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Week 9 - FVM Literature Review

FINITE VOLUME METHOD Finite Volume Method is the numerical method that solves a governing equation in its integral form, that mean the variable properties are conserved over a volume rather than a point as in case of Finite Difference Method. Finite Volume method(FVM) uses integral form of governing whereas the Finite…

MATLAB

Amol Patel
updated on 07 Jul 2021

FINITE VOLUME METHOD

Finite Volume Method is the numerical method that solves a governing equation in its integral form, that mean the variable properties are conserved over a volume rather than a point as in case of Finite Difference Method.

Finite Volume method(FVM) uses integral form of governing whereas the Finite Difference method(FDM) uses differential form of the governing equations. The discretised equation in the FVM has a clear physical interpretation while in the FDM is not that clearly interpreted.

The Finite Volume Method has the following three steps:

Step 1: Grid Generation

to create a grid in FVM we divide the entire domain into discrete control volumes. The nodal points will be at the centre of this control volumes. The general nodal point is identified using 'P' and the sorrounding point will be named as 'W','E','N','S' according to their positon at west, east, north , south of the node P respectively.

Step 2: Discretisation

to find the discretised equation at the nodal point we need to integrate the governing over hte control volume.

Step 3: Solution for Equations

after finding the discretised equations at every nodal point we get a system of linear equation that can be solved to obtain the distribution of the properties.

Example:

Now let us take an example of 1D heat conduction equation to understand the FVM more clearly

We willbe considering a problem of a source free heat conduction in an insulated rod whose end are maintained at constant temperature 100 and 500 as shown in the figure.

the governing equation for this 1D problem will be

$\frac{d}{d x} (k \frac{d T}{d x}) = 0$ $d/(dx)(k(dT)/dx) = 0$

Now we have to calculate the steady state temperature distribution in the rod.

considering the thermal consuctivity (k = 1000W/mK) and the cross sectional area (A = 10^-2 sq m).

after dividing the rod into 5 equal control volumes ( $δ x = 0.1$ $delta x = 0.1$ )we will be discretising the equation.

considering nodal point 2 as our present node for the control volume be identifeid as P the point to its west can be identified as W and similarly point east as E now the corresponding boundaries can be identified as 'w' and 'e' for west and east boundary.

applying integration on our governing equation over the control volume

$\int_{Δ V} \frac{d}{d x} (k \frac{d T}{d x}) = (k A \frac{d T}{d x}) \underset{e}{} - (k A \frac{d T}{d x}) \underset{w}{} = 0$ $underset(DeltaV)(int) d/dx (k(dT)/dx) = (k A (dT)/dx) underset(e)() - (k A (dT)/dx)underset(w)() =0$

we can interpret this form of governing equation more easily, this is one of the advantage of the FVM.

the thermal conductivity at the boundary can be interpolated using the values at the nodes.

$k \underset{e}{} = \frac{k \underset{E}{} + k \underset{P}{}}{2}$ $k underset(e)() = (k underset(E)() + k underset(P)())/2$

$k \underset{w}{} = \frac{k \underset{W}{} + k \underset{P}{}}{2}$ $k underset(w)() = (k underset(W)() + k underset(P)())/2$

the heat flux terms can be written as

$(k A \frac{d T}{d x}) \underset{e}{} = k \underset{e}{} A \underset{e}{} (\frac{T \underset{E}{} - T \underset{P}{}}{δ x \underset{P E}{}})$ $(k A (dT)/dx)underset(e)() = k underset(e)() A underset(e)() ( (T underset(E)() - T underset(P)())/(delta x underset(PE)()))$

$(k A \frac{d T}{d x}) \underset{w}{} = k \underset{w}{} A \underset{w}{} (\frac{T \underset{P}{} - T \underset{W}{}}{δ x \underset{W P}{}})$ $(k A (dT)/dx)underset(w)() = k underset(w)() A underset(w)() ( (T underset(P)() - T underset(W)())/(delta x underset(WP)()))$

our governing equation changes to

$k \underset{e}{} A \underset{e}{} (\frac{T \underset{E}{} - T \underset{P}{}}{δ x \underset{P E}{}}) - k \underset{w}{} A \underset{w}{} (\frac{T \underset{P}{} - T \underset{W}{}}{δ x \underset{W P}{}}) = 0$ $k underset(e)() A underset(e)() ( (T underset(E)() - T underset(P)())/(delta x underset(PE)())) - k underset(w)() A underset(w)() ( (T underset(P)() - T underset(W)())/(delta x underset(WP)())) =0$

the above equation can be rearranged as follows:

$(\frac{k \underset{e}{}}{δ x \underset{P E}{}} A \underset{e}{} + \frac{k \underset{w}{}}{δ x \underset{W P}{}} A \underset{w}{}) . T \underset{P}{} = (\frac{k \underset{w}{}}{δ x \underset{W P}{}} A \underset{w}{}) . T \underset{W}{} + (\frac{k \underset{e}{}}{δ x \underset{P E}{}} A \underset{e}{}) . T \underset{E}{}$ $((k underset(e)())/(delta x underset(PE)()) A underset(e)() + (k underset(w)())/(delta x underset(WP)()) A underset(w)() ). T underset(P)() = ((k underset(w)())/(delta x underset(WP)()) A underset(w)()).T underset(W)() + ((k underset(e)())/(delta x underset(PE)()) A underset(e)()).T underset(E)()$

the thermal conductivity ( $k \underset{e}{} = k \underset{w}{} = k$ $k underset(e)() = k underset(w)() = k$ ), nodal spacing ( $δ x$ $delta x$ ) and the cross sectional area ( $A \underset{e}{} = A \underset{w}{} = A$ $A underset(e)() = A underset(w)() = A$ ) are constants so we can replace them with a coefficient .

$a \underset{P}{} . T \underset{P}{} = a \underset{W}{} . T \underset{W}{} + a \underset{E}{} . T \underset{E}{}$ $a underset(P)(). T underset(P)() = a underset(W)(). T underset(W)() +a underset(E)(). T underset(E)()$

where the coefficients are

$a \underset{W}{}$ $a underset(W)()$	$a \underset{E}{}$ $a underset(E)()$	$a \underset{P}{}$ $a underset(P)()$
$\frac{k}{δ x} A$ $k / (delta x) A$	$\frac{k}{δ x} A$ $k/ (delta x) A$	$a \underset{W}{} + a \underset{E}{}$ $a underset(W)() +a underset(E)()$

this above equation is the discretised form for the internal node 2 ,3 and 4 .

For the boundary Nodes 1 and 5 , we have to solve seperate for each boundary nodes.

for node 1 the integration of governing equation for the control volume gives

$k A (\frac{T \underset{E}{} - T \underset{P}{}}{δ x}) - k A (\frac{T \underset{P}{} - T \underset{A}{}}{\frac{δ x}{2}}) = 0$ $k A (( T underset(E)() - T underset(P)())/(delta x)) - kA ((T underset(P)() - T underset(A)())/ ((deltax)/2)) =0$

after rearranging

$(\frac{k}{δ x} A + \frac{2 k}{δ x} A) T \underset{P}{} = (\frac{k}{δ x} A) T \underset{E}{} + (\frac{2 k}{δ x} A) T \underset{A}{}$ $(k /(deltax) A + (2k)/(delta x) A) T underset(P)() = (k/(delta x) A )T underset(E)() + ((2k)/(delta x) A )T underset(A)()$

replacing with the coefficients we get

$a \underset{P}{} T \underset{P}{} = a \underset{W}{} T \underset{W}{} + a \underset{E}{} T \underset{E}{} + S \underset{u}{}$ $a underset(P)() T underset(P)() = a underset(W)() T underset(W)() + a underset(E)() T underset(E)() + S underset(u)()$

where

$a \underset{W}{}$ $a underset(W)$	$a \underset{E}{}$ $a underset(E)$	$a \underset{P}{}$ $a underset(P)$	$S \underset{P}{}$ $S underset(P)$	$S \underset{u}{}$ $S underset(u)$
0	$\frac{k A}{δ x}$ $(kA)/(delta x)$	$a \underset{W}{} + a \underset{E}{} + S \underset{P}{}$ $a underset(W)() + a underset(E)()+ S underset(P)()$	$- \frac{2 k A}{δ x}$ $-(2kA)/(delta x)$	$\frac{2 k A}{δ x} T \underset{A}{}$ $(2kA)/(delta x) T underset(A)()$

