1

CHAPTER 32

32.1 First equation

Middle equations (i = 1 to 8)

Last equation

The solution is

c0 = 76.53 c5 = 29.61c1 = 63.25 c6 = 24.62c2 = 52.28 c7 = 20.69c3 = 43.22 c8 = 17.88c4 = 35.75 c9 = 16.58

32.2 Element equation: (See solution for Prob. 31.4 for derivation of element equation.)

where

Substituting the parameter values:

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2

Assembly:

Boundary conditions:

A mass balance at the inlet can be written as:

which can be solved for

Substitute into first equation

Outlet:

Therefore, the system of equations to be solved is

Solution:

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3

c0 = 74.4 c1 = 51.08 c2 = 35.15 c3 = 24.72 c4 = 20.74

32.3 According to Fick’s first law, the diffusive flux is

where J(x) = flux at position x. If c has units of g/m3, D has units of m2/d and x is measured in m, flux has units of g/m2/d. In addition, there will be an advective flux which can be calculated as

Finite divided differences can be used to approximate the derivatives. For example, for the point at the beginning of the tank, a forward difference can be used to calculate

Thus, the flux at the head of the tank is

The remainder of the values can be calculated in a similar fashion using centered (middle nodes) and backward differences (the end node):

c dc/dx J-diff J-adv J76.44 -9.588 19.176 76.44 95.61652.47 -8.076 16.152 52.47 68.62236.06 -5.484 10.968 36.06 47.02825.05 -3.394 6.788 25.05 31.83819.09 -2.384 4.768 19.09 23.858

32.4 Segmentation scheme:

Nodes 1,1 through 5,1

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4

Collecting terms gives

Node 6,1 would be modified to reflect the no flow condition in x and the Dirichlet condition at 6,0:

The nodes along the upper edge (1,2 through 5,2) would be generally written to reflect the no-flow condition in y as

The node at the upper right edge (6,2) would be generally written to reflect the no-flow condition in x and y as

Finally, the nodes along the lower edge (1,0 through 3,0) would be generally written to reflect the no-flow condition in y as

These equations can be solved for

40 5.881926 1.096415 1.216395 3.59368 4.13942 4.22014540 5.920461 1.384706 3.128728 13.49365 14.72048 14.8512340 6.017519 2.316748 12.0638 100 100 100

32.5 For simplicity, we will use a very coarse grid to solve this problem. Thus, we place nodes as in the following diagram.

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5

oil

water

v10o

v8o

v6o v6w

v4w

v2w

v0w

x

0

A simple explicit solution can be developed by substituting finite-differences for the second derivative terms in the motion equations. This is done for the three non-boundary nodes,

These three equations have 7 unknowns (v0w, v2w, v4w, v6w, v6o, v8o, v10o). The boundary conditions at the plates effectively specify v0w = 0 and v10o = 8. The former is called a “no slip” condition because it specifies that the velocity at the lower plate is zero.

The relationships at the oil-water interface can be used to eliminate two of the remaining unknowns. The first condition states that

v6o = v6w (i)

The second can be rearranged to yield

(ii)

These, along with the wall boundary conditions can be substituted into the differential equations

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6

These equations can now be integrated to determine the velocities as a function of time. Equations (i) and (ii) can be used to determine v6o and v6w. The results are plotted below:

0

4

8

0 2 4 6 8 10

t = 1.5 s

t = 0.5 s

t = 1.0 s

32.6 Using a similar approach to Sec. 32.2, the nodal equation can be developed as:

This equation can then be written for the interior nodes and solved for

0 0 0 0 0 00 -0.24461 -0.34245 -0.34245 -0.24461 00 -0.34245 -0.48922 -0.48922 -0.34245 00 -0.34245 -0.48922 -0.48922 -0.34245 00 -0.24461 -0.34245 -0.34245 -0.24461 00 0 0 0 0 0

These results can then be used as input to the right-hand side of Eq. 32.14. The nodal equation is

If this equation is written for every interior node, the resulting simultaneous equations can be solved for

0 0 0 0 0 00 0.04436 0.06914 0.06914 0.04436 00 0.06914 0.10828 0.10828 0.06914 00 0.06914 0.10828 0.10828 0.06914 0

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7

0 0.04436 0.06914 0.06914 0.04436 00 0 0 0 0 0

32.7 Grid scheme

All nodes in the above scheme can be modeled with the following general difference equation

Node 0,0:

The external nodes can be approximated with finite differences

which can be substituted into the difference equation to give

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8

Node 1,0:

Node 2,0:

Node 0,1:

Node 1,1:

Node 2,1:

Node 0,2:

Node 1,2:

Node 2,2:

The equations can be solved simultaneously for

16.3372 17.37748 18.55022 2016.29691 17.31126 18.4117 2016.22792 17.15894 17.78532

More refined results can be obtained by using finer grid spacing.

