J4 STEAM 



The difference between the given pressure and 76 Ib. is 77.7 

 76 = 1.7 Ib. For a difference of 1.7 Ib., the change of total 

 heat is 1.7 X .25 = .425 B. T. U. Hence, for 77.7 Ib., H = 1,176.0 

 + .425=1,176.425, say 1,176.4 B. T. U. 



EXAMPLE 5. Find the volume occupied by 14 Ib. of steam at 

 30 Ib. gauge pressure. 



SOLUTION. Absolute pressure = 30+14.7 = 44.7 Ib. per sq. in. 

 From the Steam Table, 



for = 44 Ib., V = 9.484 cu. ft. 



for = 46 Ib., F = 9.097 cu. ft. 



Difference, 2 Ib., .387 cu. ft. 



.387 

 The difference for 1 Ib. is = .1935. 44.7-44 = .7 Ib. 



actual difference in pressure. . 1935 X. 7 = .135 difference in 

 volume. As the pressure increases, the volume decreases; 

 and to obtain the volume at 44.7 Ib., it is necessary to sub- 

 tract the difference .135 from the volume at 44 Ib.; thus, for p 

 = 44.7, v = 9.484 -.135 = 9.349 cu. ft. The volume of 14 Ib. 

 is 14X9.349 cu. ft. = 130.89 cu ft. 



EXAMPLE 6. Find the weight of 40 cu. ft. of steam at a 

 temperature of 25-4 F. 



SOLUTION. From column 10 of the Steam Table, the weight 

 w of 1 cu. ft. of steam at 253.98 is .07820 Ib: 254 253.98 = .02. 

 Neglecting the .02, the weight of 40 cu. ft. is therefore 

 ,07820X40-3.128 Ib. 



EXAMPLE 7. How many pounds of steam at 64 Ib. pressure, 

 absolute, are required to raise the temperature of 300 Ib. of 

 water from 40 to 130 P., the water and steam being mixed 

 together? 



SOLUTION. The number of heat units required to raise 1 Ib. 

 from 40 to 130 is 130 -40 = 90 B. T. U. Actually, a little 

 more than 90 would be required but the above is near enough 

 for all practical purposes. Then, to raise 300 Ib. from 40 

 to 130 requires 90X300 = 27,000 B. T. U. This quantity of 

 heat must necessarily come from the steam. Now, 1 Ib. of steam 

 at 64 Ib. pressure gives up, in condensing, its latent heat of 

 vaporization, or 906.2 B. T. U.; but, in addition to its latent 

 heat, each pound of steam on condensing must give up an 



