STEAM 889 



bined. Hence is deduced the following composition of the weight of one cubic foot of 

 steam would have at the temperature of 32 Fahr., and pressure of one atmosphere 

 (or 147 Ibs. on the square inch), if steam were a perfect gas, and if it could exist at 

 the pressure and temperature stated. 



Data from the Experiments of Regnault. 



Half a cubic foot of oxygen at the pressure of one atmosphere b. 



and temperature, 32 ....... 0-044628 



1 cubic foot of hydrogen ....... 0-005592 



1 cubic foot of steam in the ideal state of perfect gas, at one . 

 atmosphere and 32 ........ 0-050220 



If steam were a perfect gas, the weight of a cubic foot could be calculated for any 

 given pressure and temperature by the following formula : 

 Weight of a cubic foot = 0-05022 Ib. x pressure in atmosphere = 



493-2 



For example, at one atmosphere of pressure, and 212, the weight of a cubic foot of 

 steam would be : 



JQO.OO 



0-05022 x 3r_f= 0-03679 Ib. 



But steam is known not to be a perfect gas ; and its actual density is greater than that 

 which is given by the preceding formula, though to what extent is not known by 

 direct experiment. The most probable method of indirectly determining the density 

 of steam, is by computation from the latent heat of evaporation, from which it appears 

 that at one atmosphere and 212, the weight of a cubic foot of steam is probably 

 0'03679 Ib. The greatest pressure under which steam can exist at a given temperature 

 is called the pressure of saturation for steam of a given temperature. The tempera- 

 ture is called the boiling point of water under the given pressure. The pressure of 

 saturation is the only pressure at which steam and liquid water can exist together in 

 the same vessel at a given temperature. 



It becomes necessary to understand correctly the method of determining fixed 

 temperatures by certain phenomena taking place at them. Thus ice begins to melt 

 at a point, which we call the freezing point, marked 32 upon the scale devised by 

 Fahrenheit (see THERMOMETER), and we determine the boiling point of water to be 

 212 on the same scale, under the average atmospheric pressure of 14'7 Ibs. on the 

 square inch; 2116'4 Ibs. on the square foot; 29'992 inches of the column of mercury. 

 At this latter point water ceases to be liquid, and becomes vaporiform. From 32 to 

 212, all the heat which has been poured into the water has effected no change 

 of physical condition, but the higher temperature being reached, a new condition is 

 established, and steam is produced ; this steam then beginning to act according to 

 certain fixed laws. 



A cubic inch of water evaporated under the ordinary atmospheric pressure is converted 

 into a cubic foot of steam. 



A cubic inch of water evaporated under the atmospheric pressure gives a mechanical 

 force equal to what would raise a ton weight one foot high. 



These are the effects produced at 212 under the above-named pressure. 



Careful experiments have determined, within very small Umits of error, the 

 following facts : Steam under pressure of 35 Ibs. per square inch, and at the tem- 

 perature of 261, exerts a force equal to a ton weight raised one foot; iinder the 

 pressure of 15 Ibs. and at the temperature of 213, it is 2,086 Ibs., or about seven per 

 cent, less ; and under 70 Ibs. and at 306 it is 2,382 Ibs., or nearly six and a half per 

 cent, more than a ton raised a foot. It is sufficient for all practical purposes to assume 

 that each cubic inch evaporated, whatever be the pressure, develops a gross mechanical 

 effort equivalent to a ton weight raised one foot. 



As a given power is produced by a given rate of evaporation, to determine this the 

 following rules are applicable : 



To produce the force expressed by one horse-power, the evaporation per minute 

 must develop a mechanical force equal to 33,000 Ibs., or about 15 tons raised 1 foot 

 high. Fifteen cubic inches of water would accordingly produce this effect, which, 

 without evaporation, would be equivalent to 900 cubic inches per hour. To find, 

 therefore, the gross power developed by a boiler, it would be only necessary to divide 

 the number of cubic inches of water evaporated per hour by 900. If, therefore, 

 to 900 cubic inches be added the quantity of water per hour necessary to move the 

 engine itself, independently of its load, we shall obtain the quantity of water per hour 



