Page 338 - Bird R.B. Transport phenomena

322 Chapter 10 Shell Energy Balances and Temperature Distributions in Solids and Laminar Flow

Coolant Aluminum Fig. 10B.3. Temperature distribution in a cylindrical fuel-
cladding rod assembly.
Nuclear
fuel rod

(c) From the temperature profile, obtain an expression for the heat flux at the surface. Equate
this result to the heat flux given by "Newton's law of cooling" and show that a dimensionless
heat transfer coefficient (known as the Nusselt number) is given by

Nu = Щ- = 2 (10B.1-1)

in which D is the sphere diameter. This well-known result provides the limiting value of Nu
for heat transfer from spheres at low Reynolds and Grashof numbers (see §14.4).
(d) In what respect are the Biot number and the Nusselt number different?
10B.2. Viscous heating in slit flow. Find the temperature profile for the viscous heating problem
shown in Fig. 10.4-2, when given the following boundary conditions: at x = 0, T = T ; at x = b,
o
q = 0.
x
Г-T
Answer: o = 1 - ' " 2 '
10B.3 Heat conduction in a nuclear fuel rod assembly (Fig. 10B.3). Consider a long cylindrical nu-
clear fuel rod, surrounded by an annular layer of aluminum cladding. Within the fuel rod
heat is produced by fission; this heat source depends on position approximately as

S» = S 1j £)] (10B.3-1)

Here S n0 and b are known constants, and r is the radial coordinate measured from the axis of
the cylindrical fuel rod. Calculate the maximum temperature in the fuel rod if the outer sur-
face of the cladding is in contact with a liquid coolant at temperature T L . The heat transfer co-
efficient at the cladding-coolant interface is h L, and the thermal conductivities of the fuel rod
and cladding are k F and k c.
Re
-
Answer: 7>, max - T L = ^ ^ + 1 + ^ ln
4k F V V 2k c 2/\R ch L RF
10B.4. Heat conduction in an annulus (Fig. 10B.4).
(a) Heat is flowing through an annular wall of inside radius r 0 and outside radius r v The
thermal conductivity varies linearly with temperature from k 0 at T o to к at T v Develop an ex-
л
pression for the heat flow through the wall.
(b) Show how the expression in (a) can be simplified when (r } — r )/r is very small. Interpret
o
o
the result physically.
Answer: (a) Q = 2irL(T - T )(^l)(ln ^ " ' ; (b) Q = 2nr L^^j0^j
0 1 o