Fundamentals of Transport Phenomena ChE 715

1. Fundamentals of Transport Phenomena
ChE 715
Lecture 16
Start Ch 3
•Non-dimensionalizing
S li•Scaling
•Order of Magnitude
Spring 2011

2. Scaling & order-of-magnitude analysis
What is the purpose of scaling?
How do you scale terms?
Why non dimensionalize terms?
S l t t di i li
Why non-dimensionalize terms?
How to non-dimensionalize terms?
Scale terms to non-dimensionalize
Non-dimensional terms can provide estimate /effect
of terms, variables, physical phenomena
– can eliminate or retain terms base on this
Scaling/non-dimensionalizing can reduce order of
problem, e..g, 2-D problem to 1-D problem
“Properly scaled” non-dimensional terms ~1

3. Scaling and Order-of-Magnitude Analysis
Order of magnitude equality (for this course): “~”
If x ~ y then, roughly speaking,
0 3 30.3y < x < 3y
Scaling:
b di i l i bl l hLet x be a dimensional variable (e.g., length);
Δx is the approximate range of x values.
Let L scale for x (or Δx);Let L = scale for x (or Δx);
Define as a scaled, dimensionless variable.˜x =
x
L
What are the length and temperature/concentration
scales in the previous example problems?scales in the previous example problems?

4. Scaling and Order-of-Magnitude Analysis
Estimates of Derivatives
First derivative:
∂C ΔC C0
First derivative:
Dimensionless:
∂x
~
Δx
~ 0
L
∂Ψ
∂˜
~
ΔΨ
Δ˜
~1
∂x Δx
∂2
C ΔC Δx C0Second derivative:
Dimensionless:
∂ C
∂x2
~
ΔC Δx
Δx
~
C0
L2
∂2
Ψ
~1 If the scales are correct.
∂˜x2
1
Correct scaling should reduce each dimensionless term to order 1g

5. Getting estimate of terms
Example: Heated solid immersed in fluid
How to scale in this case?
2L
L = minimum distance from the centerline to the surface
TC = centerline temperature
TS = surface temperature
T fl id (ambient) temperat reT∞ = fluid (ambient) temperature

6. Getting estimate of terms
Example: Heated solid inmmersed in a fluid
Surface boundary condition: −k n⋅∇T( )S
= h TS −T∞( )y ( )S
( )
Scaling: − n⋅∇T( )S
~
TC −TS
LL
Interpretation of Biot number: Bih ≡
hL
k
~
TC −TS
TS −T∞

7. Getting estimate of terms
Example: Heated solid inmmersed in a fluid
solid resistance
~
fluid resistance
C ST ThL
Bi
k T T
−
≡ =
fluid resistanceSk T T∞−
For Bi>>1, we have Ts = T∞
This is something we mentioned earlier, but now you can see how the
temperatures are equal!temperatures are equal!

8. Is the scaling correct?
CA = CA,0
Classic reaction-diffusion problem:
Homogeneous reaction rate, –rA = k1CAL
Impermeable solid
BCs?
x=L 0A
A
x L
dC
D
dx
− =
Impermeable solid
x=0 CA = CA,0
2
1
2
A
A
A
d C k
C
dx D
=
x Ldx =
, ; A
AO
Cx
Let
L C
η ψ= =
The problem is dimensionless.
Is it scaled properly?
22
2 2 1
2
;
k L
D
∂
ϕ ϕ
∂η
Ψ
= Ψ =
Ψ(0) =1; ′Ψ (1) = 0B C ’s: Ψ(0) =1; Ψ (1) = 0B.C. s:
No! the differential eqn has
terms greater than ~1

9. Is the scaling correct?
CA = CA,0
Classic reaction-diffusion problem:
Homogeneous reaction rate, –rA = k1CAL
Impermeable solidImpermeable solid
22
k L∂ Ψ
, ; A
AO
Cx
Let
L C
η ψ= =
, Z= m
Let ϕ η
2 2 22
2 2 1
2
;
k L
D
∂
ϕ ϕ
∂η
Ψ
= Ψ =
Ψ(0) =1; ′Ψ (1) = 0B.C.’s:2
∂ Ψ
2 2
2
2 2
m
Z
∂ ∂
ϕ
∂η ∂
∴ =
So, what is the proper length scale?
2 2
2
m
Z
∂
ϕ
∂
−Ψ
= Ψ
Set m=1:
2
2
Z
∂
∂
Ψ
= Ψ
Z=
=
m
x
ϕ η
Z∂
1
=
/D k
Z is a dynamic length scale!

