No: Date: a) 0.2 mm = o.2 xlo°m -3 200 xlo ms 200um. 3. E1- 0.365 ( log (200) -1) : 0.5368 z 0.537 mf b) 2. 3-1 me m. d. th00 0.1768 0.177m. L= (0-177m) 3.822 m 1-0.637 C) F. %3D 6(3-1) 20x29 12.34...

please verify my answers.

That is a fluidized bed reactor containing 15000 kg of sand (2700 kg/m3 density and approximate 0.2 mm in diameter) is experiencing an air flow at 300 ºC. If the entering air is at 20 atm and its viscosity can be assumed to be 0.021 mPa.s.

Determine:

a) The minimal porosity for fluidization if the reactor diameter is 2 m.

b) The minimal height of fluidized bed,

c) The pressure drop,

d) The minimal fluidization.

Extracted text: No: Date: a) 0.2 mm = o.2 xlo°m -3 200 xlo ms 200um. 3. E1- 0.365 ( log (200) -1) : 0.5368 z 0.537 mf b) 2. 3-1 me m. d. th00 0.1768 0.177m. L= (0-177m) 3.822 m 1-0.637 C) F. %3D 6(3-1) 20x29 12.34 K9/m² (0.0828)(573) (-AP)= 3.822 (2700-12.34)-0.537) (9.8) = 46609.25e Moderointe


Extracted text: Date: こ1ち。 Vmf + 1.75 2. 3 mf [l-0. 537)(2.1x105) -3 (0.537)8 (0.2xlo°)² (2700-12.34)(9.8)=150 12.34 (Vmp) Ymf + 1.75 (0.5年)02d6) 26339.07= 23G455042 Vmp 697268.92( Vmf ) 235455.424mf +677268.97 ( Vmt)?=26339.07 2354 65.42 Vmp s0 +697268.99(vmp)²-2633 9.07 0.0 886 %3D or -0.426 メ Vmf= 0.0886v mf=0.09 m/s Prdp = l0.58 5- 2.1xlo?