Example 4.3 Heat Conduction with Flux Boundary Conditions
In the previous two examples, the temperature (dependent variable) at x = 0 was specified. The same technique can be applied for eh case where the derivative of the dependent variable is known at the boundary x = 0 (flux boundary conditions). Consider the transient heat conduction problem in a slab.[4] The governing equation in dimensionless form is
=α
u(x,0) = 0
(0,t) = - k and u(∞,t) is defined (4.1.3)
The flux boundary condition has to be considered while taking the Laplace transform. Equation (4.1.3) is solved in Maple below:
| > | restart:with(linalg):with(inttrans):with(plots): |
| > | eq:=diff(u(x,t),t)=alpha*diff(u(x,t),x$2); |
| (1) |
| > | u(x,0):=0; |
| (2) |
| > | bc1:=diff(u(x,t),x)=-k; |
| (3) |
| > | bc2:=u(infinity,t)=defined; |
| (4) |
| > | eqs:=laplace(eq,t,s): |
| > | eqs:=subs(laplace(u(x,t),t,s)=U(x),eqs); |
| (5) |
| > | bc1:=laplace(bc1,t,s); |
| (6) |
The boundary condition in the Laplace domain is:
| > | bc1:=subs(laplace(u(x,t),t,s)=U(x),bc1); |
| (7) |
| > | bc1:=subs(x=0,convert(bc1,D)); |
| (8) |
| > | U(x):=rhs(dsolve({eqs,bc1},U(x))); |
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(9) |
| > | eqc:=coeff(U(x),exp(1/alpha^(1/2)*s^(1/2)*x)); |
| (10) |
| > | _C2:=solve(eqc,_C2); |
| (11) |
| > | U(x):=eval(U(x)); |
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(12) |
| > | u:=invlaplace(U(x),s,t); |
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(13) |
The dimensionless temperature profile is given by:
| > | u:=convert(u,erfc); |
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(14) |
| > | pars:={alpha=0.001,k=1}; |
| (15) |
Plots are made for particular values of parameters:
| > | plot3d(subs(pars,u),x=1..0,t=300..0,axes=boxed,title="Figure Exp. 4.1.7.",labels=[x,t,"u"],orientation=[-60,60]); |
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| > | plot([subs(pars,t=1,u),subs(pars,t=10,u),subs(pars,t=100,u),subs(pars,t=200,u)],x=0..1,axes=boxed,title="Figure Exp. 4.1.8.",thickness=5,labels=[x,"u"],legend=["t=1","t=10","t=100","t=200"]); |
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| > |