Tutorial On Scale Up.docx

(a)

Tutorial on Scale-Up The following expression for a fully baffled stirred bioreactor with a single impeller has been derived by dimensionless analysis: 𝑁𝑝 = 𝑐(𝑁𝑅𝑒 ) 𝜒

where

Np

is power number

N Re c

is Reynold’s number

𝜒

is an exponent

is a constant dependent on vessel geometry

Plots of power number against Reynolds number on log-log coordinates for various bioreactor configurations based on Figure 1 can be constructed from experimentally obtained data as in Figure 2, with the value of the exponent 𝜒 changing with flow regimes. Explain the significance of these curves in terms of scaling up a fermentation process in geometrically similar bioreactors in each of laminar, transient, and turbulent regions.

Figure 1

(b)

(5 marks)

A fermentation process in a pilot-scale stirred-tank bioreactor with four baffles and two impellers each comprising of a disc with six vertical flat-blade impellers with dimensions as below was found to produce the maximum yield when the impeller speed was N = 0.27 rps.

Dt (Fermenter diameter)

=

1.00 m

Da (Impeller diameter)

=

0.33 m

E (Height of lowest impeller)

=

0.33 m

L (Length of blade)

=

0.0825 m

W (Width of blade)

=

0.066 m

J (Baffle width)

=

0.10 m

H (Distance between impellers)

=

1.00 m

The fermentation broth fulfilled the criteria for a Newtonian fluid throughout the fermentation, and has the following characteristics: Density Viscosity

  1300 kgm 3   0.03 kgm 1s 1

The following relationships apply:

𝑁𝑝 = 𝜌𝑁𝑃30𝐷5

Power number

𝑎

𝑃0 = 𝑐𝜌𝑁 3 𝐷𝑎5 𝑃0 = 𝑐𝜇𝑁 2 𝐷𝑎3 2

𝑎 𝑁𝑅𝑒 = 𝜌𝑁𝐷 𝜇

where

(turbulent flow)

(laminar flow)

Reynold’s number

 = liquid density

 = liquid viscosity N = impeller rotational speed

𝐷𝑎 = impeller diameter

Po = power consumption for un-gassed fluid You are required to scale-up this fermentation process using the method of ‘constant power per unit volume’ to a geometrically similar production-scale

bioreactor 50 times its volume. Estimate the impeller speed of the productionscale bioreactor.

A log-log graph paper is provided in the Appendix 1.

(25 marks)

Figure 2.

Solution (a) This typical power curve can be divided up into 3 clearly defined regions, as shown in Fig. 2 and described in Table 1 below:

Table 1: Regions in a Power Curve

( )

Region

Reynolds number

Laminar flow

<101

-1

Transition

101 - 104

Variable

Turbulent flow

>104

0

Slope of plot

The Reynolds number is thus a criterion for the degree of turbulence within the system and may be used to reveal the flow regime which exist. (3 Marks) If the experimentally determined value of the exponent  are substituted into equation 1, the following expressions results:

P  cN 3 D 5 P  cN 2 D 3

(turbulent flow) (laminar flow)

These two equations are widely used to predict the effects of operating variables upon power consumption in scale-up of fermentation based on un-gassed fluids. (2 Marks) (b)

From the dimensions given, we can determine, which curve on the plot of Np against Rei that corresponds to curve A.

Check if the system is in the transition flow region: Calculate the value of the Reynolds number (NRe) on the small scale under the defined optimal conditions.

N i Da2 Rei =  Where Ni = impeller speed of agitation in rps = 0.27 rps.

