Chemical and Mechanical Design of Fluidized Drier used for Ammonium Sulphate Drying

Based on approximation, simple steps are used to know how to design a fluidized bed drier, of course we can give complete design but some useful information can help a bit. The below equation is called as Ergen’s equation. This equation is generally used for packed beds. Since in fluidized bed there is no pressure drop, we can use the below equation in fluidized bed regime. And the equation is

[(150*µ *VOM/ Øs2 Dp2)(1 – ε M)/ (ε M)3 ] + {(1.75 * ρ * VOM2)/ Ø s D p}* (1/ (ε M) 3}= g(ρs – ρ)

  • µair at 1400C                                       =          0.023 × 10-3 kg/ m. sec
  • Ø s                                                           =          0.9 (for nearly spherical particles)
  • Characteristic diameter          =          D p = 1.25 mm = 1.25 × 103 m
  • Mean void fraction, ε M                           =          0.43
  • Acceleration due to gravity, g          =          9.8 m/sec2
  • ρs = wet particle density                     =          1860.72 kg/m3
  • ρ air at 1400C                                                 =          0.9 kg/m3

Therefore on substituting the above values in the equation, we obtain

  • 19542.65VOM + 17608.512 VOM2    =          18226.236

Upon solving,

  • VOM = 0.6039 m/sec

Therefore minimum fluidization velocity = 0.6039 m/sec

Let volumetric flow rate of air be q air,

  • q air                                                        = m air/ ρ air
  • ρ                                                             = (1 × 29)/(0.0821 × 413)  = 0.855 kg/m3
  • q air                                                        = 997.813 / 0.855 = 1167.032 m3/hr  = 0.324 m3 / sec
  • Operating velocity, V              = 1.5 VOM (as per industrial data) = 1.5 × 0.6039 = 0.90585 m/sec
  • Therefore,
  • Area,                           A         = q air / V = 0.324/ 0.90585   = 0.35768 m2
  • Diameter                                = √ (A × 4 / П)   = 0.6748 m

Fluidized drier Length can be obtained by two approaches:

Fluidized Drier length calculation METHOD 1:

  • ρ s                         =          (mass of solid) / (volume of the solids) =     [W × (7/60)] / [(П/4) D 2. LS (1 – ε M)]
  • Where W         =          2373 kg/hr
  • LS                          =          0.7299m
  • LS (1 – ε M)         =          Lf (1 – ε f), Lf is fluid bed height.
  • ε M                       =          0.43
  • ε f                        =          0.6 to 0.7

Let it be 0.65

  • 0.7299(1-0.43) = Lf (1-0.65)
  • Lf                            =          1.1887 m

Fluidized Drier length calculation METHOD 2:

  • Ґ                      =          (volume) / (volumetric flow rate)

Where Ґ is space time

  • 7 / 60              =          [(П / 4) D2 L] / (2452.88 / 1860.72)
  • L × 0.2712     =          0.116
  • LS                           =          0.427 m
  • LS (1 – ε M)       =          Lf (1 – ε f)
  • Lf                            =          0.6954 m

We have calculated Lf using two methods. We consider METHOD 1 since the length that’s calculated using METHOD 2 will have porosity less than 0.60.

Therefore,

  • Minimum fluidization velocity      =          0.6039 m/sec
  • Diameter of fluid bed drier               =          0.6748 m
  • Static height of bed, Ls                          =          0.7299 m
  • Height of fluidized bed                        =          1.1887 m
Mechanical Design of Fluid Bed Drier

Thickness of the column,

PC = [(2.42 ×E) / (1 – µ2) (3/4)] × [(t / D0) (5/2) / {(L / D0) – (0.45) (t / D0) (1/2)}]

