Document ppoZXN7jrbp00xN964GoMVLg6

American Society of Heating and Ventilating Engineers Guide, 1932 Table 1. Flow of Steam in Pipes P = loss in pressure in pounds. d = inside diameter of pipe in inches. L = length of pipe in feet. D = weight of 1 cu. ft. of steam. W = pounds of steam per hour. P = 0.0000000367 ( 1 + ^ ) Col. 1 Pressure Loss 5220IN Ounces Pips Size Actual Nominal Internal Diameter Internax Area or Pipe Sq. Inches Col. 2 Steam Press. bt Gage Col. 3 Length or Pipe in Feet Col. 4 /loo y-r 0.25 65.28 i 1.049 0.864 0.536 -1.0" 0.187 20 2.240 0.50 92.28 IK 1.380 1.496 1.178 -0.5" 0.190 40 1.580 1.00 . 130.5 IK 1,610 2.036 1.828 0.0 0.193 60 1.290 2 184.6 2 2.067 3.356 3.710 0.3 0.195 80 1.120 ,3 226.0 VK 2.469 4.788 6.109 1.3 0.201 100 1.000 ' 4 261.0 3 3.068 7.393 11.183 2.3 0.207 120 0.912 5 291.8 3K 3.548 9.887 16.705 5.3 0.223 140 0.841 6 319.7 4 4.026 12.730 23.631 10.3 0.248 160 0.793 7 345.3 4K 4.506 15.947 32.134 15.3 0.270 180 0.741 8 369,1 5 5.047 20.006 43.719 20.3 0.290 200 0.710 10 412.7 6 6.065 28.886 71.762 30.3 0.326 250 0.632 12 452.0 7 7.023 38.743 106.278 40.3 0.358 300 0.578 14 488.3 8 7.981 50.027 149.382 50.3 0.388 350 0.538 16 522.0 9 8.941 62.786 201.833 60.3 0.415 400 ' 0.500 20 583.6 10 10.020 78.854 272.592 75.3 0.452 450 0.477 24 639.3 12 12.000 113.098 437.503 100.3 0.507 500 0.447 28 690.5 14 13.250 137.880 566.693 125.3 0.557 600 0.407 32 738.2 16 15.250 182.655 816.872 150.3 0.603 700 0.378 40 825.4 Column 1 X 2 X 3 X 4 lb, of steam 175.3 0.645 800 0.354 per hour that will flow through a straight 48 904.1 pipe for a given condition. 200.3 0.685 900 0.333 80 1167.2 -- 1.3 lb. press. -- 100 ft. equivalent length: 160 1650.7 130.5 X 3.710 X 0.201 Xl 97.2 lb. per hour. 97.2 X 4b = 388.8 sq. ft. equivalent radiation. 1000 1200 0.316 0.289 320 480 2334.5 2859.1 Table 1 does not allow for entrained water in.low-pressure steam, condensation in covered pipe and roughness in commertial pipe as found in practice. 1500 2000 0.258 0.224 Pounds per square inch gage 2.04 in. Vacuum, Mercury Column. . . bThe factor 4 is the approximate equivalent in square feet of steam radiation of 1 lb. of steam per hour. 132 ^ Chapter 9--Steam Heating Systems and Pipe Sizes . so that the steam flowing in any pipe may be calculated by multiplying together the proper factors in each column as shown in the example at the bottom of the table. Table 2 is a basic table giving the theoretical capacities of pipe in square feet of direct cast-iron radiation (Based on x/i lb steam per hour per square foot) and the resulting velocity in feet per second for various pressure drops in ounces per 100 ft length of pipe with an initial steam pressure of 1 lb gage. This table was compiled from the values given in Table 1. In using Tables 1 or 2 the total pressure drop figured should never equal or exceed the initial pressure. Example 1. In a 3-iri.; pipe, what pressure drop is required in ounces per 100 ft of length of pipe to supply steam to 2014 sq ft of equivalent heating surface? The initial steam pressure is 1 lb gage. Solution. In Table 2, column for 3-in. pipe, find that steam for 2014 sq ft of equivalent heating surface will be supplied at 1 lb initial pressure and a pressure drop of 3 oz per 100 ft length of run. Table 3 is to be used with Table 2 for calculating the capacity of a steam pipe, for other initial pressures and lengths when the capacity is known for 1-lb pressure and 100-ft length. To determine the capacity of any pipe for an initial pressure other than 1 lb, multiply the capacity given in Table 2 by the pressure factor in Column 2, Table 3, opposite the re quired pressure indicated in Column 1. Example 3. What is the capacity of a 100-ft, 4-in. pipe with 2-lb initial pressure and pressure drop of 1 oz? Solution. From Table 2, find 2457, the capacity of the 4-in. pipe with 1-lb initial pressure and 1-oz pressure drop. Multiplying 2457 by 1.03, the constant for 2-lb initial pressure (Column 2 of Table 3), gives 2531 as the capacity of the 4-in. pipe with 2-lb initial pressure and a pressure drop of 1 oz per 100-ft length. To determine the capacity for any length other than 100 ft, multiply the capacity given in Table 2 by the length factor in Column B, Table 3, opposite the required length in Column A. Example 3. What is the capacity of a 140-ft, 4-in. pipe with an initial pressure of 1 lb and a pressure drop of 2 oz in the 140 ft? Solution. From Table 2 it is found that the capacity of a 100-ft, 4-in. pipe with 1-lb initial pressure and 2-oz pressure drop, is 3475 sq ft. Multiplying this value by 0.841 the constant for a 140-ft length as given in Table 3 gives 2922 which is the capacity for the given conditions. Example i. What is the capacity of a 140-ft, 4-in. pipe with 2-lb initial pressure and a pressure drop of 2 oz? Solution. From Table 2 find. 3475, the capacity of the 4-in. pipe with 1-lb initial pressure and 2-oz pressure drop. Multiplying 3475 by 1.03, the constant for 2-lb initial pressure (see Column 2, Table 3), and this by 0.841, the constant for 140-ft length (see Column B, Table 3), gives 3010 as the capacity of the 4-in. pipe with 2-lb initial pressure and a pressure drop of 2 oz in the 140-ft length. Should lengths other than those given in Column A, Table 3, and under length of pipes in feet in Table 1 be desired the constant may be obtained from the formula in Column 4, Table 1, and used the same as the con stants from Table 3. Example 5. What would be the constant for 2500 ft of pipe to be used either in Tables 1 or 3? Solution. The constant to be used = = 0.2. y 2500 133