Hi
My code below solves colebrook eqn for friction factor using newton raphson method. The issue is that I dont know how to manipulate an input variable (e.g Rynolds no.) to see the resulting f factor ( x in this code）.
My function;
function nr(y,x0,n,varargin)
x = x0
f = matlabFunction(y)
y1= diff(y)
f1= matlabFunction(y1)
for c = 1:n
x = x - f(x)/f1(x)
k = [x f(x)];
disp(k)
end
my script
format long
Re = 10000;
y = (x.^(-1/2))+(0.86*log(((0.0001)/3.7)+((2.51/Re)*x.^(-1/2))));
nr(y,0.01,10);
If for example I input Re = linspace(10000,60000,10) I get an error "matrix dimension must agree"
Additionally, I would like to make a matrix with each Re and final friction factor value (x)
Current output
>> samplescribt
x =
0.010000000000000
f =
function_handle with value:
@(x)log(1.0./sqrt(x).*2.51e-4+2.702702702702703e-5).*(4.3e1./5.0e1)+1.0./sqrt(x)
y1 =
- 1/(2*x^(3/2)) - 199095708787547171/(1844674407370955161600*x^(3/2)*(4630132762501097/(18446744073709551616*x^(1/2)) + 1/37000))
f1 =
function_handle with value:
@(x)1.0./x.^(3.0./2.0).*(-1.0./2.0)-(1.0./x.^(3.0./2.0).*1.0793e-4)./(1.0./sqrt(x).*2.51e-4+2.702702702702703e-5)
x =
0.018957804377517
0.018957804377517 1.851220793655044
x =
0.027612285982097
0.027612285982097 0.447238114020280
x =
0.031211167390416
0.031211167390416 0.037929950196298
x =
0.031575183433487
0.031575183433487 0.000314699305965
x =
0.031578254339019
0.031578254339019 0.000000021974257
x =
0.031578254553479
0.031578254553479 0.000000000000002
x =
0.031578254553479
0.031578254553479 0
x =
0.031578254553479
0.031578254553479 0
x =
0.031578254553479
0.031578254553479 0
x =
0.031578254553479
0.031578254553479 0
>>
Thanks
