Cooling of Cylinders in a Stream of Air 

 and therefore It per cm. length 



= 4<(0 4 -0o 4 ), 



A = area in sq. cms. 

 I = length in cms. 

 t =time in sees. 



The values obtained are given in Table I. 

 Table I. 



123 



E (cals. per cm. length). 



Diameter of Cylinder. 



4 calories. 



•43 cm. 



1 24 „ 



•81 „ 



28-7 „ 



1-93 „ 



75 „ 



5-06 „ 



2343 „ 



15*5 ., 



115 



Stream-line Section. 



It has been assumed that the temperature of the outside 

 surface of the cooling cylinder was the same as the tempe- 

 rature of the steam. This is practically true, as can be 

 shown by calculating the value of 2 — 6 1 from 



Hi 

 (0 2 -0i)- 27rr £T' 



2 — 6i being the drop of temperature in the tube, 



H the largest value of the heat which passes across the 



tube per cm. length in time T, 

 r the mean radius = 15'4 cm., 

 d the thickness = m 2 cm., 

 k the conductivity of copper. 



The value of 6 2 — Qi is *03° C, so that the temperature of 

 the outside of cylinder can be taken to be 100° 0. 



The quantity of water condensed depends on the density 

 and the temperature of the air, so to obtain uniform results 

 a correction was applied, and the quantity of water was 

 reduced to a standard temperature and density. 



Results. 

 Experiments were performed with five copper cylinders 

 of diameters '43, *81, 1'93, 5'06, 15*5 cm. respectively, and 

 the results obtained for the loss of heat in calories per cm. 

 length (H) and the velocity (V) in metres per sec. are given 

 in Table IL 



K 2 



