Trying to solve for roots

I am trying to solve for the velocity in this problem. I have attached a photo of my work on paper, but I was trying to set it up in matlab using bisection method, but it calls error when I try to input the large term the includes velocity. I have attached my work on paper and my code. One other thing is my work on paper, in the denominator there should be two parentheses on the right, signifying that the entire denominator is being squared. When I run it, it says my parenthesis are wrong, but I thought I did them correctly. Please help.

% Bisection Method Code #2

max_error=.001;
itr_max=100;

g=9.8
L=80
D=.1
Km=6.5
Re=4826       %coefficient of v in reynolds number
e=.00015

f = @(x) .625-(x^2/(2*g))*(Km+(L/D)*1/(-1.8log((e/(D*3.7))^1.11+6.9/(Re*x)))^2)        ; %function

xl=0;   %lower bound
xu=100;   %upper bound
xr0=0;  %initial guess

for i = 1:itr_max
    fxl=f(xl);
    fxu=f(xu);
    
    xr=(xl+xu)/2;
    
    fxr=f(xr);
    
    err=abs((xr-xr0)/xr)*100;
    ferr=abs((fxr-f(xr0))/fxr)*100;
    xr0=xr;
    
    fprintf('i=%d \t xl=%f \t xu=%f \t fxl=%f \t fxu=%f \t xr=%f \t fxr=%f \t er=%f\n',i,xl,xu,fxl,fxu,xr,fxr,err)
   
    if err < max_error
        break;
    end    
    
    test1=fxr*fxl;
    test2=fxr*fxu;
    
    if test1<0
        xu=xr;
    elseif test2<0
        xl=xr;
    end
end        

Answer 1

[Cris LaPierre]

MATLAB does not do implied multiplicaton. You are getting the error in your first image because you have

f = @(x) .625-(x^2/(2*g))*(Km+(L/D)*1/(-1.8log((e/(D*3.7))^1.11+6.9/(Re*x)))^2); %function

% ^^ implied multiplication

Specify the operation to take place between -1.8 and log to fix this error.

f = @(x) .625-(x^2/(2*g))*(Km+(L/D)*1/(-1.8*log((e/(D*3.7))^1.11+6.9/(Re*x)))^2); %function

% ^

Your code runs without error once that change is made. Just be aware that, in MATLAB, log is the natural log, or ln. If you want log base 10, that is log10.

% Bisection Method Code #2

max_error=.001;
itr_max=100;

g=9.8;
L=80;
D=.1;
Km=6.5;
Re=4826;       %coefficient of v in reynolds number
e=.00015;

f = @(x) .625-(x^2/(2*g))*(Km+(L/D)*1/(-1.8*log((e/(D*3.7))^1.11+6.9/(Re*x)))^2); %function

xl=0;   %lower bound
xu=100;   %upper bound
xr0=0;  %initial guess

for i = 1:itr_max
    fxl=f(xl);
    fxu=f(xu);
    
    xr=(xl+xu)/2;
    
    fxr=f(xr);
    
    err=abs((xr-xr0)/xr)*100;
    ferr=abs((fxr-f(xr0))/fxr)*100;
    xr0=xr;
    
    fprintf('i=%d \t xl=%f \t xu=%f \t fxl=%f \t fxu=%f \t xr=%f \t fxr=%f \t er=%f\n',i,xl,xu,fxl,fxu,xr,fxr,err)
   
    if err < max_error
        break;
    end    
    
    test1=fxr*fxl;
    test2=fxr*fxu;
    