Da (Impeller diameter) = 0.33 m   1300 kgm 3

Density

  0.03 kgm 1s 1

Viscosity Therefore, Rei =

N i Da2 1300  0.27  0.332 = 1,274.13   0.03

The value of Reynolds no. indicates that the flow is in the transition region. The following procedure can be used for scale-up in the transition region: (ii)

Find the corresponding value of the power number at small scale (N P1) by interpolating the power curve. From the power curve, the corresponding value of NP1 is 4.8. With two impellers, NP1 = 2 × 4.8 = 9.6

(iii)

Calculate power of the small scale from power number equation:

P1  N P1 N13 D15

 P = 9.6 × 1300 × 0.273 × 0.335 = 0.96 W (iv) Calculate the required power consumption on the large scale:

V P2  P1 *  2  V1 (v)

  = 0.96 * 50 = 48 W 

For the large fermenter, calculate values of the Reynolds number over a range of different impeller speed :

 N 2 D22   N Re      First, what is the value of

D22 ?

 Di For geometrically similar vessels,  1  Di  2

1

  V1  3     V2  

1

1 V 3  D   2  Di1  503  0.33  1.2m  V1 

Now, calculate values of the Reynolds number over a range of different impeller speeds:

Ni(2)

Rei(2)

0.025

1560

0.050

3120

0.075

4680

0.100

6240

0.125

7800

0.150

9360

0.175

10920

0.200

12480

Transition Region

Turbulent Region

Presumably the speed of agitation is between 0.1 and 0.17 because they fall in the transition region. (vi)

Find the values of the power numbers corresponding to these Reynolds numbers by interpolating the power curve. Ni(2)

(vii)

Rei(2)

Np

0.025

1560

4.8

0.050

3120

5.4

0.075

4680

5.8

0.100

6240

6.0

0.125

7800

6.0

0.150

9360

6.0

0.175

10920

6.0

0.200

12480

6.0

Calculate the power consumption on the large scale at each selected value of impeller speed from the normal power number equation: P2  N P 2 N 23 D25 Since the large scale is geometrically the same with the small scale, hence with 2 impellers, NP(2) = 2 × NP (obtained from the curve). Therefore,

Ni(2)

Np

2Np



0.025

4.8

9.6

1300

1.2 0.0000156

2.488

0.485

-1.60

-0.31

0.05

5.4

10.8

1300

1.2

0.000125

2.488

4.367

-1.30

0.64

0.075

5.8

11.6

1300

1.2

0.000422

2.488

15.830

-1.12

1.20

0.1

6

12

1300

1.2

0.001000

2.488

38.818

-1.00

1.59

0.125

6

12

1300

1.2

0.001953

2.488

75.816

-0.90

1.88

0.15

6

12

1300

1.2

0.00338

2.488

131.010

-0.82

2.12

0.175

6

12

1300

1.2

0.005

2.488

208.039

-0.76

2.32

0.107334

6

12

1300

1.2

0.001

2.488

48.000

-0.97

1.68

(vii)

Ni(2)3

Da(2)

Da(2)5

P2

log Ni(2)

log (P2)

Plot power consumption against impeller speed on log-log coordinates for the large fermenter. Or use Goal Seek in Excel to determine what is the value of Impeller speed that corresponds to P2 = 48 W.

From Goal Seek, we shall obtain the Ni2 = 0.107 rps. Or Solve using the normal calculation, by substituting the values of P2 as 48 W into the power no equation. P2  N P 2 N 23 D25 48  12.0  1300  N 23  1.25 48  38816 N 23 N 23  1.237  10 3 N 2  3 1.237  10 3  0.107 rps

From the curve of P2 against Ni(2), at P2 =48 W, the value of Ni(2) was found to be 0.107rps. And from the plot of log P2 against log Ni(2), at log 48 = 1.68, the value of log Ni(2) was found to be -0.96. Therefore, the value of Ni(2) is 10-0.96 = 0.107 rps.

4.8

Re = 1,274.13

QUESTION 2

(PO2, PO6, CO3, C3, C3, C5)

Two geometrically similar stirred tanks with a flat-blade turbine impeller of the following dimensions are to be operated at 30C as pilot-scale and production-scale aerobic fermenters.

