Finite Volume Method Advanced Numerical Analysis by Md.Al-Amin

Finite Volume
Method
Course Title:
Advanced Numerical Analysis
Course no: 0541 12 Math 5101
1
8/31/2023 Mathematics Discipline

Presented by
Md. Al-Amin
Student ID: MSc-231206
On behalf of
Group- 04
Mathematics Discipline
Khulna University, Khulna 9208
2

Limitations
Limitations
of FVM
Solve
Solve
problems
Basic concept
Why we need FVM
(Finite volume
method)
Process
Working rules

We get solution
for specific grid
points
Basic concept
Finite difference
method
Finite Element
method
Finite Volume
method
We get solution for any
points in the interval. Profile
assumptions needed to
complete analysis
Profile assumptions
just need to evaluate
the integrals. It follows
conservation law.
4

5
.
❖ Discretization
The process of converting a continuous problem or equation into a discrete one that can be solved
using numerical methods. The process of discretization involves dividing the domain of the continuous
function into a set of smaller regions or intervals, called discretization cells or element.
▪ Discretization of the domain: Dividing the slab into ‘n’ number of domain
▪ Element: Each domain is called as element
▪ Finite Element Mesh: The collection of element
▪ Nodes: The elements are connected at points called as nodes
▪ Uniform mesh: When elements are of same dimension

Finite volume refers to the small
volume surrounding each node point on
a mesh. FVM is a numerical method
or an approximate method for
representing and evaluating ordinary or
partial differential equations in the form
of algebraic equations.
.
Similar to FDM or FEM values are
calculated at discrete places on a
meshed geometry.
In FVM volume integrals in a
partial differential equation that
contains a divergence term
converted to surface integral
using the Gauss’s divergence
theorem.
FVM follows conservation law:
The flux entering and leaving is
identical.
FINITE VOLUME METHOD (FVM)

STEP 01
Grid generation: Divide the domain
into a number of finite sized sub
domain (control volumes)
represented by finite number of grid
points.
STEP 02
Discretization: Integrating the
governing equation over control
volume to yield a discretized
equation at its nodal point.
STEP 03
Solution of equations: consider
profile assumptions for the
dependent variable to express the
result in terms of algebraic
equations which are now easier
to solve.
Important steps involving FVM for solving a differential equation are as follows:

8
.
Problem 01: Explain finite volume method to solve the one dimensional steady state diffusion equation for the
property T:
𝑑
𝑑𝑥
(𝐾
𝑑𝑇
𝑑𝑥
)+S=0, where K is the diffusion coefficient and S is the source term.
Boundary value of T at points A and B are prescribed.
For one-dimensional heat conduction in a rod, K is the thermal conductivity of the material, T is temperature and S is
the heat generation due to an electrical current passing through the rod.
Solution:
Given the governing differential equation
𝑑
𝑑𝑥
(𝐾
𝑑𝑇
𝑑𝑥
)+S=0
STEP 01
Grid generation: Divide the domain into a number of finite sized sub domain (control volumes) represented by
finite number of grid points.

9
.
Step 1: Grid generation:
Using the universal nomenclature procedure:
Divide the domain into 5 co domain(control volume) using nodal points
Center of control volume for the respective grid points
Boundary of the control volume because we have to implement the boundary conditions
Isolated part of any control volume from the domain and try to describe its features

10
Disappearing isolated part’s indication
P
Central nodal point is denoted as P
E
Immediate next east neighbor denoted as E
W
Immediate next west neighbor denoted as W
𝑑
𝑑𝑥
(𝐾
𝑑𝑇
𝑑𝑥
)+S=0 1 Dimensional Diffusion equation
so we just need W and E
𝜕
𝜕𝑥
𝐾
𝜕𝑇
𝜕𝑥
+
𝜕
𝜕𝑦
𝐾
𝜕𝑇
𝜕𝑦
+ S = 0 2 Dimensional Diffusion equation
so we need W, E, N and S
Δ𝑥 = 𝛿𝑥𝑤𝑒
West face is w and east face is e
𝛿𝑥𝑃𝑒
𝛿𝑥𝑃𝐸
𝛿𝑥𝑊𝑃
𝛿𝑥𝑤𝑃
w e
Particular length of control volume
So, here phases of the control volume are
denoted as lower case alphabets and
grid points are denoted by upper case
alphabets.
Here, our 1st step to generate grid is
completed.

