c p T u (2 r )dr
0
Substituting (9.8) for the mass flow rate, and assuming constant specific heat,
ro
T u rdr
Tm
0
ro
u rdr
0
This can be simplified by substituting the mean velocity, equation (9.8):
ro
Tm
9.4
2
u m ro2
T u r dr
(9.9)
0
Universal Velocity Profile
9.4.1 Results from Flat Plate Flow
It was shown that universal velocity profile in a pipe is very similar to that over a flat
plate (see Fig. 8.13), especially when the plate is exposed to a zero or favorable pressure
gradient
A pipe flow friction factor model was used to analyze flow over a flat plate using the
momentum integral method (see Section 8.4.3)
The characteristics of the flow near the wall of a pipe are not influenced greatly by the
curvature of the wall of the radius of the pipe, so, invoking the two-layer model that we
used to model flow over a flat plate:
Viscous sublayer
u
(8.54)
y
Law of the Wall
1
u
ln y
(8.58)
B
Continuous wall law models by Spalding and Reichardt were applied to pipe flow
and discussed in Section 8.4.2
For pipe flow, the wall coordinates are a little different than for flat plate flow, so the ycoordinate is
y
ro
r
(9.10)
This yields the y+ wall coordinate:
y
r0
(ro
r
r )u *
(9.11)
The velocity wall coordinate is the same as before:
u
u
(8.49)
u*
The friction velocity is the same
u*
o
/
(8.46)
The friction factor is based on the mean flow velocity instead of the free stream velocity:
Cf
o
(1 / 2) u m2
(9.12)
Therefore, the friction velocity can be expressed as:
u*
um C f / 2
9.4.2 Development in Cylindrical Coordinates
Since the velocity profile data for pipe flow matches that of flat plate flow, it allowed us
to develop expressions for universal velocity profiles solely from flat plate (Cartesian)
coordinates
Developing expressions for universal velocity profiles using cylindrical coordinates is
developed after revealing important issues and insights
Assuming fully-developed flow, the left side of the x-momentum equation (9.4) goes to
zero, leaving
1
r
r r
1 p
x
(9.13)
Rearranging and integrating yields an expression for shear stress anywhere in the flow
(r )
r p
2 x
(9.14)
C
The constant C is zero since the velocity gradient (and hence the shear stress) is expected
to go to zero at r = 0
Evaluating (9.14) at r and ro and taking the ratio of the two gives
(r)
r
ro
o
(9.15)
Equation (9.15) shows that the local shear is a linear function of radial location and
raises the following important issue:
Equation (9.15) is a linear shear profile depicted in Fig. 9.2
This expression contradicts how the fluid is expected to behave near the wall (recall
the Couette Flow assumption led us to the idea that τ is approximately constant in the
direction normal to the wall for the flat plate)
Yet, in Section 8.4 that the universal velocity profile that resulted from this
assumption works well for flat plate flow as well as pipe flow
This is reconciled as follows
The near-wall region over which we make the Couette flow assumption covers a
very small distance
Assume that, in that small region vary close to the wall of the pipe, the shear is
nearly constant:
0
The Couette assumption approximates the behavior near the pipe wall as
(
M
)
u
r
o
constant
(9.16)
Experimental data shows that near the wall, velocity profiles for flat plate and pipe flow
are essentially the same and suggests that the near-wall behavior is not influenced by the
outer flow, or even the curvature of the wall (see Fig. 8.13)
9.4.3 Velocity Profile for the Entire Pipe
From the previous development the velocity gradient (and the shear stress) is expected to
be zero at the centerline of the pipe; none of the universal velocity profiles developed
thus far behave this way
Reichardt attempted to account for the entire region of the pipe by suggesting the
following model for eddy viscosity:
y
6
M
r
1
ro
r
1 2
ro
2
(9.17)
This leads to the following expression for the velocity profile
1
u
where Reichardt used
ln y
0.40 and B
The slope of this equation is zero at r
of the pipe
1.5 1 r / ro
1 2(r / ro ) 2
B
(9.18)
5.5
0 , so the behavior matches that at the core
The viscous sublayer is not accounted for
As r
9.5
r0 , equation (9.18) reduces to the Law of the Wall form, eq. (8.56)
Friction Factor for Pipe Flow
9.5.1 Blasius Correlation for Smooth Pipe
Blasius developed a purely empirical correlation for flow through a smooth pipe using
dimensional analysis and experimental data:
0.0791 Re D1 / 4 ,
Cf
4000 Re D
where the friction factor is based on the mean flow velocity C f
105
o
(9.19)
/(1 / 2) u m2
Though less accurate and versatile than later correlations, this lead to the 1/7th Power
Law velocity profile
9.5.2 The 1/7th Power Law Velocity Profile
A crude but simple approximation for the velocity profile in a circular pipe was
discovered by Prandtl and von Kármán, leading from the Blasius correlation
Formulation
Recasting the Blasius correlation in terms of wall shear stress
o
1
2
u m2
0.0791
2ro u m
1/ 4
Then rearranging:
o
0.03326 u m7 / 4 ro 1 / 4
1/ 4
Assume a power law can be used to approximate the velocity profile:
(a)
u
u CL
q
y
ro
(b)
with uCL representing the centerline velocity
Assume that the mean velocity in the flow can be related to the centerline velocity as
u CL
(c)
(const)u m
Substituting (b) and (c) for the mean velocity in (a) yields
y
u
ro
o
(const)
o
(const) u 7 / 4 y (
1/ q 7 / 4
ro 1/ 4
1/ 4
This is simplified:
7 / 4 q ) ( 7 / 4 q 1 / 4)
ro
1/ 4
(d)
Prandtl and von Kármán argued that the wall shear stress is not a function of the size
of the pipe, so the exponent on r0 should be zero
Setting the exponent to zero, the value of q must be equal to 1/7, leading to the
classic 1/7th power law velocity profile:
u
uCL
y
ro
1/ 7
(9.20)
Experimental data shows that this profile adequately models velocity profile through
a large portion of the pipe and is frequently used in models for momentum and heat
transfer (recall Section 8.4.3)
Limitations
Accurate for a narrow range of Reynolds numbers: (104 to 106)
Yields an infinite velocity gradient at the wall
Does not yield a gradient of zero at the centerline
Nikuradse (a student of Prandtl’s) measured velocity profiles in smooth pipe over a wide
range of Reynolds numbers, and reported that the exponent varied with Reynolds
number:
u
u CL
y
ro
n
(9.21)
Table 9.1 lists Nikuradse’s measurements, including measurement of pipe friction factor
of the form
Cf
C
Re1D/ m
(9.22)
Nikuradse’s results show that the velocity profile becomes fuller as the mean velocity
increases
9.5.3 Prandtl’s Law for Smooth Pipe
A more theoretical model for friction factor is developed by employing the universal
velocity profile
Formulation
Beginning with the Law of the Wall, equation (8.58), substituting the wall
um C f / 2
coordinates u and y , as well as the friction velocity u *
o /
yields
u
um
2
Cf
1
ln
yu m
Cf
B
2
(9.23)
Assume that the equation holds at any value of y; evaluate the expression at the
centerline of the duct, y ro D / 2 , where u uCL , by substituting these and the
Reynolds number:
u CL
um
2
Cf
1
ln
Cf
Re D
2
B
2
(9.24)
By looking at the Law of the Wall, one can obtain a functional relationship for the
friction factor, though unfortunately a relationship for uCL / u m is unknown
To evaluate the mean velocity, u m , substitute the velocity profile (8.58) into an
expression for the mean velocity, equation (9.10):
ro
1
um
ro
2
u (2 r )dr
ro2
ro2
0
u (ro
y )dy
(9.25)
0
where y ro r
Performing the integration, the mean velocity becomes:
um
u
*
ln
ro u *
Cf
1
1
B
3
2
(9.26)
Making substitutions again
um
um
2
ln
Re D
2
Cf
2
B
3
2
(9.27)
u m cancels out of the equation; however uCL does not appear either, so the above
expression can be used directly to find an expression for C f by rearranging and
substituting
0.40 and B
5.0
1
Cf /2
2.44 ln Re D C f / 2
0.349
The expression is not yet complete, as it was assumed that the Law of the Wall is
accurate everywhere and ignores the presence of a viscous sublayer or wake region
The constants are adjusted to fit the experimental data, yielding
1
2.46 ln C f / 2 D C f / 2
Cf /2
0.29,
Re D
4000
(9.28)
This is Prandtl’s universal law of friction for smooth pipes; it is also known as the
Kármán-Nikuradse equation
Despite the empiricism of using a curve fit to obtain the constants in (9.28), using a
more theoretical basis to develop the function has given the result a wider range of
applicability than Blasius’s correlation
Equation (9.28) must be solved iteratively for C f
A simpler, empirical relation that closely matches Prandtl’s is
Cf
2
0.023Re D1 / 5 ,
3 104
Re D
106
(9.29)
This equation is also suitable for non-circular ducts with the Reynolds number
calculated using the hydraulic diameter
9.5.4 Effect of Surface Roughness
Roughness shifts the universal velocity profile downward (see Fig. 8.16 and Section
8.4.4)
The velocity profile in the logarithmic layer can be written as
1
u
ln y
B
B
where B is the shift in the curve, which increases with wall roughness k
The behavior also depends on the type of roughness, which ranges from uniform
geometries like rivets to random structures like sandblasted metal
Equivalent sand-based roughness model:
1
f 1/ 2
2.0 log10
Re D f 1 / 2
1 0.1(k / D) Re D f 1 / 2
0.8
(9.30)
where it is common to use the Darcy friction factor
f
(9.31)
4C f
If the relative roughness k / D is low enough, it doesn’t have much of an effect on
the equation
Scaling shows that roughness is not important if (k / D) Re D
10
If (k / D)Re D 1000 , the roughness term dominates in the denominator, and the
Reynolds number cancels
Friction is no longer dependent on the Re D
Colebrook and White developed the following formula for commercial pipes:
1
f
1/ 2
2.0 log10
k/D
3.7
2.51
Re D f 1 / 2
(9.32)
Representative roughness values presented in Table 9.2
This function appears in the classic Moody chart (Fig. 9.3)
9.6
Momentum-Heat Transfer Analogies
Analogy method is applied to pipe flow to the case of constant heat flux boundary
condition
Though an analogy cannot be made for the case of a constant surface temperature,
resulting models approximately hold for this case as well
Formulation
For hydrodynamically fully developed flow the x-momentum equation (9.4) becomes
1 dp
dx
1
r(
r r
M
1
r(
r r
H)
)
u
r
(9.33a)
The energy equation reduces to:
u
T
x
T
r
(9.33b)
Recall that an analogy is possible if the momentum and energy equations are
identical, so
Note that in pipe flow the pressure gradient is non-zero, although constant with
respect to x.
To ensure an analogy, the left side of (9.33b) must then be constant
For thermally fully developed flow and a constant heat flux at the wall, the shape
of the temperature profile is constant with respect to x, leading to
T
x
constant
Analogy appears to be possible; however, boundary conditions must also match
Boundary conditions are:
At r
At r
0:
du (0)
dr
0,
T (0)
r
u (ro )
0, T (ro ) Ts ( x)
(9.34a)
0
r0 :
du (ro )
dr
T (ro )
r
, k
o
(9.34b)
(9.34c)
qo
where qo is assumed to be into the flow (in the negative r-direction)
By normalizing the variables as:
U
u
,
um
T Ts
, X
Tm Ts
x
, and R
L
r
,
ro
it can be shown that both governing equations and boundary conditions are identical
in form
9.6.1 Reynolds Analogy for Pipe Flow
Assuming that
(Pr = 1) and H
M ( Prt = 1), the same assumptions used to
develop Reynold’s analogy for a flat plate, then the governing equations (9.33a) and
(9.33b) are identical
Following the same procedure as in the original derivation, the Reynolds analogy is
essentially identical for pipe flow:
St D
qo
u m c p (Ts
Cf
Tm )
2
, Pr
1
or
St D
Nu D
Re D Pr
Cf
2
(9.35)
Note that in this case the Stanton number is defined in terms of the mean velocity and
bulk temperature, as is the wall shear stress: o 12 C f u m2
9.6.2 Adapting Flat Plate Analogies to Pipe Flow
Other flat plate analogies can be adapted to pipe flow, with modifications
Example: von Kármán analogy
For pipe flow, conditions at the edge of the boundary layer are approximated by
conditions at the centerline of the pipe:
V
u CL and T
TCL
These substitutions affect the friction factor, which translates to:
Cf
o
1
2
2
u CL
Following the development exactly as before, the result is almost identical:
qo
uCL c p (Ts
Cf /2
TCL )
1 5
Cf
(9.36)
5Pr 1
6
( Pr 1) ln
2
The left side of (9.36) and the friction factor are expressed in terms of centerline
variables instead of the more common and convenient mean quantities u m and Tm
This is corrected:
qo
u m c p (Ts
u m (Ts
Tm ) u CL (Ts
Tm )
TCL )
C f / 2 u m / u CL 2
u
1 5 m
u CL
Cf
