623 
that have been examined experimentally, the conduction of heat 
throngh the wall is very great, however, because the wall is not 
very thick, and consists of a substance (mostly glass) that conducts 
heat pretty well. Consequently the influence of the resistance to heat 
of the wall of the tube is slight, and the following approximation 
may be used. When the current of heat in the wall is supposed to 
be radial, and when d and A, represent the thickness resp. the 
conductivity of this wall, a quantity of heat given by: 
1,0 
a 
flows through the wall per unit of time and surface. 
In this @ is the temperature of the substance on the inner side 
of the wall of the tube. 
When a is the radius of the interior width of the tube, we get 
the boundary conditions: 
He CAN aes PON fy 
ae hm ag Tan. 
Both members of this equation express the current of heat per 
unit of time and surface. 
In order to solve the differential equation (6) with the conditions 
(8), (9) and (10), we seek a particular solution, which is a product 
of two factors, one that depends on 2, (X,), and one that depends 
on r (R). When we substitute: 
On NAR, 
in (6), we may write for this equation: 
PR, HR, dX, AX, 
dries dr a, dz, daz,” 
rR, ae Xe 
As in this relation the first member depends only on 7 and the 
second member only on w,, both members are constant, e.g. — C. 
Then the following equations are obtained for X, and R,: 
at dh 5 
EEn En dr. + C Ji B; . 5 Oy 7 . e (11a) 
aX v, dX 
ee wes OK Oene Sew eee aie VID) 
dr? a, dr 
The solution of (41a) which remains finite for 7 = 0, is the Brssur. 
function of order zero: 
Et = Se WE). ae econ eae 
40* 