11
Given the governing differential equation 𝑑
𝑑𝑥
(𝐾
𝑑𝑇
𝑑𝑥
)+S=0
Step 2: Discretization:
STEP 02
Discretization: Integrating the
governing equation over control
volume to yield a discretized
equation at its nodal point.
So, we have,
‫׬‬𝑤
𝑒 𝑑
𝑑𝑥
(𝐾
𝑑𝑇
𝑑𝑥
)dV+ ‫׬‬𝑤
𝑒
𝑆 𝑑𝑉 = 0
or, 𝐴𝐾
𝑑𝑇
𝑑𝑥 𝑒
− 𝐴𝐾
𝑑𝑇
𝑑𝑥 𝑤
+𝑆∆𝑉 = 0
Here we have got a corresponding equation but this is still differential equation. To get corresponding
algebraic expression we must have a profile variation for T as a function of x. So that we can
calculate dT/dx.
So, the problem arising to some assumptions on how T varies with x.

12
P E
W w e
So, we have to make a choice
I. So, we have no other constraints; it could have been piece wise constant profiles of T
P E
W w e
For, constant profiles of T. The result of dT/dx =0
So, this is not a valid choice.
So, what could be the next possible choice?
II. Piece wise linear non constant profile between control volume phases.
Constant is linear but here consider non constant.
This is just a sketch to understand the concept don’t bother with the arbitrary value.
We need to calculate dT/dx at e and w. At e and w dT/dx is
discontinuous.
We will get left hand derivative for deep blue line and right
hand derivative for blue line. These two derivatives are
different as these are different lines. This implies, K dT/dx is
discontinuous. Physically means heat flux is not continuous.
But at a phase heat flux must be continuous. So, this is
also not a acceptable profile assumption.

13
P E
W w e
III. Piece wise linear constant profile between the grid points.
We need to calculate dT/dx at e and w. At e and w dT/dx is
continuous. So, this is the valid profile assumption.
Will it work?
So, as T is a linear function of x
𝐴𝐾
𝑑𝑇
𝑑𝑥 𝑒
− 𝐴𝐾
𝑑𝑇
𝑑𝑥 𝑤
+𝑆∆𝑉 = 0
or, 𝐴𝑒𝐾𝑒
𝑇𝐸−𝑇𝑃
𝛿𝑥𝑃𝐸
− 𝐴𝑤𝐾𝑤
𝑇𝑃−𝑇𝑊
𝛿𝑥𝑊𝑃
+𝑆∆𝑉 = 0
The source term S maybe a function of the
dependent variable. In such case the FVM
approximates the source term by means of
a linear form: 𝑆∆𝑉 = 𝑆𝑢 + 𝑆𝑝𝑇𝑝
or, 𝐴𝑒𝐾𝑒
𝑇𝐸−𝑇𝑃
𝛿𝑥𝑃𝐸
− 𝐴𝑤𝐾𝑤
𝑇𝑃−𝑇𝑊
𝛿𝑥𝑊𝑃
𝐴𝑒𝐾𝑒
𝛿𝑥𝑃𝐸
+ 𝑆𝑢 + 𝑆𝑝𝑇𝑝 =0
𝑜𝑟,
𝐴𝑒𝐾𝑒
𝛿𝑥𝑃𝐸
+
𝐴𝑤𝐾𝑤
𝛿𝑥𝑊𝑃
− 𝑆𝑝 𝑇𝑃 =
𝐾𝑒𝐴𝑒
𝛿𝑥𝑃𝐸
𝑇𝐸 +
𝐾𝑤𝐴𝑤
𝛿𝑥𝑊𝑃
𝑇𝑊 + 𝑆𝑢
Now, we can symbolically write this as
𝑜𝑟, 𝑎𝑃𝑇𝑃 = 𝑎𝐸𝑇𝐸 + 𝑎𝑊𝑇𝑊 + 𝑆𝑢
Step 3: Solution of the equation:
Now the discretized equations of the above form will
modified to incorporate boundary conditions. The
resulting system of algebraic equations are now easier
to solve.

14
.
Problem 02:
Solution:
Consider the problem of source-free heat conduction in an insulated rod of length 𝑳 = 𝟎. 𝟓 𝒎 and
whose ends are maintained at constant temperatures of 1000C and 5000C respectively. Then using
the finite volume method to find the steady state temperature distribution in the rod. Given that the
thermal conductivity 𝒌 = 𝟏𝟎𝟎𝟎
𝑾
𝒎.𝑲
and the cross-sectional area 𝑨 = 𝟏𝟎 × 𝟏𝟎−𝟑
𝒎𝟐
. Also
compare your numerical results with the analytical solution.
The steady state heat conduction equation with no heat source is given by
𝑑
𝑑𝑥
𝑘
𝑑𝑇
𝑑𝑥
= 0 ………………………………………… (1)
Also given one dimensional problem is sketched in the following figure (1)
.
A B
TA=1000C TB=5000C
𝑳 = 𝟎. 𝟓 𝒎
Fig. 1: Given rod with boundary condition

