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
Ts
Transfer
•
• Surface temperature is too high.
How to reduce it?
(1) Use a fan
(2) Change the fluid
(3) Increase surface area
q′s′
V∞
T∞
+
Fig. 1.1
−
1
• Conclusion: Three important factors in convection
(1) fluid motion
(2) fluid nature
(3)
surface
geometryof the role of fluid motion in convection:
• Examples
• Fanning to feel cool
• Stirring a mixture of ice and water
• Blowing on the surface of coffee in a cup
• Orienting a car radiator to face air flow
1.3 Focal Point in Convection Heat Transfer
Determination of temperature distribution in a
moving fluid
T = T ( x, y, z, t )
(1.1)
2
1.3 Fourier’s Law of Conduction
A (T − T )
q∝
L
A (Tsi − Tso )
qx = k
L
si
Tso
Tsi
so
A
x
qx
dx
(1.2)
k = thermal conductivity
0
L
x
Fig. 1.2
• Valid for:
(1) steady state
(2) constant k
(3) one-dimensional conduction
3
• Reformulate to relax restrictions. Consider element dx
T ( x ) − T ( x + dx )
T ( x + dx ) − T ( x )
q =k A
=−k A
dx
dx
dT
(1.3)
qx = − k A
dx
q′x′ = Heat flux
qx
(1.4)
q′x′ =
A
dT
(1.5)
q ′x′ = − k
dx
x
Generalize (1.5):
∂T
q ′x′ = − k
,
∂x
∂T
q ′y′ = − k
,
∂y
∂T
q ′z′ = − k
∂z
(1.6)
4
(1) Why negative sign?
(2) k ≠ constant
(3) Find T(x,y,z,t), use (1.6) to obtain q′′
(4) Changing fluid motion changes T(x,y,z,t)
1.5 Newton's Law of Cooling
q′s′ ∝ (Ts − T∞ )
q′s′ = surface flux
Ts = surface temperature
q′s′ = h(Ts − T∞ )
(1.7)
5
• Eq. (1.7) is Newton's law of cooling
• h is called the heat transfer
coefficient
h = f (geometry, motion, properties, ∆T )
(1.8)
1.6 The Heat Transfer Coefficient h
• Is h a property?
• Does h depend on temperature
distribution?
Apply Fourier’s law
∂ T ( x ,0 , z )
q′s′ = − k
∂y
(1.9)
Combine (1.7) and (1.9)
∂T ( x ,0, z )
∂y
h = −k
(Ts − T∞ )
(1.10)
Temperature distribution in needed to determine h
6
Ts•
• Apply Newton’s law to the bulb:
q ′s′
Ts = T∞ +
h
(1.11)
• Increase V∞ to increase h and lower Ts
qs′′
V∞
T∞
Fig. 1.1
+
Table 1.1 Typical values of h
h ( W/m 2 − o C )
Process
Free convection
Gases
5-30
Liquids
20-1000
Forced convection
Gases
20-300
Liquids
50-20,000
Liquid metals
5,000-50,000
Phase change
2,000-100,000
Boiling
5,000-100,000
Condensation
−
7
1.7 Differential Formulation of Basic Laws
• Three basic laws:
conservation of mass,
momentum, and energy
y
v
u
• Formulation
w
• Differential
• Integral
• Finite difference
• Key assumption: continuum
x
z
1.8 Mathematical Background
Fig. 1.4
r
(a) Velocity Vector V
r
V = ui + v j + wk
(1.12)
8
(b) Velocity Derivative
r
∂V ∂u ∂v
∂w
i+
j+
k
=
∂x ∂x
∂x
∂x
(1.13)
(c) The Operator ∇
Cartesian:
∂
∂
∂
∇≡
i+
j+ k
∂x
∂x
∂x
(1.14)
Cylindrical:
∂
1 ∂
∂
∇ ≡ ir +
iθ + i z
∂r
r ∂θ
∂z
(1.15)
Spherical: ∇ ≡
∂
1 ∂
1
∂
ir +
iθ +
iφ
∂r
r ∂θ
r sin θ ∂φ
(1.16)
9
(d) Divergence of a Vector
r
r ∂u ∂v ∂w
div .V ≡ ∇ ⋅ V = + +
∂x ∂y ∂z
(1.17)
(e) Derivative of the Divergence
r
∂
∂ ∂u ∂v ∂w
(∇ ⋅ V ) = + +
∂x
∂x ∂x ∂y ∂z
(1.18)
or
or
r
∂
∂
(∇ ⋅V ) = ∇ ⋅ (u i + v j + w k )
∂x
∂x
r
r
∂
∂V
(∇ ⋅ V ) = ∇ ⋅
∂x
∂x
(1.19)
10
(f) Gradient of Scalar
∂T
∂T
∂T
Grad T = ∇T =
i+
j+
k
∂x
∂y
∂z
(1.22)
(g) Total Differential and Total Derivative
f = flow field dependent variable such as u, v, p, etc.
Cartesian Coordinates:
f = f ( x, y, z, t )
(a)
Total differential of f:
∂f
∂f
∂f
∂f
df = dx + dy + dz + dt
∂x
∂y
∂z
∂t
or
df Df ∂f dx ∂f dy ∂f dz ∂f
+
+
+
≡
=
dt Dt ∂x dt ∂y dt ∂z dt ∂t
(b)
11
But
dx
dy
dz
= u,
= v,
=w
dt
dt
dt
(c)
Substitute (c) into (b)
df Df
∂f
∂f
∂ f ∂f
+w +
=
=u
+v
dt Dt
∂x
∂ z ∂t
∂y
Total derivative:
df Df
=
dt Dt
Convective derivative:
∂f
∂f
∂f
u
+w
+v
∂x
∂y
∂z
Local derivative:
∂f
∂t
(1.21)
(d)
(e)
12
Apply to velocity component u. Set f = u
du Du
∂u
∂u
∂u ∂u
=
+w +
=u
+v
dt Dt
∂x
∂z ∂t
∂y
(1.22)
(1.22) represents
∂u
∂u
∂u
= convective acceleration in the x-direction
u +v +w
∂x
∂y
∂z
(f)
∂u
= local acceleration
∂t
(g)
Cylindrical coordinates : r ,θ , z
dv r Dv r
∂v r v θ ∂v θ v θ2
∂v r ∂v r
= vr
+vz
=
+
−
+
dt
Dt
∂r
r ∂θ
r
∂z
∂t
(1.23a)
13
dv θ Dv θ
=
dt
Dt
∂v θ v θ ∂v θ v r v θ
∂v θ ∂v θ
= vr
+vz
+
+
+
∂r
r ∂θ
r
∂z
∂t
dv z Dv z
∂v z ∂v z
∂v z v θ ∂v z
= vr
+vz
=
+
+
dt
Dt
∂r
r ∂θ
∂z
∂t
(1.23b)
(1.23c)
• Total derivative of temperature:
set f = T in (1.21)
dT DT
∂T
∂ T ∂T
∂T
=u
+v
+w
+
=
dt
Dt
∂x
∂y
∂z ∂t
(1.24)
14
1.9 Problem Solving Format
Solve problems in stages:
(1)
(2)
(3)
(4)
Observations
Problem Definition
Solution Plan
Plan Execution
(i) Assumptions
(ii) Analysis
(iii) Computations
(iv) Checking
(5) Comments
15
1.10 Units
SI units
Length (L): meter (m)
Time (t): second (s)
Mass (m): kilogram (kg)
Temperature (T): kelvin (K)
• Celsius and kelvin scales
T(oC) = T(K) - 273.15
(1.25)
Derived units:
• Force: newton (N)
One newton = force to accelerate one kilogram
one meter per sec per sec:
Force = mass × acceleration
N = kg . m /s2
16
• Energy: joules (J)
One joule = energy due to a force of one newton
moving a distance of one meter
J = N. m = kg . m2 /s2
• Power: watts (W)
One watt = one joule per second
W = J/s = N. m/s = kg . m2 /s3
17
Example 1.1: Heat Loss from Identical Triangles
• Surface temperature: Ts
• Variable h:
C
h( x ) =
x
q1
• Determine:
q2
(1)
•
•
•
Observations
Newton’s law givesq
h varies with x
Integration is required
(2) Problem Definition.
Determine dq for dx of each triangle
18
(3) Solution Plan.
Apply Newton's law to element and integrate
(4) Plan Execution
(i) Assumptions
• steady state
•
one• dimensional
uniform T∞
• uniform Ts
• negligible radiation
(ii) Analysis
Apply Newton’s law
dq = h( x )(Ts − T∞ )dA
(a)
C
h=
x
(b)
19
Triangle 1:
dA1 = y1 ( x )dx
(c)
Triangle 2:
dA2 = y 2 ( x )dx
(d)
Geometry:
H
y1 ( x ) = ( L − x )
L
H
y2 ( x) =
x
L
(e)
(f)
(e) into (c), (f) into (d):
H
dA1 = ( L − x )dx
L
(g)
H
dA2 = xdx
L
(h)
(b) and (g) into (a), integrate
20
H L− x
q1 = ∫ dq1 = ∫ C (Ts − T∞ )
dx
1/ 2
0
L x
L
q1 = ( 4 / 3)C (T − T∞ ) HL1 / 2
s
Similarly
(i)
H x
q2 = ∫ dq2 = ∫ C (T − T∞ )
dx
1
/
2
0
L x
L
s
q 2 = ( 2 / 3)C (T − T∞ ) HL1 / 2
s
Ratio of (i) and (j)
q1
=2
q2
(j)
(k)
(iii) Checking
Dimensional check:
(b) gives units of C:
C = W/m 3/2 − o C
21
(i) gives units of q1
q1 = C(W/m3/2-oC)( Ts - T∞ )(oC)H(m)L1/2 (m1/2) = W
Qualitative check:
q1 > q2 because base of 1 is at x = 0 where h = ∞ .
(5) Comments
• Orientation is important.
Same area triangles but different q
• Use same approach for other geometries
22
CHAPTER 2
DIFFERENTIAL FORMULATION
OF THE BASIC LAWS
2.1 Introduction
• Solutions must satisfy 3 fundamental laws:
conservation of mass
conservation of momentum
conservation of energy
• Differential formulation: application of basic laws
to differential element
1
2.2 Flow Generation
(i) Forced convection: by mechanical means(fan, blower,
nozzle, jet, etc.)
(ii) Free (natural) convection: due to gravity and density
change
2.3 Laminar vs. Turbulent Flow
u
turbulent
(a)
u
laminar
t
Fig. 2.1
Laminar: No random fluctuations
Turbulent: Random fluctuations
(b)
t
2
Transition from laminar to turbulent:
Transition Reynolds number, depends on
• flow geometry
• surface roughness
• pressure gradient
• etc.
Flow over flat plate: ≈ 500,000
Flow through tubes: ≈ 2300
2.4 Conservation of Mass:
The Continuity Equation
3
2.4.1 Cartesian Coordinates
m& y +
y
dy
dx
x
(a)
∂y
dy
∂m& x
m& x +
dx
∂x
m& x
dy
∂m& y
dx
m& y
(b)
Fig. 2.2
Rate of mass added to element Rate of mass remove from element =
Rate of mass change within element
(2.1)
4
Assume continuum, use Fig. 2.2b, and (2.1)
∂m& x
m& x + m& y + m& z − m& x +
dx −
x
∂
∂m& y
∂m& z ∂m
m& y + ∂y dy + m& z + ∂z dz = ∂ t
(a)
Express (a) in terms of density and velocity
m& = ρVA
(b)
m& x = ρ udydz
(c)
m& y = ρ v dxdz
(d)
m& z = ρ wdxdy
(e)
Apply (b) to element
5
Mass m of element
m = ρ dxdydz
(f)
(c)–(f) into (a)
∂ρ ∂
∂
∂
+ (ρ u) + (ρ v ) + (ρ w ) = 0
∂ t ∂x
∂y
∂z
(2.2a)
• (2.2a) is the continuity equation
Alternate forms:
∂ρ
∂ρ
∂ρ
∂ρ
∂u ∂v ∂ w
+u
+ +v
+w
+ρ +
+
=0
∂t
∂x
∂y
∂z
∂x ∂x ∂ x
(2.2b)
r
Dρ
+ ρ ∇ ⋅V = 0
Dt
(2.2c)
or
6
or
r
∂ρ
+∇ ⋅ ρV = 0
∂t
Special case: constant density (incompressible fluid)
(2.2d)
Dρ
=0
Dt
(2.2c) becomes
r
∇ ⋅V = 0
(2.3)
2.4.2 Cylindrical Coordinates
z
(r,z,θ )
•
r
y
x
θ
Fig. 2.3
7
∂ρ 1 ∂
1 ∂
∂
(ρ rv r ) +
(ρ v θ ) + (ρ v z ) = 0
+
∂ t r ∂r
r ∂θ
∂z
(2.4)
2.4.3 Spherical Coordinates
z
(r,φ, θ )
φ
θ
•
r
x
y
Fig. 2.4
(
)
∂ρ 1 ∂
1
1 ∂
∂
(
ρ r 2v r +
ρ vθ sinθ ) +
ρ v φ = 0 (2.5)
+
2
∂ t r ∂r
r sinθ ∂θ
r sinθ ∂φ
(
)
8
Example 2.1: Fluid in Angular Motion
•
•
•
•
Shaft rotates inside tube
Incompressible fluid
No axial motion
Give the continuity equation
Solution
r
ω
θ
shaft
(1) Observations
• Cylindrical coordinates
•
No variation in axial and angular
directions
Incompressible
(2)•Problem
Definition.fluid
Simplify the 3-D continuity
(3) Solution Plan.
Apply the continuity equation in cylindrical coordinates
9
(4) Plan Execution
(i) Assumptions
• Incompressible
• No axial motion
• Shaft and tube are concentric (axisymmetric, no
angular variation)
(ii) Analysis. Start with (2.4):
∂ρ 1 ∂
1 ∂
∂
(ρ rv r ) +
(ρ v θ ) + (ρ v z ) = 0
+
∂ t r ∂r
r ∂θ
∂z
(2.4)
Simplify
∂ρ
ρ
=0
Incompressible fluid: is constant,
∂t
No axial velocity: v z = 0
∂
Axisymmetric:
=0
∂θ
10
(2.4), gives
∂
(rv r ) = 0
∂r
(a)
Integrate
rv r = C
(b)
C = constant of integration
Boundary condition:
v r ( r o ,θ ) = 0
Use (b)
C =0
(c)
(b) gives
vr = 0
(d)
(iii) Checking
Dimensional check: Each term in (2.4) has units of
density per unit time.
(5) Comments
11
2.5 Conservation of Momentum:
The Navier-Stokes Equation of Motion
2.5.1 Cartesian Coordinates
• Momentum is a vector
quantity
y
dz
dy
• Newton’s law of motion:
3 components
dx
• Apply Newton’s law to
element, Fig. 2.5
x
z
r
r
∑ δ F = (δ m )a
Fig. 2.5
(a)
12
r
a = acceleration of the element
r
δ F = external force on element
δ m = mass of the element
∑ δ Fx = (δ m )a x
(b)
δ m = ρ dxdydz
(c)
du Du
∂u
∂u
∂u ∂u
ax =
=
+w +
=u +v
∂x
∂y
∂z ∂t
dt Dt
(d)
x-direction:
Mass δ m
Total acceleration a x
(c) and (d) into (b)
Du
dxdydz
∑ δ Fx = ρ
Dt
(e)
13
External x-forces:
(i) Body force (gravity)
(ii) Surface force
Total forces
∑ δ Fx = ∑ δ Fx )body + ∑ δ Fx )surface
(f)
∑ δFx )body = ρg x dxdydz
(g)
Gravity force:
Surface forces:
σ xx = normal stress on surface dydz
14
τ yx = shearing (tangential) stress on surface dxdz
τ zx = shearing (tangential) stress on surface dxdy
Summing up x-forces, Fig. 2.6
∂σ xx ∂τ yx ∂τ zx
dxdydz
+
+
∑ δFx )surface =
∂y
∂z
∂x
Substituting (f), (g) and (h) into (e)
x-direction:
∂σ xx ∂τ yx ∂τ zx
Du
ρ
= ρ gx +
+
+
Dt
∂x
∂y
∂z
Similarly, for y and z-directions
(h)
(2.6a)
y-direction:
∂τ xy ∂σ yy ∂τ zy
Dv
=ρ gy +
+
+
ρ
∂x
Dt
∂z
∂y
(2.6b)
15
z-direction:
∂τ xz ∂τ yz ∂σ zz
Dw
= ρ gz +
+
+
ρ
Dt
∂z
∂x
∂y
(2.6c)
Unknowns in (2.6), 13:
u, v, w, ρ , σ xx , σ yy , σ zz ,τ xy , τ yx ,τ xz , τ zy , τ zx ,τ yz
However
τ xy = τ yx , τ xz = τ zx , τ yz = τ zy
(i)
Reduce number of unknowns:
Use empirical relations called the constitutive equations
τ xy = τ yx
∂v ∂u
= µ +
∂x ∂ y
(2.7a)
τ xz = τ zx
∂w ∂u
= µ
+
∂x ∂z
(2.7b)
16
∂v ∂w
τ yz = τ zy = µ +
∂z ∂y
r
∂u 2
σ xx = − p + 2 µ − µ∇ ⋅ V
∂x 3
r
∂v 2
σ yy = − p + 2 µ − µ∇ ⋅ V
∂y 3
r
∂w 2
σ zz = − p + 2 µ
− µ∇ ⋅ V
∂z 3
• Fluids obeying (2.7) are Newtonian fluids
Substitute (2.7) into (2.6 )
r
Du
∂p ∂ ∂u 2
ρ
= ρg x − + µ 2 − ∇ ⋅ V +
Dt
∂x ∂ x ∂x 3
∂ ∂u ∂v ∂ ∂w ∂u
µ + + µ +
∂y ∂ y ∂x ∂ z ∂x ∂z
(2.7c)
(2.7d)
(2.7e)
(2.7f)
(2.8x)
17
r
Dv
∂p ∂ ∂v 2
ρ
= ρg y − + µ 2 − ∇ ⋅ V +
Dt
∂y ∂y ∂y 3
(2.8y)
∂ ∂ v ∂ w ∂ ∂u ∂ v
µ + + µ +
∂z ∂z ∂y ∂x ∂y ∂x
r
Dw
∂p ∂ ∂w 2
ρ
− ∇ ⋅V +
= ρg z − + µ 2
Dt
∂z ∂z ∂z 3
∂ ∂w ∂u ∂ ∂v ∂w
µ + + µ +
∂x ∂x ∂z ∂y ∂z ∂y
(2.8z)
NOTE:
• Eqs. (2.8) are the Navier-Stokes equations of motion
• Unknowns are 6: u, v, w, p, ρ , µ
• Restrictions: continuum and Newtonian fluid
18
Vector form of (2.8x), (2.8y) and (2.8z)
r
r
r
r 2
DV
r
r 4
= ρg − ∇p + ∇ (µ∇ ⋅ V ) + ∇ (V ⋅ ∇µ ) − V∇ µ +
ρ
3
Dt
r
r
r
∇µ × (∇ × V ) − (∇ ⋅ V )∇µ − ∇ × (∇ × µV )
(2.8)
Simplified cases:
(i) Constant viscosity
∇µ = 0
(j)
r
r
r
r
2r
∇ × (∇ × µV ) = ∇ (∇ ⋅ µV ) − ∇ ⋅ ∇µV = µ∇ (∇ ⋅ V ) − µ∇ V
(k)
and
(j) and (k) into (2.8)
r
r
DV
r
r 1
2r
ρ
= ρg − ∇p + µ∇ (∇ ⋅ V ) + µ∇ V
Dt
3
(2.9)
19
Eq. (2.9) is valid for: (1) continuum, (2) Newtonian (3)
constant viscosity.
(ii) Constant viscosity and density
Continuity equation (2.3)
(2.3) into (2.9)
r
∇ ⋅V = 0
r
DV
r
r
2r
ρ
= ρg − ∇ p + µ ∇ V
Dt
(2.3)
(2.10)
Eq. (2.10) is valid for: (1) continuum, (2) Newtonian (3)
constant viscosity (4) constant density
The 3-components of (2.10):
20
x-direction:
∂u
∂u
∂u
∂u
ρ + u + v + w =
∂y
∂z
∂x
∂t
∂ 2u ∂ 2u ∂ 2u
∂p
ρg x − + µ 2 + 2 + 2
∂x
∂y
∂z
∂x
(2.10x)
∂ 2w ∂ 2w ∂ 2w
∂p
ρg z − + µ 2 + 2 + 2
∂z
∂y
∂z
∂x
(2.10y)
∂ 2v ∂ 2v ∂ 2v
∂p
ρg y − + µ 2 + 2 + 2
∂y
∂y
∂z
∂x
(2.10z)
z-direction:
∂w
∂w
∂w
∂w
+u
+v
+w
ρ
=
∂x
∂y
∂z
∂t
y-direction:
∂v
∂v
∂v
∂v
ρ + u + v + w =
∂x
∂y
∂z
∂t
21
2.5.2 Cylindrical Coordinates
Assumptions: Continuum, (2) Newtonian fluid, (3) constant
viscosity and (4) constant density.
r-direction:
∂v r v θ ∂v r v θ 2
∂v r ∂v r
−
+vz
+
=
+
ρ v r
∂r
r ∂θ
r
∂z
∂t
2
2
∂ 1 ∂
∂p
1 ∂ v r 2 ∂v θ ∂ v r
( rv r ) + 2
− 2
+
ρg r − + µ
2
2
∂r
∂
r
r
∂
r
∂
θ
r ∂θ
r
∂z
(2.11r)
θ -direction:
∂v θ ∂v θ
∂v θ v θ ∂v θ v r v θ
+
+vz
ρ v r
−
+
=
r ∂θ
r
∂r
∂z
∂t
2
2
∂ 1 ∂
∂
v
∂
vθ
v
∂
1 ∂p
1
2
r
θ
ρgθ −
+ µ
( rv θ ) + 2
+ 2
+
2
2
r ∂θ
∂
r
r
∂
r
∂
θ
r ∂θ
r
∂z
(2.11 θ )22
z-direction:
∂ v z ∂v z
∂v z v θ ∂ v z
+ vz
+
ρ v r
+
=
∂t
∂r
r ∂θ
∂z
1 ∂ ∂v z 1 ∂ 2 v z ∂ 2 v z
∂p
ρg z − + µ
+
r
+ 2
2
2
∂z
∂z
r ∂r ∂r r ∂θ
(2.11z)
2.5.3 Spherical Coordinates
Assumptions: Continuum, (2) Newtonian fluid, (3) constant
viscosity and (4) constant density.
r-direction:
2
2
∂v
v
v
v
+
v
∂
v
∂
v
∂
v
φ
θ
φ
r −
+ r=
ρ v r r + θ r +
∂r
r ∂θ r sinθ ∂φ
r
∂t
2
∂p
2
2 ∂v θ 2v θ cotθ
2 ∂v φ
−
− 2
ρgr − + µ ∇ v r − 2 v r − 2
2
∂r
r
r ∂θ
r
r sinθ ∂φ
(2.12r)
23
θ -direction:
2
∂v
v
v
v θ ∂v θ
v r vθ
φ ∂v θ
φ cot θ ∂v θ
θ
+
−
+
=
+
−
ρ vr
∂r
∂t
r ∂θ r sin θ ∂φ
r
r
2
vθ
1 ∂p
2 ∂v r
2 cosθ ∂v φ
+ µ ∇ v θ +
−
−
ρgθ −
2
2
2
2
2
r ∂θ
r ∂θ r sin θ r sin θ ∂φ
(2.11 θ )
φ -direction:
∂v φ v θ
ρ v r
+
∂r
r
1
ρgφ −
r sinθ
∂v φ
∂θ
+
vφ
∂v φ
r sinθ ∂φ
+
vφ v r
r
+
vθ v φ
r
∂v θ
cotθ +
∂t
=
∂p
+
∂φ
vφ
2
2
2 cosθ ∂v θ
∂v r
µ ∇ v φ − 2 2 + 2 2
+ 2 2
r sin θ r sin θ ∂φ r sin θ ∂φ
(2.11φ )
24
Where ∇ 2 is
2
1
∂
∂
1
1
∂
∂
∂
∇2 = 2 r 2 + 2
sin θ
+ 2
∂θ r sin 2 θ ∂φ 2
r ∂r ∂r r sin θ ∂θ
(2.13)
Example 2.2: Thin Liquid Film Flow Over an
Inclined Surface
• Incompressible
y
• Parallel streamlines.
u
• Write the Navier-Stokes
equations
0
g
x
(1) Observations
θ
• Flow is due to gravity
• Parallel streamlines: v = 0
• Surface pressure is uniform (atmospheric)
• Cartesian geometry
25
(2) Problem Definition.
Simplify the x and y components of the Navier-Stokes
equations
(3) Solution Plan.
Start with the Navier-Stokes equations in Cartesian
coordinates and simplify for this case
(4) Plan Execution
(i) Assumptions
•
•
•
•
•
•
Newtonian
steady state
flow is in the x-direction only
constant properties
uniform ambient pressure
parallel streamlines
26
(ii) Analysis
Start with (2.10x ) and (2.10y)
∂u
∂u
∂u
∂u
ρ + u + v + w =
∂x
∂y
∂z
∂t
∂ 2u ∂ 2u ∂ 2u
∂p
ρ g x − + µ 2 + 2 + 2
∂x
∂y
∂z
∂x
(2.10x)
∂v
∂v
∂v
∂v
+u +v
+w =
∂y
∂z
∂x
∂t
ρ
∂ 2v ∂ 2v ∂ 2v
∂p
ρg y − + µ 2 + 2 + 2
∂y
∂y
∂z
∂x
(2.10y)
Gravitational acceleration:
g x = g sinθ
,
g y = − g cosθ
(a)
27
Simplifications:
∂u ∂ v
=
=0
∂t ∂t
Axial flow (x-direction only):
Steady state:
∂
w= =0
∂z
Parallel streamlines:
v =0
(b)
(c)
(d)
(a)-(d) into (2.10x) and (2.10y)
and
∂ 2u ∂ 2u
∂u
∂p
ρ u = ρ g sinθ −
+ µ
+
2
2
∂x
∂x
∂
x
y
∂
(e)
∂p
0 = − ρg cosθ −
∂y
(f)
(f) is thy y-momentum equation
28
Simplify (e) using continuity (2.3)
r ∂u ∂ v ∂ w
∇ ⋅V =
+
+
=0
∂x ∂y ∂ z
(c) and (d) into (g)
∂u
=0
∂x
(h) into (e)
2
(g)
(h)
∂p
∂ u
ρ g sinθ −
+ µ 2= 0
∂x
∂y
(i)
p = −( ρ g cosθ ) y + f ( x )
(j)
Integrate (f)
f(x) = “constant” of integration
At free surface, y = H , pressure is uniform equal to p∞ .
Set y = H in (j)
f ( x ) = p∞ + ρ gH cosθ
(k)
29
(k) into (j)
Different (k)
(m) into (i)
p = ρ g( H − y) cosθ + p∞
(l)
∂p
=0
∂x
(m)
ρ g sinθ + µ
d 2u
dy
2
=0
(n)
This is the x-component
(iii) Checking
Dimensional check: Each term in (f) and (n) must have
same units:
ρ g cos θ = (kg/m3)(m/s2) = kg/m2-s2
∂ p N/m2 N kg − m/s2
2
2
=
=
kg/m
s
= 3=
∂y
m
m
m3
30
ρg sin θ = kg/m2-s2
d 2u
m/s
2-s2
µ
=
(kg/m-s)
=
kg/m
d y2
m2
Limiting check: For zero gravity fluid remains stationary.
Set g = 0 in (n) gives
d 2u
dy
2
=0
(o)
Solution to (o): u = 0, ∴ fluid is stationary
(5) Comments
• Significant simplifications for: For 2-D incompressible,
parallel flow
• The flow is 1-D since u depends on y only
31
2.6 Conservation of Energy: The Energy Equation
2.6.1 Cartesian Coordinates
y
dz
dy
dx
x
z
Fig. 2.5
Energy can not be created or destroyed
Apply to element dxdydz
32
A
Rate of change of
internal and kinetic
energy of element
B
=
C
Net rate of heat
addition by conduction
Net rate of internal and kinetic
energy transport by convection
_
+
D
Net rate of work done by
element on surroundin gs
(2.14)
• Express each term in (2.14) in terms of temperature
(Appendix A)
• Explain physical significance of each term
• Result is called the energy equation
• Assumptions
33
• Continuum
• Newtonian
• Negligible nuclear, electromagnetic and radiation energy
(1) A = Rate of change of internal and kinetic energy of
element
• Internal energy of element depends on temperature
(thermodynamic)
• Kinetic energy of element depends on velocity (flow
field)
∂
A=
ρ ( uˆ + V 2 / 2) dxdydz
∂t
[
]
(A-1)
34
(2) B = Net rate of internal and kinetic energy by convection
• Internal energy convected through sides with mass flow.
Depends on temperature
• Kinetic energy convected through sides of element with
mass flow. Depends on velocity
B=−
{ [(uˆ + V 2 / 2) ρ V ] }dxdydz
∇
•
(A-2)
(3) C = Net rate of heat addition by conduction
• Conduction at each surface depends on temperature
gradient
• Apply Fourier’s law (1.6)
C = −(∇ q′′ ) dxdydz
•
(A-3)
35
(4) D = Net rate of work done by the element on the
surroundings
Rate of work = force × velocity
36
• 18 surface forces (Fig. 2.6)
• 3 body forces (gravity)
• Total 21 forces at 21 velocities
r r
∂
D = − ρ (V ⋅ g )dxdydz − ( uσ xx + v τ xy + wτ xz ) +
∂x
∂
∂
( uτ yx + v σ yy + wτ yz ) + ( uτ zx + v τ zy + wσ zz ) dxdydz
∂y
∂z
(A-7)
Substitute (A-1), (A-2), (A-3) and (A-7) into (2.14)
∂
1 2
1 2 r
ρ uˆ + V = −∇ ⋅ uˆ + V ρV
2
∂t
2
r r ∂
− ∇ ⋅ q′′ + ρ (V ⋅ g ) + ( uσ xx + v τ xy + wτ xz ) +
∂x
∂
∂
( uτ yx + v σ yy + wτ yz ) + ( uτ zx + v τ zy + wσ zz )
∂y
∂z
(A-8)
37
Simplify using:
• Fourier’s law (1.6)
• Continuity equation (2.2)
• Momentum equations (2.6)
• Constitutive equations (2.7)
• Thermodynamic relations for û and ĥ
DT
Dp
+ µΦ
ρ cp
= ∇ ⋅ k∇T + β T
Dt
Dt
(2.15)
where
β = coefficient of thermal expansion (compressibility)
• β is a property
1 ∂ρ
(2.16)
β =−
ρ ∂T p
Φ = dissipation function (energy due to friction)
38
∂u 2 ∂v 2
∂w 2
Φ = 2 + +
+
∂y
∂x
∂z
2
∂u ∂v 2 ∂v ∂ w 2
∂
∂
w
u
+ + +
+ −
+
∂z ∂y
∂y ∂x
∂x ∂z
2
2 ∂ u ∂ v ∂w
+ +
3 ∂x ∂y ∂z
(2.17)
• Φ is Important in high speed flow and for very viscous
fluids
2.6.2 Simplified Form of the Energy Equation
(a) Cartesian Coordinates
• Use (2.15)
• Assumptions leading to (2.15):
39
• Continuum
• Newtonian
• Negligible nuclear, electromagnetic and radiation
energy transfer
• Special cases
(i) Incompressible fluid
β =0
and
c p = cv = c
DT
= ∇ ⋅ k∇T + µ Φ
ρc p
Dt
(ii) Incompressible constant conductivity fluid
(2.18) is simplified further constant k:
(2.18)
40
or
DT
= k∇ 2T + µ Φ
ρc p
Dt
(2.19a)
∂ 2T ∂ 2T ∂ 2T
∂T
∂T
∂T
∂T
+ µΦ
+u
+v
+w
+
+
ρ cΡ
= k
∂x 2 ∂y 2 ∂z 2
∂x
∂y
∂z
∂t
(2.19b)
(iii) Ideal gas
p
ρ=
RT
(2.20)
(2.20) into (2.16)
1 ∂ρ
1 p
1
=
β =− =
2
ρ ∂T p ρ RT
T
(2.21) into (2.15)
DT
Dp
= ∇ ⋅ k∇T +
+ µΦ
ρ cp
Dt
Dt
Using continuity (2.2c) and (2.20)
r
DT
= ∇ ⋅ k∇ T − p∇ ⋅ V + µ Φ
ρ cv
Dt
(2.21)
(2.22)
(2.23)
41
(b) Cylindrical Coordinates
Assume:
• Continuum
• Newtonian fluid
• Negligible nuclear, electromagnetic and radiation energy
transfer
• Incompressible fluid
• Constant conductivity
∂ T v θ ∂T
∂T
∂T
+ vr
+
ρcP
+vz
=
∂r
r ∂θ
∂z
∂t
1 ∂ ∂T 1 ∂ 2T ∂ 2T
k
+ 2 + µΦ
r
+ 2
2
∂z
r ∂r ∂r r ∂θ
(2.24)
where
42
2
2
2
∂v z
1 ∂v θ v r
∂v r
Φ = 2
+ + 2
+
+ 2
r
∂r
r ∂θ
∂z
2
2
1 ∂v z ∂v 0
∂v r ∂ v z
∂v θ v θ 1 ∂v r
−
+
+
+
+
+
r r ∂0
∂z
∂r
∂r
r ∂0
∂z
2
(2.25)
(c) Spherical Coordinates
Assume:
• Continuum
• Newtonian fluid
• Negligible nuclear, electromagnetic and radiation energy
transfer
• Incompressible fluid
• Constant conductivity
43
v 0 ∂ T k ∂ 2 ∂T
∂T
∂T v o/ ∂T
=
ρ c p
+
+
+ vr
r
+
2
r ∂o/ r sin φ ∂0 r ∂r ∂r
∂r
∂t
2
∂
T
∂
T
1
1
k
+ µΦ
sin o/
+ 2
2
2
2
∂o/ r sin o/ ∂0
r sin o/ ∂o/
(2.26)
where
2
2
2 ∂v
v
cot
φ
1 φ vr
1 ∂v 0 v r
φ
∂v
+
+
Φ = 2 r +
+
+
+
∂r
r ∂φ
r
r sin φ ∂θ
r
r
∂ vφ
r
∂r r
2
1 ∂v r
+
r ∂φ +
2
sin φ ∂ vθ
1 ∂v θ
1 ∂v r
∂ vθ
+
r
+
+
r ∂φ r sin φ
r sin φ ∂φ
r
sin
φ
∂
θ
r
∂
r
2
(2.27)
44
Example 2.3: Flow Between Parallel Plates
• Axial flow with dissipation
• Assume:
• Newtonian
• Steady state
• Constant density
• Constant conductivity
• Parallel streamlines
• Write the energy equation
(1) Observations
• Parallel streamlines: v = 0
• Incompressible, constant k
• Include dissipation
• Cartesian geometry
45
(2) Problem Definition
Determine the energy equation for parallel flow
(3) Solution Plan
Start with the energy equation for constant ρ and k in
Cartesian coordinates and simplify
(4) Plan Execution
(i) Assumptions
• Newtonian
• Steady state
• Axial flow
• Constant ρ and k
• Negligible nuclear, electromagnetic and radiation
energy transfer
• Parallel streamlines.
46
(ii) Analysis. Start with energy equation (2.19b)
∂ 2T ∂ 2T ∂ 2T
∂T
∂T
∂T
∂T
+ µΦ
+u
+v
+
+
+w
ρ cΡ
= k
2 ∂y 2 ∂z 2
∂x
∂y
∂z
∂t
∂x
(2.19b)
where
∂ u 2 ∂v 2
∂w 2
Φ = 2 + +
+
∂y
∂x
∂z
∂u ∂v 2 ∂v ∂w 2
∂ w ∂u 2
+ + +
+ −
+
∂z ∂ y
∂y ∂x
∂x ∂z
2
2 ∂ u ∂v ∂ w
+ +
3 ∂x ∂y ∂z
However
Steady state:
∂T
=0
∂t
(2.17)
(a)
47
Axial flow:
Parallel flow:
∂
w= =0
∂z
v =0
(b)
(c)
(a)-(c) into (2.19b) and (2.17)
∂ 2T ∂ 2T
∂T
ρ cΡ u
= k 2 + 2 + µΦ
∂x
∂y
∂x
2
2
2 ∂u 2
∂u
∂u
Φ = 2 + −
3 ∂x
∂x
∂y
Further simplification: use continuity (2.3)
r ∂u ∂v ∂ w
∇ ⋅V =
+
+
=0
∂ x ∂ y ∂z
(b) and (c) into (f) gives
∂u
=0
∂x
(d)
(e)
(f)
(g)
48
2
∂u
(g) into (e)
Φ =
∂y
(h) into (d) gives the energy equation
(h)
2
2
2
∂ T ∂ T
∂T
∂u
(i)
= k 2 + 2 + µ
ρ cΡ u
∂x
∂y
∂y
∂x
(iii) Checking
Dimensional check: Each term in (i) has the same
units of W/m 3
Limiting check: For no fluid motion, energy equation reduces
to pure conduction. Set u = 0 in (i)
∂ 2T ∂ 2T
+ 2 =0
2
∂y
∂x
49
(5) Comments
• In energy equation (i), properties c p , k , ρ and µ
represent fluid nature
• Velocity u represents fluid motion
• Last term in (i) represents dissipation, making (i) nonlinear
2.7 Solutions to the Temperature Distribution
Governing equations: continuity (2.2), momentum (2.8)
and energy (2.15)
50
TABLE 2.1
Basic law
No. of
Equations
Unknowns
p
ρ µ k
u
v
w
1
u
v
w
Momentum
3
u
v
w
Equation of State
1
T
p
ρ µ
ρ
Viscosity relation
µ = µ ( p, T )
1
T
p
µ
1
T
p
Energy
1
Continuity
Conductivity relation
k = k ( p, T )
TT
ρ
p
k
• Solution consideration: Table 2.1 Equation of state
gives c p and cv and β
51
(1) General case: variable properties
• 8 unknowns: T, u, v, w, p, ρ , µ , k , 8 eqs. (yellow box)
• 8 eqs. solved simultaneously for 8 unknowns
• Velocity and temperature fields are coupled.
(2) Special case 1: constant k and µ
• 6 Unknowns: T, u, v, w, p,ρ , 6 eqs., see blue box
• 6 eqs. solved simultaneously for 6 unknowns
(3) Special case 2: constant k , µ and ρ
• 5 unknowns: T, u, v, w, p, 5 eqs., see red box
• However, 4 unknowns: u, v, w, p, 4 eqs., give flow field,
see white box
• Velocity and temperature fields are uncoupled
52
2.8 The Boussinesq Approximation
• Free convection is driven by density change
• Can’t assume ρ = constant
• Alternate approach: the Boussinesq approximation
• Start with N-S equations for variable ρ
r
r
DV
1
r
2r
ρ
= pg − ∇p + µ∇ (∇ ⋅ V ) + µ∇ V
Dt
3
• Assume:
(1) ρ = ρ ∞ in inertia term
(2.9)
r
(2) ρ = ρ ∞ in continuity, ∴ ∇ .V = 0
(2.9) becomes
r
DV
r
2r
ρ∞
= ρg − ∇p + µ∇ V
Dt
(a)
53
( )∞ Reference state (far away from
r object) where
r
DV∞
2r
(b)
V∞ = uniform,
= ∇ V∞ = 0
Dt
Apply(a) at infinity , use (b)
r
(c)
ρ ∞ g − ∇p∞ = 0
Subtract (c) from
r (a)
DV
r
2r
(d)
ρ
= ( ρ − ρ ∞ )g − ∇ ( p − p∞ ) + µ∇ V
ρ
Dt
(3) Express ( ρ − ρ ∞ ) in term of temperature difference.
Introduce β
1 ∂ρ
(2.16)
β =−
ρ ∂T p
Assume, for free convection ρ (T , p ) ≈ ρ (T )
1 dρ
β ≈−
ρ ∞ dT
(e)
54
For small ∆T , is linear ρ (T )
1 ρ − ρ∞
β ≈−
ρ ∞ T − T∞
(f)
∴
ρ − ρ ∞ = − β ρ ∞ (T − T∞ )
Substitute (2.28) into (d)
r
DV
1
r
2r
= − β g (T − T∞ ) −
∇ ( p − p∞ ) + v ∇ V
Dt
ρ∞
(2.28)
(2.29)
• Simplification leading to (2.29) is called the Boussinesq
approximation
• This eliminates density as a variable
• However, momentum and energy are coupled
55
2.9 Boundary Conditions
(1) No-slip condition
At surface, y = 0
r
V ( x ,0, z , t ) = 0
(2.30a)
or
u( x ,0, z , t ) = v ( x ,0, z , t ) = w ( x ,0, z , t ) = 0
(2) Free stream condition
• Far away from object, assume uniform velocity
• Example:
Uniform u at y = ∞ :
u( x , ∞ , z , t ) = V∞
Uniform temperature:
T ( x , ∞ , z , t ) = T∞
(2.30b)
(2.31)
(2.32)
(3) Surface thermal conditions
56
(i) Specified temperature
T ( x ,0, z , t ) = Ts
(ii) Specified heat flux
∂ T ( x , 0, z , t )
−k
= qo′′
∂y
(2.33)
(2.34)
Example 2.4: Heated Thin Liquid Film Flow
Over an Inclined Surface
• Axial flow by gravity, thin
y
film
• Uniform plate
temperature To
• Uniform flux qo′′ at free
surface
u
qo′′
To g
x
θ
• Write the velocity and thermal boundary conditions
57
(1) Observations
• No slip condition at inclined plate
• Free surface is parallel to the inclined plate
• Specified temperature at plate
• Specified flux at free surface
• Cartesian geometry
(2) Problem Definition.
Write the boundary conditions at two surfaces for u, v and T
(3) Solution Plan
• Select an origin and coordinate axes
• Identify the physical flow and thermal conditions at the
two surfaces
• Express conditions mathematically
58
(4) Plan Execution
(i) Assumptions
• Constant film thickness
• Negligible shearing stress at free surface
• Newtonian fluid.
(ii) Analysis.
Origin and coordinates as shown
(1) No slip condition at the inclined surface
u( x ,0) = 0
v ( x ,0) = 0
(2) Parallel streamlines
v ( x, H ) = 0
(3) Negligible shear at free surface: for Newtonian
fluid use (2.7a)
(a)
(b)
(c)
59
∂v ∂u
τ xy = τ yx = µ +
∂x ∂ y
Apply (2.7a) at the free surface, use (c)
∂u( x , H )
=0
∂y
(4) Specified temperature at plate:
T ( x ,0) = To
(5) Specified heat flux at the free surface:
∂ T ( x , 0, z , t )
−k
= −qo′′
∂y
(iii) Checking
(2.7a)
(d)
(e)
(f)
Dimensional check: Each term of (f) has units of flux.
(5) Comments
60
• Must select origin and coordinates
• Why negative heat flux in (f)?
2.10 Non-dimensional Form of the Governing
Equations: Dynamic and Thermal
Similarity Parameters
• Rewrite equations in dimensionless form to:
• Identify governing parameters
• Plan experiments
• Present results
• Important factors in solutions
• Geometry
• Dependent variables: u, v, w, p, T
• Independent variables: x, y, z, t
61
• Constant quantities: p∞ , T∞ , Ts , V∞ , L, g
• Fluid properties: c p , k, β, µ, ρ
• Mapping results: dimensional vs. dimensionless
2.10.1 Dimensionless Variables
• To non-dimensionalize variables: use characteristic
quantities g, L, Ts , T∞ , V∞
• Define dimensionless variables
r
r∗ V
∗ ( p − p∞ )
V =
p =
V∞
ρ ∞V∞ 2
x
x =
L
∗
z
z =
L
r
g
g∗ =
g
y
y =
L
∗
T∗ =
∗
(T − T∞ )
(Ts − T∞ )
V∞
t =
t,
L
∗
(2.35)
62
∴
∇=
∂
∂ ∂
∂
∂
∂
1
+ + =
+
+
= ∇*
∂x ∂y ∂z L∂x ∗ L∂y ∗ L∂z ∗ L
2
∂2
∇ =
∂x
2
+
∂2
∂y
2
+
∂2
∂
∂
∂
=
+ 2 ∗2 + 2 ∗2
2 ∗2
∂z
L ∂x
L ∂y
L ∂z
1 ∗2
=
∇
2
L
2
V∞ D
D
D
=
=
∗
Dt D( Lt / V∞ ) L Dt ∗
(2.36a)
(2.36b)
(2.36c)
2.10.2 Dimensionless Form of Continuity
(2.35), (2.36) into (2.2c)
Dρ
Dt *
r*
+ ρ ∇ ⋅V = 0
(2.37)
63
2.10.3 Dimensionless Form of the NavierStokes Equations of Motion
(2.35), (2.36) into (2.29)
r*
DV
Gr * r *
1 *2 r
* *
= − 2T g −∇ P +
∇ V*
*
Re
Dt
Re
(2.38)
Re and Gr are dimensionless parameters (numbers)
βg (Tw − T∞ )L3 , Reynolds number
Gr ≡
2
(2.39)
ρV∞ L V∞ L
, Grashof number
Re ≡
=
v
µ
(2.40)
v
64
2.10.4 Dimensionless Form of the Energy
Equation
Two special cases:
(i) Incompressible, constant conductivity
(2.35), (2.36) into (2.19)
DT ∗
Dt ∗
1
Ε ∗
∗2 ∗
=
∇ T +
Φ
RePr
Re
Pr and E are dimensionless parameters
c pµ
µ/ρ
v , Prandtl number
=
=
Pr =
k
k / ρc p ∝
V∞ 2
Ε=
, Eckert number
c p (Ts − T∞ )
(2.41a)
(2.42)
(2.43)
65
(2.35), (2.36) into (2.17) gives dimensionless
dissipation function Φ *
u* 2 v * 2
∂ ∂
*
Φ = 2 * + * + ⋅ ⋅ ⋅
∂ x ∂y
(ii) Ideal gas, constant conductivity and viscosity
(2.35), (2.36) into (2.22)
*
DT *
1
Dp
Ε *
*2 *
=
∇ T +Ε * +
Φ
*
RePr
Re
Dt
Dt
(2.44)
(2.41b)
2.10.5 Significance of the Governing Parameters
Governing equations (2.37), (2.38), (2.41) are governed by
4 parameters: Re, Pr, Gr and E:
T * = f ( x * , y* , z * .t * ; Re , Pr , Gr , E )
(2.45)
66
NOTE
• Significance of parameters
• Reynolds number : viscous effect
• Prandtl number : property, heat transfer effect
• Grashof number : buoyancy effect (free convection)
• Eckert number: viscous dissipation: high speed flow and
very viscous fluids
• Dimensional form: solution depends on
• 6 quantities: p∞ , T∞ ,Ts ,V ,L, g
∞
• 5 properties c p , k, β, µ, and ρ affect the solution
• Dimensionless form: solution depends on
• 4 parameters: Re, Pr, Gr and E
67
• Special cases:
• Negligible free convection: eliminate Gr
• Negligible dissipation eliminate E.
