Applied Transport Theory: Similarity Solution Derivation Task

This assignment solution focuses on deriving a similarity solution for the spreading of a thin film along a horizontal surface using applied transport theory. The problem involves dimensionless variables and neglects surface tension effects, governed by a given partial differential equation. The solution includes determining the relationship between constants α and β, using a conservation of mass argument to further define these constants, and subsequently writing down and solving the ordinary differential equation for f(ξ) subject to specified boundary conditions. The derivation utilizes techniques such as converting the PDE to an ODE, applying conservation principles, and employing chain rule transformations to solve the differential equation and determine the final solution for h(x,t).

2018
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TITLE: APPLIED TRANSPORT THEORY
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If we neglect the effects of surface tension, the governing equation for the spreading of a thin
film along a horizontal surface is (in dimensionless variables)
∂ h
∂ t = ∂
∂ x ( 1
3 h3 ∂ h
∂ x )
Part 1
Derive a similarity solution of the form
h=tα f ( ξ ) where ξ= x
tβ
(a) Determine the relationship between the constants α ∧β
(b) Using a conservation of mass argument, determine another relationship between the
constants α∧β, and thus solve for α ∧β
(c) Use the values of α∧β, write down the ordinary differential equations for f ( ξ )
(d) Solve the differential equation from part c subject to the boundary conditions f=0,
df
dξ =0 , ξ → ∞
SOLUTION
Part a
The similarity solution exists for a given spreading of a thin film to convert the PDE
equation to an ODE for one to obtain the roots,
∂ h
∂ t = ∂
∂ x ( 1
3 h3 ∂ h
∂ x )
Let k = 1
3 , therefore,
∂ h
∂ t = ∂
∂ x ( k h3 ∂ h
∂ x )
Simplifying further,
k
4
∂
∂ x (4 h ∂ h4
∂ x )= 1
12 . ∂2
∂ x2 ( ∂h4
∂ x )
12 ∂ h
∂t = ∂2
∂ x2 ( h4 )
Using the values of h we find,
h=tα f ( ξ ) where ξ= x
tβ
1

12 ∂
∂ t (tα f ( ξ ) )= ∂2
∂ x2 ( tα f ( ξ ) )4
But , ξ= x
t β
12 ∂
∂ t ( tα f ( x
tβ )) = ∂2
∂ x2 ( tα f ( x
tβ ) )
4
Part b
Using the conservation of mass argument,
12 ∂ h
∂t = ∂2
∂ x2 ( h4 )
12 ht =hxx
4
12 ∂ h
∂t =h3 ∂2 h
∂ x2
The operator X =ξ ( x , y ) ∂ x +η( x , y)∂ y generates a point symmetry to determine the
equation,
X |2| ( y!11− y−k ) ¿ y' '− y−k=0
Where,
ζ j ( x , y , y' , … , y j )= d ζ j−1
dx − y j dξ
dx j=1,2,3 …
ζ 2 ¿ y' ' = y−k
1 + kη y−k−1=0
η=3 a1 y
ξ=ao +a1 ( k +1 ) x
The solution yields,
y ( x ) =h ( x , t ) =a2 ( a1 ( k +1 ) x
3
k+ 1 ) k ≠−1
Where a2 is a constant of integration
2

a2= ( 2 a2
2 ( k −1 ) 2 k )
− ( 1
k +1 )
y ( λ)=α
a1=α
1
3 ( k+1 )
( 3 ( k −1 ) (2 k ) )
1
3 − ( k +1 ) λ
To obtain β∧γ in terms of α,
β= 3
2
3 α
1−2 k
3 ( k−2 )
1
3
( 2 k−1 )
2
3
k ≠ 1
2 , 2 ,…
Part c
Finding the ODE function
f ( ξ )
12 ht =h3 hxx
general solution , H X =h ts , H T =h xr
Let our ξ=g , t= p
h ( x ,t )=t−r g ( x ts ) =t−r g ( p )
2 kr h' ' + ( 4 k+ p2 ) h'=0
So that,
h' ( p )=c2 e∫ 4 k + p2
2kp dp
¿ c2 e∫ ( 2
p + p
2 k ) dp
¿ c2 e−(2 ln p+ p2
4 k )
¿ c2
P2 e
− p2
4 k
Hence,
3

h ( ∞ ) −h ( p ) =∫
p
∞ C2
s2 e
−s2
4 k ds
therefore,
h ( p )=M −c2 ( 1
p e−( p2
4 k )− 1
2 k ∫
p
∞
e
− s2
4 k ds )
g ( p )= ( M +C1 ) p−C2 (e
−p 2
4 k −( p
2 k )∫
P
∞
e
−s2
4 k ds )
Part d
Solving the differential equation.
The boundary conditions are f=0, df
dξ =0 , ξ → ∞
Using chain rule,
H X=h ts , H T =h xr
H X=aγ ut tT =aγ −β ut
When γ−β =γ−2 α … PDE under stretching transformation
β=2 α , γ is arbitrary
As a result, aγ u ( aα x ,aβ t )
g ( p )= ( M +C1 ) p−C2 (e
−p 2
4 k −( p
2 k )∫
P
∞
e
−s2
4 k ds )
M +C1 =0 , as f =0
g ( p )=−C2 (e
− p2
4 k − ( p
2k )∫
P
∞
e
−s2
4 k ds )
In addition,
4

N=g' ( 0 ) = C2
2 k (∫
P
∞
e
−s2
4 k ds ) , df
dξ =0
df
dξ =0 , N = c2
√ k ∫
0
∞
e−z 2
dz
N= c2
√k . √ π
2
N= C2
2 √ π
k
h ( x ,t )=−tα
(2 N √ k
π ) [e−( x2
4 kt )− ( x
2k √t )∫
x
√t
∞
e−( x2
4 kt ) ds
]
5