KTH ROYAL INSTITUTE OF TECHNOLOGY

School of Engineering Sciences

Course: Applied Numerical Methods

Course code: SF2520

Authors: 1) Andreas Angelou

2) Vasileios Papadimitriou

3) Dionysios Zelios

Lab name: Model of a Heat Exchanger

Date of submission: 13/01/2014

CONTENTS

Part A ..3

Part B4 Part C5 Part D..6 Part E...8 Appendix..10

References.11

Model of a Heat Exchanger

In this exercise a hyperbolic partial differential equation modeling a heat exchanger is

studied. A fluid of temperature T(x,t) is flowing with constant speed u in a pipe. Outside the

pipe there is a cooling medium that keeps a constant low temperature Tcool. The

temperature of the fluid in the pipe is initially cool, i.e., T=Tcool but within a short time

period hot fluid enters the pipe. The task is to study how the temperature T(x,t) of the fluid

in the pipe depends on x and t.

The following PDE model is given:

1( ) 0cool

T Tk T T

x u t

, 0 x L

with the initial condition ( ,0) coolT x T and the boundary condition

(0, )

( )sin( ) ,0 0.5

,0.5 4

sin( ( 4)) , 4

hot

T t hot

hot

Tcool T Tcool t t

t

T Tcool t t

The length of the heat exchanger is L=3. The heat exchanger parameter is k=0.1, the velocity

u=1, the cooling temperature Tcool=5 and the hot temperature is Thot=100.

Part A

We want to show that our problem can be scaled to ( ) 0u u

u

, 0 1 .

Starting from our initial differential equation and using the following change of variables

5T u , 3x L , 3t , we have:

(5 ) (5 )0.1(5 5) 0

(3 ) (3 )

u uu

1 1

0.1( 1) 03 3

u uu

0.3( 1) 0u u

u

, with 0 1

The initial condition is then transformed: ( ,0) 1u , 0 and the boundary condition

becomes: (0, )

100 51 ( )sin( )

,0 0.55

100 / 5 ,0.5 4

sin( ( 4)) , 4

u

hotT Tcool

(0, )

1 19sin( ) ,0 0.5

20 ,0.5 4

20 sin( ( 4) , 4

u

Part B

We discretize the scaled version of the problem initially with the method of lines (MoL). We

introduce grid points i=ih, i=0,1,n, n=1. We approximate u

at the point i with 1i iu u

dx

.

1 0.3 ( 1)i i iidx

u u uu

1

1(1 0.3 ) ] 0.3i

i idx

dx

uu u

The initial condition is u(0)=1 and the boundary condition is

(0, )

1 19sin(3 ) ,0 0.5

20 ,0.5 4

20 sin( ( 4) , 4

u

The PDE problem it then approximated by a system of ODEs: ( )du

Au bd

,

where A is

(1 0.3 ) 0 0 0 0

1 (1 0.3 ) 0 0 0 0

0 1 (1 0.3 ) 0 0 01*

0 0 1 0 0

0 0

0 0 0 1 (1 0.3 )

dx

dx

dxA

dx

dx

b is

0.3 ( )

0.3

( ) 0.3

0.3

b

and u is

1

1

(0) 1

1

u

We already know that the theoretical stability result based on eigenvalue analysis as

presented for ODE systems in chapter 3 of the book is not correct for the hyperbolic PDE

model problem. The reason for this is that the eigenvalues of A are all equal, which implies

an initial polynomial growth of the solution of the difference equation.

We therefore use von Neumann analysis based on Fourier analysis to decide if our ODE is

stable or not.

We conclude that the time-stepsize should be approximately 0.2 times the space-stepsize in

order to have a stable system.

Part C

We solve the ODE system in part B with the explicit Euler method, constant time step t.

We investigate numerically how the largest constant time step tmax, preserving the stability

of the numerical solution, depends on the stepsize h defined in part B.

We find tmax for N=10,20 and 40.

N tmax

10 0.002

20 0.0008

40 0.0004

Solving the PDE system in part B, we expect to have a hyperbolic graph which shows that the

temperature is always increasing, since there is no boundary condition at the end of the pipe

that affects it.

We also show graphically the unstable numerical solution for the case N=20, when

tmax< t=0.05.

.

Part D

We work out a difference formula of Lax-Wendroff type that solves the PDE problem in part

A.