15
.
.
𝑳 = 𝟎. 𝟓 𝒎
Fig. 2: Grid generation
Let us divide the length of the rod into five equal control volumes as shown in figure (2)
.
TA=1000C TB=5000C
P E
W
𝛿𝑥𝑃𝑒
𝛿𝑥𝑤𝑃
w e
Since the length of the rod 𝑳 = 𝟎. 𝟓 𝒎 , so the node spacing 𝛿𝑥 =
0.5
5
= 0.1𝑚
The grid consists of five nodes. For each one of the nodes 2, 3 and 4 temperature values to the east and west are
available as nodal values. Consequently, discretized equation (1) can be readily written for the control volumes
surrounding these nodes:
𝐾𝑒
𝛿𝑥𝑃𝐸
𝐴𝑒 +
𝐾𝑤
𝛿𝑥𝑊𝑃
𝐴𝑤 𝑇𝑝 = (
𝐾𝑤
𝛿𝑥𝑊𝑃
𝐴𝑤 +
𝐾𝑒
𝛿𝑥𝑃𝐸
𝐴𝑒)𝑇𝑒 ……………………………… (2)

16
.
Here the thermal conductivity (𝑲𝒆 = 𝑲𝒘 = 𝑲 = 𝟏𝟎𝟎𝟎), node spacing (𝛿𝑥 = 𝛿𝑥𝑊𝑝 = 𝛿𝑥𝑃𝑒 = 0.1 𝑚) and
cross-sectional area (𝑨𝒆 = 𝑨𝒘 = 𝑨 = 𝟏𝟎 × 𝟏𝟎−𝟑) are constants.
Therefore the discretized equation for nodal points 2, 3 and 4 is
𝑎𝑤𝑇𝑝 = 𝑎𝑤𝑇𝑤 + 𝑎𝑒𝑇𝑒 …………………………. (3)
Where 𝑎𝑊 =
𝐾𝑤
𝛿𝑥𝑊𝑃
𝐴𝑤 =
1000
0.1
× 10 × 10−3= 100
𝑎𝐸 =
𝐾𝑒
𝛿𝑥𝑃𝐸
𝐴𝑒 =
1000
0.1
× 10 × 10−3
= 100
And 𝑎𝑃 = 𝑎𝑊 + 𝑎𝐸 = 100 + 100 = 200
Here 𝑆𝑢 and 𝑆𝑝 are zero in this case
since there is no sourse term in the
governing equation (1)
Nodes 1 and 5 are boundary nodes, and therefore requires special attention. Then for node 1, we have the
discretized equation as
𝑎𝑃𝑇𝑃 = 𝑎𝐴𝑇𝐴 + 𝑎𝐸𝑇𝐸 …………………………. (4)
Where 𝑎𝐴 =
𝐾𝐴
𝛿𝑥/2
=
1000×10×10−3
0.1/2
= 200
And 𝑎𝐸 = 100 so that 𝑎𝑃 = 𝑎𝐴 + 𝑎𝐸 = 200 + 100 = 300

17
.
Similarly for node 5, the discretized equation is
𝑎𝑃𝑇𝑝 = 𝑎𝑊𝑇𝑊 + 𝑎𝐵𝑇𝐵 ………………………….. (5)
Where 𝑎𝑊 = 100 as before
𝑎𝐵 =
𝐾𝐴
𝛿𝑥/2
=
1000×10×10−3
0.1/2
= 200
So that 𝑎𝑃 = 𝑎𝐵 + 𝑎𝑊 = 200 + 100 = 300
Using (3), (4), and (5), the system can be written as follows
From equation (1)
For node 1:
300𝑇𝑃 = 200𝑇𝐴 + 100𝑇𝐸 Here 𝑇𝑝 = 𝑇1
Or, 300𝑇1 = 200 × 100 + 100 × 𝑇2
Or 300𝑇1 − 100𝑇2 = 20000 …………………. (6)
From equation (3)
For node 2:
200𝑇𝑃 = 100𝑇𝑊 + 100𝑇𝐸
Or, 200𝑇2 = 100𝑇1 + 100𝑇3
Or, −100𝑇1 + 200𝑇2 − 100𝑇3 = 0 …………….…. (7)
Similarly for nodes 3 and 4, we can write
−100𝑇2 + 200𝑇3 − 100𝑇4 = 0 …………...………. (8)
−100𝑇3 + 200𝑇4 − 100𝑇5 = 0 ……………………. (9)
And finally for node 5:
300𝑇𝑃 = 100𝑇𝑊 + 200𝑇𝐵
Or, 300𝑇5 = 200𝑇4 + 100 × 500
Or, −100𝑇4 + 300𝑇5 = 1000 … … … .……………… (10)

