FJMS

Far East Journal of Mathematical Sciences (FJMS)
© 2016 Pushpa Publishing House, Allahabad, India
Published Online: March 2016
http://dx.doi.org/10.17654/MS099070969
Volume 99, Number 7, 2016, Pages 969-981 ISSN: 0972-0871
Received: May 13, 2015; Revised: October 1, 2015; Accepted: November 4, 2015
2010 Mathematics Subject Classification: 76A02.
Keywords and phrases: heat transfer, finite element method, Stefan condition, stream function.
Communicated by K. K. Azad
THREE-DIMENSIONAL MATHEMATICAL MODELING
OF HEAT TRANSFER BY STREAM FUNCTION
AND ITS NUMERICAL SOLUTION
Mohammad Hassan Mohammadi
Institute of Mathematics
National Academy of Sciences of Republic of Armenia
Armenia
e-mail: mohamadi.mh.edu@gmail.com
Abstract
This paper states the mathematical modeling of heat transfer in the
Garnissage furnace in two dimensional case, and the modeling is based
on the stream function. We invoke the three conservation laws of
physics, i.e. the mass, the momentum, and the energy conservation
laws to derive the continuity equation, the Navier-Stokes equations,
and the energy equation. The Stefan condition will be applied to
demonstrate the free boundary between solid and liquid phase. First
we express the system by stream functions, and then we convert it to
the variational formulation (weak formulation), and after that we
perform the finite element method to achieve the numerical solution of
the system.
1. Introduction
Mathematical modeling of heat transfer is vital tool to analyze the
behavior of fluids in the furnaces, and it helps researchers to describe the

Mohammad Hassan Mohammadi970
conditions in the environment of furnaces, and even it is applied by designers
to optimize the furnaces plan in the construction process [1-3]. On the other
hand this approach has low cost and more exactness, when the other physical
ways are much more expensive, then the mathematical modeling is famous
and popular between the specialists [4, 5].
In the Garnissage furnace the process of melting starts by imposing the
heat to get the batch blanket, and electrical boosters increase the temperature
of the materials and lead to changing the phase from the solid to the liquid.
The boundary between the solid and the liquid is moving during the phase
changing, and we put on the Stefan condition to stating the geometry of the
free boundary [6-8]. Also we will exploit the mass, momentum, and energy
conservation laws for the moving fluid in the tank to get continuity, Navier-
Stokes, and energy conservation equations.
We express the system in the form of stream functions, and this
representation is simple to handle, because it makes the possibility to state
the Navier-Stokes equations just by one equation. The transport system with
four nonlinear differential equations would be converted to the variational
version by using convenient test functions that are continuous enough and
have small supports. Then we discrete the domain and replace all the
variables by the approximate values, and finally we enforce the finite
element method to transfer the nonlinear system of differential equations to
gain the nonlinear system of equations. To derive the numerical solution of
the system every classical method can be performed, and we recommend the
Newton’s method to get the linear system of equations.
2. The Continuity Equation
We start the modeling process by some elementary definitions and
lemmas.
Definition 2.1. Suppose that ( )0,, vu=V is the velocity vector field of
two-dimensional flows such that
,ψ×∇=V

Three-dimensional Mathematical Modeling of Heat Transfer … 971
where ( ),,0,0 ψ=ψ then we call ψ the stream function, we compute the
velocity components as
,
y
u
∂
ψ∂
=
.
x
v
∂
ψ∂
−=
Lemma 2.1. Assume that ( ),1
0 Ω∈ψ H then the stream function ψ
satisfies in the continuity equation.
Proof. Replace the stream function ψ into the continuity equation, then
we will get as
,0
22
=
∂∂
ψ∂
−
∂∂
ψ∂
=⎟
⎠
⎞
⎜
⎝
⎛
∂
ψ∂
∂
∂
−⎟
⎠
⎞
⎜
⎝
⎛
∂
ψ∂
∂
∂
=
∂
∂
+
∂
∂
=⋅∇
yxyxxyyxy
v
x
u
V
that is the stream function ψ automatically satisfies in the continuity
equation. By this property we will not handle the continuity equation in our
transport system of equations.
3. The Navier-Stokes Equations
In the second stage of modeling [5, 6] we set the stream function ψ into
the momentum conservation law (the Navier-Stokes equations), before
replacing we will convert the system of Navier-Stokes equations to the single
equation, then for the Newtonian fluid the Navier-Stokes equations as
( ) ( ) ,gp
t
ρ+∇μ⋅∇+−∇=⎟
⎠
⎞
⎜
⎝
⎛ ∇⋅+
∂
∂
ρ VVV
V
(3.1)
where μ is the dynamic viscosity. For the steady flow the dynamic viscosity
is constant, then
( ) .gp
t
ρ+Δμ+−∇=⎟
⎠
⎞
⎜
⎝
⎛ ∇⋅+
∂
∂
ρ VVV
V
(3.2)
Let the two-dimensional flow in plane, and then we will have ( ),0,, vu=V

