ME 353 Heat Transfer 1
M.M. Yovanovich
P1MT91.MWS
Problem 1 of Midterm Exam, October 1991.
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A spherical shell of radii r1 and r2 > r1, and thermal conductivity
k. Inner boundary at r1 is subjected to uniform heat flux qinpp,
while the outer boundary at r2 is convectively cooled by a fluid
whose temperature is Tf and the heat transfer coefficient is h0.
The conbined heat transfer coefficient consists of the convection
and radiation conponents hrad and hconv. The fluid and surrounding
temperatures are equal.
Steady-state, constant properties, source-free.
Use the notation given in the problem which is the same as the text.
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> restart:
a) Derive differential equation and boundary conditions by means of
heat balances over differential control volumes for the spherical shell
of inner and outer radii r1 and r2, and thermal conductivity k.
The inner boundary has incident flux qipp, and the outer surface is
convectively cooled by heat transfer coefficient ho and a fluid
whose temperature is Tf.
> q[r]:= - k*A*diff(T(r), r); A:= 4*Pi*r^2;
> q[r + dr]:= q[r] + diff(q[r], r)*dr;
> eq:= expand((q[r] - q[r + dr])/(k*A*dr)) = 0;
> qin:= qinpp*A1;
> q:= - k*A1*diff(T(r), r); A1:= 4*Pi*r1^2;
> bc1:= expand((qin - q)/A1) = 0;
> q_Convection:= h[0]*A2*(T(r) - T[f]);
> q:= - k*A2*diff(T(r), r); A2:= 4*Pi*r2^2;
> bc2:= expand((q - q_Convection)/(k*A2)) = 0;
Summary of part a) in conventional form.
> ode:= eq;
> bc1:= diff(T(r), r) = - qinpp/k;
> bc2:= diff(T(r), r) = - (h[0]/k)*(T(r) - T[f]);
b) Obtain the temperature distribution in the plane wall.
> sol:= dsolve(ode, T(r));
> dersol:= diff(rhs(sol), r);
>
bc1:=
subs(r = r1, dersol) = - qinpp/k;
>
bc2:=
subs(r = r2, dersol) =
- (h[0]/k)*(subs(r = r2, rhs(sol)) - T[f]);
> consts:= solve({bc1, bc2}, {_C1, _C2}):
> assign(consts);
> temp:= expand(sol);
b) Show that the dimensionless temperature phi = (T(r) - Tf)/(qinpp*r1/k)
depends on the Biot number, Bi = h0*r2/k and radii ratio r1/r2.
> phi:= expand((rhs(temp) - T[f])/(qinpp*r1/k));
> phi:= subs(h[0] = (Bi/r2)*k, phi);
c) By means of the thermal resistance concept obtain the solution
for phi(r) directly, and obtain expressions for the boundary temperatures
phi(r1) and phi(r2).
> #restart:
> Temp:= T[f] + q1*(Rf2 + Rs);
>
#Rs is the solid resistance of the shell
#from r = r to r = r2.
>
Rf2:= 1/(h[0]*A2);
Rs:= 1/(4*Pi*k)*(1/r - 1/r2);
q1:= qinpp*A1;
> Temp:= expand(Temp);
The temperature distribution is identical to that obtained formally.
We can use the expression for phi to find phi(r1) and phi(r2).
> phi:= phi;
> phi_r1:= subs(r = r1, phi);
> phi_r2:= subs(r = r2, phi);
d) Determine the heat transfer rate at the outer surface r = r2
using three methods:
i) Fourier's rate equation, Q = - k A dT/dr,
ii) Newton's cooling law, Q = h A (Ts - Tf),
iii) Overall thermal energy balance, Qout = Qin,
because Qgen = 0, Qstorage = 0.
> Q_D1:= expand(subs(r = r2, - k*A2*diff(rhs(temp), r)));
> Q_D2:= expand(subs(r = r2, h[0]*A2*(rhs(temp) - T[f])));
>
Q_D3:= qinpp*A1;
All three methods give the same result.