Can't solve integrate with unknown value

Im doing calculations in isothermal reactors and need to solve an integral with an unknown value (xa):

My code:

fun = @(xa) ( 1 / (k * CA0^2 * (1 - xa)^2 )

solve ( t == CA0 * integral(fun , 0 , xa ) , xa )

Where t is known, CA0 is known , k is known and xa is my unknown

If I solve by hand I get the right equation:

xa = k*CA0*t / (1 + k*CA0*t)

And can easy find my xa, it is not possible to find the right xa with my code.

Can you get: xa = k*CA0*t / (1 + k*CA0*t) from Matlab or an equation similar, which can find my xa?

Values:

CA0 = 2500, t = 13.57 and k = 1.40 * 10^-4

the xa should be 0.826

Answer 1

[Walter Roberson]

syms k CA0 xa t XA 
fun = 1 / (k * CA0^2 * (1 - xa)^2 )
sol = solve(t == int(fun,xa,0,XA), XA, 'ReturnConditions',true)

Then

sol.XA

provided that sol.conditions is true

[Frank Hansper]

Thanks for your suggestion - but when I use the code to generate a equation from it (xa1 = 1 - 1/(k*t*CA0^2 + 1)). It gives xa = 0.999, which is wrong.

[Walter Roberson]

It gives a symbolic answer involving variables that have not been assigned a valuable. It gives a formula not a numeric value. You would need to substitute values for the symbolic variables to get a numeric value. We would need know what values you are substituting

[Torsten]

If your commands are exactly the same as Walter's three lines of code, this must be a bug.

Best wishes

Torsten.

[Frank Hansper]

Argh, your code is correct, Walter Roberson! but you forgot to include the CA0 term right after the integral.

Right code:

syms k CA0 xa t XA 

fun = 1 / (k * CA0^2 * (1 - xa)^2 )

sol = solve(t == (CA0) * int(fun,xa,0,XA), XA, 'ReturnConditions',true)

sol.XA

Thanks guys, for helping me.

One last thing, can you elaborate on 'ReturnConditions',true, I aint familiar with this?

[Walter Roberson]

When ReturnConditions is true, then the Conditions field of the returned struct will be a conditional expression. The solution that was given in the other variables is only valid when the conditional expression is true.

As a simple example, if the solution to something was 1/x then the Conditions field returned might be x ~= 0 -- that the solution 1/x just is not valid when x == 0

[Frank Hansper]

I see, thanks for the explanation and your time.

I have nothing more at the moment and consider this theard answered.

[Karan Gill]

Frank, regarding "ReturnConditions", did you try looking up the documentation for solve at https://www.mathworks.com/help/symbolic/solve.html ?

[Frank Hansper]

Yes and I must say, I had a brain fart - I thought Matlab would return the solution with ''it's'' conditions of the solution. ex. Matlab returns what it uses symbolic and what is uses numeric. So I found it unnecessary at first.

Answer 2

[Torsten]

Use "int" instead of "integral" for symbolic calculations.

Best wishes

Torsten.