18
Then the system using equation (6) to (10) can be written
as
3𝑇1 − 𝑇2 = 200 ……………..……………… (11)
−𝑇1 + 2𝑇2 = 𝑇3 = 0 …………..…………… (12)
−𝑇2 + 2𝑇3 − 𝑇4 = 0 .…………………….. (13)
−𝑇3 + 2𝑇4 − 𝑇5 = 0 …………………….. (14)
−𝑇4 + 3𝑇5 = 1000 ………………………. (15)
Now (11) + (12)*3 ⟹ 5𝑇1 − 3𝑇3 = 200 ….. (16)
(13)*5 + (16) ⟹ 7𝑇3 − 5𝑇4 = 20.…….. (17)
(14)*7 + (17) ⟹ 7𝑇4 − 7𝑇5 = 200 … (18)
(15)*9 + (18) ⟹ 20𝑇5 = 9200
⟹ 𝑇5 = 460
∴ 18 ⟹ 𝑇4 = 380
17 ⟹ 𝑇3 = 300
16 ⟹ 𝑇2 = 220
Finally , 11 ⟹ 𝑇1 = 140
Then the steady state temperature at the nodal points are
𝑇1 = 140, 𝑇2 = 220, 𝑇3 = 300, 𝑇4 = 380,
𝑇5 = 460

19
Analytical Solution:
Since 𝑘 = 1000, the given equation becomes
𝑑
𝑑𝑥
𝑘
𝑑𝑇
𝑑𝑥
= 0
⟹ 1000
𝑑2
𝑇
𝑑𝑥2 = 0
⟹
𝑑2𝑇
𝑑𝑥2 = 0
Which gives after twice time integration as
𝑇 𝑥 = 𝐶1𝑥 + 𝐶2 …………………… (19)
Where 𝐶1 and 𝐶2 are arbitrary constants
Applying the boundary condition 𝑇 0 =100
And 𝑇 0.5 = 500 in (19), we set
𝑇 0 = 100 = 𝐶1 ∗ 0 + 𝐶2 ⟹ 𝐶1 = 800
i.e. 𝑇 𝑥 = 800𝑥 + 100 ……………………. (20)
which is the exact solution
The comparison between the finite volume solution and the exact
solution is given below:
Table-1
The above table shows that the exact solution and numerical results
coincides.
Node no. Distance,
x(m)
FVS ES Percentage of
error
1 0.05 140 140 0
2 0.15 220 220 0
3 0.25 300 300 0
4 0.35 380 380 0
5 0.45 460 460 0

20
.
Problem 03:
Solution:
Fig. 1: Given plate with boundary condition
Consider a large plate of thickness L=2 cm=0.02m with constant thermal conductivity k=0.5 w/m.k and uniform
heat generative q = 1000 kw/m3. The faces A and B of the plate are kept at temperature 1000C and 2000C
respectively. Assume that the dimensions in the y-axis z-direction are so large that temperature gradients are
significant in the x-direction on y. Then using the finite volume method to calculate the steady state temperature
distribution. Also compare the numerical results with the analytical solution.
For the present problem the heat conduction equation (with heat source) is
𝑑
𝑑𝑢
𝑘
𝑑𝑇
𝑑𝑢
+ 𝑞 = 0 … … … … … … … … … … (1)
All the given problem is sketched in the figure(1):
L
A B
TA
TB
Here,
𝑻𝑨 = 𝟏𝟎𝟎°𝑪
𝑻𝑩 = 𝟐𝟎𝟎°𝑪
L = 0.02 m
The domain is divided into 5 control volume giving 𝛿𝑥 =
.02
5
= 0.004𝑚

21
Fig. 2: Grid generation
.
𝑳 = 𝟎. 𝟎𝟐 𝒎
.
TA=1000C TB=2000C
P E
W
𝛿𝑥𝑃𝑒
𝛿𝑥𝑤𝑃
w e
𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟, 𝐴𝑤 = 𝐴𝑒 = 𝐴 = 1
𝐻𝑒𝑟𝑒, 𝐾𝑒 = 𝐾𝑤 = 𝐾 = .5 Τ
𝑤 𝑚. 𝑘
𝑎𝑛𝑑, 𝑞 = 1000 Τ
𝑘𝑤 𝑚3
Now using finite volume method, the discretized equation which is valid for
nodes 2, 3, and 4. is
𝑎𝑝𝑇𝑝 = 𝑎𝑤𝑇𝑤 + 𝑎𝐸𝑇𝐸 + 𝑆𝑈……………………………..(2)
Where,
𝑎𝑤 =
𝐾𝐴
𝛿𝑥
=
0.5 × 1
.004
= 125
𝑎𝐸 =
𝐾𝐴
𝛿𝑥
=
0.5 × 1
.004
= 125
𝑆𝑢 = 𝑞𝐴𝛿𝑥 = 1000 × 103
× .004 × 1 = 4000
And 𝑎𝑃 = 𝑎𝑊 + 𝑎𝐸 = 125 + 125 = 250

