Stoichiometry Problem - Material Balance, 17

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

In a textile industry, it is desired to prepare 24% caustic soda solution by weight. Because of very high heat of dissolution of caustic soda in water, the above solution is prepared by two-step process. First, caustic soda is dissolved in the correct quantity of water in a dissolution tank to prepare 50% (by weight) solution. After dissolution and cooling is complete, this solution is taken to a dilution tank where some more water is added for producing 24% by weight caustic soda solution. Assuming no evaporation loss of water in dissolution tank, calculate the weight ratio of water fed to the dissolution tank to bypass water to the dilution tank. Also, calculate the total weight of water used in the process.

Photo by Math Principles in Everyday Life

Solution:

The given word problem is about the preparation of caustic soda solution in two steps which involves the principles of Stoichiometry. The solid NaOH is mixed with some water to make a caustic soda solution and the rest of water is bypassed to dilute the caustic soda solution into a lower concentration. 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 label further the flow diagram as follows

Photo by Math Principles in Everyday Life

Basis: 100 kg of Solid NaOH

Let w = be the total amount of water used in the process
      x = be the amount of water fed to Dissolution Tank
      y = be the amount of water bypassed and fed to Dilution Tank
      z = be the amount of 50% NaOH Solution
      p = be the amount of 24% NaOH Solution

Overall Material Balance before Dissolution Tank:

Overall Material Balance for Dissolution Tank:

Material Balance of NaOH at Dissolution Tank:

Substitute the value of z to the second equation, we have

Overall Material Balance for Dilution Tank:

Substitute the value of z to the above equation, we have

Material Balance of NaOH at Dilution Tank:

Substitute the value of z to the above equation, we have

The value of y will be equal to 

Substitute the values of x and y to the first equation, we have

Therefore, 

The total weight of water used in the process is 316.667 kg.  

Stoichiometry Problem - Material Balance, 4

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

An evaporator is fed with 15,000 kg/hr of a solution containing 10% NaCl, 15% NaOH, and the rest water. In the operation, water is evaporated and NaCl is precipitated as crystals. The thick liquor leaving the evaporator contains 45% NaOH, 2% NaCl, and the rest water. Calculate:

(a) kg/hr of water evaporated,
(b) kg/hr of salt precipitated, and
(c) kg/hr of thick liquor.

Solution:

The given word problem is about the evaporation of caustic liquid into water vapor, salt precipitate, and thick liquor 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 problem, it is better to draw the flow diagram as follows

Photo by Math Principles in Everyday Life

Basis: 15,000 kg/hr of weak liquor


Let x = be the amount of water evaporated
      y = be the amount of thick liquor
      z = be the amount of salt precipitated


Overall Material Balance of Evaporator:

Material Balance of NaOH:

Material Balance of NaCl:

Substitute the values of y and z to the first equation which is the Overall Material Balance of Evaporator, we have

Therefore,


Amount of Water Evaporated = 8,600 kg/hr
Amount of Salt Precipitated = 1,400 kg/hr
Amount of Thick Liquor = 5,000 kg/hr

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