Plotting a function given a range of inputs

I need to plot a line of the Moody chart given a relative roughness and range of Reynolds numbers but I have no clue how to write a plot code. For simplicity, lets say I want to plot y=8x+10 with a range of 2<x<20. How would I go about doing this? Thank you.

Answer 1

[Elias Gule]

First define your vector x as:

 dx = 0.01;
 x = 2:dx:20; % where dx in the increment from one x value to the other.

OR

N = 40; % number of samples
x = linspace(2,20,N);

then define y as:

y = 8*x + 10;

Then to plot:

plot(x,y);

For more help please look at the VERY USEFUL Matlab documentation. In the command window just type:

doc plot

[Vatsal Dhamelia]

clear;

clc;

pi = acos(-1);

Um = 1.1; % Fluid Velicity amplitude

D = 0.11; % cylinder diamter

L = 1; % cylinder length

m = 50; % cylinder mass

rho = 1024; % fluid density

K = 200; % spring stiffness

c = 100; % damping constant

CA = 1; % added mass coefficient

CD = 1.8; % drag coefficient

KC = 2:2:20;%KC number

T = KC*D/Um ; % period of oscillatory flow ;

omega = 2*pi/T; % angular frequency of the flow

md = rho*pi*(D*D)/4*L; % added mass coefficient

Ap = D*L; % projected area

dt = T/40; % time step

ndt = T/dt*5; % %total step to be calculated

X(1:ndt+1) = 0; % save displacement X from step 0 to step ndt

V(1:ndt+1) = 0;

Pa(1:ndt+1) = 0;

time(1:ndt+1) = (0:ndt)*dt;

for n=1:ndt

% to calculate kx1 and kv1

ta = time(n);

aw = (omega *Um*cos(omega *ta))-((2*omega*Um/3)*cos(2*omega*ta)); % acceleration of the fluid

Vw = (Um*sin(omega*ta))-((Um/3)*sin(2*omega*ta)); % velocity of the fluid

Va = V(n);

Xa = X(n);

t1 = CA*md*aw;

t2 = 0.5*rho*CD*Ap*abs(Vw-Va)*(Vw-Va);

t3 = -K*Xa-c*Va;

kx1 = V(n);

kv1 = (t1+t2+t3)/(m+CA*md);

% to calculate kx2 and kv2

ta = time(n)+dt/2;

aw = (omega *Um*cos(omega *ta))-((2*omega *Um/3)*cos(2*omega*ta)); % acceleration of the fluid

Vw = Um*sin(omega*ta)-(Um/3)*sin(2*omega*ta); % velocity of the fluid

Va = V(n)+0.5*dt*kv1;

Xa = X(n)+0.5*dt*kx1;

t1 = CA*md*aw;

t2 = 0.5*rho*CD*Ap*abs(Vw-Va)*(Vw-Va);

t3 = -K*Xa-c*Va;

kx2 = V(n)+0.5*dt*kv1;

kv2 = (t1+t2+t3)/(m+CA*md);

% to calculate kx3 and kv3

ta = time(n)+dt/2;

aw = (omega *Um*cos(omega *ta))-((2*omega *Um/3)*cos(2*omega*ta)); % acceleration of the fluid

Vw = Um*sin(omega*ta)-(Um/3)*sin(2*omega*ta); % velocity of the fluid

Va = V(n)+0.5*dt*kv2;

Xa = X(n)+0.5*dt*kx2;

t1 = CA*md*aw;

t2 = 0.5*rho*CD*Ap*abs(Vw-Va)*(Vw-Va);

t3 = -K*Xa-c*Va;

kx3 = V(n)+0.5*dt*kv2;

kv3 = (t1+t2+t3)/(m+CA*md);

% to calculate kx4 and kv4

ta = time(n)+dt;

aw = (omega *Um*cos(omega *ta))-((2*omega *Um/3)*cos(2*omega*ta)); % acceleration of the fluid

Vw = Um*sin(omega*ta)-(Um/3)*sin(2*omega*ta); % velocity of the fluid

Va = V(n)+0.5*dt*kv3;

