Stoichiometry Problem - Material Balance, 6

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

A sample of coal is found to contain 63% carbon and 24% ash on weight basis. The analysis of refuse after combustion shows 7% carbon and the rest ash. Calculate the percentage of the original carbon unburnt in the refuse. 

Solution:

The given word problem is about the combustion of a coal in a furnace which involves the principles of Stoichiometry. When you add the percent compositions of a coal, it will not be equal to 100%. The rest of the composition of a coal is either water or other substances that are mostly combustible. During the combustion of a coal, the components like carbon, water, ash, and other substances are converting into flue gas and refuse. The flue gas is a mixture of carbon dioxide, carbon monoxide, water vapor, oxygen, nitrogen, and other oxide gases. The refuse is mostly ash but sometimes there are unburnt carbon or coal in the refuse. 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

Since the amount of coal, flue gas, and refuse are not given in the problem, let's assign a basis of 100 kg of coal in the problem.

Basis: 100 kg of coal

Let x = be the amount of refuse
      y = be the amount of flue gas

Material Balance of Ash:

Therefore,

Stoichiometry Problem - Material Balance, 5

Category: Chemical Engineering Math, Algebra

"Published in Newark, California, USA"

An evaporator system concentrating a weak liquor from 5% to 50% solids handles 100 kg of solid per hour. If the same system is to concentrate a weak liquor from 4% to 35%, find the capacity of the system in terms of solids that can be handled per hour assuming water evaporation capacity to be same in both the cases. 

Solution: 

The given word problem is about the evaporation of a liquid from weak to thick liquor which involves the principles of Stoichiometry. There are two evaporators in the given problem with different cases but the water evaporation capacity is the same for both cases. 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

Case I:

Basis: 100 kg/hr of solid handling capacity of the evaporator

Let x₁ = be the amount of weak liquor
      y₁ = be the amount of thick liquor
      z₁ = be the amount of water evaporated

Amount of solids in weak liquor:

Amount of solids in thick liquor:

Overall Material Balance of Evaporator:

For Case II, the water evaporation capacity is the same as with Case I. And so, z₁ = z₂ = 1800 kg/hr of water evaporated.  

Photo by Math Principles in Everyday Life

Case II:

Basis: 1800 kg/hr of water evaporated

Let x₂ = be the amount of weak liquor
      y₂ = be the amount of thick liquor
      z₂ = be the amount of water evaporated

Overall Material Balance of Evaporator:

Material Balance of Solids:

 

Substitute the value of x₂ to the first equation, we have

Hence, the amount of weak liquor is

Therefore, the solid handling capacity of the evaporator is

or

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