AM5360 ASSIGNMENT 2 UNSTEADY HEAT CONDUCTION : BTCS SCHEME Shameem Shereef NA13B040 Problem Definition A bar of length 1 is heated such that on end of bar has temperature of 1 unit and other end 0...

AM5360
ASSIGNMENT 2
UNSTEADY HEAT CONDUCTION : BTCS SCHEME

Shameem Shereef

NA13B040

Problem Definition
A bar of length 1 is heated such that on end of bar has temperature of 1 unit and other end 0 units. Initially , at t=0 the entire bar has temperature 0 units.

The Governing PDE is :

Considering alpha =1unit
Unsteady Heat Conduction is an example for parabolic PDE. It is an Initial Boundary Value problem,which needs an initial state as well as boundary conditions for all time.
Initial Condition
At t=0: T=0 for 0= x = 1 T(x=0)=1 T(x=1)=0

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Boundary Condition :
T(x=0,t)=1 T(x=1,t) =0 ?t>0
Discretisation
BTCS (Backward-Time-Centered-Space) scheme in which a first-order approximation is used for the time derivative, but a second order one for the spatial one. Using the a finite-difference notation, the BTCS scheme is then expressed as

The discretization error is 1st order in time and second order in space
A von Neumann stability analysis shows that BTCS Scheme is unconditionally stable
.BTCS is an implicit scheme

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Where a	= c =- a	*	t/(?x)2	b	= 2a	*	t/(?x)2	+ 1
i	i	i

where matrix Ais a tridiagonal matrix. Thomas algorithm is the Gaussian elimination algorithm
tailored to solve this type of sparse system.

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Total no of multiplications and divisions in Thomas algorithm is 5n-8 which is significantly smaller than the n³/3 operations (approximately) necessary for carrying out the Gaussian elimination and backward sweep for a dense matrix.
Thomas algoritm is a direct method for solving system of linear eqns.

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Results

1. t =0.1 s ,

time _implicit
Nx	(in s)	??t/( x)^2
11	XXXXXXXXXX	10
21	XXXXXXXXXX	40
31	XXXXXXXXXX	90
41	XXXXXXXXXX	160
51	XXXXXXXXXX	250
101	XXXXXXXXXX	1000

We can observe that increasring no of nodes produces only very small increase in computationial time . The computational Time is of order of 10 ms, in this case

1. t =0.01 s ,

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Nx	time _implicit (in s)	??t/( x)^2
11	XXXXXXXXXX	1
21	XXXXXXXXXX	4
31	XXXXXXXXXX	9
41	XXXXXXXXXX	16
51	XXXXXXXXXX	25
101	XXXXXXXXXX	100

The computational Time is of order of 100 ms , in this case 3 . t =0.01 s ,

time _implicit (BTCS)	time _explicit (FTCS)
Nx	(in s)	(in s)	??t/( x)^2

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AM5360	NA13B040
11	XXXXXXXXXX	0.7443	0.1
21	XXXXXXXXXX	XXXXXXXXXX	0.4
31	0.9047	unstable	0.9
41	XXXXXXXXXX	unstable	1.6
51	XXXXXXXXXX	unstable	2.5
101	XXXXXXXXXX	unstable	10

FTCS Scheme is observed to be unstable for ??t/( x)^2 = 0.9, 1.6,2.5,10 as expected from Von Neumann Stability analysis .whereas BTCS schme yield a stable solution for all values of ??t/( x)^2 , prooving, it is unconditionally stable.
FTCS computational time is almost 10 time grater than that of BTCS scheme for Nx=21,

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4. Computational Time decreases according to a power law with delta t
Computational Time (for BTCS)= XXXXXXXXXX * t^ XXXXXXXXXX

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Appendix : Matlab Code

• AM5360 Assignment 2 : Na13b040

• This code can be used to solve the temperature profile upto t=20 s, using both FTCS and BTCS Scheme %The Computational Time upto t=20s is calcuated

dt=input('Enter? dt:');?

Nx=input('Enter? Nx ,the no of gridpoints : ');?

k=input('Enter? 1 for implicit (BTCS),else 0 for explicit(FTCS)scheme : ');?

alpha=1;
L=1;
dx=L/(Nx-1);
gamma=alpha*dt/(dx)^2

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if?k==1
%implicit method (BTCS Scheme) tic
t=0;
T_old=zeros(Nx,1); %Initial? Condition while?t
t=t+dt; T_new=zeros(Nx,1); %Boundary Conditions T_new(1)=1; T_new(Nx)=0;
% setting TDM : a_n *T_n+1+ b_n * T_n+1 + c_n * T_n+1 = d_n a=-1*gamma*ones(1,Nx-3);
c=a;
b=(1+2*gamma)*ones(1,Nx-2); %? dominant diagonal d=T_old(2:Nx-1);

d(1)=d(1)+gamma*T_new(1); d(Nx-2)=d(Nx-2)+gamma*T_new(Nx);
%Solving the TDM using Thomas Algorithm :
% Triangulation : Forward sweep with Normalization c(1)=c(1)/b(1);
d(1)=d(1)/b(1); for?i=2:Nx-3
c(i)=c(i)/(b(i)-a(i-1)*c(i-1));

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end for?i=2:Nx-2
d(i)=(d(i)-a(i-1)*d(i-1))/(b(i)-a(i-1)*c(i-1));
end
% Backward Sweep T_new(Nx-1)=d(Nx-2); for?i=Nx-3:-1:1
T_new(i+1)=d(i)-c(i)*T_new(i+2);
end T_old=T_new;
end
toc
else
%Explicit method FTCS tic
t=0;
T_old=zeros(Nx,1); %Initial? Condition while?t
t=t+dt; T_new=zeros(Nx,1); %Boundary Condition T_new(Nx)= 0; T_new(1) = 1;
% Time Marching for?i= 2:Nx-1
T_new(i)= (1-2*gamma)*T_old(i)+gamma*(T_old(i-1)+T_old(i+1));

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end
T_old=T_new;
end
toc
end
Published with MATLAB® R2014a
References

• http://nptel.ac.in/courses/ XXXXXXXXXX/downloads/sections/module4/section4.pdf

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