22
Also the discretized equation for node 1 is
𝑎𝑝𝑇𝑝 = 𝑎𝐴𝑇𝐴 + 𝑎𝐸𝑇𝐸 + 𝑆𝑈……………………….(3)
Where, 𝑎𝐴 =
𝐾𝐴
Τ
𝛿𝑥 2
= 2 ×
𝐾𝐴
𝛿𝑥
=
2×.5×1
.004
= 250
𝑎𝐸 =
𝐾𝐴
𝛿𝑥
=
.5 × 1
.004
= 125
𝑎𝑃 = 𝑎𝐴 + 𝑎𝐸 = 250 + 125 = 375
Similarly, for node 5, the discretized equation is
𝑎𝑝𝑇𝑝 = 𝑎𝑤𝑇𝑤 + 𝑎𝐵𝑇𝐵 + 𝑆𝑈…………………………….(4)
Where,
𝑎𝐵 =
𝐾𝐴
𝛿𝑥/2
=
2 × 0.5 × 1
0.004
= 250
𝑎𝑊 =
𝐾𝐴
𝛿𝑥
=
0.5 × 1
0.004
= 125
And 𝑎𝑃 = 𝑎𝐵 + 𝑎𝑊 = 250 + 125 = 375

23
Now using these equation as
375 𝑇1 − 125 𝑇2 = 25000 + 4000 = 29,000
−125 𝑇1 + 250 𝑇2 − 125 𝑇3 = 4,000
−125 𝑇2 + 250𝑇3 − 125𝑇4 = 4,000
−125𝑇3 + 250𝑇4 − 125𝑇5 = 4,000
−125𝑇4 + 375𝑇5 = 50000 + 4000 = 54,000
⇒ 3𝑇1 − 𝑇2 = 232
−𝑇1 + 2𝑇2 − 𝑇3 = 32
−𝑇2 + 2𝑇3 − 𝑇4 = 32
−𝑇3 + 2𝑇4 − 𝑇5 = 32
−𝑇4 + 3𝑇5 = 432
Applying the Gaussian elimination method,
We get
3𝑇1 − 𝑇2 = 232
5𝑇2 − 3𝑇3 = 32
7𝑇3 − 5𝑇4 = 32
9𝑇4 − 7𝑇5 = 32
20𝑇5 = 432
Then using backward substitution, we get
𝑇5 = 230, 𝑇4 = 258, 𝑇3 = 254, 𝑇2 = 218, 𝑇1 = 150
Which are required steady state temperature
distribution at noded point.

24
Analytical solution:
The analytical solution to this problem ray be obtained by integrating
equation twice with repeat to x and by subsequent application of the
boundary conditions given
𝑇 𝑥 =
𝑇𝐵 − 𝑇𝐴
𝐿
+
𝑞
2𝑘
𝐿 − 𝑥 𝑥 + 𝑇𝐴 … … … … … … … … … … 5
The comparison between the finite volume solution and the exact
solution is shown in table given below
Node
No.
Distance, x(m) FVS ES Percentage
error
1 0.02 150 146 2.73
2 0.006 218 214 1.86
3 0.01 254 250 1.60
4 0.014 258 254 1.57
5 0.018 230 226 1.76
0.0 0.4 0.8 1.2 1.6 2.0
50
100
150
200
250
300
exact
Numerical
Distance(cm)
Temperature(℃)
Fig 3: comparison of the numerical result with the analytical solution
From table and fig it can be seen that the agreement is very
good.
Here,𝑻𝑨 = 𝟏𝟎𝟎°𝑪, 𝑻𝑩 = 𝟐𝟎𝟎°𝑪, L = 0.02 m, k=0.5 w/m, q = 1000 kw/m3 = 1000 *1000 w/m3

25
.
Problem on one dimensional convection-diffusion equation :
Solution:
Finite Volume Method for
Convection-Diffusion Problem
In problems, where fluid flow plays a significant role. We must account for the effects of Convection. Diffusion
always occurs alongside convection in nature. So here we explain the finite volume method to predict combined
convection and diffusion.
Describe the finite volume method to solve the steady convection-diffusion equation
𝑑
𝑑𝑥
𝑃 𝑢 𝑄 =
𝑑
𝑑𝑥
(⎾
𝑑𝑄
𝑑𝑥
) for a property 𝑄,
Where the one-dimensional flow field 𝑢 satisfy the continuity equation
𝑑(𝑃𝑢)
𝑑𝑥
= 0
Given that,
𝑑
𝑑𝑥
𝑃 𝑢 𝑄 =
𝑑
𝑑𝑥
(⎾
𝑑𝑄
𝑑𝑥
) … … … 1
and
𝑑(𝑃𝑢)
𝑑𝑥
= 0 … … … (2)
We consider the one-dimensional control volume as shown in fig-1. A general nodal point is identified by 𝑷 and its
west and east node are identified by 𝑾 and E. The west side and east side is referred to by 𝒘 and 𝒆 respectively. The
control volume width is ∆𝒙 = 𝜹𝒙𝒘𝒆. Also the distance between 𝑤 and 𝑃, 𝑃 and 𝑒 are shown in fig-1.