2
( Pr 1) ln
5Pr 1
6
where
C f is defined in terms of the mean velocity C f
o
2
/ 12 u m
The terms q o / u m c p (Ts Tm ) collectively are the Stanton number for
pipe flow
Simplifying yields von Kármán Analogy for pipe flow:
St D
T s Tm
Ts TCL
C f / 2 u m / u CL
1 5
um
u CL
Estimates for the ratios u m / uCL and Ts Tm / Ts
definition of mean temperature, equation (9.9):
Cf
2
( Pr 1) ln
(9.37)
5 Pr 1
6
TCL are found using the
th
u m and Tm are estimated using the 1/7 Law profiles, which for a circular pipe are:
y
ro
u
u CL
1/ 7
(9.20)
Similar to (8.111) for a flat plate:
T Ts
TCL Ts
y
ro
1/ 7
(9.38)
Substituting these models into (9.8) and (9.9), it can be shown that
um
u CL
0.817
Tm Ts
TCL Ts
(9.39)
0.833
(9.40)
9.6.3 Other Analogy-Based Correlations
A simple correlation for turbulent flow in a duct is based on the Colburn analogy:
Starting equation (8.96), and using equation (9.27) for the friction factor, one finds
St D
Nu D
0.023Re D1/ 5 Pr
2/3
0.023Re D4 / 5 Pr 1 / 3
(9.41)
The Dittus-Boelter correlation is an empirical correlation based on the Colburn analogy:
Nu D
where n = 0.4 for heating ( Ts
0.023Re D4 / 5 Pr n
(9.42)
Tm ) and n = 0.3 for cooling
Simplicity and the fact that this analogy compares well with experimental data make
it popular
In recent years its accuracy, along with that of the Colburn analogy, have been
challenged
Models by Petukhov and the Gnielinski correlation (see Section 9.8) are preferred for
their improved accuracy and range of applicability
Analogies remain a common way to model heat transfer in pipes; there are models
developed specifically for pipe flows
9.7
Algebraic Method Using Uniform Temperature Profile
As was done for the flat plate, the universal temperature and velocity profiles can be
used to estimate the heat transfer in a circular duct
Formulation
The Nusselt number for flow in a duct is defined as
hD
k
Nu D
qo D
(Ts Tm )k
(9.43)
To invoke the universal temperature profile, the definition of T , equation 8.102, is
used to define the mean temperature as
Tm
(Ts
c p u*
Tm )
qo
(Ts
Tm )
c pum C f / 2
qo
(9.44)
For duct flow, the friction velocity u * is defined in terms of the mean velocity, so
substituting the above into (9.43) for q o and invoking the definitions of the Reynolds
and Prandtl numbers:
Nu D
Re D Pr C f / 2
(9.45)
Tm
There are several ways to proceed from here:
One approach: evaluate Tm using a dimensionless version of (9.33):
ro
Tm
2
u m ro 2
T u (ro
y ) dy
(9.46)
0
Substituting appropriate universal temperature and velocity profiles into
(9.46) and integrating must be done numerically
A second approach yields a simpler closed-form solution
Rewrite the original Nusselt number relation (9.43) as follows:
Nu D
qo D
(Ts
(Ts Tm )k (Ts
TCL )
TCL )
where TCL is the centerline temperature
Substituting the definition of T
denominator:
Nu D
for the centerline temperature in the
Re D Pr C f / 2 (Ts TCL )
(Ts Tm )
TCL
(9.47)
The universal temperature profile, equation (8.118), is used to evaluate TCL :
TCL
Prt
ln yCL 13Pr 2 / 3
7
(9.48)
As was done for the flat plate, substitute the Law of the Wall velocity profile
(8.59) for ln y CL :
u CL
1
ln y CL
(9.49)
B
Substituting these into the Nusselt number relation:
Re D Pr C f / 2
Nu D
Prt (u CL
B) 13Pr
Expressions are needed for u CL and (Ts
2/3
TCL ) /(Ts
For the centerline velocity, use the definition of u
u CL
u CL
u*
u CL
um
(Ts TCL )
7 (Ts Tm )
(9.50)
Tm ) :
for pipe flow
2
Cf
(9.51)
Mean velocity and temperature are needed, so, the 1/7th power law is used to
avoid the complexity of the logarithmic velocity and temperature profiles
From the previous section, the 1/7th power law yields:
um
u CL
0.817 and
Tm Ts
TCL Ts
0.833
Using the definition of Stanton number, St D NuD /( Re D Pr) , and selecting
Prt = 0.9 and B = 5.0, equation (9.50) is rearranged:
St D
Cf /2
0.92 10.8 Pr 2/3
0.89 C f / 2
(9.52)
A first approximation suggests this model be limited to ReD < 1 x 105
The ultimate test is to compare this model to experimental data
9.8
Other Correlations for Smooth Pipes
Petukhov evoked Reichardt’s model for eddy diffusivity and velocity profile (9.15, 9.16)
to obtain:
St D
Cf /2
1.07 12.7 Pr 2/3 1 C f / 2
,
0.5
10
4
Pr
ReD
2000
5 106
(9.53)
This compares well to experimental data over a wide range of Prandtl and Reynolds
numbers
The following model was used for the friction factor:
Cf
2
(2.236 ln Re D
4.639)
2
(9.54)
Note the similarity between Petukhov’s relation (9.53) and the algebraic result, equation
(9.52)
In 1976, Gnielinski modified Petukhov’s model slightly, extending the model to include
lower Reynolds numbers:
Nu D
( Re D 1000) PrC f / 2
1 12.7 Pr 2/3 1 C f / 2
,
0.5
3
3 10
Pr
ReD
2000
5 106
(9.55)
Pethkhov’s friction model can be used in (9.55) for the friction factor
For the above models, properties should be evaluated at the film temperature
These correlations are reasonable for channels with constant surface temperature as well
as constant heat flux; the flows are relatively insensitive to boundary conditions
9.9
Heat Transfer in Rough Pipes
The effects of roughness on the heat transfer from flat plates was discussed in Section
8.5.6, and much of the same physical intuition applies to flow in channels
Norris presents the following empirical correlation for flow through circular tubes:
Nu
Nu smooth
Cf
C f ,smooth
n
,
Cf
C f ,smooth
4
(9.56)
where n 0.68Pr 0.215
A correlation like Colebrook’s (9.30) could be used to determine the rough-pipe friction
factor
The behavior of this relation reflects what is expected physically:
The Prandtl number influences the effect of roughness
For very low Prandtl fluids the roughness plays little role in heat transfer
The influence of roughness size is limited
The effect of increasing roughness vanishes beyond (C f / C f ,smooth) 4 , hence a
maximum is reached
Though roughness enhances heat transfer, it increases friction, which increases pumping
costs, though the increase of friction due to roughness also reaches a limiting value
The application of roughness to increase heat transfer requires benefits to be weighed
against increasing costs
CHAPTER 10
CORRELATION EQUATIONS:
FORCED AND FREE CONVECTION
10.1 Introduction
A key factor in convection is the heat the heat transfer coefficient h.