T * = f ( x * , y* , z * , t * ; Re , Pr )
(2.46)
• Significance of (2.45) and (2.46):
Geometrica lly similar bodies have
the same velocity and temperature
solution if the parameters are the same
• Use (2.45) to:
• Plan experiments
• Carry out numerical computations
• Organize presentation of results
68
2.10.6 Heat Transfer Coefficient: The Nusselt
Number
− k ∂T ( x ,0, z )
h=
(T − T∞ )
∂y
(1.10)
s
Express in dimensionless form: use (2.30)
* ( x* ,0 , z * )
hx
∂
T
= − x*
k
∂ y*
• Local Nusselt number Nu x
hx
Nu x =
k
(2.47)
(2.48)
• Average Nusselt number Nu
hL
Nu =
k
(2.49)
69
where
Recall
L
h = ∫ h( x )dx
L 0
(2.50)
T * = f ( x * , y* , z * .t * ; Re, Pr , Gr , E )
(2.45)
Nu x = f ( x * ; Re , Pr , Gr , E )
(2.51)
1
Thus
Special case: negligible buoyancy and viscous dissipation
(2.52)
Nu x = f ( x * ; Re , Pr )
For free convection with negligible dissipation we obtain
Nu x = f ( x * ; Gr , Pr )
For the average Nusselt number
Nu =
hL
= f ( Re , Pr ,Gr , E )
k
(2.53)
(2.54)
70
Example 2.5: Heat Transfer Coefficient for Flow
over Cylinders
Two experiments, different cylinders, same fluid:
Experiment # 1
D1 = 3 cm
Experiment # 2
V1 = 15 m/s
V2 = 98 m/s
D2 = 5 cm
h 2 = 144 W/m2-oC
h1 = 244 W/m2-oC
Compare results with correlation equation
Nu D =
hD
= C ReD0.6 Pr n
k
(a)
Are experimental data accurate?
(1) Observations
• Compare data for h1 and h2 correlation (a)
• h appears in definition of Nu
71
• Fluid, C and n are unknown, (a) does not give h
• Use (a) to determine ratio h1 / h2
(2) Problem Definition.
Determine h1 / h2 using data and correlation (a)
(3) Solution Plan.
Apply correlation (a) equation to determine h1 / h2 and
compare experimental data
(4) Plan Execution
(i) Assumptions
• Correlation (a) is valid for both experiments
• Fluid properties are constant
(ii) Analysis
Use Re D =
VD
ν
into (a)
72
hD
VD
= C
k
ν
Solve for h
h=
0.6
Pr n
C k V 0.6 Pr n
ν
0.6
(b)
(c)
0.4
D
Apply (c) to the two experiments
h1 =
and
h2 =
Take ratio of (d) and (e)
CkV10.6 Pr n
ν
0.6
D1
(d)
0.4
CkV2 0.6 Pr n
ν 0.6 D2 0.4
h1 V1
=
h2 V2
0.6
D2
D1
(e)
0.4
(f)
73
(iii) Computations
Substitute data for V1 , V2 , D1 and D2 into (f)
h1 15(m s )
=
h2 98 (m s )
0.6
5(cm )
3(cm )
0.4
= 0 .4
(g)
Experimental data for ratio h1 / h2
W
244
2o
h1
m - C = 1.69
=
h2 144 W
2o
m - C
The two results are not the same
(h)
Conclusion: Incorrect experimental data
(iv) Checking
Dimensional check: units of (f) are correct
74
Limiting check: If V1 = V2 and D1 = D2 , then h 1 = h2 .
This is confirmed by (f)
Qualitative check: If V is increased, h should increase.
This is substantiated by (c).
(5) Comments
• Critical assumption: correlation (a) applies to both
experiments
• Analysis suggests an error in the experimental data
• More conclusive check can be made if C, n and fluid are
known
2.11 Scale Analysis
Procedure to obtain approximate results
(order of magnitue) without solving equations
75
Example 2.6: Melting Time of Ice Sheet
• Ice sheet thickness L
• At freezing temperature T f
• One side is at To > T f
xi
L
x
solid
liquid T f
• Other side is insulated
• Conservation of energy at
the melting front:
0
dx i
∂T
−k
=ρL
∂x
dt
xi = melting front location
To
(a)
L = latent heat of fusion
• Use scale analysis to determine total melt time
76
(1) Observations
• Entire sheet melts when xi = L
• Largest temperature difference is To − T f
• Time is in equation (a)
• Scaling of equation (a) should be useful
(2) Problem Definition
Determine the time t = t o when xi ( t ) = L
(3) Solution Plan
Apply scale analysis to equation (a)
(4) Plan Execution
(i) Assumptions
• Sheet is perfectly insulated at x = L
• Liquid phase is stationary
77
(ii) Analysis
Equation (a) is approximated by
∆ xi
∆T
−k
=ρL
∆x
∆t
Select scales for variables in (a)
Scale for ∆T :
Scale for ∆ x :
Scale for ∆ x i :
Scale for ∆t :
Substitute into (a)
(b)
∆T ∼ (To − T f )
∆x ∼ L
∆ xi ∼ L
∆t ∼ t o
k
(To − T f )
L
Solve for melt time to
to ∼
L
∼ρ L
to
ρ L L2
k(To − T f )
(c)
78
(iii) Checking
Dimensional check: Each term in (c) has units of time:
to =
ρ ( kg/m 3 )L( J/kg ) L2 ( m 2 )
o
o
k ( W/m- C)(To − T f )( C)
=s
Limiting check:
(1) If L is infinite, melt time is infinite. Set L = ∞ in (c)
gives t o = ∞
(2) If thickness is zero, melt time should vanish. Set L = 0
in (c) gives to = 0
Qualitative check:
Expect t o to:
Directly proportional to mass, L and L, and
Inversely proportional to k (To − T f )
This is confirmed by solution (c)
79
(5) Comments
• t o is estimated without solving governing equations
• Exact quasi-steady solution
to =
ρ L L2
2k(To − T f )
(d)
• Scaling answer is within a factor of 2
80
CHAPTER 3
EXACT ONE-DIMENSIONAL SOLUTIONS
3.1 Introduction
• Temperature solution depends on velocity
• Velocity is governed by non-linear Navier-Stokes eqs.
• Exact solution are based on simplifications governing
equations
3.2 Simplification of the Governing Equations
Simplifying assumptions:
(1) Laminar flow
(2) Parallel streamlines
(3.1)
v =0
1
(3.1) into continuity for 2-D, constant density fluid:
∴
∂u
= 0 , everywhere
∂x
∂ 2u
=0
∂x 2
(3) Negligible axial variation of temperature
∂T
= 0 , everywhere
∂x
(3.4) is valid under certain conditions. It follows that
(3.2)
(3.3)
(3.4)
∂ 2T
(3.5)
=0
2
∂x
(4) Constant properties: velocity and temperature fields are
uncoupled (Table 2.1, white box)
2
TABLE 2.1
Basic law
No. of
Equations
Unknowns
p
ρ µ k
u
v
w
1
u
v
w
Momentum
3
u
v
w
Equation of State
1
T
p
ρ µ
ρ
Viscosity relation
µ = µ ( p, T )
1
T
p
µ
1
T
p
Energy
1
Continuity
Conductivity relation
k = k ( p, T )
TT
ρ
p
k
Similar results are obtained for certain rotating flows.
3
Fig. 3.2:
• Shaft rotates inside sleeve
• Streamlines are concentric circles
• Axisymmetric conditions, no axial
variations
∂T
=0
∂θ
∴
∂ 2T
∂θ 2
(3.6)
=0
(3.7)
3.3 Exact Solutions
3.3.1 Couette Flow
• Flow between parallel plate
• Motion due to pressure drop and/or moving plate
• Channel is infinitely long
4
Example 3.1: Couette Flow with Dissipation
• Very large parallel
plates
• Incompressible fluid
• Upper plate at To moves with velocity Uo
• Insulate lower plate
• Account for dissipation
• Laminar flow, no gravity, no pressure drop
• Determine temperature distribution
(1) Observations
• Plate sets fluid in motion
• No axial variation of flow
• Incompressible fluid
• Cartesian geometry
5
(2) Problem Definition.
Determine the velocity and temperature distribution
(3) Solution Plan
• Find flow field, apply continuity and Navier-Stokes
equations
• Apply the energy to determine the temperature distribution
(4) Plan Execution
(i) Assumptions
• Steady state
• Laminar flow
• Constant properties
• Infinite plates
• No end effects
• Uniform pressure
• No gravity
6
(ii) Analysis
Start with the energy equation
∂T
∂T
∂T
∂T
+u
+v
+w
ρ cΡ
=
∂y
∂z
∂x
∂t
∂ 2T ∂ 2T ∂ 2T
k 2 + 2 + 2 + µΦ
∂x
∂z
∂y
Φ is dissipation
2
∂u 2 ∂ v 2
∂
w
Φ = 2 + + +
∂y
∂z
∂x
∂u ∂v 2 ∂v ∂ w 2
∂w ∂u 2
+ + +
+ −
+
∂z ∂ y
∂x ∂z
∂y ∂ x
2
2 ∂u ∂v ∂w
+
+
3 ∂ x ∂ y ∂z
(3.19b)
(3.17)
7
Need u, v and w. Apply continuity and the Navier-Stokes
equations
Continuity
∂ρ
∂ρ
∂ρ
∂ρ
∂ u ∂v ∂ w
+u
+ +v
+w
+ρ +
+
=0
∂t
∂z
∂x
∂y
∂x ∂y ∂z
(2.2b)
Constant density
Infinite plates
∂ρ ∂ ρ ∂ρ ∂ρ
=0
=
=
=
∂ t ∂x ∂y ∂z
(a) and (b) into (2.2b)
Integrate (c)
(a)
∂
∂
=w=0
=
∂x ∂z
(b)
∂v
=0
∂y
(c)
v = f ( x)
(d)
8
f ( x ) is “constant” of integration
Apply the no-slip condition
v ( x ,0 ) = 0
(d) and (e) give
Substitute into (d)
(e)
f ( x) = 0
v =0
(f)
∴ Streamlines are parallel
To determine u we apply the Navier-Stokes eqs.
∂u
∂u
∂u
∂u
ρ + u + v + w =
∂x
∂y
∂z
∂t
∂ 2u ∂ 2u ∂ 2u
∂p
ρg x − + µ 2 + 2 + 2
∂x
∂y
∂z
∂x
(2.10x)
9
Simplify:
Steady state
∂u
=0
∂t
(g)
No gravity
gx = 0
(h)
Negligible axial pressure variation
∂p
=0
∂x
(b) and (f)-(i) into (2.10x) gives
d 2u
Solution to (j) is
dy 2
=0
u = C1 y + C 2
Boundary conditions
u(0) = 0 and u( H ) = U o
(i)
(j)
(k)
(l)
10
(k) and (I) give
(m) into (k)
C1 = Uo
and
C2 = 0
u y
=
Uo H
(m)
(3.8)
Dissipation: (b) and (f) into (2.17)
Use (3.8) into (n)
∂u
Φ =
∂y
Φ=
2
2
Uo
2
(n)
(o)
H
Steady state: ∂T / ∂t = 0
Infinite plates at uniform temperature:
∂T ∂ 2T
= 2 =0
∂x ∂x
11
Use above, (b), (f) and (o) into energy (2.10b)
k
Integrate
T =−
B.C.
d 2T
+µ
dy 2
2
µU o
2
2kH
dT(0)
−k
=0
dy
U o2
H2
(p)
=0
y 2 + C3 y + C4
and
T ( H ) = To
(q)
(r)
B.C. and solution (q) give
C3 = 0 and C4 = To +
µUo2
(s) into (q)
T − To
µU o2
k
1
y 2
= 1 − 2
2
H
(s)
2k
(3.9)
12
Fourier’s law gives heat flux at y = H
(3.9) into the above
dT ( H )
q′′( H ) = − k
dx
q ′′( H ) =
µU o2
H
(iii) Checking
Dimensional check: Each term in (3.8) and (3.9) is
dimensionless. Units of (3.10) is W/m2
(3.10)
Differential equation check: Velocity solution (3.8) satisfies
(j) and temperature solution (3.9) satisfies (p)
Boundary conditions check: Solution (3.8) satisfies B.C. (l),
temperature solution (3.9) satisfies B.C. (r)
13
Limiting check: (i) Stationary upper plate: no fluid motion.
Set Uo = 0 in (3.8) gives u(y) = 0
(ii) Stationary upper plate: no dissipation, uniform
temperature To, no surface flux. Set Uo = 0 in (o), (3.9) and
(3.10) gives Φ = 0, T(y) = To and q′′( H ) = 0
(iii) Inviscid fluid: no dissipation, uniform temperature To.
Set µ = 0 in (3.9) gives T(y) = To
(iv) Global conservation of energy: Frictional energy is
conducted through moving plate:
W = Friction work by plate
q′′(H ) = Heat conducted through plate
W = τ ( H )U o
(t)
where
14
τ (H ) = shearing stress
(3.8) into (u)
(v) and (t)
du( H )
τ (H ) = µ
dy
(u)
Uo
τ (H ) = µ
H
(v)
W =
µU o2
H
(w)
(w) agrees with (3.10)
(4) Comments
• Infinite plate is key assumption. This eliminates x as a
variable
• Maximum temperature: at y = 0 Set y = 0 in (3.9)
T (0) − To =
µU o2
2k
15
3.3.2 Hagen-Poiseuille Flow
•Problems associated with axial flow in channels
• Motion due to pressure drop
• Channel is infinitely long
Example 3.2: Flow in a Tube at Uniform
Surface Temperature
• Incompressible
fluid flows in a
long tube
• Motion is due to
pressure gradient ∂p / ∂z
• Surface temperature To
• Account for dissipation
• Assuming axisymmetric laminar flow
16
• Neglecting gravity and end effects
• Determine:
[a] Temperature distribution
[b] Surface heat flux
[c] Nusselt number based on [ T ( 0) − To]
(1) Observations
• Motion is due to pressure drop
• Long tube: No axial variation
• Incompressible fluid
• Heat generation due to dissipation
• Dissipated energy is removed by conduction at the
surface
• Heat flux and heat transfer coefficient depend on
temperature distribution
17
• Temperature distribution depends on the velocity
distribution
• Cylindrical geometry
(2) Problem Definition.
Determine the velocity and temperature distribution.
(3) Solution Plan
• Apply continuity and Navier-Stokes to determine flow field
• Apply energy equation to determine temperature
distribution
• Fourier’s law surface heat flux
• Equation (1.10) gives the heat transfer coefficient.
(4) Plan Execution
18
(i) Assumptions
• Steady state
• Laminar flow
• Axisymmetric flow
• Constant properties
• No end effects
• Uniform surface temperature
• Negligible gravitational effect
(ii) Analysis
[a] Start with energy equation (2.24)
∂ T v θ ∂T
∂T
∂T
+ vr
+
+vz
=
∂r
r ∂θ
∂z
∂t
1 ∂ ∂T 1 ∂ 2T ∂ 2T
k
+ 2 + µΦ
r
+ 2
2
∂z
r ∂r ∂r r ∂θ
ρ c P
(2.24)
19
where
2
2
2
∂v z
1 ∂v θ v r
∂v r
Φ = 2
+ + 2
+
+ 2
r
∂r
r ∂θ
∂z
2
2
1 ∂v z ∂v 0
∂v r ∂ v z
∂v θ v θ 1 ∂v r
−
+
+
+
+
+
r r ∂0
∂z
∂r
∂r
r ∂0
∂z
• Need v r, v θ and v z
• Flow field: use continuity and Navier-Stokes eqs.
1 ∂
∂
∂ρ 1 ∂
(ρ rv r ) +
(ρ v θ ) + (ρ v z ) = 0
+
∂ t r ∂r
r ∂θ
∂z
Constant ρ
∂ρ ∂ρ ∂ρ ∂ρ
=
=
=
=0
∂ t ∂r ∂θ ∂z
Axisymmetric flow
∂
vθ =
=0
∂θ
2
(2.25)
(2.4)
(a)
(b)
20
Long tube, no end effects
∂
=0
∂z
(c)
d
(r v r ) = 0
dr
(d)
r v r = f (z )
(e)
(a)-(c) into (2.4)
Integrate
f ( z ) is “constant” of integration. Use the no-slip B.C.
v ( ro , z ) = 0
(e) and (f) give
f (z) = 0
Substitute into (e)
∴ Streamlines are parallel
vr = 0
(f)
(g)
v z Determine : Navier-Stokes eq. in z-direction
21
∂ v z ∂v z
∂v z v θ ∂ v z
+ vz
+
ρ v r
+
=
∂t
∂r
r ∂θ
∂z
1 ∂ ∂v z 1 ∂ 2 v z ∂ 2 v z
∂p
ρg z − + µ
+
r
+ 2
2
2
∂z
∂z
r ∂r ∂r r ∂θ
Simplify
Steady state:
No gravity:
(2.11z)
∂
=0
∂t
gr = g z = 0
(b), (c) and (g)-(i) into (2.11z)
∂p
1 d dv z
−
+µ
=0
r
∂z
r dr dr
∴ v z depends on r only, rewrite (3.11)
∂p
1 d dv z
=µ
r
= g( r )
∂z
r dr dr
(h)
(i)
(3.11)
(j)
22
Integrate
p = g( r )z + C o
Apply Navier-Stokes in r-direction
(k)
∂v r v θ ∂v r v θ 2
∂
v
∂
v
r
r
=
−
+vz
+
+
ρ v r
r
r
r
z
t
∂
∂
∂
∂
θ
2
2
∂ 1 ∂
∂p
1 ∂ v r 2 ∂v θ ∂ v r
( rv r ) + 2
− 2
+
ρg r − + µ
2
2
∂r
∂
r
r
∂
r
∂
θ
r ∂θ
r
∂z
(2.11r)
(b), (g) and (i) into (2.11r)
Integrate
∂p
=0
∂r
p = f (z)
(l)
(m)
f ( z ) = “constant” of integration
• Equate two solutions for p: (k) and (m):
23
∴
p = g(r )z + Co = f ( z)
g(r) = C
C = constant. Use (o) into (j)
Integrate
integrate again
Two B.C. on : v z
∂p
1 d dv z
=µ
r
=C
∂z
r dr dr
(n)
(o)
(p)
dv z 1 d p 2
r
r + C1
=
dr 2µ d z
1 dp 2
vz =
r + C1 ln r + C2
4µ d z
dv z (0)
= 0,
dr
(q) and (r) give C1and C2
v z (ro ) = 0
(q)
(r)
24
C1 = 0
,
1 dp 2
C2 =
ro
4µ d z
Substitute into (q)
1 dp 2
vz =
( r − ro2 )
4µ d z
For long tube at uniform temperature:
∂T ∂ 2 T
= 2 =0
∂z ∂z
(b), (c), (g), (h) and (s) into energy (2.24)
1 d dT
k
r
+ µΦ = 0
r dr dr
(b), (c) and (g) into (2.25)
dv z
Φ =
dr
(3.12)
(s)
(t)
2
Substitute velocity solution (3.11) into the above
25
2
1 d p 2
Φ =
r
2µ d z
(u) in (t) and rearrange
(u)
2
Integrate
d dT
1 d p 3
r
r
=−
dr dr
4kµ d z
(3.13)
2
1 d p 4
T =−
r + C 3 ln r + C 4
64kµ d z
Need two B.C.
dT (0)
= 0 and T ( ro ) = To
dr
2
(v) and (w) give
1 d p 4
C 3 = 0 , C 4 = To +
ro
64kµ d z
Substitute into (v)
ro4
d p
T = To +
64kµ d z
2
4
r
1 −
4
r
o
(v)
(w)
(3.14a)
26
In dimensionless form:
T − To
4
r
1 −
=
2
4
4
r
ro d p
o
64kµ d z
[b] Use Fourier’s
dT ( ro )
q ′′( ro ) = − k
dr
(3.14) into above
2
3
ro d p
q ′′( ro ) =
16 µ d z
[c] Nusselt number:
hD 2hro
Nu =
=
k
k
(1.10) gives h
dT ( ro )
k
h=−
[T (0) − To ] dr
(3.14a) into (y)
(3.14b)
(3.15)
(x)
(y)
27
(z) into (x)
2k
h=
ro
Nu = 4
(z)
(3.16)
(iii) Checking
Dimensional check:
• Each term in (3.12) has units of velocity
• Each term in (3.14a) has units of temperature
• Each term in (3.15) has units of W/m2
Differential equation check: Velocity solution (3.12)
satisfies (p) and temperature solution (3.14) satisfies (3.13)
Boundary conditions check: Velocity solution (3.12) satisfies
B.C. (r) and temperature solution (3.14) satisfies B.C. (w)
Limiting check:
28
(i) Uniform pressure ( dp / dz = 0 ): No fluid motion.
Set dp / dz = 0 in (3.12) gives v z = 0
(ii) Uniform pressure ( dp / dz = 0 ): No fluid motion, no
dissipation, no surface flux. Set dp / dz = 0 in (3.15) gives
q′′( ro ) = 0
(iii) Global conservation of energy:
Heat leaving tube = Pump work
Pump work W for a tube of length L
(z-1)
p1 = upstream pressure
p2 = downstream pressure
= flow rate
29
(3.12) into the above, integrate
(z-2)
(z-1) and (z-2)
π ro4 dp
W =−
( p1 − p2 )
8µ dz
Work per unit area W ′′
(z-3) into the above
(z-3)
W
W ′′ =
2π ro L
ro3 dp ( p1 − p2 )
W ′′ = −
16 µ dz
L
( p1 − p2 )
dp
However
=−
L
dz
Combine with (z-4)
ro3 dp 2
W ′′ =
16 µ dz
(z-4)
30
This agrees with (3.15)
(5) Comments
• Key simplification: long tube with end effects. This is
same as assuming parallel streamlines
• According to (3.14), maximum temperature is at center r
=0
• The Nusselt number is constant independent of Reynolds
and Prandtl numbers
31
3.3.3 Rotating Flow
Example 3.3: Lubrication Oil Temperature
in Rotating Shaft
• Lubrication oil between shaft
and housing
• Angular velocity is ω
• Assuming laminar flow
• Account for dissipation
• Determine the maximum temperature
rise in oil
(1) Observations
• Fluid motion is due to shaft rotation
• Housing is stationary
32
• No axial variation in velocity and temperature
• No variation with angular position
• Constant ρ
• Frictional heat is removed at housing
• No heat is conducted through shaft
• Maximum temperature at shaft
• Cylindrical geometry
(2) Problem Definition.
Determine the velocity and temperature distribution of oil
(3) Solution Plan
• Apply continuity and Navier-Stokes eqs. to determine flow
field
• Use energy equation to determine temperature field
• Fourier’s law at the housing gives frictional heat
33
(4) Plan Execution
(i) Assumptions
• Steady state
• Laminar flow
• Axisymmetric flow
• Constant properties
• No end effects
• Uniform surface temperature
• Negligible gravitational effect
(ii) Analysis
• Energy equation governs temperature
∂ T v θ ∂T
∂T
∂T
+ vr
+
ρcP
+vz
=
∂r
r ∂θ
∂z
∂t
1 ∂ ∂T 1 ∂ 2T ∂ 2T
k
+ 2 + µΦ
r
+ 2
2
∂z
r ∂r ∂r r ∂θ
(2.24)
34
where
2
2
2
∂v z
1 ∂v θ v r
∂v r
Φ = 2
+ + 2
+
+ 2
r
∂r
r ∂θ
∂z
2
2
1 ∂v z ∂v 0
∂v r ∂ v z
∂v θ v θ 1 ∂v r
−
+
+
+
+
+
r r ∂0
∂z
∂r
∂r
r ∂0
∂z
2
(2.25)
Need flow field v r , v θ and v z
• Apply continuity and Navier-Stokes to determine flow
field
∂ρ 1 ∂
1 ∂
∂
(ρ rv r ) +
(ρ v θ ) + (ρ v z ) = 0
+
(2.4)
∂ t r ∂r
r ∂θ
∂z
Constant ρ
∂ρ ∂ρ ∂ρ ∂ρ
(a)
=
=
=
=0
∂ t ∂r ∂θ ∂z
Axisymmetric flow
∂
(b)
=0
∂θ
35
Long shaft:
(a)-(c) into (2.4)
Integrate
∂
vz =
=0
∂z
d
(r v r ) = 0
dr
r vr = C
Apply B.C. to determine C
v r ( ro ) = 0
(e) and (f) give C = 0
Use (e)
vr = 0
(c)
(d)
(e)
(f)
(g)
∴ Streamlines are concentric circles
Apply the Navier-Stokes to determine v θ
36
∂v θ ∂v θ
∂v θ v θ ∂v θ v r v θ
−
+vz
+
ρ v r
+
=
∂t
∂r
r ∂θ
r
∂z
2
∂ 1 ∂
1 ∂p
2 ∂v r ∂ 2 v θ
1 ∂ vθ
ρgθ −
+ µ
( rv θ ) + 2
+ 2
+
2
2
r ∂θ
r ∂θ
r ∂θ
∂z
∂ r r ∂r
(2.11 θ )
For steady state:
∂
=0
∂t
Neglect gravity, use (b),(c), (g), (h) into (2.11 θ )
d 1 d
(
r
v
)
θ =0
dr r dr
Integrate
B.C. are
C1
C2
vθ =
r+
2
r
v θ ( ri ) = ω ri
v θ ( ro ) = 0
(h)
(3.17)
(i)
(j)
37
(j) gives C1 and C 2
C1 = −
(k) into (i)
2ω ri2
ro2
− ri2
C2 =
ω ri2 ro2
ro2
− ri2
v θ ( r ) ( ro / ri ) 2 ( ri / r ) − ( r / ri )
=
ω ri
( ro / ri ) 2 − 1
(k)
(3.18)
Simplify energy equation (2.24) and dissipation function
(2.25). Use(b), (c), (g), (h)
1 d dT
k
r
+ µΦ = 0
r dr dr
and
dv 0 v θ
−
Φ =
r
dr
(l)
2
(3.18) into above
38
2
1
Φ =
2
4
1
−
(
r
/
r
)
r
i
o
Combine (m) and (l)
2ω ri2
d dT
µ
r
=−
dr dr
k
(m)
2
2ω ri
1
1 − ( r / r ) 2 r 3
i
o
2
Integrate(3.19) twice
2
2
1
µ 2ω ri
T (r ) = −
2 + C 3 ln r + C 4
2
4k 1 − ( ri / ro ) r
Need two B.C.
dT(ri )
T (ro ) = To and
=0
dr
(n) and (o) give C 3 and C 4
µ
(n)
(o)
2
1
C3 = −
2
2
2k 1 − ( ri / ro ) ri
2ω ri2
(3.19)
39
µ
C 4 = To +
2
4k 1 − ( ri / ro )
Substitute into (o)
2ω ri2
2
or
2
1
2
2 + 2 ln ro
ro ri
µ 2ω ri
2
2
T ( r ) = To +
( ri / ro ) − ( ri / r ) + 2 ln( ro / r )
4k 1 − ( r / r ) 2
i o
(3.20a)
T ( r ) − To
µ
2
[
]
= ( ri / ro ) 2 − ( ri / r ) 2 + 2 ln( ro / r )
2
4k 1 − ( ri / ro )
Maximum temperature at r = ri
2ω ri
µ
2
(3.20b)
2
T ( ri ) − To =
1
+
(
r
/
r
)
i o + 2 ln( ro / ri )
2
4k 1 − ( ri / ro )
(3.21)
2ω ri
[
Use Fourier’s law to determine frictional energy per
unit length q′( ro )
]
40
(3.20a) in above
(iii) Checking
dT ( ro )
q′( ro ) = −2π ro k
dr
(ω ri ) 2
q′( ro ) = 4π µ
2
1 − ( ri / ro )
(3.22)
• Each term in solutions (3.18) and (3.20b) is dimensionless
• Equation (p) has the correct units of W/m
Differential equation check:
• Velocity solution (3.18) satisfies (3.17) and temperature
solution (3.20) satisfies (3.19)
Boundary conditions check:
• Velocity solution (3.18) satisfies B.C. (j) and temperature
solution (3.20) satisfies B.C. (o)
41
Limiting check:
• Stationary shaft: No fluid motion. Set ω = 0 in (3.18) gives
vθ = 0
• Stationary shaft: No dissipation, no heat loss Set ω = 0 in
(3.22) gives q′( ro ) = 0
• Global conservation of energy:
Heat leaving housing = shaft work
Shaft work per unit length
W ′ = −2π riτ ( ri )ω ri
τ ( ri ) = shearing stress
dv 0 v θ
τ ( ri ) = µ
−
r r = ri
dr
(3.18) into the above
τ ( ri ) = −2
( ro / ri ) 2 µω ri
2
( ro / ri ) − 1
(p)
(q)
(r)
42
Combining (p) and (r)
W ′ = 4π µ
(ω ri ) 2
1 − ( ri / ro )
2
(s)
This is identical to surface heat transfer (3.22)
(5) Comments
• The key simplifying assumption is axisymmetry
• Temperature rise due to frictional heat increase as
the clearance s decreased
• Single governing parameter: ( ri / ro )
43
CHAPTER 4
BOUNDARY LAYER FLOW
APPLICATION TO EXTERNAL FLOW
4.1 Introduction
• Boundary layer concept (Prandtl 1904): Eliminate
selected terms in the governing equations
• Two key questions
(1) What are the conditions under which
terms in the governingequations can
be dropped?
(2) What terms can be dropped ?
1
• Answer: By two approaches
• Intuitive arguments
• Scale analysis
4.2 The Boundary Layer Concept:
Simplification of Governing Equations
4.2.1 Qualitative Description
2
Under certain conditions the action of viscosity
is confined to a thin region near the surface
called the viscous or velocity boundary layer
Under certain conditions thermal interaction between
moving fluid and a surface is confined to a thin region
near the surface called the thermal or temperature
boundary layer
• Conditions for viscous boundary layer:
(1) Slender body without flow separation
(2) High Reynolds number ( Re > 100)
• Conditions for thermal boundary layer:
3
(1) Slender body without flow separation
( 2) High product of Reynolds and Prandtl numbers
( RePr > 100)
ρV∞ L c p µ ρ c pV∞ L
Peclet Number = Pe = RePr =
=
(4.1)
k
k
µ
(1) Fluid velocity at surface vanishes
(2) Rapid changes across BL to V∞
(3) Rapid changes temperature across BL from Ts to T∞
(2) Boundary layers are thin:
For air at 10 m/s parallel to 1.0 m long plate, δ = 6 mm
at end
(3) Viscosity plays negligible role outside the viscous BL
(4) Boundary layers exist in both forced and free convection
flows
4
4.2.2 The Governing Equations
Simplified case:
Assumptions:
(1) steady state
(2) two-dimensional
(3) laminar
(4) constant properties
(5) no dissipation
(6) no gravity
Continuity:
∂u ∂v
+
=0
∂x ∂ y
(2.2)
5
x-direction:
∂u
∂u
∂u
∂u
ρ + u + v + w =
∂y
∂z
∂x
∂t
∂ 2u ∂ 2u ∂ 2u
∂p
ρg x − + µ 2 + 2 + 2
∂x
∂y
∂z
∂x
(2.10x)
y-direction:
∂w
∂w
∂w
∂w
+u
+v
+w
ρ
=
∂x
∂y
∂z
∂t
∂ 2w ∂ 2w ∂ 2w
∂p
ρg z − + µ 2 + 2 + 2
∂z
∂y
∂z
∂x
(2.10y)
Energy:
∂ 2T ∂ 2T
∂T
∂T
+v
ρ cΡ u
= k 2 + 2
∂y
∂y
∂x
∂x
(2.19)
6
4.2.3 Mathematical Simplification
4.2.4 Simplification of the Momentum Equations
(i) Intuitive Arguments
Two viscous terms in (2.10x):
2
2
∂ u ∂ u
+ 2
2
∂x
∂y
is one smaller than the other?
7
Insect dilemma: Too windy at position 0, where to go?
Move to position 4!
Conclusion:
Changes in u with respect to y are more
pronounced than changes with respect to x
2
2
• Neglect
2
∂ u
∂ u
<< 2
2
∂x
∂y
(4.2)
∂ u in (2.10x)
∂x 2
Pressure terms in (2.10x) and (2.10y):
• Slender body
• Streamlines are nearly parallel
• Small vertical velocity
8
∂p
≈0
∂y
∴
(4.3)
p depends on x only, i.e. p ≈ p(x)
∂p dp dp∞
≈
≈
∂x dx dx
(4.4)
(4.2) and (4.4) into (2.10x) gives:
Boundary layer x-momentum equation
∂u
∂u
1 dp∞
∂ 2u
u +v
=−
+ν 2
∂x
∂y
ρ ∂x
∂y
(4.5)
• Continuity equation (2.2) and the x-momentum boundary
layer equation (4.5) contain three unknowns: u, v, and p∞
9
• p∞ is pressure at edge of BL (y = δ), obtained from
solution of inviscid flow outside BL
(ii) Scale Analysis
• Use scaling to arrive at BL approximations.
• Assign a scale to each term in an equation
Slender body
Free stream velocityV∞
Length L
BL thickness δ
Postulate:
δ
L
<< 1
(4.6)
10
If (4.6) is valid, we pose three questions:
(1) What terms in the governing equations can be dropped?
(2) Is normal pressure gradient negligible compared to axial
pressure gradient?
(3) Under what conditions is (4.6) valid?
Assign scales:
u ∼ V∞
(4.7a)
y ∼δ
(4.7b)
x∼L
(4.7c)
Apply (4.7) to continuity (2.2)
∂v
∂u
=−
∂x
∂y
11
Using (4.7)
Solving for v
V∞
∼
L
δ
v
v ∼ V∞
δ
L
Conclusion: v << V∞
Order of magnitude of inertia and viscous terms
x-momentum equation (2.10x)
• First inertia term:
∂u
V∞
u ∼ V∞
∂x
L
• Second inertial term:
(4.7d)
(a)
∂u
V∞
v
∼v
∂y
δ
12
Use (4.7d)
∂u
V∞
v
∼ V∞
∂y
L
(b)
Conclusion: 2 inertia terms are of the same order
Examine 2 viscous terms in (2.10x)
• First viscous term:
2
• Second viscous term:
Conclusion:
∴ Neglect
∂ u
V∞
2 ∼
∂x
L2
(c)
∂ 2u
V∞
2 ∼
∂y
δ2
(d)
∂ 2u
∂ 2u
<< 2
2
∂x
∂y
(4.2)
∂ 2 u / ∂x 2 in (2.10x)
13
Examine 2 viscous terms in (2.10y)
2
2
∂ v
∂ v
2 <<
∂x
∂y 2
(4.8)
Simplify (2.10x) and (2.10y) Using (4.2) and (4.8)
∂u
∂u
1 ∂p
∂ 2u
u +v
=−
+ν 2
∂x
∂y
ρ ∂x
∂y
(4.9x)
1 ∂p
∂v
∂v
∂ 2v
u +v
=−
+ν 2
∂x
∂y
ρ ∂y
∂y
(4.9y)
This answers first question
• Second question: pressure gradient
Scale
∂p
∂p
and
∂x
∂y
14
Balance axial pressure with inertia in (4.9x)
∂p
∂u
∼ ρu
∂x
∂x
Scale using (4.7)
(e)
∂p
V∞2
∼ρ
∂x
L
Balance pressure with inertial in (4.9y)
∂p
V∞2 δ
∼ρ
∂y
L L
(f)
Compare (e) and (f) using (4.6)
∂p
∂p
<<
∂y
∂x
(4.10)
15
Since
p = p( x , y )
∂p
∂p
dp = dx + dy
∂x
∂y
or
dp ∂p (∂p / ∂y ) dy
=
1+
dx ∂x (∂p / ∂x ) dx
Scale
dy
dx
dy δ
∼
dx L
(4.11)
(g)
(e)-(g) into (4.11)
dp ∂p
2
=
1 + (δ / L)
dx ∂x
dp ∂p
≈
dx ∂x
[
Invoke(4.6)
]
(h)
(i)
16
Conclusion
Boundary layer pressure depends on x only.
Variation with y is negligible
∴ Pressure p(x) inside BL = pressure
p∞ ( x ) at edge
p( x , y ) ≈ p∞ ( x )
∴
(j)
∂p dp∞
≈
∂x dx
(4.12)
∂u
∂u
1 dp∞
∂ 2u
u +v
=−
+ν 2
∂x
∂y
ρ dx
∂y
(4.13)
(4.12) into (4.9x)
(4.13) is x-momentum eq. for BL flow. Result is based on key
assumption that δ / L << 1.
17
• Third question: condition for validity of (4.6)
δ
L
<< 1
(4.6)
Balance inertia with viscous force in (4.13)
Inertia:
∂u
V∞
u ∼ V∞
∂x
L
(a)
Viscous:
∂ 2u
V∞
ν 2 ∼ν 2
∂y
δ
(b)
Equate
Rearrange
V∞2
V∞
∼ν 2
L
δ
δ
L
∼
ν
V∞ L
(4.14a)
18
or
δ
1
∼
L
ReL
where
ReL =
δ
L
<< 1 when
V∞ L
ν
(4.14b)
(4.15)
Re L >> 1
Generalized (4.14)
δ
1
∼
x
Rex
(4.16)
19
4.2.5 Simplification of the Energy Equation
Simplify (2.19)
∂ 2T ∂ 2T
∂T
∂T
ρ cΡ u
+v
= k 2 + 2
∂y
∂y
∂x
∂x
(2.19)
(i) Intuitive Arguments
Two conduction terms in (2.19):
∂ 2T ∂ 2T
+ 2
2
∂x
∂y
is one smaller than the other?
20
Insect dilemma: Too hot at position 0, where to go?
Move to position 2!
Conclusion:
Changes in T with respect to y are more
pronounced than with respect to x
∂ 2T
∂ 2T
<< 2
2
∂x
∂y
(4.17)
21
∂ 2T
• Neglect
in (2.19):
∂x 2
∂T
∂T
∂ 2T
+v
u
=α 2
∂x
∂y
∂y
(4.18)
(4.18) is the boundary layer energy equation.
(ii) Scale Analysis
• Use scaling to arrive at BL approximations
• Assign a scale to each term in an equation
Slender body
Free stream velocity V∞
Free stream temperature T∞
Length L
BL thickness δ t
22
Postulate:
δt
L
<< 1
(4.19)
If (4.19) is valid, we pose two questions:
(1) What terms in (2.19) can be dropped?
(2) Under what conditions is (4.19) valid?
• Answer first question
Assign scales:
y ∼ δt
(4.20)
∆T ∼ Ts − T∞
(4.21)
x∼L
(4.7b)
Scales for u and v depend on whether δ t is larger or smaller
than δ .
23
Two cases, Fig. 4.4:
Case (1): δ t > δ
u ∼ V∞
(4.22)
Scaling of continuity:
v ∼ V∞
δt
(4.23)
L
Scales for convection terms in (2.19):
24
Use (4.7b) and (4.20-4.23)
and
∂T
∆T
u
∼ V∞
∂x
L
∂T
∆T
v
∼ V∞
L
∂y
(a)
(b)
Conclusion: the two terms are of the same order
Scale for conduction terms:
2
and
∂ T ∆T
∼ 2
2
∂x
L
∂ 2T ∆T
∼ 2
2
∂y
δt
δt
Compare (c) with (d), use :
<< 1
L
(c)
(d)
25
∴
∂ 2T
∂ 2T
<< 2
2
∂x
∂y
(e)
∴ Energy equation simplifies to
∂T
∂T
∂ 2T
u
+v
=α 2
∂x
∂y
∂y
(4.18)
Second question: Under what conditions is (4.19) valid?
δt
L
<< 1
(4.19)
Balance between convection and conduction:
∂T
∂ 2T
u
∼α 2
∂x
∂y
Scaling
∆T
∆T
V∞
∼α 2
L
δt
26
or
δt
L
∼
or
α
V∞ L
δt
k
∼
L
ρc pV∞ L
or
δt
Conclusion:
1
∼
L
PrReL
δt
L
<< 1 when
PrReL >> 1
(4.24)
(4.25)
Define Peclet number Pe
Pe = PrReL
Example: For Pe = 100,
δt
L
(4.26)
∼ 0.1
27
• When is δ t > δ ?
• Take ratio of (4.24) to (4.14b)
1
δt
∼
Pr
δ
(4.27)
∴ Criterion for : δ t > δ
δ t > δ when
Pr << 1
(4.28)
Case (2): δ t < δ Fig. 4.4
V∞
u
δt
δ
28
• u within the thermal boundary layer is smaller than free
stream velocity
• Similarity of triangles
Scaling of continuity
δt
u ∼ V∞
δ
(4.29)
δ t2
v ∼ V∞
Lδ
(4.30)
Use (4.29), (4.30) and follow procedure of case (1):
conclusion:
(1) The two terms are of the same order
(2) Axial conduction is negligible compared to normal
conduction
Second question: Under what conditions is (4.19) valid?
29
Balance between convection and conduction:
2
∂T
∂ T
u
∼α 2
∂x
∂y
Use (4.29) for u, scale each term
V∞
δ t ∆T
∆T
∼α 2
δ L
δt
or
3
( δ t / L)
However
δ
k
δ
∼
ρ c pV∞ L L
1
∼
L
Re L
(f)
(4.14b)
Substitute into (f)
30
δt
L
1
∼
Pr
1/3
(4.31)
Re L
Conclusion:
δt
<< 1 when Pr1/3 ReL >> 1
L
• When is δ t < δ ?
