"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
