Problem 1: ME 303 Spring 1998

M.M. Yovanovich

S98MTPROB1.MWS

Solution for Problem 1.

> restart:

(a) What are the S.I. units of given the units of ?

> units_nu:= solve(m/s/m^2 - 1/nu*m/s/s, nu);

The units of the terms of the given PDE are . Since the units of are

the units of the transport property are .

(b) Find the steady-state velocity distribution.

The steady-state velocity distribution is obtained from the ODE:

> odess:= diff(u(y),y,y) = 0;

> solss:= dsolve(odess, u(y));

The boundary conditions on the steady-state ODE and its solution are and .

> bc1:= subs(y = 0, rhs(solss)) = U;

> bc2:= subs(y = H, rhs(solss)) = 0;

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

> assign(consts):

> solss:= solss;

The steady-state velocity distribution is linear in as expected.

(c) By means of the steady-state velocity distribution and Newton's Law of Viscosity find the relation between the wall shear and the parameters: .

> tau[w]:= -mu*Diff(rhs(solss),y);

> tau[w]:= value(%);

(d) If the units of , what are the units of ?

> units_of_mu:= solve(N/m^2 - mu*(m/s)/m = 0, mu);

(e) Let in the given PDE and obtain the SLP: Y'' + Y = 0, in the interval , with homogeneous BCS: and .

> restart:
PDE:= diff(u(y,t),y,y) - diff(u(y,t),t)/nu = 0;

> u(y,t):= v(y) + w(y,t);

> PDEsep:= PDE;

The given PDE can be separated in an ODE which represents the steady-state solution, and another homogeous PDE which can be separated by means of the substitution into two related ODEs.

> pde1:= op(1,PDEsep);
pde2:= op(2,pde1) + op(3,pde1) = 0;

> w(y,t):= Y(y)*tau(t);

> pde3:= expand(pde2/w(y,t));

The PDE has been separated. We can find the related ODEs by setting the first term to and the second term to to get the separated ODEs.

> odeY:= diff(Y(y),y,y) + lambda^2*Y(y) = 0;

> odet:= diff(tau(t),t) + lambda^2*nu*tau(t) = 0;

The solutions of the separated ODEs are:

> solodeY:= dsolve(odeY, Y(y));

> solodet:= dsolve(odet, tau(t));

The boundary conditions on the equation and its solution come from the conditions on the equation.

Since and , we have which requires that . Since cannot be , then .

Since and , we have which requires . Since cannot be , then we have .

(f) What eigenfunctions and eigenvalues satisfy the SLP problem?

> bc1:= simplify(subs(y = 0, rhs(solodeY))) = 0;

> bc2:= simplify(subs(y = H, rhs(solodeY))) = 0;

From the first BC we see that which removes the cosine function from the solution.

The second BC requires that or . The first choice leads to the trivial solution , and therefore it is rejected. The second option requires that .

For this SLP the eigenfunctions are the sines: and the eigenvalues are , where , etc .