[scilab-Users] No reaction
Jaundre Venter
jaundreventer at gmail.com
Wed Oct 5 22:24:08 CEST 2011
Thank you very much Adrien.
always nice if someone can explain to you where your problems are and why.
Thanks
Was there any other problems you saw that i have to be aware of?
On Wed, Oct 5, 2011 at 5:38 PM, Adrien Vogt-Schilb <vogt at centre-cired.fr>wrote:
> Hi
>
> When you use ode, it's ok, if say, dx(1) depends on dx(4).
> but you still have say that to scilab properly, something like:
>
> function dx = f(t,x)
> dx(6)=((F+Fab+Floss)*(x(2))), // culture Volume V
> dx(1)=(((mu)*(x(1)))-(((x(1))/(x(6)))*((dx(6))))*(CO)*(x(2)))), //biomass
> concentration X
> and so one
>
> note that because i had to know dx(6) to compute dx(1) i just computed
> dx(6) before dx(1): no problem. and note that i used x(2). The idea of the
> ode is to compute dx from x!
>
> make sure you understand that using dx_6 instead of dx(6), your ODE solver
> is not updating dx_6 at each time step, it is using the initial and only
> dx_6 forever. That's why your last varaibles do not move, somehow their
> speeds are never updated.
> for instance, dx(6)=((F+Fab+Floss)*(HION)), // culture Volume V is
> constance in time (i guess)
>
>
>
> On 05/10/2011 15:00, Jaundre Venter wrote:
>
> Hi Adrien
>
> i am new to SCILAB! I just want to say that.
>
> Yes dx_1 is equal to dx1 but the only reason why i have programmed it like
> that is becasue the ODE's looks as follows - (see word file attached that
> can explain the ODE"s better. with regards to the HION and CO it actually
> refers to [H+] and CO2 as you said. the only reason why n multiplied HION
> and CO wit hsome ODE's is because some does have a influence on some ODE's.
>
> This is the first time i am working with SCILAB thus i am struggling to
> understand how SCILAB wants the code so that all 9 ODE's are shown and so
> that the ODE's that is having a effect on other does happen. I though if you
> refer to dx(1) for example in a other ODE it means that SCILAb will know the
> dx(1) has a influence on the other ODE.
>
> The main goal of my assesment is to deliver similar results obtained from
> MATLAB on SCILAB. all i got was how the grpahs should look like and the
> ODE's.
>
> On Wed, Oct 5, 2011 at 2:43 PM, Adrien Vogt-Schilb <vogt at centre-cired.fr>wrote:
>
>> Hi
>>
>> Try to isalote your problem
>> if i understood well, the following code
>>
>>
>> // initial values
>> x0=[0.1, 1e-5, 0, 15, 1.16, 100,0,297,0.5]';
>> t=0:0.005:400;
>> y=ode(x0, 0, t, f);
>>
>> returns y such that sum(y(6:9,:)>x0) == 0 ?
>> if this is true, we do not need the plots to solve the problem
>> can you check that ?
>>
>> I believe the f function is erroneous.
>> It seems that dx_1 should be equal to dx(1) at each time step, and that
>> HION should be equal to x(2) at each time step, etc.
>>
>> in other terms, some of your phisical variables seem to be represented by
>> to variables (i am guessing HION=[H+] and x(2)=[H+] also) but scilab does
>> not have any chance to know that.
>> if my guess is right, you have to rewrite the f function in a way that
>> eliminates all references to HION, dx_1, dx_6 and so on
>>
>>
>> On 05/10/2011 14:27, Jaundre Venter wrote:
>>
>> Hi all
>>
>> Can someone please explain to me the following:
>>
>> I am busy with a project of simulation the production of penicillin in a
>> bio reactor. Now i have 9 ODE's which i want to simulate.
>>
>> now for some reason the last three graphs i am getting doesn't show any
>> response what so ever. i am using the following code.
