生物过程反应原理_Chapter 7 modeling and Scale up

Chapter 7. Scale-Up 8.1 Similitude For the optimum design of a production-scale fermentation system(prototype), we must translate the data on a small scale(model)to the large scale. The fundamental requirement for scale-up is that the model and prototype should be similar to each other Two kinds of conditions must be satisfied to ensure similarity between model and prototype They are: (1)Geometric similarity of the physical boundaries

Chapter 7. Scale-Up 8.1 Similitude For the optimum design of a production-scale fermentation system (prototype), we must translate the data on a small scale (model) to the large scale. The fundamental requirement for scale-up is that the model and prototype should be similar to each other. Two kinds of conditions must be satisfied to ensure similarity between model and prototype. They are: (1) Geometric similarity of the physical boundaries:

The model and the prototype must be the same shape and all linear dimensions of the model must be related to the corresponding dimensions of the prototype by a constant scale factor (2) Dynamic similarity of the flow fields: The ratio of flow velovities of corresponding fluid particles is the same in model and prototype as well as the ratio of all forces acting on corresponding fluid particles. When dynamic similarity of two flow fields with geometrically similar flow patterns The first requirement is obvious and easy to accomplish but the second is difficult to understand and also to accomplish and needs explanation. For example, if force that may act on a fluid element in a fermenter during agitation are the viscosity force Fv, drag force on impeller D, and gravity force FG each can be expressed with

The model and the prototype must be the same shape, and all linear dimensions of the model must be related to the corresponding dimensions of the prototype by a constant scale factor. (2) Dynamic similarity of the flow fields: The ratio of flow velovities of corresponding fluid particles is the same in model and prototype as well as the ratio of all forces acting on corresponding fluid particles. When dynamic similarity of two flow fields with geometrically similar flow patterns. The first requirement is obvious and easy to accomplish, but the second is difficult to understand and also to accomplish and needs explanation. For example, if force that may act on a fluid element in a fermenter during agitation are the viscosity force FV, drag force on impeller FD, and gravity force FG,each can be expressed with

characteristic quantities associated with the agitating system. According to Newtons equation of viscosity. viscosity force is (9.1) Where du/dy is velocity gradient and A is the area on which the viscosity force acts. For the agitating system the fluid dynamics involved are too complex to calculate a wide range of velocity gradients present However, it can be assumed that the average velocity gradient is proportional to agitation speed n and the area A is to D2, which results (9.2) he drag force Fp can be characterized in an agitating system as

characteristic quantities associated with the agitating system. According to Newton’s equation of viscosity, viscosity force is (9.1) Where du/dy is velocity gradient and A is the area on which the viscosity force acts. For the agitating system, the fluid dynamics involved are too complex to calculate a wide range of velocity gradients present. However, it can be assumed that the average velocity gradient is proportional to agitation speed N and the area A is to D2 I , which results. (9.2) The drag force FD can be characterized in an agitating system as A dy du FV = ( ) 2 FV  NDI

o D OC (9.3) Since gravity force FG is equal to mass m times gravity constant g, F OC pDig (9.4) G The summation of all forces is equal to the inertial force as ∑F=F+FD+FG=F1∝pDN2(9.) Then dynamic similarity between a model(m) and a prototype(p)is achieved if (FYm(Fpm (fm(fu) 9.6) (FP(FDp (FP (F

(9.3) Since gravity force FG is equal to mass m times gravity constant g, (9.4) The summation of all forces is equal to the inertial force FI as, = = (9.5) Then dynamic similarity between a model(m) and a prototype(p) is achieved if = = = (9.6) FG DI g 3   F FV + FD + FG 4 2 FI  DI N V P V m F F ( ) ( ) D P D m F F ( ) ( ) G P G m F F ( ) ( ) I P I m F F ( ) ( ) D N P F I mo D 

Or in dimensionless forms F P D D (9.7) P G The ratio of inertial force to viscosity force is 7 DN DD/N FV UNDi -Re(9.8

Or in dimensionless forms: = = (9.7) = The ratio of inertial force to viscosity force is = = = (9.8) P V I F F ( ) m V I F F ( ) P D I F F ( ) P G I F F ( ) m D I F F ( ) m G I F F ( ) V I F F 2 4 2 I I ND D N    DI N 2 i NRe