Instead of determining h we determine the Nusselt number Nu, which a
dimensionless heat transfer coefficient.
Whenever it is difficult or not possible to determine the Nusselt number analytically,
we search for a correlation equation which gives the Nusselt number.
Correlation equations are usually based on experimental data.
This chapter gives correlation equations for:
(1) External forced convection over plates, cylinders, and spheres.
(2) Internal forced convection through channels.
(3) External free convection over plates, cylinders and spheres.
(4) free convection in enclosures.
10.2 Experimental Determination of Heat Transfer Coefficient h
Newton's law of cooling is used to determine h
experimentally:
h
qs
Ts T
(10.1)
Measure surface temperature Ts , surface heat
flux q s , and free stream temperature T , and use
(10.1) to determine h.
Example: Determining q s .
In Fig. 10.1 the
cylinder is heated electrically, current and voltage
give power (heat) dissipated, heat flux q s is power
divided by surface area
qs
ΔV
Ts
T
i
V
Fig. 10.1
Experimental data is correlated in terms of dimensionless variables and parameters.
Example: Forced convection for constant properties and no dissipation, local Nusselt
number is correlated as
Nu x = f ( x* ; Re, Pr )
10.3 Limitations and Accuracy of Correlation Equations
(2.52)
2
All correlation equations have LIMITATIONS. They MUST be carefully noted.
Examples of limitations:
Geometry: An equation for each configuration.
Range of parameters, such as the Reynolds, Prandtl and Grashof numbers, for which
a correlation equation is valid, are determined by the availability of data and/or the
extent to which an equation correlates the data.
10.4 Procedure for Selecting and Applying Correlation Equations
There are many, many correlation equations. Each is for a specific application and
is valid under specified conditions. Presenting all correlation equations and
discussing their applications and limitations is not the most effective and efficient
approach to studying and learning the material. Instead, we will describe a
systematic procedure for searching, identifying and selecting correlation equations.
Some of the common applications will be presented as examples. Selecting
correlation equations for applications not discussed in this chapter follow the same
procedure described below.
(1) Identify the geometry under consideration. Is it flow over a flat plate, over a cylinder,
through a tube, or through a channel?
(2) Identify the classification of the heat transfer process. Is it forced convection, free
convection, external flow, internal flow, entrance region, fully developed region,
boiling, condensation, micro-gravity?
(3) Determine if the objective is finding the local heat transfer coefficient (local Nusselt
number) or average heat transfer coefficient (average Nusselt number).
(4) Check the Reynolds number in forced convection. Is the flow laminar, turbulent or
mixed?
(5) Identify surface boundary condition. Is it uniform temperature or uniform flux?
(6) Examine the limitations on the correlation equation to be used. Does your problem
satisfy the stated conditions?
(7) Establish the temperature at which properties are to be determined. For external flow
properties are usually determined at the film temperature T f
Tf
(Ts
T )/2
(10.2)
and for internal flow at the mean temperature Tm . However, there are exceptions that
should be noted.
(8) Use a consistent set of units in carrying out computations.
(9) Compare calculated values of h with those listed in Table 1.1. Large deviations from
the range of h in Table 1.1 may mean that an error has been made.
3
10.5 External Forced Convection Correlations
10.5.1 Uniform Flow over a Flat Plate: Transition to Turbulent Flow
Boundary layer flow over a semi-infinite flat plate.
Mixed flow: laminar and turbulent. Transition or critical Reynolds number Re xt :
V xt
Rext
5 105
(10.3)
V ,T
xt
(1) Plate at Uniform Surface Temperature.
The local heat transfer coefficient is
determined from the local Nusselt
number.
In the turbulent region, x
Valid for:
Fig.10.2
8.2
Fig.
xt
hx
k
Nu x
x
turbulent
transition
laminar
0.0296 ( Rex ) 4 / 5 ( Pr )1 / 3
(10.4a)
flat plate, constant Ts
5 105 Rex 107
0.6 Pr 60
properties at Tf
(10.4b)
Average heat transfer coefficient for mixed flow: consider both laminar and
turbulent:
xt
L
h
1
L
1
L
h( x)dx
0
L
hL ( x)dx
h t ( x)dx
(10.5)
xt
0
Use (4.72) for hL (x) and (10.4a) for ht (x ) , integrate
h
k
0.664 ( Rext )1 / 2
L
0.037 ( Re L ) 4 / 5
( Rext )
4/5
( Pr )1 / 3
(10.7a)
Limitations: see limitations on the respective correlations for the Nusselt numbers.
Expressed in terms of the average Nusselt number Nu L :
hL
k
Nu L
0.664 ( Re xt )1 / 2
0.037 ( Re L ) 4 / 5
( Re xt ) 4 / 5
( Pr )1 / 3
(10.7b)
(2) Plate at Uniform Surface Temperature with an Insulated Leading Section.
Turbulent flow
Nu x
hx
k
0.0296( Re x )
1 ( xo /x )
V
4/5
1/3
( Pr )
9 / 10 1 / 9
(10.8)
T
t
insulation
xt
xo
Ts
Fig. 10.3
x
4
Limitations: see (10.4b) and review limitations on (4.72)
(3) Plate with Uniform Surface Flux.
Nu x
hx
k
0.030( Re x ) 4/5 Pr 1 / 3
(10.9)
Properties T f (Ts T ) / 2, where Ts is the
average surface temperature.
10.5.2 External Flow Normal to a Cylinder
T
V
Average Nusselt number
Fig. 10.5
N uD
hD
k
0.3
0.62Re1D/ 2 Pr 1 / 3
1
0.4
Pr
2 / 3 1/ 4
1
ReD
282,000
5/8 4/5
(10.10a)
Valid for:
flo
w
n
o
rm
a
ltoc
y
lin
d
e
r
PeRe
Pr0
.2
D
p
ro
p
e
rtie
sa
tT
f
(10.10b)
For Pe < 0.2, use
NuD
hD
k
1
0.8237 0.5 ln Pe
(10.11a)
Valid for:
flo
w
n
o
rm
a
ltoc
y
lin
d
e
r
PeRe
Pr0
.2
D
p
ro
p
e
rtie
sa
tT
f
(10.11b)
The above gives examples of correlation equations and their limitations.