26
.
.
.
P E
W
𝛿𝑥𝑃𝑒
𝛿𝑥𝑤𝑃
w e
Fig 1: A control volume around node 𝑃
Now integration over the control volume of the equation (1) gives
‫׬‬𝑐𝑣
𝑑
𝑑𝑥
𝑃 𝑢 𝑄 𝑑𝑣 =
𝑑
𝑑𝑥
‫׬‬𝑐𝑣
𝑑
𝑑𝑥
(⎾
𝑑𝑄
𝑑𝑥
)𝑑𝑣 … … … (3)
For a vector Ԧ
𝑎 the Gauss’s divergence theorem states that ‫׬‬𝑐𝑣
𝑑𝑖𝑣 Ԧ
𝑎 𝑑𝑣 = ‫׬‬𝐴
ෝ
𝑛. Ԧ
𝑎 𝑑𝐴 … … … (4)
Where ො
𝑛 is the unit outward normal vector to the surface element 𝑑𝐴.
Thus, applying the Gauss’s divergence theorem (4), the equation (3) can be written as follows:
න
𝐴
ො
𝑛. 𝑃 𝑢 𝑄 𝑑𝐴 = න
𝐴
ො
𝑛. (⎾
𝑑𝑄
𝑑𝑥
)𝑑𝐴
or, (𝑃 𝑢 𝐴𝑄)𝑒− 𝑃 𝑢 𝐴𝑄 𝑤 = ⎾𝐴
𝑑𝑄
𝑑𝑥 𝑒
− ⎾𝐴
𝑑𝑄
𝑑𝑥 𝑤
… … … 5

27
And integration of continuity equation (2) yields
න
𝑐𝑣
𝑑(𝑝𝑢)
𝑑𝑥
dv = 0
or, ‫׬‬𝐴
ො
𝑛. 𝑃 𝑢 𝑑𝐴 = 0
or, (𝑃 𝑢 𝐴)𝑒−(𝑃 𝑢 𝐴)𝑤= 0 … … … (6)
or, 𝐹𝑒𝐴𝑒 − 𝐹𝑤𝐴𝑤 = 0
or, 𝐹𝑒𝐴 − 𝐹𝑤𝐴 = 0
or, 𝐹𝑒 − 𝐹𝑤 = 0
To obtain discretized equation for convection-diffusion problem we must approximate terms in equation (5). It
is convenient to define two variables F and D to represent the convective mass flux per unit area and diffusion-
conductance at cell faces:
𝐹 = 𝑃𝑢 and 𝐷 =
⎾𝑒
𝛿𝑥
… … … (7)
The cell face values of the variables F and D can be written as
𝐹𝑤 = (𝑃𝑢)𝑤 , 𝐹𝑒 = (𝑃𝑢)𝑒 … … … 8𝑎
and 𝐷𝑤=
⎾𝑤
𝛿𝑥𝑊𝑃
, 𝐷𝑒 =
⎾𝑒
𝛿𝑥𝐸𝑃
… … … (8b)
Now assume that 𝐴𝑤 = 𝐴𝑒 = A,

28
so, we can divide the left and right sides of equation (5) by area A. Also the diffusive flux terms are evaluated as
(⎾𝐴
𝑑𝑄
𝑑𝑥
) 𝑒= 𝐹𝑒𝐴𝑒
𝑄𝐸 − 𝑄𝑃
𝛿𝑥𝑃𝐸
… … … 9𝑎
And (⎾𝐴
𝑑𝑄
𝑑𝑥
) 𝑒= 𝐹𝑤𝐴𝑤
𝑄𝑃−𝑄𝑊
𝛿𝑥𝑊𝑃
… … … 9𝑏
Using (8) and (9), the integral convection-diffusion equation (5) can now be written as
𝐹𝑒𝑄𝑒 − 𝐹𝑤𝑄𝑤 = 𝐷𝑒 𝑄𝐸 − 𝑄𝑝 − 𝐷𝑤 𝑄𝑝 − 𝑄𝑊 … … … 10
and the integral continuity equation (6) can be written as
𝐹𝑒 − 𝐹𝑤 = 0 … … … (11)
We also assume that the velocity field is ‘somehow known’ which takes care of the values of 𝐹𝑒 and 𝐹𝑤. In order to solve
the equation (10) we need to calculate the hand post property Q at the e and w faces. Using the central differencing
scheme and for a uniform grid, we can write the cell faces values of property Q as 𝑄𝑒 =
𝑄𝑃+𝑄𝐸
2
… … … 12𝑎
And 𝑄𝑤 =
𝑄𝑊+𝑄𝑃
2
… … … 12𝑏
Now substituting the above expression (12a) and (12b) into equation (10) we obtain
𝐹𝑒
2
𝑄𝑝 + 𝑄𝑒 −
𝐹𝑤
2
𝑄𝑤 + 𝑄𝑝
= 𝐷𝐸 𝑄𝑒 − 𝑄𝑃 − 𝐷𝑤 𝑄𝑃 − 𝑄𝑤 … … … (13)

