200B_HW4

2. c) Evaluate the velocity field numerically with = 1, T=1

() = ||/

The Fourier Transform of the velocity above is:

() = 2

1 + ()2

We set out integration limit so the value of this function becomes very small (10-9).

2

1 + ()2= 109 = 44,721

Because we have a significant high upper integration bound, our step also needs to be refined, thus:

= 0.000001

Figure 1- Problem 2c) velocity distribution profile for various t

2. c) Evaluate the wall shear stress numerically, and plot it for = , = , = .

Figure 2-Problem 2.d) Shear stress and velocity for

3. b) Evaluate numerically the velocity profile for the plate velocity

() = (

)

2

For this velocity profile, we need to set the upper bound of integration following the same procedure

used in the lecture notes (page 61), thus:

() = (

2 )

2

= 109

=9.1

The integration step was chosen to be 0.001 for a good accuracy.

4.b) Plot the temperature solution for k=1, To=1.

Figure 3- Problem 4.b) Temperature distribution over an infinite rod

MATLAB code:

%%MAE 200B HW4 %%Laura Novoa %% Problem 2 clc clear all close all

%Solves for the velocity distribution of a semi-infinite Newtonian fluid %over an infinite plate with velocity u0(t)=exp[-(abs(t)*T]

%Kinematic viscosity T = 1; nu=1.0;

%Time and Y vectors t=T*[-5:1:5];

y=[0:0.05:10];

%Integration limit Omega=sqrt((2*T-(1e-9))/(1e-9)*T^2);

%Omega increment domega=0.000001*Omega; % since we increased integration limits by a factor of

10,

%Resulting omega vector omega=[0:domega:Omega];

%Plotting velocity profile

%================ Start Time and Y loops ============== for i=1:length(t) for j=1:length(y)

%Fourier transform: integration versus omega u(i,j)=0; for k=1:length(omega) beta=sqrt(0.5*omega(k)/nu); uo = 2*T/(1+(omega(k).*T).^2); exp1=exp(-beta*y(j)); oscillation=cos(omega(k)*t(i)-beta*y(j)); u(i,j)=u(i,j)+domega*uo*exp1*oscillation; end end end %================ End time and Y loops ==============

%Scale the answer u=u*T/(pi);

%Legend text for i=1:length(t) legtext{i}=['t / T = ', num2str(t(i)/T)]; end

%Plot colors c(1,:)=[0.0 ,0.0 ,0.0]; c(2,:)=[0.0 ,0.0 ,1.0]; c(3,:)=[0.0 ,0.7 ,0.0]; c(4,:)=[1.0 ,0.0 ,1.0]; c(5,:)=[1.0 ,0.0 ,0.0]; c(6,:)=[1.0 ,0.5, 0.0]; c(7,:)=[0.5, 1.0, 1.0]; c(8,:)=[1.0 ,0.5, 0.5]; c(9,:)=[0.5 ,1.0, 0.5];

%Line plots for different times close figure('Position',[150,150,400,500],'Color','w') subplot('Position', [0.14,0.11,0.81,0.85]) for i=1:length(t)

plot(u(i,:), y, 'linewidth', 1.5) % , 'color', c(i,:)) hold on end hold off %axis([0,1,0,max(y)]) axis 'auto' set(gca,'fontsize',14) xlabel('{\it u/U}','FontName', 'Times','FontAngle', 'normal','fontsize',18); ylabel('{\it y}','FontName', 'Times','FontAngle', 'normal','fontsize',18); legend(legtext, 1) text(0.1, 9.5, ['{\it T} = ', num2str(T)], 'FontName', 'Times',

'fontsize',20)

%% Plotting shear stress

%run the other section first %increasing time resolution for plotting t = T*[-5:0.01:5]; rho = 1; mu = nu*rho;

