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
y
• Momentum is a vector
quantity
dy
• Newton’s law of motion:
3 components
• Apply Newton’s law to
element, Fig. 2.5
dz
dx
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
To g
x
qo′′
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
liquid T f
0
To
• Other side is insulated
• Conservation of energy at
the melting front:
solid
dx i
∂T
−k
=ρL
∂x
dt
xi = melting front location
(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