HEAT FROM SMALL CYLINDERS IN A STREAM OF FLUID. 
379 
assuming the flux u(£) to be constant over the boundary, 
writing u (^) = u 0 , 
U = u 0 /3 0 . 
From (18) we have, 
.(19) 
As a result of the high heat-conductivity of the cylinder in the experiments carried 
out the temperature 0 O of the cylinder may be considered constant over its entire 
boundary, an assumption justified by a calculation carried out in Section 13. There 
will therefore be a discontinuity in the temperature over the boundary; we assume 
that the temperature of the stream in contact with the cylinder becomes finally equal 
to that of the cylinder at the point /3 0 where it leaves the boundary. The integral in 
(17) must be divided into two parts in order to make the argument n(x — g) in the 
function K 0 |n(cc — £)| positive; we then obtain 
2t tkQ ( x ) = u 0 
e n{x —^)} f e n( ^~ x) K 0 {n (i—x)} d£ 
J x 
Making the substitution u = n (x — g) in the first integral and v = n(g— x) in the 
second, we may write the above equation in the form 
2 t tk6 (x) = {u 0 /n) e“K 0 (u) du + e~ v K 0 (v) dv 
'n(Po-z) 
( 20 ) 
Writing 6 = 6 0 when x — /% in the above equation and making use of (19) we find for 
the heat-loss of the cylinder per unit length the expression 
H = 27 rK0 o n/3 o 
•nPo 
e u K 0 (u) du 
( 21 ) 
where 6 0 denotes the temperature of the cylinder above that of the stream at a great 
distance. The discontinuity of temperature occurring over the boundary may easily 
be calculated at any point from (20) ; the maximum discontinuity occurring at x = 0 
is easily seen to be given by 
e “K 0 (u) du 
•nft, 
e“K 0 ( u ) du 
( 22 ) 
Section 5. Numerical Evaluation of the Functions Employed in the 
Preceding Section. 
Before we can proceed to evaluate the functions occurring in the preceding section 
we proceed to derive the convergent and asymptotic expansions of the function 
F (x) = j e“K 0 (u) du. Expanding e u in powers of u, and making use of the expansion 
Jo 
(10), we obtain on integrating term by term the convergent series applicable for 
small values of the variable x, 
(l/a?)j e u K 0 (u)du = (l +\x + ^xl 2 + f£%x i + ...) — (y + log-^c)(l +%x + \x 2 + f^x*+...). (23) 
3 c 2 
