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FLOW IN PACKED BEDS FLUID FRICTION IN POROUS MEDIA PACKED TOWERS • Packed towers are finding applications in adsorption, absorption, ion-exchange, distillation, humidification, catalytic reactions, regenerative heaters etc., • Packing is to provide a good contact between the contacting phases. • Based on the method of packing, Packings are classified as (a) Random packings (b) Stacked packings • The packings are made with clay, porcelain, plastics or metals. Packing Void fraction, ε Berl saddle 0.6 - 0.7 Intalox saddle 0.7 - 0.8 Rasching ring 0.6 - 0.7 Pall ring 0.9 - 0.95 Principal requirements of a tower packing • It must be chemically inert to the fluids in the tower. • It must be strong without excessive weight. • It must contain adequate passages for the contacting streams without excessive pressure drop. • It must provide good contact between the contacting phases. • It should be reasonable in cost. FLUID FRICTION IN POROUS MEDIA • In this approach, the packed column is regarded as a bundle of crooked tubes of varying cross sectional area • The theory developed for single st. tubes is used to develop the results of bundle of crooked tubes….. – Laminar flow – Turbulent flow – Transition flow Laminar flow in packed beds • Porosity (void fraction) is given by ε = (volume of voids in the bed / (total volume of bed ) EMPTY TOWER VELOCITY • Superficial velocity ‘vs’ = (Q / Apipe) • Interstitial velocity ‘vI’ = (Q / ε Apipe) Velocity based on the area actually open to the flowing fluid vI vs i.e., AREA AVAILABLE FOR FLOW = A ε • In a packed bed consider a (non-circular CSA) set of crooked tubes • rH = (cross sectional area of channel) / (wetted perimeter of channel) • Multiply and divide by LENGTH of bed =rH = (A ε ) L / (wetted perimeter) L • =rH = ε (volume of bed) / (Total wetted surface area of solids) • To find wetted surface area……. • Total wetted surface area of solids = (no. of spherical particles) x (surface area of one particle) and we know…. No. of particles = (volume of bed) (1- ε) / (volume of one particle) volume fraction (vol.bed) rH (vol.bed)(1 ) surfaceareaof one particle vol.of oneparticle vol.of oneparticle 1 surfaceareaof one particle D 3p 6 1 D 2 p rH Dp 1 6 so, modified Re ynoldsno. N Re, p Deq v I 4 Dp vs 6 1 ERGUN defined NRe,p without the constant term (4/6) for PACKED BED N Re, p 1 D p vs 1 for La min ar flow( Hagen Poiseuilleeqn) Lv P 32 2 I Deq vs L 32 2 D p 4 6 1 v s (1 ) 2 P 72 2 L Dp 3 By several experiments it has been found that the constant value should be 150 KOZNEY-CARMANN EQN. v s (1 ) 2 P 150 2 L Dp 3 Only if NRe,p < 10 (LAMINAR) Turbulent flow in packed beds we use the general equation for friction factor PDeq 2 f f v P s 2 2 Lv I2 Dp L 4 2 6 1 v s2 (1 ) 3f Dp 3 By several expts it has been found that for turbulent flow, the ‘ 3f ’ should be replaced by a value 1.75 P 1.75 v s2 (1 ) L Dp 3 BURK- PLUMMER EQN. Only if NRe,p > 1000 (TURBULENT) FOR TRANSITION REGION… 2 2 150 v ( 1 ) 1 . 75 v P s s (1 ) 2 3 L Dp Dp 3 ERGUN EQN. if NRe,p between 10 and 1000 (TRANSITION) PROB….. • Calculate the pressure drop of air flowing at 30ºC and 1 atm pressure through a bed of 1.25 cm diameter spheres, at a rate of 60 kg/min. The bed is 125 cm diameter and 250 cm height. The porosity of the bed is 0.38. The viscosity of air is 0.0182 cP and the density is 0.001156 gm/cc. Data: • Mass flow rate of Air = 60 kg/min = 1 kg/sec • Density of Air () = 0.001156 gm/cc = 1.156 kg/m3 • Viscosity of Air () = 0.0182 cP = 0.0182 x 10-3 kg/(m.sec) • Bed porosity () = 0.38 • Diameter of bed (D)= 125 cm = 1.25 m • Length of bed (L) = 250 cm = 2.5 m • Dia of particles (Dp)= 1.25 cm = 0.0125 m • Volumetric flow rate = mass flow rate / density = 1 / 1.156 = 0.865 m3/sec • Superficial velocity Vo = 0.865 / ( (/4) D2 ) = 0.865 / ( (/4) 1.252 ) = 0.705 m/sec • NRe,P = 0.0125 x 0.705 x 1.156 / (0.0182 x 10-3 x ( 1- 0.38 ) ) = 903…….Transition region • We shall use Ergun equation to find the pressure drop. • p = 2492.92 N/m2 Pressure Drop in Regenerative Heater • A