VOLUME AND WEIGHT OF AIR. The volume of air and its weight per cubic foot change with the temperature. The final volume may be computed by the following: Rule.-Reduce both the original and final temperatures to absolute temperatures. Multiply the original volume by the final absolute temperature and divide by the original absolute temperature. The quotient will be the final volume. Or, let Voriginal volume; V1 = final volume; Toriginal absolute temperature, V= EXAMPLE.-What will be the volume of 400 cu. ft. of air, having a temperature of 150°, when it is cooled to 10°? SOLUTION.-Applying the above rule, 400 (460 +10) 308.19 cu. ft. Ans. 460 + 150 The final weight of a given volume of air may be com. puted from the following: Rule.-Multiply the original weight by the original absolute temperature, and divide the product by the final absolute temper• ature. The quotient will be the final weight. Or, letting W original weight; W1 = final weight; WT Wi T EXAMPLE.-A chimney of 1 sq. ft. area and 120 ft. high is filled with hot air at a temperature of 450°; the temperature of the atmosphere is 60°; what is the difference in the weight of the air before and after it is heated? V T = = T SOLUTION.-The volume of the air is 120 cu. ft. The original weight is (see preceding table) 120 X .076389.1656 lb., and the absolute temperature is 60+ 460 = 520°. Then, apply. ing the last rule, 9.1656 X 520 The change in weight = 9.1656-5.2375 3.9281 lb. Ans. 5.2375 lb. VOLUME, DENSITY, AND PRESSURE OF AIR, (D. K. Clark.) 180 200 210 212 220 Pressure at Constant Volume Pounds per Sq. 12.96 13.86 14.08 14.36 14.70 14.92 15.21 15.49 15.77 16.05 16.33 16.61 16.89 17.19 17.50 17.76 18.02 18.58 18.86 18.92 19.14 Comparative Pres sure .881 .943 .958 .977 1.000 1.015 1.034 1.054 1.073 1.092 1.111 1.130 1.149 1.168 1.187 1.206 1.226 1.264 1.283 1.287 1.302 DEFINITIONS. The process of changing water (or other liquids) into vapor by means of heat is called evaporation or vaporization. When steam is in contact with the water from which it is generated it is called saturated steam. Steam, if not in contact with water, may be heated like air or any other gas until its temperature is higher than the boiling point. Steam in this condition is said to be superheated. The specific heat of superheated steam is .4805, or, say, .48 for ordinary purposes. STEAM TABLES. Whenever the pressure of saturated steam is changed, there are other properties that change with it. These properties are the following: 1. The temperature of the steam, or, what is the same thing, the boiling point. 2. The number of B. T. U. required to raise a pound of water from 32° (freezing) to the boiling point corresponding to the given pressure. This is called the heat of the liquid. 3. The number of B. T. U. required to change the water at the boiling temperature into steam at the same temperature. This is called the latent heat of vaporization, or, simply, the latent heat. 4. The number of heat units required to change a pound of water at 32° to steam of the required temperature and pressure. This is called the total heat of vaporization, or, simply, the total heat. It is plain that the total heat is the sum of the heat of the liquid and the latent heat. That is, total heat heat of liquid latent heat. 5. The specific volume of the steam at the given pressure; that is, the number of cubic feet occupied by a pound of steam of the given pressure. 6. The density of the steam; that is, the weight of 1 cubic foot of the steam at the given pressure. All the above properties are different for different pressures. For example, if steam boils under atmospheric pressure, the temperature is 2120; the heat of the liquid is 180.531 B. T. U.; the latent heat, 966.069 B. T. U.; the total heat, 1,146.6 B. T. U. A pound of steam at this pressure occupies 26.37 cu. ft., and a cubic foot of the steam weighs about .037928 lb. When the pressure is 70 lb. per sq. in. above vacuum, the temperature is 302.7740; the heat of the liquid is 272.657 B. T. U.; the latent heat is 901.629 B. T. U.; the total heat is 1,174.286 B. T. U. A pound of the steam occupies 6.076 cu. ft., and a cubic foot of the steam weighs .164584 lb. These properties have been determined by direct experiment for all ordinary steam pressures. They are given in the table of the properties of saturated steam. EXPLANATION OF THE TABLE. Column 1 gives the pressures from 1 to 300 lb. These pressures are above vacuum. The steam gauges fitted on steam boilers register the pressure above the atmosphere. That is, if the steam is at atmospheric pressure, 14.7 lb. per sq. in., the gauge registers 0. Consequently, the atmospheric pressure must be added to the reading of the gauge to obtain the pressure above vacuum. In using the table, care must be taken not to use the gauge pressures without first adding 14.7 lb. per sq. in. Pressures registered above vacuum are called absolute pressures. The pressures given in column 1 are absolute. Absolute pressure per square inch = gauge pressure per square inch + 14.7. Column 2 gives the temperature of the steam when at the pressure shown in column 1. Column 3 gives the heat of the liquid. It will be noticed that the values in column 3 may be obtained approximately by subtracting 320 from the temperature in column 2. If the specific heat of water were exactly 1.00, it would, of course, take exactly 212-32 = 180 B. T. U. to raise a pound of water from 32° to 2120. But experiment shows that the specific heat of water is slightly greater than 1.00 when the temperature of the water is above 62°, and it therefore takes 180.531 B. T. U. to raise a pound of water from 32° to 2120. Column 4 gives the latent heat of vaporization, which is seen to decrease slightly as the pressure increases. Column 5 gives the total heat of vaporization. The values in column 5 may be obtained by adding together the corresponding values in columns 3 and 4. Column 6 gives the weight of a cubic foot of steam in pounds. As would be expected, the steam becomes denser as the pressure rises, and weighs more per cubic foot. Column 7 gives the number of cubic feet occupied by 1 pound of steam at the given pressure. It will be noticed that the corresponding values of columns 6 and 7 multiplied together always produce 1. Thus, for 31.3 pounds pressure, gauge, .11088 X 9.018 = 1.000, nearly. Column 8 gives the ratio of the volume of a pound of steam at the given pressure, and the volume of a pound of water at 39.2°. The values in column 8 may be obtained by dividing 62.425, the weight of a cubic foot of water at 39.20, by the numbers in column 6. EXAMPLES ON THE USE OF THE STEAM TABLE. EXAMPLE 1.-Calculate the heat required to change 5 lb. of water at 32° into steam at 92 lb. pressure above vacuum. SOLUTION. From column 5, the total heat of 1 lb. at 92 lb. pressure is 1,180.045 B. T. U. 1,180.045 X 5 = 5,900.225 B. T. U. EXAMPLE 2.-How many heat units are required to raise 8 lb. of water from 32° to 250° F.? 219.261.293 SOLUTION.-Looking in column 3, the heat of the liquid of 1 lb. at 250.2939 is 219.261 B. T. U. heat of liquid for 250°. = 1,861.228 B. T. U. 218.968 Then, for 8 lb. it is B. T. U. 218.968 X 8 EXAMPLE 3.-How many foot-pounds of work will it require to change 60 lb. of boiling water at 80 lb. pressure, absolute, into steam of the same pressure? = SOLUTION.-Looking under column 4, the latent heat of vaporization is 895.108; that is, it takes 895.108 B. T. U. to change 1 lb. of water at 80 lb. pressure into steam of the same pressure. Therefore, it takes 895.108 X 60 = 53,706.48 B. T. U. to perform the same operation on 60 lb. of water. 53,706.48778 41,783,641.44 ft.-lb. EXAMPLE 4.-Find the volume occupied by 14 lb. of steam at 30 lb., gauge pressure. SOLUTION. 30 lb., gauge pressure - 30 +14.7 = 41.7, absolute pressure. The nearest pressure in the table is 44 lb., and the volume of a pound of steam at that pressure is 9.403 cu. ft. The volume of a pound at 46 lb. pressure is 9.018 cu. ft. 9.4039.018 = .385 cu. ft., the difference in volume for a .385 difference in pressure of 2 lb. .1925 cu. ft., the difference in volume for a difference in pressure of 1 lb. .1925 × .7 = .135 cu. ft., the difference in volume for a difference in pressure of .7 lb. Therefore, 9.403.135 = 9.268 cu. ft. is the volume of 1 lb. of steam at 44.7 lb. pressure. The .135 cu. ft. |