I need to find the value of Ta and Tc for two simultaneous equations to get a numerical answer using NSolve. I have tried using FindMinimum and NMinimize and still could not solve the problem. Here is my code:
qe = 1.6*10^-19;
Ah = 0.0025; (*m^2*)
AL = 0.025; (*m^2*)
ALoss = 2.5*10^-4; (*m^2*)
ATi = 2.5*10^-4; (*m^2*)
h = 6.626*10^-34; (*J.s*)
kB = 1.38*10^-23; (*J/K*)
Krd = 1.2*10^6; (*A/m^2.K^4*)
\[Epsilon]in = 0.18;
\[Epsilon] = 0.9;
Th = 1500; (*K*)
Tl = 300; (*K*)
T0 = 300; (*K*)
Uh = 1.1022*10^4; (*W/m.K^4*)
Ul = 2.2045*10^4; (*W/m.K^4*)
Uloss = 1.1022*10^4; (*W/m.K^4*)
\[Sigma] = 5.67*10^-8; (*W/m.K^4*)
\[Epsilon]in = 0.18;
wfc = 1.0; (*eV*)
V = 1.0; (*V*)
k1 = (ATi*\[Sigma]*\[Epsilon]in)/(Uh*Ah) + (\[Epsilon]*\[Sigma])/Uh;
k2 = (2*kB*Krd*ATi)/(qe*Uh*Ah);
k3 = (Krd*ATi)/(qe*Uh*Ah);
k4 = 1;
k5 = (ATi*\[Sigma]*\[Epsilon]in)/(Uh*Ah);
k6 = Th + (\[Epsilon]*\[Sigma]*Th^4)/Uh;
k7 = (ATi*\[Sigma]*\[Epsilon]in)/(Ul*AL) + (\[Epsilon]*\[Sigma])/Ul;
k8 = (2*kB*Krd*ATi)/(qe*Ul*AL);
k9 = (Krd*ATi)/(qe*Ul*AL);
k10 = 1;
k11 = (ATi*\[Sigma]*\[Epsilon]in)/(Ul*AL);
k12 = Tl + (\[Epsilon]*\[Sigma]*Tl^4)/Ul;
eqn1 = k1*Tc^4 + k2*Tc^3*Exp[-(wfc*qe + qe*V)/(kB*Tc)] +
k3*(wfc*qe + qe*V)*Tc^2*Exp[-(wfc*qe + qe*V)/(kB*Tc)] + Tc -
k4*Ta^4 - k2*Ta^3*Exp[-(wfc*qe)/(kB*Ta)] -
k3*(wfc*qe + qe*V)*Ta^2*Exp[-(wfc*qe)/(kB*Ta)] - k5;
eqn2 = k6*Ta^4 + k7*Ta^3*Exp[-(wfc*qe)/(kB*Ta)] +
k8*(wfc*qe)*Ta^2*Exp[-(wfc*qe)/(kB*Ta)] + Ta - k9*Tc^4 -
k7*Tc^3*Exp[-(wfc*qe + V*qe)/(kB*Tc)] -
k8*(wfc*qe)*Tc^2*Exp[-(wfc*qe + qe*V)/(kB*Tc)] - k10;
sol = NSolve[{eqn1 == 0, eqn2 == 0}, {Ta, Tc}, Reals]
The value of Ta should be close to 300 and Tc is around 1500.
qe = 1.6*10^-19;
. $\endgroup$ – Roman Jun 11 at 21:22FindRoot
is more similar toNSolve
thanFindMinimum
. Have you tried it? $\endgroup$ – Chris K Jun 11 at 21:29Plot3D[eqn1, {Ta, 200, 400}, {Tc, 1000, 2000}]
andPlot3D[eqn2, {Ta, 200, 400}, {Tc, 1000, 2000}]
to see that neithereqn1
noreqn2
are close to zero within this region. $\endgroup$ – Jack LaVigne Jun 12 at 1:56