CHE311-crystallization-calc-procedure

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CHE311-2021 Separation processes Evaporative crystallizer design 1 T HERMODYNAMIC MODEL (M ARGULES 1- PARAMETER ) Table 1 – solubility data Solubility mol frac. x 20c T, Celsius 20 The value of x 20 is the solubility of the salt at 20C, or sometimes 25C, is often reported in various databases – it tends to be the more reliable value, because it is the most measured. Note that most times the solubility data comes as a wt% or as g solute/100 g solvent. You will need MWsolute =________________ g/mol and MW solvent=_______________ g/mol to obtain the molar fraction required in the table. You will also need the melting point of the solute Tm = __________ C , and a literature or assumed value of the heat of dissolution ΔH sol =_________ kJ/mol . This value will later be used as an adjusted or fitted – the actual ΔH sol is really a function of the composition. Considering the solid-liquid equilibrium, with the Margules 1-parameter (component 1 being the solute): x 1 *exp (A(1-x 1 )^2)= exp(-(ΔH sol /R)*(1/T-1/Tm)) , and the solubility at 20 C, then the value of A is: A= -((ΔH sol /R)*(1/T-1/Tm)+ Ln(x 20c ))/(1-x 20c )^2 The saturation temperature (at the solid solubility limit can be calculated) for a given x1 T s-l, sat, C = (-(Ln(x 1 ) + A(1-x 1 )^2)/(ΔH sol /R) +1/(Tm C +273))^-1 -273 The table can then be augmented to: Solubility mol frac. x 20c T, Celsius 20 T s-l, sat, C At this point, the value of ΔH sol can be optimized to produce a better match between T and T s-l from the table. 2 P ROPERTY BLOCK Page 1 of 8

A Antoine B Antoine C Antoine Antoine for solvent (water in this case) 4.6543 1435.264 -64.848 ΔH sol (adjusted) kJ/mol Tm= Celsius A(Margules) = Enthalpy of formation (kJ/mol) at 25C cp (J/molK) MW (g/mol) Density, kg/m^3 viscosity, Pa Solvent(l) -285.8 (for water) 75 (water) 18 1000 1.00E-03 Solvent(g) -241.8 (for water) 35 (water) -- -- -- Solute(aq) -- -- -- -- Solute(s) -- The best sources of data for ΔH sol , ΔH f and cp for numerous solutes is (a) CRC handbook of Chemistry and Physics. The second possible resource, also very useful is the second chapter of Perry’s handbook . I believe you can download this chapter for free from the UofT online resource for Perry’s handbook. One note of caution when using Perry for heat of dissolution -> the values are reported with a sign opposite to the convention used in the energy balances, and it comes in kcal/mol. For example, for (NH 4 ) 2 SO 4 , the heat of solution is reported as -2.75. The value we would use is +(2.75)*(4.186) kJ/mol. Also, Chemspider has data of Tm, density and solubility at room temperature. NIST webbook has information on Cp and Tm for some components. The Antoine constants for the solvent could also be found in the NIST database. 3 P ROCESS F LOWCHART Page 2 of 8 Filter F-01 Feed (S1) Vapour (S2) Product (S3) Heater (HX-01) Recycle fluid (S4) Recycle suspension (S5) Mixer (M-01) H l D (S6) (S7) D o

Figure 1. Flowchart for a DTB evaporative crystallizer, including filtering unit. Stream S1 (feed) in Figure 1 brings the solute in dissolved form, typically close to its solubility limit at the temperature of feed, but sometimes it can be at a concentration substantially lower than the solubility. The lower the feed concentration, the more solvent that needs to be evaporated in the crystallizer. S1 is mixed with the supersaturated solution from the filter (S4) and the internal recirculation line (S5). S5 is necessary to maintain the crystallizer bed agitated, and it can be used to melt the fine particles. The mixed stream (S6) is fed to the heater (HX-01), which introduces the energy necessary to evaporate the solvent. The biggest challenge with HX-01 is that we need to prevent the fluid from increasing the temperature too much because it can induce crystallization in the exchanger, which can foul HX-01. In the DTB, the hot stream (S7) is fed directly to the internal draft tube were the mixer M-01 is used to maintain the system in suspension and well mixed in the tank. The bottom of the tank has an elutriation leg from where the product suspension is withdrawn to the filter F-01, from where a filtered, assumed dry product (S3), and the supersaturated solution (S4) are obtained. 4 P ROCESS CONDITIONS BLOCK Table 2. Operating conditions T feed Celsius Page 3 of 8