>>
>> dx(1)=(((mu)*(x(1)))-(((x(1))/(x(6)))*((dx_6)))*(CO)*(HION)), //biomass
>> concentration X
>> dx(2)=((z*(((mu)*(x(1)))-(((F)*(x(1)))/(x(6)))))+(QQ)), //hydrogen ion
>> concentration H+
>> dx(3)=((((mupp)*(x(1)))-((K)*(x(3)))-((x(3))/(x(6)))*(dx_6))*(HION)),
>> //Penicilin concentration P
>> dx(4)=((-((mu)/(Yxs))*(x(1)))-(((mupp)/(Yps))*(x(1)))-((mx)*(x(1)))+((Fsf)/(x(6)))-((x(4)/(x(6)))*(dx_6))),
>> //Substrate concentration S
>> dx(5)=(-(((mu)/(Yxo))*(x(1)))-(((mupp)/(Ypo))*(x(1)))-(((mo))*(x(1)))+((kla)*(cll-(x(5))))-(((x(5))/(x(6)))*(dx_6))),
>> //dissolved oxygen
>> dx(6)=((F+Fab+Floss)*(HION)), // culture Volume V
>> dx(7)=(((rq1)*(dx_1)*(x(6)))+(rq2)*(x(1))*(x(6))), //Heat generation Qrxn
>> dx(8)=((((F)/(sf))*(Tf-(x(8))))+(1/((x(6))*(pcp)))*(QT)), // Temperature
>> T
>> dx(9)=(((a1)*(dx_1))+((a2)*(x(1)))+(a3)), // CO2 evolution, CO2
>> endfunction
>>
>> now when i ask for plotting the graphs i am using the following.:
>>
>> // initial values
>> x0=[0.1, 1e-5, 0, 15, 1.16, 100,0,297,0.5]';
>> t=0:0.005:400;
>> y=ode(x0, 0, t, f);
>>
>> // the plots of each variable
>> da.title.text="BIOMASS CONCENTRATION"
>> da.x_label.text="Time, hours";
>> da.y_label.text="X,g/l ";
>> scf(1);clf; //Opens and clears figure 1
>> plot(t,y(1,:))
>>
>> da.title.text="HYDROGEN ION H+ CONCENTRATION"
>> da.y_label.text="H+,mol/l ";
>> scf(2);clf; //Opens and clears figure 2
>> plot(t,y(2,:))
>>
>> da.title.text="PENICILLIN CONCENTRATION"
>> da.y_label.text="P,g/l ";
>> scf(3);clf; //Opens and clears figure 3
>> plot(t,y(3,:))
>>
>> da.title.text="SUBSTRATE CONCENTRATION"
>> da.y_label.text="S,g/l ";
>> scf(4);clf; //Opens and clears figure 4
>> plot(t,y(4,:))
>>
>> da.title.text="DISSOLVED OXYGEN CONCENTRATION"
>> da.y_label.text="C_l,g/l ";
>> scf(5);clf; //Opens and clears figure 5
>> plot(t,y(5,:))
>>
>> da.title.text="CULTURE VOLUME"
>> da.y_label.text="V,l";
>> scf(6);clf; //Opens and clears figure 6
>> plot(t,y(6,:))
>>
>> da.title.text="HEAT OF REACTION"
>> da.y_label.text="Qrxn,cal";
>> scf(7);
>> clf; //Opens and clears figure 7
>> plot(t,y(7),:)
>>
>> da.title.text="TEMPERATURE"
>> da.y_label.text="T,Kelvin";
>> scf(8);
>> clf; //Opens and clears figure 8
>> plot(t,y(8),:)
>>
>> da.title.text="CO2 EVOLUTION"
>> da.y_label.text="CO2,mmol/l/";
>> scf(9);
>> clf; //Opens and clears figure 9
>> plot(t,y(9),:)
>>
>> Am i doing something wrong? before the ODE's i have just programmed the
>> initial values and constants :
>>
>> funcprot(0);
>> function dx = f(t,x)
>> K1=1.0e-10 //mol/l
>> K2=7.0e-05 //mol/l
>> Kx=0.15 // Contois saturation constant, g/l
>> Kox=2e-02 // oxygen limitation constant
>> mux=0.092 // maitenance coefficient on subsrate
>> p=3 //constant
>> Kp=0.0002 // inhibition constant
>> Kop=2e-02 // oxygen limitation constant
>> K=0.04 // Penicillin hydrolysis constant, per h
>> Yxs=0.45 // yield constant,g biomass/g glucose = dimensionless
>> Yps=0.90 // yield constant, g pinicillin/ g glucose = dimensionless
>> mx= 0.014 // Maintenance coefficient on substrate, per h
>> Yxo=0.04 // yield constant, g biomass/g oxygen = dimensionless
>> Ypo=0.20 // yield constant, g penicillin/g oxygen= dimensionless
>> mo= 0.467 // maintenance coefficient of oxygen, per h
>> mup=0.0005 // specific rate of penicilline production (per h)
>> sf=600 // Feed substrate concentration, g/l
>> kla=23 // function of agitation power input and oxugen flow rate,
>> dimensional
>> cll=1.16 // dissolved oxygen concentration, g/l
>> Cab=3 // concentrations in both solutions
>> Fa=5 // acid flow rate, l/h !!