32.8 The fluxes can be determined using finite divided differences as

dh/dx1.04029 1.10651 1.31126 1.449781.01435 1.05740 1.34437 1.588300.93102 0.77870 0.62638

dh/dy

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0.04029 0.06623 0.13852 0.000000.05464 0.10927 0.38245 0.000000.06898 0.15232 0.62638

dh/dn1.04107 1.10849 1.31855 1.449781.01582 1.06303 1.39771 1.588300.93357 0.79345 0.88583

(radians)0.03871 0.05978 0.10525 0.000000.05381 0.10297 0.27716 0.000000.07396 0.19317 0.78540

(degrees)2.21777 3.42509 6.03034 0.000003.08314 5.90002 15.88014 0.000004.23765 11.06765 45.00000

Velocity-5.205E-04 -5.542E-04 -6.593E-04 -7.249E-04-5.079E-04 -5.315E-04 -6.989E-04 -7.942E-04-4.668E-04 -3.967E-04 -4.429E-04

32.9 Because of the equi-spaced grid, the domain can be modeled with simple Laplacians. The resulting solution is

25 40 40 3010 21.87149 24.04033 20 1510 13.44564 14.28983 12.63401 10 7.510 7.62124 7.039322 6.246222 5.311556 55 0 0 0 0 2.5

32.10 A convenient segmentation scheme can be developed as

Simple Laplacians reflecting the boundary conditions can be developed and solved for

99.56378 99.49429 99.13184 97.64677 91.45524 81.4052 7099.63328 99.64076 99.69315 100 93.3845 82.08277 7099.6878 99.74233 100 100 100 83.54139 70

99.63328 99.64076 99.69315 100 93.3845 82.08277 70

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10

99.56378 99.49429 99.13184 97.64677 91.45524 81.4052 70

32.11 The system to be solved is

which can be solved for x1 = 2, x2 = 3, x3 = 6, and x4 = 6.75.

32.12 The system to be solved is

which can be solved for x1 = 8, x2 = 12, x3 = 13.33333, x4 = 16, and x5 = 18.

32.13 Substituting the Crank-Nicolson finite difference analogues to the derivatives

into the governing equations gives the following finite difference equations:

Substitute for the end point boundary conditions to get the end point finite difference equations. Substitute the first order Crank Nicolson analogues to the derivatives

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11

into the midpoint boundary condition and get

where and are fictitious points located in the opposite side of the midpoint from their half. Write out the two finite difference equations from above for the point i = L (the midpoint) and then combine these two equations with the midpoint boundary condition to obtain the midpoint finite difference equation:

%PDE Parabolic Problem - Transient Heat conduction in a composite rod% u[xx]=u[t] 0<x<0.5% r(u[xx])=u[t] 0.5<x<1% BC u(0,t)=1 u(1,t)=1% u[x]=r(u[x]) x=0.5% IC u(x,0)=0 0<x<1% i=spatial index, from 1 to imax % R = no. of x points (R=21 for 20 dx spaces)% n=time index from 1 to N% N = no. of time steps,% Crank-Nicolson Formulation R=41; %(imax must be odd for point L to be correct)N=69; % last time step = nmax+1L=(R-1)/2+1; % L = midpoint of point no. (for R=41, L=21) % Constantsr=0.01;dx=1/(R-1); dx2=dx*dx;dt=dx2; % Setting dt to dx2 for good stabilility and results% Independent space variablex=0:dx:1;% Sizing matricesu=zeros(R,N+1); t=zeros(1,N+1);a=zeros(1,R); b=zeros(1,R); c=zeros(1,R); d=zeros(1,R);ba=zeros(1,R); ga=zeros(1,R);up=zeros(1,R);% Boundary Conditions at t=0u(1,1)=1;u(R,1)=1;% Time step loop% n=1 represents 0 time, next time = n+1t(1)=0;for n=1:N t(n+1)=t(n)+dt; % Boundary conditions & Constants u(1,n+1)=1; u(R,n+1)=1; dx2dt=dx2/dt; % coefficients b(2)=-2-2*dx2dt; c(2)=1; d(2)=(2-2*dx2dt)*u(2,n)-u(3,n)-2; for i=3:L-1 a(i)=1;