    if test1<0
        xu=xr;
    elseif test2<0
        xl=xr;
    end
end        

i=1 	 xl=0.000000 	 xu=100.000000 	 fxl=0.625000 	 fxu=-5023.050705 	 xr=50.000000 	 fxr=-1262.750501 	 er=100.000000
i=2 	 xl=0.000000 	 xu=50.000000 	 fxl=0.625000 	 fxu=-1262.750501 	 xr=25.000000 	 fxr=-318.703577 	 er=100.000000
i=3 	 xl=0.000000 	 xu=25.000000 	 fxl=0.625000 	 fxu=-318.703577 	 xr=12.500000 	 fxr=-80.759644 	 er=100.000000
i=4 	 xl=0.000000 	 xu=12.500000 	 fxl=0.625000 	 fxu=-80.759644 	 xr=6.250000 	 fxr=-20.370447 	 er=100.000000
i=5 	 xl=0.000000 	 xu=6.250000 	 fxl=0.625000 	 fxu=-20.370447 	 xr=3.125000 	 fxr=-4.877890 	 er=100.000000
i=6 	 xl=0.000000 	 xu=3.125000 	 fxl=0.625000 	 fxu=-4.877890 	 xr=1.562500 	 fxr=-0.845006 	 er=100.000000
i=7 	 xl=0.000000 	 xu=1.562500 	 fxl=0.625000 	 fxu=-0.845006 	 xr=0.781250 	 fxr=0.223444 	 er=100.000000
i=8 	 xl=0.781250 	 xu=1.562500 	 fxl=0.223444 	 fxu=-0.845006 	 xr=1.171875 	 fxr=-0.230346 	 er=33.333333
i=9 	 xl=0.781250 	 xu=1.171875 	 fxl=0.223444 	 fxu=-0.230346 	 xr=0.976562 	 fxr=0.016848 	 er=20.000000
i=10 	 xl=0.976562 	 xu=1.171875 	 fxl=0.016848 	 fxu=-0.230346 	 xr=1.074219 	 fxr=-0.101702 	 er=9.090909
i=11 	 xl=0.976562 	 xu=1.074219 	 fxl=0.016848 	 fxu=-0.101702 	 xr=1.025391 	 fxr=-0.041163 	 er=4.761905
i=12 	 xl=0.976562 	 xu=1.025391 	 fxl=0.016848 	 fxu=-0.041163 	 xr=1.000977 	 fxr=-0.011841 	 er=2.439024
i=13 	 xl=0.976562 	 xu=1.000977 	 fxl=0.016848 	 fxu=-0.011841 	 xr=0.988770 	 fxr=0.002583 	 er=1.234568
i=14 	 xl=0.988770 	 xu=1.000977 	 fxl=0.002583 	 fxu=-0.011841 	 xr=0.994873 	 fxr=-0.004609 	 er=0.613497
i=15 	 xl=0.988770 	 xu=0.994873 	 fxl=0.002583 	 fxu=-0.004609 	 xr=0.991821 	 fxr=-0.001008 	 er=0.307692
i=16 	 xl=0.988770 	 xu=0.991821 	 fxl=0.002583 	 fxu=-0.001008 	 xr=0.990295 	 fxr=0.000788 	 er=0.154083
i=17 	 xl=0.990295 	 xu=0.991821 	 fxl=0.000788 	 fxu=-0.001008 	 xr=0.991058 	 fxr=-0.000110 	 er=0.076982
i=18 	 xl=0.990295 	 xu=0.991058 	 fxl=0.000788 	 fxu=-0.000110 	 xr=0.990677 	 fxr=0.000339 	 er=0.038506
i=19 	 xl=0.990677 	 xu=0.991058 	 fxl=0.000339 	 fxu=-0.000110 	 xr=0.990868 	 fxr=0.000115 	 er=0.019249
i=20 	 xl=0.990868 	 xu=0.991058 	 fxl=0.000115 	 fxu=-0.000110 	 xr=0.990963 	 fxr=0.000003 	 er=0.009624
i=21 	 xl=0.990963 	 xu=0.991058 	 fxl=0.000003 	 fxu=-0.000110 	 xr=0.991011 	 fxr=-0.000053 	 er=0.004812
i=22 	 xl=0.990963 	 xu=0.991011 	 fxl=0.000003 	 fxu=-0.000053 	 xr=0.990987 	 fxr=-0.000025 	 er=0.002406
i=23 	 xl=0.990963 	 xu=0.990987 	 fxl=0.000003 	 fxu=-0.000025 	 xr=0.990975 	 fxr=-0.000011 	 er=0.001203
i=24 	 xl=0.990963 	 xu=0.990975 	 fxl=0.000003 	 fxu=-0.000011 	 xr=0.990969 	 fxr=-0.000004 	 er=0.000601

[David Ziehl]

Thank you very much for the reply. My code works now.

Regards,

David