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.  

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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

Maximum Minimum Problem, 8

Category: Differential Calculus, Analytic Geometry, Algebra

"Published in Newark, California, USA"

Find the shortest distance from the point P(2, 0) to a point on the curve y² - x² = 1, and find the point on the curve closest to P.

Solution:

To illustrate the problem, it is better to draw the figure and sketch the graph of the curve (hyperbola) as well as the given point in Rectangular Coordinate System as follows

Photo by Math Principles in Everyday Life

As you can see from the figure above that there will be two distances of a point to a curve because the curve is hyperbola and it is symmetrical with x axis and y axis. The other point on the curve is a point of tangency.  The distance of a point to the point of tangency is the perpendicular distance of a point to the tangent line. To understand more the problem, label further the above figure as follows

Photo by Math Principles in Everyday Life

The distance of two points is

If the first point is P(2, 0), then the above equation becomes

If the second point is on the curve, then the above equation becomes

Next, we need to eliminate y at the above equation. If the equation of a curve is

,then the value of y will be equal to 

Substitute the value of y to the first equation which is the distance of two points, we have

Take the derivative of the above equation with respect to x, we have

Set dd/dx = 0 because we want to minimize the distance of a point to a curve.

Substitute the value of x to the equation of the curve in order to get the value of y, we have

Therefore, the point on the curve is

The shortest distance of a point to a curve is