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12

b(i)=-2-2*dx2dt; c(i)=1; d(i)=-u(i-1,n)+(2-2*dx2dt)*u(i,n)-u(i+1,n); end a(L)=2; b(L)=-2*(1+r)-4*dx2dt; c(L)=2*r; d(L)=-2*u(L-1,n)+(2*(1+r)-4*dx2dt)*u(L,n)-2*r*u(L+1,n); for i=L+1:R-2 a(i)=r; b(i)=-2*r-2*dx2dt; c(i)=r; d(i)=-r*u(i-1,n)+(2*r-2*dx2dt)*u(i,n)-r*u(i+1,n); end a(R-1)=r; b(R-1)=-2*r-2*dx2dt; d(R-1)=-r*u(R-2,n)+(2*r-2*dx2dt)*u(R-1,n)-2*r; % Solution by Thomas Algorithm ba(2)=b(2); ga(2)=d(2)/b(2); for i=3:R-1 ba(i)=b(i)-a(i)*c(i-1)/ba(i-1); ga(i)=(d(i)-a(i)*ga(i-1))/ba(i); end % Back substitution step u(R-1,n+1)=ga(R-1); for i=R-2:-1:2 u(i,n+1)=ga(i)-c(i)*u(i+1,n+1)/ba(i); end dt=1.1*dt;end % end of time step loop% Plot% Storing plot value of u as up, at every 5 time steps% j=time index% i=space indexfor j=5:5:N+1 for i=1:R up(i)=u(i,j); end plot(x,up) hold onendgridtitle('u[xx]=u[t] 0<x<0.5; r(u[xx])=u[t] 0.5<x<1; u(0,t)=1, u(1,t)=1, u(x,0)=0; u[x]=r(u[x]) x=0.5')xlabel('x - ND Space')ylabel('u - ND Temperature')hold offgtext('r=0.01')

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13

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4u[xx]=u[t] 0<x<0.5; r(u[xx])=u[t] 0.5<x<1; u(0,t)=1, u(1,t)=1, u(x,0)=0; u[x]=r(u[x]) x=0.5

x - ND Space

u -

ND

Tem

pera

ture

r=1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4u[xx]=u[t] 0<x<0.5; r(u[xx])=u[t] 0.5<x<1; u(0,t)=1, u(1,t)=1, u(x,0)=0; u[x]=r(u[x]) x=0.5

x - ND Space

u -

ND

Tem

pera

ture

r=0.1

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14

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1u[xx]=u[t] 0<x<0.5; r(u[xx])=u[t] 0.5<x<1; u(0,t)=1, u(1,t)=1, u(x,0)=0; u[x]=r(u[x]) x=0.5

x - ND Space

u -

ND

Tem

pera

ture

r=0.01

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1u[xx]=u[t] 0<x<0.5; r(u[xx])=u[t] 0.5<x<1; u(0,t)=1, u(1,t)=1, u(x,0)=0; u[x]=r(u[x]) x=0.5

x - ND Space

u -

ND

Tem

pera

ture

r=0.001

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15

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.2

0.4

0.6

0.8

1

1.2

1.4u[xx]=u[t] 0<x<0.5; r(u[xx])=u[t] 0.5<x<1; u(0,t)=1, u(1,t)=1, u(x,0)=0; u[x]=r(u[x]) x=0.5

x - ND Space

u -

ND

Tem

pera

ture r=0

32.14% PDE Parabolic Problem – Heat conduction in a rod% u[xx]+u[yy]=u[t]% BC: u(0,y,t)=0 u(l,y,t)=l% BC: u(x,0,t)=0 u(x,l,t)=l% IC: u(x,y,0)=0 0<=x<1 0<=y<1% Crank-N1colson Formulation% ADI Solution Method% Intermediate value of u stored as ul% MATLAB version with multi-dimensional arrays% i=spatial index in x-direction, from 1 to R % j=spatial index in y-direction, from 1 to S % n=time index from 1 to NR=21; % last x-pointS=21; % last y-pointN=20; % last time step = N+1% Constantsdx=1/(R-1);dx2=dx*dx; dy=1/(S-1);dy2=dy*dy;dxdy=dx2/dy2;dydx=dy2/dx2;dt=dx2; %setting dt to dx2 vfor good stability and results% Independent space variablesx=0:dx:1;y=0:dy:1;% Sizing matricesu=zeros(R,S,N); u1=zeros(R,S);% u(1,j,n)=present time,u1=first pass intermediate results% u(i,j,n+1)=next time stept=zeros(1,N+1);a=zeros(1,R); b=zeros(1,R); c=zeros(1,R); d=zeros(1,R);ba=zeros(1,R); ga=zeros(1,R);