  • Young’s modulus,   E         = 190 × 103 MN / m2
  • Poisson’s ratio,         µ          = 0.305
  • Let the thickness‘t’ be 4 mm         =          0.004m
  • Outer diameter D0           =          D i + (2 ×t)  =          0.6748 + (2 × 0.004) =          0.6828 m
  • Total drier height                              =          1.5 Lf = 1.7831 m
  • L                                   =          (3 / 4) × (total height)  =          1.3373 m
  • PC                               =          675377.94 N / m2 =          6.665 atm
  • PC / 4 = 6.665 / 4                 =          1.666 atm

As 1.666 atm is greater than operating pressure which is 1 atm, 4 mm thickness is feasible.  Thickness of the hemispherical head,

t h  = 4.4 RC √ [3(1-µ)2]   √(P / 2 E)

Where,

  • RC                  =          crown radius = 0.6748 m
  • µ                      =          0.305
  • E                     =          190 × 103 ×106 N /m2
  • P                     =          101325 N /m2

Therefore

  • t h                     =          1.8456 mm
  • Thickness of the hemispherical head                  = 1.8456 mm

 Thickness of fluidized drier conical bottom,

  • th                      =          (P × D) / (2fJcosα)
  • α                      =          half apex angle of the cone          =           600
  • P                     =          design pressure       = 1 atm           =          101325 N/m2
  • f                       =          design stress at operating temperature   =          P ×D / (2tcosα)   =          17093527.5 N/m2
  • D                     =          Diameter of the plate = inner diameter = 0.6748 m
  • J                      =          joint efficiency          = 0.9

 

Therefore, thickness of conical bottom = 4.4 mm

Hemispherical head is attached by lap welding and conical head using butt welding. At the junction of the conical head and cylindrical shell, localized stresses are produced. In order to intensity of these localized stresses in the case of conical heads where the angle exceeds 300, provision has to be made to attach it to the shell, by a knuckle having inside radius not less than 10% of the internal radius of the shell.

Auto CAD 2D drawing of ammonium sulphate fluidized drier

Simple model Auto CAD drawing of fluidized bed drier


DESIGN FLUIDIZED DRIER SUPPORTS:

  • Diameter of the vessel                    =          0.6748 m
  • Height of the vessel                        =          1.7831 m
  • Clearance from vessel bottom
  • of foundation            `                       =          0.3m
  • Weight of the vessel with
  • contents                                             =          dead weight + weight of the contents=    1000 + 2453=    3453 kg=    3453 × 9.8=    33839.4 N

From literature, for the above force

  • Number of supports                         =          3
  • Maximum compressive load, P      =          33839.4 / 3 =          11279.8 N
  • From data in a literature, for diameter 0.6748 m
  • a                      =          150 – (2 × 4 ×10-3)=          149.992 mm
  • B                     =          150 mm

Thickness of the fluidized drier base plate,

Average pressure on the

  • plate, P av                                        =          P/ (a × B)  =          11279.8 / (149.992 × 150)  =          0.50135 N/ mm2
  • Stress, f                                           =          0.5 P av (B2 / T1 2) (a 4 / [B4 + a4])
  • We have, f                                         =          157.5 N/ mm2

Where T1 is the thickness of the base plate.

  • Therefore T1                                                  =          5.0065mm

T1 can be taken as 6 mm.

Thickness of the web plate,

Bending moment on each plate  =    (P / 2) [(A – D0) / 2] × 100  =  (11279.8 / 2) [(0.8 – 0.6828) / 2] ×100 = 33049.8 N cm

Stress on the edge of f   =  [(bending moment) / (T2 × a × a)] ×(1/cos 45) × (1/100) => 157.5 = [(33049.8/ (T2 × 14.9992 × 14.9992)] × (1/0.707) × (1/100) =>T= 0.027 m = 2.7 mm

Therefore T2, web plate thickness can be taken between 3 to 4 mm.

In chemical design we consider mostly the flow rate and pressure of air required to the know volume and mass of the ammonium sulphate and optimum dimension of the drier. Thickness of supports and it accessories position is dealt in mechanical design section. Design of the fluidized drier effectively drawn in AUTO-CAD software which can be used for fabrication.