Biochem Proj 8.3 and Additl on Chap9

8.3) You need to design a filter for a 10,000 gallon fermenter that will be aerated at a rate of 535 ft 3/min (at 20oC and 1 atm) . The bacterial count in the air air is 80 per ft3. The average size of the t he bacteria is 1μm with density of 1.08 g/cm3. You are going to use glass fibers (Dc= 15 μm) with packing density α=0.03. The cross-sectional area of the filter w ill be designed to give a superficial air velocity vo of 5 ft/s. a. What depth of the filter would you recommend to prevent contamination? b. how is the answer in (a) c hanged if vo is decreased to 1 ft/s? Explain the re sults.

Given: @20oC & 1atm

Vt= 10,000 gal Q= 8000 ft3/d

ρ= 1.2 X10-3 g/cm3

Cno= 1000/ft3

μ= 1.8X10-4

Dp= 1 μm= 1.5X10-3 cm α= 0.03

vo= 8ft/s Required: a) B b) B if vo is decreased to 1 ft/s Solution: V=

ft/s   −. = 8.2744ft/s  .  = 251.445 cm/s 

).   . Dcvρ = (.) Nre= = 2.5145  . λ=

.X− 29/88.314107293 = 6.4995X10-6 cm g  √ 29/88.31410 .. X−m 

   .X−(1.257 + 0.04 −.. CF =1+ )= 1.1634  Nst=

. .. .().  ..  = 0.6501

Nimp=0.075 (0.6501)1.2=0.0447 K=

  = 1/15 .

 +         ]= 3.9426x10-3 Nint= [1 ln1     .−.   + D B=

.(.) -7 .  = 2.7729x10

Npe=

.. =1361619.856 77

Ndif =1.3(1361619.856)2/3 + 0.7(1/15)2= 3.2169x10-3 nc= 1-(1-0.0447)(1-3.2169x10-3)(1-3.9426x10-3)= 0.0515 Assume T= 1 day

 =8000 ft3/day (1 day) (1000/ft3) = 8000000   = − (   . 0.0515  . −. B=11.7567 cm b) if vo=1ft/s V=

ft/s   −. = 8.2744ft/s  .  = 31.4307 cm/s 

Dcvρ = (.).7   .= 0.3143 Nre=  . λ=

.X− 7 -6 g  √ 29/88.31410 293 = 6.4995X10 cm .. X−m 

   .X−(1.257 + 0.04 −.. CF =1+ )= 1.1634 

..().7  ..  = 0.0813

Nst=

Nimp=0.075 (0.0813)1.2=3.6912x10-3 K=

  = 1/15 .

 +         -3 Nint=    ]= 1.3476x10 .−. [1 ln1   + D B=

.(.) -7 .  = 2.7729x10

Npe=

..7 = 170024.3427 77

Ndif =1.3(170024.3427) 2/3 + 0.7(1/15)2= 4.0023x10-3 nc= 1-(1-1.3476x10-3)(1-3.6912x10-3)(1-4.0023x10-3)= 9.0160x10-3 Assume T= 1 day

 =8000 ft3/day (1 day) (1000/ft3) = 8000000   = − (   . 9.0160x103  . −. B=67.1548 cm The Decrease on upstream velocity caused the combined collection efficiency to decrease and a decrease in collection efficiency requires a larger value of B (depth) in order to complete sterilization.

Chapter 9 (notes ni maam) The specific oxygen demands and critical oxygen concentrations for typical microbial plants and animal cell cultures are listed below Cell culture

qo

Concentration, mmol/L

Escherichia Coli

0.5 mmol(gdw)-1h-1 

0.0082

Vitisvinifera (grape)

0.6 mmol(gdw)-1h-1

0.055

Chinese Hamster Ovary

3.0x10-10 mmol(gdw)-1h-1 

0.020

a.Estimate the kLa requirement to achieve cell concentrations of 25 gram dry weight/ L for E. Coli and V. Vinifera and 3.0x109 for CHO cells, while maintaining dissolved oxygen concentration above critical. The oxygen solubility in the media used for the cultures is 7.2x10-3 kg/m3 b. The relationship between kLa and the power input to a 1-m3 stirred bioreactor is kLa

 (

0.5

,

compare the bioreactor power requirements for culture of the three different cell types under the conditions described in a. Solution: a.)

for e coli: q= 8.5mmol/h (25/L)=292.65 mmol/L.h v.vinifera: q= 0.6 mmol/h (25/L)=15 mmol/L.h

CHO: q= 3x10-10mmol/h (3x109/L)=0.9 mmol/L.h Oxygen Solubilty: 7.3x10-3 kg/m3 292.5 mmol/L.h= [7.2 x10-3 kg/m3 (1kmol/32kg)(1x106 mmol/kmol)-0.0002)kla kLa= 980.1661/hr 15 mmol/L.h= [7.2 x10-3 kg/m3 (1 kmol/32kg)(1x106 mmol/kmol)-0.055)kla kLa= 88.2353/hr 0.9 mmol/L.h= [7.2 x10-3 kg/m3 (1kmol/32kg)(1x106 mmol/kmol)-0.02)kla kLa= 4.3902/hr

b.) kLa= k



0.5

PT E.C.= 960725.5836 m3/hr2 K2 PT v.v. = 7765.4682 m3/hr2 K2 PT CHO= 19.2739 m3/hr2 K2 PT E.C.> PT v.v.> P T CHO