(4.32)
• Take ratio of (4.31) to (4.14b)
δt
1
∼ 1/3
δ
Pr
(4.33)
∴ Criterion for : δ t < δ
δ t < δ when Pr 1/3 >> 1
(4.34)
31
4.3 Summary of Boundary Layer Equations
for Steady Laminar Flow
Assumptions:
(1) Newtonian fluid
(2) two-dimensional
(3) negligible changes in kinetic and potential energy
(4) constant properties
• Assumptions leading to boundary layer model
(5) slender surface
(6) high Reynolds number (Re > 100)
(7) high Peclet number (Pe > 100)
32
• Introduce additional simplifications:
(8) steady state
(9) laminar flow
(10) no dissipation (Φ = 0)
(11) no gravity and
(12) no energy generation ( q′′′ = 0 )
Governing boundary layer equations:
Continuity:
∂u ∂v
+
=0
∂x ∂y
(2.2)
1 dp∞
∂u
∂u
∂ 2u
u +v
=−
+ν 2
∂x
∂y
ρ dx
∂y
(4.13)
x-Momentum:
33
Energy:
∂T
∂T
∂ 2T
+v
=α 2
u
∂x
∂y
∂y
(4.18)
Note the following:
(1) Continuity is not simplified for boundary layer flow
(2) Pressure in (4.13) is obtained from inviscid solution
outside BL. Thus (2.2) and (4.13) have two
unknowns: u and v
(3) To include buoyancy, add ρ β g(T − T∞ )to right (4.13)
(4) Recall all assumptions leading the 3 equations
34
4.4 Solutions: External Flow
• Streamlined body in an infinite flow
• Examine thermal interaction
• Need temperature distribution T
• Temperature depends on velocity distribution
• For constant properties, velocity distribution is
independent of temperature
4.4.1 Laminar Boundary Layer Flow over
Semi-infinite Flat Plate: Uniform Surface
Temperature
35
• Plate is at temperature Ts
• Upstream temperature is T∞
• Upstream velocity uniform and parallel
• For assumptions listed in Section 4.3 the continuity,
momentum and energy are given in (2.2), (4.13) and (4.18)
• Transition from laminar to turbulent at:
Ret = V∞ xt /ν ≈ 500,000
36
(i) Velocity Distribution
Find:
• Velocity distribution
• Boundary layer thickness δ ( x )
• Wall shearing stress τ o ( x )
(a) Governing equations and boundary conditions:
Continuity and x-momentum:
∂u ∂v
+
=0
∂x ∂y
2
∂u
∂u
1 dp∞
∂ u
u +v
=−
+ν 2
∂x
∂y
ρ dx
∂y
(2.2)
(4.13)
The velocity boundary conditions are:
u( x ,0) = 0
v ( x ,0 ) = 0
(4.35a)
(4.35b)
37
u( x , ∞ ) = V∞
(4.35c)
u(0, y ) = V∞
(4.35d)
(b) Scale analysis: Findδ ( x )and τ o ( x )
Result of Section 4.2.4:
δ
1
∼
x
Re x
Wall stress τ o :
τ xy = τ yx
∂v ∂u
= µ +
∂x ∂y
(4.16)
(2.7a)
At wall y = 0 , v ( x ,0) = 0
∂u( x ,0)
τo = µ
∂y
(4.36)
Scales for u and y :
38
u ∼ V∞
x∼L
(4.7a)
(4.7c)
(4.36) is scaled using (4.7)
τo ∼ µ
Use (4.16) for δ
V∞
δ
V∞
τo ∼ µ
Rex
x
(a)
(b)
Friction coefficient C f :
τo
Cf =
(1 / 2) ρ V∞2
(4.37a)
Use (b) for τ o
1
Cf ∼
Rex
(4.37b)
39
(c) Blasius solution: similarity method
• Solve (2.2) and (4.13) for the u and v
• Equations contain 3 unknowns: u, v, and p∞
• Pressure is obtained from the inviscid solution outside BL
Inviscid solution:
• Uniform inviscid flow over slightly curved edge BL
• Neglect thickness δ
• Model: uniform flow over a flat plate of zero thickness
• Solution:
u = V∞ , v = 0, p = p∞ = constant
(4.38)
Thus the pressure gradient is
dp∞
=0
dx
(4.39)
40
(4.39) into (4.13)
∂u
∂u
∂ 2u
u +v
=ν 2
∂x
∂y
∂y
(4.40)
• (4.40) is nonlinear
• Must be solved simultaneously with continuity (2.2)
• Solution was obtained by Blasius in 1908 using
similarity transformation:
Combine x and y ito a single variable η ( x , y )
V∞
η( x , y ) = y
νx
(4.41)
u
Postulate that
depends on η ( x , y ) only
V∞
41
u df
=
V∞ dη
(4.42)
f = f (η) to be determined
NOTE:
(1) Including V∞ / ν in definition of η , is for convenience
only
(1) η ( x , y ) in (4.41) is arrived at by formal procedure
Continuity (2.2) gives v:
∂v
∂u
=−
∂x
∂y
Multiplying by dy, integrate
v = −∫
∂u
dy
∂x
(a)
42
Use (4.41) and (4.42) to express dy and ∂u / ∂x in terms of
the variable η
V∞
dy =
dη
νx
Chain rule:
(b)
∂u du dη
=
∂x dη dx
Use (4.41) and (4.42) into above
2
∂u
V∞ d f
=−
η
2
∂x
2 x dη
(c)
(b) and (c) into (a)
v
1 ν
=
V∞ 2 V∞ x
∫
2
d f
η dη
2
dη
43
Integration by parts gives
v
1 ν df
− f
=
η
V∞ 2 V∞ x dη
(4.43)
• Need function f (η ) , use momentum equation
First determine ∂u / ∂y and ∂ 2 u / ∂y 2
∂u du dη
d 2 f V∞
=
= V∞
∂y dη dy
dη 2 ν x
(d)
d 3 f V∞
∂ 2u
= V∞
2
dη 3 ν x
∂y
(e)
(4.42), (4.43) and (c)-(e) into (4.40)
44
d3 f
d2 f
2 3 + f (η ) 2 = 0
dη
dη
(4.44)
Partial differenti al equations are transformed
into an ordinary differenti al equation
NOTE: x and y are eliminated in (4.44)
• Transformation of boundary conditions
df (∞ )
=1
dη
(4.45a)
f ( 0) = 0
(4.45b)
df (0)
=0
dη
(4.45c)
45
df (∞ )
=1
dη
(4.45d)
Equations (4.44) is third order.
How many boundary conditions?
• Difficulty: (4.44) is nonlinear
• Solution by power series (Blasius)
• Result: Table 4.1
46
Table 4.1 Blasius solution [1]
V∞
η=y
vx
0.0
0.4
0.8
2.4
2.8
3.2
3.6
4.0
4.4
4.8
5.0
5.2
5.4
5.6
f
df
u
=
d η V∞
d3 f
dη 3
0.0
0.02656
0.10611
0.92230
1.23099
1.56911
1.92954
2.30576
2.69238
3.08534
3.28329
3.48189
3.68094
3.88031
0.0
0.13277
0.26471
o.72899
0.81152
0.87609
0.92333
0.95552
0.97587
0.98779
0.99155
0.99425
0.99616
0.99748
0.33206
0.33147
0.32739
0.22809
0.18401
0.13913
0.09809
0.06424
0.03897
0.02187
0.01591
0.01134
0.00793
0.00543
47
• Find δ ( x ) wall stress τ o ( x )
• Define δ as the distance y from the plate where
u/V∞ = 0.994, Table 4.1 gives
δ = 5.2
or
νx
V∞
δ
5.2
=
x
Re x
Scaling result:
δ
(4.46)
1
∼
x
Re x
(4.16)
∂u( x ,0)
τo = µ
∂y
(4.36)
• Wall stress τ o : use
48
(d) into (4.36), use Table 4.1
2
τ o = µ V∞
V∞ d f ( 0)
V∞
= 0.33206 µV∞
2
ν x dη
νx
(4.47)
Friction coefficient C f : (4.47) into (4.37a)
0.664
Cf =
Re x
(4.48)
Scaling result:
1
Cf ∼
Rex
(4.37b)
49
(ii) Temperature Distribution
• Isothermal semi-infinite plate
• Determine: δ t , h(x) and Nux
• Need temperature distribution
(a) Governing equation and boundary conditions
Assumption: Listed in Section 4.3
50
Energy equation
∂T
∂T
∂ 2T
+v
=α 2
u
∂x
∂y
∂y
(4.18)
The boundary condition are:
T ( x ,0) = Ts
(4.49a)
T ( x , ∞ ) = T∞
(4.49b)
T (0, y ) = T∞
(4.49c)
(b) Scale analysis: δ t , h(x) and Nux
From Section 4.2.5: Set L = x (4.24) and (4.31)
Case (1): δ t > δ ( Pr <<1)
δt
1
∼
x
PrRe x
(4.50)
51
Case (2): δ t < δ (Pr >>1)
δt
x
1
∼
1/3
Pr
(4.51)
Re x
Heat transfer coefficient h(x)
∂ T ( x ,0 )
∂y
h = −k
Ts − T∞
(1.10)
Use scales of (4.20) and (4.21) into above
h∼
k
δt
(4.52)
Where δ t is given by (4.50) and (4.51).
52
Case (1): δ t > δ ( Pr <<1), (4.50) into (4.52)
x
h∼
PrRe x ,
k
Local Nusselt number Nux
(4.53) into (4.54)
for Pr <<1
hx
Nux =
k
Nux ∼ PrRe x
,
for Pr <<1
(4.53)
(4.54)
(4.55)
Case (2): δ t << δ ( Pr >>1). Substituting (4.51) into (4.52)
k 1/3
h ∼ Pr
Re x , for Pr >>1
x
(4.56)
Nusselt number:
Nux ∼ Pr
1/3
Re x
, for Pr >>1
(4.57)
53
(c) Pohlhausen’s solution: T(x,y), δ t , h(x), Nux
• Energy equation (4.18) is solved analytically
• Solution by Pohlhausen (1921) using similarity
transformation
• Defined θ
(4.58) into (4.18)
T − Ts
θ=
T∞ − Ts
(4.58)
2
B.C.
∂θ
∂θ
∂θ
u
+v
=α 2
∂x
∂y
∂y
(4.59)
θ ( x ,0) = 0
θ ( x ,0 ) = 1
(4.60a)
θ ( x ,0 ) = 1
(4.60c)
(4.60b)
54
• Solve (4.59) and (4.60) using similarity
• Introduce transformation variable η
V∞
η( x , y ) = y
νx
(4.41)
Assume
θ ( x , y ) = θ (η )
• Blasius solution gives u and v
u df
=
V∞ dη
(4.42)
v
1 ν df
− f
=
η
V∞ 2 V∞ x dη
(4.43)
(4.41)-(4.43) into (4.59) and noting that
55
∂θ dθ ∂η
η dθ
=
=−
∂x dη ∂x
2 x dη
∂θ dθ ∂η
V∞ dθ
=
=
∂ y dη ∂ y
ν x dη
∂ 2θ V∞ d 2θ
=
2
ν x dη 2
∂y
(4.59) becomes
2
d θ Pr
dθ
=0
f (η )
+
2
2
dη
dη
(4.61)
Result:
Partial differenti al equation is transformed
into an ordinary differenti al equation
56
NOTE:
(1) One parameter: Prandtl number Pr
(2) (4.61) is linear, 2nd order ordinary D.E.
(3) f (η ) in (4.61) represents the effect motion
Transformation of B.C.:
θ (∞ ) = 1
θ ( 0) = 0
θ (∞ ) = 1
Solution: Separate variables, integrate twice, use B.C.
(4.62) (Details in Appendix)
Pr
∞
2
d f )
dη
2
η dη
θ (η ) = 1 −
Pr
∞
d 2 f
dη
2
0
dη
(4.62a)
(4.62b)
(4.62c)
∫
(4.63)
∫
57
Surface temperature gradient:
dθ (0)
=
dη
[0.332]
∞
∫η
Pr
Pr
(4.64)
d 2 f
2 dη
dη
• Integrals are evaluated numerically
d2 f
•
is obtained from Blasius solution
2
dη
• Results are presented graphically in Fig. 4.6
58
1.0
100 10
1 0.7(air )
0.8
T − Ts
T∞ − Ts
0.1
0.6
0.4
Pr = 0.01
0.2
0
2
4
6
8
10
12
14
V∞
η=y
νx
Fig. 4.6 Pohlhausen's solution
• Determine: δ t , h(x) and Nux
• Fig. 4.6 gives δ t . At y = δ t , T ≈ T∞ , or
59
T − Ts
θ=
≈ 1 , at y = δ t
T∞ − Ts
(4.65)
• Fig. 4.6 shows that δ t ( x ) depends on Pr
• Local heat transfer coefficient h(x): use (1.10)
where
∂ T ( x ,0 )
∂y
h = −k
Ts − T∞
(1.10)
∂T ( x ,0) dT dθ (0) ∂η
=
∂y
dθ dη ∂ y
Use (4.41) and (4.58) into above
V∞ dθ (0)
∂ T ( x ,0 )
= (T∞ − Ts )
∂y
ν x dη
60
Substitute into (1.10)
V∞ dθ ( 0)
h( x ) = k
ν x dη
(4.66)
• Average heat transfer coefficient:
1
h=
L
L
∫ h( x )dx
(2.50)
0
Use (4.66) and integrate
k
dθ ( 0 )
h =2
Re L
dη
L
(4.67)
• Local Nusselt number: (4.66) into (4.54)
dθ ( 0 )
Nu x =
Re x
dη
(4.68)
61
• Average Nusselt number:
dθ ( 0 )
NuL = 2
Re L
dη
• Total heat transfer rate qT :
(4.69)
Plate length L and width W. Apply Newton’s law
qT =
∫
L
h( x )(Ts − T∞ )W dx =
0
L
(Ts − T∞ )W
or
∫ h( x )dx = (T
s
− T∞ )WL h
0
qT = (Ts − T∞ ) A h
s
(4.70)
Heat transfer coefficient and Nusselt number
dθ (0)
depend on surface temperature gradient
dη
62
dθ (0)
•
depends on Pr
dη
• It is determined from (4.64)
Table 4.2
dθ ( 0 )
Pr
dη
0.001
0.0173
• Values in Table 4.2
• Approximate values of
0.01
0.0516
0.1
0.140
dθ ( 0 )
are given by:
dη
0.5
0.259
0.7
0.292
1.0
0.332
7.0
0.645
10.0
0.730
15.0
0.835
50
1.247
100
1.572
1000
3.387
63
dθ ( 0 )
1/ 2
= 0.564 Pr
,
dη
dθ ( 0 )
1/ 3
= 0.332 Pr
,
dη
Pr < 0.05
(4.71a)
0.6 < Pr < 10
(4.71b)
dθ ( 0 )
= 0.339 Pr 1 / 3 ,
dη
Pr >10
(4.71c)
• Compare with scaling:
• Two cases: Pr << 1 and Pr >> 1
Combine (4.71a) and (4.71c) with (4.68)
Nux = 0.564 Pr 1 / 2 Re x , for Pr < 0.05
Nux = 0.339 Pr
1/ 3
Re x , for Pr > 10
(4.72a)
(4.72c)
64
Scaling results:
Nux ∼ PrRe x
Nux ∼ Pr
1/3
,
Re x
for Pr <<1
(4.55)
, for Pr >>1
(4.57)
• Fluid properties: Evaluated at the film temperature T f
Ts + T∞
Tf =
2
(4.73)
65
4.4.2 Applications: Blasius Solution,
Pohlhausen’s Solutions and Scaling
• Three examples
Example 4.1: Insect in Search of Advice
• Air at 30oC,
V∞ = 4 m/s
• Insect at 0
• Determine velocity u at locations 0, 1, 2, 3, 4.
• Is insect inside BL?
(1) Observations.
• External forced convection boundary layer problem
66
• Changes in velocity between 1 and 3 should be small
compared to those between 2 and 4
• Location 4 should have the lowest velocity
• If the flow is laminar Blasius applies
• The flow is laminar if Reynolds number is less than 500,000
(2) Problem Definition. Determine u at the five locations
(3) Solution Plan.
• Check the Reynolds number for BL approximations and
if the flow is laminar
• If laminar, use Blasius solution, Table 4.1, to determine
u and δ
(4) Plan Execution
67
(i) Assumptions. All assumptions leading to Blasius
solution: These are:
• Newtonian fluid
• steady state
• constant properties
• two-dimensional
• laminar flow (Rex < 5×
×105)
• viscous boundary layer flow (Rex > 100)
• (7) uniform upstream velocity
• flat plate
• negligible changes in kinetic and potential energy
• no buoyancy (β = 0 or g = 0)
68
(ii) Analysis
Re x =
V∞ x
ν
(a)
V∞ = upstream velocity = 4 m/s
−6
ν = kinematic viscosity = 16.01 × 10 m2 /s
Transition Reynolds number:
Re xt = 5×
×105
(b)
• Laminar flow if Rex < Re xt
• Viscous BL approximations are valid for
Re x > 100
(c)
At x = 151 mm:
4( m/s )0.151(m )
Re x =
= 37,726
−6
2
16.01 × 10 (m /s )
69
∴BL flow is laminar. Use Blasius solution
Determine
V∞
η=y
νx
5.2
δ
=
x
Re x
(d)
(4.46)
(iii) Computations.
• Calculate η at each location, use Table 4.1 to find u/V∞.
Results:
location
x (m)
y (m)
η
u/V∞
u(m/s)
0
1
2
3
4
0.150
0.151
0.150
0.149
0.150
0.002
0.002
0.003
0.002
0.001
2.581
2.573
3.872
2.59
1.291
0.766
0.765
0.945
0.768
0.422
3.064
3.06
3.78
3.072
1.688
70
• Use (4.46) to determine δ at x = 0.151m and Re x = 37,726
5.2
5 .2
x=
0.151(m ) = 0.004 m = 4 mm
δ=
Re x
37,726
Thus the insect is within the boundary layer
(iv) Checking. Dimensional check:
Equations (a) and (d) are dimensionally correct
Qualitative check: u at the five locations follow expected
behavior
(5) Comments.
• The insect should move to location 4
• Changes in u with respect to x are minor
• Changes in u with respect to y are significant
71
• What is important for the insect is the magnitude of the
velocity vector V = (u2 + v2)1/2 and not u. However, since
v << u in boundary layer flow, using u as a measure of
total velocity is reasonable
Example 7.2: Laminar Convection over a Flat
Plate
• Water
• V∞ = 0.25 m/s
• T∞ = 35°C
• Ts= 85°C
• L = 75 cm
72
[a] Find equation for δ t ( x )
[b] Determine h at x = 7.5 cm and 75 cm
[c] Determine qT for a plate 50 cm wide
[d] Can Pohlhausen's solution be used to q′′ at the trailing
end of the plate if its length is doubled?
(1) Observations
• External forced convection over a flat plate
• δ t ( x ) increases with x
• Newton’s law of cooling gives q′′and qT
• h( x ) decreases with x
• Pohlhausen's solution is applies laminar flow and all
other assumptions made
• Doubling the length doubles the Reynolds number
73
(2) Problem Definition. Determine temperature distribution
(3) Solution Plan
• Compute the Reynolds and Peclet numbers to establish if
this is a laminar boundary layer problem
• Use Pohlhausen's solution to determine δ t , h(x), q′′and qT
(4) Plan Execution
(i) Assumptions. All assumptions leading to Blasius
solution: These are:
• Newtonian fluid
• two-dimensional
• negligible changes in kinetic and potential energy
• constant properties
• boundary layer flow
74
• steady state
• laminar flow
• no dissipation
• no gravity
• no energy generation
• flat plate
• negligible plate thickness
• uniform upstream velocity V∞
• uniform upstream temperature T∞
• uniform surface temperature Ts
• no radiation
75
(ii) Analysis and Computations
• Are BL approximations valid? Calculate the Reynolds and
Peclet. Condition:
Re x > 100 and Pe = Re x Pr > 100
V∞ x
Re x =
(a)
ν
Transition Reynolds number: Re
t
Re x = 5 × 105
(b)
T f = (Ts + T∞ ) / 2
(c)
Properties at T f
Ts = 85oC
T∞ = 35oC
T f = (85+ 35)(oC)/2 = 60oC
76
k = 0.6507 W/m-oC
Pr = 3.0
ν = 0.4748 × 10−6 m2/s.
at x = 7.5 cm Re x and Pe are
Re x =
V∞ x
ν
0.25(m/s )0.075(m )
4
=
= 3.949 × 10
−6
2
0.4748 × 10 (m / s )
Pe = Re x Pr = 3.949 × 104 × 3 = 11.85 × 104
∴ BL approximations are valid, flow is laminar
Pohlhausen's solution is applicable.
[a] Determine δ t : At y = δ t , T ≈ T∞
(
T∞ − Ts )
≈1
θ (ηt ) =
(T∞ − Ts )
77
From Fig. 4.6: Value of ηt at θ (ηt ) = 1and Pr = 3 at is
approximately 2.9
ηt ≈ 2.9 = δ t V∞ /ν x
or
ηt
2.9
2.9
=
=
x
V∞ ν x
Re x
(d)
[b] Heat transfer coefficient:
dθ ( 0 )
:
dη
V∞ dθ ( 0)
h( x ) = k
ν x dη
dθ ( 0 )
= 0.332 Pr 1 / 3 ,
dη
0.6 < Pr < 10
(4.66)
(4.71b)
78
Pr = 3
dθ ( 0 )
1/ 3
= 0.332 (3) = 0.4788
dη
Substituting into (4.66) for x = 0.075 m
W
h = 825.5 2 o
m − C
At x = 0.75 m
W
h = 261 2 o
m − C
[c] Heat transfer rate:
qT = (Ts − T∞ ) A h
s
(4.70)
L = length of plate = 75 cm =0.75 m
W = width of plate = 50 cm = 0.5 m
79
k
dθ ( 0 )
h =2
Re L
L
dη
(4.67)
ReL = 3.949× 105 . Substitute into the above
W
h = 522.1 2 o
m − C
Substitute into (4.70)
qT = 9789 W
[d] Doubling the length of plate:
Re2 L = 2 (3.949 × 105) = 7.898 × 105
∴ Re2 L > Ret
Flow is turbulent, Pohlhausen's solution is not applicable
80
(iii) Checking. Dimensional check:
Reynolds number is dimensionless and that units of h
and are h correct
Qualitative check: As x is increased h decreases
Quantitative check: Computed values of h are within the
range of Table 1.1
(5) Comments
• Check Reynolds number before applying Pohlhausen's
solution
• Velocity boundary layer thickness δ is given by
δ
5.2
=
x
Re x
(4.46)
Compare (d) with equation (4.46):
δt < δ
81
Example 7.3: Scaling Estimate of Heat Transfer
Rate
Use scaling to determine the total heat transfer rate for
conditions described in Example 7.2
(1) Observation
•Newton’s law gives heat transfer rate
• The heat transfer coefficient can be estimated using scaling
(2) Problem Definition.
Determine the heat transfer coefficient h
(3) Solution Plan.
Apply Newton’s law of cooling and use scaling to determine h
(4) Plan Execution
82
(i) Assumptions
• Newtonian fluid
• two-dimensional
• negligible changes in kinetic and potential energy
• constant properties
• boundary layer flow
• steady state
• no dissipation
• no gravity
• no energy generation
• no radiation
(ii) Analysis. Application of Newton’s law of cooling gives
qT = (Ts − T∞ ) A h
s
(4.70)
83
A = surface area = LW, m2
h = average heat transfer coefficient, W/m2-oC
L = length of plate = 75 cm =0.75 m
qT = total heat transfer rate from plate, W
Ts = surface temperature = 85oC
T∞ = free stream temperature = 35oC
W = width of plate = 50 cm = 0.5 m
h by (1.10)
∂ T ( x ,0 )
∂y
h = −k
Ts − T∞
(1.10)
k = thermal conductivity = 0.6507 W/m-oC
84
Follow analysis of Section 4.41, scale of h for Pr >>1
k 1/3
h ∼ Pr
Re x , for Pr >>1
x
V∞ x
and Pr = 3
Re x =
(4.56)
ν
Set h ∼ h , x = L, A = WL and substitute (4.56) into (4.70)
qT ∼ (Ts − T∞ )W k Pr
1/3
(a)
Re L
(iii) Computations
Re L = 3.949 × 105
Substitute into (a)
1/3
qT ∼ (85 − 35)( o C) 0.5(m ) 0.6507( W/m − o C) 3
394900
qT ∼ 14740 W
85
Using Pohlhausen’s solution gives qT = 9789 W
(iv) Checking.
Dimensional Check:Solution (a) is dimensionally correct
(5) Comments.
Scaling gives an order of magnitude estimate of the heat
transfer coefficient. In this example the error using scaling
rate is 50%
86
4.4.3 Laminar Boundary Layer Flow over
Semi-infinite Flat Plate:
Variable Surface Temperature
• Consider uniform flow over plate
• Surface temperature varies with x as:
Ts ( x ) − T∞ = Cx
n
(4.72)
87
• C and n, constants
• T∞ is free stream temperature
• Determine T ( x , y ) , h( x ) , Nu x and qT
• Assumptions: summarized in Section 4.3
(i) Velocity Distribution
• For constant properties velocity is independent of the
temperature distribution
• Blasius solution is applicable:
u df
=
V∞ dη
(4.42)
v
1 ν df
− f
=
η
V∞ 2 V∞ x dη
(4.43)
88
V∞
η( x , y ) = y
νx
(4.41)
(ii) Governing Equations for Temperature Distribution
Based on assumptions OF Section 4.3:
∂T
∂T
∂ 2T
u
+v
=α 2
∂x
∂y
∂y
(4.18)
Boundary condition
T ( x ,0) = Ts = T∞ + Cx
n
(4.73a)
T ( x , ∞ ) = T∞
(4.73b)
T (0, y ) = T∞
(4.73c)
89
(iii) Solution
• Solution to (4.18) is by similarity transformation
Define :θ
Assume
T − Ts
θ=
T∞ − Ts
(4.58)
θ ( x , y ) = θ (η )
(4.75)
Use (4.41)-(4.43), (4.58), (4.72), (4.75), energy (4.18)
transforms to (Appendix C)
d 2θ
df
Pr
dθ
+ nPr (1 − θ ) +
f (η )
=0
2
dη
2
dη
dη
B.C. (4.73):
θ (∞ ) = 1
θ ( 0) = 0
θ (∞ ) = 1
(4.76)
(4.76a)
(4.76b)
(4.76c)
90
• Note: Two B.C. coalesce into one
Heat transfer coefficient and Nusselt number:
Use (1.10)
∂ T ( x ,0 )
∂y
h = −k
Ts − T∞
(1.10)
where
∂T ( x ,0) dT dθ (0) ∂η
=
∂y
dθ dη ∂ y
Use (4.41),(4.58) and (4.72) into the above
∂ T ( x ,0 )
n V∞ dθ ( 0)
= −Cx
∂y
ν x dη
Substitute into (1.10)
91
V∞ dθ (0)
h( x ) = k
ν x dη
(4.78)
• Average heat transfer coefficient: Use (2.50)
1
h=
L
L
∫ h( x )dx
(2.50)
0
Substitute (4.78) into (2.50) and integrate
k
dθ ( 0 )
h =2
Re L
L
dη
(4.79)
• Local Nusselt number: (4.78) into (4.54)
dθ ( 0 )
Nu x =
Re x
dη
(4.80)
• Average Nusselt number:
92
dθ ( 0 )
NuL = 2
Re L
dη
(4.81)
Heat transfer coefficient and Nusselt number
dθ (0)
depend on surface temperature gradient
dη
(ii) Results:
• Equation (4.76) subject to boundary conditions (4.77) is
solved numerically
• Solution depends on two parameters: the Prandtl
number Pr and the exponent n in (4.72) dθ (0) / dη is
presented in Fig. 4.8 for three Prandtl numbers.
93
2.0
30
dθ (0)
dη
10
1.0
Pr = 0.7
0
0.5
1.0
n
1.5
dθ (0) for plate with varying surface temperature
Fig. 4.8
dη
Ts ( x ) − T∞ = C x n
4.4.3 Laminar Boundary Layer Flow over a
Wedge: Uniform Surface Temperature
94
• Symmetrical flow over
a wedge of angle β π
• Uniform surface
temperature
• Uniform upstream
velocity, pressure and
temperature
• Both pressure and velocity
outside the viscous BL vary with distance x along wedge
• For assumptions of Section 4.3, the x-momentum eq. is
2
∂u
∂u
1 dp∞
∂ u
u +v
=−
+ν 2
∂x
∂y
ρ dx
∂y
(4.13)
C is a constant and m describes wedge angle:
95
m=
β
2− β
(4.83)
dp∞
• Apply (4.13) at edge of BL to determine
:
dx
• Flow is inviscid
• v =ν = 0
• u = V∞ ( x )
1 dp∞
∂V∞
−
= V∞
ρ dx
∂x
Substitute into (4.13)
∂u
∂u
∂V∞
∂ 2u
u +v
= V∞
+ν 2
∂x
∂y
∂x
∂y
(4.84)
96
The B.C. are
u( x , ∞ ) = V∞ ( x ) = Cx
m
(4.84a)
u( x ,0) = 0
(4.84b)
v ( x ,0 ) = 0
(4.84c)
(i) Velocity Solution:
• By similarity transformation (follow Blasius approach)
• Define a similarity variable η :
V∞ ( x )
C ( m −1) / 2
=y
x
η( x , y ) = y
νx
ν
(4.86)
• Assume u(x, y) to depend on η :
u
dF
=
V∞ ( x ) dη
(4.87)
97
Continuity (2.2), (4.86) and (4.87) give v
m +1
1 − m dF
v = −V∞ ( x )
F−
η
xV∞ ( x ) 2
1 + m dη
ν
(4.88)
Substitute (4.82) and (4.86)-(4.88) into (4.84)
3
2
2
d F m +1 d F
dF
F
−m +m =0
+
3
2
2
dη
dη
dη
(4.89)
This is the transformed momentum equation B. C. (4.85)
transform to
dF (0)
=0
dη
F ( 0) = 0
dF (∞ )
=1
dη
(4.89a)
(4.89b)
(4.89c)
98
Note the following regarding (4.89) and (4.90):
• x and y do not appear
• Momentum eq. (4.89) is 3rd order non-linear
• Special case: m = β = 0 represents a flat plate
• Setting m = 0 in (4.89) and (4.90) reduces to Blasius
problem (4.44) & (4.45), F (η ) = f (η )
• (4.89) is integrated numerically
• Solution gives F (η ) and dF / dη . These give u and v
(ii) Temperature Solution:
Energy equation:
2
∂θ
∂θ
∂θ
u
+v
=α 2
∂x
∂y
∂y
(4.59)
99
Boundary conditions:
θ ( x ,0) = 0
θ ( x ,0 ) = 1
(4.60a)
θ ( x ,0 ) = 1
(4.60c)
T − Ts
θ=
T∞ − Ts
(4.58)
(4.60b)
where
• Same energy equation and B.C. as the flat plate.
• Is temperature distribution the same?
• Equation (4.59) is solved by similarity transformation.
Assume:
θ ( x , y ) = θ (η )
(4.75)
where
100
V∞ ( x )
C ( m −1) / 2
=y
x
η( x , y ) = y
νx
ν
(4.86)
Substitute (4.86)-(4.88) and (4.75) into (4.59) and (4.60)
2
d θ Pr
dθ
( m + 1)F (η )
=0
+
2
2
dη
dη
θ ( 0) = 0
θ (∞ ) = 1
θ (∞ ) = 1
(4.91)
(4.92a)
(4.92b)
(4.92c)
• Partial differential equations is transformed into
ordinary equation
• Two governing parameters: Prandtl number Pr and the
wedge size m
• (491) a linear second order equation requiring two B.C.
101
• F (η ) in (4.91) represents effect of fluid motion
• B.C. (4.60b) and (4.60c) coalesce into a single condition
• Special case: m = β = 0 represents flat plate. Set m = 0
in (4.91) reduces to Pohlhausen’s problem (4.61)
Solution: (Details in Appendix B)
• Separate variables in (4.91)
• Integrate twice
• Applying B.C. (4.92), gives
(m + 1) Pr η
exp
F
(
)
d
d
−
η
η
η
∫η
∫
0
2
θ (η ) = 1 −
∞
(m + 1) Pr η
F (η )dη dη
∫0 exp −
∫
0
2
∞
(4.93)
102
dθ (0)
Temperature gradient at surface
:
dη
• Differentiate (4.93), evaluate at η = 0
dθ ( 0 )
(m + 1) Pr
F (η )dη dη
= ∫ exp −
∫
0
dη
2
0
∞
η
−1
(4.94)
• F (η ) is given in the velocity solution
• Evaluate integrals in (4.93)&(4.94) numerically
dθ (0)
• Results for
and F ′′( 0) are in Table 4.3
dη
103
dθ ( 0 )
Table 4.3 Surface temperature gradient
and
dη
velocity gradient F ′′(0) for flow over an isothermal wedge
m
wedge angle
. πβ
F ′′(0)
0
0
0.111
0.333
1.0
dθ ( 0 ) / dη
at five values of Pr
0.7
0.8
1.0
5.0
10.0
0.3206
0.292
0.307
0.332
0.585
0.730
π / 5 (36o)
0.5120
0.331
0.348
0.378
0.669
0.851
π / 2 (90o)
0.7575
0.384
0.403
0.440
0.792
1.013
1.2326
0.496
0.523
0.570
1.043
1.344
π
(180o)
• Use Table 4.3 to determine h( x ) and Nu x
104
where
∂ T ( x ,0 )
∂y
h = −k
Ts − T∞
(1.10)
∂T ( x ,0) dT dθ (0) ∂η
=
∂y
dθ dη ∂ y
Use (4.58),(4.75) and (4.86) into above
∂ T ( x ,0 )
V∞ ( x ) dθ (0)
= (T∞ − Ts )
∂y
ν x dη
Substitute into (1.10)
V∞ ( x ) dθ (0)
h( x ) = k
ν x dη
(4.95)
Local Nusselt number: substitute (4.95) into (4.54)
105
dθ ( 0 )
Nu x =
Re x
dη
where
Re x =
xV∞ ( x )
ν
(4.96)
(4.97)
• Key factor in determining h( x ) and Nu x :
dθ (0)
Surface temperature gradient is
, listed in Table 4.3.
dη
106
CHAPTER 5
APPROXIMATE SOLUTIONS:
THE INTEGRAL METHOD
5.1 Introduction
• Why approximate solution?
• When exact solution is:
• Not Available
• Complex
1
• Requires numerical integration
• Implicit
• Approximate solution by integral method:
• Advantages: simple, can deal with
complicating factors
• Used in fluid flow, heat transfer,
mass transfer
2
5.2 Differential vs. Integral Formulation
Differential Formulation
Conservati on laws are applied to
an infinitesm al element dx × dy
• Result: Solutions are exact
3
Integral Formulation
Conservation laws are satisfied
in an average sense
• Result: Solutions are approximate
4
5.3 Integral Method Approximation:
Mathematical Simplification
• Number of independent variables are reduced
• Reduction in order of differential equation
5.4 Procedure
(1) Integral formulation of the basic laws
• Conservation of mass
• Conservation of momentum
• Conservation of energy
5
(2) Assumed velocity and temperature profiles
• Satisfy boundary conditions
• Several possibilities
• Examples: Polynomial, linear, exponential
• Assumed profile has an unknown parameter
or variable
(3) Determination of the unknown parameter
or variable
• Conservation of momentum gives the
unknown variable in the assumed velocity
6
• Conservation of energy gives the unknown variable in
the assumed temperature
5.5 Accuracy of the Integral Method
• Accuracy depends on assumed profiles
• Accuracy is not sensitive to form of profile
• Optimum profile: unknown
7
5.6 Integral Formulation of the Basic Laws
5.6.1 Conservation of Mass
• Boundary layer flow
• Porous curved wall
• Conservation of mass for element δ × dx :
8
dme
mx
δ
dm x
mx +
dx
dx
dmo
Fig. 5.3
dm x
m x + dmo + dme = m x +
dx
dx
dm x
dme =
dx − dmo
dx
(a)
9
dm e = mass from the external flow
dm o = mass through porous wall
m x = mass from boundary layer rate entering
element at x
One-dimensional mass flow rate:
m = ρVA
(b)
Apply (b) to porous side, assume that injected is
the same as external fluid
dmo= ρ vo Pdx
(c)
10
P = porosity
ρ = density
Applying (b) to dy
dm x = ρ udy
Integrating
δ ( x)
mx =
∫ ρ udy
(d)
0
(c) and (d) into (a)
11
d
dme =
dx
ρ udy dx − ρ v o Pdx
δ ( x)
∫0
(5.1)
5.6.2 Conservation of Momentum
Apply momentum theorem to the element δ × dx
∑ Fx = M x (out ) − M x (in)
(a)
∑ Fx = External x-forces on element
12
M x (in) = x-momentum of the fluid entering
M x (out)= x-momentum of the fluid leaving
dp
( p + )dδ
2
pδ
δ
d
pδ + ( pδ )dx
dx
τ o (1 − P )dx
Fig. 5.4 Forces
13
V∞ ( x )dme
Mx
dM x
Mx +
dx
dx
Fig. 5.4 x - Momentum
Fig. 4 and (a):
dp
d
pδ + p + dδ − pδ − ( pδ )dx − τ o (1 − p )dx =
2
dx
dM x
M
+
dx − M x − V∞ ( x )dme
x
dx
14
p = pressure
V∞ = velocity at the edge of the boundary layer
τ o = wall stress
δ ( x)
∫
ρ u2dy
(b)
∂u( x ,0 )
τo = µ
∂y
(c)
Mx =
0
and
15
(b) and (c) into (a)
∂ u ( x ,0 )
dp
(
)
−δ
=
− µ 1− P
∂y
dx
d
dx
δ (x)
δ (x)
∫
∫
ρ u 2 dy − V ∞ ( x ) d
0
dx
ρ udy − V ∞ ( x )ρ P v o
0
• x-momentum of dmo ?
• Shear force on slanted surface?
• (5.2) applies to laminar and turbulent flow
16
• Curved surface: V∞ ( x ) and p( x )
• (5.2) is the integral formulation of conservation
of momentum and mass
• (5.2) is a first order O.D.E. with x as the
independent variable
Special Cases:
(i) Case 1: Incompressible fluid
dp dp ∞
≈
dx dx
(4.12)
For boundary layer flow
17
∂u
∂u
1 dp∞
∂ 2u
=−
u +v
+ν 2
ρ ∂x
∂x
∂y
∂y
Apply (4.5) at y = δ where
u = V∞
dp dp∞
dV∞
= − ρ V∞ ( x )
≈
dx dx
dx
(5.3) into (5.2), constant
(4.5)
(5.3)
ρ
d V∞
∂ u( x , 0 )
=
− ν (1 − P )
δ V∞ ( x )
dx
∂y
d
dx
δ ( x)
δ ( x)
∫
∫
d
u dy − V∞ ( x )
dx
0
2
0
(5.4)
udy − V∞ ( x ) Pv o
18
(ii) Case 2: Incompressible fluid and impermeable
flat plate
Flat plate, (5.3)
dV ∞ dp dp ∞
=0
=
≈
dx
dx
dx
(d)
Impermeable plate
v o = 0, P = 0
(e)
(d) and (e) into (5.4)
∂u( x ,0 )
d
v
= V∞
∂y
dx
δ (x)
∫0
d
udy −
dx
δ (x)
∫0
u 2dy
(5.5)
19
5.6.3 Conservation of Energy
Neglect:
(1) Changes in kinetic and potential energy
(2) Dissipation
(3) Axial conduction
20
Conservation of energy element δ t × dx
dEe
δt
dE x
Ex +
dx
dx
Ex
dx
dEc
dEo
Fig. 5.6
dE x
E x + dEc + dEo + dEe = E x +
dx
dx
21
Rearranging
dE x
dEc =
dx − dEe − dEo
dx
Fourier’s law
∂T ( x ,0 )
dEc = − k (1 − P )
dx
∂y
(b)
Energy with mass dm e
dEe = c pT∞ dme
22
Use (5.1) for dme
d
dEe = c pT∞
dx
ρ udy dx − c pT∞ ρ v o Pdx
δ ( xt )
∫0
(c)
Energy of injected mass
dE o = ρ c pTo v o pdx
(d)
Energy convected with fluid
Ex =
δ t ( x)
∫0
ρ c p uT dy
(e)
23
(b)-(e) into (a)
∂T ( x ,0 ) d
− k (1 − P )
=
∂y
dx
d
c pT∞
dx
δ t ( x)
∫
ρ c p uTdy −
0
δ t ( x)
∫
(5.6)
ρ udy − ρ c p v o P(To − T∞ )
0
Note
(1) (5.6) is integral formulation of conservation
energy
(2) (5.6) is a first order O.D.E. with x as the
independent variable
24
Special Case: Constant properties and
impermeable flat plate
Simplify (5.6)
−α
∂ T ( x ,0 ) d
=
∂y
dx
δ t ( x)
∫
u(T − T∞ )dy
(5.7)
0
25
5.7 Integral Solutions
5.7.1 Flow Field Solution:
Uniform Flow over a Semi-Infinite Plate
• Integral solution to Blasius problem
• Integral formulation of momentum (5.5)
26
∂u( x ,0 )
d
= V∞
v
∂y
dx
δ (x)
∫0
d
udy −
dx
δ (x)
∫0
u2dy
(5.5)
• Assumed velocity profile u( x , y )
u( x , y ) =
N
∑
an ( x ) y n
(5.8)
n=0
Example, a third degree polynomial
u( x , y ) = a0 ( x ) + a1 ( x ) y + a2 ( x ) y 2 + a3 ( x ) y 3
(a)
27
Boundary conditions give a n
(1) u( x ,0 ) = 0
(3) ∂u( x , δ ) ≅ 0
∂y
(2) u( x , δ ) ≅ V∞
2
∂
u( x ,0 )
(4)
=0
2
∂y
Note:
(1) B.C. (2) and (3) are approximate. Why?
(2) B.C. (4): set
y = 0 in the x-component of the
Navier equations, (2.10x)
(3) 4 B.C. and (a) give:
28
a 0 = a 2 = 0 , a1 =
3 V∞
1 V∞
, a3 =
2 δ
2δ 3
u 3 y 1 y
= −
V∞ 2 δ 2 δ
3
(5.9)
Assumed velocity is in terms of a
single unknown variable δ ( x )
(5.9) into (5.5), evaluate integrals
3
1
39
2 dδ
vV∞ =
V∞
2
δ 280
dx
(b)
29
(b) is first order O.D.E. in δ ( x )
140 v
δ dδ =
dx
13 U ∞
Integrate, use B.C. δ (0) = 0
140 v x
∫0 δ dδ =
∫0 dx
13 U ∞
δ
Evaluate integrals
δ
280 / 13 4.64
=
=
x
Re x
Re x
(5.10)
30
(5.10) into (5.9) gives u( x , y )
Friction coefficient C f :Use (4.36) and (4.37a)
∂u
µ ( x ,0 )
3v
τo
∂y
Cf =
=
=
2
2
V∞δ ( x )
ρV∞ / 2
ρV∞ / 2
Use (5.10) to eliminate δ ( x )
Cf =
0.646
Re x
(5.11)
31
Accuracy: Compare (5.10) and (5.11) with Blasius solution
δ
5.2
=
,
x
Re x
Blasius solution
(4.46)
0.664
,
Cf =
Re x
Blasius solution
(4.48)
Note:
(1) Both solutions have same form
(2) Error in δ ( x ) is 10.8%
32
(3) Error in C f is 2.7%
(4) Accuracy of C f is more important than δ ( x )
5.7.2 Temperature Solution and Nusselt Number:
Flow over a Semi-Infinite Plate
(i) Temperature Distribution
• Flat plate
• Insulated section x o
33
• Surface at T s
• Laminar, steady, two-dimensional,
constant properties boundary layer flow
• Determine: δ t , h( x ) , Nu( x )
• Must determine: u( x , y ) and T ( x , y ) (2)
Integral formulation of conservation of energy
∂T ( x ,0 ) d
−α
=
∂y
dx
δ ( x)
∫
ρ c p u(T − T∞ )dy
(5.7)
0
34
• Integral Solution to u( x , y ) and δ ( x ) :
u 3 y 1 y
= −
V∞ 2 δ 2 δ
3
δ
280 / 13 4.64
=
=
x
Re x
Re x
(5.9)
(5.10)
Assumed temperature
T ( x, y ) =
N
∑
bn ( x ) y n
(5.12)
n=0
35
Let
T ( x , y ) = b0 ( x ) + b1 ( x ) y + b2 ( x ) y 2 + b3 ( x ) y 3
(a)
Boundary conditions give bn ( x )
(1) T ( x ,0 ) = Ts
(2) T ( x , δ t ) ≅ T∞
(3) ∂T ( x , δ t ) ≅ 0
∂y
(4) ∂ 2T ( x ,0 )
=0
2
∂y
36
Note
(1) B.C. (2) and (3) are approximate. Why?
(2) B.C. (4): set y = 0 in the x-component of
the energy equation (2.19)
(3) Four B.C. and (a) give:
3
1 b = 0,
b0 = Ts , b1 = (T∞ − Ts ) , 2
δt
2
1
1
b3 = − (T∞ − Ts ) 3
2
δt
37
Therefore
3 y 1 y3
T ( x , y ) = Ts + (T∞ − Ts )
−
3
2 δ t 2 δ t
(5.13)
(5.9) and (5.13) into (5.7) and evaluating the integral
3 T∞ − Ts d
= (T∞ − Ts )V∞δ
α
2
δt
dx
4
3 δ t 2
3 δ t
−
20 δ 280 δ
(5.14)
• (5.14) is simplified for Pr > 1
δt
δ
< 1 , for
Pr > 1
(5.15)
38
Last two term in (5.14):
3 δ t
280 δ
δt
3
<<
δ
20
2
Simplify (5.14)
2
α
d
δ t
10 = V∞
δ
δt
dx δ
Use (5.10) for
(b)
δ
280 ν x
δ =
13 V ∞
(c)
39
(c) into (b)
3
2
δt
δ t d δ t 13 1
+ 4 x
=
δ
δ dx δ 14 Pr
δt
Solve (d) for . Let
δ
δt
r =
δ
(d)
3
(e)
(e) into (d)
4 dr 13 1
r+ x =
3 dx 14 Pr
(f)
40
Separate variables and integrate
3
δt
− 3 / 4 13 1
r = = C ( x)
+
14 Pr
δ
(g)
C = constant. Use boundary condition on δ t
δ t ( xo ) = 0
(h)
13 1 3 / 4
C=−
xo
14 Pr
(i)
Apply (h) to (g)
41
(i) into (h)
δ t 13 1 xo
=
1 −
δ 14 Pr x
Use (c) to eliminate
δ
3 / 4 1 / 3
(5.16)
280 vx
13 V∞
(5.17a)
in (5.16)
13 1 x
δt =
1 − o
14 Pr x
1/ 3
3 / 4
or
x
o
1
−
= 1/3
1/2
x Pr Re x x
δt
4.528
1/ 3
3/ 4
(5.17b)
42
Re x is the local Reynolds
Re x =
V∞ x
(5.18)
ν
(ii) Nusselt Number
Local Nusselt number:
hx
Nu x =
k
(j)
∂T ( x ,0)
−k
∂y
h=
Ts − T∞
(k)
h is given by
43
Use temperature solution (5.13) in (k)
3k
h( x ) =
2δt
(5.19)
Use (5.17a) to eliminate δ t in (5.19)
3/4
k xo
h( x ) = 0.331 1 −
x x
−1 / 3
Pr 1/3 Re x 1/2
(5.20)
1/2
(5.21)
Substitute into (j)
xo
Nux = 0.331 1 −
x
3 / 4 −1 / 3
Pr
1/3
Re x
44
(iii) Special Case: Plate with no Insulated Section
set xo = 0 in above solution
1/ 3
δ t 13 1
=
δ 14 Pr
δt
x
=
0.975
Pr 1 / 3
4.528
=
Pr
1/3
Re x
1/2
(5.22)
(5.23)
45
k 1/3
h( x ) = 0.331 Pr Re x 1/2
x
(5.24)
Nu = 0.331 Pr 1/3 Re x 1/2
(5.25)
x
Accuracy of integral solution:
(1) for Pr = 1, δ t / δ = 1 . Set
Pr = 1 in (5.22)
δt
= 0.975
δ
Error is 2.5%
(2) Compare with Pohlhausen’s solution. For Pr > 10
Nu x = 0.339 Pr 1 / 3 Re x , for Pr > 10
Error is 2.4%
(4.72c)
46
Example 5.1:
Laminar Boundary Layer Flow over a Flat Plate:
Uniform Surface Temperature
Use a linear profiles. Velocity:
u = a0 + a1 y
(a)
Boundary conditions
(1) u( x ,0) = 0 ,
u = V∞
(2) f u( x , δ ) ≈ V∞
y
(b)
δ
47
Apply integral formulation of momentum, (5.5).
