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;

[Maple Math]

[Maple Math]

> q[r + dr]:= q[r] + diff(q[r], r)*dr;

[Maple Math]

> eq:= expand((q[r] - q[r + dr])/(k*A*dr)) = 0;

[Maple Math]

> qin:= qinpp*A1;

[Maple Math]

> q:= - k*A1*diff(T(r), r); A1:= 4*Pi*r1^2;

[Maple Math]

[Maple Math]

> bc1:= expand((qin - q)/A1) = 0;

[Maple Math]

> q_Convection:= h[0]*A2*(T(r) - T[f]);

[Maple Math]

> q:= - k*A2*diff(T(r), r); A2:= 4*Pi*r2^2;

[Maple Math]

[Maple Math]

> bc2:= expand((q - q_Convection)/(k*A2)) = 0;

[Maple Math]

Summary of part a) in conventional form.

> ode:= eq;

[Maple Math]

> bc1:= diff(T(r), r) = - qinpp/k;

[Maple Math]

> bc2:= diff(T(r), r) = - (h[0]/k)*(T(r) - T[f]);

[Maple Math]

b) Obtain the temperature distribution in the plane wall.

> sol:= dsolve(ode, T(r));

[Maple Math]

> dersol:= diff(rhs(sol), r);

[Maple Math]

> bc1:=
subs(r = r1, dersol) = - qinpp/k;

[Maple Math]

> bc2:=
subs(r = r2, dersol) =
- (h[0]/k)*(subs(r = r2, rhs(sol)) - T[f]);

[Maple Math]

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

> assign(consts);

> temp:= expand(sol);

[Maple Math]

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));

[Maple Math]

> phi:= subs(h[0] = (Bi/r2)*k, phi);

[Maple Math]

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);

[Maple Math]

> #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;

[Maple Math]

[Maple Math]

[Maple Math]

> Temp:= expand(Temp);

[Maple Math]

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;

[Maple Math]

> phi_r1:= subs(r = r1, phi);

[Maple Math]

> phi_r2:= subs(r = r2, phi);

[Maple Math]

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)));

[Maple Math]

> Q_D2:= expand(subs(r = r2, h[0]*A2*(rhs(temp) - T[f])));

[Maple Math]

> Q_D3:= qinpp*A1;

[Maple Math]

All three methods give the same result.