29
This can be rearranged to give
[(𝐷𝑊 −
𝐹𝑊
2
) + (𝐷𝑒 +
𝐹𝑒
2
)]𝑄𝑃 =(𝐷𝑊 +
𝐹𝑊
2
)𝑄𝑊 + (𝐷𝑒 −
𝐹𝑒
2
)𝑄𝐸
Or, [(𝐷𝑊 +
𝐹𝑊
2
) + 𝐷𝑒 −
𝐹𝑒
2
+ (𝐹𝑒 − 𝐹𝑤)]𝑄𝑃 =(𝐷𝑊 +
𝐹𝑊
2
)𝑄𝑊 + (𝐷𝑒 −
𝐹𝑒
2
)𝑄𝐸 … … … (14)
Now identifying the coefficient 𝜑𝑊 and 𝜑𝐸 as 𝑎𝑊 and 𝑎𝐸 , the central differencing expressions for the
discretized convection diffusion equations are
𝑎𝑃𝜑𝑃 = 𝑎𝑊𝜑𝑊 + 𝑎𝐸𝜑𝐸 (15)
Where
𝑎𝑊 = 𝐷𝑊 +
𝐹𝑤
2
,
𝑎𝐸 = 𝐷𝑒 +
𝐹𝑒
2
and 𝑎𝑃 = 𝑎𝑊 + 𝑎𝐸 + (𝐹𝑒 − 𝐹𝑤) (16)
Then to solve a one-dimensional convection diffusion problem we write discretized equations of the form
(15) for all grid nodes. This yields a set of algebraic equations that is solved to obtain the distribution of the
transport property 𝜑 .

30
.
Problem 02:
Solution:
Fig. 1: Grid generation for discretization
A property 𝜑 is transported by means of convection and diffusion through one-dimensional domain. The governing
equation is
𝑑
𝑑𝑥
ρ𝑢𝜑 =
𝑑
𝑑𝑥
(Γ
𝑑𝜑
𝑑𝑥
) with the boundary conditions are 𝜑0 = 1 at 𝑥 = 0 and 𝜑𝐿 = 0 at 𝑥 = 𝐿 .
Then using the finite volume method with five equally spaced cells and central differencing scheme for convection
and diffusion to calculate the distribution of 𝜑 as a function of x when 𝑢 = 0.1 𝑚/𝑠, length 𝑳 = 𝟎. 𝟏 𝒎, 𝜌 =
1.0 𝑘𝑔/𝑚3
and Γ = 0.1𝑘𝑔/𝑚𝑠. Also compare the results with the analytical solution
𝜑 − 𝜑0
𝜑𝐿 − 𝜑0
=
𝑒− 𝜌𝑢𝑥/Γ −1
𝑒− 𝜌𝑢𝐿/Γ −1
The method of solution is demonstrated using the simple grid in Fig. 1 . The domain is divided into five control
volumes giving 𝛿𝑥 =
1.0
5
= 0.2m . Here 𝐹 = 𝜌𝑢, 𝐹𝑒 = 𝐹𝑤 = 𝐹 and 𝐷𝑒 = 𝐷𝑤 = 𝐷 everywhere. The boundaries are
denoted by A and B.
.
.
x=0, 𝝋=1
P E
W
𝛿𝑥𝑃𝑒
𝛿𝑥𝑤𝑃
w e
x=L, 𝝋=0

