Question

Particles having size of 0.1mm, a shape factor of 0.8 and a density of 1650 kg/m' are to be fluidized using air at 25°C (p=2.374 SI, u=1.45x10ʻSI) and 202.65kPa (abs pressure). The bed's height at minimum fluidizing conditions is Lmf=1.6m. The bed diameter is 0.5m and the bed height when the particles are at rest is Lo=1m. 3.1. What is the mass of particles to be used ? 3.2. Calculate the minimum void fraction of the fluidized bed? 3.3. Calculate the pressure drop at minimum fluidizing conditions ? 3.4. Calculate the minimum velocity for fluidization? (hint : compute (a,b,c) of the equation a.V2 + b. Vm + c = 0 found in the appendix) 3.5. If the expansion of the bed is to insure that e' is between 1.1 Eny and 1.3 Enf, What should be the expansion velocity v'?

Answer 1

Answer:- The Giren data is, Particle size DP = 0.1mm Shape factor =0.8 density of Partide pp = 1650 kg/mo Density of aio Pf -d 374 kg/m3 Viscosity of ait M = 1.45x105 Pas minimum Huidization bed length Lmp -1.60 Bed diameter D-0.5 Bed Icngth when pasticicos asc at fast 20-IM -> we need sphesicity for this problem, - spheticity = do = shape 58 = 1.25 factor 3.1) what is the mass of pasticks to be used mass - density X VOLUME so mass of each pastiche me - fe хve Now Vp = a(OpJ3 -(0.8)*(0.1X103 je vp = 8X10" ma

Per 1650 kg/me mp = Pp xe mp = 1650 x8x1063 kg mp = 1.80 x109 kg void traction of the froidized bed 342) Calculate the minimum Total mass of pousticle 3 betose fividization Tota1 mass of particico affeo fiudization -> (m) belosc = (m) after Pp (V) belose Pp (after) Allength) setoo = A (length) after 20(1-E) - Lot (1-Emt) Hest & is void traction when Pasticks we are at sast there is nothing mentioned io question Go We ate assuming it to be 3000 Go to = Lmt (1-Emt) . Substituting, 1=1.6(1-Emb) Emp 1 G Ent = 0.375]

3.3 calculate the possuo doop at minimum fiudizina Condition. POxsovoc drop can be -> At the minimum fiuidization conditions diocchy found by using the formula AP = (1-Emp) imf (Pp-7)x9 sp - (1-0.375) 16 (1650 - 2:374)*9.8 Ise- 16146.7348 pa 34) Calculate the minimum velocity foo fluidization formua known as Hess, we are using the most geonesas 'Goguo equation" Eo quo equation ;- 11.45 pg Lut , C1- Em lumi (150 MLmt x (1 Emp) Jump + ( Pop (Emtje Jump DP = ( Do De Em ? Reassanging the terms: - 1 175 Pt. Lont - 150% 297 x (1 -EmpJump-Op 30 (1-Emt) Jums + ( op Cempios OSDP (Em 3 ) Poop Cemtjo in -> Hesc, a, b, c ate accooding to hiot a 1.75 X .374x1. 125 X0.1 X163 (0.37583

Ta 620253.037) - 150x145x105 x 16 x (1.0.375² Clasje (0.18159 (0.3753 b= 1649777 178 - = C = -1614617348 now, aiunt + bime +c=0 -> Here, we are using a scientific calculator anc positik soot and on negative soot we will delenity take positik tool that is obvious I vint -9.46x10°mis. 3.5) It the expansion of the bed is to insure that εis between them and 1.3emt. What should be the expansion Velocity v? → As ' is in beteren HEMA and IJEME. so we take & = 1.28EMA &' - 1. 2X01375 [ε =0457

-> Now, According to theory, As & temt it means length of fluidiaed bed is incocased, but pocosuoc doop semains constant after seaching Holidation Conditions in the fiudization so, using the Juation ارد AP = (1-EJL (Pp - PfJg (1) Eogun Suation, ne sourL (1-EJ" 175 p (v) xL x 1-6 SP RP 8 OBDP & > Equating both the duation we get, 150 MV! x (1-s'] + 1.7504 x (Uy to : 9 (Pp-PH) as ope (€) QSDP 150x1.45X10 X (-0.45), v't /1.75X2:374 1-0.45) 1(orje Cras y + x[0.1 X1033 Cougis ) * (C125) *xlovlx1ājCous je lag 59.8(1650 - 2.374) -> Buol 64.6091) v' + (36 4729-7663) (v')*=1614617348 (361729. 7663) (V+)* +(840164.6091) 0'-16146: 7348= 0 -7 saving esing scientific calculatos, eux act n's 0.0190 mts