as the velocity vector field. We represent the Navier-Stokes equations as
,
1
xgu
x
p
y
u
v
x
u
u +Δϑ+
∂
∂
ρ
−=
∂
∂
+
∂
∂
(3.3)
,
1
ygv
y
p
y
v
v
x
v
u +Δϑ+
∂
∂
ρ
−=
∂
∂
+
∂
∂
(3.4)
where
ρ
μ
=ϑ is the kinematic viscosity.
Now we differentiate equation (3.3) respect to y and equation (3.4)
respect to x and after subtracting we attain
y
v
x
v
x
v
u
x
v
x
u
y
u
v
y
u
y
v
yx
u
u
x
u
y
u
∂
∂
∂
∂
−
∂
∂
−
∂
∂
∂
∂
−
∂
∂
+
∂
∂
∂
∂
+
∂∂
∂
+
∂
∂
∂
∂
2
2
2
22
( ) ( ).
2
v
x
u
yyx
v
v Δ
∂
∂
ϑ−Δ
∂
∂
ϑ=
∂∂
∂
− (3.5)
Now we set the stream function ψ into the equality (3.5) and we establish
( ) ( ) .2
ψΔϑ=ψΔ
∂
∂
∂
ψ∂
−ψΔ
∂
∂
∂
ψ∂
yxxy
(3.6)
4. The Heat Equation
Assume that the flow is steady and incompressible, then the energy
conservation equation is as
,
1
S
cc
k
y
v
x
u
ρ
+θΔ
ρ
=
∂
θ∂
+
∂
θ∂
(4.1)
now we set the stream function ψ into equation (4.1) and we gain
.
1
S
cc
k
yxxy ρ
+θΔ
ρ
=
∂
θ∂
∂
ψ∂
−
∂
θ∂
∂
ψ∂
(4.2)
5. Variational Formulation of Transport Equations
We will handle the finite element method to solve numerically the system
of transport equations. For this aim we indicate the variational formulation of

transport equations in the stream function form [9, 10]. Assume that ∈η
( ),1
0 ΩH now multiply equation (3.6) by η, and integrate, then
( ) ( )∫∫ ∫∫Ω Ω
⎟
⎠
⎞
⎜
⎝
⎛ η
∂
ψ∂
ψΔ
∂
∂
−⎟
⎠
⎞
⎜
⎝
⎛ η
∂
ψ∂
ψΔ
∂
∂
dxdy
xy
dxdy
yx
( )∫∫Ω
ψηΔϑ= .2
dxdy (5.1)
We rewrite the integral equality (4.1) by using the integration by parts and
we have
∫∫Ω
=⎟
⎠
⎞
⎜
⎝
⎛ ηΔϑ+
∂
ψ∂
∂
η∂
−
∂
ψ∂
∂
η∂
ψΔ .0dxdy
xyyx
(5.2)
We continue the process by the same technic for energy equation (4.2), and
after integration we get
∫∫ ∫∫Ω Ω
⎟
⎠
⎞
⎜
⎝
⎛ η
∂
θ∂
∂
ψ∂
−⎟
⎠
⎞
⎜
⎝
⎛ η
∂
θ∂
∂
ψ∂
dxdy
yx
dxdy
xy
∫∫ ∫∫Ω Ω
⎟
⎠
⎞
⎜
⎝
⎛ η
ρ
+⎟
⎠
⎞
⎜
⎝
⎛ θηΔ
ρ
= ,dxdy
c
S
dxdy
c
k
(5.3)
now we perform the integration by parts (Green’s first identity) for equation
(5.3) and get
∫∫ ∫∫Ω Ω ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
η∂
∂
θ∂
+η
∂∂
θ∂
ψ+⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
η∂
∂
θ∂
η
∂∂
θ∂
ψ− dxdy
xyyx
dxdy
yxyx
22
( )∫∫ ∫ ∫∫Ω Ω∂ Ω
η
ρ
+η⋅θ∇
ρ
+η∇⋅θ∇
ρ
−= ,dxdy
c
S
c
k
dxdy
c
k
xn (5.4)
in this part [14] we put the Stefan condition [ ] txx nλ=⋅λ−=⋅θ∇ +
− nwn
into the integral equation (5.4) and we have
∫∫ ∫∫Ω Ω
η∇⋅θ∇
ρ
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
η∂
∂
θ∂
−
∂
η∂
∂
θ∂
ψ− dxdy
c
k
dxdy
xyyx
∫ ∫∫Ω∂ Ω
η
ρ
+ηλ
ρ
+ ,dxdy
c
S
n
c
k
t (5.5)