Our initial scaled differential equation is 0.3( 1) 0u u

u

(1)

0.3( 1)u u

u

(2)

2 2 2

2 20.3 0.6 0.09( 1)

d u u u d u d u duu

d d dt d d

(3)

The Taylor expansion at the point (i,k):

2 21 3

2( )

2

k kk k i t ii i t t

u h uu u h O h

d

By developing our PDE (1) in terms of Taylor expansion and using (2) and (3) equations, we

have:

2 21 ( 0.3[ 1]) [ 0.6 0.09( 1)]

2

k k kk k k ki t i ii i t i i

u h u uu u h u u

(4)

Using the central difference approximations:

1 1

2

k k k

i i i

x

u u u

h

(6)

2

1 1

2 2

2k k k ki i i i

x

u u u u

h

(5)

We have: 2 2 2 2 2 2 2

1

1 12 2 2

0.09 0.6 0.6 0.09(1 0.3 ) ( ) ( ) 0.3

2 2 2 4 2 2 4 2

k k k kt t t t t t t t ti i t i i t

x x x x x x x

h h h h h h h h hu u h u u h

h h h h h h h

Hence:

2 2

1 2

2 2

2 2

2 2

3 2

2

4

0.6

2 2 4

0.091 0.3

2

0.6

2 2 4

0.090.3

2

t t t

x x x

t tt

x

t t t

x x x

tt

h h hc

h h h

h hc h

h

h h hc

h h h

hc h

Part E

We program the Lax-Wendroff formula in part D as a Matlab program and we run it for

h=0.1 and t=0.002 and we present the result in a 3D plot.

Appendix % Matlab code, Lab 6 % Andreas Angelou, Vasileios Papadimitriou, Dionysios Zelios

clear clc close all N = 10; %total spacesteps dx = 1/N; %space stepsize T=10; %total time (seconds) dt = 0.01; %time stepsize A = zeros(T/dt,N); %setting up matrix A A(1,1) = -(1/dx)-0.3; for i = 2:N A(i,i) = -(1/dx)-0.3; A(i,i-1) = (1/dx); end for k=1:N u0(k)=1; end U=zeros(T/dt,N); %setting the solution matrix U(1,:)=u0; b = zeros(T/dt,1); %setting the boundary matrix a =1; b(1) =0.3+a; %initiating the boundary matrix for t = 2:(T/dt) b(t) = 0.3; end for t = 1:(T/dt) %working on the inner points for i = 1:N if t*dt < (0.5/3) && t*dt > 0 a = 1+ 19*sin(3*pi*t); else if t*dt >(4/3) a = 20 + sin(pi*(3*t-4)); else a=20; end end b(1) = 0.3+a; U(t+1,i) = U(t,i)+dt*(A(t,i)*U(t,i)+b(t)); end end mesh(U) xlabel('Space(scaled)') ylabel('Time(scaled)') zlabel('Temperature(scaled)') title('Unstable solution')

% Lax Wendroff clear clc close all N = 10; %total spacesteps dx = 1/N; %space stepsize T=10; %total time (seconds) dt = 0.002; %time stepsize for k=1:N

u0(k)=1; end U=zeros(T/dt,N); %setting the solution matrix U(1,:)=u0; b = zeros(T/dt,1); %setting the boundary matrix a =1; b(1) =a; %initiating the boundary matrix for t = 1:(T/dt) %working on the inner points for i=2:N-1 if t*dt < (0.5/3) && t*dt > 0 a = 1+ 19*sin(3*pi*t); else if t*dt >(4/3) a = 20 + sin(pi*(3*t-4)); else a=20; end end b(1) = a; U(t+1,i) = U(t,i-1)*((dt/(2*dx))-((dt^2)/(2*(dx)^2))-

(0.6*(dt)^2/(4*dx)))+U(t,i)*(1-

(0.3*dt)+((dt)^2/(dx)^2)+(0.09*(dt)^2/2))+U(t,i+1)*(-dt/(2*dx)-

((dt)^2/(2*(dx)^2))+(0.6*(dt)^2/(4*dx)))+0.3*dt-(0.09*(dt)^2)/2; end % U(t+1,N) = 2*U(t,N-1)*((dt/(2*dx))-((dt^2)/(2*(dx)^2))-

(0.6*(dt)^2/(4*dx)))+U(t,N-2)*(1-

(0.3*dt)+((dt)^2/(dx)^2)+(0.09*(dt)^2/2))+0.3*dt-(0.09*(dt)^2)/2; U(t,N) = 2*U(t,N-1)-U(t,N-2); end mesh(U) xlabel('Spacesteps') ylabel('Timesteps') zlabel('Temperature(/5)')

References

Introduction to computation and modeling for differential equations,

Lennart Edsberg