Cor r elation equations for other configur ations will be listed without details.
Limitations and conditions on their use should be noted.
10.5.3 External Flow over a Sphere
10.6 Internal Forced Convection Correlations
5
10.6.1 Entrance Region: Laminar Flow through Tubes at Uniform Surface
Temperature
(1) Fully Developed Velocity, Developing Temperature: Laminar Flow.
(2) Developing Velocity and Temperature: Laminar Flow. this case is given by [5, 7]
10.6.2 Fully Developed Velocity and Temperature in Tubes: Turbulent Flow
(1) The Colburn Equation
(2) The Gnielinski Equation
10.6.3 Non-circular Channels: Turbulent Flow
10.7 Free Convection Correlations
10.7.1 External Free Convection Correlations
(1) Vertical Plate: Laminar Flow, Uniform Surface Temperature.
(2) Vertical Plates: Laminar and Turbulent, Uniform Surface Temperature.
(3) Vertical Plates: Laminar Flow, Uniform Surface Heat Flux.
(4) Inclined Plates: Laminar Flow, Uniform Surface Temperature.
(i) Heated upper surface or cooled lower surface
(ii) Heated lower surface or cooled upper surface
(6) Vertical Cylinders.
(7) Horizontal Cylinders.
(8) Spheres.
10.7.2 Free Convection in Enclosures
(1) Vertical Rectangular Enclosures.
(2) Horizontal Rectangular Enclosures.
(3) Inclined Rectangular Enclosures.
10.8 Other Correlations
Keep in mind that this chapter presents correlation equations for very limited
processes and configurations.
There are many other correlation equations for topics such as:
Condensation
Boiling
High speed flow
Jet impingement
Dissipation
Liquid metals
Heat transfer enhancements
Finned geometries
Irregular geometries
Micro-gravity
Non-Newtonian fluids
Etc.
Consult textbooks, handbooks and journals.
CHAPTER 11
CONVECTION IN MICROCHANNELS
11.1 Introduction
11.1.1 Continuum and Thermodynamic Hypothesis.
Previous chapters are based on two fundamental assumptions:
(1) Continuum: Navier-Stokes equations, and the energy equation are applicable
(2) Thermodynamic equilibrium: No-velocity slip and no-temperature jump at
boundaries.
Validity criterion: The Knudsen number:
Kn
(1.2)
De
is the mean free path.
Continuum: valid for:
Kn
0.1
(1.3a)
0.001
(1.3b)
No-slip, no-temperature jump:
Kn
11.1.2 Surface Forces
Surface area to volume ratio increases as channel size is decreases.
Surface forces become more important as channel size is reduced.
11.1.2 Chapter Scope
Classification
Gases vs. liquids
Rarefaction
Compressibility
Velocity slip and temperature jump
Analytic solutions: Couette and Poiseuille
flows
Table 11.1
gas
11.2 Basic Considerations
11.2.1 Mean Free Path
Ideal gas:
(11.2)
RT
p 2
Properties if various gases; Table 11.1
7
R
Air
Helium
J/kg K kg/m3
287.0
2077.1
Hydrogen 4124.3
Nitrogen
296.8
Oxygen
259.8
1.1614
0.1625
0.08078
1.1233
1.2840
10
kg/s m
m
184.6
199.0
89.6
178.2
207.2
0.067
0.1943
0.1233
0.06577
0.07155
2
11.2.2 Why Microchannels?
The heat transfer coefficient increases as channel size is decreased.
Examine fully developed flow through tubes and note the effect of diameter
h
3.657
k
D
(11.3)
11.2.3 Classification. Based on Knudsen number
Kn
0.001
continuum, no slip flow
(11.4)
0.001 Kn 0.1 continuum, slip flow
0.1 Kn 10
transition flow
10
Kn
free molecular flow
11.2.4 Macro and Microchannels
Macrochannels
Continuum and thermodynamic equilibrium model applies.
No-velocity slip and no-temperature jump.
Microchannels
Failure of macrochannel theory and correlation.
Distinguishing factors: two and three dimensional effects, axial conduction,
dissipation, temperature dependent properties, velocity slip and temperature
jump at the boundaries and the increasing dominant role of surface forces.
11.2.5 Gases vs. Liquids
Mean free paths of liquids are much smaller than those of gases.
Onset of failure of thermodynamic equilibrium and continuum is not well defined for
liquids.
Surface forces for liquids become more important.
Liquids are almost incompressible while gases are compressible.
11.3 General Features
Rarefaction: Knudsen number effect.
Compressibility: large channel pressure drop, changes in density (compressibility).
Dissipation: Increased viscous effects.
11.3.1 Flow Rate
No-velocity slip: Fig. 11.3a
Velocity slip: Fig. 11.3b
Flow rate Q : Macrochannel theory
underestimates flow rate:
(a)
(b)
Fig. 11.3
3
Qe
Qt
1
(11.5)
w
(4.37a)
11.3.2 Friction Factor
Friction coefficient C f
Cf
(1 / 2) u m2
Friction factor f
f
1D
2 L
p
(11.6)
u m2
Fully developed laminar flow in macrochannels:
f Re
(11.7)
Po
Po is known as the Poiseuille number
Macrochannel theory does not predict Po. The following ratio is a measure of
prediction error
Po
Po
e
C*
(11.8)
t
Po appears to depend on the Reynolds number.
Both increase and a decrease in C are reported.
11.3.3 Transition to Turbulent flow
Macrochannels
Ret
uD
2300
(6.1)
Microchannels: reported transition Reynolds numbers ranged from 300 to 16,000
11.3.4 Nusselt number
Macrochannels:
Fully developed laminar flow: constant Nusselt number, independent of
Reynolds number.
Microchannels: Macrochannel theory does not predict Nu. The following ratio is a
measure of reported departure from macrochannel prediction
0.21
( Nu ) e
( Nu ) t
100
(11.9)
11.4 Governing Equations
In the slip-flow domain, 0.001 Kn 0.1 , the continuity, Navier Stokes equations,
and energy equation are valid.
Important effects: Compressibility, axial conduction, and dissipation.
4
11.4.1 Compressibility
Compressibility affects pressure drop, Poiseuille number and Nusselt number.
