samiyare

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12 years, 133 days

 

Amir

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These are questions asked by samiyare

Hi,

I have the a code with some parameters including

Nr= 0, 50, 100

Ha=0, 5, 10

EPSILONE= 0, 0.5, 1

Phiavg= 0.02, 0.06, 0.1

0.1<NBT<10

I can give the solution for higher values of 5<NBT<10 and there is no problem. However, As I reduce the values of NBT, the convergence of the problem is hard. for some values of parameters I cannot find the solution. for example:

Nr=Ha=0

EPSILONE=1

Phiavg=0.06

NBT=0.3

 

I would be most grateful if you can tel me how change the algorithm to find the solution in the range of all parameters.

Many thanks for your attentions in advance

The code has been attached

code_7-8-2014_(1).mw

 

Amir

Hi,

according to my previous question

http://www.mapleprimes.com/questions/201435-ODE-With-Constraint

I wrote the following code. at first, the code solve the equation for f and when it slves that, I want to solve Theta in such a way that use the values of f in previous calculation. I use the command 'known' but i couldnt find thesolution

I would be most grateful if you could help me in this problem

Thanks for your attentions in advance


restart; # Notice that Restart (capital R) has no effect (to catch that use semicolon, not colon)
a:=0.13:
b:=0.41:
reynolds:=1.125*10^7;  
Eq1:=diff(f(x),x$3)+diff(f(x),x$2)*f(x)+b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)^2*x^2,x$1);
Eq2:=diff(g(x),x$3)+diff(g(x),x$2)*g(x)+c*a^2*sqrt(2*reynolds)*diff(diff(g(x),x$2)^2*x,x$1);
eq1:=isolate(Eq1,diff(f(x),x,x,x));
eq2:=subs(g=f,isolate(Eq2,diff(g(x),x,x,x)));
EQ:=diff(f(x),x,x,x)=piecewise(x<c*0.1,rhs(eq1),rhs(eq2));


c:=75:
;
Q:=proc(pp2) local res,F0,F1,F2;
print(pp2);
if not type(pp2,numeric) then return 'procname(_passed)' end if:
res:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=pp2},numeric,output=listprocedure);
F0,F1,F2:=op(subs(subs(res),[f(x),diff(f(x),x),diff(f(x),x,x)])):
F1(c)-1;
end proc;


fsolve(Q(pp2)=0,pp2=(0..102));
se:=%;
res2:=dsolve({EQ,f(0)=0,D(f)(0)=0,(D@@2)(f)(0)=se},numeric,output=listprocedure);
G0,G1,G2:=op(subs(subs(res2),[f(x),diff(f(x),x),diff(f(x),x,x)])):

plots:-odeplot(res2,[seq([x,diff(f(x),[x$i])],i=0..2)],0..2); #This plots from and past 0.1*c
pr:=1;
prt:=0.89;

Eq11:=diff(theta(x),x$2)+pr*diff(theta(x),x$1)*f(x)+pr/prt*b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(theta(x),x$1)*x^2,x$1);
Eq22:=diff(g(x),x$2)+pr*diff(g(x),x$1)*f(x)+pr/prt*a^2*c*sqrt(2*reynolds)*diff(diff(f(x),x$2)*diff(g(x),x$1)*x^1,x$1);
eq11:=isolate(Eq11,diff(f(x),x,x));
eq22:=subs(g=theta,isolate(Eq22,diff(g(x),x,x)));
EQT:=diff(theta(x),x,x)=piecewise(x<c*0.1,rhs(eq11),rhs(eq22));


QT:=proc(pp3) local res3,theta0,theta1;
print(pp3);
if not type(pp3,numeric) then return 'procname(_passed)' end if:
res3:=dsolve({EQT,theta(0)=1,D(theta)(0)=pp3,known=f},numeric,output=listprocedure);
theta0,theta1:=op(subs(subs(res),[theta(x),diff(theta(x),x)])):
theta0(c);
end proc;

fsolve(QT(pp3)=0,pp3=(0..200));
res3(0);



Amir

Hi

 I have the following ODEs

Restart:
a:=0.13:
b:=0.41:
reynolds:=1.125*10^8:
Eq1:=diff(f(x),x$3)+diff(f(x),x$2)*f(x)+b^2*sqrt(2*reynolds)*diff(diff(f(x),x$2)*f(x)*x^2,x$1);
Eq2:=diff(g(x),x$3)+diff(g(x),x$2)*g(x)+c*a^2*sqrt(2*reynolds)*diff(diff(g(x),x$2)*x,x$1);
f(0)=0;
D(f(0)):=0;
# continuity condition  
g(0.1*c)=f(0.1*c):
D(g(0.1*c))=D(f(0.1*c)):
(D@2)(g(0.1*c))=(D@2)(f(0.1*c)):

the value of c is unknown which must be obtained via D(g(c)):=1;

 How can I solve it?