10. Revisit Heated Wire– can we determine scaling factor?
Constant temperature bath, T∞; heat transfer coefficient, h
Constant heat generation rate, HVg V
What is the scaling factor for temperature?
Bi t b f h t t f
hR
BiBiot number for heat transfer: Bi
k
=

11. Heated Wire Problem
( )
T h
T T
r k
∞
∂
= − −
∂
r=R0V
k d dT
r
r dr dr
H⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
Let r=0 0
T
r
∂
=
∂
; ; Bi=
T
T
T r hR
R k
η∞
Θ =
Δ
−
≡ −
1k T d dθ⎛ ⎞Δ
2
2
1
4 2
V Vr
T T R R
R
H H
k h∞
⎡ ⎤⎛ ⎞
− − +⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
=
2
1
0V
k T d d
R d d
H
θ
η
η η η
⎛ ⎞Δ
+ =⎜ ⎟
⎝ ⎠
2
1
0V Rd d Hθ
η
⎛ ⎞
+⎟ =⎜
Exact soln from before
4 2Rk h⎝ ⎠⎢ ⎥⎣ ⎦
0
d d Tk
η
η η η
+⎟
⎠ Δ⎜
⎝
2
V
T
RH
∴Δ =
2
2
1 1
1
4V
T T r
R RH Bi
∞
⎡ ⎤− ⎛ ⎞
− +⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
=
k k
⎣ ⎦
Defined first time we did problem:
T T
c
T T
T T
∞
∞
−
Θ ≡ −
−
So, not the correct scaling?

12. Heated Wire Problem
2
2
1 1
1
4V
T T r
R RH Bi
k
∞
⎡ ⎤− ⎛ ⎞
− +⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
=
2
1
1
T T r∞
⎡ ⎤− ⎛ ⎞
⎢ ⎥⎜ ⎟
0V
k d dT
r
r dr dr
H⎛ ⎞
+ =⎜ ⎟
⎝ ⎠
L t l d d t d i ti
k
Let Bi >>1
2
1
4V R RH
k
∞ ⎛ ⎞
−⎢ ⎥⎜ ⎟
⎝ ⎠⎢ ⎥⎣ ⎦
=
Let us compare scaled and exact derivatives
2
V
V
R
RdT
dr R
H
Hk
k
∼ ∼ Same order of magnitude
dr R k
(exact, for maxm value, i.e, r=R)
2
V RdT
dr
H
k
= −
We can do the same for the second derivative

13. Long Fin Problem
Constant temperature (T0)
2W
L
2W
x
Heat transfer coefficient h, ambient temperature T∞.
L
z
“Thin Fin” approximation: assume γ = L/W >> 1.
Develop the dimensionless transport problem in terms of γ and .W
hW
Bi ≡p p p γ W
k
How does this approach simplify the problem when BiW << 1?
What does scaling tell us about the temperature profile in the x and z-direction?

14. Long Fin Problem
Energy Eq. for the fin:
2 2
2 2
0
d T d T
dx dz
+ =
BCs: [ ]( , ) ( , )
T h
x L T x L T
k
∞
∂
= − −
∂
T(x,0) = T0
[ ]( ) ( )
z k
∞
∂
(0 ) 0
T
z
∂
[ ]( ) ( )
T h
W z T W z T
∂
( , ) 0
(0, ) 0z
x
=
∂
[ ]( , ) ( , )W z T W z T
x k
∞= − −
∂
Can we reduce the dimensionality of the problem ?Can we reduce the dimensionality of the problem ?

