Applications of the Formulas 



equal (x x' ); thus we may obtain the value of M, in the two 

 cases that we have considered by replacing in the general formu- 

 las -~ by -= - . 



Thus the general formula relative to the first case, assuming 

 one and then two air spaces, becomes: 



M= _ Q(T-) 



M= _ Q(T-o) 



f e i .e' i e" 

 \C + Q ^C ^~Q^C" 



If the walls were all of the same material and of the same 

 thickness e, and n in number, we would have 



Assuming a continuous wall of the same total thickness, the 

 quantity of heat transmitted would be: 



M'= Q (T ~ 0) Q(T-O) 



and we would have 



M' 



876. If we assume the air spaces and the walls each to be 

 .79" thick, and the walls to be of brick, we would have Q- =1.14 

 (7=4.84 and the ratio: 



M_ 2-f.i85 (in ij 

 W'~2-\-.iS5 n+ (n i ) 

 making n successively equal to 



i 2 3 4 5 10 



we find for M^-M' 



i -75 -64 -57 -53 -43 



