Can't solve integrate with unknown value

2016 11 21
2 个回答
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更新时间：2016 11 24
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0 个投票

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

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 采纳的回答

Walter Roberson
Walter Roberson 2016-11-22

1 个投票

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

8 个评论
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Frank Hansper
Frank Hansper 2016-11-22
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
Walter Roberson 2016-11-22
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
Torsten 2016-11-22
If your commands are exactly the same as Walter's three lines of code, this must be a bug.
Best wishes
Torsten.
Frank Hansper
Frank Hansper 2016-11-22
编辑：Frank Hansper 2016-11-22
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
Walter Roberson 2016-11-23
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
Frank Hansper 2016-11-23
I see, thanks for the explanation and your time.
I have nothing more at the moment and consider this theard answered.
Karan Gill
Karan Gill 2016-11-23
Frank, regarding "ReturnConditions", did you try looking up the documentation for solve at https://www.mathworks.com/help/symbolic/solve.html ?
Frank Hansper
Frank Hansper 2016-11-24
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.

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更多回答（1 个）

Torsten
Torsten 2016-11-22

0 个投票

Use "int" instead of "integral" for symbolic calculations.
Best wishes
Torsten.

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