Design Calculation (Sizing) Of A Crystallizer
Problem Statement And Given Data:
Given
the following data and information, it is required to design a forced
circulation crystallizer of the type shown in Figure 1 operating under vacuum.
 |
| Figure 1 Forced circulation (FC) evaporation crystallizer. |
Feed
(an aqueous solution) rate, Qi = 15 m3/h, feed concentration, Ci =
200 kg/m3 solution; feed temperature = 55oC; average
density of the solution = 1100 kg/m3 and average specific heat =
0.90 kcal/kg.oC; operating pressure = 100 mm Hg (660 mm Hg vacuum);
boiling point elevation of the saturated solution = 13oC;
saturation concentration at the crystallization temperature = 250 kg/m3;
magma density allowed, MT = 350 kg crystal/m3 solution;
crystal growth rate determined experimentally under the conditions of the
crystallizer, G = 4.67 x 10-8 m/s; crystal density = 1700 kg/m3;
desired dominant crystal size LD = 0.8 mm; heat of vaporization of
water at the temperature of the crystallizer = 570 kcal/kg; superheat allowed
in the heat exchanger = 3oC; overall heat transfer coefficient in
the heat exchanger U = 1000 kcal/m2.h.oC; low pressure
steam is available for the heat exchanger at 3 kg/cm2 gauge, latent
heat of condensation = 510 kcal/kg.
Solution:
Material balance (basis 1 hour operation)
Solution in = 15 m3 ; solute in = (15 m3)(200 kg/m3) =
3000 kg
Water
in = (15)(1100 - 200) = 13500 kg
Magma
density = 350 kg per cubic meter solution
= 350 kg crystal per
250 kg solute in the solution.
Crystals
produced = [(350)/(350 + 250)] x 3000 kg = 1750 kg.
Solute leaving with mother liquor = 3000 kg - 1750 kg = 1250 kg.
Volume
of solution (clear mother liquor) leaving = 1250 kg / (250 kg/m3,
solubility) = 5 m3
Mass
of solvent (water) leaving = (5 m3)(1100 - 250) kg/m3 =
4250 kg
Mass
of water evaporated = water in with feed – water out with mother liquor
= 13500 – 4250
= 9250 kg
Volume
of slurry leaving per hour = volume of solution + volume of crystals leaving
= 5 m3 + [1750 kg/ (1700 kg/m3)] = 6.03 m3
per hour
Crystallizer Volume
Dominant
size of the product, LD = 0.8 mm = 3GTH ; G = 4.67 x 10-8
m/s (given).
Required
holding time, TH = LD/3G = (8 x 10-4
m)/(3)(4.67 x 10-8 m/s) = 5710 s = 1.6 h
Volume
of suspension in the crystallizer at any time = (6.03 m3/h)(1.6 h) =
9.65 m3
This
is the “working volume” of the crystallizer (note that the suspension holdup in
the pipe line and in the heat exchanger tubes has been neglected). Add 60% to
account for vapour bubbles and froth.
Effective suspension volume in the
crystallizer = (9.65 m3)(1.6) = 15.44 m3
Select
a 3 m diameter vessel (this is to be
checked and changed if necessary) with a conical bottom. Take a cone angle of
45 degrees for the conical bottom.
Volume
of the conical part (radius = depth = 1.5 m) of the tank = (π/3)(1.5)2(1.5)
= 3.53 m3
Volume
of the cylindrical part = 15.44 – 3.53 = 11.91 m3 ; height =
11.91/(π)(1.5)2 = 1.68 m
Add
1.25 m space above the boiling liquid for disengagement of the entrained
droplets.
Total
length of the cylindrical part of the tank = 1.68 m + 1.25 m = 2.93 m say 3 m
Now
check the assumed diameter of the tank.
Absolute
pressure in the vapour space = 760 – 660 = 100 mm Hg = 0.1316 atm
B.P
of water = 52oC. B.P elevation = 130C = B.P of the
solution = 52 + 13 = 65oC = 338 k.
Density
of vapour (steam) at this temperature and pressure, ρv = (18)(0.1316)/(0.0821)(338)
= 0.0854 kg/m3
Volumetric
rate of vapour generation = (9250 kg/h)/(0.0854 kg/m3) =
A
Souders-Brown type equation (Equation 1) is used to determine the allowable velocity
of the vapour without risking entrainment.
 |
| Equation 1 |
For
evaporation under vacuum with a demister, a conservative value of Cv
= 0.04 m/s is used.
Allowable
velocity, uv = 0.04 [(1100
/ 0.0854)0.5 ] = 4.54 m/s
Area
required for evaporation = (1.083 x 105 m3/h)/(3600)(4.54 m/s) = 6.63
m2
Cross-section
of the tank = 7.07 m2. Hence a tank of 3 m diameter is suitable.
Energy balance and heat exchanger area
The
feed liquor enters at 55oC; feed rate = 15 m3/h =
(15)/(1100 kg/m3) = 1.65 x 104 kg/h
Take
the B.P of the liquor, 65oC, as the reference temperature.
Required
heat input to raise the liquor temperature to 65oC,
= (1.65 x 104)(0.9)(65
- 55) = 148,500 kcal
Heat
required for evaporation of water = (9250 kg/h)(510 kcal/kg) = 5.272 x 106
kcal/h
(Superheat
of the vapour produced is neglected)
Heat
of crystallization = (1750 kg/h)(30 kcal/kg) = 5.25 x 104 kcal/h
(absorption)
Total
heat input required, Qh = 1.485 x 105 + 5.272 x 106 + 5.25
x 104 = 5.473 x 106 kcal/h
Heating
steam supplied at 3 kg/cm2 gauge (saturated) = steam temperature =
143 oC
Latent
heat = 510 kcal/kg = steam rate = (5.473 x 106 kcal/h)/(510
kcal/kg) = 10,730 kg/h
Heat exchanger area
Temperature
driving force, ∆T1 = 143 – 65 = 78oC; ∆T2 = 143 – 68 = 75oC;
(∆T)m = 76.5oC
Area
of the heat exchanger = Qh/U.(∆T)m = (5.473 x 106)/(1000)(76.5)
= 71.5 m2
An
outline design of the crystallizer is given in Fig 2 below.
 |
| Figure 2 An outline diagram of the crystallizer. |
Read crystallizer selection and design considerations articles for better understanding.
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