11.4.2 Axial Conduction
Axial conduction is neglected in macrochannels for Peclet numbers greater than 100.
Microchannels typically operate at low Peclet numbers. Axial conduction may be
important.
Axial conduction increases the Nusselt number in the velocity-slip domain.
11.4.3. Dissipation
Dissipation becomes important when the Mach number is close to unity or larger.
11.5 Velocity Slip and Temperature Jump Boundary Conditions
In microchannels fluid velocity is not the same as surface velocity. The velocity slip
condition is
2
u ( x,0)
u
u ( x,0) u s
(11.10)
n
u
u(x,0) = fluid axial velocity at surface
u s surface axial velocity
x = axial coordinate
n = normal coordinate measured from the surface
u = tangential momentum accommodating coefficient
Gas temperature at a surface differs from surface temperature:
T ( x,0) Ts
2
T
T
2
1
T ( x,0)
Pr
n
(11.11)
T(x,0) = fluid temperature at the boundary
Ts = surface temperature
c p / cv , specific heat ratio
T
= energy accommodating coefficient
u
and
T
are assume equal to unity.
(11.10) and (11.11) are valid for gases.
11.6 Analytic Solutions: Slip Flows
Consider Couette and Poiseuille flows.
Applications: MEMS.
Thermal boundary conditions: Uniform surface temperature and uniform surface heat flux.
Examine the effects of rarefaction and compressibility.
5
11.6.1 Assumptions
(1) Steady state
(2) Laminar flow
(3) Two-dimensional
(4) Slip flow regime (0.001 < Kn < 0.1)
(5) Ideal gas
(6) Constant viscosity, conductivity, and specific heats
(7) Negligible lateral variation of density and pressure
(8) Negligible dissipation (unless otherwise stated)
(9) Negligible gravity
(10) The accommodation coefficients are assumed equal to unity,
u
T
1 .0 .
11.6.2 Couette Flow with Viscous Dissipation: Parallel Plates with Surface Convection
Stationary lower plate, moving upper plate.
Insulated lower plate, convection at the
upper plate.
y
ho T
us
Determine:
(1) Velocity distribution
(2) Mass flow rate
(3) Nusselt number
x
u
H
Fig. 11.6
Flow Field
x-component of the Navier-Stokes equations for compressible, constant viscosity
flow (2.9), simplifies to
d 2u
(11.12)
0
dy 2
Boundary conditions: apply (11.10)
du( x,0)
dy
u( x,0)
u ( x, H )
du( x, H )
dy
us
(g)
(h)
Solution
u
us
1
(y
1 2 Kn H
Kn)
(11.14)
Mass Flow Rate. The flow rate, m, for a channel of width W is
H
m W
u dy
0
(11.15)
6
(11.14) into (11.15)
m
us
2
WH
(11.16)
Macrochannels flow m o
mo
WH
us
2
(11.17)
Thus
m
mo
(11.18)
1
Nusselt Number. Defined as
2 Hh
k
Nu
(l)
Heat transfer coefficient h:
T (H )
y
Tm Ts
k
h
Substitute into (l)
Nu
T (H )
y
2H
Tm Ts
(11.19)
k thermal conductivity of fluid
T fluid temperature function (variable)
Tm fluid mean temperature
Ts
plate temperature
NOTE:
Fluid temperature at the surface, T ( x, H ), is not equal to surface temperature Ts .
Surface temperature is unknown in this example
Relation between T ( x, H ) and T s is given by the temperature jump condition:
Ts
2
1
T ( x, H )
T ( x, H )
Pr
y
(11.20)
uT dy
(11.22)
Mean temperature Tm
Tm
H
2
us H
0
Temperature distribution: Energy equation simplifies to
2
k
T
y2
0
(11.23)
7
Dissipation function
:
2
u
y
(11.24)
(11.24) into (11.23)
d 2T
du
k dy
dy 2
2
(11.25)
Boundary conditions
dT (0)
dy
k
dT ( H )
dy
(m)
0
ho (Ts T )
Use (11.20)
k
dT ( H )
dy
ho T ( x, H )
2
1
T ( x, H )
T
Pr
n
(n)
Solution: Use (11.14) for u, substitute into (11.25), solve and use boundary
conditions (m) and (n)
kH
H2
2 Kn 2
T
y2
H
T
(11.26)
2
ho
2
1 Pr
us
k H (1 2 Kn )
2
Nusselt number: Use (11.26) to formulate Ts ,
(11.19) Nu
(p)
dT ( H )
and Tm , substitute into
dy
8(1 2 Kn )
8
8 (1 2 Kn ) Kn
1
Kn
3
1
Pr
(11.27)
NOTE:
The Nusselt number is independent of Biot number.
The Nusselt number is independent of the Reynolds number. This is also the case
with macrochannel flows.
The Nusselt number depends on the fluid (Pr and ).
Nusselt number for macrochannel flow, Nu o : set Kn
Nu o
Thus
8
0 in (11.27)
(11.28)
8
Nu
Nu o
1 2 Kn
8 (1 2 Kn ) Kn
1
Pr
8
1
Kn
3
(11.29)
11.6.3 Fully Developed Poiseuille Channel Flow: Uniform Surface Flux
Inlet and outlet pressures are p i and p o
Surface heat flux: q s
Determine:
qs
y
H/2
(1) Velocity distribution
(2) Pressure distribution
(3) Mass flow rate
(4) Nusselt number
H/2
x
qs
Poiseuille flow in microchannels differs
from macrochannels as follows:
Fig. 11.7
Streamlines are not parallel.
Lateral velocity component v does not vanish.
Axial velocity changes with axial distance.
Axial pressure gradient is not linear.
Compressibility and rarefaction are important.
Assumptions. See Section 11.6.1. Additional assumptions:
(11) Isothermal flow.
(12) Negligible inertia forces.
2
(13) The dominant viscous force is
u
y2
.
Flow Field. Determine the axial velocity distribution.
