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Chemistry Mathematics Equation Solving Wolfram Language
Jonathan Witt

Solve the following simple system of equations?

Jonathan Witt
Posted 14 days ago
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Hi I can't get this simple system to solve. I want to know a, b, and c which are flow rates and I know the mole fractions, but mathematica won't solve for it. Please help. 

Solve[yCO == (0.015 a + 0.05 b)/(a + b + c) && 
  yCO2 == (0.07*a)/(a + b + c) && yN2 ==  1 - yCO - yCO2, {a, b, c}]
POSTED BY: Jonathan Witt
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The equation

yN2 == 1 - yCO - yCO2

adds no information relevant to estimating a, b, or c so you really have two equations and three unknowns.

However, as an example you can solve for a and b in terms of c, yCO, and yCO2:

Solve[yCO == (0.015 a + 0.05 b)/(a + b + c) && yCO2 == (0.07*a)/(a + b + c), {a, b}]

{{a -> -((1.42857 c yCO2)/(-0.1 + 2. yCO + 1. yCO2)), 
  b -> (0.428571 (-4.66667 c yCO + 1. c yCO2))/(-0.1 + 2. yCO + 1. yCO2)}}

A slightly cleaner solution would be to "rationalize" all of the coefficients:

Solve[yCO == ((15/1000) a +(5/100) b)/(a + b + c) && yCO2 == ((7/100)*a)/(a + b + c), {a, b}]

{{a -> -((100 c yCO2)/(7 (-1 + 20 yCO + 10 yCO2))), 
  b -> -((10 c (14 yCO - 3 yCO2))/(7 (-1 + 20 yCO + 10 yCO2)))}}
POSTED BY: Jim Baldwin
Answer

Adding option MaxExtraConditions -> All to Solve helps:

Solve[Rationalize[
  yCO == (0.015 a + 0.05 b)/(a + b + c) && 
   yCO2 == (0.07*a)/(a + b + c) && yN2 == 1 - yCO - yCO2, 0], {a, b, 
  c}, MaxExtraConditions -> All]

 (* {{a -> ConditionalExpression[0, 
     yCO2 == 0 && yCO == 0 && yN2 == 1 && c != 0], 
   b -> ConditionalExpression[0, 
     yCO2 == 0 && yCO == 0 && yN2 == 1 && c != 0]}, {a -> 
    ConditionalExpression[0, 
     yCO2 == 0 && yCO + yN2 == 1 && yCO != 0 && b != 0], 
   c -> ConditionalExpression[(b - 20 b yCO)/(20 yCO), 
     yCO2 == 0 && yCO + yN2 == 1 && yCO != 0 && b != 0]}, {b -> 
    ConditionalExpression[-((3 a)/10), 
     yCO == 0 && yCO2 + yN2 == 1 && yCO2 != 0 && a != 0], 
   c -> ConditionalExpression[-((7 (-a + 10 a yCO2))/(100 yCO2)), 
     yCO == 0 && yCO2 + yN2 == 1 && yCO2 != 0 && a != 0]}, {b -> 
    ConditionalExpression[(a (14 yCO - 3 yCO2))/(10 yCO2), 
     yCO + yCO2 + yN2 == 1 && yCO2 != 0 && yCO != 0 && a yCO != 0], 
   c -> ConditionalExpression[(
     3 a - 200 a yCO + (a (14 yCO - 3 yCO2))/yCO2 - (
      20 a yCO (14 yCO - 3 yCO2))/yCO2)/(200 yCO), 
     yCO + yCO2 + yN2 == 1 && yCO2 != 0 && yCO != 0 && a yCO != 0]}}*)
POSTED BY: Mariusz Iwaniuk
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