Mass Transfer (Stoffaustausch) Fall Semester 2014

Transcription

1 Ma Tranfer (Stoffautauch) Fall Semeter 4 Tet 5 Noember 4 Name: Legi-Nr.: Tet Duration: 45 minute Permitted material: NOT permitted: calculator copy of Culer book Diffuion ( nd or rd edition) printout of the lecture cript, without note on the exercie heet ( page) ummary exercie (alo handwritten on ummary), notebook, mobile phone

2 Problem (5 point) A mall cented alt ball i floating at the bottom of a ery large bathtub filled with water. The alt aturation concentration at the urface i contant, but the diameter of the alt ball decreae oer time. The concentration profile i at teady tate. a) Draw a ketch of the problem and draw the alt concentration profile at t =. (6 point) b) Find the concentration profile, tarting from the generalized ma balance equation. Aume here no conection and tate all further aumption made. Calculate the flux at the ball urface. (9 point) c) Calculate the ma tranfer coefficient. Jutify your choice of appropriate ma tranfer coefficient correlation. What i the contribution of pure diffuion? Wa the aumption of neglecting conection in (b) reaonable? ( point) d) Conidering conection, how long will it take to reduce the ball diameter to half it initial alue? Aume a contant ma tranfer coefficient and do not forget that the ball radiu i a function of time. (5 point) Data: C at=6.4* -4 mol/ Diffuion coefficient of alt in water = 6.5* -5 / Denity of alt =.7 g/ Denity of water = 998. kg/m Denity of water aturated with alt =.7 kg/m Schmidt number (Sc) = Initial Diameter of alt ball = Molecular ma of alt = g/mol

3 Problem (5 point) In water purification, wate water ( o C) contaminated with tetrachlorobiphenyl (PCB, C H 6Cl 4, ee Fig. ) i fed at L/min through a packed bed of pherical actiated carbon particle with a diameter of.5 (Fig. ). The packed bed ha a height of L = 8 and a diameter of. PCB i adorbed immediately upon contact with actiated carbon. (a) Calculate the diffuion coefficient of PCB (C H 6Cl 4) in water. Aume PCB molecule are pherical. (4 point) (b) Calculate the ma tranfer coefficient for the purification ytem. If you hae not oled (a), make a reaonable aumption for the diffuion coefficient to proceed. ( point) V Az (c) Derie a differential equation from the ma balance oer uing the ma tranfer coefficient, where A i the cro ectional area of the packed bed. (6 point) (d) Calculate the required pecific exchange area (= urface area of the actie carbon/bed olume) to remoe 99% of PCB. ( point) Data: Vicoity of o C:.9 g/( Denity of o C: 998. kg/m ) Figure. Tetrachlorobiphenyl (PCB, C H 6Cl 4) z = C PCB V r x d Reactor z = L=8 Figure. Schematic image of the packed bed

4 Solution a) Draw a ketch of the problem and draw the alt concentration profile. c bal l r(t) r i water c at = 6.4* -4 mol/ r i =.5 c = b) Find the concentration profile and calculate the flux at the ball urface. Start from the generalized ma balance equation. Aume here no conection and tate all further aumption made. r General ma balance in pherical coordinate: dc dt + c r r + θ c r θ + φ c r in θ φ = D [ c r r (r r ) + r in θ θ (in θ c θ ) + c r in θ φ ] + r Aumption: ) No conection: r = θ = φ = ) Steady tate: dc dt = ) Symmetry: = and = θ φ 4) No reaction: r = Simplified ma balance: = D r r (r c r ) Integrate, with a and b a contant: c r = a r c = a r + b Uing the boundary condition c (r = ) = and c (r = R) = c,at c (r = ) = b = b = c (r = R) = a R + b = a R = c,at a = Rc,at 4

5 c (r) = R r c,at The flux i calculated a j(r) = D dc dr = D R r c,at The flux at the phere urface (r=r) i o j(r) r=r = D R c,at Plug in the alue: j(r) r=r = = [ mol ] c) Calculate the ma tranfer coefficient. Jutify your choice of appropriate ma tranfer coefficient correlation. What i the contribution of pure diffuion? Wa the aumption of neglecting conection in (b) reaonable? i) The phyical ituation i a free conection ( ρ = =.5 g/ > 9 g/ ) around a olid phere, thu from table 8..: kd D =. +.6 (d ρg ρ ) 4 ( D ) The ma tranfer coefficient can be calculated a: k = D d [. +.6 ρg 4 (d ρ ) ( D ) ] Recognize that the Schmidt number i defined a Sc =, o that the kinematic icoity D can be calculated: = Sc D = = 6.5 Plug in the numerical alue: k = =.5 [. +.6 ( ( ) (6.5 ) ) ii) The contribution of diffuion i k diffuion = D d. 6.5 ( 6.5 5) ] 5

6 Plug in the alue: k diffuion =. =. 4 / The diffuion contribute only to k diffuion % = 6.4% of the total ma tranfer k coefficient. Thu conection i important and hould not be neglected. d) Still conidering conection, how long will it take to reduce the ball diameter to half it initial alue? Aume contant ma tranfer coefficient. The diolution rate i written a a ma balance on the ball: d (m Mw) dt = k A (c,at c ) Deelop the term dependent on the ball radiu: d(v ρ MW) = k 4πr (c dt,at c ) d ( 4 πr ρ MW) 4π ρ MW dt dr = k 4πr (c,at c ) dt = k 4πr (c,at c ) 4π ρ MW r dr dt = k 4πr (c,at c ) dr dt = k MW ρ (c,at c ) = k MW ρ c,at Integrate (initial diameter: at t=, r= r i) r t dr = k MW r i ρ c,at (r r i ) = k MW ρ dt c,at t r = r i k MW ρ c,at t 6

7 t = (r i r) ρ MW k c,at Time to reduce the diameter to half it initial alue: r = r i t = (r i r i) ρ k MW c at = Plug in the numerical alue: t = r i ρ k MW c at = 7.9 hr 7

8 Solution a) D k T f k T 6R B B (a-) where f i the friction coefficient of the olute, k B i the Boltzmann contant, μ i the olent icoity, and R i the olute radiu which can be calculated auming that the hape of the PCB molecule i pherical: R V PCB (a-) 4 VPCB A (a-) Thu, V PCB 47.48Å 8 (a-4) R.895Å The diffuion coefficient i: 6 g.8 9.5K kt B D K 6.6 6R g (a-5) b) Ma tranfer coefficient for packed bed: k d.4 / D.7 (b-) where : uperficial elocity without packing; : kinematic icoity of the water; D: diffuion coefficient of PCB in water; d: particle diameter of actiated carbon Superficial elocity of the bed: Q L/ min / 6. A dreactor 5 Thu, (b-) 8

9 k.4 / d D.7.4 / c) Ma balance oer a differential length Δz: i A c A c k V c c (c-) z zz where V Az, c c zz z k c c z i (c-) z dc dz k c c i (c-) d) At the urface of actie carbon, c i : dc k c (d-) dz Integration:.c c L k dc c dz (d-) Note that z : c c and z L (8 ) : c. c. 9

10 Thu, ln c. c c k L z (d-).c ln c k L (d-4) Sole for. ln() ln() 8.6 k L (d-5)