Axial component of the Navier-Stokes equations
p
x
2
u
0
y2
(c)
Boundary conditions
u ( x,0)
y
0
(e)
u ( x, H / 2)
y
(f)
H 2 dp
y2
1 4 Kn ( p ) 4 2
8 dx
H
(11.30)
u ( x, H / 2)
Solution
u
Must determine pressure distribution and lateral velocity v. Continuity for compressible
flow:
9
u
x
v
y
(h)
0
Use ideal gas law in (h)
pv
y
(i)
pu
x
(11.30) into (i)
H2
dp
y2
p (1 4 Kn ( p ) 4 2 )
8
x dx
H
( p v)
y
(j)
Boundary conditions on v
v(x,0)
(k)
0
v( x, H / 2)
(l)
0
Multiply (j) by dy, integrate
y
H2
dp
p
8
x dx
d ( p v)
0
y
(1
4 Kn ( p) 4
0
y2
H2
)
dy
(m)
0
(n)
Evaluate the integrals, solve for v, and use (l)
x
p
4 y3
3 H3
dp
y
1 4 Kn( p)
dx
H
y H /2
Introduce Knudsen number
Kn
Evaluate (n) at y
H
H
2
1
p
RT
(11.33)
H / 2, substitute (11.33) into (n) and integrate
1 2
p
6
H
2 RT p
Cx
(o)
D
where C and D are constants of integration. The solution to this quadratic equation is
2
p( x)
3
2 RT
H
18 RT
H2
6Cx
(p)
6D
Boundary conditions on p
p (0)
pi , p ( L )
po
(q)
Use (q) to find C and D, substitute into (p) and use the definition of Knudsen number
p ( x)
po
6 Kn o
6 Kn o
pi
po
2
(1
pi2
2
po
) 12 Kn o (1
Mass Flow Rate. The flow rate m for a channel of width W is
pi x
)
po L
(11.35)
10
H/ 2
m
2W
(s)
u dy
0
Use (11.30), (11.35) and the ideal gas law
WH 3
p 6
12 RT
H
m
2
RT
dp
dx
(11.38)
Using (11.35) to formulate the pressure gradient, substituting into (11.38), assuming
constant temperature ( T To ), and rearranging, gives
m
1 W H 3 po2
24 LRTo
pi2
1 12 Kn o (
2
po
pi
po
1)
(11.39)
For macrochannel
mo
1 W H 3 po2
12 LRTo
pi
po
1
(11.40)
1 12 Kno
(11.41)
Taking ratio
m
mo
1
2
pi
po
NOTE:
m in microchannels is very sensitive to channel height H.
(11.39) shows the effect of rarefaction and compressibility.
Nusselt Number. Follow Section 11.6.2
Nu
2H qs
k (Ts Tm )
(v)
T s is given by (11.11)
Ts
T ( x, H / 2)
2
1
T ( x, H / 2)
Pr
y
(11.42)
Tm is given by
H /2
uT dy
Tm
0
H /2
udy
0
Temperature distribution. Solve the energy equation.
Additional assumptions:
(11.43)
11
(14) Axial velocity distribution is approximated by the solution to the isothermal case.
(15) Negligible dissipation,
0
2
T / y2
(16) Negligible axial conduction, 2T / x 2
(17) Negligible effect of compressibility on the energy equation, u / x
(18) Nearly parallel flow, v 0
Energy equation: (2.15) simplifies to
T
c pu
x
2
k
T
v/ y
0
(11.44)
y2
Boundary conditions
T ( x,0)
y
(w)
0
and
k
T ( x, H / 2)
y
(x)
qs
Assume:
(19) Fully developed temperature. Define
T ( x, H / 2) T ( x, y )
T ( x, H / 2) Tm ( x)
(11.45)
(y)
(11.46)
0
(11.47)
Fully developed temperature:
Thus
x
Equations (11.45) and (11.46) give
dT ( x, H / 2)
dx
T
x
( y)
dT ( x, H / 2)
dx
dTm ( x)
dx
0
(11.48)
The heat transfer coefficient h, is given by
T ( x, H / 2)
y
Tm ( x) Ts ( x)
k
h
(y)
Use (11.42) and (11.45) into (y)
h
k[T ( x, H / 2) Tm ( x)] d ( H / 2)
Ts ( x) Tm ( x)
dy
Newton’s law of cooling:
h
Equating the above with (11.49)
qs
Ts ( x) Tm ( x)
(11.49)
12
qs
d ( H / 2)
dy
T ( x, H / 2) Tm ( x)
(11.50)
constant
Combining this with (11.48), gives
dTm ( x)
dx
dT ( x, H / 2)
dx
T
x
(11.51)
Conservation of energy for the element in Fig. 11.8 gives
2q sWdx mc pTm
mc p Tm
qs
dTm
dx
dx
Simplify and eliminate m
dTm
dx
m
2q s
= constant
c pum H
Tm
Tm
(11.52)
dx
dTm
dx
dx
qs
(11.52) into (11.51)
dTm ( x)
dx
dT ( x, H / 2)
dx
T
x
Fig. 11.8
2q s
c pum H
(11.53) into (11.44)
2
2q s u
kH u m
T
y2
(11.54)
where u m is given by
um
2
H
H /2
(cc)
udy
0
(11.30) into (cc)
um
H 2 dp
1 6 Kn
12 dx
(11.55)
(11.30) and (11.55)
u
um
6
1
1 6 Kn 4
Kn
y2
(11.56)
H2
(11.56) into (11.54)
2
T
y2
qs 1
12
1 6 Kn kH 4
Kn
y2
(11.57)
H2
Integrating twice and use (w)
T ( x, y )
12q s
1 1
(
(1 6 Kn )kH 2 4
Kn) y 2
y4
12H 2
g ( x)
(11.58)
13
To determine g(x), find Tm using two methods.
First method: Integrate (11.52)
Tm
x
2q s
c pum H
dTm
Tmi
Tm (0)
dx
0
Tmi
(11.59)
2q s
x Tmi
c pum H
(11.60)
Evaluating the integrals
Tm ( x)
Second method: use definition in (11.43). (11.30) and (11.58) into (11.43)
Tm ( x)
3q s H
k (1 6 Kn)
2
13
Kn
40
( Kn) 2
13
560
(11.61)
g ( x)
(11.60) and (11.61) give g(x)
g ( x)
Tmi
2q s
x
c pum H
3q s H
k (1 6 Kn )
Surface temperature Ts ( x, H / 2) :
3q s H
1
Ts ( x )
Kn
k (1 6 Kn ) 2
2
5
48
( Kn ) 2
2
13
Kn
40
qs H
Kn
1 kPr
13
560
(11.62)
(11.63)
g ( x)
Nusselt number: (11.61) and (11.63) into (v)
Nu
2
3
1
Kn
(1 6 Kn) 2
5
48
1
( Kn) 2
(1 6 Kn)
(11.64)
13
Kn
40
(i) The Nusselt number is an implicit function of
x since Kn is a function p which is a function of
x.
(iii) The effect of temperature jump on the
Nusselt number is represented by the last term
in the denominator of (11.64).
