# Question: Operation of Batch Mixers: A viscous aqueous solution {density of the liquid pL = 1500 kgfmi viscosity of the liquid pi; = 0.01 Pas) containing spherical sucrose particles of initial radius HI, = 6 mm is agitated in a tank of diameter (if = 2 m ﬁlled to a height HT = 2 m. A six-bladed turbine with a diameter of the impeller (ii = 0.5 m rotates at a rotational speed N = 90 rpm. -Free Course Hero Question Answer.

Question Description: *Operation of Batch Mixers: A viscous aqueous solution {density of the liquid pL = 1500 kgfmi viscosity of the liquid pi; = 0.01 Pas) containing spherical sucrose particles of initial radius HI, = 6 mm is agitated in a tank of diameter (if = 2 m ﬁlled to a height HT = 2 m. A six-bladed turbine with a *diameter of the impeller (ii = 0.5 m rotates at a rotational speed N = 90 rpm. The tank is baffled, and the system corresponds to the so-called standard turbine conﬁguration. The tank can be considered large enough that the bulk concentration of sucrose is negligible. For this conﬁguration, Kulov et a1. (Chem. Eng. Commun, 21, 259 (1933)) suggested a correlation for Sherwood number Sh for mass transfer to particles in stirred tanks as 4 if! Sh *- ‘d = 0.2673c1a’4393-“N1″ (ﬂ) _ I)” P Yd!- 1Where Reynolds number He, Schmidt number Sr: and power number Np are defined as R N dig S VL NP* "L 3.43 P FLNgdiS Where, “1.1 kinematic viscosity of the liquid V: liquid volume in the vessel P: mixing power a] [6 points] What is the expected mass transfer coefficient kc (in mfs] between the agitated liquid and the solid particles immersed in the liquid? For this conﬁguration, M; = 6. The diffusion coefficient of the dissolved solute in the liquid I)” is 3 x 10’1″ mzfs. b) [9 points] Assuming pseudo-steady operation. find the time to dissolve 10% of the initial mass of sucrose. The density and solubility in water of sucrose are given as p3 = 1600 lag/m3 and CE = 1300 kgg’mS.*

**Course Hero** **Answer & Explanation**:

(a) k_{c} = 1.95 * 10^{-3} m/s

Since all parameters are in SI units so rotational speed has also been converted in SI unit

(b) time= 95.477 seconds

Image transcription text

(b) Mass balance Post BurstAre N VdCs Kea (( -C’s ) (CS = At equilibrium des = It HT ( ( * – CS ) Since Ps = 1600 kg /m] After 10% of initial mass dissolve Cs = 160 kg / m 3 = 1 . 95 X 10 – 3 (C* – Cs ) 160 de ( 1800 – C3 ) ( 1 . 95 x 10- 3 ) O CS = 0 1 160 – In ( 1800 – ( s ) = 1. 85 x 10- 3 O N – lu / 1800 – 160 1800 1. 95 x 10-3 += 95.47 7 seconds 66