Which is the Reynolds number. Similarly F PD,N (9.9) F Pmo /D, 10 2 F PD;N D, N Fr (9.10) D g Dynamic similarity is achieved when the values of the nondimensional parameters are the same at geometrically similar locations x7/x7 Rei p l丿m (NPP=(N Pm (NErP(NFr)

Which is the Reynolds number. Similarly, = = = (9.9) = = = (9.10) Dynamic similarity is achieved when the values of the nondimensional parameters are the same at geometrically similar locations. (NRei )P = (NRei )m (NP )P = (NP )m (9.11) (NFr )P = (NFr )m D I F F P D N D N mo I I / 4 2  mo I P N D 3 5  NP 1 G I F F D g D N I I 3 4 2   g DI N 2 NFr

Therefore, using dimensionless parameters for the correlation of data has advantages not only for the consistency ofunits, but also for the scale-up purposes However, it is difficult, if not impossible, to satisfy the ynamIc similarity when more than one dimensionless group is involved in a system, which creates the needs of scale-up criteria. The following example addresses this problem Example 9.1 The power consumption by an agitator in an unbaffled vessel can be expressed as no PNDZ N2DI DND

Therefore, using dimensionless parameters for the correlation of data has advantages not only for the consistency of units, but also for the scale-up purposes. However, it is difficult, if not impossible, to satisfy the dynamic similarity when more than one dimensionless group is involved in a system, which creates the needs of scale-up criteria. The following example addresses this problem. ------------------------------------------------------ Example 9.1 The power consumption by an agitator in an unbaffled vessel can be expressed as = 3 5 I mo N D P  ( , ) 2 2 g ND N D f I I  

Can you determine the power consumption and impeller speed of a 1000 gallon fermenter based on the findings of the optimum condition from a geometrically similar one gallon vessel? If you cannot, can you scale up by using a different fluid system? Solution Since vv=1000. the scale ratio is (D) P=100013 =10 (9.12) (D) To achieve dynamic similarity, the three . ionless numbers for the prototype and the model must be equal as follows P VD)=( nmo nmo ON (9.13)

Can you determine the power consumption and impeller speed of a 1000 gallon fermenter based on the findings of the optimum condition from a geometrically similar one￾gallon vessel? If you cannot, can you scale up by using a different fluid system? Solution: Since Vp /Vm=1000, the scale ratio is, = 1000 1/3 = 10 (9.12) To achieve dynamic similarity, the three dimensionless numbers for the prototype and the model must be equal, as follows: = (9.13) I m I P D D ( ) ( ) P I mo N D P ( ) 3 5  m I mo N D P ( ) 3 5 

OND N P (9.14) N-D N-D (9.15) If you use the same fluid for the model and the prototype Pppm and Hp- Hm. Canceling out the same physical properties and substituting Eq (8. 12 ) to Eq (8.13)yields mo P ,=10(P m0) (9.16 The equality of the reynolds number requires Np=0.0INm (9,17)

P NDI ( ) 2   = (9.14) = (9.15) If you use the same fluid for the model and the prototype, ρp =ρm and μp=μm. Canceling out the same physical properties and substituting Eq.(8.12) to Eq. (8.13) yields (Pmo )P = 105 ( Pmo )m (9.16) The equality of the Reynolds number requires NP = 0.01Nm (9.17) m NDI ( ) 2   P I g N D ( ) 2 m I g N D ( ) 2 3 ( ) m P N N

On the other hand, the equality of the Froude number requires N 10 (9.18) Which is conflicting with the previous requirement for the equality of the reynolds number. Therefore, it is impossible to satisfy the requirement of the dynamic similarity unless you use different fluid systems fp fm and p unto satisfy eqs. 8. 14)and( 8.15),the following relationship must hold m31.6p (9.19)

On the other hand, the equality of the Froude number requires = (9.18) Which is conflicting with the previous requirement for the equality of the Reynolds number. Therefore, it is impossible to satisfy the requirement of the dynamic similarity unless you use different fluid systems. Ifρp ρm and μp μm, to satisfy Eqs.(8.14) and (8.15), the following relationship must hold. = (9.19) NP Nm 10 1   m ( )   P ( ) 31.6 1  