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. No:Date:a)0.2 mm = o.2 xlo°m-3200 xlo ms 200um.3.E1- 0.365 ( log (200) -1) : 0.5368 z 0.537mfb)2.3-1mem.d.th000.1768 0.177m.L= (0-177m)3.822 m1-0.637C)F.%3D6(3-1)20x2912.34 K9/m²(0.0828)(573)(-AP)= 3.822 (2700-12.34)-0.537) (9.8) = 46609.25eModerointe Date:こ1ち。Vmf+ 1.752.3mf[l-0. 537)(2.1x105)-3(0.537)8 (0.2xlo°)²(2700-12.34)(9.8)=15012.34 (Vmp)Ymf + 1.75(0.5年)02d6)26339.07= 23G455042 Vmp697268.92( Vmf )235455.424mf +677268.97 ( Vmt)?=26339.072354 65.42 Vmp s0+697268.99(vmp)²-2633 9.070.0 886%3Dor-0.426メVmf= 0.0886vmf=0.09 m/sPrdp= l0.585-2.1xlo?

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Answered: please verify my answers. That is a…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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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.

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

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?

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