15. Long Fin Problem
Let us look at BC at the top surface:
[ ]( , ) ( , )
T h
W z T W z T
x k
∞
∂
= − −
∂
Estimating the order of magnitude and rearranging:
[ ]
[ ]
(0, ) ( , )
~
( , )
T z T W z hW
Bi
T W z T k∞
−
=
−
For Bi << 1, temp variation in x direction negligible! So T=T(z) good approx.
Replace local value with cross-sectional average (at const. z):p g
0
1
( ) ( , )
W
T z T x z dx
W
≡ ∫

16. Long Fin Problem
Remember energy eq.:
2 2
2 2
0
d T d T
dx dz
+ =
Averaging each term over the cross-sectional average in the energy eq.:
2
2
00
1 1
( )
x WW
x
T T h
dx T T
W x W x Wk
=
∞
=
∂ ∂
= = − −
∂ ∂∫
2 2 2
2 2 2
0 0
1 1
W W
T T
dx Tdx
W z z W z
⎛ ⎞∂ ∂ ∂
= =⎜ ⎟
∂ ∂ ∂⎝ ⎠
∫ ∫
Energy eq. is now:
2
2
( ) 0
T h
T T
Wk
∞
∂
− − =
∂ 2
( )
z Wk
∞
∂

17. Long Fin Problem
Averaging BCs involving z:
0(0)T T=
( ) ( )
dT h
L T L T⎡ ⎤= − −⎣ ⎦
Define dimensionless temperature and z:
( ) ( )L T L T
dz k
∞⎡ ⎤⎣ ⎦
0
( )
T T
T T
ζ ∞
∞
−
Θ =
−
z
W
ζ =
W more representative
than length for thin
objects.
Temperature
is scaled.
Dimensionless Energy eq:
2
Bi
∂ Θ
Θ
Dimensionless BCs
(0) 1Θ = ( ) ( )Biγ γ
∂Θ
= Θ L
2
Bi
ζ
= Θ
∂
(0) 1Θ = ( ) ( )Biγ γ
ζ
= − Θ
∂
Are eqns. properly scaled?
L
W
γ =

18. Long Fin Problem
Scaling z coord:
m is
undetermined
constant
m
Z Bi ζ=
Dimensionless Energy Eq. becomes:
constant
2
∂ Θ
Choose m = ½ then:
2
2
0m
Bi Bi
Z
∂ Θ
− Θ =
∂
Choose m = ½, then:
Scale for z is1/ 2
1/2 h
Z Bi z
Wk
ζ
⎛ ⎞
= = ⎜ ⎟
⎝ ⎠
1/2
h
Wk
⎛ ⎞
⎜ ⎟
⎝ ⎠
Wk⎝ ⎠ Competition between heat
conduction and heat loss

19. Long Fin Problem
BC at top with scaled coord
is then:
∂Θ 1/2
( ) ( )Bi
Z
∂Θ
Λ = − Θ Λ
∂
1/2
h⎛ ⎞h
L
Wk
⎛ ⎞
Λ ≡ ⎜ ⎟
⎝ ⎠
where
If Bi 0 : And BCs are:If Bi 0 :
2
2
0
d
dZ
Θ
− Θ = ( ) 0
d
dZ
Θ
Λ =(0) 1Θ =
Solving:
( ) cosh tanh sinhZ Z ZΘ = Λ( ) cosh tanh sinhZ Z ZΘ = − Λ

20. Long Fin Problem
( ) cosh tanh sinhZ Z ZΘ = − Λ
( ) 1ZΘ →
Limiting cases:
“Short” fin (Λ << 1 or ) :
1/2
Wk
L
⎛ ⎞
<< ⎜ ⎟ ( 0)Λ →( ) 1ZΘ →Short fin (Λ << 1 or ) :L
h
<< ⎜ ⎟
⎝ ⎠
( 0)Λ →
Short fin essentially isothermal
“Long” fin (Λ >> 1) : ( ) Z
Z e−
Θ → ( )Λ → ∞
Long fin reaches ambient temperatureLong fin reaches ambient temperature
well before the tip