>> Fb=5 // base flow rate, l/h !!
>> delta_t=0.01 // time step in digital PID controller - arbitrary
>> value!!!
>> z=10e-5 // constant
>> F=0.042 // feed substrate flow rate l/h
>> T0=273 // temperature at freezing, K
>> Tv=373 // temperature at boiling
>> T=298 // feed temp of substrate
>> h=(2.5e-4) // constant
>> Floss=(x(6)*(h)*(exp(5)*((T-T0)/(Tv-T0))))
>> Fab=Fa+Fb // volume increase due to influx of acid Fa and base Fb
>> Fsf=((sf)*(F))
>> kg= 7e-3 // Arrhenius constant for growth
>> kd=10e33 // Arrhenius constant for cell death
>> Eg= 5100 // Activation energy for growth, cal/mol
>> Ed= 50000 // Activation energy for cell death, cal/mol
>> R= 1.987 // gas constant, cal/mol k
>> T= 297 // Temperature
>> RT= R*T
>> alpa= 70 // constant in Kla
>> betha= 0.4 // constant in Kla
>> Pw= 30 // Agitation power input, W
>> fg= 8.6 // Flow rate of oxygen
>> V=100 // Volume
>> QE= ((kg*exp(-(Eg/RT)))-(kd*exp(-(Ed/RT))))
>> kla= alpa*((sqrt(fg)*(Pw/x(6)))^betha)
>> mu
>> =(((mux)/(1+((K1)/(x(2)))+((x(2))/(K2))))*((x(3))/(((Kx)*(x(1)))+(x(3))))*((x(5))/(((Kox)*(x(1)))+(x(5))))*(QE))
>> // Specific growth rate
>> mupp =
>> ((mup)*((x(4))/((Kp)+(x(4))+(x(4)^2)/(K1)))*((x(5)^p)/((Kop)*(x(1)))+(x(5)^p)))
>> // Specific penicillin production rate
>> B =(((1e-14/x(2)-x(2))*x(6)-Cab*(Fa+Fb)*delta_t)/(x(6)+(Fa+Fb)*delta_t))
>> QQ =((-B+sqrt(B^2+4e-14))/2-(x(2)))*(1/delta_t)
>> dx_6 = (F+Fab+Floss) //Culture Volume V
>> dx_1 = (((mu)*(x(1)))-((x(1))/(x(6)))*(dx_6)) //biomass concentration X
>> rq1 = 60 // yield of heat generation, cal/g biomass
>> rq2 = 1.6783e-4 // Constant, cal/g biomass h
>> Tf = 296 // substrate feed temperature, Kelvin
>> a = 1000 // heat transfer coefficient of cooling/heating liquid, cal/h
>> degree C
>> b = 0.60 // constant
>> Fc=0.1 // Cooling water flow rate, not sure about value, l/h
>> pcCpc = 1/2000 // Density times heat capacity of cooling liquid, per l
>> degree C
>> pcp = 1/1500 // density times heat capacity of medium
>> QT = ((x(7)-(((a)*(Fc^b+1))/((Fc)+((a)*(Fc^b))/2*pcCpc))))
>> a1=0.143 // constant relating CO2 to growth, mmol CO2/g biomass
>> a2=4e-7 // Constant relating CO2 to mainteneance energy, mmol CO2/g
>> biomass h
>> a3=1e-4 // Constant relating CO2 to penicillin production, mmol CO2/l h
>> CO= (((a1)*(dx_1))+((a2)*(x(1)))+(a3)), // CO2 evolution, CO2
>> HION=((z*(((mu)*(x(1)))-(((F)*(x(1)))/(x(6)))))+(QQ))
>>
>> Thanks.
>>
>>
>>
>> --
>> Adrien Vogt-Schilb (Cired)
>> Tel: (+33) 1 43 94 *73 77*
>>
>
>
>
> --
> Adrien Vogt-Schilb (Cired)
> Tel: (+33) 1 43 94 *73 77*
>
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