Result:
δ
x
=
12
Re x
(5.26)
Temperature profile:
T = b0 + b1 y
(c)
Boundary conditions
(1) T ( x ,0) = Ts ,
(2) T ( x , δ t ) ≈ T∞
T = Ts + (T∞ − Ts )
y
δt
(d)
48
Apply integral formulation of energy, (5.7).
Result:
Nu x = 0.289 Pr
1/3
[
Re x 1 − ( xo / x )
3/4
]
− 1/3
(5.27)
Special case: x o = 0
Nu x = 0.289 Pr 1/3 Re x
Comments
(i) Linear profiles give less accurate results
than polynomials.
(ii) More accurate prediction of Nusselt number
than viscous boundary layer thickness.
49
5.7.3 Uniform Surface Flux
• Flat plate
• Insulated section of length x o
• Plate is heated with uniform flux q ′s′
• Determine h( x ) and Nux
50
q′′ = h( x ) [Ts ( x ) − T∞ ]
(a)
s
or
h( x ) =
qs′′
Ts ( x ) − T∞
Nusselt number:
q′′ x
Nux =
k [Ts ( x ) − T∞ ]
(b)
s
Apply conservation of energy to determine Ts ( x )
∂T ( x ,0 ) d
−α
=
∂y
dx
∫
δt
u(T − T∞ )dy
(5.7)
0
51
Integral solution to u( x , y ) :
u
3 y 1 y
= −
V∞ 2 δ 2 δ
3
(5.9)
Assume temperature profile:
T = b0 + b1 y + b2 y 2 + b3 y 3
(c)
Boundary conditions:
∂T ( x ,0 )
= q′s′
(1) − k
∂y
(2) T ( x , δ t ) ≅ T∞
∂T ( x , δ t )
(3)
≅0
∂y
2
∂
T ( x ,0 )
(4)
=0
2
∂y
52
2
1 y 3 q′s′
T ( x , y ) = T∞ + δ t − y +
2
3δt k
3
(5.29)
set y = 0 to obtain Ts ( x )
2 q′s′
Ts ( x ) = T ( x ,0) = T∞ +
δt
3k
(5.30)
5.30 into (b)
3 x
Nu x =
2 δ t (x)
(5.31)
53
Must determine δ t . Substitute (5.9) and (5.29)
into (5.7)
δt
d
α = V∞
dx
0
∫
3 y 1 y3 2
1 y 3
−
δ t − y +
dy
3
2
3δt
2 δ 2 δ 3
(d)
Evaluating the integrals
3
d 2 1 δ t
1 δ t
= δ t
−
V∞ dx 10 δ 140 δ
α
(e)
For Pr > 1 , δ t / δ < 1
3
1 δt
1 δt
<<
140 δ
10 δ
(f)
54
(f) into (e)
d δ t3
10
=
V∞ dx δ
α
Integrate
α
δ t3
10
x=
+C
V∞
δ
(g)
δ t ( xo ) = 0
(h)
Boundary condition:
Apply (h) to (g)
C = 10
α
V∞
(i)
xo
55
(i) into (g)
α
δ t = 10 ( x − xo )δ
V∞
1/ 3
(j)
Use(5.10) to eliminate δ in (j)
α
280 280 / 13
δ t = 10 ( x − xo )
13
Re x
V∞
x
1/ 3
or
3.594 xo
1−
= 1/3
x Pr Re x
x
δt
1/3
(5.32)
Surface temperature: (5.32) into (5.30)
q′′ x
Ts ( x ) = T∞ + 2.396 s 1 − o
k
x
1/3
x
Pr
1/3
Re1/2
x
(5.33)
56
Nusselt number: (5.32) into (5.31)
x
Nu x = 0.417 1 − o
x
−1/3
Pr 1/3 Re 1/2
x
(5.34)
Special case: x o = 0
q′s′
x
Ts ( x ) = T∞ + 2.396
k Pr 1/3 Re1/2
x
(5.35)
Does Ts ( x ) increase or decrease with distance x?
Nusselt number:
Nu x = 0.417 Pr 1/3 Re1/2
x
(5.36)
Exact solution:
Nu x = 0.453 Pr 1/3 Re1/2
x
(5.37)
57
Example 5.2:
Laminar Boundary Layer Flow over a Flat Plate:
Variable Surface Temperature
• Specified surface temperature
Ts ( x ) = T∞ + C x
58
• Determine the local Nusselt number
(1) Observations
• Determine u( x , y ) and T ( x , y )
• Variable Ts ( x )
• Constant properties: T ( x , y ) is independent of u( x , y )
(2) Problem Definition. Determine u( x, y) and T ( x , y )
(3) Solution Plan
• Start with the definition of Nux
• Apply the integral method to determine T ( x , y )
59
(4) Plan Execution
(i) Assumptions
(1) Steady state
(2) Constant properties
(3) Two-dimensional
×105)
(4) Laminar flow (Rex < 5×
(5)
(6)
(7)
(8)
(9)
Viscous boundary layer flow (Rex > 100)
Thermal boundary layer (Pe > 100)
Uniform upstream velocity and temperature
Flat plate (9)
Negligible changes in kinetic and potential energy
60
(10) Negligible axial conduction
(11) Negligible dissipation
(12) No buoyancy (b = 0 or g = 0)
(ii) Analysis
Nu x =
hx
k
(a)
(1.10) gives h
∂T ( x ,0)
∂y
h=
Ts ( x ) − T∞
−k
(1.10)
To determine T ( x , y ) use (5.7)
61
∂T ( x ,0 ) d
−α
=
∂y
dx
δ t ( x)
∫
u(T − T∞ )dy
(5.7)
0
(5.9) gives u(x,y)
u
3 y 1 y
= −
V∞ 2 δ 2 δ
3
(5.9)
where
280 / 13
280 xν
x=
δ =
13 V∞
Re x
(5.10)
Assume
T ( x , y ) = b0 ( x ) + b1 ( x ) y + b2 ( x ) y 2 + b3 ( x ) y 3
(a)
62
Boundary conditions:
(1) T ( x ,0 ) = Ts ( x )
(2) T ( x , δ t ) ≅ T∞
(3) ∂T ( x , δ t )
≅0
∂y
(4) ∂ 2T ( x ,0 )
=0
2
∂y
3 y 1 y3
−
T ( x, y ) = Ts ( x) + [T∞ − Ts ( x)]
3
2 δ t 2 δ t
(b) into (1.10)
3k
h( x ) =
2 δt
(b)
(c)
63
(c) into (a)
3 x
Nux =
2δt
(d)
• Use (5.7) to determine δ t
• (5.9) and (b) into (5.7), evaluating the integral
3 Ts ( x ) − T∞
=
α
2
δt
d
[Ts ( x ) − T∞ ]V∞δ
dx
(e)
4
3 δ t 2
3 δt
−
20 δ 280 δ
64
For Pr > 1
δt
δ
(5.15)
for Pr > 1
< 1,
Thus
4
3 δt
3 δt
<<
280 δ
20 δ
2
Simplify (e)
d
α
10 [Ts ( x ) − T∞ ] = V∞ [Ts ( x ) − T∞ ]δ
dx
δt
2
δt
δ
(f)
However
Ts ( x ) − T∞ = C x
(g)
65
(5.10) and (g) into (f)
d
13 V∞ 2
α
10 [C x ] = V∞ C x
δt
dx
280 νx
δt
Simplify
3/2
280 α
[ν / V∞ ] x dx = δ t2 dδ t
5
13 ν
Boundary condition on δ t
(h)
:
(i)
δ t ( 0) = 0
Integrate (h) using (i)
1/3
δ t = [10 280 / 13 ] ( Pr )−1/3 (ν x / V∞ )1/2
(j)
(j) into (d)
Nu x = 0.417 Pr 1/3 Re1/2
x
(5.38)
66
(3) Checking
• Dimensional check
• Boundary conditions check
(4) Comments
(i) (5.38) is identical with (5.36) for uniform flux
q′s′
x
Ts ( x ) = T∞ + 2.396
k Pr 1/3 Re1/2
x
(5.35)
Rewrite (5.35)
Ts ( x ) = T∞ + C x
(ii) Use same procedure for other specified Ts ( x )
67
CHAPTER 6
HEAT TRANSFER IN CHANNEL FLOW
6.1 Introduction
• Important factors:
(1) Laminar vs. turbulent flow
transition Reynolds number Re D t is
Re Dt =
where
uD
ν
≈ 2300
(6.1)
D = tube diameter
u = mean velocity
ν = kinematic viscosity
1
(2) Entrance vs. fully developed region
Based on velocity and temperature distribution: two regions:
(i) Entrance region
(ii) Fully developed region
(3) Surface boundary conditions
Two common thermal boundary conditions:
(i) Uniform surface temperature
(ii) Uniform surface heat flux
(4) Objective
Depends on the thermal boundary condition:
(i) Uniform surface temperature. Determine: axial variation of
(1) Mean fluid temperature
(2) Heat transfer coefficient
(3) Surface heat flux
2
(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
• Uniform inlet velocity V i and temperature Ti
• Developing boundary velocity and thermal boundary layers
• Two regions:
r
(1) Entrance region
(2) Fully developed region
Vi
6.2.1 Velocity Field
(1) Entrance Region (Developing Flow,
0 ≤ x ≤ Lh )
x
uc
u
δ
Lh
fully developed
Fig. 6.1
3
• Hydrodynamic entrance region
• Length Lh : hydrodynamic entrance length
• Streamlines are not parallel ( v r ≠ 0)
• Core velocity uc = uc ( x )
(increasing or decreasing with x?)
• Pressure p = p( x ) , ( dp / dx < 0 )
• δ < D/2
r
Vi
x
uc
δ
Lh
(2) Fully Developed Flow Region
u
fully developed
Fig. 6.1
x > Lh : fully developed flow
Streamlines are parallel ( v = 0)
r
For 2 -D, constant ρ : ∂ u / ∂ x = 0
4
6.2.2 Temperature Field
• Entrance Region (Developing
Temperature, 0 ≤ x ≤ Lt ):
r
• Thermal entrance region
Ts
Tc
T
• Length Lt : Thermal entrance length Vi
• Core temperature Tc is uniform
(T c = T i )
Ti
δt < D / 2
x
Ts
δt
Lt
fully
developed
Fig. 6.2
(2) Fully Developed Temperature Region
• x ≥ Lt fully developed temperature
• T = T ( r , x ) or ∂ T / ∂ x ≠ 0
• Dimensionless temperature φ is invariant with x ( ∂φ / ∂x = 0 )
5
6.3 Hydrodynamic and Thermal Entrance Lengths
Lh and Lt are determined by:
(1) Scale analysis
(2) Analytic or numerical methods
6.3.1 Scale Analysis
(1) Hydrodynamic Entrance Length Lh
• Scaling of external flow:
δ
1
∼
x
Re x
r
(4.16)
Vi
Apply (4.16) to flow in tube: at x = Lh ,
δ ∼D
D
∼
Lh
1
Re Lh
x
uc
u
δ
Lh
fully developed
Fig. 6.1
(a)
6
Express Re L in terms Re D
h
Re Lh
(b) into (a)
=
u Lh u D Lh
L
=
= Re D h
v
v D
D
1/ 2
Lh / D
Re D
(2) Thermal Entrance Length Lt
~1
(6.2)
Scale for u: u ~ u (all Prandtl numbers)
Scale of δ t : For external flow
δ t ~ LRe L−1/2 Pr −1/2
(4.24)
Apply (4.24) for flow in tube: L = Lt , δ t ∼ D
−
D ~ L t Ret 1/2 Pr
−1/2
7
(a)
Express Re L in terms Re D
t
Re Lt
u Lt u D Lt
Lt
=
= Re D
=
v
v D
D
(b)
(b) into (a)
1/2
Lt / D
Re D Pr
~1
Lt
~ Pr
Lh
(6.3)
(6.4)
6.3.2 Analytic/Numerical Solutions: Laminar Flow
(1) Hydrodynamic Entrance Length Lh
Lh
= C h Re D
De
(6.5)
8
De = equivalent diameter
4 Af
De =
P
A f = flow area
Table 6.1
Entrance length coefficients
and
Uniform
surface
flux
Uniform surface
temperature
0.056
0.043
0.033
0.090
0.066
0.041
2
0.085
0.057
0.049
4
0.075
0.042
0.054
0.011
0.012
0.008
Geometry
a
b
a
Compare with scaling:
b
1/ 2
C t [1]
Ct
Ch
P = perimeter
C h = coefficient Table 6.1
Lh / D
Re D
Ch
a
~1
(6.2)
b
Rewrite (6.5)
1/2
Lh / D
Re D
1/2
= (C h )
(a)
9
Example: Rectangular channel, a / b = 2, , Table 6.1 gives C h = 0.085
Substitute into (a)
1/ 2
Lh / D
Re D
1/ 2
= (0.085 )
= 0.29
(b)
(2) Thermal Entrance Length Lt
Lt
= Ct PrReD
De
(6.6)
C t is given in Table 6.1
Compare with scaling
1/2
Lt / D
Re D Pr
~1
(6.3)
Rewrite (6.6):
10
1/2
Lt / D
Re D Pr
= C t1/2
(b)
Example: Rectangular channel, a / b = 2, , Table 6.1 gives Ct = 0.049 gives
1/2
Lt / D
Re D Pr
1/ 2
= (0.049 )
(c)
= 0.22
Turbulent flow: Experimental results:
• Lh and Lt are shorter than in laminar flow
Rule of thumb :
Ln
10 <
< 60
D
Ln
40 <
< 100
D
(6.7a)
(6.7b)
11
6.4 Channels with Uniform Surface Heat Flux q ′s′
L
When surface heat flux is uniform
Surface temperature is variable
Tmi
x Tm ( x )
• Section length L
• Inlet temperature: Tmi = Tm ( 0)
• Surface flux q ′s′
q′s′
Fig. 6.3
Determine:
(1) Total heat transfer
(2) Mean temperature variation Tm ( x )
(3) Surface temperature variation Ts ( x )
12
Total heat:
q s = q ′s′ As = q ′s′ P x
(6.8)
As = surface area
P = perimeter
Conservation of energy:
Assumptions:
(1) Steady state
(2) No energy generation
(3) Negligible changes in kinetic and potential energy
(4) No axial conduction
q s = q ′s′ P x = mc p [Tm ( x ) − Tmi ]
or
Tm ( x ) = Tmi
m = mass flow rate
q ′s′ P
+
x
mc p
(6.9)
13
c p = specific heat
(6.9) applies to any region and any flow
(laminar, turbulent or mixed)
Use heat transfer analysis to determine
surface temperature Ts ( x )
Newton’s law of cooling
q ′s′ = h( x )[Ts ( x ) − Tm ( x )]
Solve for Ts ( x )
q ′s′
Ts ( x )= Tm ( x ) +
h( x )
Use (6.9) to eliminate Tm ( x )
14
Ts ( x ) = Tm i
Px
1
+ q′s′
+
c
h
x
(
)
m
p
(6.10)
h(x) is needed in (6.10) to determine Ts ( x )
To determine h(x):
(1) Laminar or turbulent flow?
(2) Entrance or fully developed region?
Example 6.2: Maximum Surface Temperature
• Water flows through tube
• Mean velocity = 0.2 m/s
o
• Tmi = 20 C
15
•
L
Tmo = 80 o C
• D = 0.5 cm
Tmi
• Uniform surface heat flux =
0.6 W/cm2
x
q′s′
D
Tm ( x )
0
q′s′
Ts ( x )
• Fully developed flow at outlet
• Nusselt number for laminar fully developed flow
NuD =
hD
= 4.364
k
(A)
Determine the maximum surface temperature
(1) Observations
• Uniform surface flux
•
Ts = Ts ( x ) , maximum at the outlet
16
• Laminar or turbulent flow? Check
Re D
• Is outlet fully developed? Check Lh and
• Uniform Nusselt number (h is constant)
Lt
• Length of tube section is unknown
(2) Problem Definition
(i) Determine L
(ii) Determine
T s (L)
(3) Solution Plan
(i) Apply conservation of energy
(ii) Compute Re D
(iii) Calculate Lh and Lt
(iv) Apply uniform flux analysis
( v ) If applicable use (A) to determine h
17
(4) Plan Execution
(i) Assumptions
• Steady state
• Constant properties
• Axisymmetric flow
• Uniform surface heat flux
• Negligible changes in kinetic and potential energy
• Negligible axial conduction
• Negligible dissipation
(ii) Analysis
Conservation of energy:
π DLq"s = mc p (Tmo − Tmi )
18
(a)
c p = specific heat, J/kg-oC
D = tube diameter = 0.5 cm = 0.005 m
L = tube length, m
m = mass flow rate, kg/s
Tmi = mean temperature at the inlet = 20oC
Tmo = mean temperature at the outlet = 80oC
q ′s′ = surface heat flux = 0.6 W/cm2 = 6000 W/m2
From (a)
Conservation of mass:
mc p (Tmo − Tmi )
L=
π Dq′′
(b)
m = (π / 4) D2 ρ u
(c)
s
where
19
u = mean flow velocity = 0.2 m/s
ρ = density, kg/m3
Surface temperature: Apply (6.10)
Ts ( x ) = Tm i
Px
1
+ q′s′
+
mc p h( x )
(6.10)
h = local heat transfer coefficient, W/m2-oC
P = tube perimeter, m
Ts ( x ) = local surface temperature, oC
x = distance from inlet of heated section, m
Perimeter P:
P=πD
(d)
Maximum surface temperature: set x = L in (6.10)
20
Ts ( L) = Tm i
PL
1
+ q ′s′
+
mc p h( L)
Determine h(L): Is flow laminar or turbulent? Compute
Re D =
(e)
Re D
uD
(f)
ν
Properties Tm
Tmi + Tmo
Tm =
2
(g)
( 20 + 80)( o C)
T =
= 50o C
2
For water:
c p = 4182 J/kg-oC
k = 0.6405 W/m-oC
21
Pr = 3.57
-6
ν = 0.5537× 10 m2/s
3
ρ = 988 kg/m
Use (g)
Re D =
0.2(m/s)0.005(m)
−6
2
0.5537 × 10 (m /s)
Compute Lh and
= 1806 , laminar flow
Lt using (6.5) and (6.6)
Lh
= C h Re D
De
Lt
= C t PrRe D
De
C h = 0.056 (Table 6.1)
C t = 0.043 (Table 6.1)
(6.5)
(6.6)
22
Lh = 0.056 × 0.005 (m) × 1806 = 0.506 m
Lt = 0.043 × 0.005 (m) × 1806 × 3.57 = 1.386 m
Is L smaller or larger than
Lh and Lt ?
Compute L using (b). Use (c) to compute m
m = 988(kg/m3) 0.2(m/s)π (0.005)2(m2)/4 = 0.00388kg/s
Substitute into (b)
L=
0.00388(kg/s) 4182(J/kg− o C)(80 − 20)( o C)
2
4
2
2
= 10.33 m
0.005(m) 0.6 (W/cm )10 (cm /m )
L is larger than both Lh and Lt .
Flow is fully developed at the outlet
23
Equation (A) is applicable
NuD =
hD
= 4.364
k
(A)
(iii) Computations. Apply (A)
0.6405(W/m − o C)
h(L) = 4.364
= 559 W/m2-oC
0.005(m)
Use (e)
o
2
Ts (L) = 20 C + 6000(W/m )
+
o
2 o
0.00388(kg/s)4182(J/kg− C) 559(W/m − C)
π 0.005(m)10.43(m)
1
Ts ( L) = 90.7o C
(iv) Checking. Dimensional check:
Quantitative checks: (1) Alternate approach: apply Newton’s law at outlet
24
q ′s′ = h ( L )[T s ( L ) − Tmo
]
(i)
solve for T s ( L)
T s ( L) = Tmo
2
4
2
2
q ′s′
0.6(W/cm
)
×
10
(cm
/m
)
+ = 80 (oC) +
= 90.7oC
h
559(W/m 2 − o C)
(2) Compare value of h with Table 1.1
Limiting check: For Tmi
= Tmo , L = 0. Set Tmi = Tmo in (b) gives L = 0.
(3) Comments.
• In laminar flow local h depends on local flow condition: entrance vs.
fully developed
• Check
Re D to determine:
(i) If flow is laminar or turbulent
(ii) Entrance or fully developed
25
6.5 Channels with Uniform Surface Temperature
When surface temperature is uniform,
surface heat flux is variable
• Surface temperature: Ts
Ts
m
• Inlet temperature: Tmi = Tm (0)
• Section length: L
(2) Total heat q s
dx
dqs
Determine
(1) Mean temperature variation Tm ( x )
Tm ( x )
x
Tmi
dTm
Tm +
dx
dx
Tm
dx
Fig. 6.4
26
(3) Surface flux variation q′s′(x)
Analysis
Apply conservation of energy to element dx
Assumptions
(1) Steady state
(2) No energy generation
(3) Negligible changes in kinetic and potential energy
(4) No axial conduction
dq s = m c p dTm
(a)
Newton's law:
dq s = h( x )[Ts − Tm ( x )]Pdx
27
Combine (a) and (b)
dTm
P
=
h( x )dx
Ts − Tm ( x ) m c p
(c)
Integrate from x = 0 ( Tm = Tm (0) = Tmi ) to x ( Tm = Tm ( x ) )
Tm ( x ) − Ts
P
ln
=
−
T
−
T
m cp
mi
s
x
∫0 h( x )dx
Must determine h(x). Introduce h
1
h=
x
x
∫0 h( x )dx
(6.12)
(6.12) into (6.11)
28
Tm ( x ) = Ts + (Tmi
Ph
− Ts ) exp[−
x]
mcp
(6.13)
(6.13) applies to any region and any flow
(laminar, turbulent or mixed)
To determine h(x):
(1) Is flow laminar or turbulent flow?
(2) Entrance or fully developed region?
Total heat: Apply conservation of energy:
q s = m c p [Tm ( x ) − Tmi ]
29
Surface flux: Apply Newton’s law:
q ′s′ ( x ) = h( x )[Ts − Tm ( x )]
(6.15)
Properties: At mean of inlet and outlet temperatures
Example 6.3: Required Tube Length
• Air flows through tube
• Uniform surface temperature,
r
Ts = 130o C
• Mean velocity = 2 m/s,
u0
• Tmi = 35o C
• D = 1.0 cm
Ts
Ts
D
x
u
L
30
• Nusselt number for laminar fully developed flow
hD
NuD =
= 3.657
k
(A)
Determine: tube length to raise temperature to Tmo = 105o C
(1) Observations
• Laminar or turbulent flow? Check Re D
• Uniform surface temperature
• Uniform Nusselt number (h is constant) for fully
developed laminar flow
• Length of tube is unknown
31
(2) Problem Definition. Determine tube length needed to raise
temperature to specified level
(3) Solution Plan.
• Use uniform surface temperature analysis
• Compute Re D . Laminar or turbulent?
(4) Plan Execution
(i) Assumptions
• Steady state
• Fully developed flow
• Constant properties
32
• Uniform surface temperature
• Negligible changes in kinetic and potential energy
• Negligible axial conduction
• Negligible dissipation
(ii) Analysis
Tm ( x ) = Ts + (Tmi − Ts ) exp[−
Ph
x]
mcp
(6.13)
c p = specific heat, J/kg − o C
h = average h, W/m 2 − o C
m = flow rate, kg/s
P = perimeter, m
33
Tm(x) = mean temperature at x, o C
Tmi = mean inlet temperature = 35 o C
Ts = surface temperature = 130 o C
x = distance from inlet, m
Apply (a) at the outlet (x = L), solve for L
m cp
Ts − Tmi
L=
ln
P h Ts − Tmo
(a)
Tmo = outlet temperature = 105 o C
Properties: at Tm
Tm = Tmi + Tmo
2
(b)
34
P=πD
(c)
D2
m =π
ρu
4
(d)
D = inside tube diameter = 1 cm = 0.01 m
u = mean flow velocity = 2 m/s
ρ = density, kg/m3
For fully developed laminar flow
hD
NuD =
= 3.657
k
(e)
h = heat transfer coefficient, W/m 2 − o C
k = thermal conductivity of air, W/m − o C
35
k
h = h = 3.657 , for laminar fully developed
D
Compute: Reynolds number
uD
Re D =
(f)
(g)
ν
Use (b)
o
(
35
+
105
)(
C)
Tm =
= 70 o C
2
Properties:
c p = 1008.7 J/kg− o C
k = 0.02922 W/m − o C
Pr = 0.707
ν = 19.9 × 10 − 6 m 2 /s
ρ = 11.0287 kg/m 3
36
Use (f)
Re D = 2(m/s)0.01(m) = 1005 , flow is laminar
−6
2
19.9 × 10 (m /s)
(iii) Computations
P = π 0.01(m) = 0.03142 m
(0.01)2 (m 2 )
m =π
1.0287(kg/m 3 )2(m/s) = 0.0001616 kg/s
4
o
0.02922(W
/
m
−
C) = 10.69 W/m 2 − o C
h = 3.657
0.01(m)
Substitute into (a)
L=
0.0001616(kg/s)1008.7(J/kg −o C )
2 o
0.03142(m)10.69(W/m − C)
ln
(130 − 35)(o C)
o
(130 − 105)( C)
= 0.65 m
37
(iv) Checking. Dimensional check
(i) L = 0 for Tmo = Tmi . Set Tmo = Tmi in (a) gives L = 0
(ii) L = ∞ for Tmo = Ts . Set Tmo = Ts in (a) gives L = ∞
Quantitative checks: (i) Approximate check:
(h)
Energy added at the surface = Energy gained by air
Energy added at surface ≈ hπ DL(Ts − Tm )
(i)
Energy gained by air = mc p (Tmo − Tmi )
(j)
(j) and (k) into (i), solve for L
L=
m c p (Tmo − Tmi )
(k)
h πD(Ts − Tm )
38
L=
0.0001616(kg/s)1008.7(J/kg− o C)(105 − 35)( o C)
2
o
o
10.69(W/m − C)π (0.01)(m)(130 − 70)( C)
= 0.57 m
(ii) Value of h is low compared with Table 1.1. Review solution.
Deviation from Table 1.1 are expected
(5) Comments. This problem is simplified by two conditions:
fully developed and laminar flow
39
6.6 Determination of Heat Transfer Coefficient
h( x ) and Nusselt Number Nu D
r
q′s′
Two Methods:
ro
(1) Scale analysis
0
(2) Analytic or numerical solutions
Tm
6.6.1 Scale Analysis
Ts
Fig. 6.5
Fourier’s law and Newton’s law
∂T ( ro , x )
−k
∂r
h=
Tm − Ts
(6.16)
40
Scales:
r ~ δt
(a)
∂T ( ro , x ) Tm − T s
~
δt
∂r
(b)
(a) and (b) into (6.16)
k
h~
(Tm − Ts )
δt
Tm − Ts
or
h~
k
δt
(6.17)
41
Nusselt number:
hD
NuD =
k
(6.17) into the above
NuD ~
D
δt
(6.18)
Entrance region: δ t < D , NuD > 1
Special case: fully developed region
δ t ( x) ~ D
(6.18) gives
NuD ~ 1 (fully developed)
(6.19)
42
δ t in the entrance region: For all Pr
−1 / 2
−1 / 2
x
Pr
Re
δt ~
x
(4.24)
(4.24) into (6.18)
D 1/ 2 1/ 2
Nu D ~ Pr Re x
x
(c)
Express in terms of Re D
ux
uD x
x
Re x =
=
= Re D
ν
ν D
D
(d)
(d) into (c)
1/2
D
NuD ~
x
Pr 1/2Re1/2
x
(6.20a)
43
or
NuD
1/ 2
PrRe
x/D
~1
(6.20b)
44
6.6.2 Basic Considerations for the Analytical
Determination of Heat Flux, Heat Transfer
Coefficient and Nusselt Number
(1) Fourier’s law and Newton’s law
∂T ( x , ro )
∂r
Define dimensionless variables
(a)
q′s′ = − k
T − Ts ,
x/D
θ ≡
ξ=
Re D Pr
Ti − Ts
v ∗x
vx
vr
∗
=
, vr =
u
u
,
,
r
R=
ro
Re D =
uD
ν
(6.21)
45
(6.21) into (a)
k
∂ 0(ξ ,1)
q′s′(ξ ) = (Ts − Ti )
ro
∂R
(6.22)
Newton’s law
q"s
h(ξ ) =
Tm − Ts
(6.23)
Combine (6.22) and (6.23)
k (Ts − Ti ) ∂θ (ξ ,1)
k 1 ∂θ (ξ ,1)
h(ξ ) =
=−
ro (Tm − Ts ) ∂R
ro θ m (ξ ) ∂R
(6.24)
Dimensionless mean temperatureθ m :
Tm − Ts
0m ≡
Ti − Ts
(6.25)
46
Nusselt number:
h(ξ ) D h(ξ )2ro
Nu(ξ ) =
=
k
k
(6.26)
(6.24) into (6.26)
− 2 ∂0 (ξ ,1)
Nu(ξ ) =
0 m (ξ ) ∂R
(6.27)
Determine: q ′s′ (ξ ), h(ξ ) and: Nu(ξ )
Find θ (ξ , R ) . Apply energy equation
(2) The Energy Equation
Assumptions
• Steady state
47
• Laminar flow
• Axisymmetric
• Negligible gravity
• Negligible dissipation
• Negligible changes in kinetic and potential energy
• Constant properties
1 ∂ ∂T ∂ T
∂T
∂T ∂ T
+ vz
ρc p v r
+
= k
r
+ 2
∂z
∂r ∂θ
r ∂r ∂r ∂z
2
(2.24)
Replace z by x, express in dimensionless form
4 ∂ ∂0
1
∂0
∗ ∂0
∗ ∂0
vx
+ 2 Re D Pr v r
=
R +
∂ξ
∂R R ∂R ∂R ( Re D Pr )2 ∂ξ 2
2
(6.28)
48
Pe = Re D Pr , Peclet number
(2.29)
• Third term: radial conduction
• Fourth term: axial conduction
• Neglect axial conduction for:
Pe = PrRe D ≥ 100
(6.30)
Simplify (6.28)
v ∗x
∂θ
4 ∂ ∂θ
∗ ∂θ
+ 2 Re D Pr v r
=
R
∂ξ
∂R R ∂R ∂R
(6.31)
(3) Mean (Bulk) Temperature Tm
Need a reference local temperature. Use Tm ( x )
49
mc pTm =
where
m=
∫0
ro
∫0
ro
ρ c p v xT 2πrdr
(a)
ρ v x 2πrdr
(b)
(b) into (a), assume constant properties
ro
v x T r dr
∫
Tm = 0 r
∫0 v x rdr
(6.32a)
o
In dimensionless form:
Tm − Ts
θm =
=
Ti − Ts
1
∫0
1
v ∗x RdR
∫0
v ∗xθ RdR
(6.32b)
50
6.7 Heat Transfer Coefficient in the Fully
Developed Temperature Region
6.7.1 Definition of Fully Developed Temperature
Profile
Fully developed temperature:
Ts ( x ) − T ( r , x )
φ=
Ts ( x ) − Tm ( x )
Let
x / d > 0.05 Re D Pr
(6.33)
Definition:
For fully developed temperature
φ
is independent of x
51
Therefore
φ = φ (r )
(6.34)
From (6.34)
∂φ
=0
∂x
(6.33) and (6.35):
∂φ
∂ Ts ( x ) − T ( r , x )
=0
=
∂x ∂x Ts ( x ) − Tm ( x )
(6.35)
(6.36a)
Expand and use (6.33)
dTs ∂T
dTs dTm
− φ (r )
=0
−
−
dx
dx ∂x
dx
(6.36b)
52
6.7.2 Heat Transfer coefficient and Nusselt number
∂T ( ro , x )
−k
∂r
h=
Tm − Ts
(6.16)
Use(6.33) to form ∂T ( ro , x ) / ∂r , substitute into (6.16)
do/ (ro )
h = −k
dr
(6.37)
Conclusion:
The heat transfer coefficient in the fully developed region
is constant regardless of boundary condition
53
Nusselt number:
hD
dφ ( ro )
NuD =
= −D
h
dr
(6.38)
Entrance region scaling result:
NuD ~ 1 (fully developed)
(6.19)
Scaling of fully developed region:
scale for ∂T ( ro , x ) / ∂r
∂T ( ro , x ) Ts − Tm
~
∂r
D
Substitute into (6.16)
k
h~
D
(6.39)
54
(6.39) into (6.38)
NuD ~ 1 (fully developed)
(6.40)
6.7.3 Fully Developed Region for Tubes at
Uniform Surface flux
r
• Uniform flux
• Determine
• Ts ( x )
• h
Newton’s law:
u
0
q′s′
T
D
x
q′s′
Fig. 6.6
q′s′ = h [Ts ( x ) − Tm ( x )]
(a)
55
Ts ( x ) and Tm ( x ) are unknown
q ′s′ and h are constant
(a) gives:
[Ts ( x ) − Tm ( x )] = constant
Differentiate (b)
(b)
dT s dTm
=
dx
dx
(c)
∂T dT s
=
∂x
dx
(d)
(c) into (6.36b)
Combine (c) and (d)
∂T dTs dTm (for constant q ′′)
s
=
=
∂x
dx
dx
(6.41)
56
To determine h form (6.16) :
Determine: T ( r , r ), Tm ( x ) and Ts ( x )
Conservation of energy for dx
dTm
′
′
qs Pdx + mc pTm = mc p Tm +
dx
dx
Simplify
dTm q ′s′ P
=
dx
mc p
(6.42)
= constant
q′s′
(6.42) into (6.41)
∂T dTs dTm q ′s′ P
= constant
=
=
=
∂x
dx
dx mc p
(6.43)
m
dTm
Tm +
dx
dx
Tm ( x )
dx
Fig. 6.7
57
Conclusion: T ( x , r ), Tm ( x ) and T s ( x ) are linear with x
Integrate(6.43)
q′s′P
Tm ( x ) =
x + C1
mc p
(e)
Tm (0) = Tmi
(f)
C 1 = constant
Boundary condition:
Apply (e) to (f)
C1 = Tmi
(e) becomes
Tm ( x ) = Tmi
q′s′P
+
x
mc p
(6.44)
58
Determine T ( r , x ) and Ts ( x )
Apply energy equation (2.23) in the fully developed region
Assumptions
• Negligible axial conduction
• Negligible dissipation
• Fully developed, v r = 0
ρ c pv x
∂T k ∂ ∂T
=
r
∂x r ∂r ∂r
(6.45)
r2
v x = 2u 1 −
2
r
o
(6.46)
59
However
m = π ro2 ρ u
P = 2π ro
equation (g) becomes
4q′s′ r 2 k ∂ ∂T
r
1 − 2 =
ro ro r ∂r ∂r
(6.47)
Boundary conditions:
∂ T ( 0, x )
=0
∂r
Integrate (6.47)
∂T ( ro , x )
k
= q′s′
∂r
4 r 2 r 3
∂T
q′s′ − 2 = kr
+ f (x)
ro 2 4ro
∂r
(6.48a)
(6.48b)
(h)
60
f ( x ) = “constant” of integration
Boundary condition (6.48a)
f ( x) = 0
(h) becomes
∂T 4q′s′ r r 3
=
− 2
∂r kro 2 4ro
Integrate
4q′s′
T (r , x ) =
kro
r 2 r4
− 2 + g( x )
4 4ro
(6.49)
g ( x ) = “constant” of integration
• Boundary condition (6.48b) is satisfied
• Use solution to Tm ( x ) to determine g ( x )
61
Substitute(6.46) and (6.49) into (6.32a)
7 roq′s′
Tm ( x ) =
+ g( x )
24 k
(6.50)
Two equations for Tm ( x ) : (6.44) and (6.50). Equating
g ( x ) = Tmi
7 roq′s′ Pq′s′
−
+
x
24 k
mc p
(6.51)
(6.51) into (6.49)
T ( r , x ) = Tmi
4q′s′
+
kro
r 2
r 3 7 roq′s′ Pq′s′
−
+
x (6.52)
−
2
mc p
4 16ro 24 k
Surface temperature Ts ( x ) : set r = ro in (6.52)
62
Ts ( x ) = Tmi
11 roq′s′ Pq′s′
+
x
+
24 k
mc p
(6.53)
T ( r , x ), Tm ( x ) and Ts ( x ) are determined
(6.44), (6.52) and (6.53) into (6.33)
24 1
φ (r ) = 1 −
11 ro2
2 r 4 24 Pq ′s′
7
x+ x
r − 2 +
11
4ro 11 mc p
(6.54)
Differentiate(6.54) and use (6.38)
48
NuD =
= 4.364 , laminar fully developed
11
(6.55)
63
Comments:
• (6.55) applies to:
• Laminar flow in tubes
• Fully developed velocity and temperature
• Uniform surface heat flux
• Nusselt number is independent of Reynolds and Prandtl
numbers
• Scaling result:
Nu D ~ 1
(6.40)
64
6.7.4 Fully Developed Region for Tubes at
Uniform Surface Temperature
• Fully developed
• Uniform surface temperature Ts
Determine: Nu D and h
Assumptions:
• Neglect axial conduction
• Neglect dissipation
• Fully developed: v r = 0
Energy equation (2.24):
ρ c pv x
∂ T k ∂ ∂T
=
r
∂x r ∂ r ∂ r
(6.45)
65
Boundary conditions:
∂ T ( 0, x )
=0
∂r
T ( ro , x ) = Ts
Axial velocity
r2
v x = 2u 1 − 2
ro
(6.56a)
(6.56b)
(6.46)
Eliminate ∂T / ∂x in equation (6.45). Use (6.36a)
∂φ
∂ Ts ( x ) − T ( r , x )
=0
=
∂x ∂x Ts ( x ) − Tm ( x )
(6.36a)
66
for Ts ( x ) = Ts, (6.36a) gives
∂T Ts − T ( r , x ) dTm
=
∂x Ts − Tm ( x ) dx
(6.57)
(6.46) and (6.57) into (6.45)
r 2 Ts − T ( r , x ) dTm k ∂ ∂T
ρ c p u 1 − 2
=
r
r ∂r ∂ r
ro Ts − Tm ( x ) dx
(6.58)
Result: Solution to (6.58) by infinite power series:
NuD = 3.657
(6.59)
67
6.7.5 Nusselt Number for Laminar Fully
Developed Velocity and Temperature
in Channels of Various Cross Sections
• Analytical and numerical solutions
• Results for two classes of boundary conditions:
(1)Uniform surface flux
(2)Uniform surface temperature
• Nusselt number is based on the equivalent diameter
De =
4Af
(6.60)
P
• Results: Table 6.2
68
69
• Compare with scaling
NuD ~ 1 (fully developed)
(6.40)
Example 6.4: Maximum Surface Temperature
in an Air Duct
q′s′
• 4 cm × 4 cm square duct
• Uniform heat flux = 590 W/m 2
Tmo
o
• Heating air from 40 C to 120o C
u
• u = 0.32 m/s
Tmi
L
• No entrance effects (fully developed)
70
Determine: Maximum surface temperature
(1) Observations
• Uniform surface flux
• Variable Surface temperature, Ts ( x ) , maximum at outlet
• Compute the Reynolds number
• Velocity and temperature are fully developed
• The heat transfer coefficient is uniform for fully developed
flow
• Duct length is unknown
• The fluid is air
71
(2) Problem Definition
(i) Find the required length
(ii) Determine surface temperature at outlet
(3) Solution Plan
(i) Apply conservation of energy
(ii) Compute the Reynolds
(iii) Apply constant surface solution
(iv) Use Table 6.2 for h
(4) Plan Execution
(i) Assumptions
• Steady state
72
• Constant properties
• Uniform surface flux
• Negligible changes in kinetic and potential energy
• Negligible axial conduction
• Negligible dissipation
(ii) Analysis
Conservation of energy
P L q′s′ = m c p (Tmo − Tmi )
(a)
c p = specific heat, J/kg − o C
L = channel length, m
m = mass flow rate, kg/s
73
P = perimeter, m
q ′s′ = surface heat flux = 590 W/m 2
Tmi = 40 o C
Tmo = 120 o C
Solve (a) for L
L=
mc p (Tmo − Tmi )
(b)
P q ′′
Find m and P
m = ρ S 2u
(c)
P = 4S
(d)
S = duct side = 0.04 m
u = mean flow velocity = 0.32 m/s
74
ρ = density, kg/m 3
(c) and (d) into (b)
L=
ρ S u c p (Tmo − Tmi )
(e)
4 q′′
Surface temperature: Use solution (6.10)
Ts ( x ) = Tmi
Px
1
+
+ q′s′
m
c
h
(
x
)
&
p
(f)
h( x ) = local heat transfer coefficient, W/m 2 − o C
Ts ( x ) = local surface temperature,o C
x = distance from inlet, m
75
Maximum surface temperature at x = L
Ts ( L) = Tmi
4L
1
+
+ q′s′
ρ
S
u
c
h
(
L
)
p
(g)
Determine h(L): Compute the Reynolds number
Re De =
u De
(h)
ν
De = equivalent diameter, m
ν = kinematic viscosity, m 2 /s
2
A
S
De = 4 = 4
=S
P
4S
(i)
76
(i) into (h)
Re De =
uS
(j)
ν
Properties: At Tm
Tm = Tmi + Tmo
2
Tm
(k)
(40 + 120)( o C)
= 80 o C
=
2
Properties:
c p = 1009.5 J/kg − o C
k = 0.02991 W/m − o C
Pr = 0.706
77
ν = 20.92× 10− 6 m2/s
ρ = 0.9996 kg/m3
(j) gives
Re De
0.32(m/s)0.04(m)
=
= 611.9 ,
−6
2
20.92 × 10 (m /s)
laminar flow
(6.55) and Table 6.2
Nu De
h=h
h De
=
k
=
3.608
k
h = 3.608
De
(l)
(m)
78
(iii) Computations. Use (e)
3
o
o
0.9996(kg/m ) 0.04(m)0.32(m/s) 1009.5(J/kg- C)(120 − 40)( C)
L=
= 0.4378
2
(4) 590(W/m )
m
Use (m)
0.02991(W/m − o C)
h(L) = h = 3.608
= 2.7 W/m 2 − o C
0.04(m)
Substitute into (g)
Ts ( L) = 40( o C ) +
590( W/m
2
4(0.4378)(m)
1
)
0.9996(kg/m 3 ) 0.04(m)0.32 (m/s) 1009.5(J/kg − o C) + 2.7(W/m 2 − o C)
79
Ts ( L) = 338.5 oC
(iv) Checking. Dimensional check:
Quantitative checks: (1) Alternate approach to determining :
Newton’s law at outlet
q′s′ = h [Ts ( L) − Tmo ]
(n)
T s ( L) Solve for
2
′s′
590(W/m
)
q
o
o
T s ( L) = Tmo =
120
(
C)
+
=
338.5
C
=
2 o
2.7(W/m − C)
h
(2) Compare h with Table 1.1
Limiting check: L =0 for Tmo = Tmi . Set Tmo = Tmi
into (e) gives L = 0
80
(5) Comments
(i) Maximum surface temperature is determined by the
heat transfer coefficient at outlet
(ii) Compute the Reynolds number to establish if the flow
is laminar or turbulent and if it is developing or fully
developed
81
6.8 Thermal Entrance Region: Laminar Flow
Through Tubes
6.8.1 Uniform Surface Temperature:
Graetz Solution
• Laminar flow through tube
r
Ts
• Velocity is fully developed
• Temperature is developing
u
• No axial conduction (Pe > 100)
• Uniform surface temperature Ts
T
Ti
0
x
δt
Fig.6.8
Velocity:
vr = 0
(3.1)
82
1 dp 2
vz =
( r − ro2 )
4µ d z
(3.12)
Rewrite (3.1)
vz
= = 2[1 − R 2 ]
u
(3.1) and (6.61) into (6.31)
v*z
1
1 ∂ ∂θ
2 ∂θ
1− R
=
R
2
∂ξ R ∂R ∂R
(
)
(6.61)
(6.62)
Boundary conditions:
∂θ (ξ ,0)
=0
∂R
θ (ξ ,1) = 0
θ (0, R ) = 1
83
Solution summary: Assume a product solutions
(a)
θ (ξ , R ) = X (ξ )R (R)
(a) into (6.62), separating variables
dX n
+ 2λ2n X n = 0
dξ
(b)
d 2Rn 1 dRn
2
2
2
+
(
1
−
R
)
R
λ
+
n
n =0
2
2
R dR
dR
(c)
• λ n = eigenvalues obtained from the boundary conditions
• Solution X n (ξ ) to (b) is exponential
• Solution Rn ( R ) to (c) is not available in terms of
simple tabulated functions
84
Solutions to (b) and (c) into (a)
∞
θ (ξ , R ) = ∑ C nRn ( R ) exp(−2λ2nξ )
(6.64)
n=0
C n = constant
Surface flux:
k
∂0(ξ ,1)
q′s′(ξ ) = (Ts − Ti )
ro
∂R
(6.22)
(6.64) gives
dRn (1)
∂θ (ξ ,1) ∞
Cn
=
exp(−2λ2nξ )
dRn
∂R
n=0
∑
(d)
Define
C n dR n (1)
Gn = −
2
dR
(e)
85
(e) into (6.22)
∞
2k
q′s′ (ξ ) = − (Ts − Ti ) Gn exp(−2λ2nξ )
ro
n=0
∑
(6.65)
Local Nusselt number: is given by
− 2 ∂0 (ξ ,1)
Nu(ξ ) =
0 m (ξ ) ∂R
(6.27)
(d) gives ∂θ (ξ ,1) / ∂R
Mean temperature θ m (ξ ): (6.61) and (6.64) into (6.32b),
integrate by parts and use(e)
∞
θ m (ξ ) = 8∑
Gn
2
λ
n=0 n
exp(−2λ2nξ )
(6.66)
86
(d), (e) and (6.66) into (6.27)
∞
∑
Nu(ξ ) =
Gn exp(−2λnξ )
n=0
∞
2
2
Gn
∑ λ2
n=0
(6.87)
exp(−2λ2nξ )
n
Average Nusselt number: For length x
h (ξ ) D
Nu(ξ ) =
k
Two methods for determining h (ξ ) :
(f)
(1) Integrate local h(ξ ) to obtain h (ξ )
(2) Use (6.13)
87
Ph
Tm ( x ) = Ts + (Ti − Ts ) exp [−
x]
mcp
(6.13)
Solve for h
mcp
Tm ( x ) − Ts
ln
h =−
Px
Ti − Ts
(g)
(g) into (f), use m = ρ u π D 2 / 4 P = π D and definitions of
ξ , Re D and θ m in (6.21) and (6.25)
1
Nu(ξ ) = − lnθ m (ξ )
4ξ
(6.68)
• Need λ n and G n to compute q ′s′ (ξ ), θ m (ξ ), Nu (ξ ) and Nu (ξ )
• Table 6.3 gives λ n and G n
• (6.67) and (6.68) are plotted in Fig. 6.9 as Nu(ξ ) and Nu88(ξ )
89
Average Nu
Nusselt number
Local Nu
x/D
ReD Pr
Local and average Nusselt number for
tube at uniform surface temepratu re [4]
ξ=
Fig. 6.9
90
Comments
(1) Nu D and NuD decrease with distance from entrance
(2) At any location ξ Nu D > NuD
(3) Asymptotic value (at ξ ≈ 0.05 ) for Nu D and NuD
is 3.657. Same result of fully develop analysis
Nu(∞ ) = 3.657
(6.69)
(4) Properties at
Tmi + Tmo
Tm =
2
(6.70)
(5) Solution by trial and error if Tmo is to be determined
91
Example 6.5: Hot Water Heater
• Fully developed velocity in tube
• Developing temperature
• Uniform inlet temperature Ti = 25 o C
r
• Diameter = 1.5 cm
• Length = 80 cm
• Flow rate = 0.002 kg/s
• Heat water to 75 o C
Determine: Surface temperature
u
Ts
Ti
0
x
δt
(1) Observations
• Uniform surface temperature
92
• Compute Reynolds number: Laminar or turbulent flow?