then
∫∫ ∫∫Ω Ω
η∇⋅θ∇
ρ
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
η∂
∂
θ∂
−
∂
η∂
∂
θ∂
ψ− dxdy
c
k
dxdy
xyyx
∫∫ ∫∫Ω Ω
η
ρ
+
∂
η∂
ρ
λ
+ .dxdy
c
S
tc
k
(5.6)
6. Discretization of the Domain
To follow the finite element method [15-18] we construct the finite-
dimensional subspace ( )Ω⊂ 1
0HVh which consists of test functions of the
form
( )
⎪
⎩
⎪
⎨
⎧
<<⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
−=φ
,otherwise;0
,1,1;
1
1
exp
1
1
exp
, 22
yx
yxyx (6.1)
with the shape
Now we redefine the problem as finding hhh V∈θψ , such that hh θψ ,
satisfy in the integral equations

,0∫∫Ω
=⎟
⎠
⎞
⎜
⎝
⎛ ηΔϑ+
∂
ψ∂
∂
η∂
−
∂
ψ∂
∂
η∂
ψΔ dxdy
xyyx h
hhhh
h (6.2)
( )∫∫ ∫∫Ω Ω
η∇⋅θ∇
ρ
−=⎟
⎠
⎞
⎜
⎝
⎛
∂
η∂
∂
θ∂
−
∂
η∂
∂
θ∂
ψ− dxdy
c
k
dxdy
xyyx hh
hhhh
h
∫∫ ∫∫Ω Ω
η
ρ
+
∂
η∂
ρ
λ
+ ,dxdy
c
S
tc
k
h
h (6.3)
for every .hh V∈η Suppose that
( ),dim hNVh =
and
{ ( )},...,,,, 321 hNh spanV φφφφ=
where the basis functions ( ),, yxiφ ( ),...,,2,1 hNi = have small support.
Now we put on the approximate solutions ,hψ hθ in terms of the basic
functions ( ),, yxiφ thus we can write
( ) ( )
( )
∑=
φ=ψ
hN
i
iih yxUyx
1
,,, (6.4)
( ) ( )
( )
∑=
φ=θ
hN
i
iih yxVyx
1
,,, (6.5)
where ,, ii VU ( ),...,,3,2,1 hNi = are to be determined. We replace hψ
from (6.4) into the integral equation (6.2) and we get
( )( )( )
∫∫ ∑ ∑ ∑Ω
= = =
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
φΔϑ+
∂
φ∂
∂
φ∂
−
∂
φ∂
∂
φ∂
φΔ
hN
i
hN
j
hN
j
k
j
j
kj
j
k
ii dxdy
x
U
yy
U
x
U
1 1 1
,0
( ),...,,3,2,1 hNk = (6.6)
and the equality (6.6) can be restated as

( )( )
∑ ∑ ∫∫= =
Ω ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
φ∂
∂
φ∂
−
∂
φ∂
∂
φ∂
φΔ
hN
i
hN
j
kjkj
iji dxdy
yxxy
UU
1 1
( )
∑ ∫∫=
Ω
=φΔφΔϑ+
hN
i
kii dxdyU
1
,0 (6.7)
and finally we will derive the nonlinear system for ( )hNkji ...,,3,2,1,, =
as
( )( )( )
∑ ∑ ∑= = =
=+
hN
i
hN
j
hN
i
ikiijkji bUaUU
1 1 1
,0 (6.8)
where
∫∫Ω ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
∂
φ∂
∂
φ∂
−
∂
φ∂
∂
φ∂
φΔ= ,dxdy
yxxy
a kjkj
iijk
∫∫Ω
φΔφΔϑ= .dxdyb kiik
Now we will complete the discretization process and we replace hθ from
(6.5) into the energy conservation equation (6.3) and we get
( ) ( )
∫∫ ∑∑Ω
==
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
φ∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
φ∂
−
∂
φ∂
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
φ∂
ψ− dxdy
xy
V
yx
V
j
hN
i
i
i
j
hN
i
i
ih
11
( )
∫∫ ∫∫ ∫∫∑Ω Ω Ω
=
φ
ρ
+
∂
ξ∂
ρ
λ
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
φ∇⋅φ∇
ρ
−= ,
1
dxdy
c
S
tc
k
dxdyV
c
k
j
j
hN
i
jii
(6.9)
where ( )tyx ,,ξ=ξ is the convenient test function with small support and it
is defined as
( )tyx ,,ξ
⎪
⎩
⎪
⎨
⎧
<<<⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
−
⎟
⎠
⎞
⎜
⎝
⎛
−=
.otherwise;0
,1,1,1;
1
1
exp
1
1
exp
1
1
exp
222
tyx
yxt

