partial derivatives system of differential equations in multiple unknowns

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EldaEbrithil
EldaEbrithil 2020년 3월 17일
댓글: EldaEbrithil 2020년 3월 18일
Hi all
i have some problem with pdepe : "Unable to perform assignment because the size of the left side is 4-by-1 and the size of the right side is 4-by-4."
I have 4 differential equation, one of these has two indipendent variables. These equations are connected also in terms of BC
r = (0:0.01: 1);
t = (0:0.01:2);
m = 1;
sol = pdepe(m,@pdefun,@pdeic,@pdebc,r,t);
Th = sol(:,:,1);
Tg = sol(:,:,2);
Tsurfintchamber=sol(:,:,3);
Qdot=sol(:,:,4);
function [c,effe,s] = pdefun(r,t,T,dTdr)
Th=T(1);
Tg=T(2);
Tsurfintchamber=T(3);
Qdot=T(4);
y1=((hcoefficientinternalcyclefinal*Abobbin.*(T(1)-T(2)))-...
(hcoefficientoutcyclefinal*Ainternalsurfacechamber.*(T(2)-T(3))))./(densityxenonhotgas*Accycle2*Lbobbin*cvxenon);
y2=(1/((Nuout*thermalconductivity/((2*Rintchambercycle2final)-...
(2*Rextbobbincycle2final)))*(Rintchambercycle2final)))/(2*pi*Lbobbin);
y3=((1/((Nuout*thermalconductivity/((2*...
Rintchambercycle2final)-(2*Rextbobbincycle2final)))*(Rintchambercycle2final)))+...
((log(0.0129./(Rintchambercycle2final)))/ThermalconductivityAISI304)+((log((0.0129+0.005)./0.0129))/PROMALIGHT1200thermalconductivity)+...
((log((0.0129+0.005+Thicknessoutercylinder)./(0.0129+0.005)))/ThermalconductivityAISI304));
y4=(Power/(((pi/4)*Dtantalum^2)*Lstraightheater))+hcoefficientinternalcyclefinal*dTdr(1);
c=[densitytantalum*specificheattantalum,0,0, 0;0,1,0,0;0,-1,1,y2;0,-(10/100)*2*pi*Lbobbin, 0, y3];
effe=[thermalconductivitytantalum;0;0;0]*dTdr(1);
s=[y4;y1;0;0];
end
function T0 = pdeic(r)
Tcoldgas=300;%kelvin
T0 = [Tcoldgas; Tcoldgas;Tcoldgas;0];
end
function [pL,qL,pR,qR] = pdebc(rL,TL,rR,TR,t)
hcoefficientinternalcyclefinal=21.697199999999990;
pL = [TL(1)-900; 0;0;0];
qL = [0; 0;0;0];
pR = [hcoefficientinternalcyclefinal*(TR(1)-TR(2)); 0;0;0];
qR = [-1;0;0;0];
end
  댓글 수: 3
EldaEbrithil
EldaEbrithil 2020년 3월 17일
the equations are in this form c(x,t,u,Du/Dx) * Du/Dt = x^(-m) * D(x^m * f(x,t,u,Du/Dx))/Dx + s(x,t,u,Du/Dx)
Initial conditions:
T0 = [Tcoldgas; Tcoldgas;Tcoldgas;0];
boundary conditions are in this form:
p(x,t,u) + q(x,t) * f(x,t,u,Du/Dx) = 0

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