• Compute Lh and Lt : Can they be neglected?
(2) Problem Definition
(i) Determine Ts
(ii) Determine h
(3) Solution Plan
(i) Apply uniform surface temperature results
(ii) Compute the Reynolds number: Establish if problem is
entrance or fully developed
(iii) Use appropriate results for Nusselt number
(4) Plan Execution
(i) Assumptions
• Steady state
93
• Constant properties
• Uniform surface temperature
• Negligible changes in kinetic and
potential energy
• Negligible axial conduction
• Negligible dissipation
(ii) Analysis
Uniform surface temperature
Tm ( x ) = T s + (Tmi − T s ) exp[−
Ph
x]
mcp
h = average heat transfer coefficient, W/m 2 − o C
m = flow rate = 0.002 kg/s
(6.13)
94
Tmi = mean inlet temperature = 25 o C
Tmo = mean outlet temperature = 75 o C
Apply (6.13) at outlet (x = L) and solve for Ts
Ts =
1
Tmi − Tm ( L) exp( Ph L / mc p )
1 − exp( Ph L / mc p )
[
Properties: at Tm
]
(a)
Tm = Tmi + Tmo
2
Perimeter P
P=πD
D = diameter = 1.5 cm = 0.015 m
95
Determine h : Compute the Reynolds
Re D =
u=
uD
ν
4m
(e)
ρ π D2
Properties: at Tm
c p = 4182 J/kg-oC
o
+
(
20
80
)(
C)
Tm =
= 50 o C
2
k = 0.6405 W/m-oC
Pr = 3.57
ν = 0.5537×10-6 m2/s
96
ρ = 988 kg/m3
Use (e)
u=
4(0.002)(kg/s)
3
2
2
988(kg/m )π (0.015) (m )
= 0.01146 m/s
Use (d) gives
Re D =
0.01146(m/s)0.015(m)
−6
2
0.5537 × 10 (m /s)
= 310.5 , laminar flow
Determine Lh and Lt
Lh
= C h Re D
D
Lt
= C t PrRe D
D
(6.5)
(6.6)
97
C h = 0.056 (Table 6.1)
C t = 0.033 (Table 6.1)
(6.5) and (6.6)
Lh = 0.056 × 0.015 (m) × 310.5 = 0.26 m
Lt = 0.033 × 0.015 (m) × 310.5 × 3.57 = 0.55 m
• Lh and Lt are not negligible, tube length L = 0.8 m
• Use Graetz solution Fig. 6.9 or Table 6.4
Compute ξ
x/D
ξ=
Re D Pr
(f)
98
Nusselt number Nu gives h
k
h = Nu
D
(g)
(iii) Computation. Evaluating ξ at x = L
0.8(m)/).015(m )
= 0.0481
ξ=
310 .5 × 3.57
At ξ = 0.481 Fig. 6.9 gives
Nu ≈ 4.6
Substitute into (g)
0.6405(W/m− o C)
h=
4.6 = 196.4W/m 2 − o C
0.015(m)
99
Equation (a) gives Ts
Compute the exponent of the exponential in (a)
Ph L π (0.015)(m)( 196.4(W/m 2 − o C)0.8(m)
=
= 0.88524
o
mc p
0.002(kg/s)4182(J/kg− C)
Substitute into (a)
Ts =
[
]
1
25( o C) − 75( o C)exp(0.88524) = 110.1o C
1 − exp(0.88524)
(iv) Checking. Dimensional check:
Limiting checks:
(i) For Tmi = T ( L) (no heating) Ts should be equal to Tmi .
Set Tmi = T (L) in (a) gives Ts = Tmi
100
(ii) If L = 0, Ts should be infinite. Set in L = 0 (a) gives Ts = ∞
Quantitative checks:
(i) Approximate check:
Energy added at the surface = Energy gained by water
(h)
Let
Tm = average water temperature in tube
Energy added at surface = hπ DL(Ts − Tm )
(i)
Energy gained by water = mc p (Tmo − Tmi )
(j)
(j) and (k) into (i), solve for Ts
101
Ts = Tm +
m c p (Tmo − Tmi )
(k)
h π DL
(k) gives
Ts = 50(o C) +
0.002(kg/s)4182(J/kg−o C)(75 − 25)(o C)
196.4(W/m2 −o C)π (0.0155(m)(0.8)(m)
= 106.5o C
(ii) Compare computed h with Table 1.1
(5) Comments
• Small error is due to reading Fig. 6.9
• Fully developed temperature model:
Nu D = 3.657, gives h = 156.3 W/m 2 − o C
102
6.8.2 Uniform Surface Heat Flux
• Repeat Graetz entrance problem with uniform
surface heat
• Fully developed inlet velocity
• Laminar flow through tube
• Temperature is developing
• No axial conduction (Pe > 100)
r
Energy equation: Same as
for Graetz problem
u
T
Ti
0
x
q′s′
D
δt
q′s′
Fig. 6.10
103
1
1 ∂ ∂θ
2 ∂θ
1− R
=
R
2
∂ξ R ∂R ∂R
(
)
(6.62)
Boundary conditions:
∂θ (ξ ,0)
)=0
∂R
∂θ (ξ ,1)
q′s′ro
=
∂R
k (Ti − Ts )
θ ( 0, R ) = 1
(6.71a)
(6.71b)
(6.71c)
Analytic solutions: Based on separation of variables
(1) Local Nusselt number
104
hx 11 1
Nu(ξ ) =
An exp(−2 β nξ )
= −
k
48 2 n =1
∞
∑
−1
2
(6.72)
2
β
Table 6.6 lists eigenvalues n and constant An
(2) Average Nusselt number
hx 11
Nu(ξ ) =
= −
k 48
2 −1
1
1 − exp(−2 β n ξ )
An
2
2 n =1
2β nξ
∞
∑
(6.73)
Limiting case:
Fully developed: Set ξ = ∞ (6.72) or (6.73)
−1
11
Nu(∞ ) = = 4.364
48
Same as fully developed result (6.55).
(6.74)
105
Graphical results: Fig. 6.11
106
Average Nu
Nusselt number
Local Nu
ξ=
x/D
ReD Pr
Fig. 6.11 Local and average Nusselt number for
tube at uniform surface heat flux [4]
107
CHAPTER 7
FREE CONVECTION
7.1 Introduction
• Applications:
Solar collectors
Pipes
Ducts
Electronic packages
Walls and windows
7.2 Features and Parameters of Free Convection
(1) Driving Force: Natural
1
Requirements: (i) Acceleration field, (ii) Density gradient
Temperature gradient → density gradient
(2) Governing Parameters: Two:
(i) Grashof number GrL
GrL =
β g(Ts − T∞ )L3
ν
2
(7.1)
β = coefficient of thermal expansion (compressibility), a property
β=
1
T
for ideal gas
(2.21)
2
(ii) Prandtl number Pr
Pr =
c pµ
k
Rayleigh number Ra L
β g(Ts − T∞ ) L3
β g(Ts − T∞ ) L3
Ra L = GrL Pr =
Pr =
2
ν
να
(7.2)
(3) Boundary Layer: laminar, turbulent or mixed
Criterion:
Ra x > 104
(4) Transition: Laminar to Turbulent Flow
Criterion for vertical plates: Transition Rayleigh number
3
Rax t ≈ 109
(5) External vs. Enclosure free convection:
(i) External: over:
Vertical surfaces
Inclined surfaces
Horizontal cylinders
Spheres
(ii) Enclosure: in:
4
Rectangular confines
Concentric cylinders
Concentric spheres
(6) Analytic Solution
Requires simultaneous integration of
continuity, momentum and energy
7.3 Governing Equations
Approximations:
(1) Density is constant except in gravity forces
5
(2) Boussinesq approximation: relate density change to temperature
change
(3) Negligible dissipation
Assume:
Steady state
Two-dimensional
Laminar
Continuity:
∂u ∂v
+
=0
∂x ∂y
(7.1)
x-momentum:
6
∂u
∂u
1 ∂
∂ 2u ∂ 2u
u +v
= β g (T − T∞ ) −
( p − p∞ ) + v ( 2 + 2 )
∂x
∂y
ρ ∞ ∂x
∂x
∂y
(7.2)
y-momentum:
1 ∂
∂v
∂v
∂ 2v ∂ 2v
u +v
=−
( p − p∞ ) + v ( 2 + 2 )
∂x
∂y
ρ ∞ ∂y
∂x
∂y
(7.3)
Energy:
∂ 2T ∂ 2T
∂T
∂T
u
+v
= α 2 + 2
∂x
∂y
∂y
∂x
(7.4)
NOTE:
(1) Gravity points in the negative x-direction
(2) Flow and temperature fields are coupled
7
7.3.1 Boundary Layer Equations
• Velocity and temperature boundary layers
• Apply approximation used in forced convection
y-component of the Navier-Stokes equations reduces
to
∂
( p − p∞ ) = 0
∂y
(a)
External flow: Neglect ambient pressure variation in x
∂
( p − p∞ ) = 0
∂x
(b)
Furthermore, for boundary layer flow
8
∂ 2u
∂x
2
<<
∂ 2u
∂y
2
(c)
x-momentum:
∂u
∂u
∂ 2u
u +v
= + βg (T − T∞ ) + v 2
∂x
∂y
∂y
(7.5)
Neglect axial conduction
∂ 2T
∂x
2
<<
∂ 2T
∂y
2
(d)
Energy: (7.4) becomes
∂T
∂T
∂ 2T
u
+v
=α 2
∂x
∂y
∂y
(7.6)
9
7.4 Laminar Free Convection over a Vertical Plate:
Uniform Surface Temperature
•
Vertical plate
•
Uniform temperature Ts
•
Infinite fluid at temperature T∞
Determine: velocity and temperature distribution
7.4.1 Assumptions
(1) Steady state
(2) Laminar flow
(3) Two-dimensional
10
(1)
(2)
(3)
(4)
(5)
(9)
Constant properties
Boussinesq approximation
Uniform surface temperature
Uniform ambient temperature
Vertical plate
Negligible dissipation
7.4.2 Governing Equations
Continuity:
∂u ∂v
+
=0
∂x ∂y
(7.1)
x-momentum:
11
∂u
∂u
∂ 2u
u
+v
= + βg (T − T ∞ ) + v
∂x
∂y
∂y 2
(7.5)
Energy:
∂θ
∂θ
∂ 2θ
u
+v
=α 2
∂x
∂y
∂y
(7.7)
where θ is a dimensionless temperature defined as
T − T∞
θ=
Ts − T∞
(7.8)
7.4.3 Boundary Conditions
Velocity:
(1) u( x ,0) = 0
(2) v ( x ,0) = 0
12
(3) u( x , ∞ ) = 0
(4) u(0, y ) = 0
Temperature:
(5) θ ( x ,0) = 1
(6) θ ( x , ∞ ) = 0
(7) θ ( 0, y ) = 0
7.4.4 Similarity Transformation
• Transform three PDE to two ODE
• Introduce similarity variable η ( x , y )
y
η( x, y) = C
x
1/ 4
(7.8)
13
βg (Ts − T∞ )
C=
4v 2
1/ 4
(7.9)
(7.9) into (7.8)7
Gr
η= x
4
1/ 4
y
x
(7.10)
Local Grashof number:
Grx =
β g(Ts − T∞ )x 3
ν
2
(7.11)
Let
0 ( x , y ) = 0 (η )
(7.12)
Stream function ψ satisfies continuity
14
u=
∂ψ
∂y
(7.14)
Using Blasius solution as a guide, ψ for this problem is
1
4
Grx
ψ == 4v
ξ (η )
4
(7.15)
ξ (η ) is an unknown function
Introduce (7.15) into (7.13) and (7.14)
Grx dξ
u = 2v
x dη
(Grx )1 / 4 dξ
v = 1/ 4
η
−
3
ξ
dη
x
( 4)
v
(7.16)
(7.17)
15
Combining (7.5), (7.7), (7.8), (7.12), (7.16), (7.17)
3
d ξ
dη
2
2
3
+ 3ξ
d 2θ
dη
2
dξ
−
2
+θ = 0
2
dη
dη
d ξ
+ 3 Prξ
dθ
=0
dη
(7.18)
(7.19)
NOTE
• x and y are eliminated (7.18) and (7.19)
• Single independent variable
η
Transformation of boundary conditions:
Velocity:
16
dξ (0)
(1)
=0
dη
(2) ξ (0) = 0
dξ (∞ )
(3)
=0
dη
(4)
dξ (∞ )
=0
dη
Temperature:
(1) θ ( 0) = 1
(2) θ (∞ ) = 0
(3) θ (∞ ) = 0
17
NOTE:
(1) Three PDE are transformed into two ODE
(3) Five BC are needed
(4) Seven BC are transformed into five
(5) One parameter: Prandtl number.
7.4.5 Solution
• (7.18) and (7.19) are solved numerically
• Solution is presented graphically
• Figs. 7.2 gives u(x,y)
• Fig. 7.3 gives T(x,y)
18
19
20
7.4.5 Heat Transfer Coefficient and Nusselt Number
Start with
∂T ( x ,0)
−k
∂y
h=
Ts − T∞
(7.20)
Express in terms of θ and η
− k dT dθ (0) ∂η
h=
Ts − T∞ dθ dη ∂y
Use (7.8) and (7.10)
− k Gr x
h=
x 4
1/ 4
dθ ( 0 )
dη
(7.21)
21
Define: local Nusselt number:
hx
Grx
Nu x =
= −
k
4
1/ 4
dθ (0)
dη
(7.22)
Average heat transfer coefficient
1 L
h = ∫ h( x )dx
L 0
(2.50)
(7.21) into (2.50), integrate
4 k GrL
h=−
3 L 4
1/ 4
dθ (0)
dη
(7.23)
Average Nusselt number
22
hL
4 k GrL
Nu L =
=−
k
3 L 4
1/ 4
dθ (0)
dη
Table 7.1 [1,2]
(7.24)
Pr
dθ ( 0 )
dη
•
•
•
•
•
Is key factor in solution
Depends on Prandtl number
Values are listed in Table 7.1
Obtained from numerical solution
d 2ξ (0)
dη
2
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
_ dθ ( 0 )
dη
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 2ξ ( 0)
dη 2
0.9862
0.676
0.6741
0.6421
0.5713
0.4192
0.2517
0.1450
in Table 7.1 gives surface velocity
gradient and shearing stress
23
Special Cases
Nu x = 0.600 ( PrRa x )1/4 ,
Nu x = 0.503 ( PrGrx )1/4 ,
Pr → 0
Pr → ∞
(7.24)
(7.25)
Example 7.1: Vertical Plate at Uniform Surface
Temperature
• 8cm × 8cm vertical plate in air at 10 oC
• Uniform surface temperature = 70 oC
• Determine the following at x = L = 8 cm:
(1) u at y = 0.2 cm , (2) T at y = 0.2 cm
(3) δ , (4) δ t
24
(5) Nu L , (6) h(L)
(7) q ′s′ ( L ) , (8) qT
Solution
(1) Observations
• External free convection
• Vertical plate
• Uniform surface temperature
•
Check Rayleigh number for laminar flow
• If laminar, Fig. 7.2 for u(x,y) and Fig. 7.3 T(x,y) and δ t
• Determine local Nu and h at x = L
(2) Problem Definition
25
Determine flow and heat transfer characteristics for free convection
over a vertical flat plate at uniform surface temperature.
(3) Solution Plan
• Laminar flow? Compute Rayleigh number
• If laminar, use Figs. 7.2 and 7.3.
• Use solution for Nu and h
(4) Plan Execution
(i) Assumptions
(1) Newtonian fluid
(2) Steady state
(3) Boussinesq approximations
(4) Two-dimensional
26
(5) Laminar flow ( Rax < 109 )
(6) Flat plate
(7) Uniform surface temperature
(8) No dissipation
(9) No radiation.
(ii) Analysis and Computation
Compute the Rayleigh number:
β g(Ts − T∞ ) L3
Ra L =
να
(7.2)
g = gravitational acceleration = 9.81 m/s 2
L = plate length = 0.08 m
27
Ra L = Rayleigh number at the trailing end x = L
Ts = surface temperature = 70o C
T∞ = ambient temperature =10o C
α = thermal diffusivity, m 2 /s
β = coefficient of thermal expansion = 1 / T f K -1
ν = kinematic viscosity, m 2 /s
Properties at T f
Ts + T∞ (70 + 10)o C
= 40o C
Tf =
=
2
2
k = thermal conductivity = 0.0271 W/m − o C
Pr = 0.71
ν = 16.96 × 10 −6 m 2 /s
28
16.96 × 10 −6 m 2 /s
=
= 23.89 × 10 −6 m 2 /s
α=
Pr
0.71
ν
β=
1
40o C + 273.13
= 0.0031934 K -1
Substituting into (7.2)
Ra L =
0.0031934 ( K −1 )9.81(m/s 2 )(70 − 10)(K)(0.08) 3 (m 3 )
16.96 × 10 − 6 (m 2 /s)23.89 × 10 − 6 (m 2 /s)
= 2.3752 × 10 6
Thus the flow is laminar
(1) Axial velocity u:
Grx dξ
u = 2v
x dη
(7.10)
29
Fig. 7.2:
dξ
vs. η
dη
Grx
η=
4
1/ 4
y
x
(a)
Ra L 2.3792 × 106
Grx = GrL =
=
= 3.351 ×
Pr
0.71
Use (a), evaluate η at x = 0.08 m and y = 0.002 m
3 .351 × 10
η =
4
6
1/4
0 .0032 ( m )
= 1 .21
0 . 08 ( m )
Fig. 7.2, at η = 1.21 and Pr = 0.71, gives
30
dξ
x
=u
≈ 0.27
dη
2ν Grx
Solve for u
u=
2ν GrL
0.27
L
= 0.27
2(16.96 × 10
−6
2
6
)(m /s) 3.351 × 10
= 0.2096 m/s
0.08(m)
(1) Temperature T: At η = 1.21 and Pr = 0.71, Fig. 7.3 gives
T − T∞
θ =
≈ 0.43
Ts − T∞
T ≈ T∞ + 0.43(Ts − T∞ ) = 10(o C) + 0.43(70 − 10)(o C) = 35.8o C
(3) Velocity B.L. thickness
δ:
31
At y = δ , axial velocity u ≈ 0, Fig. 7.2 gives
GrL
η( x , δ ) ≈ 5 =
4
Solve for δ
1/ 4
δ
L
1/ 4
4
δ = 5(0.08)(m )
6
3.37321 × 10
= 0.0132 m = 1.32 cm
(4) Temperature B.L. thickness δ t :
At
y = δ t , T ≈ T∞ , θ ≈ 0. Fig. 7.3 gives
GrL
η( x , δ t ) ≈ 4.5 =
4
1/ 4
δ
L
32
Solve for δ t
δ t = 4.5(0.08)(m )
1/4
4
6
3.3511 × 10
= 0.0119 m = 1.19 cm
(5) Local Nusselt number:
hx
Grx
Nu x =
= −
k
4
1/4
dθ ( 0 )
dη
(7.22)
dθ ( 0 )
Table 7.1 gives
at Pr = 0.71
dη
dθ ( 0 )
= −0.5018
dη
Nusselt Number: Use (7.22), evaluate at x = L = 0.08 m
33
NuL =
hL
k
GrL
=−
4
1/ 4
6 1 / 4
3.351 × 10
= 0.5018
dη
4
dθ ( 0 )
= 15.18
(6) Local heat transfer coefficient: at x = L = 0.08 m
k
0.0271(W/m− o C)
h( L) = NuL =
15.18 = 5.14 W/m 2 − o C
L
0.08(m)
(7) Heat flux: Newton’s law gives
q ′s′ = h (T s − T∞ ) = 5 .14 ( W/m − o C ) ( 70 − 10 )( o C) = 308.4 W/m
2
(8) Total heat transfer:
qT = hA(Ts − T∞ )
34
4 k GrL
h=−
3 L 4
4 (0.0271)(W/m − C) 3.351 × 10
o
h=−
3
0.08(m)
4
1/4
dθ (0)
dη
6 1/4
( −0.5018 ) =
6.86 W/m 2 − o C
qT = 6.86(W/m 2 − o C)0.08(m)0.08(m)(70 − 10)( o C) = 2.63 W
(iii) Checking
Dimensional check:
Units for
u, T , δ , δ t , Nu, h, q ′′ and qT are consistent
Quantitative check:
35
(i) h is within the range given in Table 1.1
(ii) δ
> δ t . This must be the case
(5) Comments
(i) Magnitudes of u and h are relatively small
(ii) h( x ) < h ( x )
7.5 Laminar Free Convection over a Vertical
Plate: Uniform Surface Heat Flux
• Vertical Plate
• Uniform surface heat flux
• Infinite fluid at temperature T∞
• Assumptions: Same as Section 7.4
36
• Governing equations: Same as Section 7.4
• Boundary conditions: Replace uniform surface temperature with
uniform surface flux
∂ T ( x ,0 )
−k
= q ′s′
∂y
• Surface temperature is variable and unknown: Ts ( x )
• Objective: Determine Ts ( x ) and
Nu x
• Solution: Similarity transformation (Appendix F)
• Results:
(1) Surface temperature
37
ν (q ′s′ )
Ts ( x ) − T∞ = − 5
x
4
β gk
2
θ (0) = constant, depends on Pr,
4
1/ 5
θ ( 0)
given in Table 7.2
Table 7.2 [4]
(2) Nusselt number
β gq ′s′ 4
Nu x = − 2 x
5ν k
1/ 5
(7.27)
1
θ ( 0)
(7.28)
Correlation equation for θ ( 0) [5]
4 + 9 Pr + 10 Pr
θ ( 0) = −
2
5 Pr
1/2
Pr
θ ( 0)
0.1
1.0
- 2.7507
10
- 0.76746
100
- 0.46566
- 1.3574
1/ 5
(7.29)
38
Example 7.2: Vertical Plate at Uniform Surface Flux
•
8cm × 8cm vertical plate in air at 10 oC
• Uniform surface flux = 308.4 oC
• Determine the following at x = 2, 4, 6 and 8 cm
(1) Surface temperature
(2) Nusselt number
(3) Heat transfer coefficient
Solution
(1) Observations
• External free convection
• Vertical plate
39
• Uniform surface heat flux
• Check Rayleigh number for laminar flow
• If laminar:
Equation (7.27) gives Ts ( x )
Equation (7.28) gives
Nux
Nux gives h(x)
(2) Problem Definition
Determine Ts ( x ) ,
in free convection
Nux and h(x) for uniformly heated vertical plate
(3) Solution Plan
• Laminar flow? Compute Rayleigh number
40
• Use (7.27) for Ts ( x )
• Use (7.28) for
•
Nux
Nux gives h(x)
(4) Plan Execution
(i) Assumptions
(10) Newtonian fluid
(11) Steady state
(12) Boussinesq approximations
(13) Two-dimensional
9
(14) Laminar flow ( Ra x < 10 )
(15) Flat plate
41
(16) Uniform surface heat flux
(17) No dissipation
(18) No radiation
(ii) Analysis
Rayleigh number:
β g(Ts − T∞ ) L3
Ra L =
να
(7.2)
Surface temperature:
ν (q ′s′ )
Ts ( x ) − T∞ = − 5
x
4
β gk
2
4
1/ 5
θ ( 0)
(7.27)
Nusselt number:
42
β gq ′s′ 4
Nu x = − 2 x
5ν k
1/ 5
1
θ ( 0)
(7.28)
Heat transfer coefficient
k
h( x ) = Nu x
x
(a)
Determine θ ( 0) : Table 7.2 or equation (7.29)
4 + 9 Pr + 10 Pr
θ ( 0) = −
2
5 Pr
1/ 2
1/ 5
(7.29)
(iii) Computations
Properties at T f
43
Ts + T∞
Tf =
2
(b)
where
Ts ( L ) + T s ( 0 )
Ts =
2
(c)
Ts (L) is unknown. Use iterative procedure:
(1) Assume Ts (L)
(2) Compute T f
(3) Find properties at T f
(4) Use (7.27) to calculate Ts (L)
(5) Compare with assumed value in step (1)
(6)
Repeat until (1) and (5) agree
44
o
Assume Ts ( L) = 130 C
Ts ( L) + Ts (0) (130 + 10)( o C)
Ts =
=
= 70o C
2
2
Ts + T∞ ( 70 + 10)o C
Tf =
= 40o C
=
2
2
o
Properties of air at 40 C :
k = 0.0271 W/m − o C
Pr = 0.71
ν = 16.96 × 10−6 m 2 /s
45
16.96 × 10 −6 m 2 /s
α=
=
= 23.89 × 10 −6 m 2 /s
Pr
0.71
ν
1
β= o
= 0.0031934 K -1
40 C + 273.13
Substitute into (7.2)
0.0031934 ( K −1 )9.81(m/s 2 )(130 − 10)(K)(0.08) 3 (m 3 )
Ra L =
16.96 × 10 − 6 (m 2 /s)23.89 × 10 − 6 (m 2 /s)
= 4.7504 × 10 6
•
Flow is laminar
Use (7.9) to determine θ (0)
46
4 + 9(0.71) + 10(0.71)
θ ( 0) = −
2
5(0.71)
1/ 2
1/ 5
= −1.4928
Use (7.27) to compute Ts (L)
o
Ts (L) = 10( C) −
1/ 5
6
2
4
2
4
4
8
−
(16.96× 10 ) (m /s )(308.4) (W /m )
o
5
(0.1)(m)
(
−
1
.
4928
)
=
63
.
6
C
2
4 4 4 4
0.0031934(1/K)9.81(m/s )(0.0271) (W /m −K
• Computed Ts (L) is lower than assumed value
• Repeat procedure until assume Ts (L) ≈ computed Ts (L)
Result: Ts ( L) = 63.2o C
47
Thus,
T f = 23.3o C
Properties of air 23.3o C
k = 0.02601 W/m − o C
Pr = 0.7125 25
ν = 15.55 × 10− 6 m 2 /s
α = 21.825 × 10− 6 m 2 /s
β = 0.003373 K -1
(7.29) gives θ ( 0)
θ ( 0) = −1.4913
(7.27) gives Ts ( x )
48
T s ( L ) = 10 ( o C ) −
(15.55 × 10 ) (m /s )(308.4) (W /m )
5
2
4
4
4
4
0.003373
(1/K)9.81
(m/s
)(0.02601)
(W
/m
−
K
−6
2
4
2
4
4
8
1/5
(x1)(m)
( − 1 .4913 )
(7.28) gives Nu x
0.003373(1/K)9.81(m/s )308.4(W/m ) 4 4
Nux = −
(x) (m )
−6
2
o
5(15.55 × 10 )(m /s)0.02601(W/m − C)
2
2
1/ 5
1
− 1.4913
(e)
(a) gives h(x)
49
k
0.02601(W/m − o C)
h( L) = NuL =
NuL
L
x (m)
(f)
Use (d)-(f) to tabulate results at x = 0.02, 0.04, 0.06 and 0.08 m
x(m) Ts ( x ) (o C) Nux
0.02
0.04
0.06
0.08
50.3
56.3
60.2
63.2
5.88
10.24
14.16
17.83
h( x )( W/m 2 − o C)
7.65
6.66
6.14
5.795
q′s′ ( W/m 2 )
308.3
308.4
308.2
308.3
(iii) Checking
50
Dimensional check:
Units for Ts , Nu and h are consistent.
Quantitative check:
(i) h is within the range given in Table 1.1
(ii) Compute q ′s′ using Newton’s law gives uniform surface flux
(5) Comments
(i) Magnitude of h is relatively small
(ii) Surface temperature increases with distance along plate
7.6 Inclined Plates
Two cases:
51
•
•
•
Hot side is facing downward, Fig. 7.5a
Cold side is facing upward, Fig. 7.5b
Flow field: Same for both
• Gravity component = gcos θ
•
Solution: Replace g by gcos θ in
solutions of Sections 7.4 and 7.5
•
Limitations
θ ≤ 60o
7.7 Integral Method
52
• Obtain approximate solutions to free
convection problems
• Example: Vertical plate at uniform surface
temperature
7.7.1 Integral Formulation of
Conservation of Momentum
•
Simplifying assumption
δ = δt
(a)
valid for Pr ≈ 1
Apply momentum theorem in x-direction to the element δ × dx
53
∑ Fx
= M x (out ) − M x (in )
∑ Fx :
•
•
•
Mx +
Wall stress
Pressure
Gravity
(b)
dM x
dx
dx
pδ +
d
( pδ )dx
dx
dy
τ odx
dx
M x dW ρ gdxdy
pδ
((pp++dp
dp//22))ddδδ
Fig. 7.7
dM x
dp
d
pδ + p + dδ − pδ − ( pδ )dx − τ o dx − = M x +
dx − M x
2
dx
dx
(c)
Simplify
− δ dp − τ o dx − dW =
dM x
dx
dx
(d)
Wall stress:
54
τo = µ
∂u( x ,0 )
∂y
(e)
Weight:
• Variable density
• Integrate weight of element dx × dy
dW = dx
δ
∫0 ρ gdy
(f)
x-momentum:
• Constant density
δ ( x)
Mx = ρ
∫0
u2dy
(g)
Substitute(e), (f) and (g) into (d)
55
δ
δ
∂u( x ,0 )
dp
d
−µ
−δ
− ∫ ρgdy = ρ ∫ u2dy
∂y
dx 0
dx 0
(h)
Combine pressure and gravity terms (B.L. flow)
dp dp∞
= −ρ∞ g
≅
dx
dx
(i)
Rewrite
δ
dp
δ
= − ρ ∞ gδ = − ∫ ρ ∞ gdy
dx
0
(j)
(j) into (h)
∂u( x ,0 )
−µ
+g
∂y
δ
∫0
d
( ρ ∞ − ρ )dy = ρ
dx
δ
∫0
u 2dy
(k)
Density change:
56
ρ ∞ − ρ = ρβ (T − T∞ )
(2.28)
(2.28) into (k), assume constant ρ β
∂u( x ,0)
−ν
+βg
∂y
δ
∫0
d
(T − T∞ )dy =
dx
δ
∫0
u2 dy
(7.30)
NOTE:
(1)
no shearing force on the slanted surface
(2)
(7.30) applies to laminar and turbulent flow
(3)
(7.30) is a first order O.D.E.
7.7.2 Integral Formulation of Conservation of Energy
Assume:
57
•
•
•
•
No changes in kinetic and potential energy
Neglect axial conduction
Neglect dissipation
Properties are constant
∂T ( x ,0 )
d
−α
=
∂y
dx
NOTE:
δ ( x)
∫0
u(T − T∞ )dy
(7.31)
Integral formulation of energy is the same for
free convection and for forced convection
7.7.3 Integral Solution
58
• Vertical plate
• Uniform temperature
Ts
• Quiescent fluid at uniform temperature T∞
Solution Procedure for Forced Convection:
(1) Velocity is assumed in terms of δ ( x )
59
(1) Momentum gives δ ( x )
(2) Temperature is assumed in terms δ t ( x )
(3) Energy gives δ t ( x )
Solution Procedure for Free Convection:
Since we assumed δ = δ t ,
either momentum or energy gives δ t .
To satisfy both,
introduce another unknown in assumed velocity
60
Assumed Velocity Profile
u( x , y ) = a 0 ( x ) + a 1 ( x ) y + a 2 ( x ) y 2 + a 3 ( x ) y 3
(a)
Boundary conditions:
(1) u( x ,0) = 0
(2) u( x , δ ) ≅ 0
(3)
(4)
∂u( x , δ )
≅0
∂y
∂ 2 u( x ,0)
∂y 2
=−
βg
(Ts − T∞ )
ν
• Setting y = 0 in the x-component, (7.2), to obtain condition (4)
• 4 boundary conditions give an
61
(a) gives
β g(Ts − T∞ )
y y2
δ y 1 − 2 +
u=
2
2
δ δ
4ν
Rewrite as
2
β g(Ts − T∞ ) 2 y
y
δ 1 −
u=
2
δ δ
4ν
(b)
Introduce a second unknown in (b). Define
β g(Ts − T∞ ) 2
δ
uo ( x ) =
2
4ν
(c)
(b) becomes
62
y
y
u = uo ( x ) 1 −
δ δ
2
(7.32)
Assumed Temperature Profile
T ( x , y ) = b0 ( x ) + b1 ( x ) y + b2 ( x ) y 2
(d)
Boundary Conditions:
(1) T ( x ,0) = Ts
(2) T ( x , δ ) ≅ T∞
(3) ∂T ( x , δ ) ≅ 0
∂y
≅
• 3 boundary conditions give bn
63
(d) becomes
y
T ( x , y ) = T∞ + (Ts − T∞ ) 1 −
δ
2
(7.33)
Heat Transfer Coefficient and Nusselt Number
∂ T ( x ,0 )
−k
∂y
h=
Ts − T∞
(7.20)
(7.33) into (7.20)
2k
h=
δ ( x)
(7.34)
Local Nusselt number:
64
Nu x =
hx
x
=2
k
δ ( x)
(7.35)
Determine δ ( x )
Solution
Use momentum: substitute (7.32) and (7.33) into (7.30)
−ν
uo
δ
δ
+ β g (Ts − T∞ )
∫0
2
y
d uo2
1 − δ dy = dx 2
δ
δ
∫0
4
y
y 1 − dy
δ
2
(e)
Evaluating the integrals
[ ]
u
1 d 2
1
uo δ = βg (Ts − T∞ ) δ − v o
105 dx
3
δ
(7.36)
Use energy: substitute(7.32) and (7.33) into (7.31)
65
d uo
2α (Ts − T∞ ) = (Ts − T∞ )
dx δ
δ
1
δ ( x)
∫0
4
y
y 1 − dy
δ
(f)
Evaluating the integrals
1 d
[uoδ ] = α 1
60 dx
δ
(7.37)
• The two dependent variables are δ ( x ) and uo ( x )
• (7.36) and (7.37) are two simultaneous first order O.D.E.
• Assume a solution of the form
uo ( x ) = Ax m
(7.38)
δ ( x ) = Bx n
(7.39)
• Determine the constants A, B, m and n
66
• Substitute (7.38) and (7.39) into (7.36) and (7.37)
2m + n 2 2m + n−1 1
A
A Bx
= βg (To − T∞ )Bx n − vx m − n
105
3
B
m+n
1 −n
m
+
n
−
1
ABx
=α x
210
B
(7.40)
(7.41)
To satisfy (7.40) and (7.41) at all values of x ,
the exponents of x in each term must be identical
(7.40) requires
2m + n − 1 = n = m − n
(g)
Similarly, (7.41) requires that
m + n − 1 = −n
(h)
67
Solve (g) and (h) for m and n gives
1
1
m = , n=
4
2
(i)
Introduce (i) into (7.40) and (7.41)
1 2
1
A
A B = βg (To − T∞ )B − v
85
3
B
(j)
1
1
AB = α
280
B
(k)
Solve (j) and (k) for A and B
−1 / 2
20 −
A = 5.17ν Pr +
21
β g(Ts − T∞ )
ν2
1/ 2
(l)
and
68
1/ 4
20
B = 3.93 Pr -1/2 Pr +
21
β g(Ts − T∞ )
ν2
−1 / 4
(m)
Substitute (i) and (m) into (7.39)
20 1
= 3.93
+ 1
x
21 Pr
δ
1/ 4
( Ra x ) −1 / 4
(7.42)
Introduce(7.42) into (7.35)
−1 / 4
−
20 1
Nu x = 0.508
+ 1
21
Pr
( Ra x )1 / 4
(7.43)
7.7.4 Comparison with Exact Solution for Nusselt Number
Exact solution:
Gr
Nu x = − x
4
1/ 4
dθ (0)
dη
(7.22)
69
Rewrite above as
Grx
4
−1 / 4
−1 / 4
−
20 1
Nu x = 0.508
+ 1
21
Pr
(4 Pr )1 / 4
(7.45)
Rewrite integral solution (7.43) as
Grx
4
−1 / 4
Nu x = −
dθ (0)
dη
(7.44)
• Compare right hand side of (7.45) with − dθ ( 0) / dη of exact solution
(7.44)
• Comparison depends on the Prandtl number
70
• Table 7.2 gives results
Table 7.3
Limiting Cases
Pr
(1) Pr → 0
Nu x exact = 0.600 ( PrRa x )1/4 , Pr → 0 (7.25a)
Nu x integral = 0.514( PrRa x )1 / 4 , Pr → 0 (7.46a)
(2) Pr → ∞
Nu x exact = 0.503 ( Ra x )1/4 ,
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
_ dθ (0)
dη
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
Pr → ∞
Nu x integral = 0.508( Ra x )1 / 4 , Pr → ∞
20 1
0.508
+ 1
21 Pr
−1 / 4
( 4 Pr )1/4
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
1.2665
4.0390
(7.25b)
(7.46b)
71
NOTE:
(1) Error ranges from 1% for Pr → ∞ to 14% for Pr → 0
(2) We assumed δ = δ t ( Pr = 1 ). Solution is reasonable for all Prandtl
numbers
72
CHAPTER 8
CONVECTION IN EXTERNAL TURBULENT FLOW
8.1 Introduction
• Common physical phenomenon, but complex
• Still relies on empirical data and rudimentary conceptual drawings
• Tremendous growth in research over last 30 years
y
V∞
uo
u(y)
turbulent wake behind body
turbulent jet
V∞
uo
turbulent wake behind
smokestack
1
8.1.1 Examples of Turbulent Flows
(i) Mixing Processes
(ii) Free Shear Flows
(iii) Wall-Bounded Flows
• Varying shape of instantaneous velocity profile
• Instantaneous velocity fluctuation
u
y
u′(y,t)
u(y,t)
u(y,t)
u
Instantaneous velocity
profile
u
velocity profiles
Instantaneousu(t)
velocity fluctuation
t
fluctuating component
2
Turbulent velocity profile vs. laminar
• Time averaged
u(y)
y
δturb
turbulent
δlam
laminar
u
Turbulent flows can enhance performance
• Turbulators
• Dimpled golf balls
3
8.1.2 The Reynolds Number and the Onset of Turbulence
• Osborne Reynolds (1883) first identifies laminar and turbulent regimes
• Reynolds number:
Re D =
uD
ν
(8.1)
• Internal flow: critical flow number is Rec = uD / ν ≈ 2300
• Flow over semi-infinite flat plate is Rec = V∞ xt / ν ≈ 500, 000
Why the Reynolds number predicts the onset of turbulence
• Reynolds number represents the ratio of inertial to viscous forces
o Inertial forces accelerate a fluid particle
o Viscous forces slow or damp the motion of the particle
• At low velocity, viscous forces dominate
o Infinitesimal disturbances damped out
o Flow remains laminar
4
• At high enough fluid velocity, inertial forces dominate
o Viscous forces cannot prevent a wayward particle from motion
o Chaotic flow ensues
Turbulence near wall
• For wall-bounded flows, turbulence initiates near the wall
5
8.1.3 Eddies and Vorticity
• An eddy is a particle of vorticity, ω,
r
ω = ∇ ×V
r
(8.2)
• Eddies typically form in regions of velocity gradient.
• Vorticity can be found from Eqn. (8.2) to be
∂v ∂ u
ωz =
−
∂x ∂y
A Common View of Eddy Formation
• Eddy begins as a disturbance near the wall
• Vortex filament forms
• Stretched into horseshoe or hairpin vortex
• Lifting phenomenon
6
7
8.1.4 Scales of Turbulence
• Largest eddies break up due to inertial forces
• Smallest eddies dissipate due to viscous forces
• Richardson Energy Cascade (1922)
8
Kolgomorov Microscales (1942)
• Attempt to estimate size of smallest eddies
Re −3/ 4
v / u Re −1/4
τ / t Re −1/ 2
η/l
(8.3a)
(8.3b)
(8.3c)
• Important impacts:
o There is a vast range of eddy sizes, velocities, and time scales
in a turbulent flow. This could make modeling difficult.
o The smallest eddies small, but not infinitesimally small.
Viscosity dissipates them into heat before they can become
too small.
o Scale of the smallest eddies are determined by the scale of
the largest eddies through the Reynolds number.
Generating smaller eddies is how the viscous dissipation is
increased to compensate for the increased production of
turbulence.
9
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 by viscosity into heat.
• Turbulent flows are not only dissipative, but also dispersive through
the advection mechanism.
8.1.6 Analytical Approaches
• Considering small eddies, is continuum hypothesis still valid?
o The smallest eddies: approximately 2 × 10−5 m
10
o Mean free path of air at atmospheric pressure is on the order
of 10−8 m three orders of magnitude smaller
o Continuum hypothesis OK
• Are numerical simulations possible?
o Direct Numerical Simulation (DNS) a widespread topic of
research
o However, short time scales and size range of turbulence a
problem
o Still have to rely on more traditional analytical techniques
Two Common Idealizations
• Homogeneous Turbulence: Turbulence, whose microscale motion,
on average, does not change from location to location and time to time.
• Isotropic Turbulence: Turbulence, whose microscale motion,
on average, does not change as the coordinate axes are rotated.
11
8.2 Conservation Equations for Turbulent Flow
8.2.1 Reynolds Decomposition
• Turbulent flow seems well-behaved on average.