The integral equality (6.9) is written as
( )
∑ ∫∫=
Ω ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
φ∇⋅φ∇
ρ
+
∂
φ∂
∂
φ∂
ψ+
∂
φ∂
∂
φ∂
ψ−
hN
i
ji
ji
h
ji
hi dxdy
c
k
xyyx
V
1
∫∫ ∫∫Ω Ω
φ
ρ
+
∂
ξ∂
ρ
λ
= ,dxdy
c
S
tc
k
j
j
(6.10)
then we will convert the equality (6.10) to the linear system
( )
( )
∑=
==
hN
i
jiij hNjdVc
1
,...,,3,2,1, (6.11)
where
∫∫Ω ⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
φ∇⋅φ∇
ρ
+
∂
φ∂
∂
φ∂
ψ+
∂
φ∂
∂
φ∂
ψ−= ,dxdy
c
k
xyyx
c ji
ji
h
ji
hij
∫∫ ∫∫Ω Ω
φ
ρ
+
∂
ξ∂
ρ
λ
= .dxdy
c
S
tc
k
d j
j
j
Thus by applying the systems of equations (6.8) and (6.11) we restate the
transport system problem as:
Finding ( ( )) ( )hN
hN RUUU ∈...,,, 21
and ( ( )) ( )hN
hN RVVV ∈...,,, 21 such that
( )
( )( )( )
∑ ∑ ∑= = =
==+
hN
i
hN
j
hN
i
ikiijkji hNkbUaUU
1 1 1
,...,,3,2,1,0 (6.12)
( )
( )
∑=
==
hN
i
jiij hNjdVc
1
....,,3,2,1,
As we see in the system (6.12) in the first part, we have the nonlinear
system of ( )hN unknowns ( ),...,,, 21 hNUUU and ( )hN equations, then we

will perform the Newton’s method to find the unknowns ( )....,,, 21 hNUUU
In the second part, we have the linear system of ( )hN unknowns ...,,, 21 VV
( ),hNV and ( )hN equations, then we can execute every classical numerical
methods to solve this linear system.
7. The Newton’s Method
In the final stage of the process of numerical method we show that how
the Newton method must be used to solve the nonlinear system (6.8), to start
the approach define the functions
( ( ))
( )( )( )
∑ ∑ ∑= = =
+=
hN
i
hN
j
hN
i
ikiijkjihNk bUaUUUUUF
1 1 1
21 ,...,,,
( ),...,,3,2,1 hNk = (7.1)
then the iterative process is
( ) ( ) ( ) ( ( ) ) ,...,,,,
000 210
11 T
hN
nnnn
UUU=−= −+
UUFUDFUU
where
( ( )) ,...,,, 21
T
hNUUU=U
and the Jacobian matrix DF is defined
( )
( )
( ) ( ) ( )
( )
,
21
2
2
2
1
2
1
2
1
1
1
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
⎛
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
=
hN
hNhNhN
hN
hN
U
F
U
F
U
F
U
F
U
F
U
F
U
F
U
F
U
F
DF
and finally the value vector F is
( ( ) ( ) ( )( )) ....,,, 21
T
hN UFUFUF=F

8. Summary and Conclusion
In this work, we prepared the approach to get the mathematical modeling
of heat transfer in the Garnissage furnace and we exerted the process of
finding its numerical solution. We invoked the physical conservation laws,
that is the mass, the momentum, and the energy conservation laws, to earn
the continuity, the Navier-Stokes, and the heat equation, and the Stefan
condition was used to model the free boundary between the solid and liquid
phase. The approach was continued by deriving the new version of equations
in the stream function system before we wanted to transfer the system into
the weak formulation.
Since the finite element method have been chosen to achieve the
numerical solution of the transport system, we got the system of equations in
the variational system by handling the sufficient test functions. The process
was completed by discretization of the domain and applying the approximate
value instead of variables. At last we reached the system with nonlinear and
linear equations (6.12),
( )
( )( )( )
∑ ∑ ∑= = =
==+
hN
i
hN
j
hN
i
ikiijkji hNkbUaUU
1 1 1
,...,,3,2,1,0
( )
( )
∑=
==
hN
i
jiij hNjdVc
1
,...,,3,2,1,
and finally we suggested the Newton’s method to solve the nonlinear part of
the system.
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