Half-Space Neumann Problems

M.M. Yovanovich

HSNEUMANN1.MWS

Neumann half-space solution. Applications to several different problems where one of the several system parameters are found given the remaining parameters.

This requires a numerical inversion of the Neumann solution when one of the parameters are not given. This worksheet shows the usefulness of a Computer Algebra System such as Maple.

> restart:

> Neumann:= T - Ti - 2*q0/k*sqrt(alpha*t)*exp(-x^2/(4*alpha*t))+ (q0*x)/k*erfc(x/(2*sqrt(alpha*t)));

Solve for temperature .

> case1:= (Ti=300, k = 20, alpha=10e-6, t=200, q0=2e4, x=10/1000);

> T1:= evalf(solve(subs(case1, Neumann), T), 6);

Solve for time .

> case2:= (Ti=300, T = 380, k = 20, alpha=10e-6, q0=2e4, x=10/1000);

> t2:= evalf(fsolve(subs(case2, Neumann), t),6);

Solve for incident heat flux .

> case3:= (Ti=300, T = 380, k = 20, alpha=10e-6, t= 300, x=10/1000);

> q03:= evalf(fsolve(subs(case3, Neumann), q0),6);

Solve for the thermal diffusivity .

> case4:= (Ti=300, T = 380, k = 20, q0 = 16054, t= 300, x=10/1000);

> alpha4:= evalf(fsolve(subs(case4, Neumann), alpha),6);

> case5:= (Ti=300, T = 380, k = 20, q0 = 16054, t= 300, alpha=10e-6);

Solve for the position in half-space .

> x5:= evalf(fsolve(subs(case5, Neumann), x),6)*1000;

Solve for the thermal conductivity .

> case6:= (Ti=300, T = 390, x = 30/1000, q0 = 8e4, t= 300, alpha=10e-6);

> k6:= evalf(fsolve(subs(case6, Neumann), k),6);

The above six examples illustrate how the Neumann solution can be used to find the unknown parameters when the remaining parameters are known.