Application of LINEAR ALGEBRIC EQUATION in Chemical Engineering

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APPLICATION OF LINEAR ALGEBRIC EQUATION IN CHEMICAL ENGINEERING Prepared By: Lakhlani Nishith

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Here,One of many Applications of Linear Algebraic Equation is explained with Example.

Transcript of Application of LINEAR ALGEBRIC EQUATION in Chemical Engineering

Page 1: Application of LINEAR ALGEBRIC EQUATION in Chemical Engineering

APPLICATION OF LINEAR ALGEBRIC EQUATION IN CHEMICAL ENGINEERING

Prepared By: Lakhlani Nishith

Page 2: Application of LINEAR ALGEBRIC EQUATION in Chemical Engineering

Linear Algebraic EquationThe 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.

Page 3: Application of LINEAR ALGEBRIC EQUATION in Chemical Engineering

Application In Chemical EngineeringOne 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.

Page 4: Application of LINEAR ALGEBRIC EQUATION in Chemical Engineering

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

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Example-1Suppose 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.

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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.

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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

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Example-2Problem

By examination we can see that the following equation is not balanced.

CH4 + O2 CO2+H2O

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Cont..Solution

First assign variables to each of the unknown coefficients in the equation which gives us:

wCH4 + xO2 yCO2 + zH2O

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Cont..CARBON w = y There is 1 carbon

atom in the w term and 1 in the y term.

HYDROGEN 4w = 2zThere 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.

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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).

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Cont..This gives us the following values for our

variables.

w = 1/2zx = 1zy = 1/2z

above equations we calculate the values of our 4 variables to be:

w x y z1 2 1 3

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Cont..Replace these values as the coefficients to

our original equation.

CH4 + 2O2 CO2 + 2H2O

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