112 On Radiation from a Cylindrical Wall. 



Provided these conditions are satisfied, the energy received 

 by s" from S" will be 



77 ^A(T ] 4 -T 2 4 ). 



From this result the problem can be readily solved. With 

 the same figure as on page 360 and using the same notation, 

 consider any point V on the disk. It is receiving radiation 

 from a surface ot* temperature T\ occupying the solid angle 

 between the two cones having V as vertex and the circles M 

 and K as bases. The disk therefore is receiving from the 

 inner walls an amount of energy equal to that which it 

 would receive from a circular disk K, less what it would 

 receive from a circular disk M, both disks having the same 

 temperature and emissivity as the walls of the cylinder. 



Construct a sphere of which the circle K and the receiving- 

 disk are small circles. The radius of this sphere is 



v 



Pi (at-V -x^y. 



4^ 2 



By substitution of the spherical areas of the two disks in the 

 formula, the result obtained after reduction is that the energy 

 received by the disk from a disk K is: 



1T<J 



2-[v / (^i 2 + « 2 + Vf-kaW-^ + a 2 + 6 2 )] x (T^-T, 4 ). 



Similarly, the energy received by the disk from a disk 

 at M would be 



™[ ^{tf + a * + b*y + ±a*b*- fe 2 + a 2 + 6 2 )](T 1 4 -T 2 4 ;. 



Therefore the energy received by the disk from the walls 

 of the cylinder is 



~ (1\ 4 - T 2 4 )[ V(^ 2 + a 2 + b 2 ) 2 -4a 2 b 2 



- v/ (i 2 2 -r- a 2 -f 6 a ) 2 - 4a 2 T 2 - x x 2 + x 2 2 ] . 



This is the result obtained on page 364. 



In the same manner an expression can be obtained for the 

 energy received by a circular disk from a surface bounded by 

 any number of circles, provided that each of these circles is 

 co-spherical with the disk and conditions (2) and (3) stated 

 previously are satisfied. 



This result can be generalized for the case where the radiating 



