Se ha denunciado esta presentación.
Utilizamos tu perfil de LinkedIn y tus datos de actividad para personalizar los anuncios y mostrarte publicidad más relevante. Puedes cambiar tus preferencias de publicidad en cualquier momento.

Application of linear algebric equation in chemical engineering

4.080 visualizaciones

Publicado el

Here.two simple applications of linear algebraic equations is explained with example.

Publicado en: Ingeniería
  • Sé el primero en comentar

Application of linear algebric equation in chemical engineering

  2. 2. Linear Algebraic Equation  The definition of a linear equation is an algebraic equation in which each term has an exponent of one.  The graphing of the equation results in a straight line.  One or more variables in the equation.
  3. 3. Application In Chemical Engineering  One of the most important organizing principles in chemical engineering is the Conservation of Mass.  The principle of mass conservation, states that for any closed system the mass of the system must remain constant over time.
  4. 4. Cont..  independent of any chemical and physical changes taking place within the system.  For stable condition(i.e., steady-state) it can be represented as: Input = output
  5. 5. Example-1  Suppose we are performing a mass balance for a conservative substance (i.e., one that doesn’t increase or decrease due to chemical transformation) in a reactor, we would have to quantify the rate mass flows into the reactor through the two inflow pipes and out of the reactor through outflow pipe.
  6. 6. Cont..  For pipe 1,product flow rate Q1=2 m3/min , Q2= 1.5 m3/min and concentrate C1=25 mg/m3 C2= 10 mg/m3 ; therefore, the rate at which mass flows into reactor through pipe 1 and pipe 2 accordingly.
  7. 7. Cont..  Because of Steady state of Reactor  Input = Output , according to that Q1C1 + Q2C2 = Q3C3 50 + 15 = 3.5 C3 C3 =18.6 mg/m3
  8. 8. Example-2 Problem  By examination we can see that the following equation is not balanced. CH4 + O2 CO2+H2O
  9. 9. Cont.. Solution  First assign variables to each of the unknown coefficients in the equation which gives us: wCH4 + xO2 yCO2 + zH2O
  10. 10. Cont.. CARBON w = y There is 1 carbon atom in the w term and 1 in the y term. HYDROGEN 4w = 2z There are 4 Hydrogen atoms in the w term and 2 in the z term. OXYGEN 2x = 2y + z There are 2 oxygen atoms in the x term, 2 in the y term, and 1 in the z term.
  11. 11. Cont..  Rewrite the linear equations in standard form to get a homogeneous system of equations with 4 variables: w-y = 0 4w-2z = 0 2x-2y-z = 0  Create a matrix for the above systems of equations augmented with zero’s (left) and perform the Gauss-Jordan elimination method to reduce the matrix(right).
  12. 12. Cont..  This gives us the following values for our variables. w = 1/2z x = 1z y = 1/2z  above equations we calculate the values of our 4 variables to be: w x y z 1 2 1 3
  13. 13. Cont..  Replace these values as the coefficients to our original equation. CH4 + 2O2 CO2 + 2H2O