(iv) The Nusselt for no-slip, Nu o , is determined
by setting Kn 0 in (11.64)
2
1
Kn
1 Pr
9
NOTE:
(ii) Unlike macrochannels, the Nusselt number
depends on the fluid, as indicated by Pr and
in (11.64).
13
560
8
7
Nu
6
5
4
0.04
0
0.08
0.12
Kn
Fig. 11.9 Nusselt number for air flow between
parallel plates at unifrorm surface
heat flux for air, = 1.4, Pr = 0.7,
u
T 1
14
Nu o
140
17
(11.65)
8.235
(v) Rarefaction and compressibility have the effect of decreasing the Nusselt number.
11.6.4 Fully Developed Poiseuille Channel Flow: Uniform Surface Temperature
Assumptions: same as the uniform flux
case.
The velocity, pressure, and mass flow
rate, are the same as for uniform flux.
Surface boundary condition is different.
Must determine temperature distribution
Ts
y
H/2
x
H/2
Fig. 11.10
Temperature Distribution and Nusselt
Number. Newton’s law of cooling the
Nusselt number for this case is given by
Nu
2 Hh
k
2H
T ( x, H / 2)
Tm ( x) Ts
y
Ts
(11.66a)
Energy equation: Include axial conduction
c pu
2
T
x
k(
T
2
2
y2
x
T
)
(11.67a)
Boundary conditions:
T ( x,0)
y
2
T ( x, H / 2) Ts
(11.68a)
0
H
T ( x, H / 2)
Kn
1 Pr
y
T (0, y )
(11.69a)
Ti
(11.70a)
T ( , y ) Ts
(11.71a)
Axial velocity is given by (11.56)
u
um
6
1
1 6 Kn 4
Kn
y2
(11.56)
H2
Dimensionless variables
T Ts
,
Ti Ts
y
, Re
H
x
,
H RePr
2 um H
, Pe
RePr
(11.72)
Use (11.56) and (11.72), into (11.66a)-(11.71a)
Nu
2
m
( , / 2)
(11.66)
15
6
(1
1 6 Kn 4
Kn
2
( Pe) 2
( ,0)
( ,1 / 2)
2
2
1
)
2
2
(11.67)
2
(11.68)
0
1
Kn
1 Pr
( ,1 / 2)
(11.69)
(0, ) 1
(11.70)
( , ) 0
(11.71)
Solution: method of separation of variables
8
Pe = 0
1
5 8
Result: Fig. 11.11.
NOTE:
7
Nu
The Nusselt number decreases as the Knudsen
number is increased.
Axial conduction
number.
increases
the
6
Nusselt
5
No-slip (Kn = 0) and negligible axial
conduction (Pe
):
Nu o
7.5407
0
0.08
0.04
0.12
Kn
Fig. 11.11 Nusselt number for flow between
parallel plates at uniform surface
temperature for air, Pr = 0.7,
,
1 .4 , u
T 1 [14]
(11.73)
11.6.5 Fully Developed Poiseuille Flow in Microtubes: Uniform Surface Flux
This problem is identical to
Poiseuille flow between
parallel plates at uniform flux
presented in Section 11.6.3.
qs
r
z
Determine the following:
(1) Velocity distribution
(2) Nusselt number
qs
Fig. 11.12
Assumptions. See Section 11.6.3.
Flow Field. Following the analysis of Section 11.6.3. Use cylindrical coordinates.
Results:
r
ro
16
ro2 dp
1 4 Kn
4 dz
vz
vz
v zm
p( z )
po
8Kn o
2
ro4 po2
16 LRTo
m
2
(11.77)
pi2
(1
pi z
)
po L
) 16 Kn o (1
2
po
pi2
pi
1
16
Kn
(
1)
o
po
po2
ro4 p o2
8 LRT
mo
(11.74)
ro2
1 4Kn (r / ro ) 2
1 8Kn
pi
po
8Kn o
r2
( pi
po
(11.78)
(11.79a)
1)
(11.79b)
Nusselt Number. Define
2ro h
k
Nu
Nu
(d)
2 ro q s
k (Ts Tm )
(e)
Results
T (r , z )
Tm
g ( z)
Tmi
q s ro
2
14
Kn
3
16Kn 2
1 r4
4 ro2
7
24
k (1 8Kn)
2q s
q s ro
z
16Kn 2
2
c p ro v z m
k (1 8Kn )
Ts (ro , z )
Nu
qs
(1 4 Kn ) r 2
(1 8 Kn ) k ro
4q s ro
3
Kn
k (1 8Kn )
16
4
(11.92)
g ( z)
(11.95)
g ( z)
14
Kn
3
7
24
(11.96)
q s ro
Kn g ( z )
1 kPr
(11.97)
2
4
( Kn 3 )
(1 8Kn )
16
1
(1 8Kn ) 2
16Kn
2
14
Kn
3
7
24
Nusselt number variation with Knudsen number for air, with
is plotted in Fig. 11.14.
No-slip Nusselt number, Nu o , is obtained by setting Kn
Nu o
48
11
4.364
4
1
Kn
1 Pr
(11.98)
1.4 and Pr
0.7,
0 in (11.98)
(11.99)
18
11.6.6 Fully Developed Poiseuille Flow in Microtubes: Uniform Surface Temperature
The uniform surface flux of Section 11.6.5
is repeated with the tube maintained at
uniform surface temperature Ts .
r
r
z
Temperature Distribution and Nusselt Number
Fig. 11.15
Ts
Same flow field as the uniform surface
flux case of Section 11.6.5
Follow the analysis of Section 11.6.4. and use the flow field of Section 11.6.5.
Dimensionless variables
T Ts
,
Ti Ts
z
, R
2ro RePr
2 um ro
r
, Re
ro
, Pe
RePr
(11.106)
Nusselt number, energy equation, and boundary conditions in dimensionless form
Nu
2
m
1 4 Kn R 2
2(2 16Kn)
(0. )
R
(1, )
2
0
Kn
1 Pr
(1, )
R
1
(R )
R R
R
(11.100)
(2 Pe) 2
(11.101)
2
4.5
(11.102)
(1, )
R
2
1
Pe = 0
1
5
4.0
(11.103)
3.5
(R,0) 1
(11.104)
(R, ) 0
(11.105)
Nu
3.0
Solution: By separation of variables.
2.5
Results: Fig. 11.16.
2.0
0
0.12
0.08
Kn
Fig. 11.16 Nusselt number for flow through tubes
at uniform surface temperature for air,
Pr = 0.7,
1.4, u T 1 , [14]
0.04
ro