3.10 – Practice Problem 2¶
An evaporation process is often employed to obtain dry salts in a mixture. The equipment to dry the salt is the evaporator and a crystallizer. The crystallizer filters out the solid and leaves behind liquid which is often recycled to maximize yield. A diagram of this process is modeled exactly in example 4.5-2. For us, we are looking to separate NaCo3 from water. the Feed is a 19.6 wt% of NaCO3, the crystallizer filters out solid NaCO3 and a bit of the NaCO3 solution at 40 wt% NaCO3. The solid to solution ratio is 10kg:1kg. The filtrate is also a 40 wt% solution and is recycled back to the evaporator. 45% of the water is evaporated by the evaporator. The evaporator has a maximum capacity of 175 kg/s.
Assume the process is operating at maximum capacity. Draw and label a flowchart and do the degree-of-freedom analysis for the overall system, the recycle–fresh feed mixing point, the evaporator, and the crystallizer. Then write in an efficient order (minimizing simultaneous equations) the equations you would solve to determine all unknown stream variables. In each equation, circle the variable for which you would solve, but don’t do the calculations.
Calculate the maximum production rate of solid \(K SO\), the rate at which fresh feed must be supplied to achieve this production rate, and the ratio kg recycle/kg fresh feed.
Calculate the composition and feed rate of the stream entering the crystallizer if the process is scaled to 75% of its maximum capacity.
The wet filter cake is subjected to another operation after leaving the filter. Suggest what it might be. Also, list what you think the principal operating costs for this process might be.
Use an equation-solving computer program to solve the equations derived in part (a). Verify that you get the same solutions determined in part (b).
a)¶
We can draw label and specify the flow diagram in 2 ways. First, here is the diagram that should have been drawn and labelled with the information given.
Figure 1.
When we label our unknowns, we can either label all the unknowns based on the NaCO3 salt as the basis. This is what it would look like:
Figure 1.
Every \(m\) denotes a mass flow rate in the system with m3 being the only non-salt variable. Every \(x\) denotes the mass fraction of the salt in the mass flow.
We can also label the unknowns \(m\) as both water and NaCO3. If a wt% is given then there is only 1 unknown labelled.
Figure 1.
Both way of labelling the diagram are correct, however; the latter is better in the long run, this will be explained later. As such, the latter diagram will be used for analysis.
The degree of freedom analysis:¶
overall balance: 2 Unknowns (m1,m7,m9) -2 equations(mass balance, NaCO3 balance) -1 relationship (10:1 solid:liquid) DOF = 0 (m4 is a known value assuming it is running at max capacity)
mixing balance 4 Unknowns (m1,m2,m3,m8) -2 equations (mass balance, NaCO3 balance) DOF = 2
crystallizer balance 4 Unknowns (m5,m6,m7,m8) -2 equations (mass balance, NaCO3 balance) DOF = 2
evaporation balance 4 Unknowns (m2,m3,m5,m6) -2 equations (mass balance, NaCO3 balance) -1 constraint( percent evaporation) DOF = 1 (m4 is a known value assuming it is running at max capacity)
As clear by the DOF analysis the overall balance should be solved first, then the evaporation balance and the other variables should become clear which ones to solve.
Overall balance at max capacity:¶
the units are omitted for clarity, do not forget them on a test
Mass balance:
NaCO3 balance:
Relationship:
The resulting numbers (in kg/s) are :
\(m_1 = 220.8\)
\(m_7 = 41.6\)
\(m_9 = 4.2\)
Evaporation + mixer balance at max capacity:¶
Since we have the relationship of water evaporated, the H2O balance has 1 less degree of freedom than the NaCO3 balance. By isolating the H2O mass balance, we can use the evaporation and the mixer balance to solve the for the recycle.
H2O balance on the mixer:
Percent evaporated:
The resulting numbers (in kg/s) are :
\(m_3 = 388.888\)
\(m_8 = 350.8\)
m6 could have been solved by using the balance on the evaporator and the H2O balance but there would have been a dead end there since m5 cannot be found with m6. At this point, all variables can be solved for using the crystallizer balance.
Recycle ratio:¶
Part b¶
Crystallizer at maximum capacity¶
Solving for m6 (H2O):
Summing streams m5 and m6 results with: \(396.6\)
Finally, reducing the capacity by 75%: \(297.45\) where 71.6% is water.
Part c¶
Drying is the step most likely to come after crystallization. The principal costs are the evaporation, the cooling from the crystallizer and the heating from the drying process.
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