I am trying to use the trapezoidal rule to compute the flow rate of fluid through a pipe.

clear

% Velocity distribution function of the pipe
v = 3.9*(1-r./r0).^n;

% Area = pi*r^2
% dA = 2*pi*r dr
% Q = V dA dr

% Value of exponent in velocity distribution function
n = 1/5; 

% Value of radius in the pipe (cm)
r0 = 6; 

% Range over which the integral is to be performed
r = [0,6];

% Now using the trapz method
Q1 = 2.0*pi*trapz(r,v)

It spits out an error after the initialization of the velocity distribution function saying it doesn't recognize the variable 'r', but I don't understand because I stated r as a variable below that.

[VBBV]

Define equation below vector r

[VBBV]

clear
% Velocity distribution function of the pipe
% Area = pi*r^2
% dA = 2*pi*r dr
% Q = V dA dr
% Value of exponent in velocity distribution function
n = 1/5; 
% Value of radius in the pipe (cm)
r0 = 6; 
% Range over which the integral is to be performed
r = [0,6];
v = 3.9*(1-r./r0).^n;
% Now using the trapz method
Q1 = 2.0*pi*trapz(r,v)

Answer 1

[Alan Stevens]

I think it needs to be more like this:

% Velocity distribution function of the pipe
% Area = pi*r^2
% dA = 2*pi*r dr
% Q = V dA dr
% Value of exponent in velocity distribution function
n = 1/5; 
% Value of radius in the pipe (cm)
r0 = 6; 
% Range over which the integral is to be performed
dr = r0/1000;
r = 0:dr:r0;
v = 3.9*(1-r./r0).^n;
% Now using the trapz method
Q1 = 2.0*pi*trapz(v)*dr

Q1 = 122.5092

% True value
vfn = @(r) 3.9*(1-r/r0).^n;
Q2 = 2*pi*integral(vfn,0,r0)

Q2 = 122.5221

[Torsten]

To get the volume flow rate, the correct integral should be

Vdot = 2*pi*integral_{r=0}^{r=r0} r*v(r ) dr

not

Vdot = 2*pi*integral_{r=0}^{r=r0} v(r ) dr

You can already see this when you consider the unit of Vdot, namely m^3/s (the last integral has m^2/s as unit).