A fluidized-bed dryer is to be sized to dry 5,000 kg/h (dry basis) of spherical polymer beads that...

Question

A fluidized-bed dryer is to be sized to dry 5,000 kg/h (dry basis) of spherical polymer beads that are closely sized to 1 mm in diameter. The beads will enter the dryer at 25°C with a moisture content of 80% (dry basis). The drying medium will be superheated steam, which will enter the bed at 250°C. The pressure in the vapor space above the bed will be 1 atm. A fluidization velocity of twice the minimum will be used to obtain bubbling fluidization. The bed temperature, exit solids temperature, and exit vapor temperature can all be assumed to be 100°C. The beads are to be dried to a moisture content of 10% (dry basis). The void fraction of the bed before fluidization is 0.47. The specific heat of the dry polymer is 1.15 kJ/kg · K, while the density is 1,500kg/m3. Batch fluidization experiments show that drying will all take place in the falling-rate period, as governed by diffusion according to (18-92), where 50% of the moisture is evaporated in 150 s. Bed expansion is expected to be about 20%. Determine the dryer diameter, average bead residence time, and expanded bed height. Is the dryer size reasonable? If not, what changes in operation could be made to make the size reasonable? In addition, calculate the entering superheated-steam flow rate and the necessary heat-transfer rate.

Robert · Accepted Answer

ID -376285 
 Fluidised bed  dryer- 
  Mass  flow  rate  of spherical  polymer beads  = 5000 kg  /hr  =  5000 /60 *60  =1.39  kg  /s 
 D = diameter   of beads     = 1  mm  =  1 /1000 =0.001  m 
 Radius    of beads  =  0.001/2=  0.0005  m 
 Initial  temperature  = t = 25  oc   , initial  moisture    =80 % =80/100  =0.8 
  Super heated  steam   =  T  = 250  oc   , P = 1   atm ,   fluidization   velocity   = 2 * V min   
 Final    moisture  =   10%  = 10 /100  =0.1  s 
  Void  fraction   =  Vf  = 0.47 
  Specific    heat  =   1.15  kj /kg –k 
  Density  of    polymer beads  = 1500  kg/m3  ,   drying  falling  rate   period  =150  sec  , during  
fludisation   ,  moisture  content  = 50 % = 0.5  , bed  expansion = 20 %  = 20 /100=0.2 
  Area  of beads  = A  =  4∏*r^2  = 4*22/7*(0.0005)^2   = 0.0000031 m2 
  Volume  of beads  = V  =  4/3*∏*r^3  = 4/3*22/7*(0.0005)^3  = 5.24*10^-10 m3 
R = RATE OF DRYING  =  --dm /2*S* dt  ----( 1) 
  Dm = dm  =2S *s*ρs*dx   ----(2)
 Here  S  =  AREA OF CROSS SECTION  OF  SOLID ,s  = thick  ness  of  solid  , ρs  = density of solid 
 Combining  eq  (1) and (2) – 
R  =   - s*ρs*dx/dt    ----(3) 
  Tt =  total  drying  time  =  s*ρs∫ ( limit  x1  -x2) ,dx /R  -----( 4) 
    Tc   = constant drying period = s*ρs *∫(  limit  x1-x2) .dx /Rc  =  s*ρs (x1-x2) /Rc   ----(5) 
   When  the   rate of drying   is linear  in  X  during   a  falling   rate  period   -- 
  R = a*X + b   ---( 6) 
 Tf  = s*ρs/a *∫( limit R1-R2) . d R /R   = s*ρs/a .ln R1/R2   ----- (7) 
 HERE  ,

A fluidized-bed dryer is to be sized to dry 5,000 kg/h (dry basis) of spherical polymer beads that are closely sized to 1 mm in diameter. The beads will enter the dryer at 25°C with a moisture content...

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