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Why does Mathematica not solve this simple integral (using units)?

GROUPS:
I am trying to find the volume of a PFR reactor. For this purpose I have written the rate expression as a function of conversion, and stuffed it in to an integral. I am using the AutomaticUnits package by Jon McLoone.

The problem is that when I execute the equation containing the integral, Mathematica starts working, but returns no result. I abort the operation after waiting what I would call suffcient time :-)  My setup is seen below:
 rA[X_] := k*((1 - X)/(1 + 2* X)*P/(R  T) - (
       X/(1 + 2 *X)*P/(R  T)*((2*X)/(1 + 2 *X) *P/(R  T))^2)/
       kc) /. {k -> 0.0440238  Minute^-1,
     kc -> 0.025  Mol^2* Liter^-2} //. {P -> 10  Atmosphere,
    T ->  127  Kelvin + 273.15  Kelvin,
    R -> 0.08205746  (Liter*Atmosphere)/(Kelvin*Mol)}
 
 
 V == FA0*\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \( .9*0.512618\)]\(
\*SuperscriptBox[\(rA[X]\), \(-1\)] \[DifferentialD]X\)\)  /. {FA0 ->
   2.5  Mol Minute^-1}

I have investigated it for a long time now, and have found out the following:
  • If I remove the Units, the expression Evaluates as it is supposed to.
  • If I keep the units, and take the reciprocal of 1/rA   (   so just rA   ), it evaluates. Not with the correct answer of course.
  • I have done the exact same operations once before, but with a much smaller function rA, which worked!
I am pretty much on bare ground, and hope someone can help me by pointing out my mistakes. I am afraid the answer is obvious, but I have stared too much at the same two lines of code, and have hence become immune toward finding a solution.

I am using Mathematica 9.

Best Regards

Simon P.
POSTED BY: Simon Pedersen
Answer
6 months ago
What do you get as a result?   I hope I decyphered the linear syntax of your second input correctly.
 In[1]:= $Version
 Out[1]= "9.0 for Mac OS X x86 (64-bit) (January 24, 2013)"
 
 In[2]:= rA[X_] := k*((1 - X)/(1 + 2*X)*
        P/(R T) - (X/(1 + 2*X)*P/(R T)*((2*X)/(1 + 2*X)*P/(R T))^2)/
        kc) /. {k -> 0.0440238 Minute^-1,
     kc -> 0.025 Mol^2*Liter^-2} //. {P -> 10 Atmosphere,
    T -> 127 Kelvin + 273.15 Kelvin,
    R -> 0.08205746 (Liter*Atmosphere)/(Kelvin*Mol)}


In[3]:= V == FA0*Integrate[ 1/rA[x], {x, 0, 0.9*0.512618}] /. {FA0 -> 2.5 Mol Minute^-1}

Out[3]= V == 270.514 Liter

Do you mean "V=" at the end? 
POSTED BY: Bruce Miller
Answer
6 months ago
Hello Bruce,

Thank you for your reply!

You have decyphered it all correctly. I am sorry you had to do that. I use the *dintt* method in Mathematica, and so it pasted it oddly. 

When I remove the units in the function rA and calculate, I get the same result as you in a matter of seconds (270.514 Liters), which is what I would expect. Nevertheless, when I execute the "V==" with units in rA, Mathematica just keeps running, without finding anything. So after waiting a long time, I just abort. Naturally I would like to calculate the expression with units. Add why does it help reciprocating the reciprocal ( 1/rA )?

What is funny, we run the same operating system, and the same version of Mathematica. Thus I do not get why it does that. Did you use the AutomaticUnits package when you executed?

Best Regards

Simon P.
POSTED BY: Simon Pedersen
Answer
6 months ago
Seems to work fine using the built-in units support in Mathematica 9:
 In[2]:= rA[X_] := k*((1 - X)/(1 + 2*X)*P/(R T) - (X/(1 + 2*X)*P/(R T)*((2*X)/(1 + 2*X)*P/(R T))^2)/
        kc) /. {k -> Quantity[0.0440238, 1/"Minutes"],
     kc -> Quantity[0.025, "Moles"^2/"Liters"^2]} //. {P ->
     Quantity[10, "Atmospheres"], T -> Quantity[127 + 273.15, "Kelvins"],
    R -> Quantity[0.08205746, "Liters" "Atmospheres"/("Kelvins" "Moles")]}
 
 In[2]:= V = Quantity[2.5, "Moles"/"Minutes"] Integrate[1/rA[x], {x, 0, 0.9*0.512618}]
 
 Out[3]= Quantity[0.270514, ("Meters")^3]

In[4]:= UnitConvert[V, "Liters"]

Out[4]= Quantity[270.514, "Liters"]
POSTED BY: Ilian Gachevski
Answer
6 months ago
Hello Ilian,

So I assume we can conclude that it is an inexplicability in the AutomaticUnits package?
POSTED BY: Simon Pedersen
Answer
6 months ago
I had not heard of the AutomaticUnits package until your note. 
In my version, Mathematica treats all these "units" as undefined symbols.
Most conveniently cancel.

The AutomaticUnits package I found at
   http://library.wolfram.com/infocenter/MathSource/7655/
is three years old.  There could be incompatibilities with Mathematica 9.
POSTED BY: Bruce Miller
Answer
6 months ago