Xa = X(n)+0.5*dt*kx3;

t1 = CA*md*aw;

t2 = 0.5*rho*CD*Ap*abs(Vw-Va)*(Vw-Va);

t3 = -K*Xa-c*Va;

kx4 = V(n)+dt*kv3;

kv4 = (t1+t2+t3)/(m+CA*md);

% next step values

X(n+1) = X(n)+(dt/6)*(kx1+2*kx2+2*kx3+kx4);

V(n+1) = V(n)+(dt/6)*(kv1+2*kv2+2*kv3+kv4);

Power = (CA*md*aw+0.5*rho*CD*Ap*abs(Vw-V(n+1))*(Vw-V(n+1)))* (V(n+1));

Pa(n+1) = Power;

end

%---------------------------------------------------------------------

% the code below is to use the Euler method

% Xe and Ve are the displacement and the velocity from the Euler Method

Xe(1:ndt+1) = 0; % save displacement X from step 0 to step ndt

Ve(1:ndt+1) = 0;

PaE(1:ndt+1) = 0;

for n=1:ndt

% to calculate kx1 and kv1

ta = time(n);

aw = (omega *Um*cos(omega *ta))-((2*omega *Um/3)*cos(2*omega*ta)); % acceleration of the fluid

Vw = Um*sin(omega*ta)-(Um/3)*sin(2*omega*ta); % velocity of the fluid

t1 = CA*md*aw;

t2 = 0.5*rho*CD*Ap*abs(Vw-Ve(n))*(Vw-Ve(n));

t3 = -K*Xe(n)-c*Ve(n);

Xe(n+1) = Xe(n)+dt*Ve(n);

Ve(n+1) = Ve(n)+dt*(t1+t2+t3)/(m+CA*md);

PowerE =(CA*md*aw+0.5*rho*CD*Ap*abs(Vw-Ve(n+1))*(Vw-Ve(n+1)))* (Ve(n+1));

PaE(n+1) = PowerE;

end

%average of power

Pam(1:ndt+1) = 0;

PamE(1:ndt+1) = 0;

for n = 1:ndt

Pam(n) = mean (Pa);

PamE(n) = mean (PaE);

end

% write the results in to a file output.txt

fileID = fopen('output.txt','w');

for n=1:ndt+1

fprintf(fileID,'%12.6e %12.6e %12.6e %12.6e %12.6e\r\n', ...

time(n),X(n),V(n),Xe(n),Ve(n));

end

fclose(fileID);

% draw the displacement of the cylinder

figure (01);

plot(time(1:ndt),X(1:ndt),'-r');

xlabel('time (s)'); %honrizontal axis's label is 'x'

ylabel('X (m)'); %vertical axis's label is 'T'

hold;

plot(time(1:ndt),Xe(1:ndt),'-b');

legend ('RK method', 'Euler method');

% draw the velocity of the cylinder

figure (02)

plot(time(1:ndt),V(1:ndt),'-r');

xlabel('time (s)'); %honrizontal axis's label is 'x'

ylabel('V (m/s)'); %vertical axis's label is 'T'

hold;

plot(time(1:ndt),Ve(1:ndt),'-b');

legend ('RK method', 'Euler method');

figure (03)

plot(time(1:ndt),Pa(1:ndt),'-r');

xlabel('time (s)'); %honrizontal axis's label is 'x'

ylabel('Power (W)'); %vertical axis's label is 'T'

hold;

plot(time(1:ndt),PaE(1:ndt),'-b');

legend ('RK method', 'Euler method');

figure (04)

plot(time(1:ndt),Pam(1:ndt),'-r');

xlabel('time (s)'); %honrizontal axis's label is 'x'

ylabel('power(W)'); %vertical axis's label is 'T'

hold;

plot(time(1:ndt),PamE(1:ndt),'-b');

legend ('RK method', 'Euler method');

[Vatsal Dhamelia]

can you please help me fo plot KC vs Pam

Answer 2

[Star Strider]

One possibility:

x = linspace(2, 20, 25);
y = 8*x + 10;

figure(1)
plot(x, y, '-pg')
grid
xlabel('x')
ylabel('y')
title('Moody Chart')

See the documentation for the various functions for details on their uses.