• Reynolds Decomposition: Separate velocity, properties into timeaveraged and fluctuating components:
g = g + g′
(8.4)
• Time-averaged component is determined by:
g=
1
τ
g ( t )dt
∫
τ
(8.5)
0
• Time average of fluctuating component is zero:
g′ =
1
τ
g ′( t )dt = 0
∫
τ
(8.6)
0
12
Average Identities:
• For two variables a = a + a′ and b = b + b′
a =a
(a )
2
= (a )
2
ab = ab + a ′b′
a+b=a +b
∂a
=0
∂t
(8.7a)
ab = ab
(8.7b)
(8.7c)
aa ′ = 0
(8.7d)
(8.7e)
a = ( a ) + ( a′ )
(8.7g)
∂a ∂a
=
∂x ∂x
(8.7h)
(8.7i)
∂a
=0
∂t
(8.7j)
2
2
2
(8.7f)
8.2.2 Conservation of Mass
13
• By identities (8.7a) and (8.6):
∂ρ ∂ ( ρ u ) ∂ ( ρ v ) ∂ ( ρ w )
+
+
=0
+
∂t
∂x
∂y
∂z
(2.2a)
• Assume incompressible, two-dimensional flow. Substituting the
Reynolds-decomposed velocities u = u + u′ and v = v + v ′,
∂ ( u + u′ ) ∂ ( v + v ′ )
+
=0
∂y
∂x
(a)
∂u ∂u′ ∂v ∂v ′
+
=0
+
+
∂x ∂ x ∂y ∂ y
(b)
• Expanding,
• Time-average the equation:
∂ u ∂ u′ ∂ v ∂ v ′
+
=0
+
+
∂x ∂ x ∂ y ∂y
(c)
• Then, simplify each term by invoking identity (8.7h):
14
∂ u ∂ u′ ∂ v ∂ v ′
+
=0
+
+
∂ x ∂x ∂ y ∂y
(d)
• By identities (8.7a) and (8.6),
∂ u ∂v
=0
+
∂x ∂y
(8.8)
8.2.3 Conservation of Momentums
• The x and y momentum equations are given by:
∂ 2u ∂ 2u ∂ 2u
∂u
∂u
∂u
∂u
∂p
ρ + u + v + w = ρ gx − + µ 2 + 2 + 2
∂x
∂y
∂z
∂y
∂z
∂x
∂t
∂x
(2.10x)
∂ 2v ∂ 2v ∂ 2v
∂v
∂v
∂v
∂v
∂p
ρ + u + v + w = ρ gy − + µ 2 + 2 + 2
∂y
∂z
∂y
∂z
∂x
∂y
∂t
∂x
(2.10y)
15
• Simplifying for steady, 2D flow, no body forces:
∂ 2u ∂ 2u
∂u
∂u
∂u
∂p
ρ +u +v = − +µ 2 + 2
∂x
∂y
∂x
∂y
∂t
∂x
(8.10x)
∂ 2v ∂ 2v
∂v
∂v
∂v
∂p
ρ +u +v = − + µ 2 + 2
∂y
∂y
∂x
∂y
∂t
∂x
(8.10y)
• For the x-momentum equation, the terms u(∂u / ∂x ) and v (∂u / ∂y ) can
be replaced by the following relations, derived from the product rule of
derivation:
∂ u ∂u 2
∂u
u
=
−u
(a)
∂x ∂ x
∂x
∂u ∂ ( uv )
∂v
v
=
−u
∂y
∂y
∂y
(b)
• Substitute (a) into the x-momentum equation (8.10x):
2
∂u ∂u
∂ 2u ∂ 2u
∂u ∂ ( uv )
∂v
∂p
ρ +
+
−u =−
+ µ 2 + 2
−u
∂y
∂y
∂x
∂y
∂x
∂x
∂t ∂x {
{
a
b
(c)
16
• Note that terms marked
and
in the above can be combined as:
∂u ∂v
−u
+ = 0 by continuity
∂x ∂y
(d)
• Thus, the x-momentum equation reduces to:
∂u ∂u 2 ∂ ( uv )
∂ 2u ∂ 2u
∂p
+ µ 2 + 2
ρ +
+
=−
∂x
∂y
∂y
∂t ∂x
∂x
(8.11)
• Following Reynolds decomposition and averaging,
∂ 2u ∂ 2u
∂u
∂p
∂ ( u′ ) 2
∂ u′ v ′
∂u
ρu
+v
+ µ 2 + 2 −ρ
−ρ
(8.12x)
=−
∂
x
∂
y
∂
x
∂
x
y
∂
x
∂
y
∂
∂ 2v ∂ 2v
∂v
∂v
∂p
∂ u′v ′
∂ ( v ′ )2
ρu
+v
+ µ 2 + 2 − ρ
−ρ
(8.12y)
=−
∂y
∂y
∂y
∂x
∂y
∂x
∂x
8.2.4 Conservation of Energy
17
• For incompressible flow, negligible heat generation, constant properties,
the energy equation is given by
∂T
∂T
∂T
∂T
+u
+v
+w
ρcp
∂x
∂y
∂z
∂t
∂ 2T ∂ 2T ∂ 2T
= k 2 + 2 + 2
∂y
∂z
∂x
+ µΦ
(2.19b)
• The energy equation reduces to:
∂T
∂T
∂T
ρcp
+u
+v
∂x
∂y
∂t
∂ 2T ∂ 2T
= k 2 + 2
∂y
∂x
(8.13)
• Following Reynolds decomposition and time averaging, Eqn. (8.13)
becomes:
∂ 2T ∂ 2T
∂T
∂T
+v
ρcp u
= k 2 + 2
∂y
∂y
∂x
∂x
• NOTE TWO NEW TERMS
(
)
(
∂ u′ T ′
∂ v ′T ′
− ρcp
− ρcp
∂x
∂y
)
(8.14)
18
8.2.5 Summary of Governing Equations for Turbulent Flow
• Continuity:
∂ u ∂v
=0
+
∂x ∂y
(8.8)
• x-momentum:
∂ 2u ∂ 2u
∂u
∂p
∂ ( u′ ) 2
∂ u′ v ′
∂u
+v
+ µ 2 + 2 − ρ
−ρ
ρu
(8.12x)
=−
∂y
∂x
∂x
∂y
∂y
∂x
∂x
• y-momentum:
∂ 2v ∂ 2v
∂v
∂p
∂v
+v
+µ 2 + 2
ρu
=−
∂y
∂y
∂y
∂x
∂x
∂ u′v ′
∂ ( v ′ )2
−ρ
− ρ
∂x
∂y
(8.12y)
• Energy:
∂ 2T ∂ 2T
∂T
∂T
+v
ρcp u
= k 2 + 2
∂y
∂y
∂x
∂x
(
)
(
∂ u′ T ′
∂ v ′T ′
− ρcp
− ρcp
∂x
∂y
)
(8.14)
19
8.3 Analysis of External Turbulent Flow
8.3.1 Turbulent Boundary Layer Equations
(i) Turbulent Momentum Boundary Layer Equation
• Consider a flat plate in turbulent flow.
• Assume boundary layer is thin:
δ
1
L
(8.15)
• Following the same arguments as for the laminar boundary layer, the
following scalar arguments are made:
u V∞
(8.16a)
x
L
(8.16b)
y
δ
(8.16c)
20
• It can be shown that the viscous dissipation terms in (8.12x) compare
as follows:
∂ 2u
∂x 2
∂ 2u
∂y 2
(8.17)
• Also, the pressure gradient in the y-direction is negligible:
∂p
≈0
∂y
(8.18)
• The pressure gradient in the x-direction can be expressed as:
∂p dp dp∞
=
=
∂x dx dx
(8.19)
Simplifying the Fluctuation Terms:
• Fluctuation Terms:
∂ ( u′ )
∂x
2
∂ u′ v ′
and
∂y
• If fluctuation terms are the result of eddies, one could argue that
there is no preferred direction to the fluctuations:
21
u′
v′
(8.20)
u′v ′
(8.21)
or
( u′ )
2
• Using scale analysis:
First Fluctuation Term:
∂ ( u′ )
∂x
Second Fluctuation Term:
• Since δ / L
2
( u′ )
L
∂ u′v ′
∂y
2
(a)
u′v ′
( u′ )
δ
δ
2
(b)
1, we conclude that:
∂ ( u′ )
∂x
2
∂ u′v ′
∂y
(8.22)
• The x-momentum equation for the turbulent boundary layer reduces to:
22
∂u
dp
∂ 2u
∂ u′v ′
∂u
ρu
+v
+µ 2 −ρ
=−
∂y
dx
∂y
∂y
∂x
(8.20)
(ii) Turbulent Energy Equation
• Scaling arguments for the thermal boundary is:
x
L
(8.16b)
y
δt
(8.24)
Ts − T∞
(8.25)
∂ 2T
∂y 2
(8.26)
∆T
• Then:
∂ 2T
∂x 2
• Fluctuation terms:
ρcp
(
∂ u′T ′
∂x
) and ρ c ∂ ( v′T ′ )
p
∂y
• Assuming there is no preferred direction to the fluctuations:
u′ v ′
(8.20)
23
or
u′T ′
v ′T ′
(8.27)
∂ ( v ′T ′ )
∂y
(8.28)
• We can then show that:
∂ ( u′T ′ )
∂x
• The energy equation then reduces to:
(
∂ v ′T ′
∂T
∂ 2T
∂T
ρcp u
+v
= k 2 − ρcp
∂y
∂y
∂y
∂x
)
(8.29)
8.3.2 Reynolds Stress and Heat Flux
• Can write the x-momentum and energy boundary layer equations as:
∂u
dp ∂ ∂u
∂u
′
′
ρu
+v
+ µ
− ρu v
=−
∂y
dx ∂y ∂y
∂x
(8.30)
∂T
∂T ∂ ∂T
ρcp u
+v
− ρ c p v ′T ′
=
k
∂ y ∂ y ∂y
∂x
(8.31)
24
• Fluctuating term in (8.30) “looks” like a shear stress
• Fluctuating term in (8.31) “looks” like a heat flux
• Consider a particle “fluctuation” imposed on some average velocity profile
• ρ u′v ′ is called the turbulent shear stress or the Reynolds stress
•ρ c p v ′T ′is called the turbulent heat flux or the Reynolds heat flux
8.3.3 The Closure Problem of Turbulence
• Turbulent boundary layer equations:
• Continuity:
∂ u ∂v
=0
+
∂x ∂y
(8.8)
25
∂u
dp ∂ ∂u
∂u
′
′
• x-momentum: ρ u
+ µ
− ρu v
+v
=−
∂y
dx ∂y ∂y
∂x
• Energy:
∂T
∂T ∂ ∂T
′
′
+v
− ρcp v T
ρcp u
= k
∂ y ∂y ∂y
∂x
(8.30)
(8.31)
• Boundary conditions:
u ( x , 0) = 0
v ( x , 0) = 0
u ( x , ∞ ) = V∞
u (0, y ) = V∞
T ( x , 0) = Ts
T ( x , ∞ ) = T∞
T (0, y ) = T∞
(8.31a)
(8.31b)
(8.31c)
(8.31d)
(8.31e)
(8.31f)
(8.31g)
• Also have, outside the boundary layer:
dp dp∞
=
dx dx
dV∞
1 dp∞
V∞
=−
dx
ρ dx
(8.32)
(8.33)
26
• Leaves us with three equations (8.8), (8.30) and (8.31), but five unknowns:
u , v , T , u′v ′ and v ′T ′
• This is the closure problem of turbulence.
8.3.4 Eddy Diffusivity
• Customary to model the Reynolds stress as
− ρ u′v ′ = ρε M
∂u
∂y
(8.34)
• ε M is called the momentum eddy diffusivity. ρε M is often referred to as
eddy viscosity.
• Similarly, we can model the Reynolds heat flux as
− ρ c p v ′T ′ = ρ c pε H
∂T
∂y
(8.35)
• ε H is called the thermal eddy diffusivity. ρ c pε H is often referred to as
eddy conductivity.
• We can then write the boundary layer momentum and energy equations as
27
∂u
∂u ∂
∂u
u
+v
=
(ν + ε M )
∂x
∂y ∂ y
∂y
(8.38)
∂T
∂
∂T
∂T
u
+v
=
(α + ε H )
∂x
∂y ∂y
∂y
(8.39)
• The terms in brackets represent the apparent shear stress and apparent
heat flux, respectively:
τ app
∂u
= (ν + ε M )
∂y
ρ
−
′′
qapp
ρcp
= (α + ε H )
∂T
∂y
(8.40)
(8.41)
8.4 Momentum Transfer in External Turbulent Flow
8.4.1 Modeling Eddy Diffusivity: Prandtl’s Mixing Length Theory
• Simplest model by Boussinesq: constant ε M
o does not allow u′v ′ to approach zero at the wall
28
• Prandtl (1925): used analogy to kinetic theory of gases
• Define the mixing length l as the distance the particle travels towards the
wall as the result of a fluctuation.
• The velocity fluctuation u′ that results can be approximated from a
Taylor series as
∂u
dy
∂y
− uinitial
u′ ≈ uinitial +
(a)
u′ = u final
(a)
• Thus,
u′
∂u
l
∂y
(b)29
• If we assume, as we have before, that fluctuations have no preferred
direction, then u′ v ′, and so
v′
∂u
l
∂y
(c)
• One could argue, then, that the turbulent stress term − u′v ′ is of the
following scale:
2
− u′v ′
( u′ ) ( v ′ )
∂u
l2
∂
y
(d)
• Finally, we can solve Equation (8.34), for the eddy viscosity:
εM
− u′ v ′
=
∂u / ∂y
l2
∂u
∂y
(8.42)
• Prandtl proposed the following model for the mixing length,
l =κy
(8.43)
• Leading to Prandtl’s mixing-length model:
εM
∂u
=κ y
∂y
2
2
(8.44)
30
8.4.2 Universal Turbulent Velocity Profile
• One way to solve momentum is to assume a velocity profile, then use
approximate methods to solve integral momentum (like in Chap. 5)
(i) Large-Scale Velocity Distribution: “Velocity Defect Law”
• First Step: normalize variables: u / V∞ vs. y / δ
• Doesn’t collapse curves with varying friction
Velocity Defect Law
• Introduce coefficient of friction:
τo
Cf =
( 1 / 2 ) ρV∞ 2
(8.45)
• Second, define a friction velocity as:
u* ≡ τ o / ρ
(8.46)
u* = V∞ C f / 2
(8.47)
31
• Define velocity defect:
u − V∞
u*
(8.48)
• This works, but doesn’t provide
enough detail near the wall.
(ii) Wall Coordinates
• Dimensional analysis suggests the
following wall coordinates:
u
u ≡ *
u
+
(8.49)
(iii) Near-Wall Profile: Couette Flow Assumption
• Very close to the wall, scaling analysis suggests:
∂
∂u
(ν + ε M ) 0
Near the Wall:
∂y
∂y
τ app
∂u
= (ν + ε M )
constant
∂y
ρ
(8.50)
(8.51)
32
• This result is similar to Couette flow: Couette Flow Assumption
• Resulting curve:
33
• What do we do with this? We can use (8.51) to develop an expression for
the velocity profile.
• First, we need to express (8.51) in terms of the wall coordinates u +
+
and y . Substituting their definitions, it can be shown that:
ε M ∂u +
1 + ν ∂y + = 1
(8.52)
• And after rearranging and integrating,
+
y
u+ =
dy +
∫ (1 + ε
0
M
(8.53)
/ν )
(iv) Viscous Sublayer
• Very close to the wall, viscous forces dominate,
ν
εM
• Couette Flow Assumption (8.52) reduces to:
∂u +
=1
+
∂y
• Integrating, with boundary condition u + = 0 at y + = 0
u + = y + , (0 ≤ y + ≤ 7)
(8.54)
34
• This relation compares well to experimental data from y + ≈ 0 to 7
which we call the viscous sublayer.
(v) Fully Turbulent Region: “Law of the Wall”
• Further away from the wall, turbulent fluctuations dominate, ε M
• Couette Flow Assumption (8.52) becomes
ε M ∂u +
=1
+
ν ∂y
ν
(8.55)
• Substitute Prandtl’s mixing length, and wall coordinates:
ε M = κ 2 ( y+ )
2
• Substitute into Equation (8.55),
κ
2
(y )
+
2
∂u
+
∂y
+
∂u +
ν +
∂y
(8.56)
2
=1
• Solve for the velocity gradient,
∂u +
1
=
∂y + κ y +
(8.57)
35
• Finally, integrate the above to obtain
u+ =
1
κ
ln y + + B
(8.58)
• This is sometimes referred to as the Law of the Wall.
• The constant κ is called von Karman’s constant, and experimental
measurements show that κ ≈ 0.41.
• The constant of integration B can be estimated by noting that the
viscous sublayer and the Law of the Wall region appear to intersect at
+
+
roughly y = u ≈ 10.8 . Using this as a boundary condition, the
integration constant is found to be β ≈ 5.0.
• Thus, 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
• van Driest’s continuous law of the wall
o van Driest proposed a mixing length model of this form:
l = κ y (1 − e− y/ A )
(8.60)
o van Driest used this equation with (8.42) and (8.51) to obtain:
36
τ app
2 ∂u
2 2
− y/ A
= ν + κ y ( 1 − e
)
∂y
ρ
(8.61)
• Transforming (8.61) into wall coordinates, and solving for ∂y + / ∂u +
one can obtain:
∂u +
=
+
∂y
2
2
1+ 1+κ y
+2
(1 − e
− y + / A+
)
2
(8.62)
• Spalding’s Law:
y + = u + + e − κβ
2
3
+
+
+
κ
u
κ
u
(
)
(
)
eκ u − 1 − κ u+ −
−
2
6
(8.63)
o Works for flat plate and pipe flow
• Reichardt’s Law, applied frequently to pipe flow:
y + −0.33 y +
− y+ / X
u = ln ( 1 + κ y ) + C 1 − e
e
−
κ
X
+
1
+
(8.64)
(vii) Effect of Pressure Gradient
37
• 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,” and the region y > 350 is
commonly referred to as the wake region, where the velocity profile
deviates even from the overlap region.
• 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, which
helps explain why the models
developed here apply as well to
pipe flow.
38
8.4.3 Approximate Solution for Momentum Transfer: Momentum Integral
Method
(i) Prandtl - von Karman Model
• Consider a flat, impermeable plate exposed to incompressible,
zero-pressure-gradient flow
• The integral momentum equation reduces to equation (5.5),
ν
∂u ( x , 0)
d
= V∞
∂y
dx
δ ( x)
∫
0
udy −
d
dx
δ (x)
∫
u 2 dy
(5.5)
0
• Applies to turbulent flows as well – without modification – if we look at
the behavior of the flow on average, and we interpret the flow properties
as time-averaged values.
Estimate of Velocity Profile
• The integral method requires an estimate for the velocity profile
in the boundary layer.
• Prandtl and von Kármán both used a crude but simple model
for the velocity profile using prior knowledge about pipe flow.
39
• Using the Blasius model for the shear at the wall of a circular pipe,
Prandtl [18] and von Kármán [16] each showed that the velocity
profile in the pipe could be modeled as
y
u
=
uCL ro
1/ 7
• This is the well-known 1/7th Law velocity profile, discussed
further in Chapter 9.
• Why base a velocity profile for flat plate on pipe flow? The velocity data
for pipe flow and flat plate flow (at zero or favorable pressure gradient)
have essentially the same shape, so the use of this model to describe flow
over a flat plate is not unreasonable.
• To apply the 1/7th law to flat plate, we approximate ro as the edge of
the boundary layer δ , and approximate uCL as V∞ . Then,
u y
=
V∞ δ
1/ 7
(8.65)
Model for Wall Shear
• LHS of Integral Momentum is an expression for wall shear; uses
assumed velocity profile.
40
• Problem: our assumed profile goes to infinity as y approaches zero.
• To avoid this dilemma, Prandtl and von Kármán again looked to
pipe flow knowledge
• They adapted the Blasius correlation for pipe flow friction factor to find an
expression for the wall shear on a flat plate
• Recasting the Blasius correlation terms of the wall shear and the tube
radius, they obtained
τo
V∞δ
0.02333
=
=
ν
2
ρV∞ 2
Cf
−1/ 4
(8.67)
• This is used in the LHS of the momentum integral relation.
Example 8.2: Integral Solution for Turbulent Boundary Layer
Flow over a Flat Plate
Consider turbulent flow over a flat plate, depicted in Fig. 8.8. Using
the 1/7th law velocity profile (8.65) and the expression for friction
factor (8.67), obtain expressions for the boundary layer thickness
and friction factor along the plate.
41
(1) Observations. The solution parallels that of Chapter 5 for laminar
flow over a flat plate.
(2) Problem Definition. Determine expressions for the boundary layer
thickness and friction factor as a function of x.
(3) Solution Plan. Start with the integral Energy Equation (5.5),
substitute the power law velocity profile (8.65) and friction factor
(8.67), and solve.
(4) Plan Execution.
(i) Assumptions. (1) Boundary layer simplifications hold, (2)
constant properties, (3) incompressible flow, (4) impermeable flat plate.
(ii) Analysis.
• Substitute 1/7th power law velocity profile into the Mom. Int. Equation:
δ (x)
1/ 7
δ ( x)
d
y
2 y
∫0 V∞ δ dy − dx ∫0 V∞ δ
• Dividing the expression by V∞ 2 , and collecting terms,
δ (x)
y 1/ 7 y 2/7
τo
d
=
− dy
2
∫
dx 0 δ
ρV∞
δ
τo
d
= V∞
ρ
dx
2/ 7
dy
(a)
(b)
42
• After integrating,
τo
7 dδ
=
ρV∞ 2 72 dx
(8.68)
• Now substituting (a) into the wall shear expression (8.67),
V∞δ
0.02333
ν
−1/4
7 dδ
=
72 dx
(c)
• Then, separating variables and integrating,
4 5/ 4
δ
45
72 V∞
= 0.02333
7 ν
−1/4
x+C
(8.69)
• To complete the solution, a boundary condition is needed.
• Can assume that δ ( x ) is zero at x = 0, which ignores the initial laminar
boundary layer region
• However crude the
assumption, we find that the
results of this analysis
compare well to experimental
data.
43
• With the boundary condition established, the integration constant C
equals zero. Then solving (8.69) for δ ( x ) ,
V∞ x
δ ( x ) = 0.3816
ν
δ 0.3816
=
x Re x 1/5
or
−1/5
(d)
x
(8.70)
• Finally, solve for the friction factor. Substituting (8.70) into (8.67),
V∞ x
0.3816V∞
Cf
ν
= 0.02333
2
ν
Which reduces to
Cf
0.02968
=
Re x 1/ 5
2
−1/ 5
x
−1/ 4
(e)
(8.71)
(5) Checking. Equations (8.70) and (8.71) are both dimensionless, as
expected.
44
(6) Comments. Note that, according to this model, the turbulent
boundary layer δ/x varies as Re x −1/ 5, as does the friction factor C f . This
−1/ 2
is contrast to laminar flow, in which δ/x and C f vary as Re x .
(ii) Newer Models
• One 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 considered to be of limited applicability even for
pipe flow.
White’s Model
• White [14] uses the Law of the Wall velocity profile (8.59) to model the
wall shear.
• First, substituting the definitions of u + and y + , as well as u* , into
the Law of the Wall expression (8.59),
u
V∞
yV
2
= 2.44 ln ∞
ν
Cf
Cf
2
−1/ 4
+ 5.0
• In theory, 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 = δ ) = V∞ . Then, the above can be expressed as
45
Cf
+ 5.0
= 2.44 ln Reδ
2
Cf / 2
1
(8.72)
• Still a difficult relation to use, but a simpler curve fit over a range of
4
7
values from Reδ ≈ 10 to 10 gives
(8.73)
C f ≈ 0.02 Reδ −1/ 6
δ
• We can now use this expression to estimate the wall shear in the integral
method.
• For the velocity profile, the 1/7th power law is still used.
• It can be shown that the solution to the momentum integral equation in
this case becomes
δ
0.16
(8.74)
=
1/7
x Reδ
and
C f 0.0135
(8.75)
=
2
Reδ 1/ 7
• Equations (8.74) and (8.75) replace the less accurate Prandtl-von
Kármán correlations, and White recommends these expressions for
general use.
46
Kestin and Persen’s Model
• Perhaps a more accurate correlation would result if we use one
of the more advanced velocity profiles to estimate the wall shear,
as well as to replace the crude 1/7th power law profile.
• Kestin and Persen used Spalding’s law of the wall for the
velocity profile and shear stress.
• The resulting model is extremely accurate, but cumbersome.
White [20] modified the result to obtain the simpler relation
0.455
Cf = 2
ln ( 0.06 Re x )
(8.76)
• White reports that this expression is accurate to within 1% of Kestin
and Persen’s model.
(iii) Total Drag
• The total drag is found by integrating the wall shear along the
entire plate. Assuming the presence of an initial laminar flow region,
xcrit
FD =
∫ (τ )
o lam
0
L
wdx +
∫ (τ )
o turb
wdx
(8.76)
xcrit
47
1
1
2
• Dividing by ρV∞ A = ρV∞ 2 wL , the drag coefficient C D is:
2
2
1
CD =
L
xcrit
∫
0
C f , lam dx + ∫ C f ,turb dx
xcrit
L
(8.78)
• Substituting Eqn. (4.48) for laminar flow and using White’s model
(8.75) for turbulent flow, we obtain with some manipulation,
CD =
0.0315 1477
−
1/ 7
Re L
Re L
(8.79)
• Assume xcrit = 5 × 10 5
8.4.4 Effect of Surface Roughness on Friction Factor
• The interaction between the already complex turbulent flow and the
complex, random geometric features of a rough wall is the subject of
advanced study and numerical modeling.
• However, with crude modeling and some experimental study we can
gain at least some physical insight.
• Define k as the average height of roughness elements on the surface.
In wall coordinates:
+
*
k = ku / ν
48
• Experiments show that for small values of k + (less than approximately
5), the velocity profile and friction factor are unaffected by roughness
• For k + > 10 or so, however, the roughness extends beyond the viscous
sublayer, and the viscous sublayer begins to disappear, likely due to
the enhanced mixing in the roughness provided.
• Beyond k + > 70 viscous effects are virtually eliminated, and the flow
is referred to as fully rough. Beyond this value of roughness, the shape
of the velocity profile changes very little. Consequently, we might expect
that once the surface is fully rough, increasing the roughness would not
change the friction factor.
49
8.5 Energy Transfer in External Turbulent Flow
• Not surprisingly, energy transfer is also greatly complicated under
turbulent flow.
• We found in Chapter 2 that the heat transfer for flow over a
geometrically similar body like a flat plate (neglecting both buoyancy
and viscous dissipation) could be correlated through dimensionless
analysis by
Nux = f ( x * , Re , Pr )
(2.52)
• Turbulence introduces two new variables into the analysis: the
momentum and thermal eddy diffusivities, ε M and ε H .
• One way to deal with these new terms is to introduce a new
dimensionless parameter: Turbulent Prandtl Number
εM
Prt =
εH
Approaches to Analyzing Turbulent Heat Transfer
• Find a mathematical analogy between heat and mass transfer
• Develop a universal temperature profile, similar to how we
developed a universal velocity profile.
(8.81)
50
o 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.
•There are more advanced methods, like numerical solutions to the
boundary layer flow, which we will forgo in this text. We will instead
remain focused on some of the more traditional methods, which are the
basis of the correlations commonly in use.
8.5.1 Momentum and Heat Transfer Analogies
• Osborne Reynolds first discovered a link between momentum and
heat transfer in 1874 while studying boilers.
• He theorized that the heat transfer and the frictional resistance in a pipe
are proportional to each other.
• This is a significant and bold assertion! If we can measure or predict the
friction along a wall or pipe, we can determine the heat transfer simply
by using a multiplying factor. This approach would allow us to solve
for the heat transfer directly, avoiding the difficulty of solving the energy
equation.
51
(i) Reynolds Analogy
• Consider parallel flow over a flat plate.
• The pressure gradient dp/dx is zero, and the boundary layer momentum
and energy equations (8.38) and (8.39) reduce to
∂u
∂u ∂
∂u
u
+v
=
(ν + ε M )
∂x
∂y ∂ y
∂y
(8.82a)
∂T
∂
∂T
∂T
u
+v
=
(α + ε H )
∂x
∂y ∂y
∂y
(8.82b)
• The boundary conditions are
u ( y = 0) = 0, T ( y = 0) = Ts
u ( y → ∞ ) = V∞ , T ( y → ∞ ) = T∞
(8.83a)
(8.83b)
• Notice that equations (8.82a&b) and their respective boundary
conditions are very similar; if they were identical, their solutions –
the velocity and temperature profiles – would be the same.
Normalizing the Variables
• Select the following variables,
52
T − Ts
u
v
x
y
U=
, V=
,θ=
, X = and Y =
V∞
V∞
T∞ − Ts
L
L
• The boundary layer equations become
∂U
1 ∂
∂U
∂U
U
+V
=
(ν + ε M )
∂X
∂Y V∞ L ∂Y
∂Y
∂θ
∂θ
1 ∂
∂θ
U
+V
=
α
+
ε
(
H )
∂X
∂Y V∞ L ∂y
∂Y
(8.84a)
(8.84b)
• With boundary conditions:
U (Y = 0) = 0, θ (Y = 0) = 0
U (Y → ∞ ) = 1, θ (Y → ∞ ) = 1
(8.85a)
(8.85b)
• Normalizing the variables has made the boundary conditions identical.
• The boundary layer equations (8.84) can then be made identical if
ν + ε M =α + ε H which is possible under two conditions.
1. The kinematic viscosity and thermal diffusivity are equal:
ν = α (Pr = 1)
(8.86)
2. The eddy diffusivities are equal:
53
ε M = ε H (Prt = 1)
(8.87)
• We can provide some justification for this assumption by arguing that
the same turbulent mechanism—the motion and interaction of fluid
particles—is responsible for both momentum and heat transfer.
Reynolds made essentially the same argument, and so Equation (8.87) by
itself is sometimes referred to as Reynolds’ analogy.
• The analogy is now complete, meaning that the normalized velocity and
temperature profiles, U ( X , Y ) and V ( X , Y ) are equal.
Developing the Analogy
• Begin by writing the ratio of the apparent heat flux and shear stress
(equations. 8.40 and 8.41),
54
′′ / ρ c p
qapp
τ app / ρ
α + ε H ) ∂T / ∂y
(
=−
(ν + ε M ) ∂u / ∂y
(8.88)
• Imposing the two conditions ν = α (8.86) and ε M = ε H (8.87),
substituting the dimensionless variables yields
′′
qapp
τ app
=
c p ( Ts − T∞ ) ∂θ / ∂Y
V∞ ∂U / ∂Y
(8.89)
• Since the dimensionless velocity and temperature profiles are identical,
their derivatives cancel.
′′ / τ app
• Another important implication of (8.89) is that the ratio qapp
is constant throughout the boundary layer. This means we can represent
this ratio by the same ratio at the wall. Equation (8.89) then becomes
qo′′
τo
=
c p ( Ts − T∞ )
V∞
• can recast this into a more convenient form by substituting
qo′′ = h Ts − T∞ and τ o = 0.5C f ρV∞ 2 into the above, and rearranging,
(
)
Cf
h
=
2
ρV∞ c p
55
• The terms on the left side can also be written in terms of the Reynolds,
Nusselt and Prandtl numbers,
Cf
Nux
St x =
( Pr = 1)
=
Re x Pr
2
(8.90)
• This is the Reynolds Analogy.
• St is called the Stanton Number.
•Note: The same analogy can also be derived for laminar flow over a
flat plate (for Pr =1).
Limitations to the Reynolds Analogy
• It is limited to Pr = 1 fluids.
o A reasonable approximation for many gases, but for most liquids
the Prandtl numbers are much greater than unity—values of up
to 700 are possible.
• Therefore the Reynolds analogy is not appropriate for liquids.
• Also doesn’t account for the varying intensity of molecular and turbulent
diffusion in the boundary layer.
(ii) Prandtl-Taylor Analogy
56
• In independent works, Prandtl [9] and Taylor [10] modified the
Reynolds analogy by dividing the boundary layer into two regions
o a viscous sublayer where molecular effects dominate:
ν
ε M and α
εH
ε M and α
εH
o a turbulent outer layer, where turbulent effects dominate:
ν
o Notice that neither of these conditions restricts us to Pr = 1 fluids.
Analogy for the Viscous Layer
• Define the viscous sublayer from y = 0 to y = y1
• The boundary conditions for this region are
u (0) = 0, T (0) = Ts
u ( y1 ) = u1 , T ( y1 ) = T1
• Define the following normalized variables:
T − Ts
u
v
x
y
U= , V = , θ=
, X=
and Y =
u1
u1
T1 − Ts
y1
y1
• Then, for the viscous sublayer, the ratio of the apparent heat flux and
apparent shear stress (Eqn. 8.86) leads to the following:
57
qo′′
Ts − T1 =
Pru1
(8.93)
τ oc p
′′ / τ app = qo′′ / τ o = constant
Where we have again noted that qapp
Analogy for the Outer Layer
• Closely resembles the Reynolds analogy, with ε M = ε H but this time
we assume that the turbulent effects outweigh the molecular effects,
equation (8.92).
• The boundary conditions for this region are
u ( y1 ) = u1 , T ( y1 ) = T1
u ( y → ∞ ) = V∞ , T ( y → ∞ ) = T∞
• The following normalized variables will make the analogy valid in this
region:
u − u1
v − u1
T − T1
x
y
U=
, V =
, θ=
, X=
and Y =
V∞ − u1
V∞ − u1
T∞ − T1
L
L
• Then, for the outer region, the ratio of the apparent heat flux and
apparent shear stress (equation 8.86) leads to
qo′′
T1 − T∞ =
(V∞ − u1 )
τ oc p
(8.94)
58
′′ / τ app is constant, so we have chosen the value at
As before, the ratio qapp
y = y1 (which, as we found for the viscous sublayer, can be represented by
qo′′ / τ o ).
• Adding (8.93) and (8.94) gives
u1
qo′′
Ts − T∞ =
V∞ ( Pr − 1 ) + 1
τ oc p
V∞
• Substituting τ o =
1
C f ρV∞ 2 into the above yields
2
Cf / 2
qo′′
St =
=
ρV∞ c p ( Ts − T∞ ) u1 Pr − 1 + 1
(
)
V∞
• The velocity at the edge of the viscous sublayer, u1, is still unknown
• Estimate u1 using the universal velocity profile. Approximate the edge
+
+
of the viscous region u = y ≈ 5
• Then, from the definition of u +
u1
2
+
u =5=
V∞ C f
59
Cf
u1
=5
V∞
2
(8.95)
• Thus, the Prandtl-Taylor analogy is
Cf / 2
Nu x
St x =
=
Re x Pr C f
5
( Pr − 1) + 1
2
(8.96)
(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. The result, developed in Appendix D, is
Nu x
St x =
=
Re x Pr
Cf / 2
Cf
5 Pr + 1
1+ 5
( Pr − 1 ) + ln
2
6
(8.97)
(iv) Colburn Analogy
• Colburn [24] proposed a purely empirical modification to the Reynolds
analogy that accounts for fluids with varying Prandtl number.
60
• He proposed the following correlation through an empirical fit of
available experimental data:
St x Pr
2/ 3
=
Cf
2
(8.98)
• The exponent (2/3) on the Prandtl number is entirely empirical, and
does not contain any theoretical basis.
• The Colburn analogy is considered to yield acceptable results for
Re x < 107 (including the laminar flow regime) and Prandtl number
ranging from about 0.5 to 60.
Example 8.3: Average Nusselt Number on a Flat Plate
Determine the average Nusselt number for heat transfer along a flat
plate of length L with constant surface temperature. Use White’s
model (8.75) for turbulent friction factor, and assume a laminar
region exists along the initial portion of the plate.
(1) Observations. This is a mixed-flow type problem, with the initial
portion of the plate experiencing laminar flow.
(2) Problem Definition. Determine an expression for the average
Nusselt number for a flat plate of length L.
61
(3) Solution Plan. Start with an expression for average heat transfer
coefficient, equation (2.50), and split the integral up between laminar
and turbulent regions.
(4) Plan Execution.
(i) Assumptions. (1) Boundary layer assumptions apply, (2) mixed
(laminar and turbulent) flow, (3) constant properties, (4)
incompressible flow, (5) impermeable flat plate, (6) uniform surface
temperature. (7) transition occurs at xc = 5*105.
(ii) Analysis.
• The average heat transfer coefficient is found from:
L
1
hL = ∫ h( x )dx
L0
(2.50)
• Splits this into laminar and turbulent regions:
x
L
1 c
hL = ∫ hlam ( x )dx + ∫ hturb ( x )dx
L 0
xc
(8.99)
•From the definition of Nusselt number, we can write the above as:
hL L
Nu L =
=
k
xc
∫
0
L
1
1
Nux ,lam dx + ∫ Nux ,turb dx
x
x
xc
(a)
62
• To find expressions for local Nusselt number, we will use the friction
factors for laminar flow, and White’s model for turbulent flow (8.75),
and apply them to Colburn’s analogy (8.98).
• The results are
Nux ,lam = 0.332 Pr 1/ 3 Re x 1/ 2
(b)
Nux ,turb = 0.0135 Pr 1/ 3 Re x 6/ 7
(c)
• Substituting these expressions into (a) gives:
Nu L =
hL L
=
k
xc
1/ 3 V∞
0.332
Pr
)
ν
∫0 (
1/ 2
L
V
+ ∫ ( 0.0135 ) Pr 1/ 3 ∞
x xc
ν
dx
6/7
dx
x 1/7
Which yields
1/ 3
Nu L = 0.664 Pr Re xc
1/ 2
7
+ ( 0.0135 ) Pr 1/ 3 Re L 6/7 − Re xc 6/ 7 (d)
6
(
)
• Finally, since Rexc = 5*105, (d) reduces to:
Nu L = ( 0.0158 Re L 6/ 7 − 739 ) Pr 1/ 3
(iii) Checking. The resulting Nusselt number correlation is
dimensionless.
(8.100)
63
(5) Checking. If the laminar length had been neglected, the resulting
correlation would be
(8.101)
Nu L = 0.0158 Re 6/7 Pr 1/ 3
L
This result also makes sense when examining the mixed-flow
correlation (8.100). If the plate is very long, such that the majority of
the plate is in turbulent flow, the second term in the parentheses
becomes negligible, leading to (8.101).
8.5.2 Validity of Analogies
• Generally valid for slender bodies, where pressure gradient does not
vary greatly from zero.
• Approximately valid for internal flows in circular pipes as well,
although other analogies have been developed specifically for internal
flow.
• Although they are derived assuming constant wall temperature, the
above correlations work reasonably well even for constant heat flux.
• To address property variation with temperature, evaluate properties
at the film temperature:
Ts + T∞
(8.102)
T =
f
2
64
Effect of the Turbulent Prandtl Number
• Analogies also that Prt = 1. Valid?
• Prt as high as 3 near wall, but 0.7–1 outside the viscous sublayer
• Prt 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.
This suggests that the analogies should be approximately valid for real
flows.
Validity of the Colburn Analogy
• Arguably, the most popular analogy is that of Colburn.
• The analogy is as primitive as the Reynolds analogy, adds no new
theoretical insight, and is in fact merely a curve-fit of experimental data.
• Why has this method maintained its usefulness over the decades?
• Easy to use
• More advanced models are based on theoretical assumptions
that are, at best, approximations.
65
o Prandtl-Taylor and von Kármán analogies assume that the viscous
sublayer and conduction sublayer are the same thickness.
• Colburn analogy backed by experimental data over a range of conditions
and fluids.
• Empiricism is sometimes better than pure theoretical arguments; the
test is the experimental data.
• Colburn analogy does have critics. Churchill and Zajic [29]
demonstrated that that the Colburn analogy under-predicts the Nusselt
number by 30-40% for fluids with Prandtl numbers greater than 7.
• Despite their shortcomings, analogies are fairly straightforward, and
facilitate the development of empirical correlations that are often
reasonably accurate and easy to use.
• Numerical solutions, on the other hand, are still difficult to obtain and
are limited in applicability.
• For these reasons, heat and mass transfer analogies remain in
widespread use, and new correlations are still being developed
often based on this technique.
8.5.3 Universal Turbulent Temperature Profile
66
(i) Near-Wall Profile
• Begin with the turbulent energy equation, (8.39).
• Akin to the Couette flow assumption, we assume that, near the wall,
the velocity component v ~ 0, as is the temperature gradient ∂T / ∂x .
Thus the left-hand-side of (8.39) approaches zero. Then,
∂
∂T
(α + ε H )
∂y
∂y
Near the Wall:
0
• This implies that the apparent heat flux is approximately constant with
respect to y,
′′
qapp
ρcp
= − (α + ε H )
∂T
∂y
constant
(8.103)
• The idea here is the same as we developed for the universal velocity
profile: we can solve the above relation for the temperature profile.
′′ / ρ c p is constant throughout this region,
• First, recognize that, since qapp
′′ with qo′′ . Then, substituting wall coordinates
we can replace qapp
u + and y + , (8.103) can be rearranged to
*
∂T ρ c p u
ν
− +
=
′′
∂y
qo
α + εH
(8.104)
67
• Define a a temperature wall coordinate as,
T + ≡ ( Ts − T )
• Then (8.104) becomes,
ρ c p u*
qo′′
∂T +
ν
=
+
α + εH
∂y
(8.105)
(8.106)
• We can now integrate the above expression:
y+
+
dy
ν
T+ = ∫
α + εH
0
(8.107)
• We will divide the boundary layer into two regions in order to
evaluate this expression.
(ii) Conduction Sublayer
• Very close to the wall, we expect molecular effects to dominate the heat
transfer; that is, α
εH
• Invoking this approximation, (8.107) reduces to:
T + = Prdy + = Pry + + C
68
• The constant of integration, C, can be found by applying the boundary
condition that T + ( y + = 0) = 0 . This condition yields C = 0, so the
temperature profile in the conduction sublayer is
T + = Pry + , (y + < y1+ )
(8.108)
+
• In the above, y1 is the dividing point between the conduction and outer
layers.
(iii) Fully Turbulent Region
• Outside the conduction-dominated region close to the wall, we expect
that turbulent effects dominate: α + ε H
y
+
+
T = T1 +
∫
y1+
ν +
dy
εH
(8.109)
• Rather than develop some new model for εH, we invoke the
turbulent Prandtl number:
Pr
εH =
t
εM
• We already have a model for εM, and will assume a constant value for Prt
• Prandtl’s mixing length theory was
ε M = κ 2 y2
∂u
∂y
(8.44) 69
• In terms of wall coordinates, we can write (8.44) as
ε M = κ 2 ( y+ )
2
∂u +
ν +
∂y
(8.110)
• The partial derivative ∂u + / ∂y +can be found from the Law of the Wall,
Equation (8.58). Substituting the above and (8.58) into (8.109),
we obtain
y+
Pr +
(8.111)
T+ =
dy
+
∫ κy
y1+
• Finally, if we assume Prt and κ are constants, then (8.110) becomes:
y+
T =
ln + ,
κ
y1
+
Prt
(y + > y1+ )
(8.112)
• Kays et al. [30] assumed Prt = 0.85 and κ = 0.41, but found that
+
y
the thickness of the conduction sublayer ( 1) varies by fluid.
• White [14] reports a correlation that can be used for any fluid with
Pr ≥ 0.7:
Pr
(8.113)
T + = t ln y + + 13 Pr 2/ 3 − 7
κ
• Prt is assumed to be approximately 0.9 or 1.0.
70
(iv) A 1/7th Law for Temperature
• As with the velocity profile, a simpler 1/7th power law relation is
sometimes used for the temperature profile:
T − Ts y
=
T∞ − Ts δ
1/ 7
(8.114)
8.5.4 Algebraic Method for Heat Transfer Coefficient
• The existence of a universal temperature and velocity profile makes
for a fairly simple, algebraic method to estimate the heat transfer. 71
• Begin with the definition of the Nusselt number, which can be expressed
using Newton’s law of cooling as:
qo′′ x
hx
Nux ≡
=
k
k ( Ts − T∞ )
(8.115)
• Invoke the universal temperature profile, T+: Using the definition of T+,
equation 8.105, we can define the free stream temperature as follows,
T∞ = ( Ts − T∞ )
+
ρ c p u*
qo′′
= ( Ts − T∞ )
ρ c pV∞ C f / 2
qo′′
(8.116)
• Substituting this expression into (8.115) for (Ts –T∞) and rearranging,
Nu x =
ρ c pV∞ C f / 2 x
kT∞ +
• Then, multiplying the numerator and denominator by ν,
Nu x =
Re x Pr C f / 2
T∞ +
(8.117)
• We can now use the universal temperature profile, Eqn. (8.113),
+
to evaluate T∞ ,
72
T∞ + =
+
Prt
κ
ln y∞ + + 13 Pr 2/ 3 − 7
(8.118)
• A precise value for y∞ is not easy to determine. However, we can make a
clever substitution using the Law of the Wall velocity profile, Eqn. (8.58).
• In the free stream, we can evaluate (8.58) as
+
u∞ =
1
κ
ln y∞ + + B
(8.119)
+
• Substituting (8.119) into (8.118) for ln y∞ , the Nusselt number relation
then becomes
Re Pr C / 2
Nu x =
x
f
Prt ( u∞ + − B ) + 13 Pr 2/ 3 − 7
• We can simplify this expression further. Using the definition of
Stanton number, St x = Nux / ( Re x Pr ) ,selecting B = 5.0 and Prt = 0.9, and
+
noting that the definition of u leads to u∞ + = 2 / C f , we can
rearrange the relation above to arrive at the final result:
St x =
Cf / 2
0.9 + 13 ( Pr
2/ 3
− 0.88 ) C f / 2
(8.120)
• Note the similarity to the more advanced momentum-heat transfer
analogies of Prandtl and Taylor (8.96) and von Kármán (8.97).
