98 STEAM 



tz = temperature of water in barrel after steam is con- 



densed; 

 L = latent heat of 1 Ib. of steam at observed pressure. 



Then, Q = - 



EXAMPLE. In a calorimetric test, the weight of cold water 

 was 420 Ib., and of steam condensed, 36 Ib. The initial 

 temperature of the cold water was 40 F., the final tempera- 

 ture was 130 F., and the average steam pressure was 60 Ib. 

 Find the quality of the steam. 



SOLUTION. Absolute pressure = 60+ 14.7 = 74.7 Ib. per sq. in. 

 Latent heat of steam at this pressure, from Steam Table, is 

 898.5. The temperature of steam at this pressure is 307.2. 

 Hence, by the formula 



-- .9714 



898.5 



That is, the boiler generates a mixture that is composed of 

 97.14% of dry steam and 2.86% of water. 



If the result found by the foregoing formula shows that Q 

 is greater than 1, the conclusion is that the steam, instead of 

 being wet, was superheated, and therefore gave up, in con- 

 densing, a greater amount of heat per pound than would have 

 been given up by 1 Ib. of dry saturated steam at the same 

 pressure. 



The barrel calorimeter must be used very carefully in order 

 to obtain accurate results. The operation should be repeated 

 once or twice before the actual test is made so as to warm up 

 the barrel. The most important observation is the tempera- 

 ture. This should be taken by a thermometer graduated to 

 fifths or tenths of a degree. The weights should be as accu- 

 rate as possible. The chief merit of the barrel calorimeter is 

 its availability. It can be rigged up in almost any situation. 



The quality of the steam having been determined, the actual 

 amount of water evaporated by a steam boiler is found by 

 multiplying the observed amount of feedwater by the quality 

 of the steam expressed decimally. 