Scale Pilot Production

Impeller diameter, Di (m) 0.24 0.8

Liquid volume, V (m 3)

Tank diameter (m)

0.17 6.28

0.50 1.6

Satisfactory results were obtained with the pilot-scale fermentor at a rotational impeller speed Ni of 1.5 s-1 and volumetric air flow rate, Q of 0.0083 m3 s-1. The density and viscosity of the broth are 1050 kgm-3 and 0.002 kg/m.s, respectively. Van’t Riet (1979) has discovered a correlation for oxygen mass transfer coefficient, k La as the scale-up criterion in relation with the following equations: −3

𝑘𝐿 𝑎 = 2.0 × 10

𝑃𝑎 0.7 0.2 ( ) 𝑣𝑠 𝑉 −0.20

𝑃𝑎 𝑄 −0.25 𝑁𝑖2 𝐷𝑖4 = 0.10 ( ) ( ) 𝑃0 𝑁𝑖 𝑉 𝑔𝑊𝑖 𝑉 2⁄3 where

Pa

=

aerated power (W)

vs

=

superficial gas velocity (m/s)

Wi

=

impeller blade width

P0

=

power consumption for unaerated bioreactor.

g

=

gravitational acceleration = 9.81 ms-2

(a) By referring to Curve 1 in Fig. 3, calculate the Reynolds number and hence determine the power consumption of aerated bioreactor at pilot scale. The following relationships apply:

𝑁𝑝 = 𝜌𝑁𝑃30𝐷5

𝑎

Power number

𝑃0 = 𝑐𝜌𝑁 3 𝐷𝑎5 𝑃0 = 𝑐𝜇𝑁 2 𝐷𝑎3 2

𝑎 𝑁𝑅𝑒 = 𝜌𝑁𝐷 𝜇

where

(turbulent flow)

(laminar flow)

Reynold’s number

 = liquid density

 = liquid viscosity N = impeller rotational speed

𝐷𝑎 = impeller diameter

Po = power consumption for ungassed fluid (7 marks)

Figure 3. The power number-Reynolds number relationship for various impellers.

(b) Calculate the superficial gas velocity, vs and estimate the oxygen mass transfer 𝑄 coefficient, kLa for pilot scale. Given that 𝜐𝑠 = 𝐴

where A = cross-sectional area of the bioreactor, Q = volumetric flowrate of air sparged into the fermentor, (5 marks)

(c) Estimate the kLa for the production scale by employing constant superficial gas velocity and constant power consumption per unit volume as scale up criteria and comment on the value obtained in comparison with the value of k La in (b).

Given that, correlation for speed of agitation in terms of volume is 2⁄ 9

𝑁2 𝑉1 =( ) 𝑁1 𝑉2

The subscript 1 and 2 refer to pilot scale and production scale, respectively. (15 marks)

Solution: (i)

𝑁𝑅𝑒 =

𝑁𝐷2 𝑎 𝜇

=

1050×1.5×0.242 0.002

= 45,360 (turbulent flow) (2 marks)

From Curve 1 in Fig. 1, Np = 5

Then power consumption for un-aerated bioreactor, 𝑃 = 𝑁𝑝 𝜌𝑁 3 𝐷𝑎5 5 × 1050 × 1.53 × 0.245 = 14 W.

(1 marks)

Then the aerated power consumption can be calculated by the given formula;

−0.20

𝑃𝑎 𝑄 −0.25 𝑁𝑖2 𝐷𝑖4 = 0.10 ( ) ( ) 𝑃 𝑁𝑖 𝑉 𝑔𝑊𝑖 𝑉 2⁄3 

𝑃𝑎 𝑃

= 0.10 (

0.10(2.35) (

1.5×0.17 0.024

9.81𝑊𝑖

Given that,

−0.25

0.0083

𝑊𝑖

)

(

1.52 ×0.244

2 9.81𝑊𝑖 ×0.173

−0.20

−0.20

)

=

)

(1 mark)

= 0.2

𝐷𝑖

(1 mark)

Therefore, Wi = 0.2Di = 0.2 × 0.24 = 0.048 m

(1 mark)

By substituting P = 14 W and Wi = 0.048 m into the above equation,



𝑃𝑎 14

= (0.235) (

0.024 9.81×0.048

−0.20

)

= 0.426

Then, Pa = 14 × 0.426 = 5.97 W.

