Mr Wilson, On Convection of Heat. 



415 



An expression for the temperature due to a line source can 

 also be got from that for a point source. For a point source at 

 the origin we have 



4<7rkr 



Integrating this along the z-axis, we get, for the temperature 

 due to a line source, 



4-7T& 



r. 



+ a> o2k 



dz = t --re 2k 



spu 

 2k ' 



dz. 



Comparing this with the result previously obtained, we con- 

 clude that 



K n 



ursp 



ir 



6 J —cc 



-f« 



dz 



2k ) 2 J _ M Jr 2 + z 2 



where r is now Jx 2 + y 2 instead of Jx 2 + y 2 + z 2 . 



We shall now consider the distribution of temperature in a 

 slab bounded by two infinite parallel planes between which it 

 slides with a uniform velocity. 



y=t 



2» 



y=o 



x O 



Let AB and CD be the two planes, and take the origin of 

 coordinates at 0, so that the plane CD is given by y = and the 

 plane AB by y = t, where t is the thickness of the slab. We 

 suppose then that v = w = and u is positive. The differential 

 equation is therefore 



d>6 , ffl _spu d6_ 



dx 2 dy 2 k dx 



Assume a solution of the form 



6= 2 C n sin — f- ) , 



n=\ \ l ' 



where C n is a function of x only. 