in the same way for node 5 we can write the equation

$k A (\frac{T \underset{B}{} - T \underset{P}{}}{\frac{δ x}{2}}) - k A (\frac{T \underset{P}{} - T \underset{W}{}}{δ x}) = 0$ $kA ((T underset(B)() - T underset(P)())/ ((deltax)/2)) - k A (( T underset(P)() - T underset(W)())/(delta x)) =0$

after rearranging

$(\frac{k}{δ x} A + \frac{2 k}{δ x} A) T \underset{P}{} = (\frac{k}{δ x} A) T \underset{W}{} + (\frac{2 k}{δ x} A) T \underset{B}{}$ $(k /(deltax) A + (2k)/(delta x) A) T underset(P)() = (k/(delta x) A )T underset(W)() + ((2k)/(delta x) A )T underset(B)()$

replacing with the coefficients we get

where

$a \underset{W}{}$ $a underset(W)$	$a \underset{E}{}$ $a underset(E)$	$a \underset{P}{}$ $a underset(P)$	$S \underset{P}{}$ $S underset(P)$	$S \underset{u}{}$ $S underset(u)$
$\frac{k A}{δ x}$ $(kA)/(delta x)$	0	$a \underset{W}{} + a \underset{E}{} + S \underset{P}{}$ $a underset(W)() + a underset(E)()+ S underset(P)()$	$- \frac{2 k A}{δ x}$ $-(2kA)/(delta x)$	$\frac{2 k A}{δ x} T \underset{A}{}$ $(2kA)/(delta x) T underset(A)()$

Now we have equations for every node and if we substitute the value of thermal conductivity , cross sectional area , nodal distance and temperature at ends A and B we get the system of algebraic equation that we can solve using MATLAB.

the set of value of various coefficients after substituitng the values becomes

node	$a \underset{W}{}$ $a underset(W)$	$a \underset{E}{}$ $a underset(E)$	$S \underset{P}{}$ $S underset(P)$	$S \underset{u}{}$ $S underset(u)$	$a \underset{P}{}$ $a underset(P)$
1	0	100	200. $T \underset{A}{}$ $T underset(A)()$	-200	300
2	100	100	0	0	200
3	100	100	0	0	200
4	100	100	0	0	200
5	100	0	200. $T \underset{B}{}$ $T underset(B)()$	-200	300

our set of algebraic equations beocmes

$300 T \underset{1}{} = 100 T \underset{2}{} + 200 T \underset{A}{}$ $300 T underset(1)() = 100 T underset(2)() + 200 T underset(A)()$

$200 T \underset{2}{} = 100 T \underset{1}{} + 100 T \underset{3}{}$ $200 T underset(2)() = 100 T underset(1)() + 100 T underset(3)()$

$200 T \underset{3}{} = 100 T \underset{2}{} + 100 T \underset{4}{}$ $200 T underset(3)() = 100 T underset(2)() + 100 T underset(4)()$

$200 T \underset{4}{} = 100 T \underset{3}{} + 100 T \underset{5}{}$ $200 T underset(4)() = 100 T underset(3)() + 100 T underset(5)()$

$300 T \underset{5}{} = 100 T \underset{4}{} + 200 T \underset{B}{}$ $300 T underset(5)() = 100 T underset(4)() + 200 T underset(B)()$

we have the values of $T \underset{A}{} = 100$ $T underset(A)() = 100$ and $T \underset{B}{} = 500$ $T underset(B)() = 500$

so after solving we the system of equation we get the temperature at each node

T1 = 140

T2 = 220

T3 = 300

T4 = 380

T5 = 460

this result can be now compared with the exact analytical solution that can be found by integrating the govrnin equation and puttting the limits .

this results are almost similar, if we find any error we can then refine our grid and try to solve the numerical solution again.

Need for Interpolation schemes :

We need interpolation schemes to find the value of properties at the boundaries of the control volumes.

If we want to find out the value of thermal conductivity at a boundary of a control volume and if it a function of temperature we will have to interpolate the values of the temperature from the neighbouring nodesto find the temperature at the boundary and that only we will be able to find the value of thermal conductivity at that boundary.

Need for Flux Limiter:

While calculations for the interpolations there might be sudden jump in the values of fluxes due to shocks in the system this can create errors and may even blow up our solutions so to lime the values of fluxes we need to provide flux limiters.