(1 mark)

ii) Cross-sectional area of the stirred tank, given that DT = 0.5 m

A=𝜋

𝐷𝑇2 4

Then, 𝜐𝑠

=𝜋

=

𝑄 𝐴

0.52 4

=

=0.196 m2

0.0083 0.196

= 0.042 𝑚/𝑠

(2 marks)

hence, −3

𝑘𝐿 𝑎 = 2.0 × 10 −3

𝑘𝐿 𝑎 = 2.0 × 10

𝑃𝑎 0.7 0.2 ( ) 𝑣𝑠 𝑉

5.97 0.7 ( ) 0.0420.2 0.17

𝑘𝐿 𝑎 = 2.0 × 10−3 × 12.07 × 0.53 = 0.0128𝑠 −1

(3 marks)

(iii) The rotational speed of agitator for production scale can be determined from,

2⁄ 9

𝑁2 𝑉1 =( ) 𝑁1 𝑉2

2⁄ 9

𝑁2 0.17 =( ) 1.5 6.28

N2 = 0.67 rps

(3 marks)

From constant power consumption for unaerated bioreactor per unit volume where,

P1 P  2 V1 V2 𝑃

14

Then P2 = 𝑉1 𝑉2 = 0.17 (6.28) = 517 𝑊 1

(2 marks)

Then the aerated power consumption can be calculated by the given formula; −0.20

𝑃𝑎 𝑄 −0.25 𝑁𝑖2 𝐷𝑖4 = 0.10 ( ) ( ) 𝑃 𝑁𝑖 𝑉 𝑔𝑊𝑖 𝑉 2⁄3

𝑄

From the constant superficial gas velocity, 𝜐𝑠 = 𝐴2 = 0.042𝑚/𝑠 2

where A2 = 𝜋

𝐷𝑇2

=𝜋

4

1.62 4

=2 m2

Therefore, Q2 = A2vs = 2 × 0.042 = 0.084 m3/s marks)



𝑃𝑎 𝑃

= 0.10 (

0.10(2.66) (

)

0.67×6.28 0.054

9.81𝑊𝑖

Given that,

−0.25

0.084

𝑊𝑖 𝐷𝑖

−0.20

(

0.672 ×0.84

2 9.81𝑊𝑖 ×6.283

)

= 0.2

Therefore, Wi = 0.2Di = 0.2 × 0.8 = 0.16 m

(2

−0.20

)

=

By substituting P = 517 W and Wi = 0.16 m into the above equation,



𝑃𝑎 517

= (0.266) (

0.054 9.81×0.16

−0.20

)

= 0.52

Then, Pa = 517 × 0.52 = 269.8 W.

(4 marks)

hence, −3

𝑘𝐿 𝑎 = 2.0 × 10 𝑘𝐿 𝑎 = 2.0 × 10

−3

𝑃𝑎 0.7 0.2 ( ) 𝑣𝑠 𝑉

269.8 0.7 ( ) 0.0420.2 6.28

𝑘𝐿 𝑎 = 2.0 × 10−3 × 13.876 × 0.53 = 0.0147𝑠 −1

(1 mark)

Percentage error between the 2 kLa values are defined by, (0.0147−0.0128) 0.0128

× 100 = 14.8%

(1 mark)

Theoretically, the two values should be similar, however, it is very difficult to achieve a constant kLa values because of the correlation (Van’t Riet) used here is very general that applies to this calculation. In addition, the aeration in larger scale will consume more power and cost that will affect the kLa values. (2 marks)