73
8.5.5 Integral Methods for Heat Transfer Coefficient
• One use for the universal temperature profile is to model the heat
transfer using the integral energy equation.
• Consider turbulent flow over a flat plate, where a portion of the leading
surface is unheated
• Can assume a 1/7th power law profile for both the velocity and
temperature (equations 8.65 and 8.114), and substitute them into the
Energy Integral Equation
• Even with the simplest of assumed profiles, the development is
mathematically cumbersome.
• A detailed development appears in Appendix E; the result of the analysis
is:
74
C f xo
Nu x
St x =
=
1 −
Re x Pr
2 x
9/10
1/ 9
(8.121)
• Applies to turbulent flow over a flat plate with unheated starting length
xo.
• Note that (8.121) reduces to the Reynolds analogy when xo = 0. This is
because the Prandtl number was assumed to be 1 as part of the derivation.
• The model has been used to approximate heat transfer for other fluids as
follows. Equation (8.121) can be expressed as
Nux =
Nu xo = 0
1 − ( x / x ) 9/10
o
1/9
(8.122)
• In this form, other models for heat transfer, like von Kármán’s analogy,
could be used to approximate Nuxo = 0 for Pr ≠ 1 fluids.
8.5.6 Effect of Surface Roughness on Heat Transfer
• We would expect roughness to increase the heat transfer, like it did for
the friction factor.
• However, the mechanisms for momentum and heat transfer are different.
75
• As roughness increases, the viscous sublayer diminishes, to such an
extent that for a fully rough surface the viscous sublayer disappears
altogether.
o The turbulent fluid elements are exchanging momentum with surface
directly (like profile or pressure drag), and the role of molecular
diffusion (i.e., skin friction) is diminished.
• Heat transfer, on the other hand, relies on molecular conduction at the
surface, no matter how rough the surface, or how turbulent the flow.
o There is no “pressure drag” equivalent in heat transfer.
o Moreover, fluid in the spaces between roughness elements is largely
stagnant, and transfers heat entirely by molecular conduction.
o The conduction sublayer, then, can be viewed as the average height
of the roughness elements.
o The stagnant regions between roughness elements effectively create
a resistance to heat transfer, and is the major source of resistance
to heat transfer [27].
• Bottom line: we can not expect roughness to improve heat transfer as
much as it increases friction.
76
• This is also means that we can not predict the heat transfer by simply
using a friction factor for rough plates along with one of the
momentum-heat transfer analogies.
What Influences Heat Transfer on a Rough Plate?
• The roughness size k
o Expect that roughness size has no influence until it extends beyond
the viscous and conduction sublayers.
o Its influence reaches a maximum beyond some roughness size
(the fully rough limit).
• The Prandtl number
o Since molecular conduction is important.
o Fluids with higher Prandtl number (lower conductivity) would be
affected more by roughness. Why? The lower-conductivity fluid
trapped between the roughness elements will have a higher resistance
to heat transfer. Also, the conduction sublayer is shorter for these
fluids, so roughness elements penetrate relatively further into the
thermal boundary layer.
o In contrast, for a liquid metal, the conduction sublayer may fully
engulf the roughness elements, virtually eliminating their influence
77
on the heat transfer.
• Kays et al. [30] develop a correlation for rough plate, which is equivalent
to
−1
Cf
0.2
+
0.44
(8.123)
St =
Pr + C k
Pr
C / 2
(
t
s )
f
2
• where k s+ = k s u* / ν is based on the equivalent sand-grain roughness ks
and C is a constant that depends on roughness geometry.
• Bogard et al. [31] showed that this model compared well with
experimental data from roughened turbine blades.
o Showed a 50% increase in heat transfer over smooth plates.
o Demonstrated that increasing roughness beyond some value showed
little increase in the heat transfer.
78
CHAPTER 9
CONVECTION IN TURBULENT CHANNEL FLOW
9.1 Introduction
• We will begin this subject with the criteria for fully developed velocity and
temperature profiles.
• Will focus most of our attention on analyzing fully developed flows.
• As in Chapter 6, our analysis is limited to general boundary conditions:
(i) uniform surface temperature, and (ii) uniform heat flux.
9.2 Entry Length
• Common rules of thumb:
Lh Le
≈ 10
≈
De De
(6.7)
o De is the hydraulic or equivalent diameter
De =
4 Af
P
o Bejan [1] recommends (6.7) particularly to Pr = 1 fluids.
1
• White [2] recommends the following approximation:
Lh
≈ 4.4 Re 1/6
De
De
(9.1)
• Latzko (see Reference 3) suggests:
Lh
≈ 0.623 Re 1/4
De
De
(9.2)
• Thermal entry length doesn’t lend itself to a simple, universallyapplicable equation, since the flow is influenced so much by fluid
properties and boundary conditions.
• The hydrodynamic entry length is much shorter for turbulent flow than
for laminar. In fact, the hydrodynamic entrance region is sometimes
neglected in the analysis of turbulent flow.
9.3 Governing Equations
• Consider flow through a circular pipe.
• Assume 2D, axisymmetric,
incompressible flow.
2
9.3.1 Conservation Equations
• Conservation of Mass:
∂u 1 ∂
+
( rvr ) = 0
∂x r ∂r
(9.3)
• x-momentum equation reduces to:
∂v r
∂u
1 dp 1 ∂
u
+ vr
=−
+
∂x
∂r
ρ dx r ∂r
∂u
r
+
ν
ε
M )
(
∂r
(9.4)
• Conservation of Energy:
∂T
∂T 1 ∂
u
+ vr
=
∂x
∂ r r ∂r
∂T
r ( α + ε H ) ∂r
(9.5)
9.3.2 Apparent Shear Stress and Heat Flux
• Similar to that of the flat plate:
τ app
∂u
= (ν + ε M )
∂r
ρ
(9.6)
′′
qapp
∂T
= − (α + ε H )
∂r
ρcp
(9.7)
3
9.3.3 Mean Velocity and Temperature
Mean Velocity
• Calculating by evaluating the mass flow rate in the duct:
ro
m = ρ um A = ∫ ρ u ( 2π r ) dr
0
• Assuming constant density:
ro
um =
ro
1
2
u
2
r
dr
urdr
=
π
( )
2 ∫
2 ∫
ro 0
π ro 0
(9.8)
Bulk Temperature
• Evaluating by integrating the total energy of the flow:
ro
∫ Turdr
Tm ≡
0
ro
∫ urdr
0
• Can be simplified by substituting the mean velocity, equation (9.8),
4
ro
Tm =
2
Turdr
2 ∫
um ro 0
(9.9)
9.4 Universal Velocity Profile
9.4.1 Results from Flat Plate Flow
• Already seen that the universal velocity profile in a pipe is very similar
to that of flow over a flat plate at zero or favorable pressure gradient.
• We even adapted a pipe flow friction factor model to analyze flow over a
flat plate using the momentum integral method.
• It is apparent, then, that 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.
• Therefore a reasonable start to modeling pipe flow is to invoke the
two-layer model that we used to model flow over a flat plate:
Viscous Sublayer:
u+ = y +
(8.54)
Law of the Wall:
+
u =
1
κ
ln y + + B
• We also have continuous wall law models by Spalding (8.63) and
Reichardt (8.64) that have been applied to pipe flow.
(8.58)
5
Wall Coordinates for Internal Flow
• Note that for pipe flow, the wall coordinates are a little different than for
flat-plate flow.
• First, the y-coordinate for pipe flow is
y = ro − r
(9.10)
• So the wall coordinate y+ is
*
r
−
r
u
(
)
y+ = r + − r + = o
o
ν
(9.11)
• The velocity wall coordinate is the same as before,
u
u ≡ *
u
+
(8.49)
• and the friction velocity is the same,
u* ≡ τ o / ρ
(8.46)
• The friction factor is based on the mean flow velocity instead of the
free-stream velocity:
τo
Cf =
(9.12)
2
( 1 / 2 ) ρ um
6
• So the friction velocity can be expressed as:
u* = um C f / 2
9.4.2 Development in Cylindrical Coordinates
• The velocity profile data for pipe flow matches that of flat plate flow
o This fact allowed us to develop expressions for universal velocity
profiles solely from flat plate (Cartesian) coordinates.
• Would we have achieved the same results if we had started from the
governing equations for pipe flow (i.e., cylindrical coordinates)?
• Assume fully developed flow. x-momentum reduces to
1 ∂ rτ
r ∂r ρ
1 ∂p
=
ρ ∂x
(9.13)
• Rearranging and integrating, we obtain an expression for the shear stress
anywhere in the flow:
r ∂p
+C
τ (r ) =
2 ∂x
(9.14)
• The constant C is zero, since we would expect the velocity gradient (and
hence the shear stress) to zero at r = 0.
7
• Evaluating (9.14) at r and ro and taking the ratio of the two gives:
τ (r ) r
=
ro
τo
(9.15)
• This result shows that the local shear is a linear function of radial location.
• But the Couette Flow assumption meant that τ is
approximately constant in the direction normal to the
wall! How do we reconcile this?
o Remember that the near-wall region over which we
make the Couette flow assumption covers a very
small distance.
o Therefore we could assume that, in that small
region vary close to the wall of the pipe, the shear
is nearly constant, τ = τo
o Thus the Couette assumption approximates the behavior near the
pipe wall as
(ν + ε M )
∂u τ o
= constant
=
∂r ρ
(9.16)
• Exp. data suggest that the near-wall behavior is not influenced by the
outer flow, or even the curvature of the wall.
8
9.4.3 Velocity Profile for the Entire Pipe
• The velocity gradient (and the shear stress) is supposed to be zero at the
centerline of the pipe.
• Unfortunately, none of the universal velocity profiles we’ve developed so
far behave this way
• Reichardt attempted to account for the entire region of the pipe. He
suggested a model for eddy viscosity:
2
r
εM κ y
r
=
1 + 1 + 2
6
ro
ν
ro
+
(9.17)
• Which leads to the following expression for the velocity profile:
1.5 ( 1 + r / ro )
+
+B
u = ln y
2
κ 1 + 2 ( 1 + r / ro )
+
1
(9.18)
• Reichardt used κ= 0.40 and B = 5.5.
• The profile does not account for the viscous sublayer, but as r →ro,
equation (9.18) does reduce to the original Law of the Wall form,
equation (8.58).
9
9.5 Friction Factor for Pipe Flow
9.5.1 Blasius Correlation for Smooth Pipe
• Based on dimensional analysis and experimental data, Blasius developed
a purely empirical correlation for flow through a smooth circular pipe:
C f ≈ 0.0791 Re D−1/4 (4000<ReD<105)
(9.19)
o The friction factor is based on the mean flow velocity,
τo
Cf =
( 1 / 2 ) ρ um 2
(9.12)
• Later correlations have proven to be more accurate and versatile, but this
correlation led to the development of the 1/7th Power Law velocity profile.
9.5.2 The 1/7th Power Law Velocity Profile
• Discovered independently by Prandtl [7] and von Kármán [8].
• Begin with the Blasius correlation, which can be recast in terms of wall
−1/4
shear stress:
τ
2
r
u
o
or
( 1 / 2 ) ρ um 2
= 0.0791 o
ν
m
τ o = 0.03326 ρ um7/4 ro−1/ 4ν 1/4
(a)
10
• Assume a power law velocity profile:
y
u
=
uCL ro
q
(b)
• Assume that the mean velocity in the flow can be related to the centerline
velocity as:
uCL = ( constant ) u
(c)
• Substituting (b) and (c) for the mean velocity in (a) yields :
y
τ o = ( const ) ρ u
ro
−1/ q
7/4
ro−1/ 4ν 1/4
• Simplifies to:
τ o = ( const ) ρ u 7/ 4 y ( −7/ 4 q ) ro(7/ 4 q −1/ 4)ν 1/ 4
(d)
• Both Prandtl and von Kármán argued that the wall shear stress is not a
function of the size of the pipe.
• Then the exponent on ro should be equal to 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,
11
y
u
=
uCL ro
1/7
(9.20)
• Experimental data show that this profile adequately models the velocity
profile through a large portion of the pipe, and is frequently used in
models for momentum and heat transfer.
• Limitations:
o Accurate for only a narrow range of Reynolds numbers (roughly,
104 to 106).
o Yields an infinite velocity gradient at the wall
o Does not yield a gradient of zero at the centerline
Nikuradse’s Improvement to the 1/7th Power Law
• Another student of Prandtl, Nikuradse [10] measured velocity profiles in
smooth pipe over a wide range of Reynolds numbers, and reported that
the exponent varied with Reynolds number,
y
u
=
uCL ro
n
(9.21)
12
• Also correlated pipe friction factor of the form
Cf =
C
Re 1/D m
(9.22)
• As one might expect, Nikuradse’s results show that the velocity profile
becomes fuller as the mean velocity increases.
9.5.3 Prandtl’s Law for Smooth Pipe
• Whereas the Blasius correlation is purely empirical, we can develop a
more theoretical model for friction factor by employing the universal
velocity profile.
13
• Begin with the Law of the Wall, equation. Substitute the wall coordinates
u+ and y+, as well as the friction velocity u* = τ o / ρ = um C f / 2 :
u
um
2
1 yum
= ln
C f κ ν
Cf
+ B
2
(9.23)
• If we assume that the equation holds at any value of y, we could
evaluate the expression at the centerline of the duct, y = ro=D/2, where
u = uCL :
uCL
um
2
1 Re D
= ln
C f κ 2
Cf
+ B
2
(9.24)
• We now have a functional relationship for the friction factor.
• However, the ratio uCL/um is still unknown.
Evaluating the Mean Velocity for Prandtl’s Law
• Goal is to integrate the Law of the Wall velocity profile (8.58) across
the pipe.
• Start with expression for the mean velocity, equation (9.10). Using the
variable substitution y = ro – r , (9.10) becomes
14
ro
um =
ro
1
2
u
(2
π
r
)
dr
=
u ( ro − y )dy
2 ∫
2 ∫
ro 0
π ro 0
(9.25)
• Then, substitute the Law of the Wall for u .
• Performing the integration, it can be shown that the mean velocity
becomes
*
1 ro u
3
um = u ln
+ B−
2
κ
ν
κ
*
(9.26)
• Or, making substitutions again for u*
um = um
C f 1 ReD
ln
2 κ 2
Cf
3
+ B−
2
2κ
(9.27)
• Warning: the term we were trying to evaluate, um, cancels out of the
expression! However, uCL doesn’t not appear either.
• We can use the above expression directly to find an expression for Cf.
Rearranging, and substituting the values κ= 0.41 and B = 5.0 gives
1
Cf / 2
(
)
= 2.44 ln ReD C f / 2 − 0.349
15
• This development ignores the presence of a viscous sublayer or a wake
region.
• Empirically, a better fit to experimental data is
1
Cf / 2
(
)
= 2.46 ln C f / 2 D C f / 2 + 0.29
(Re D > 4000)
(9.28)
• This is called Prandtl’s universal law of friction for smooth pipes.
o Sometimes referred to as the Kármán-Nikuradse equation.
• Note that, 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 Cf. A simpler, empirical
relation that closely matches Prandtl’s is:
Cf
2
≈ 0.023 ReD−1/ 5
(3 × 10−4 <Re D < 106 )
(9.29)
• This correlation is also suitable for non-circular ducts, with the
Reynolds number calculated using the hydraulic diameter.
16
9.5.4 Effect of Surface Roughness
• From our discussion of turbulent flow over a rough flat plate, we saw
that roughness shifts the universal velocity profile downward.
• We could write the velocity profile in the logarithmic layer as:
+
u =
1
κ
ln y + + B − ∆B
∆B is the shift in the curve, which increases with wall roughness k+
• The behavior of the velocity profile also depends on the geometry of the
roughness, like rivets to random structures like sandblasted metal.
• The following model is based on equivalent sand grain roughness [2],
1
f 1/ 2
ReD f 1/ 2
≈ 2.0 log10
− 0.8
1/ 2
1 + 0.1( k / D ) ReD f
(9.30)
where it is common to use the Darcy friction factor,
f = 4C f
(9.31)
Two Important Points on Surface Roughness:
1. If the relative roughness k/D is low enough, it doesn’t have much of
an effect on the equation.
17
o Scaling shows that roughness is not important if ( k / D ) Re D < 10
2. On the other hand, if , the roughness term ( k / D ) Re D > 1000
dominates in the denominator, and the Reynolds number cancels;
in other words, the friction factor is no longer dependent on the ReD.
Colebrook-White Equation
• Developed for commercial pipes,
1
f
1/ 2
k/D
2.51
= −2.0 log10
+
1/ 2
3.7
Re
f
D
(9.32)
• This function is what appears
in the classic Moody Chart
18
Moody Chart:
19
9.6 Momentum-Heat Transfer Analogies
• Development is applied to the case of a constant heat flux boundary
condition.
• Strictly speaking, an analogy cannot be made in pipe flow for the case
of a constant surface temperature. But resulting models approximately
hold for this case as well.
Development
• x-momentum equation (9.4) becomes, for hydrodynamically fully
developed flow,
1 dp 1 ∂
=
ρ dx r ∂r
∂u
r (ν + ε M ) ∂r
(9.33a)
∂T
r
α
+
ε
H )
(
∂
r
(9.33b)
• Energy equation reduces to,
u
∂T 1 ∂
=
∂x r ∂r
• Are the left-hand sides analogous?
20
o Note that in pipe flow the pressure gradient is non-zero, although
constant with respect to x. To ensure an analogy, then, the left side
of (9.33b) must then be constant.
o 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:
dT
= constant
dx
o So the analogy holds on the LHS.
Boundary Conditions
• Boundary conditions must match:
At r = 0:
At r = ro:
du (0) dT (0)
=
=0
∂r
∂r
u ( ro ) = 0, T ( ro ) = Ts ( x )
du ( ro )
∂T ( ro )
= τo, k
= qo′′
µ
dr
∂r
• If we normalize as follows:
(9.34a)
(9.34b)
(9.34c)
21
T − Ts
u
x
r
U=
,θ=
, X = and R =
um
Tm − Ts
L
ro
• We can show that both the governing equations and the boundary
conditions are identical in form.
9.6.1 Reynolds Analogy for Pipe Flow
• Assume ν = α (Pr = 1) and εM = εH (Prt = 1)
o Same assumptions used to develop Reynold’s analogy for a flat plate
• Then the governing equations (9.33a) and (9.33b) are identical.
• Follow exactly the same process that we followed for the original
derivation, we find that the Reynolds analogy is essentially identical for
pipe flow,
Cf
qo′′
St D ≡
=
ρ um c p (Ts − Tm ) 2
Cf
Nu D
St D =
=
ReD Pr
2
(Pr = 1)
(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:
τ o = 12 C f ρ um2
22
9.6.2 Adapting Flat-Plate Analogies to Pipe Flow
• We saw that Reynold’s analogy is identical for flat plate and pipe flows.
• We know that the velocity profiles near the wall are similar.
• Can we adapt other flat-plate analogies to pipe flow?
Von Kármán Analogy for Pipe Flow
• Take original von Kármán analogy, replace V∞ and T ∞ with
V∞ ≈ uCL and T∞ ≈ TCL
• These substitutions also affect the friction factor, which translates to:
τo
Cf ≈ 1
2
u
ρ
2
CL
• Following the development exactly as before, the result is almost identical:
qo′′
=
ρ uCL c p (Ts − TCL )
Cf / 2
Cf
5 Pr + 1
1+ 5
(
Pr
−
1)
+
ln
2
6
(9.36)
• Problem: the LHS and the friction factor are expressed in terms of
centerline variables instead of the more common and convenient mean
quantities um and Tm. Correct this as follows:
23
um (Ts − Tm )
qo′′
=
ρ um c p (Ts − Tm ) uCL (Ts − TCL )
(C
f
)
/ 2 ( um / uCL )
2
Cf
5 Pr + 1
( Pr − 1) + ln
2
6
um
1+ 5
uCL
• Now, Cf is again defined in terms of the mean velocity, C f = τ o / 12 ρ um2
and the terms qo′′ / ρ um c p (Ts − Tm ) are collectively the Stanton number
for pipe flow.
• Simplifying,
Ts − Tm
St D
=
Ts − TCL
(C
f
)
/ 2 ( u m / uCL )
um C f
5 Pr + 1
1 + 5
( Pr − 1) + ln
6
uCL 2
(9.37)
• This is the von Kármán Analogy for pipe flow.
Estimates for Mean Temperature and Velocity
• We can develop estimates for the ratios ( um / uCL )and ( Ts − Tm ) / ( Ts − TCL )
using the definition of mean temperature, equation (9.9).
• Estimate um and Tm using the 1/7th Law profiles, which for a circular
pipe are:
24
y
u
=
uCL ro
1/7
(9.20)
and, similar to (8.111) for a flat plate,
y
T − Ts
=
TCL − Ts ro
1/7
(9.38)
• Substituting these models into (9.8) and (9.9), we can show that:
um
= 0.817
uCL
Tm − Ts
= 0.833
TCL − Ts
(9.39)
(9.40)
9.6.3 Other Analogy-Based Correlations
• A simple correlation for turbulent flow in a duct is based on the
Colburn analogy.
• Beginning with the analogy, equation (8.96), and using equation (9.27)
for the friction factor, we obtain
25
St D = 0.023 ReD−1/ 5 Pr −2/ 3
or
NuD = 0.023 ReD4/5 Pr 1/ 3
(9.41)
• One of the most popular correlations is the Dittus-Boelter correlation,
which is an empirical correlation based on the Colburn analogy:
NuD = 0.023 ReD4/ 5 Pr n
(9.42)
o where n = 0.4 for heating (Ts > Tm) and n = 0.3 for cooling.
• Although still popular, the Colburn analogy and its derivative, the
Dittus-Boelter correlation have been challenged in recent years.
• Models such as those by Petukhov and Gnielinski correlation (see
Section 9.8) are preferred for their improved accuracy and range
of applicability.
• Other analogies have been developed specifically for pipe flows, instead
of adapting existing flat-plate models. Examples
o Reichardt [16]
o Boelter, Martinelli, and Jonassen [17]
o Churchill and Zajic [18] in 2002 (which the authors claim is to
date the most accurate model for the internal flow.)
26
9.7 Algebraic Method Using Universal Temperature Profile
• As we did for flow over a flat plate, we can use the universal
temperature and velocity profiles to estimate the heat transfer in a
circular duct.
• Begin again with the definition of the Nusselt number, which for flow
in a duct can be expressed as
qo′′D
hD
N uD ≡
=
k
(Ts − Tm )k
(9.43)
• To later invoke the universal temperature profile, we use the definition
of T+, equation 8.102, to define the mean temperature as
+
m
T = (Ts − Tm )
ρ c p u*
qo′′
= (Ts − Tm )
ρ c p um C f / 2
qo′′
(9.44)
• Recall that for duct flow, the friction velocity u* is defined in terms of
the mean velocity. Substituting this expression into (9.43) for qo′′
and invoking the definitions of the Reynolds and Prandtl numbers,
Nu D =
ReD Pr C f / 2
Tm+
(9.45)
27
• Several ways to proceed with the analysis. One approach is to evaluate
Tm+ using a dimensionless version of the mean temperature expression
ro+
(9.33):
2
(9.46)
Tm+ = + +2 T + u + ( ro+ − y + )dy +
um ro
∫
0
• Theoretically, can simply substitute appropriate universal temperature
and velocity profiles into the above and integrate. Practically, this
requires numerical integration.
• A simpler, second approach can be taken. First, we rewrite the original
Nusselt number relation (9.43) as follows,
qo′′D
(Ts − TCL )
Nu D =
(Ts − Tm )k (Ts − TCL )
where TCL is the centerline temperature.
• Then, substitute the definition of T+ for the centerline temperature in
the denominator, we obtain
ReD Pr C f / 2 (Ts − TCL )
N uD =
+
TCL
(Ts − Tm )
(9.47)
28
• We can now use the universal temperature profile, equation (8.118), to
evaluate TCL+:
+
CL
T
Prt
=
κ
+
ln yCL
+ 13 Pr 2/ 3 − 7
(9.48)
• Now, just like in our analysis for flat plate flow, we can substitute the
Law of the Wall velocity profile (8.59) for ln yCL+:
+
CL
u =
1
κ
+
ln yCL
+B
(9.49)
• Substituting these into the Nusselt number relation,
ReD Pr C f / 2
(Ts − TCL )
NuD =
(9.50)
+
2/ 3
Prt ( uCL − B ) + 13 Pr − 7 (Ts − Tm )
+
• We need expressions for uCL
and (Ts − TCL ) / (Ts − Tm ). For the
centerline velocity, we can use the definition of u+ for pipe flow:
+
CL
u
uCL uCL
= * =
u
um
2
Cf
(9.51)
29
• It appears that, if we are to complete the analysis, we will need to
evaluate the mean velocity and temperature after all. To avoid the
complexity of the logarithmic velocity and temperature profiles, we could
estimate these quantities using the much simpler 1/7th Law profiles,
which we saw in the last section yields
um
= 0.817
uCL
Tm − Ts
= 0.833
TCL − Ts
(9.39)
(9.40)
• Finally, using the definition of Stanton number, St D = NuD / ( ReD Pr ) ,
selecting Prt = 0.9 and B = 5.0, we can rearrange (9.50) obtain
St D =
Cf / 2
0.92 + 10.8 ( Pr
2/3
− 0.89 ) C f / 2
(9.52)
• Under what conditions is this equation applicable? The ultimate test
would be to compare the expression to experimental data.
• However, since we invoked the 1/7th power law, which is valid around
1×105, it might be reasonable as a first approximation to limit this model
to ReD < 1×105.
30
9.8 Other Correlations for Smooth Pipes
Petukhov’s Model
• Petukhov followed a more rigorous theoretical development, invoked
Reichardt’s model for eddy diffusivity and velocity profile (9.15, 9.16):
St D =
Cf / 2
1.07 + 12.7 ( Pr
2/3
− 1)
0.5 ≤ Pr ≤ 2000
, 4
6
C f / 2 10 < Re D < 5 × 10
(9.53)
• Compares well to experimental data over a wide range of Prandtl and
Reynolds numbers.
• Petukhov used the following model for friction factor, which he also
developed:
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). It seems as if we have captured the essential
functionality even in our modest approach.
Gnielinski’s Model
• Gnielinski modified Petukhov’s model slightly, extending the model to
include lower Reynolds numbers:
31
N uD =
0.5 ≤ Pr ≤ 2000
,
3
6
3
×
10
<
Re
<
5
×
10
Cf / 2
D
( Re D − 1000) PrC f / 2
1 + 12.7 ( Pr
2/3
− 1)
(9.55)
• Use Petukhov’s friction model in (9.55) for the friction factor.
• For all models, properties should be evaluated at the film temperature.
• As was the case with the analogy-based correlations, 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
• We’ve discussed the effects of roughness on the heat transfer from flat
plates in Section 8.5.6, and much of the same physical intuition applies
to flow in channels. Norris [21,3] presents the following empirical
correlation for flow through circular tubes:
n
Cf
Cf
Nu
=
< 4
,
Nusmooth C f , smooth
C f , smooth
where n = 0.68 Pr 0.215
(9.56)
32
• A correlation like Colebrook’s (9.30) could be used to determine the
rough-pipe friction factor.
• The behavior of this relation reflects what we expect physically.
o The Prandtl number influences the effect of roughness, and for very
low-Pr fluids the roughness plays little role in the heat transfer.
o Regardless of Prandtl number, the influence of roughness size is
limited: Norris reports that the effect of increasing roughness
vanishes beyond (C f / C f , smooth ) ≈ 4 , and so the equation reaches
a maximum.
• Although roughness enhances heat transfer, it also increases the friction.
• Neither the friction nor the heat transfer increase indefinitely with
roughness size – both reach a limiting value.
33
CHAPTER 10
CORRELATION EQUATIONS:
FORCED AND FREE CONVECTION
10.1 Introduction
• Correlation equations: Based on experimental data
• Chapter outline: Correlation equations for:
(1) External forced convection over:
Plates
Cylinders
Spheres
(2) Internal forced convection through channels
(3) External free convection over:
Plates
Cylinders
Spheres
1
10.2 Experimental Determination of Heat Transfer Coefficient h
Newton's law of cooling defines h:
q′s′
h=
Ts − T∞
∆V
(10.1)
q ′s′ = surface flux
Ts = surface temperature
T∞ = ambient temperature
Example: Electric heating
Measure: Electric power, Ts , T∞
Use (10.1) to calculate h
q′s′
−•
•
Ts
T∞
•
+
V∞
Fig. 10.1
• Form of correlation equations:
• Dimensionless: Nusselt number Is a dimensionless heat transfer
coefficient.
2
1.Example: Forced convection with no dissipation
hx
Nu x =
= f ( x * ; Re, Pr )
k
(2.52)
Use (2.52) to plan experiments and correlate data
10.3 Limitations and Accuracy of Correlation Equations
All correlation equations have limitations !
• Limitations on:
(1) Geometry
(2) Range of parameters: Reynolds, Prandtl, Grashof, etc.
(3) Surface condition: Uniform flux, uniform temperature, etc.
• Accuracy: Errors as high as 25% are not uncommon!
10.4 Procedure for Selecting and Applying Correlation Equations
(1) Identify the geometry
3
(2) Identify problem classification:
Forced convection
Free convection
External flow
Internal flow
Entrance region
Fully developed region
Boiling
Condensation
Etc.
(3) Define objective: Finding local or average heat transfer coefficient
(4) Check the Reynolds number:
(a) Laminar
(b) Turbulent
(c) Mixed
(5) Identify surface boundary condition:
(a) Uniform temperature
4
(b) Uniform flux
(6) Note limitations on correlation equation
(7) Determine properties at the specified temperature:
(a) External flow: at the film temperature T f
T f = (Ts + T∞ ) / 2
(10.2)
(b) Internal flow: at the mean temperatureTm
(c) However, there are exceptions
(8) Use a consistent set of units
(9) Compare calculated values of h with Table 1.1
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
5
Three regions:
V∞
(1) Laminar
(2) Transition
(3) Turbulent
•
T∞
x
• t
laminar
Re x=t Transition or
x
turbulent
transition
Fig. 10.2
critical Reynolds
number:
Re x t depends on: Geometry, surface finish, pressure gradient, etc.
For flow over a flat plate:
V x
Re xt = ∞ t ≈ 5 × 105
ν
• Examples of correlation equations for plates:
Laminar region, x < xt :
6
Use (4.72a) or (4.72b) for local Nusselt number to obtain local h
Turbulent region, x > xt :
Local h:
hx
Nu x =
= 0.0296( Re x )4 / 5 ( Pr )1 / 3
k
Limitations:
flat plate, constant Ts
5 × 105 < Re x < 107
0.6 < Pr < 60
properties at T f
(10.4a)
(10.4b)
Average h
x
L
1 L
1 t
h=
h( x )dx =
hL ( x )dx +
ht ( x )dx
0
0
L
L
xt
∫
∫
∫
(10.5)
7
hL = local laminar heat transfer coefficient
ht = local turbulent heat transfer coefficient
(4.72b) and (10.4a) into (10.5):
1/ 2
k
V∞
h = 0.332
L
ν
0
∫
xt
V∞
+
0
.
0296
ν
1/ 2
x
dx
dx
1/ 3
(
)
Pr
x t x1 / 5
4/5 L
∫
(10.6)
Integrate
h=
{
[
) ] } ( Pr )1 / 3
k
0.664 Re xt 1 / 2 + 0.037 ( Re L )4 / 5 − Re x t 4 / 5
L
(
)
(
Dimensionless form:
NuL =
{
(
)
[
(
) ]}( Pr )1 / 3
1/ 2
4/5
4/5
hL
= 0.664 Re x t
+ 0.037 ( Re L )
− Re x t
k
(2) Plate at uniform surface temperature
with an insulated leading section
x0=Length of insulated section
V∞
T∞
xt
0
•
insulation
(10.7b)
δt
•
xo
x
•
Ts
Fig. 10.3
8
Two cases:
• Laminar flow, x t > x o : Use (5.21) for the local Nusselt number to obtain
local h
•Turbulent flow, x t < x o : The local Nusselt number is
hx 0.0296Re 4x / 5 Pr 1 / 3
Nu x =
=
k
9 / 10 1 / 9
1 − ( xo / x )
[
(10.8)
]
(3) Plate with uniform surface flux
Two regions:
• Laminar flow, 0 < x < xt
Use (5.36) or (5.37) for the local
Nusselt number to obtain local h
V∞
T∞
0
•Turbulent flow, x > xt :
hx
Nu x =
= 0.030Re 4x / 5 Pr 1 / 3
k
xt
x
•
q′s′
Fig. 10.4
(10.9)
9
Properties at T f = (Ts + T∞ ) / 2 and Ts is the average surface temperature
10.5 External Flow Normal to a Cylinder
• For uniform surface temperature or uniform
surface flux
V∞
T∞
θ
Fig. 10.5
Nu L =
5/8
0.62 Re1D/ 2 Pr 1 / 3 Re D
1 +
1
/
4
282,000
Pr 2 / 3
hD
= 0.3 +
k
1 + (4 /
Limitations:
[
)
]
Flow norm al to cylinder
Pe = Re D Pr > 0 . 2
properties at T f
4/5
(10.10a)
(10.10b)
Pe = Peclet number = ReD Pr
10
For Pe < 0.2, use:
hD
1
NuD =
=
k
0.8237 − 0.5 ln Pe
Limitations
(10.11a)
flow normal to cylinder
Pe = Re D Pr < 0.2
properties at T f
10.5.3 External Flow over a Sphere
( )
1/4
hD
0 .4 µ
1/ 2
2/3
Nu D =
= 2 + 0.4 Re D + 0.06 Re D Pr
µs
k
[
Limitations:
]
flow over sphere
3.5 < ReD < 7.6 × 104
0.71 < Pr < 380
1< µ
(10.12a)
(10.12b)
< 3.2
µs
properties at T∞ , µ s at Ts
11
10.6 Internal Forced Convection Correlations
Chapter 7:
Analytic solutions to h for
fully developed laminar flow
Correlation equations for h in the
entrance and fully developed regions
for laminar and turbulent flows
• Transition or critical Reynolds number for smooth tubes:
Re Dt =
uD
ν
≈ 2300
(10.13)
12
10.6.1 Entrance Region: Laminar Flow Through
Tubes at Uniform Surface Temperature
• Two cases:
(1) Fully Developed Velocity, Developing Temperature: Laminar Flow
• Solution: Analytic
• Correlation of analytic
results:
Ts
T
u
FDV
•
developing
0
x
δt
u
temperature
insulation
Fig. 10.6
hD
NuD =
= 3.66 +
k
0.0668 ( D/L ) Re D Pr
2/3
{1 + 0.04 [( D/L) ReD Pr ] }
(10.14a)
13
Limitations:
entrance region of tubes
uniform surface temperature Ts
laminar flow (ReD < 2300)
fully developed velocity
developing temperature
properties at Tm = (Tmi + Tmo ) / 2
(10.14b)
(2) Developing Velocity and Temperature: Laminar flow
hD
1 / 3 µ
[
]
Nu D =
= 1.86 ( D/L) Re D Pr
µ
k
s
0.14
(10.15a)
14
Limitations:
entrance region of tube
uniform surface temperature Ts
laminar flow (ReD < 2300)
developing velocity and temperature
0.48 < Pr < 16700
0.0044 < µ µ < 9.75
s
properties at Tm , µ s at Ts
10.6.2 Fully Developed Velocity and Temperature in Tubes: Turbulent Flow
• Entrance region is short: 10-20 diameters
• Surface B.C. have minor effect on h for Pr > 1
• Several correlation equations for h:
(1) The Colburn Equation: Simple but not very accurate
Nu D =
Limitations:
4/5 1/3
hD
= 0.023Re D Pr
k
(10.16a)
15
fully developed turbulent flow
smooth tubes
ReD > 104
0.7 < Pr < 160
L /D > 60
properties at Tm
(10.16b)
• Accuracy: Errors can be as high as 25%
(2) The Gnielinski Equation: Provides best correlation of experimental
data
Nu D =
2/3
[
]
1
+
(
D
/
L
)
1/2
2/3
8 ) ( Pr − 1) ]
( f 8 )( Re D − 1000) Pr
[1 + 12.7( f
(10.17a)
• Valid for: developing or fully developed turbulent flow
16
Limitations:
2300 < ReD < 5 × 106
0.5 < Pr < 2000
0 < D/L <1
properties at Tm
(10.17b)
• The D/L factor in equation accounts for entrance effects
• For fully developed flow set D/L = 0
The Darcy friction factor f is defined as
∆p D
f =
ρ u2
2 L
(10.18)
For smooth tubes f is approximated by
f = (0.79ln Re D − 1.64) − 2
(10.19)
17
10.6.3 Non-circular Channels: Turbulent Flow
Use equations for tubes. Set D = De (equivalent diameter)
4Af
De =
P
A f = flow area
P = wet perimeter
10.7 Free Convection Correlations
x
10.7.1 External Free Convection Correlations
(1) Vertical plate: Laminar Flow, Uniform Surface
Temperature
u
Ts •
T∞
• Local Nusselt number:
g
y
Fig. 10.7
18
hx 3
Pr
Nu x =
=
k 4 2.435 + 4.884 Pr 1 / 2 + 4.953 Pr
1/ 4
( Ra x )1 / 4
(10.21a)
• Average Nusselt number:
hL
Pr
Nu L =
=
1/2
k
2.435 + 4.884Pr + 4.953Pr
1/4
( Ra L )1/4
(10.21b)
(10.21a) and (10.21b) are valid for:
Limitations:
vertical plate
uniform surface temperature Ts
laminar, 10 4 < Ra L < 10 9
0 < Pr < ∞
properties at T f
(10.21c)
19
(2) Vertical plates: Laminar and Turbulent, Uniform Surface
Temperature
1/6
h L
0.387 Ra L
Nu L =
= 0.825 +
8/27
k
1 + (0.492 /Pr ) 9/16
[
]
2
(10.22a)
Limitations:
vertical plate
uniform surface temperature Ts
laminar, transition, and turbulent
10 −1 < Ra L < 1012
0 < Pr < ∞
properties at T f
(10.22b)
(3) Vertical Plates: Laminar Flow, Uniform Heat Flux
• Local Nusselt number:
20
hx
Pr 2
*
Nux =
=
Grx
1/2
k 4 + 9Pr + 10Pr
1/ 5
(10.23)
Determine surface temperature: Apply Newton’s law:
where Grx* is defined as
q′s′
h( x ) =
Ts ( x ) − T∞
Grx*
=
(10.24)
β gq ′s′
4
x
kν 2
(10.25)
(10.24) and (10.25) into (10.23) and solve for Ts ( x ) − T∞
4 + 9 Pr 1 / 2 + 10 Pr α ν q′s′ 4
Ts ( x ) − T∞ =
( β g )( k )
Pr
x
1/ 5
(10.26a)
(10.23) and (10.26a) are valid for:
21
vertical plate
laminar, 104 < Grx* Pr < 109
uniform surface flux, q′s′
0 < Pr < ∞
• Properties in (10.26a) depend on surface temperatureTs (x) which is not
known. Solution is by iteration
(4) Inclined plates: Constant surface temperature
• Use equations for vertical plates
θ
• Modify Rayleigh number as:
β gcosθ (Ts − T∞ )
Ra x =
αv
Ts > T∞
Ts < T∞
(10.27)
(a)
g T∞
θ
(b)
Fig. 10.9
22
Limitations:
inclined plate
uniform surface temperatur e Ts
Laminar, Ra L < 109
(10.28)
0 ≤ θ ≤ 60o
(5) Horizontal plates: Uniform surface temperature:
(i) Heated upper surface or cooled lower surface
Nu L = 0.54( Ra L )1 / 4 , 2 × 104 < Ra L < 8 × 106
Nu L = 0.15( Ra L )1 / 3 , 8 × 106 < Ra L < 1.6 × 109
Limitations:
horizontal plate
hot surface up or cold surface down
all properties , except , β , at T f
β at T f for liquids , Ts for gases
(10.29b)
(10.29c)
23
(ii) Heated lower surface or cooled upper surface
Nu L = 0.27( Ra L )1 / 4 , 105 < Ra L < 1010
Limitations:
horizontal plate
hot surface down or cold surface up
all properties, except, β, at Tf
β at Tf for liquids, Ts for gases
Characteristic length L:
L=
(10.30b)
surface area
perimeter
(6) Vertical Cylinders. Use vertical plate correlations for:
D
35
>
for Pr ≥ 1
1
/
4
L (GrL )
(10.32)
(7) Horizontal Cylinders:
24
1/ 6
h D
0.387( Ra D )
= 0.60 +
Nu D =
8 / 27
k
9
/
16
1 + (0.559/Pr )
[
Limitations:
(8) Spheres
Limitations:
]
2
(10.33a)
horizontal cylinder
uniform surface temperature or flux
10 − 5 < Ra D < 1012
properties at T f
hD
Nu L =
= 2+
k
0.589( Ra D )1 / 4
[1 + (0.Pr469 ) ]
9 / 16 4 / 9
(10.34a)
sphere
uniform surface temperature or flux
Ra D < 1011
Pr > 0.7
properties at T f
25
10.7.2 Free Convection in Enclosures
Examples:
• Double-glazed windows
• Solar collectors
• Building walls
• Concentric cryogenic tubes
• Electronic packages
Fluid Circulation:
• Driving force: Gravity and unequal surface temperatures
Heat flux:
Newton’s law:
q′′ = h(Th − Tc )
(10.35)
Heat transfer coefficient h:
Nusselt number correlations depend on:
26
• Configuration
• Orientation
• Aspect ratio
• Prandtl number Pr
• Rayleigh numberRa δ
(1) Vertical Rectangular Enclosures
δ
Rayleigh number
β g ( Th − Tc )δ 3
Raδ =
Pr
2
ν
Tc
Tc
(10.36)
L
g
Several equations:
Fig. 10.10
27
hδ
Pr
Nuδ =
Raδ
= 0.18
k
0.2 + Pr
Valid for
0.29
(10.37a)
vertical rectagular enclosure
L
1<
<2
δ
10 − 3 < Pr < 10 5
Pr
Ra δ > 10 3
0 . 2 + Pr
properties at T = ( T c + T h ) / 2
hδ
Pr
Nuδ =
Raδ
= 0.22
k
0.2 + Pr
Valid for
0.28
(10.37b)
L
δ
− 0.25
(10.38a)
vertical rectagular enclosure
L
2 < < 10
δ
Pr < 10 5
10 3 < Raδ < 1010
properties at T = (Tc + Th ) / 2
(10.38b)
28
hδ
Nuδ =
= 0.046 [Raδ ]1 / 3
k
Valid for
(10.39a)
vertical rectagular enclosure
L
1 < < 40
δ
(10.39b)
1 < Pr < 20
10 6 < Raδ < 10 9
properties at T = (Tc + Th ) / 2
hδ
0.012
Nuδ =
[Raδ ]
= 0.42 [Pr ]
k
Valid for
0.25
L
δ
−0
0..3
(10.40a)
vertical rectagular enclosure
L
10 < < 40
δ
1 < Pr < 2 × 10
4
104 < Reδ < 107
properties at T = (Tc + Th ) / 2
(10.40b)
29
(2) Horizontal Rectangular Enclosures
• Enclosure heated from below
• Cellular flow pattern develops at critical Rayleigh number Ra δ c = 1708
• Nusselt number:
L
hδ
δ
Nuδ =
= 0.069[Raδ ]1 / 3 [Pr ]0.074
k
Tc
g
Th
(10.41a)
Fig. 10.11
Valid for
horizontal rectangular enclosure
heated from below
3 × 105 < Raδ < 7 × 10 7
properties at T = (Tc + Th ) / 2
(10.41b)
30
δ
(3) Inclined Rectangular Enclosures
• Applications: Solar collectors
• Nusselt number:correlations depend on:
• Inclination angle
• Aspect ratio
•Prandtl number Pr
• Rayleigh numberRa δ
Tc
g
Th
L
θ
For:
Fig. 10.12
0 o < θ < 90 o: heated lower surface, cooled upper surface
90
o
< θ < 180
o
Table 10.1
critical tilt angle
: cooled lower surface,
heated upper surface
• Nusselt number is minimum at
L/δ
θc
1
3
6
12 > 12
25o 53o 60o 67 o 70o
31
a critical angle θ c : Table 10.1
1708
hδ
Nuδ =
= 1 + 1.441 −
k
Raδ cosθ
( Raδ cosθ )1 / 3
− 1
18
*
1708(1.8 sinθ )1.6
1 −
+
Raδ cosθ
*
(10.42a)
Valid for
inclined rectangular enclosure
L / δ ≤ 12
0 < θ ≤ θc
(10.42b)
∗
set [ ] = 0 when negative
properties at T = (Tc + Th ) / 2
32
hδ
o Nuδ ( 90 )
0.25
Nuδ =
= Nuδ (0 )
(sinθ c )
o
k
Nuδ (0 )
o
θ /θ c
(10.43a)
Valid for
inclined rectangular enclosure
L / δ ≤ 12
0 < θ ≤ θc
properties at T = (Tc + Th ) / 2
Nuδ =
Valid for
hδ
= Nuδ (90o ) [sinθ ] 0.25
k
(10.43b)
(10.44a)
inclined rectangular enclosure
all L / δ
o
θ c < θ < 90
properties at T = (Tc + Th ) / 2
(10.44b)
33
[
]
hδ
Nuδ =
= 1 + Nuδ (90o ) − 1 sinθ
k
(10.45a)
Valid for
inclined rectangular enclosure
all L / δ
(10.45b)
90o < θ < 180 o
properties at T = (Tc + Th ) / 2
Do
(4) Horizontal Concentric Cylinders
5
• Flow circulation forT i > T o
Ti
• Flow direction is reversed for T i < T o .
