Integration - Algebraic Substitution, 4

Category: Integral Calculus, Algebra

"Published in Newark, California, USA"

Evaluate:

Solution:

Consider the above equation

The above equation cannot be integrated by simple integration because one of the variables has a radical sign. We need to eliminate the radical sign by substituting with another variable as follows

Let

So that the above equation becomes

but 

Therefore,

 

Integration - Partial Fractions

Category: Integral Calculus, Algebra

"Published in Newark, California, USA"

Evaluate

Solution:

Consider the given equation above 

if

then

Since the numerator is only dx, then we cannot integrate the above equation by simple integration. However, we can factor the denominator as follows

The above equation can be expressed into partial fractions as follows 

Consider

Multiply both sides of the equation by their Least Common Denominator (LCD), we have
 

 

 

  Equate x²:

Equate x:

Equate x0:

Substitute the values of A, B, and C to the original equation, we have

You can also simplify further the above equation as follows

 

Stoichiometry Problem - Material Balance, 22

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

A multiple effect evaporator system has a capacity of processing 1000 kg per day of solid caustic soda when it concentrates weak liquor from 4% to 25% by weight caustic soda. When the same plant is fed with 10% weak liquor and if it is concentrated to 50% (both on weight basis), find the capacity of the plant in terms of solid caustic soda. Assume that the water evaporating capacity to be same in both cases. 

Solution:

The given word problem is about evaporation of caustic soda solution with two cases with the same water evaporating capacity which involves the principles of Stoichiometry. The total amount of a substance in the reactants or incoming ingredients must be equal to the total amount of a substance in the final products. In short, the Law of Conservation of Mass must be followed all the time.  To illustrate the given problem, it is better to draw the flow diagram as follows

Case I:

Photo by Math Principles in Everyday Life

Basis: 1000 kg/day of solid caustic soda on Weak Liquor and Thick Liquor

Let x = be the amount of Weak Liquor
      y = be the amount of Thick Liquor
      z = be the amount of Water Vapor

Overall Material Balance of Multiple Effect Evaporator:

Material Balance of Caustic Soda:

If the amount of solid caustic soda at Weak Liquor and Thick Liquor are the same, then it follows that

We can solve for the value of x as follows

We can solve for the value of y as follows

Substitute the values of x and y to the first equation in order to solve for the value of z which is the amount of water vapor as follows

Case II:

The amount of water evaporated in Case I is the same as in Case II. 

Photo by Math Principles in Everyday Life

Basis: 21000 kg/day of Water Vapor

Let x = be the amount of Weak Liquor
      y = be the amount of Thick Liquor

Overall Material Balance of Multiple Effect Evaporator:

Material Balance of Solid Caustic Soda:

Substitute the value of x to the first equation, we have

Therefore, the capacity of the plant in terms of solid caustic soda or the amount of solid caustic soda is