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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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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.
Answered Same DayDec 31, 2021

Solution

Robert answered on Dec 31 2021
82 Votes
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 = Rc -- R’/Xc -- X’ --- (8)
Rc = rate at first critical point
R’ = rate at ...
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