• Circulation enhances thermal conductivity
q′ =
2π keff
ln( Do / Di )
(Ti − To )
To
Di
(10.46)
Fig. 10..13
34
Correlation equation for the effective conductivity keff :
Pr
*
= 0.386
Ra
k
0.861 + Pr
keff
Ra * =
[ln( Do / Di )] 4
[
δ 3 ( Di )− 3 / 5 + ( Do )− 3 / 5
δ=
Valid for
]
5
1/ 4
(10.47a)
Raδ
Do − Di
2
(10.47b)
(10.47c)
concentric cylinders
10 2 < Ra* < 107
(10.47d)
properties at T = (Tc + Th ) / 2
35
10.8 Other Correlations
The above presentation is highly abridged.
There are many other correlation equations for:
• Boiling
• Condensation
• Jet impingement
• High speed flow
• Dissipation
• Liquid metals
• Enhancements
• Finned geometries
• Irregular geometries
• Non-Newtonian fluids
• Etc.
Consult textbooks, handbooks, reports and journals
36
Heat Transfer in Microchannels
11.1 Introduction:
Applications
Cooling of microelectronics
Inkjet printer
Medical research
Micro-electro-mechanical systems (MEMS): Micro heat
exchangers, mixers, pumps, turbines, sensors and
actuators
1
11.1.1 Continuum and Thermodynamic
Equilibrium Hypothesis
Properties: (pressure, temperature, density, etc) are
macroscopic manifestation of molecular activity
Continuum: material having sufficiently large number
of molecules in a given volume to give unique values
for properties
Validity of continuum assumption: the molecularmean-free path, , is small relative to the characteristic
dimension of the system
Mean-free-path: average distance traveled by molecules
before colliding
2
Knudson number Kn:
Kn =
λ
(1.2)
De
De = characteristic length
Gases: the criterion for the validity of the continuum
assumption is:
Kn < 10 −1
(1.3a)
Thermodynamic equilibrium: depends on collisions
frequency of molecules. The condition for thermodynamic
equilibrium is:
Kn < 10
−3
(1.3b)
3
At thermodynamic equilbirium:
fluid and an adjacent surface have the same velocity and temperature:
no-velocity slip
no-temperature jump
Continuity, Navier-Stokes equations, and energy equation
are valid as long as the continuum assuption is valid
No-velocity slip and no-temperature jump
are valid as long as thermodynamic equilibrium is justified
Microchannels: Channels where the continuum assumption
and/or thermodynamic equilibrium break down
4
9.1.2. Surface Forces. Examine ratio of surface to
volume for tube:
A
πDL
4
=
=
V πD 2 L / 4 D
(11.1)
•For D = 1 m, A/V = 4 (1/m)
•For D = 1 µm, A/V = 4 x 10 6 (1/m)
• Consequence:
(1) Surface forces may alter the nature of surface
boundary conditions
(2) For gas flow, increased pressure drop results in
large density changes. Compressibility becomes
important
5
9.1.3 Chapter Scope
• Classification
• Gases vs. liquids
• Surface boundary conditions
• Heat transfer in Couette flow
• Heat transfer in Poiseuille flow
9.2 Basic Consideration
9.2.1 Mean Free Path. For gases:
λ=
µ π
p
2
RT
(11.2)
6
Table 11.1
p = pressure
R = gas constant
T = temperature
µ = viscosity
values of λ for common gases
R
gas
Air
Helium
ρ
µ × 10 7
J/kg− K kg/m3 kg/s− m
287.0
1.1614
2077.1
0.1625
Hydrogen 4124.3 0.0808
Nitrogen
296.8 1.1233
Oxygen
259.8 1.2840
184.6
199.0
89.6
178.2
207.2
λ
µm
0.067
0.1943
0.1233
0.06577
0.07155
NOTE:
• Pressure drops along a channel → λ increases
→ Kn increases
• λ is very small, expressed in terms of the micrometer,
µ = 10 − m
6
7
11.2.2 Why Microchannels?
Nusselt number: fully developed
flow through tubes at uniform
surface temperature
hD
NuD =
= 3.66
k
k
h = 3.657
D
106
water
5
10
104
air
103
(6.57)
(11.3)
102
continuum
101
100
101
102
103
104
D( µm)
Fig. 11.1
As D ↓ h ↑
sink
Application:
Water cooled microchips
flow
microchip q
Fig. 11.2
8
11.2.3 Classification
Based on the Knudsen number:
Kn < 0.001
continuum, no − slip flow
0.001< Kn < 0.1 continuum, slip flow
0.1 < Kn < 10
transition flow
10 < Kn
free molecular flow
(11.4)
Four important factors:
(1) Continuum
(2) Thermodynamic equilibrium
(3) Velocity slip
(4) Temperature jump
9
(1) Kn < 0.001: Macro-scale regime (previous chapters):
Continuum: valid
Thermodynamic equilibrium: valid
No velocity slip
No temperature jump
(2) 0.001 < Kn < 0.1: Slip flow regime:
Continuum: valid
Thermodynamic equilibrium: fails
• Velocity slip
• Temperature jump
Continuity, Navier-Stokes equations, and energy equations are valid
No-velocity slip and No-temperature jump conditions, conditions fail
Reformulate boundary conditions
10
(3) 0.1< Kn<10: Transition flow:
Continuity and thermodynamic equilibrium fail
Reformulate governing equations and boundary conditions
Analysis by statistical methods
(4) Kn>10: Free molecular flow: analysis by kinetic theory
of gases
11.2.4 Macro and Microchannels
Macrochannels: Continuum domain, no velocity slip,
no temperature jump
Microchannels: Temperature jump and velocity slip,
with or without failure of continuum assumption
11
Distinguishing factors:
(1) Two and three dimensional effects
(2) Axial conduction
(3) Viscous dissipation
(4) Compressibility
(5) Temperature dependent properties
(6) Slip velocity and temperature
(7) Dominant role of surface forces
12
11.2.5 Gases vs. Liquids
Macro convection:
• No distinction between gases and liquids
• Solutions for both are the same for the same geometry,
governing parameters (Re, Pr, Gr,…) and boundary
conditions
Micro convection:
• Flow and heat transfer of gases differ from liquids
Gas and liquid characteristics:
(1) Mean free path: λliquid << λ gas
• Continuum assumption may hold for liquids but fail
for gases
13
• Typical MEMS applications: continuum assumption is
valid for liquids
(2) Knudsen number: used as criterion for thermodynamic
equilibrium and continuum for gases but not for liquids
(3) Onset of failure of thermodynamic equilibrium and
continuum: not well defined for liquids
(4) Surface forces: liquid forces are different from gas forces
(5) Boundary conditions: differ for liquids from gases
(6) Compressibility: liquids are almost incompressible while
gases are not
(7) Flow physics: liquid flow is not well known. Gas flow is
well known
14
(8) Analysis: more complex for liquids than gases
11.3 General Features
• Flow and heat transfer phenomena change as channel
size is reduced:
Rarefaction: Knudsen number effect
Compressibility: Effect of density change due to pressure
drop along channel
Viscous dissipation: Effect of large velocity gradient
Examine: Effect of channel size on:
• Velocity profile
• Flow rate
15
• Friction factor
• Transition Reynolds number
• Nusselt number
Consider:
Fully developed microchannel gas flow as the Knudsen
number increases from the continuum through the slip
flow domain
16
11.3.1 Flow Rate
Slip flow: increased velocity and
flow rate
Qe
>1
Qt
(11.5)
(a) no-slip velocity
(b) slip velocity
Fig. 11.3
e = determined experimentally
t = from macrochannel theory or correlation equations
11.3.2 Friction Factor f
• Define friction coefficient C f
Cf =
τw
2
( 1 / 2 ) ρ um
(4.37a)
17
τ w = wall shear stress
um = mean velocity
• Fully developed flow through channels: define friction
factor f
1 D ∆p
f =
2
2 L ρ um
(11.6)
D = diameter
L = length
∆p = pressure drop
18
Macrochannels: fully developed laminar flow:
(1) f is independent of surface roughness
(2) Product of f and Reynolds number is constant for each
channel geometry:
f Re = Po
Po = Poiseuille number
(3) Po is independent of Reynolds number
Microchannels: compare experimental data, ( Po)e, with
theoretical value, ( Po) t , (macroscopic, continuum)
19
(Po)e
( Po)t
= C*
(11.8)
Conclusion:
(1) C * departs from unity:
1 << C * << 1
(2) Unlike macrochannels, Po for fully developed flow
depends on the Re
(3) Conflicting findings due to: difficulties in measurements
of channel size, surface roughness, pressure distribution,
uncertainties in entrance effects, transition, and
determination of properties
20
11.3.3 Transition to turbulent flow
Macrochannels: smooth macrotubes
Re t =
uD
ν
≈ 2300
(6.1)
Microchannels: reported transition
300 < Re t < 16,000
Factors affecting the determination of Re t :
• Variation of fluid properties
• Measurements accuracy
• Surface roughness
21
11.3.4 Nusselt number. For fully developed conditions:
Macrochannel: Nusselt number is constant
Microchannels: In general, Nusselt number is not well
established:
• Nu varies along microchannels
• Nu depends on:
• Surface roughness
• Reynolds number
• Nature of gas
• Widely different reported results:
( Nu)e
0.21 <
< 100
( Nu)t
(11.9)
22
where:
( ) e = experimental
( ) t = macrochannel theory
Factors affecting the determination of Nue :
• Variation of fluid properties
• Measurements accuracy
23
11.4 Governing Equations
Slip flow regime: 0.001 < Kn < 0.1 :
Continuity, Navier - Stokes equations, and energy equation are valid
No - velocity slip and no - temperature jump conditions fail
Reformulate boundary conditions
Factors to be considered:
• Compressibility
• Axial conduction
• Dissipation
11.4.1 Compressibility: Expressed in terms of Mach number
fluid velocity
M=
speed of sound
24
Macrochannels:
• Incompressible flow, M < 1
• Linear pressure drop
Microchannels:
• Compressible flow
• Non-linear pressure drop
• Decrease in Nusselt number
11.4.2 Axial Conduction
Macrochannels: neglect axial conduction for
Pe = Re D Pr > 100
(6.30)
25
Pe = Peclet number
Microchannels: low Peclet numbers, axial conduction may
be important, it increases the Nusselt number
11.4.3 Dissipation
Microchannels: large velocity gradient, dissipation may
become important
11.5 Slip Velocity and Temperature Jump
Boundary Conditions
Slip velocity for gases:
2 − σ u ∂u( x ,0)
u( x ,0) − us =
λ
∂n
σu
(11.10)
26
u( x ,0) = fluid axial velocity at surface
u s = surface axial velocity
x = axial coordinate
n = normal coordinate measured from the surface
σ u = tangential momentum accommodating coefficient
Temperature jump for gases
2 − σ T 2γ λ ∂T ( x ,0)
T ( x ,0) − Ts =
σ T 1 + γ Pr ∂n
(11.11)
T(x,0) = fluid temperature at the boundary
T s = surface temperature
27
γ = c p / cv , specific heat ratio
σ T = energy accommodating coefficient
NOTE
(1) Eq. (11.10) and (11.11) are valid for gases
(2) Eq. (11.10) and (11.11) are valid for Kn < 0.1
(3) σu and σT, are:
• Empirical factors
• They depend on the gas, geometry and surface
• Values range from zero (perfectly smooth) to unity
28
• Difficult to determine experimentally
• Values for various gases are approximately unity
29
11.4. 8 Analytic Solutions: Slip Flows
Two common flow types, extensive use in MEMS:
(1) Couette flow (shear driven): fluid is set in motion by
a moving surface
Examples:
100 µm
stator
rotor
stationary
ω
2µ
movable
Fig. 11.4
Fig. 11.5
30
(2) Poiseuille flow (pressure driven): fluid is set in motion by
an axial pressure gradient
Examples:
Micro heat exchangers, mixers, microelectronic heat sinks
NOTE
• No pressure drop in Couette flow
• Signifiant pressure drop in Poiseuille flow
Boundary conditions: two types:
(1) Uniform surface temperature
(2) Uniform surface heat flux
31
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
32
(10) The accommodation coefficients are equal to unity,
σ u = σ T = 1 .0
11.6.2 Couette Flow with Viscous Dissipation:
Parallel Plates with Surface Convection
y
ho T∞
us
• Infinitely large parallel plates
u
• Gas fills gap between plates
H
x
Fig. 11.6
• Upper plate: moves with velocity us
• Lower plate: stationary, insulated
• Convection at the upper plate
• Consider dissipation and slip conditions
33
Determine:
(1) Velocity distribution
(2) Mass flow rate
(3) Nusselt number
Find flow field and temperature distribution
Flow Field
• Normal velocity and all axial derivatives vanish
• Axial component of the Navier-Stokes equations, (2.9),
simplifies to
d 2u
dy 2
=0
34
Boundary conditions: use (11.10), Setσ u = 1
• Lower plate: n = y = 0 and us = 0,
(11.10) gives
du( x ,0)
u( x ,0) = λ
dy
(g)
• Upper plate: n = H – y, (9.10) gives
du( x , H )
u( x , H ) = us − λ
dy
Solution
y
u
1
(
+ Kn )
=
u s 1 + 2 Kn H
(11.14)
35
Kn is the local Knudsen number
Kn =
λ
H
(11.13)
NOTE
(1) Fluid velocity at the moving plate: set y = H in (11.14)
u( H ) 1 + Kn
<1
=
us
1 + 2 Kn
Effect of slip:
• Decrease fluid velocity at the moving plate
• Increase fluid velocity at the stationary plate
36
(2) Velocity distribution is linear
(3) Setting Kn = 0 in (11.14) gives the no-slip solution
y
u
=
us H
(k)
Mass Flow Rate m
m =W
H
∫0
ρ u dy
(11.15)
W = channel width
Neglect variation of ρ along y, (11.14) into (11.15)
37
us
m = ρ WH
2
• Flow rate is independent of the Knudsen number
• Compare with macrochannel flow rate mo
(k) into (11.15)
us
mo = ρ WH
2
(11.17)
This is identical to (11.16), thus
m
=1
mo
(11.18)
38
Nusselt Number
• Equivalent diameter for parallel plates, De = 2H
• Nusselt number
2 Hh
Nu =
k
(l)
Heat transfer coefficient h:
∂T ( H )
−k
∂y
h=
Tm − Ts
39
∂T ( H )
∂y
Nu = −2 H
Tm − Ts
(11.19)
k = conductivity of fluid
T = fluid temperature
Ts = plate temperature
NOTE
(1) Fluid temperature at the moving plate, T (x,H), is not
equal to surface temperature
(2) h is defined in terms of surface temperature Ts
40
(3) Use temperature jump, (11.11), to determine Ts
(4) For the upper plate, n =H – y, eq. (11.11) gives
2γ λ ∂T ( x , H )
Ts = T ( x , H ) +
1 + γ Pr
∂y
(11.20)
• Mean temperature Tm: defined in Section 6.6.2
mc pTm = W
H
∫0
ρ c p uT dy
(11.21)
• Neglect variation of cp and ρ along y, use (11.14)
. for u and (11.15) for m
41
2
Tm =
us H
H
∫0
uT dy
(11.22)
Determine temperature distribution:
• Use energy equation, (2.15)
• Apply above assumptions, note that axial derivatives
vanish, (2.15) gives
k
∂ 2T
∂y
2
+ µΦ = 0
(11.23)
42
(2.17) gives the dissipation function Φ which simplifies to
∂u
Φ =
∂y
2
(11.24)
(9.24) into (9.23)
2
d T
dy
2
=−
µ du
k dy
2
(11.25)
Boundary conditions
Lower plate:
dT (0)
=0
dy
(m)
43
Upper plate:
dT ( H )
−k
= ho (Ts − T∞ )
dy
Use (920) to eliminate Ts
dT ( H )
−k
= ho
dy
2γ λ ∂T ( x , H )
− T∞
T ( x , H ) + 1 + γ Pr
∂n
(n)
Use velocity solution (9.14), solve for T
2
kH
H
2γ Kn 2
β
β
2
T =− y +
H β + T∞
+
+
2
ho
2
γ + 1 Pr
β
(11.26)
44
where
µ
us
β =
k H (1 + 2 Kn )
2
(p)
Velocity solution (11.14), temperature solution (11.26)
giveTs , Tm and Nu
kHβ
Ts =
+ T∞
ho
Tm =
(u)
1
2
2γ Kn 2
1 2
2 kHβ
H
β
KnH
β
+
H β + T∞ (w)
+
+
γ + 1 Pr
1 + 2 Kn 4
3
ho
45
2
Nu =
1
2γ Kn
1 2
+ Kn +
1 + 2 Kn 4 3 γ + 1 Pr
(11.27)
Note the following regarding the Nusselt number
(1) It is independent of Biot number
(2) It is independent of the Reynolds number
(3) Unlike macrochannels, it depends on the fluid
(4) First two terms in the denominator of (11.27) represent
rarefaction (Knudsen number). The second term
represents effect of temperature jump
46
(5) Nusselt number for macrochannels, Nuo: set Kn = 0
in (11.27):
Nuo = 8
(11.28)
Ratio of (11.27) and (11.28)
Nu
1 + 2 Kn
=
Nuo 8 8γ Kn
1 + 3 Kn + γ + 1 Pr
(11.29)
NOTE: Ratio is less than unity
47
11.6.3 Fully Developed Poiseuille Channel
Flow: Uniform Surface Flux
• Pressure driven flow between parallel plates
• Fully developed velocity and temperature
• Inlet and outlet pressures are pi and po
• Uniform surface flux, q ′s′
q′s′
y
Determine:
(1) Velocity distribution
(2) Pressure distribution
(3) Mass flow rate
(4) Nusselt number
H/2
H/2
x
q′s′
Fig. 11.7
48
Note: ∂p ≠ 0
∂x
Major difference between macro and micro fully developed
slip flow:
Macrochannels: incompressible flow
(1) Parallel streamlines
(2) Zero lateral velocity component (v = 0)
(3) Invariant axial velocity ( ∂ u / ∂ x = 0)
(4) Linear axial pressure ( dp / dx = constant)
49
Microchannels: compressibility and rarefaction change
above flow pattern:
(1) None of above conditions hold
(2) Large axial pressure drop → density changes
→ compressible flow
(3) Rarefaction: pressure decreases → λ increases
→ Kn increases with x
(4) Axial velocity varies with axial distance
(5) Lateral velocity v does not vanish
(6) Streamlines are not parallel
(7) Pressure gradient is not constant
50
Assumptions
(1) Steady state
(2) Laminar flow
(3) H / R << 1
(4) Two-dimensional
(5) Slip flow regime (0.001 < Kn < 0.1)
(6) Ideal gas
(7) Constant viscosity, conductivity and specific heats
(8) Negligible lateral variation of density and pressure
(9) σ u = σ T = 1.0
51
(10) Negligible dissipation
Flow Field
Additional assumptions:
(11) Isothermal flow
(12) Negligible inertia forces:
∂u
∂v
ρ (u + v ) = 0
∂x
∂y
(13) The dominant viscous force is µ
∂ 2u
∂y 2
Navier-Stokes equations (2.9) simplify to:
∂p
∂ 2u
− +µ
=0
2
∂x
∂y
52
Boundary conditions:
Symmetry at y = 0
∂u( x,0)
=0
∂y
(e)
For the upper plate, n = H – y
∂u( x , H / 2)
u( x , H / 2) = − λ
∂y
(f)
Solution to u
H 2 dp
y2
u=−
1 + 4 Kn( p ) − 4 2
8 µ dx
H
53
For an ideal gas
Kn =
λ
H
=
µ π
H
1
RT
2
p
(11.33)
Pressure Distribution p:
To determine p(x), must determine vertical component v:
start with continuity (2.2a)
∂ρ ∂
∂
∂
+ ( ρ u) + ( ρ v ) + ( ρ w ) = 0
∂ t ∂x
∂y
∂z
Apply above assumptions
54
∂
∂
(ρ u) + (ρ v ) = 0
∂x
∂y
(h)
Use ideal gas to eliminate ρ:
p
ρ=
RT
(11.31)
(11.31) into (h), assuming constant temperature
∂
∂
( pv ) = − ( pu )
∂y
∂x
(i)
(11.30) into (i)
55
∂
H 2 ∂ dp
y2
( pv ) =
p (1 + 4 Kn( p) − 4 2 )
∂y
8 µ ∂x dx
H
(j)
Boundary conditions:
v ( x ,0 ) = 0
(k)
v ( x , H / 2) = 0
(l)
Multiply (j) by dy, integrate and using (k)
∫
y
H 2 ∂ dp
y2
d ( pv ) =
(1 + 4 Kn( p) − 4 2 ) dy
p
8 µ ∂x dx 0
0
H
y
∫
(m)
56
Evaluate the integrals
H 3 1 ∂ dp
v=
p
8 µ p ∂x dx
y 4 y 3
[1 + 4 Kn( p)] −
3
H
3
H
(11.32)
Determination of p(x): Apply boundary condition (l) to (11.32)
∂ dp
p
∂x dx
y 4 y 3
=0
[1 + 4 Kn( p)] −
3
H 3 H
y= H / 2
Express Kn in terms of pressure. Equations (11.2) and
(11.13) give
λ
µ π
1
Kn =
H
=
H
2
RT
p
(n)
(11.33)
57
Evaluate (n) at y = H/2, substitute (11.33) into (n) and
integrate
p
dp
dx
1
1 µ
+
2
π
RT
=C
3 H
p
Integrate again (T is assumed constant)
1 2 µ
p +
2π RT p = Cx + D
6
H
(o)
Solve for p
p( x ) = −3
µ
H
2π RT + 18π RT
µ2
H
2
+ 6Cx + 6 D
(p)
58
Pressure boundary conditions
p(0) = pi , p(L) = po
(q)
Apply (q) to (p)
1
µ
C=
( po − pi ) +
2πRT ( po − pi )
6L
HL
2
2
pi µ
D=
2πRT pi
+
6 H
2
Substitute into (p) and normalize by po
59
p( x)
3µ
=−
2π RT +
po
Hpo
(r)
pi x pi 6µ
pi
18µ 2 π RT pi 6µ
+
1
−
+
2
π
RT
(
1
−
)
+
+
2
π
RT
po L po2 Hpo
po
H 2 po po Hpo
2
2
2
2
Introduce outlet Knudsen number Kno using (11.2) and
(11.13)
Kno =
λ ( po )
H
=
µ
π
H po
2
RTo
(11.34)
Substitute (11.34) into (r)
2
pi
pi2
pi x
p( x )
= −6 Kno + 6 Kno +
+ (1 − 2 ) + 12 Kno (1 − p ) L
po
p
po
o
o
(11.35)
60
NOTE:
(1) Unlike macrochannel Poiseuille flow, pressure
variation along the channel is non-linear
(2) Knudsen number terms represent rarefaction effect
(3) The terms (pi/po)2 and [1- (pi/po)2](x/L) represent the
effect of compressibility
(4) Application of (11.35) to the limiting case of Kno =0
gives
p( x )
=
po
pi2
pi2 x
+ (1 − 2 )
2
po
po L
(11.36)
This result represents the effect of compressibility alone
61
Mass Flow Rate
m = 2W
H/2
∫0
ρ udy
(s)
W = channel width
(11.30) in (s)
WH 3
[1 + 6 Kn( p)] ρ dp
m=−
12µ
dx
(t)
Density ρ :
p
ρ=
RT
(11.37)
62
(11.33) gives Kn(p)
Kn( p ) =
λ ( p)
H
=
µ
π
Hp 2
RT
(11.33)
(11.33) and (11.37) in (t)
dp
WH 3
µ π
m=−
p+6
RT
12µ RT
H 2
dx
(11.38)
(11.35) into (11.38) and let T=To
1 W H 3 po2
m=
24 µ LRTo
pi2
pi
2 − 1 + 12 Kno 24( − 1)
po
po
(11.39)
63
Compare with no-slip, incompressible macrochannel case:
1 W H 3 po2 pi
mo =
− 1
12 µ LRT po
(11.40)
m 1 pi
= + 1 + 12 Kno
mo 2 po
(11.41)
Taking the ratio
NOTE
(1) Microchannels flow rate is very sensitive to H
(2) (11.39) shows effect of rarefaction (slip) and
compressibility on m
64
(3) Since pi / po > 1, (11.41) shows that neglecting
compressibility and rarefaction underestimates m
Nusselt Number
2 Hh
Nu =
k
(u)
For uniform surface flux q ′s′′
q′s′
h=
Ts − Tm
Substitute into (g)
2 H q′s′
Nu =
k (Ts − Tm )
(v)
65
Plate temperature Ts: use (11.11)
Ts = T ( x , H / 2) +
2γ λ ∂T ( x , H / 2)
1 + γ Pr
∂y
(11.42)
Mean temperature Tm:
H/2
Tm
∫
= 0
uT dy
(11.43)
H/2
∫0
udy
Need u(x,y) and T(x,y)
Velocity distribution: (11.30) gives u(x,y) for isothermal 66flow
Additional assumption:
(14) Isothermal axial velocity solution is applicable
(15) No dissipation, Φ = 0
(16) No axial conduction, ∂ 2T / ∂x 2 << ∂ 2T / ∂y 2
(17) Negligible effect of compressibility on the energy
equation
(18) Nearly parallel flow, v = 0
Energy equation: equation (2.15) simplifies to
∂T
∂ 2T
=k 2
ρ c pu
∂x
∂y
(11.44)
67
Boundary conditions:
k
∂T ( x ,0)
=0
∂y
(w)
∂T ( x , H / 2)
= q′s′
∂y
(x)
To solve (11.44), assume:
(19) Fully developed temperature
Solution: T(x,y) and Tm(x): Define
T ( x , H / 2) − T ( x , y )
φ=
T ( x , H / 2) − Tm ( x )
(11.45)
68
Fully developed temperature:
φ
is independent of x
φ = φ ( y)
(11.46)
∂φ
=0
∂x
(11.47)
Thus
(11.45) and (11.46) give
∂φ
∂ T ( x , H / 2) − T ( x, y )
=
=0
∂x ∂x T ( x , H / 2) − Tm ( x )
69
Expanding and use (11.45)
dT ( x , H / 2) ∂T
dT ( x , H / 2) dTm ( x )
− φ ( y)
−
−
= 0 (11.48)
∂x
dx
dx
dx
Determine:
∂T ( x , y ) dT ( x , H / 2)
dTm ( x )
,
and
∂x
dx
dx
Heat transfer coefficient h:
∂T ( x , H / 2)
−k
∂y
h=
Tm ( x ) − Ts ( x )
(y)
70
(11.42) gives Ts(x). (11.45) gives temperature gradient in (y)
T ( x, y ) = T ( x, H / 2) − [T ( x, H / 2) − Tm ( x )]φ
Differentiate
∂T ( x , H / 2)
dφ ( H / 2)
= −[T ( x , H / 2) − Tm ( x )]
∂y
dy
(z)
(z) into (y), use (11.42) for Ts(x)
k[T ( x , H / 2) − Tm ( x )] dφ ( H / 2)
h=−
Ts ( x ) − Tm ( x )
dy
(11.49)
71
Newton’s law of cooling:
q′s′
h=
Ts ( x ) − Tm ( x )
Equate with (11.49)
q′s′
T ( x , H / 2) − Tm ( x ) = −
= constant
dφ ( H / 2)
dy
(11.50)
Differentiate
∂T ( x , H / 2) ∂Tm ( x )
−
=0
∂x
∂x
Combine this with (11.48)
72
dT ( x , H / 2) dTm ( x ) ∂T
=
=
dx
dx
∂x
(11.51)
NOTE:
(11.51) replaces∂T with dTm in (11.44)
∂x
dx
Determine dTm :
dx
q′s′
m
Conservation of energy for element:
Tm +
Tm
dTm
dx
dx
dx
q′s′
Fig. 11.8
73
Conservation of energy for element:
dTm
2q′s′Wdx + mc pTm = mc p Tm +
dx
dx
Simplify
dTm 2Wq′s′
=
= constant
dx
mc p
(aa)
However
m = WHρ um
(bb)
(bb) into (aa)
74
dTm
2q′s′
=
= constant
dx
ρ c p um H
(11.52)
(11.52) into (11.51)
dT ( x , H / 2) dTm ( x ) ∂T
2q′s′
=
=
=
dx
dx
∂x ρ c p um H
(11.53)
(11.53) into (11.44)
∂ 2T
2q′s′ u
=
2
kH um
∂y
(11.54)
75
Mean velocity:
um
2
=
H
H /2
∫0
udy
(cc)
(11.30) gives velocity u. (11.30) into (cc)
H 2 dp H / 2
um = −
4 µ dx 0
∫
y2
1 + 4 Kn − 4 2 dy
H
Integrate
H 2 dp
[1 + 6 Kn]
um = −
12µ dx
(11.55)
76
Combining (11.30) and (11.55)
u
6 1
y2
=
+ Kn − 2
um 1 + 6 Kn 4
H
(11.56)
(11.56) into (11.54)
12
q′s′ 1
y2
=
+ Kn − 2
2
1 + 6 Kn kH 4
H
∂y
∂ 2T
(11.57)
Integrate twice
1 1
12q′s′
y4
T ( x, y) =
( + Kn) y −
+ f ( x) y + g( x)
2
(1 + 6 Kn)kH 2 4
12 H
2
(dd)
77
f(x) and g(x) are “constants” of integration
Boundary condition (w) gives
f ( x) = 0
Solution (dd) becomes
1 1
12q′s′
y4
T ( x, y) =
( + Kn) y −
+ g( x)
2
(1 + 6 Kn)kH 2 4
12 H
2
(11.58)
NOTE:
(1)Boundary condition (x) is automatically satisfied
(2) g(x) is determine by formulating Tm using two methods
78
Method 1: Integrate (11.52)
Tm
∫Tmi
2q′s′
dTm =
ρ c p um H
x
∫0 dx
where
Tm (0) = Tmi
(11.59)
Evaluate the integrals
2q′s′
Tm ( x ) =
x + Tmi
ρ c p um H
(11.60)
79
Method 2: Use definition of Tm. Substitute (11.30) and
(11.58) into (11.43)
Tm ( x ) =
H dp
−
8µ dx
H /2
∫0
1 1
y 2
12q′s′
y4
2
1 + 4 Kn − 4 2
( + Kn) y −
+ g ( x )dy
2
12 H
H (1 + 6 Kn)kH 2 4
H dp H / 2
y2
−
1 + 4 Kn − 4 2 dy
8µ dx 0
H
∫
Evaluate the integrals
3q′s′ H
13
2 13
Tm ( x ) =
( Kn) + Kn +
+ g( x)
2
40
560
k (1 + 6 Kn)
(11.61)
Equating (11.60) and (11.61) gives g(x)
2q′s′
3q′s′ H
13
2 13
g( x) = Tmi +
x−
(
Kn
)
+
Kn
+
40
560
ρc pum H
k(1 + 6Kn)2
(11.62)
80
(11.58) into (11.42) gives Ts
3q′s′ H 1
5
2γ q′s′ H
Ts ( x ) =
Kn + +
Kn + g ( x )
k (1 + 6 Kn) 2
48 γ + 1 kPr
(11.63)
The Nusselt number is given in (v)
2 H q′s′
Nu =
k (Ts − Tm )
(v)
(11.61) and (11.63) into (v)
Nu =
2
3
1
5
1
13
2γ Kn
2 13
(
)
−
Kn
Kn
+
+
+
Kn +
(1 + 6 Kn) 2
48 (1 + 6 Kn)
40
560 γ + 1 Pr
(11.64)
81
NOTE:
(1) Kn in (11.64) depends of local pressure p
(2) Pressure varies with x, Kn varies with x
(3) Unlike macrochannels, Nu is not constant
(4) Unlike macrochannels, Nu depends on the fluid
(5) No-slip Nu for macrochannel
flow, Nuo: set Kn = 0 in (11.64)
0
8
Nu
6
140
Nuo =
= 8.235
17
(11.65)
4
0
0.04
0.08
0.12
Kn
Fig. 11.9 Nusselt number for air
82
This agrees with Table 6.2
(6) Rarefaction and compressibility decrease the Nusselt
number
83
11.6.4 Fully Developed Poiseuille Channel
Flow: Uniform Surface Temperature
Repeat Section 11.6.3 with plates at uniform surface
temperature Ts
• Flow field: same for both cases:
y
Ts
(11.30)→ u(y)
H/2
(11.35) → p( x ) / po
H/2
x
(11.35) → m
Ts
Fig. 11.10
• Energy equation: (11.44) is modified to include axial
conduction
84
• Boundary conditions: different for the two cases
Nusselt number:
2 Hh
− 2H
∂T ( x , H / 2 )
Nu =
=
k
Tm ( x ) − Ts
∂y
Need T(x,y) and Tm(x)
Solution approach:
Solve the Graetz channel entrance problem and set
x → ∞ to obtain the fully developed solution
85
Axial conduction: can be neglected for:
Pe = PrRe > 100
Microchannels:
Small Reynolds → Small Peclet number → Axial
conduction is important
Include axial conduction: modify energy equation (11.44)
∂T
∂ 2T ∂ 2T
ρ c pu
= k( 2 + 2 )
∂x
∂x
∂y
(11.67a)
86
Boundary and inlet conditions:
∂T ( x ,0)
=0
∂y
T ( x , H / 2) = Ts −
2γ H
∂T ( x , H / 2 )
Kn
∂y
γ + 1 Pr
(11.68a)
(11.69a)
T ( 0, y ) = Ti
(11.70a)
T (∞, y ) = Ts
(11.71a)
87
8
Axial velocity
7
Pe = 0
1
5 ∞8
Nu
1
u
6
y2
=
+ Kn − 2
um 1 + 6 Kn 4
H
(11.56)
6
5
0
Solution
0.04
0.08
0.12
Kn
Fig. 11.11 Nusselt number for flow between
parallel plates at uniform surface
temperature for air, Pr = 0.7,
γ = 1.4 , σu = σT =1, [14]
• Use method of separation of variables
• Specialize to fully developed:
set x → ∞
Result: Fig. 11.11 shows Nu vs. Kn
88
NOTE
(1) Nu decreases as the Kn is increased
(2) No-slip solution overestimates microchannels Nu
(3) Axial conduction increases Nu
(4) Limiting case: no-slip (Kn = 0) and no axial conduction
( Pe = ∞ ) :
Nuo = 7.5407
(11.73)
This agrees with Table 6.2
Heat Transfer Rate, q s:
Following Section 6.5
89
q s = m c p [Tm ( x ) − Tmi ]
(6.14)
Tm (x) is given by
Tm ( x ) = Ts + (Tmi − Ts ) exp [−
Ph
x]
mc p
(6.13)
h , is determine numerically using (6.12)
1 x
h = ∫ h( x )dx
x 0
(6.12)
90
11.6.5 Fully Developed Poiseuille Flow in
Micro Tubes: Uniform Surface Flux
q′s′
r
r
ro
Consider:
• Poiseuille flow in
micro tube
• Uniform surface flux
z
q′s′
Fig. 11.12
• Fully developed velocity and temperature
• Inlet and outlet pressures are pi and po
91
Determine
(1) Velocity distribution
(2) Nusselt number
• Rarefaction and compressibility affect flow and heat
transfer
• Velocity slip and temperature jump
• Axial velocity variation
• Lateral velocity component
• Non-parallel stream lines
• Non-linear pressure
92
Assumptions
Apply the 19 assumptions of Poiseuille flow between
parallel plates (Sections 11.6.3)
Flow Field
• Follow analysis of Section 11.6.3
• Axial component of Navier-Stokes equations in
cylindrical coordinates:
1 ∂ ∂v z
1 ∂p
(r
)=
r ∂r
∂r
µ ∂z
(a)
v z ( r , z ) = axial velocity
93
Boundary conditions:
Assume symmetry and set σu = 1
∂v z (0, z )
∂r
=0
∂v z ( ro , z )
u( ro , z ) = − λ
∂r
(b)
(c)
Solution
ro2 dp
r2
vz = −
1 + 4 Kn − 2
4 µ dz
ro
(11.74)
94
Knudsen number
Kn =
λ
2 ro
(11.75)
Mean velocity vzm
v zm =
1
π ro2
ro
∫0
2π r v z dr
Use (11.74), integrate
ro2 dp
v zm = −
(1 + 8 Kn)
8µ dz
(11.76)
95
(11.74) and (11.76)
vz
1 + 4 Kn − ( r / ro ) 2
=2
v zm
1 + 8 Kn
Solution to axial pressure
2
p( z )
pi
pi
pi z
= −8 Kno + 8 Kno +
+ (1 − ) + 16 Kno (1 − )
po
po
po L
po
2
(11.78)
2
(11.76) and (11.78) give m
π ro4 po2
m=
16 µ LRT
pi2
pi
− 1) (11.79a)
2 − 1 + 16 Kno (
po
po
96
For incompressible no-slip (macroscopic)
π ro4 po2 pi
mo =
(
− 1)
8 µ LRT po
(11.79b)
Nusselt Number
• Follow Section 11.6.3
2ro h
Nu =
k
(d)
Heat transfer coefficient h:
q′s′
h=
Ts − Tm
97
Substituting into (d)
2 ro q′s′
Nu =
k (Ts − Tm )
(e)
Ts = tube surface temperature, obtained from temperature
jump condition (11.11)
2γ λ ∂T ( ro , z )
Ts = T ( ro , z ) +
1 + γ Pr
∂r
Mean temperature:
ro
Tm
∫
= 0
v z T r dr
ro
∫0
v z rdr
(f)
(11.80)
98
Energy equation:
ρ c pv z
∂T k ∂ ∂T
(r )
=
∂ z r ∂r ∂r
(11.81)
Boundary conditions:
∂T (0, z )
=0
∂r
(g)
∂T ( ro , z )
k
= q′s′
∂r
(h)
Define
99
T (ro , z ) − T (r , z )
φ=
T (ro , z ) − Tm ( z )
(11.82)
Fully developed temperature:
Thus
φ = φ (r )
(11.83)
∂φ
=0
∂z
(11.84)
(11.82) and (11.84) give
∂φ
∂ T (ro , z ) − T (r , z )
=
=0
∂ z ∂ z T (ro , z ) − Tm ( z )
100
Expand (11.82)
dT (ro , z ) dTm ( z )
dT (ro , z ) ∂T
−
− φ (r )
−
=0
dz
∂z
dz
dz
(11.85)
Determine:
∂T (r , z ) dT (ro , z )
dTm ( z )
and
,
∂z
dz
dz
Heat transfer coefficient h,:
∂T (ro , z )
−k
∂r
h=
Tm ( z ) − Ts ( z )
(i)
101
Rewrite (11.81)
T (r , z ) = T (ro , z ) − [T (ro , z ) − Tm ( z )]φ
Differentiate and evaluating at r = ro
∂T (ro , z )
dφ (ro )
= −[T (ro , z ) − Tm ( z )]
∂r
dr
(j)
k[T (ro , z ) − Tm ( z )] dφ (ro )
h=−
Ts ( z ) − Tm ( z )
dr
(k)
(j) into (i)
102
Newton’s law of cooling h
q′s′
h=
Ts ( z ) − Tm ( z )
Equate with (k)
q′s′
T (ro , z ) − Tm ( z ) = −
= constant
dφ (ro )
k
dr
(11.86)
Differentiate
∂T (ro , z ) ∂Tm ( z )
−
=0
∂z
∂z
103
Combine with (11.85)
dT (ro , z ) dTm ( z ) ∂T
=
=
dz
dz
∂z
(11.87)
Will use (11.87) to replace ∂T / ∂z in (11.81) withdTm / dz.
q′s′
Conservation of energy to dx
m
Tm +
Tm
dTm
dx
dx
dx
dTm
′
′
2π ro qs dz + mc pTm = mc p Tm +
dz
dz
q′s′
Fig. 11.13
104
Simplify
dTm 2π roq′s′
=
dz
mc p
(l)
m = ρπ ro v z m
(m)
dTm
2q′s′
=
dz
ρ c p rov z m
(11.88)
However
2
(m) into (l)
(11.88) into (11.87)
dT (ro , z) dTm ( z ) ∂T
2q′s′
=
=
=
dz
dz
∂z ρ c provzm
(11.89)
105
(11.89) into (11.81)
2q′s′ v z
∂ ∂T
(r ) =
r
∂r ∂r
kro v z m
(11.90)
(11.77) is used to eliminate v z / z z m in the above
q′s′
∂ ∂T
4
(r
)=
∂ r ∂r
1 + 8 Kn kro
r2
1 + 4 Kn − 2 r
ro
(11.91)
Integrate
q′s′
T (r , z ) =
(1 + 8 Kn)k ro
4
2 1r
(1 + 4 Kn) r −
+ f ( z) y + g( z)
2
4 ro
(n)
106
Condition (g) gives
f (z ) = 0
Solution (n) becomes
q′s′
T (r , z ) =
(1 + 8 Kn)k ro
4
2 1r
(1 + 4 Kn) r −
+ g(z)
2
4 ro
(11.92)
Condition (h) is automatically satisfied
Determine g(z): Use two methods to determine Tm
Method 1: Integrate (11.88)
107
Tm
∫T
mi
2q′s′
dTm =
ρ c p v z m ro
z
∫0dz
where
Tm (0) = Tmi
(11.93)
Evaluate the integral
2q′s′
Tm =
z + Tmi
ρ c pv z m ro
(11.94)
Method 2: Use definition of Tm in (11.80). Substitute (11.74)
and (11.92) into (11.80)
108
1
4q′s′
r 2
r4
2
1 + 4 Kn − 2
( + Kn) r −
+ g ( z ) r dr
2
ro (1 + 8 Kn)kro 4
16 ro
ro
2
r rdr
1 + 4 Kn − 2
ro
0
ro
∫0
Tm =
∫
Integrate
Tm =
q′s′ ro
7
2 14
16
Kn
Kn
+
+ g( z)
+
2
3
24
k (1 + 8 Kn)
(11.95)
Equate (11.94) and (11.95), solve for g(z)
2q′s′
q′s′ro
7
2 14
g(z) = Tmi +
z−
16Kn + Kn+
2
3
24
ρc provzm k(1 + 8Kn)
(11.96)
109
Use (f) and (11.92) to determineTs (ro , z )
4q′s′ro
3
4γ q′s′ro
Ts (ro , z ) =
Kn + +
Kn + g ( z )
k (1 + 8 Kn)
16 γ + 1 kPr
(11.97)
Nusselt number: (11.95) and (11.97) into (e)
Nu =
2
4
3
1
14
7
4γ 1
( Kn +
)−
16
Kn
Kn
+
+
Kn
+
2
(1 + 8 Kn)
16 (1 + 8 Kn)
3
24 γ + 1 Pr
2
(11.98)
110
Results: Fig. 11.14
• Fig. 11.14 gives Nu vs. Kn for air
• Rarefaction and compressibility 4.0
decrease the Nusselt number
• Nusselt number depends on Nu 3.0
the fluid
• Nu varies with distance
2.0
0
along channel
Fig. 11.14
• No-slip Nusselt number, Nuo,
is obtained by setting
Kn = 0 in (11.97)
48
Nuo =
= 4.364
11
0.04
0.08
0.12
Kn
Nusselt number for air flow through
tubes at unifrorm surface heat flux
(11.99)
111
This agrees with (6.55) for macro tubes
11.9.6 Fully Developed Poiseuille Flow in
Micro Tubes: Uniform Surface Temperature
• Repeat Section 11.6.5
with the tube at surface
temperature Ts
• Apply same assumptions
r
r
ro
z
Fig. 11.15
Ts
• Boundary conditions are
different
•Flow field solution is identical for the two cases
112
Nusselt number:
2ro h
− 2ro ∂T ( ro , z )
Nu =
=
k
Tm ( z ) − Ts
∂r
• Determine T(r,z) and Tm(z)
• Follow the analysis of Section 11.6.4
• Solution is based on the limiting case of Graetz tube
entrance problem
• Axial conduction is taken into consideration
• Energy equation (11.81) is modified to include axial
conduction:
113
∂T k ∂ ∂T
∂ 2T
(r
)+k
=
ρ c pv z
∂ z r ∂r ∂r
∂z 2
(11.101a)
Boundary and inlet conditions
∂T ( r ,0)
=0
∂z
(11.102a)
∂T ( ro , z )
2γ 2ro
Kn
∂r
γ + 1 Pr
(11.103a)
T ( ro , z ) = Ts −
T ( r ,0) = Ti
(11.104a)
T (r , ∞ ) = Ts
(11.105a)
114
(11.76) gives axial velocity
vz
1 + 4 Kn − ( r / ro ) 2
=2
v zm
1 + 8 Kn
(11.76)
• Solution by the method of separation of variables
• Solution is specialized for fully developed conditions at
large z
• Result for air shown in Fig. 11.16
• Neglecting axial conduction: set Pe = ∞
• Axial conduction increases the Nu
115
• Limiting case: no slip and no axial conduction: at Kn = 0
and Pe = ∞
4.5
Nuo = 3.657
4.0
(11.72)
∞
3.5
Nu
This agrees with (6.59)
• Limiting case: no slip with
axial conduction: at Kn =0
and Pe = 0:
3.0
2.5
2.0
0
0.04
0.08
0.12
Kn
Fig. 11.16 Nusselt number for flow through tubes
Nuo = 4.175
at uniform surface temperature for air
116
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