31
Here 𝐹 = 𝜌𝑢 = 1 × 0.1 = 0.1 = 𝐹𝑒 = 𝐹𝑤
𝐷𝑒 = 𝐷𝑤 = 𝐷 =
Γ
𝛿𝑥
=
0.1
0.2
= 0.5
Now the discretized problem is
𝑎𝑃𝜑𝑃 = 𝑎𝑊𝜑𝑊 + 𝑎𝐸𝜑𝐸 (1)
Which is applied only for the internal nodal points 2, 3, and 4, but control volume 1 and 5 needs special
treatment since they are adjacent to the domain boundaries.
Now the coefficients are:
𝑎𝑊 = 𝐷 +
𝐹
2
= 0.5 +
0.1
2
= 0.55
𝑎𝐸 = 𝐷 −
𝐹
2
= 0.5 −
0.1
2
= 0.45
𝑎𝑃 = 𝑎𝑊 + 𝑎𝐸 = 0.55 + 0.45 = 1.0
From equation (1)
𝜑𝑃 = 0.55𝜑𝑊 + 0.45𝜑𝐸 (2)
For the boundary node,
𝐷𝐴 = 𝐷𝐵 =
Γ
𝛿𝑥2
= 2D = 1.0
For node1 the equation (1) becomes,
𝑎𝑃𝜑𝑃 = 𝑎𝐴𝜑𝐴 + 𝑎𝐸𝜑𝐸 (3)
𝑎𝐴 = 𝐷𝐴 + 𝐹𝐴 = 1.1, 𝑎𝐸 = 0.45, 𝑎𝑃 = 1.55 so that (3) reduces
for node 1:
1.55𝜑𝑃 = 1.1 + 0.45𝜑𝐸 (4)
Also for node 5, the equation (1) becomes,
𝑎𝑃𝜑𝑃 = 𝑎𝐵𝜑𝐵 + 𝑎𝑊𝜑𝑊 (5)
𝑎𝑊 = 0.55, 𝑎𝐵 = 𝐷𝐵 = 𝐹 = 0.9
So, 𝑎𝑃 = 0.9 + 0.55 = 1.45 so that equation (5) reduces to
1.45𝜑5 = 0.55𝜑4 + 0
𝑜𝑟, 1.45𝜑5 − 0.55𝜑4 = 0 (6)

32
Then the system (2), (4), and (6) can be written as,
1.55𝜑1 − 0.45𝜑2 = 1.1
−0.55𝜑1 + 𝜑2 − 0.45𝜑3 = 0
−0.55𝜑2 + 𝜑3 − 0.45𝜑4 = 0
−0.55𝜑3 + 𝜑4 − 0.45𝜑5 = 0
−0.55𝜑4 + 1.45𝜑5 = 0
Using the Gaussian elimination method, the above system can
be written as,
1.55𝜑1 − 0.45𝜑2 = 1.1
1.3025𝜑2 − 0.6975𝜑3 = 0.605
0.9189𝜑3 − 0.5861𝜑4 = 0.3328
0.5965𝜑4 − 0.4135𝜑5 = 0.1830
0.6375𝜑5 = 0.10065
Using backward substitution we get,
𝜑5 = 0.1578, 𝜑4 = 0.4162, 𝜑3 = 0.6276, 𝜑2 = 0.8006, 𝜑1
= 0.9421
Comparison with the analytical solution:
Substitution of the given data we get the exact
solution of the problem:
𝜑 𝑥 =
2.7183 − 𝑒𝑥
1.7183
Comparison of numerical and analytical solution are
given in the Table-1 which shows reasonable
agreement.
Node Distanc
e
FVS ES Differe
nce
Percen
tage
error
1 0.1 0.9421 0.9387 -0.003 -0.36
2 0.3 0.8006 0.7963 -0.004 -0.53
3 0.5 0.6276 0.6224 -0.005 -0.83
4 0.7 0.4163 0.4100 -0.006 -1.53
5 0.9 0.1579 0.1505 -0.007 -4.91

STE
P
02
STE
P
03
STE
P
01
STE
P
05
STE
P
06
STE
P
04
Grid-dependence: The accuracy of the
FVM solution depends on the quality of
the grid used to discretize the domain.
In general, the smaller the grid size,
the more accurate the solution.
Limited accuracy: The FVM is generally
less accurate than other numerical
techniques, such as the finite element
method (FEM) or the spectral method.
This is especially true for complex
geometries or problems with high
gradients.
Difficulty with high-order derivatives:
The FVM is typically less accurate when
approximating high-order derivatives,
such as the Laplacian. This can lead to
errors in the solution, particularly in
regions of high curvature or high
gradients.
Limitations of FVM
Difficulty with unstructured grids: The
FVM is generally more suited to structured
grids, where the grid points are arranged
in a regular pattern. Unstructured grids,
where the grid points are irregularly
spaced, can be more difficult to handle with
the FVM.
Limited adaptivity: The FVM is typically
less adaptable than other numerical
methods, such as the adaptive FEM,
which can adjust the grid size and shape
to improve the accuracy of the solution in
regions of interest.
Difficulty handling nonlinear equations: The
FVM is often less effective in handling
nonlinear equations than other numerical
methods, such as the finite difference
method (FDM) or FEM.

D
W
L
W
Working
rule of FVM
Limitations
of FVM
Why we
need FVM ?
Difference among
3 methods
Output of Today’s Lecture

Thanks to all
 
I appreciate your patience in
listening to this boring class
attentively.
35

Finite Volume Method Advanced Numerical Analysis by Md.Al-Amin

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