BY JAMES TOLSON, ESQ. 95 
peri (Cxp+(OxEx T’) 
C+(QxE). 
SS 315 X24 X 14) 
Rae (es 815 Xi24): 
== EGyOde — ts 
— 
To find the amount of heat transmitted through the 
walls— 
1st.—Calculated from the temperature of the walls. 
Cx (t—t!) = 55 X74°54—15'04 = 1°366 
E 24 
2nd.—Calculated from the temperature of the surface of 
the internal wall and the temperature of the internal air. 
Q x (t’—T)=1°315 X (15°04 — 14) = 1.366. 
3rd.—Calculated from the temperature of the external 
surface of the walls and the temperature of the internal air. 
Ode 2) £0) Gans Xan ge ihe 
RO ehh in 4 Gee aay meer 
ie 55 
41th.—Calculated from the temperature of the external 
and internal air only. This is, perhaps, the most useful 
rule. 
(se xO) x aT) 
{Cx [2x A)+RI]}+{ExAxQ} 
‘568 "5.5.x I 315) X(77—14) 
ss Fare 24% 568 X 1315 ¢ 
==) 20: 
Having found the flow of heat per square foot per hour into 
a chamber of the above dimensions, it is an easy matter to 
ascertain the total amount of heat gained by leakage. As 
there is a considerable difference between the superficial 
area of the external and internal surfaces of the walls, it 
will be necessary to take the mean area. ‘This is as stated 
above—6,536 sup. feet. Multiplying 6,536 by 1°366, we 
have the total flow of heat through the wall into the 
chamber of 
8,928 units per hour. 