Thanks for your attentions in advance

Amir

Dear collegues

I wrote the following code

 


restart:
Digits := 15;
a[k]:=0;
b[k]:=7.47;
a[mu]:=39.11;
b[mu]:=533.9;
mu[bf]:=9.93/10000;
k[bf]:=0.597;
ro[p]:=3880 ;
ro[bf]:= 998.2;
c[p]:= 773;
c[bf]:= 4182;
#mu[bf]:=1;
Gr[phi]:=0; Gr[T]:=0;
#dp:=0.1;
Ree:=1;
Pr:=1;
Nbt:=cc*NBTT+(1-cc^2)*6;

#######################
slip:=0.1;         ####
NBTT:=2;           ####
lambda:=0.1;       ####
phi_avg:=0.02;    ####
#######################


eq1:=diff( (1+a[mu]*phi(eta)+b[mu]*phi(eta)^2)*diff(u(eta),eta),eta)+dp/mu[bf]+Gr[T]*T(eta)-Gr[phi]*phi(eta);
eq2:=diff((1+a[k]*phi(eta)+b[k]*phi(eta)^2)*diff(T(eta),eta),eta)+lambda*T(eta)/k[bf];
eq3:=diff(phi(eta),eta)+1/Nbt*diff(T(eta),eta);
Q:=proc(pp2,fi0) local res,F0,F1,F2,a,INT0,INT10;
global Q1,Q2;
print(pp2,fi0);
if not type([pp2,fi0],list(numeric)) then return 'procname(_passed)' end if:
res := dsolve({subs(dp=pp2,eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=fi0}, numeric,output=listprocedure,continuation=cc);
F0,F1,F2:=op(subs(res,[u(eta),phi(eta),T(eta)])):
INT0:=evalf(Int(F0(eta),eta=0..1));
INT10:=evalf(Int(F0(eta)*F1(eta),eta=0..1));
a[1]:=evalf(Int(F0(eta),eta=0..1))-Ree*Pr;;
a[2]:=INT10/INT0-phi_avg;
Q1(_passed):=a[1];
Q2(_passed):=a[2];
if type(procname,indexed) then a[op(procname)] else a[1],a[2] end if
end proc;
Q1:=proc(pp2,fi0) Q[1](_passed) end proc;
Q2:=proc(pp2,fi0) Q[2](_passed) end proc;
Optimization:-LSSolve([Q1,Q2],initialpoint=[0.3,0.0007]);




se:=%[2];
res2 := dsolve({subs(dp=se[1],eq1)=0,eq2=0,eq3=0,u(0)=slip*D(u)(0),u(1)=-slip*D(u)(1),D(T)(0)=0,D(T)(1)=1,phi(0)=se[2]}, numeric,output=listprocedure,continuation=cc);
G0,G1,G2:=op(subs(res2,[u(eta),phi(eta),T(eta)])):
TTb:=evalf(Int(G0(eta)*G2(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1))/evalf(Int(G0(eta)*(G1(eta)*ro[p]*c[p]+(1-G1(eta))*ro[bf]*c[bf] ),eta=0..1));
with(plots):
odeplot(res2,[[eta,phi(eta)/phi_avg]],0..1);
odeplot(res2,[[eta,T(eta)/TTb]],0..1);
odeplot(res2,[[eta,u(eta)/(Ree*Pr)]],0..1);

res2(1);
Nuu:=(1/TTb);
1/((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2));
(1/TTb)*(((1+a[k]*G1(1)+b[k]*G1(1)^2)/(1+a[k]*phi_avg+b[k]*phi_avg^2)));
>

I want to run the code for the value of NBTT in the range of 0.2 to 10. this code gave the results in the range of 4-10 easily. So, I used the continuation which improve the range of the results between 2-10. However, I coudnt gave the results when 0.2<NBTT<2. Would you please help me in this situation.

Also, It is to be said that the values of phi should be positive. in some ranges, I can see that phi(1) is negative. Can I place a condition in which the values phi restricted to be positive.

Thanks for your attentions in advance

Amir

Hi,

I wrote the following code which is properly run

 


restart:

# parametrs

MUR:=(1-phi)^2.5:
RhoUR:=(1-phi+phi*rho[p]/rho[f]):
RhoCPR:=(1-phi+phi*rhocp[p]/rhocp[f]):
BetaUR:=(phi*rho[p]*beta[p]+(1-phi)*rho[f]*beta[f])/(RhoUR*rho[f])/beta[f]:

dqu3:=diff(h(x),x$1)-RhoUR*BetaUR*T(x);
dqu2:=5*diff(T(x),x$2)+k[f]/k[nf]*Pr*RhoCPR*f(x)*diff(T(x),x$1);
dqu1:=5/(MUR)*diff(f(x),x$3)
+ 2*(diff(h(x),x$1)*x-h(x))
+RhoUR*(3*f(x)*diff(f(x),x$2)-diff(f(x),x$1)^2);
rho[f]:=998.2: cp[f]:=4182: k[f]:=0.597:   beta[f]:= 2.066/10000:
rho[p]:=3380: cp[p]:=773: k[p]:=36:   beta[p]:= 8.4/1000000:

k[nf]:=((k[p]+2*k[f])-2*phi*(k[f]-k[p]))/((k[p]+2*k[f])+phi*(k[f]-k[p])):
rhocp[nf]:=rho[p]*cp[p]*phi+rho[f]*cp[f]*(1-phi):
rhocp[p]:=rho[p]*cp[p]:
rhocp[f]:=rho[f]*cp[f]:

phi:=0.00:
binfinitive:=6: Pr:=7: lambda:=0:


with(plots):
pppe:=dsolve( {dqu1=0,dqu2=0,dqu3=0,T(0)=1,T(binfinitive)=0,f(0)=0,D(f)(0)=lambda,D(f)(binfinitive)=0,h(binfinitive)=0}, numeric );
-pppe(0);
print(odeplot(pppe,[x,diff(f(x),x)],0..binfinitive,color=black,numpoints=400));
print(odeplot(pppe,[[x,diff(f(x),x)]],0..binfinitive,color=black,numpoints=400));
print(odeplot(pppe,[[x,T(x)]],0..binfinitive,color=black,numpoints=400));


However, in some range of parameters, I must increase the value of binfinitive (for example binfinitive=50). however, my code is doesnt converge for higher values of 10 (at most). Can anyone change this algorithm in a way that it insensitive to the value of binfinitive?

Many thanks for your attention in advance

 

Amir

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