376 
PKOF. LOUIS VESSOT KING ON THE CONVECTION OF 
problem leads immediately to the corresponding solution in the case of a cylinder for 
which the hydrodynamical stream-lines can be calculated .( 6 ) 
Section 3. Solution of Boussinesq’s Transformed Problem. 
The complete statement of the problem just formulated may be made in terms of 
an integral equation, and, although the general solution is not yet forthcoming, an 
expression for the heat-loss per unit length may be obtained under special assumptions 
justified by a good agreement with the results of observation. The solution of the 
transformed problem requires us to solve the equation 
Pe , d 2 e 0 be 
— + —2 = 5“ 
ox cy ox 
( 5 ) 
subject to the condition that over the portion of the te-axis between x — 0 and x — /3 0 
the boundary conditions are specified. Since equation (5) is linear, we may build up 
a solution by the integration of a linear distribution of line sources. Writing 
<]> — 6e~ nx , (5) takes the more symmetrical form 
<LS . Tf 
'y 2 ' 'N 2 
ox cy 
n <p, 
( 6 ) 
( 6 ) Boussinesq, loc. cit. (‘ J. de Matt.’), p. 295. The approximate solution given by Boussinesq 
proceeds as follows: assuming the heat conductivity k small the temperature varies very slowly with ft, 
while it varies rapidly with a. If, in addition, the velocity is great enough so that the coefficient cY/k is 
small, we may neglect the term d 2 9/d/3 2 in the differential equation (2), which then reduces to the 
simple Fourier form 
c 2 9/dcc 2 = 2n 30/3/3,.. (i.) 
of which the appropriate solution is 
Q = J{2f) 
f (ft — noi 2 jv 2 ) dv, 
(ii.) 
which prescribes the temperature over the boundary a = 0 by the relation 9 = f (ft). Writing w = na, 2 /v 2 , 
we have (c9/ca .) 0 = - 2 J(2njir) j f (ft — w 2 ) dw, from which we derive from (4) the result 
H = 4k J(2nft r) [f(ft Q - oP) - /( - oP)] d« .(iii.) 
Making the further assumption that approximately 9 = 0 from ft = - oo to ft = 0, 9 = 9 0 from ft = 0 
to ft = fto, and 9 = 0 from ft = ft 0 to ft = oo, (iii.) becomes 
H = 4k J(2n/Tr) ft<?9 0 .. (iv.) 
Applying the results to the case of a circular cylinder, the hydrodynamical solution gives /3 0 = 4a, 
where a is the radius of the cylinder. Also writing c = s<r, where s is the specific heat (at constant 
volume if the fluid is a gas) per unit mass and cr is the density, we obtain finally 
H = 8 (so-K'Va/7r) a 9 {) , .(v.) 
which will be referred to as Boussinesq’s formula. Here 9 0 is the temperature of the fluid cylinder above 
that of the surrounding fluid at a great distance. It must be kept in mind that the above formula cannot 
be expected to represent reality unless k is small and the term 2soYa/i< large. 