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P feed Bar M flow(S1) kg/day Solute feed x w % T operation = Celsius (S) = G= m/s φ particle ΔT HX-01 Celsius Heat flux W/m^2 V fluid /V settling factor ΔP filter Bar L cake = M Cake porosity (ε cake ) The temperature and pressure of the feed are normally set. The production can be kg/day (M flow(s1) ). The composition of the solute in the feed, in most cases, is given as wt% (x w% ). The operating temperature (T operation ) influences the properties, but more so, the operating pressure. Higher temperatures will create higher pressures, with less vacuum required. However, high T operation can lead to unstable crystallization. T operation is often less than 80C. The supersaturation (S) affects the rate of crystallization, and the particle size (L). A recommended initial guess is S = 3E-4/x 1sat at T operation . Higher or lower (S) can be used if necessary, to optimize the size of the particle (L) and the size of the crystallizer. The growth rate (G) used in most crystallizers ranges from 1 to 10 E-8 m/s, a value of 3E-8 is typically a safe estimation. Lower G could be used with systems with (S)>1. The volume fraction of particles in the crystallizer φ particle is another optimizable parameter, but it normally ranges from 0.01 to 0.1. High φ particle could lead to fouling. The temperature increases in the exchanger ΔT HX-01 often ranges from 0.05 to 5 Celsius. Higher values are not recommended because precipitation and fouling can occur. For the rest of the parameters, see heuristic recommendations in slide set 4. 5 M ASS AND ENERGY BALANCE BLOCK Table 3, Mass and Energy balance table S1 S2 S3 S4 S5 S6 S7 T, Celsius T feed T operation T operation T operation T operation P, bar P feed P operation P operation P operation P operation P operation P operation Page 4 of 8

mol/s Solvent (l) 0 0 Solvent (g) 0 0 0 0 0 0 Solute(aq) 0 0 Solute (s) 0 0 0 0 0 0 Total, mol/s h i , kJ/mol Solvent (l) Solvent (g) Solute(aq) Solute (s) H i , kW Solvent (l) Solvent (g) Solute(aq) Solute (s) Total (H), kW T6 needs to be calculated based on the energy balance at the mixing point. T7 = T6 + ΔT HX-01 P operation is calculated using the SLV equilibrium equation: (1-x op )exp(A(1-x op )^2)P sat,solvent = P operation Where P sat,solvent is calculated using Antoine and T operation For x op => see crystallizer block For n̊ sv-l,s1 (molar flow of liquid solvent in stream S1) = M flow(S1) (1- x w %/100)/MW sv (convert to required units) For n̊ so-aq,s1 (molar flow of dissolved solute in stream S1) = M flow(S1) (x w %/100)/MW so (convert to required units) For n̊ sv-g,s2 (molar flow of gas solvent in stream S2) = n̊ sv-l,s1 assumes the all the solvent is evaporated in the crystallizer. For n̊ so-s,s3 (molar flow of solute solid in S3) = n̊ so-aq,s3 (assumes a nearly dry product production) n̊ so-aq,s4 = n̊ so-s,s3 *(MWso/ρ solid )*((1-φ particle )/φ particle )*(ρ operation /MW operation )*x op n̊ sv-l,s4 = n̊ so-s,s3 *(MWso/ρ solid )*((1-φ particle )/φ particle )*(ρ operation /MW operation )*(1-x op ) For ρ operation , MW operation => see crystallizer block For the molar flows in S5 and S6 require to incorporate mass and energy balances in the mixing point and in the heat exchanger. Before discussing those, let us discuss the enthalpy calculations and the overall energy balance. To calculate the molar enthalpies in any stream: Page 5 of 8

h sv-l,Sj = ΔHf° sv-l + cp sv-l (T sj -25°C) , please be careful with the units. h sv-g,Sj = ΔHf° sv-g + cp sv-g (T sj -25°C) , please be careful with the units. h so-aq,Sj = ΔHf° so-s +cp so-s (T sj -25°C) + ΔHsol so-s , please be careful with the units. h so-s,Sj = ΔHf° so-s +cp so-s (T sj -25°C) , please be careful with the units. To calculate the enthalpy flow of each component “i” in stream “j” : H i,Sj = h i,Sj = ΔHf° so-s n̊ i,sj To calculate the total enthalpy flow of stream “j” : H Sj = Σ i H i,Sj The overall energy balance can be used to calculate the duty of the HX-01 Duty of HX-01 = H S3 +H S2 – H S1 Energy balance around HX-01 to find n̊ S5 . Assume T S6 =T operation , and T S7 =T S6 +ΔT HX-01 . As first guess, make n̊ S5 = n̊ S4 . In that case, n̊ so-aq,s5 = n̊ s5 *x op , n̊ sv-l,s5 = n̊ s5 *(1-x op ) , we are going to neglect the solids in S5 (the baffle should have removed a significant fraction). Mass balances n̊ so-aq,s6 = n̊ so-aq,s7 = n̊ so-aq,s4 + n̊ so-aq,s5 ; n̊ sv-l,s6 = n̊ sv-l,s7 = n̊ sv-l,s4 + n̊ sv-l,s5 HX-01 energy balance: H S7 -H S6 - duty HX-01 = 0 , using goal seek to change n̊ s5 until the energy balance closes. Mixing point energy balance to find T S6 . This value is likely very close to T operation (this calculation will not change the previous calculation for n s5 ). Assume T s6 =T operation . Then balance should be H S6 -H S5 -H S4 = 0 , use goal seek to find T S6 . 6 C RYSTALLIZER DESIGN BLOCK Table 4. Crystallizer design parameters x sat (S) estimated Lm(app) (mass-average particle size) μ m x op (mol fraction in the crystallizer) τ (residence time) Hours density of feed (ρ feed ) kg/m^3 density of liquid in suspension (ρ operation ) kg/m^3 MW avg suspension liquid (MW operation ) g/mol Vflow(s1) volumetric flow of feed m^3/s Vl (crystallizer liquid volume) m^3 Dv (diameter from volume) m ρ gas density of vapour kg/m^3 υ gas,max (maximum gas superficial velocity) m/s Page 6 of 8

