"Function might be unused"

I can't seem to figure our why Matlab doesn't like my function. It keeps telling me that it is probably called incorrectly but I can't see what I did wrong here.

   Ideal Boiler Function

hfg=930

lhv=21500

p_percent=100

p=p_percent/100

MFfr=0.5191

%This function will calculate the steam mass flow rate (SMfr)

%of an actual boiler with 100% efficency (p_percent=100%)

%This funtion has 4 inputs (p,hfg,lhv,and MFfr),

%and 1 output (SMfr).

function SMfr=Ideal_Boiler(p,hfg,lhv,MFfr)

      SMfr=(MFfr*p*lhv)/hfg;

      output=SMfr;

end

disp('SMfr =');disp(SMfr)

Answer 1

[bio lim]

I don't see any problems, unless you defined the function inside your main script, and MATLAB doesn't allow that.

hfg=930;
lhv=21500;
p_percent=100;
p=p_percent/100;
MFfr=0.5191;

SMfr=Ideal_Boiler(p,hfg,lhv,MFfr)
disp('SMfr =');disp(SMfr)

SMfr =

     12.0007

SMfr =
   12.0007

[Shelby Russ]

Um. So, how do I know if it is defined in the main script? I just opened one and saved my file as the function name and it hates me for it. I'm so confused.

[bio lim]

You need a two .m files, one for your function and one for your main program.

In one of them, you declare your function as:

function SMfr=Ideal_Boiler(p,hfg,lhv,MFfr)
      SMfr=(MFfr*p*lhv)/hfg;
      output=SMfr;
end

And in your main program, you actually use the values, and call for your function:

hfg=930
lhv=21500
p_percent=100
p=p_percent/100
MFfr=0.5191

SMfr=Ideal_Boiler(p,hfg,lhv,MFfr) % calling Ideal_Boiler function

And the output is:

SMfr =
     12.0007

[Shelby Russ]

Oh my gosh, thank you so much. I would have never figured that out.

[bio lim]

My pleasure. One more thing:

In your code, if you end a line with a semi-colom, i.e., ; , MATLAB doesn't print out the results. So in your code:

disp('SMfr =');disp(SMfr)

Is unnecessary, if you just end your function call, i.e., SMfr=Ideal_Boiler(p,hfg,lhv,MFfr) without a semi-colon (Meaning it will print the result out).

Answer 2

[Kevin Agudo]

syms t
function y = f(t)
t=linspace(0,100);
f(x)=(3*t^2+1)./(t^3+50);
h=.01;
plot(t ,(f(t+h)-f(t))./h,t,(f(t+h)-2*f(t)+f(t-h))/h^2)
end

When I run the function, dont get a graph and it says "function f may not be used"

[Walter Roberson]

h=.01;
t=linspace(0,100);
plot(t ,(f(t+h)-f(t))./h,t,(f(t+h)-2*f(t)+f(t-h))/h^2)

function y = f(t)
y=(3*t.^2+1)./(t.^3+50);
end

Answer 3

[Johanan]

Adapt pend.m to build the double pendulum. A new pair of rod and bob must be defined for the second pendulum. Note that the pivot end of the second rod is equal to the formerly free end of the first rod: The (x, y) position of the free end of the second rod can be calculated by using simple trigonometry.

[Walter Roberson]

I do not understand how adapting pend.m will solve the problem of "Function might be unused" ???