3. An aqueous feed of 10,000 kg/h, saturated with Ba
at 100 C, enters a ᵒ
crystallizer that can be simulated with the MSMPR model. However, crystallization is achieved with negligible evaporation. The magma leaves the crystallizer at 20 C with crystals of the dihydrate. The crystallizer has a volume ᵒ
(vapor-space-free basis) of 2.0
. From laboratory experiments, the crystal
growth rate is essentially constant at 4x
crystals = 3.097 g/
−
m/s. Density of the dehydrate
. Density of an aqueous, saturated solution of barium
chloride at 20 C = 1.279 g/ ᵒ
. Determine the:
a) kg/h of crystals in the magma product product (11 marks) Firstly, a mass balance is performed around the crystallizer. From Table 17.5, solubility of BaCl 2: At 100°C = 58.3 58.3 g/100 g water At 20°C, solubility solubility = 35.7 g/100 g/100 g water Therefore, the feed contains: 10,000 kg/hr [58.3g/(58.3 + 100)g] = 3,680 kg/h of BaCl2 ,and 10,000 – 10,000 – 3,680 3,680 = 6,320 kg/h of water. MW of BaCl2 = 208.27 MW of BaCl2∙H2O = 244.31 Therefore, the crystals are 208.27/244.31 = 0.852 mass fraction BaCl2 Mass balance for BaCl2 around the crystallizer cr ystallizer.. Let x = kg/h of water in the crystals.
0.852 3,680 = = 6,320 6,320 (35.1007)+10.852
(1)
Solving for Equation (1), (1),
3,680 = = 2256.240.357 57 +5.7568 5.3998 = 1423.76 = 263 /ℎ = 263 2630.852⁄0.148 = 1,514 /ℎ of water in the crystals.
BaCl2 in the crystals
Therefore, the product crystals in the magma = 263 + 1,514 = 1,777 kg/h b) Predominant crystal size size in mm (7 marks) Next, the volumetric flow rate of the magma is calculated in order to compute the residence time in the crystallizer. crystallizer. The flow rate of solution in the magma = (10,000 – (10,000 – 1,777) 1,777) kg/h = 8,223 kg/h
From Perry’s Handbook, the density of
∙ 2
crystals = 3,097 kg/m3, and
the density of the solution in the magma = 1,279 kg/m3 The volumetric flow rate of the magma is: 8,223/1,279 + 1,777/3,097 = 7.00 m3/h Average residence time, t, of crystals in the crystallizer is: 2.0/7.0 = 0.286 h = 0.286 x 60 =17 minutes Growth rate:
10−
G = (4.0 x G=
) m/s
. cm/min / = 0.0024 cm/min
The predominant crystal size is given by: Lpd = 3Gt = 3(4.0 x 10-7)m/s[17(60)] s = 0.00122 m = 0.122 cm = 1.22 mm
c) Mass fraction of crystals in the size range from U.S. Standard 20 to 25 mesh. (9 marks) Estimate the mass fraction of crystals in the size range from 10 to 25 mesh. From Table 17.4, 20 mesh has an aperture of 0.850 mm = 0.085 cm 25 mesh has an aperture of 0.710 mm = 0.071 cm Calculate the dimensionless crystal size from (17-48),
= L/Gt
For 20 mesh, = 0.085/[(0.0024)(17)] = 2.08 For 25 mesh,
= 0.071/[(0.0024)(17)] = 1.74
From Table 17.9, the cumulative mass fraction of crystals smaller than size L, is:
= 11+ + 2 + 6
(2)
For L = 0.085cm, z = 2.08. Substitution into Equation (2),
= 1[1+2.08+ . + .]2.08
= 0.158
For L = 0.071cm, z=1.74. Substitution into Equation (2)
1.74 1. 7 4 = 11+1.74+ 2 + 6 1.74 = 0.099 Therefore, we have 15.8 wt% crystals smaller than 0.085 cm, and 9.9 wt% crystals smaller than 0.071cm Therefore, the mass fraction of crystals between 20 and 25 U.S mesh: 0.158
0.099 = 0.059 or 5.9 wt%
