### Bed operations fixed bed fluidized bed

```Item B: Bed Operations
packed bed & fluidized bed

填充床 (packed bed, fixed bed固定床), 流體化床
(fluidized bed) – solid phase, gas phase, liquid phase
 用途: 吸附/吸收/蒸餾 (mass transfer 質傳操作), 反應
(e.g. 燃燒/煆燒), ion-exchange bed逆洗去除雜質, ….
 關心的項目: pressure drop (flow resistance), reaction
rate (or adsorption rate, or..) (extent of reaction); mass
transfer rate (質傳速率 ~ in general related to diffusion
rate, surface area of solid); material loss (solid material);
 單操實驗: 先關心基本的壓降 (壓降不正常的後果?)
堆積物當然也可以是粉粒體, e.g. 觸媒粒子
 此圖例子為氣液相逆流, 所示堆積物多用於蒸餾
原則上影響填充床壓降者: 填充物尺寸, 堆積密度
(random packing, ordered packing, etc. 壓降與接觸面積

 單一流體時的壓降 Ergun equation: viscous force and
inertia force之和
 P
L
(1   )   u
2
 150
 d
3
2
p
 1 . 75
(1   )  u
2
 dp
3
 R/ρu2 = 4.17/Re + 0.29 (f = 64/Re for flow through a pipe)
R: drag force on unit area of packed material
 L’ 表示也有液體流過; G‘ superficial gas mass velocity
 乾溼塔操作不同
Loading point – 此時壓降斜率開始明顯增加; 氣體實際

Flooding point - 液體開始流不下去, stay on top; 氣體開

 ideal operation: 應該稍低於loading point
Generalized pressure
drop and flooding point
curves from Eckert
 主要correlate L' & G'
 流體化床: 基本上以氣固相為主; 流體與固體粒子充分

底部多為porous plate 除一開始支撐粒子外, 也可以順

固體粒子有被帶出的可能性; 一般操作都需要加裝回收

 實驗項目: 最小流體化速度, 終端速度(terminal velocity)
 Fixed bed to expanded bed to minimum fluidization to
pneumatic transport; bubbling fluidization; two-phase theory of
fluidization; slurry transport or pneumatic transport
堆積情形會影響壓降
正常操作壓降即為weight of bed/cross area = W/A
 bed height 會逐漸增加, 最後粒子會被吹走
mf and umf: minimum fluidization 對應的狀況
為精確的得到umf 宜先讓bed fluidized, 然後再重複之
 Φs sphericity 球型度 (為求簡化可以假設為1)
 For roughly spherical particles, εmf ~ 0.40-0.45
1
 s
1 . 75
 s  mf
3
(
d p u mf  f

3
mf
 14 ,
1   mf
 
2
s
3
mf
 11
3
150 (1   mf ) d p u mf  f
d p  f ( p   f )g
) 
(
) 
2 3
2
 s  mf


2
Terminal velocity 的一些經驗式
 For spheres & εm = 0.45, terminal velocity 大約是minimum
fluidization velocity的50倍, Rep > 103, 則只有7.7倍
g (  p   f )d p
2
ut 
4(  p   f ) g
2
ut  [
Re
18 
ut  [
225  f 
2
]
1/ 3
3 .1 g (  p   f ) d p

f
dp
]
1/ 2
p
 0 .4
0 . 4  Re
500  Re
 500
p
p
 2  10
5
```