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Wednesday, October 2, 2013

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


 

Tuesday, October 1, 2013

Stoichiometry Problem - Material Balance, 21

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

After a crystallization operation, the solution of calcium chloride in water contains 60 g of CaCl2 per 100 g of water. Calculate the amount of this solution necessary to dissolve 200 kg of CaCl2·6H2O crystals at a temperature of of 298K (25°C). The solubility of CaCl2 at 298K (25°C) is 819.2 g of CaCl2 per 1000 g of water. (Atomic Weights: Ca = 40, Cl = 35, O = 16, H = 1)

Solution:

The given word problem is about mixing of calcium chloride crystals with enough calcium chloride solution to dissolve the crystals completely 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

Photo by Math Principles in Everyday Life

Basis: 200 kg of CaCl2·6H2O Crystals

Let x = be the amount of CaCl2 Solution

      y = be the amount of Final Solution

Molecular Weight Data:

CaCl2 = 110
CaCl2·6H2O = 218
H2O = 18

Overall Material Balance of Mixer:


Consider CaCl2 Solution:





Consider CaCl2·6H2O Crystals:







Consider Final Solution:









Material Balance of CaCl2:







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









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







Therefore,

The amount of CaCl2 Solution is 144.1887 kg.

Monday, September 30, 2013

Stoichiometry Problem - Material Balance, 20

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

One hundred milliliters per minute of 5M NaOH solution (sp. gr. = 1.18) is mixed with 10M NaOH solution (sp. gr. = 1.37). It is desired to produce a solution containing 11.7 % mole NaOH. Calculate the required volumetric flow rate (ml/min) of 10M NaOH solution. Also, calculate the volumetric flow rate at the final solution. (Atomic Weights: Na = 23, O = 16, H = 1)

Solution:

The given word problem is about mixing of two NaOH solutions with different concentrations to produce a desired NaOH solution 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

Photo by Math Principles in Everyday Life

Basis: 100 ml/min of 5M NaOH Solution

Let x = be the volumetric flow rate of 10M NaOH in ml/min

Molecular Weight Data:

NaOH = 40
H2O = 18

Consider 5M NaOH Solution:








 
 


Consider 10M NaOH Solution:
 
 
 
 
 
 
 
 
 
 
 

The total moles of NaOH from the two solutions is equal to
 

The total moles of H2O from the two solutions is equal to


Hence, the % mole of NaOH is equal to




Material Balance of NaOH:








Therefore, the volumetric flow rate of 10M NaOH Solution is 78.2 ml/min.

The volumetric flow rate of Final Solution is 100 ml/min + 78.2 ml/min = 178.2 ml/min