SECT. VI.] HEATING OF CLOSED SPACES. 69 



be hS(b a J }, instead of being hS(b a), for we suppose that 

 the new surface of the solid and the surfaces which bound the 

 shell have likewise the same external conducibility h. It is 

 evident that the expenditure of the source of heat will be less 

 than it was at first. The problem is to determine the exact ratio 

 of these quantities. 



89. Let e be the thickness of the shell, m the fixed tempera 

 ture of its inner surface, n that of its outer surface, and K its 

 internal conducibility. We shall have, as the expression of the 

 quantity of heat which leaves the solid through its surface, 

 hS(b-a ). 



As that of the quantity which penetrates the inner surface 

 of the shell, hS (a - m). 



As that of the quantity which crosses any section whatever 



of the same shell. KS . 



e 



Lastly, as the expression of the quantity which passes through 

 the outer surface into the air, hS (n a). 



All these quantities must be equal, we have therefore the 

 following equations : 



rr 



h (n a) = (m ri), 



h(n a) = h (a m), 

 h(n-a)=h(b-a). 

 If moreover we write down the identical equation 



k(n a) = h(n a), 

 and arrange them all under the forms 

 n a = n a, 



m-n = - (n-a) 



I 



b a = n a, 

 we find, on addition, 



