Example4.6 rev1.mw
Example 4.6
Consider the transient heat conduction problem in a slab.[4] The governing equation in dimensionless form is
=α
u(x,0) = 1 (4.1.6)
(0,t) = h*u(0,t) and u(∞,t) is defined
where α is the thermal diffusivity (m2/s) and h is the heat transfer coefficient (m-1). Carslaw and Jaeger[4] presented solutions for this problem after transforming the governing equation and boundary conditions to a form convenient for similarity transformation. Equation (4.1.6) is solved in Maple below using the Laplace transform technique (note that the transformation is not necessary with this approach):
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restart:with(linalg):with(inttrans):with(plots): |
The governing equation is entered here:
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eq:=diff(u(x,t),t)=alpha*diff(u(x,t),x$2); |
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(1) |
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(2) |
Enter the boundary condition here:
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bc1:=diff(u(x,t),x)-h*u(x,t)=0; |
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bc2:=u(infinity,t)=defined; |
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(4) |
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eqs:=subs(laplace(u(x,t),t,s)=U(x),eqs); |
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(6) |
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(7) |
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bc1:=subs(laplace(u(x,t),t,s)=U(x),bc1); |
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(8) |
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(9) |
The boundary condition in the Laplace domain is:
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(10) |
U is solved as:
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U(x):=rhs(dsolve({eqs,bc1})); |
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eqc:=coeff(U(x),exp(1/alpha^(1/2)*s^(1/2)*x)); |
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coef:=solve(eqc,{_C1}); |
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(13) |
The temperature profile in the Laplace domain is:
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U(x):=evala(subs(coef,U(x))); |
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(14) |
Maple cannot find the inverse of the Laplace transform:
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The above expression is split into two terms U1 and U2 as shown below:
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u1:=invlaplace(U1,s,t); |
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(17) |
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The inverse of U2 is:
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U2:=subs({x=x1,alpha=alpha1},U2); |
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u2:=invlaplace(U2,s,t); |
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(20) |
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u2:=subs({x1=x,alpha1=alpha},u2); |
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(21) |
Finally, the temperature distribution is given by:
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(22) |
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pars:={alpha=1e-3,h=0.01}; |
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(23) |
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plot3d(subs(pars,u),x=1..0,t=500..0,axes=boxed,title="Figure Exp. 4.1.13.",labels=[x,t,"u"],orientation=[110,60]); |
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plot([subs(pars,t=10,u),subs(pars,t=100,u),subs(pars,t=200,u),subs(pars,t=500,u)],x=0..1,axes=boxed,title="Figure Exp. 4.1.14.",thickness=5,labels=[x,"u"]); |
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(24) |