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16

% Boundary conditions for n=1:N for i=1:R u(i,S,n)=1; end for j=1:S u(R,j,n)=1; endend% Intermediate valuesfor i=1:R u1(i,S)=1;endfor j=1:S u1(R,j)=1;end% Plot initial conditionsmesh(x,y,u(:,:,1))title('u[xx]+u[yy]=u[t];u(0,y,t)=0,u(1,y,t)=1,u(x,0,t)=0,u(x,1,t)=1,u(x,y,0)=0')xlabel('x-coordinate');ylabel('y-coordinate');zlabel('u-Temperature'); pause% *****************************************************% Time step loop %: n=1 represents 0 time, n+1 = next time step t(1)=0;for n=1:N t(n+1)=t(n)+2*dt; % First pass in x-direction ********************************** % first time step – intermediate value at u1(i,j) are calculated % Constants dx2dt=dx2/dt; % Coefficients for j=2:S-1 b(2)=-2-dx2dt; c(2)=1; d(2)=-dxdy*u(2,j-1,n)+(2*dxdy-dx2dt)*u(2,j,n)-dxdy*u(2,j+1,n); for i=3:R-2 a(i)=1; b(i)=-2-dx2dt; c(i)=1; d(i)=-dxdy*u(i,j-1,n)+(2*dxdy-dx2dt)*u(i,j,n)-dxdy*u(i,j+1,n); end a(R-1)=1; b(R-1)=-2-dx2dt; d(R-1)=-1-dxdy*u(i,j-1,n)+(2*dxdy-dx2dt)*u(i,j,n)-dxdy*u(i,j+1,n); % Solution by Thomas algorithm ba(2)=b(2); ga(2)=d(2)/b(2); for i=3:R-1 ba(i)=b(i)-a(i)*c(i-1)/ba(i-1); ga(i)=(d(i)-a(i)*ga(i-1))/ba(i); end % Back substitution step u1(R-1,j)=ga(R-1); for i=R-2:-1:2 u1(i,j)=ga(i)-c(i)*u1(i+1,j)/ba(i); end end % second pass in y-direction ******************************** % second tme step - final values at u(i,j,n+1) are calculated dy2dt=dy2/dt;

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17

% coefficients for i=2:R-1 b(2)=-2-dy2dt; c(2)=1; d(2)=-dydx*u1(i-1,2)+(2*dydx-dy2dt)*u1(i,2)-dydx*u1(i+1,2); for j=3:S-2 a(j)=1; b(j)=-2-dy2dt; c(j)=1; d(j)=-dydx*u1(i-1,j)+(2*dydx-dy2dt)*u1(i,j)-dydx*u1(i+1,j); end a(S-1)=1; b(S-1)=-2-dy2dt; d(S-1)=-1-dydx*u1(i-1,S-1)+(2*dydx-dy2dt)*u1(i,S-1)-dydx*u1(i+1,S-1); % Solution by Thomas algorithm ba(2)=b(2); ga(2)=d(2)/b(2); for j=3:S-1 ba(j)=b(j)-a(j)*c(j-1)/ba(j-1); ga(j)=(d(j)-a(j)*ga(j-1))/ba(j); end % Back substitution step u(i,S-1,n+1)=ga(S-1); for j=S-2:-1:2 u(i,j,n+1)=ga(j)-c(j)*u(i,j+1,n+1)/ba(j); end end % dt can be incremented at this point if desired as dt=1.1*dtend % end time step loop%*****************************************************************% Plot resultsmesh(x,y,u(:,:,10))title('u[xx]+u[yy]=u[t];u(0,y,t)=0,u(1,y,t)=1,u(x,0,t)=0,u(x,1,t)=1,u(x,y,0)=0')xlabel('x-coordinate');ylabel('y-coordinate');zlabel('u-Temperature'); pauset(10)mesh(x,y,u(:,:,20))title('u[xx]+u[yy]=u[t];u(0,y,t)=0,u(1,y,t)=1,u(x,0,t)=0,u(x,1,t)=1,u(x,y,0)=0')xlabel('x-coordinate');ylabel('y-coordinate');zlabel('u-Temperature'); pauset(20)

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18

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.