Problem 2: ME 303 Spring 1998

M.M. Yovanovich

S98MTPROB3.MWS

Solution for Problem 2.

> restart:

(a) Nondimensionalize the PDE with .

Transformation of the terms and .

> ux:= diff(u(zeta,tau),zeta)*diff(zeta(x),x);

> uxx:= diff(u(zeta,tau),zeta,zeta)*diff(zeta(x),x)^2;

> ut:= diff(u(zeta,tau),tau)*diff(tau(t),t);

Introduce into the PDE.

> zeta:=x-> x/L; diff(zeta(x),x);
tau:= t->alpha*t/L^2;
u(zeta,tau):= phi(zeta,tau)*u[ref];

Transformations of space and time terms of PDE.

> uxx:= uxx;

> ut:= ut;

Substitute into the dimensional PDE to get nondimensional PDE:

> pdend:= uxx + P/k - ut/alpha = 0;

Multiply through by to get another form of the PDE.

> pde2:= expand(L^2/u[ref]*pdend);

(b) Select the arbitrary reference temperature such that the source term becomes 1.

> u[ref]:= L^2*P/k;

(c) Specify the nondimensional steady-state ODE and the boundary conditions, and obtain the solution.

The steady-state ODE is obtained from the transient PDE by setting the transient term to zero. The two steady-state boundary conditions are obtained from the conditions:

and . They are and .

> unassign(zeta):
odes:= diff(phis(zeta),zeta,zeta) + 1 = 0;

> sol:= dsolve(odes, phis(zeta));

> bc1:= subs(zeta = 0, diff(rhs(sol),zeta)) = 0;

> bc2:= subs(zeta = 1, rhs(sol)) = 0;

> consts:= solve({bc1,bc2},{_C1,_C2});

> assign(consts):

> sol:= sol;

(d) >From the nondimensional steady-state solution determine the dimensionless heat transfer rates at the two boundaries defined as at and at .

> Qstar[0]:= -subs(zeta =0, diff(rhs(sol),zeta));

> Qstar[1]:= -subs(zeta =1, diff(rhs(sol),zeta));

>

The boundary at is adiabatic and all of the heat generated inside the system leaves through the right boundary.