%================ Start Time and Y loops ============== for i=1:length(t) %Fourier transform: integration versus omega tau(i)=0; for k=1:length(omega) beta=sqrt(0.5*omega(k)/nu); uo = 2*T/(1+(omega(k).*T).^2); oscillation=sin(omega(k)*t(i))-cos(omega(k)*t(i)); tau(i)=tau(i)+domega*beta*uo*oscillation; end end %================ End time and Y loops ==============

%Scaling the solution tau = mu/pi.*tau;



for i=1:length(t) uo(i) = exp(-abs(t(i))/T); end

%Line plots for different times close figure plot(t,uo) xlabel('{\it t}','FontName', 'Times','FontAngle', 'normal','fontsize',18); ylabel('{\it u_o and \tau_w}','FontName', 'Times','FontAngle',

'normal','fontsize',18); hold on plot(t,tau) axis 'auto' set(gca,'fontsize',14) xlabel('{\it t}','FontName', 'Times','FontAngle', 'normal','fontsize',18); ylabel('{\it \tau_w}','FontName', 'Times','FontAngle',

'normal','fontsize',18); legend('u_o','\tau_w') %legend(legtext, 1) text(0.1, 9.5, ['{\it T} = ', num2str(T)], 'FontName', 'Times',

'fontsize',20)

%% PROBLEM 3

clc clear all close all

%Solves for the velocity distribution of a semi-infinite Newtonian fluid %over an infinite plate with velocity u0(t)=exp[-(abs(t)*T]

%Kinematic viscosity T=1; nu=1; h=3;

%Time and Y vectors t=T*[0:1:10]; y=[0:0.05:10];

%Integration limit Omega=9.1/T;

%Omega increment domega=0.01*Omega;

%Resulting omega vector omega=[0:domega:Omega];

%Plotting velocity profile

%================ Start Time and Y loops ============== for n=1:length(t) for j=1:length(y)

%Fourier transform: integration versus omega u(n,j)=0; for k=1:length(omega) uo = T.*sqrt(pi).*exp(-(omega(k).*T/2).^2); uo2 = T.*sqrt(pi).*exp(-(-omega(k).*T/2).^2); beta=sqrt(0.5*omega(k)/nu); a1=real(-beta*(1+i)); a2=real(beta*(-1+i)); A1 = uo/(1-exp(2*a1*h)); B1 = uo/(1-exp(-2*a1*h)); A2 = uo2/(1-exp(2*a2*h)); B2 = uo2/(1-exp(-2*a2*h)); f=(A1.*exp(a1*y(j))+B1.*exp(-

a1*y(j))).*exp(real(i.*omega(k).*t(n)))+(A2.*exp(a2*y(j))+B2.*exp(-

a2*y(j))).*exp(real(-i*omega(k)*t(n))); u(n,j)=u(n,j)+domega.*f; end end end %================ End time and Y loops ==============

%Scale the answer u=u*1/2*pi;



%Line plots for different times close figure('Position',[150,150,400,500],'Color','w') subplot('Position', [0.14,0.11,0.81,0.85]) for i=1:length(t) plot(u(i,:), y, 'linewidth', 1.5) hold on end hold off %axis([0,1,0,max(y)]) axis 'auto' set(gca,'fontsize',14) xlabel('{\it u/U}','FontName', 'Times','FontAngle', 'normal','fontsize',18); ylabel('{\it y}','FontName', 'Times','FontAngle', 'normal','fontsize',18); legend(legtext, 1)

text(0.1, 9.5, ['{\it T} = ', num2str(T)], 'FontName', 'Times',

'fontsize',20)

%% PROBLEM 4

clc clear all close all

%Solves for the temperature distribution with initial temperature the %scaled Heaviside function

%Kinematic viscosity k=1.0; To=1;

%Time and Y vectors [X,T] = meshgrid(-2:0.05:2,0:.05:5);

%Function U=(To/2).*(1-erf(-X./sqrt(4.*k.*T))); %U=(To/2).*(erfc(-X./sqrt(4.*k.*T)));

surf(X,T,U) xlabel('x') ylabel('t') zlabel('u')

200B_HW4

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