wzelik

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When I posted this problem I used Maple 14.
I don't know which version you have.. But Maple Support a few months ago promissed me that problem will be resolved in next (15) version of Maple.

...and they did it.




Best regards

wzel
 

Could anyone confirm, that is it a bug of Maple?


wzel 

Could anyone confirm, that is it a bug of Maple?


wzel 

Original question was: What's happened when we rapidly change convection condition (for example heavy rain) for given material properties, IC and BC. On the beginning I assume that hconv=f(time), but Maple produce wrong results when h growing from 9 to 1000 W/m2K during few seconds. I tried another approach...
I needed steady state for T(0,x):=Tinit(x) (finally Tinit := (685-1573)*x/(.22)+1573), Tinit for given IC, BC and 9 W/m2K) and new IC1, BC1 (h=1000 W/m2K) for solution when heavy rain flow on the horizontal cylindrical surface.
I know that easier way exists, …but I had a model…
if I assume enough time period I reach thermal equilibrium… it was easiest for me in this case.


For me was big surprise than Maple produce wrong results for relatively simply problem. In the same time from MathCAD I got right results.

wzel

Hi, Thank you for replay!

...but I would prefer to check solution using different software, without fitting to know solution from another source.


wzel 

Hi, Thank you for replay!

...but I would prefer to check solution using different software, without fitting to know solution from another source.


wzel 

@hirnyk 

Comsol  Multiphisics 4.0

wzel 

Hi

I tried to follow up your indication, without success unfortunately.

I’m in Maple since May of this year  that is way I can’t construct right numeric solution.
It frustrated me very much.


Lets simplify the problem to steady state.
1st wall: x from 0 to x1;

2nd wall: x from x1 to x2;

Equation: diff(T(x),x,x)=0;

BC1 = T(0)=1200, #heated side

BC2 = diff(T(x2),x)=-(h/lambda2)*(T(x2)-Tamb) # cooling side

and …

BC3 = lambda1*diff(T(x1)=-lambda2*diff(T(x1),x)  #interior boundaries.


If it is no problem for you, please advise me, how to using piecewise transformation according your previous answer.

Regards
wzelik

Hi

I tried to follow up your indication, without success unfortunately.

I’m in Maple since May of this year  that is way I can’t construct right numeric solution.
It frustrated me very much.


Lets simplify the problem to steady state.
1st wall: x from 0 to x1;

2nd wall: x from x1 to x2;

Equation: diff(T(x),x,x)=0;

BC1 = T(0)=1200, #heated side

BC2 = diff(T(x2),x)=-(h/lambda2)*(T(x2)-Tamb) # cooling side

and …

BC3 = lambda1*diff(T(x1)=-lambda2*diff(T(x1),x)  #interior boundaries.


If it is no problem for you, please advise me, how to using piecewise transformation according your previous answer.

Regards
wzelik

Thank you very much for help.


Regards
wzel

Thank you very much for help.


Regards
wzel

Hi again!

Something wrong with my question?  ...or the problem is in pdsolve? :-) :-) :-).

Please for any adivice!

Regards
wzel 

Hi,

The problem is that I don’t  have idea how to set up pdsolve!
I have solved problem  for one domain. It was easy.

PDE := diff(T(x, t), t) = a*(diff(T(x, t), x, x));
a := 9.245*10^(-7); L := .3; h := 10; lambda := 3; `epsilon;` := .85; sigma := 5.6697*10^(-8);
IBC := {T(0, t) = 1000, T(x, 0) = 298, (D[1](T))(L, t) = -h*(T(L, t)-298)/lambda+`epsilon;`*sigma(T(L, t)^4-298^4)};
pds := pdsolve(PDE, IBC, numeric, timestep = 50);
p1 := pds:-plot(t = 100);
p2 := pds:-plot(t = 36000);
plots[display]({p1, p2}, title = `Heat profile at t=100 and 3600 seconds`); 

But I’m looking for solution from two or three domain, for classical multilayer transient heat transfer.
I know there are another numeric possibility, finite element, finite differences,
but I wonder if it possible with Maple.

Regards wzel
 
PS:
In the previous post I made typing error in BC2, sorry for that. 

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