regenerative heater is packed with a bed of 6 mm spheres. The cubes are poured into the cylindrical shell of the regenerator to a depth of 3.5 m such that the bed porosity was 0.44. If air flows through this bed entering at 25ºC and 7 atm abs and leaving at 200ºC, calculate the pressure drop across the bed when the flow rate is 500 kg/hr per square meter of empty bed cross section. Assume average viscosity as 0.025 cP and density as 6.8 kg/m3. • Mass flow rate of Air / unit area = 500 kg/(hr.m2) = 0.139 kg/(sec.m2) • Density of Air () = 6.8 kg/m3 • Viscosity of Air () = 0.025 cP = 0.025 x 10-3 kg/(m.sec) • Bed porosity () = 0.44 • Length of bed (L) = 3.5 m • Dia of particles (Dp)= 6 mm = 0.006 m • Superficial velocity Vs = mass flow rate per unit area / density = 0.139 / 6.8 = 0.0204 m/sec • NRe,p = 0.006 x 0.0204 x 6.8 / (0.025 x 103 x ( 1- 0.44 ) ) = 59.45 • We shall use Ergun equation to find the pressure drop. • ∆P = 46.37 N/m2 Design of Packed Tower with Berl Saddle packing • 7000 kg/hr of air, at a pressure of 7 atm abs and a temperature of 127oC is to be passed through a cylindrical tower packed with 2.5 cm Berl saddles. The height of the bed is 6 m. What minimum tower diameter is required, if the pressure drop through the bed is not to exceed 500 mm of mercury? For Berl saddles, p = (1.65 x 105 Z Vs1.82 1.85 ) / Dp1.4 where p is the pressure drop in kgf/cm2, Z is the bed height in meter, is the density in g/cc, Dp is nominal diameter of Berl saddles in cm, Vs is the superficial linear velocity in m/sec. Data: • Mass flow rate = 7000 kg/hr = 1.944 kg/sec • Height of bed (Z) = 6 m • Dp = 2.5 cm • 760 mm Hg = 1 kgf/cm2 = 1 atm • p = 500 mm Hg = (500/760) x 1 kgf/cm2 = 0.65 kgf/cm2 • Formula: • Ideal gas law: • PV = nRT • Formula given, • p = (1.65 x 105 Z Vs1.82 1.85 ) / Dp1.4 Calculations: = M(n/V) = M(P/RT) = 29 x 7 x 1.01325 x 105 / (8314 x (273 + 127) ) = 6.185 kg/m3 = 6.185 x 10-3 g/cc • p = (1.65 x 105 Z Vs1.82 1.85 ) / Dp1.4 • 0.65 = (1.65 x 105 x 6 x Vs1.82 x (6.185 x 10-3 )1.85 ) / 2.51.4 • Vs1.82 = 0.02886 • Vs = 0.1432 m/sec • Volumetric flow rate = mass flow rate/density = 1.944/6.185 = 0.3144 m3/sec • Required Minimum Diameter (D) = 1.6719 m. • Air flows thro a packed bed of powdery material of 1cm depth at a superficial gas velocity of 1cm/s. A manometer connected to the unit registers a pressure drop of 1cm of water. The bed has a porosity of 0.4. Assuming that Kozney-Carmann equation is valid for the range of study, estimate the particle size of the powder? Density of air = 1.23kg/m3 viscosity of air = 1.8x10-5 kg/m-s • Dp=1.24x10-4m Flow Rate of Water through IonExchange Column Figure shows a water softener in which water trickles by gravity over a bed of spherical ion-exchange resin particles, each 0.05 inch in diameter. The bed has a porosity of 0.33. Calculate the volumetric flow rate of water. Assume laminar flow. • Applying Bernoulli's equation from the top surface of the fluid to the outlet of the packed bed and ignoring the kinetic-energy term and the pressure drop through the support screen, which are both small, we find ……… p1 p2 1 2 1 2 v1 gh1 v2 gh2 h f 2 2 g(∆z) = hf hf = ∆p/ρ=3.7376 J/kg • Since Laminar flow, apply Kozney-Carmann equation • vs = 0.01055 m/sec • = Q = 21cm3/sec • Water trickles by gravity over a bed of particles each 1mm dia in a bed of 6cm and height 2m. The water is fed from a reservoir whose dia is much larger than that of packed bed, with water maintained at a height of 0.1m above the top of the bed. The bed has a porosity of 0.31. calculate the volumetric flow rate of water if its viscosity is 1cP Shape factor-Sphericity factor • For non-spherical particles instead of diameter an equivalent diameter is defined. • Sphericity Φs is defined as the surface-volume ratio for a sphere of dia Dp divided by the surface-volume ratio for the particle whose nominal size is Dp. • Φs = (6/Dp) / (sp/vp) • Therefore, actual dia to be used in Ergun eqn is = Φs Dp • For a non-spherical particle, Ergun eqn is given by……… 1.75 v (1 ) P 150v s (1 ) 2 2 3 3 L s D p s D p 2 2 s