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Vflow(S2) vapour volumetric flow m^3/s D υgas, max (diameter from υ gas,max ) m D (crystallizer main body diameter) m Hl (crystallizer liquid height) m Q HX-01 (Duty HX-01) kW A HX-01 (Area HX-01) m^2 V settling (settling velocity of the Lm particles) m/s Vflow (S5) (volumetric flow of S5) m^3/s A _annular (annular area for the DTB) m^2 Do (External – annular diameter for DTB) m V fluid (average velocity in the fluid) m/s ε m (power dissipation/mass) W/kg Power of M-01 W υ filtration (Filtration superficial velocity) m/s Area filter m^2 For the given value of T operation , you could use goal seek to find x sat that satisfy the equation: X sat *exp (A(1-x sat )^2)- exp(-(ΔH dissolution /R)*(1/(T operation +273)-1/(Tm+273))) = 0 There is another way to do this via trial and error in Excel –> to be shown in the video. (S)estimated = 3E-4/x sat , this could be used to inform the selection of (S) L m,app = 50μm*(S)^(-0.6) here one would use the (S) in the operating conditions x op = x sat *(1+(S)) τ = 4*L app /G ρ feed =100/(x w %/ρ solute +(100- x w %)/ρ solvent ) ρ operation = (x op* MW SO +(1- x op )*MW SV )/ (x op* MW SO /ρ SO +(1- x op )*MW SV /ρ SV-l ) MW operation = (x op* MW SO +(1- x op )*MW SV ) Vflow(S1) = M flow(S1) /ρ feed , watch out for the units. Vl = Vflow(S1)*τ/(1-φ particle ) For Dv, one assumes Hl/Dv =1, and the volume of a cylinder, such that (πDv^2/4)*Hl = πDv^3/4 = Vl , then Dv = (4Vl/π)^(1/3) For the density of the vapour leaving in S2, one assumes ideal gas, ρ gas = P operation MW sv-g /(R*T operation,K ) watch out for the units, you might find useful using R =8.314E- 5 bar*m^3/k-mol. υ gas,max = 0.0244 m/s*((ρ operation - ρ gas )/ρ gas )^0.5 Vflow(S2) = n̊ s2 *R*(T operation,K )/P operation D υgas, max = ((Vflow(S2)/υ gas,max )*4/π)^0.5 Page 7 of 8

D = max(Dv, D υgas, max ) Hl = Vl/(πD^2/4) Q HX-01 (Duty HX-01) from the mass and energy balance calculations A HX-01 (area HX-01) = Q HX-01 /heat flux V settling = g*(ρ solute- ρ operation )*(L m ^2)*(1-φ particle )^4.65/(18µ fluid ) Vflow (S5) = n̊ s5 * MW operation /ρ operation A _annular = Vflow (S5)/ V settling = π (Do^2-D^2)/4 Do = (4*A _annular / π + D^2)^0.5 V fluid = V settling *(V fluid /V settling factor) ε m = v fluid *g*(ρ particle -ρ operation /ρ operation ) Power of M-01 = ε m * Vl*(ρ operation + ρ particle *φ particle /(1- φ particle )) To obtain υ filtration , use goal seek to obtain ((1- ε cake )/ε cake ^3)*(150/(ρ operation *υ filtration *L app /(µ fluid *(1- ε cake )))+1.75)*(L cake /L app )* ρ operation * υ filtration ^2 - ΔP cake = 0 A filter = (1/0.3)*(n̊ s4 * MW operation /ρ operation )/ υ filtration . The factor 1/0.3 assumes a 30% of active area in the rotary drum. Page 8 of 8