The proportioning of the steam pipes in a heating plant is of the greatest importance, and should be carefully worked out by methods which experience has proved to be correct.

ROCOCO ORNAMENTAL THREE COLUMN PATTERN RADIATOR FOR WARMING BY HOT WATER.

There are several ways of doing this, but for ordinary conditions the following tables have given excellent results in actual practice. They have been computed from what is known as

Fig. 62.

Fig. 65

D'Arcy's formula, with suitable corrections made for actual working conditions. As the computations are somewhat complicated, only the results will be given here, with full directions for their proper use. The following table gives the flow of steam in pounds per minute for pipes of different diameters, and with varying drops in pressure between the supply and discharge ends of the pipe. These quantities are for pipes 100 feet in length; for other lengths the results must be corrected by the factors given in table XII. As the length of the pipe increases, the friction becomes greater, and the quantity of steam discharged in a given time is diminished.

Table X is computed on the assumption that the drop in pressure between the two ends of the pipe equals the initial pressure. If the drop in pressure is less than the initial pressure the actual discharge will be slightly greater than the quantities given in the table, but this difference will be small for pressures up to 5 pounds, and can be neglected as it is on the side of safety. For higher initial pressures, table XI has been prepared. This is to

Fig. 64.

 TABLE X. Diam.of Pipe. Drop in Pressure (Pounds.) 1/4 1/2 3/4 1 1 1/2 2 3 4 5 1 .44 .63 .78 .91 1.13 1.31 1.66 1.97 2.26 1 1/4 .81 1.16 1.43 1.66 2.05 2.39 3.02 3.59 4.12 1 1/2. 1.06 1.89 2.34 2.71 3.36 3.92 4.94 5.88 6.75 2 2.93 4.17 5.16 5.99 7.43 8.65 10.9 13.0 14.9 2 1/2 5.29 7.52 9.32 10.8 13.4 15.6 19.7 23.4 26.9 3 8.61 12.3 15.2 17.6 21.8 25.4 32 31.8 43.7 3 1/2 12.9 18.3 22.6 26.3 32.5 37.9 47.8 56.9 65.3 4 18.1 25.7 31.8 36.9 45.8 53.3 67.2 80.1 91.9 5 32.2 45.7 56.6 65.7 81.3 94.7 120 142 163 6 51.7 73.3 90.9 106 131 152 192 229 262 7 76.7 109 135 157 194 226 285 339 390 8 108 154 190 222 274 319 402 478 549 9 147 209 258 299 371 432 545 649 745 10 192 273 339 393 487 567 715 852 977 12 305 434 537 623 771 899 1130 1350 1550 15 535 761 942 1090 1350 1580 1990 2370 2720

be used in connection with table X as follows. First find from table X the quantity of steam which will be discharged through the given diameter of pipe with the assumed drop in pressure;

 TABLE XI. Drop inPressure in Pounds. Initial Pressure. 10 20 30 40 60 80 1/4 1.27 1.49 1.68 1.84 2.13 2.38 1/2 1.26 1.48 1.66 1.83 2.11 2.36 1 1.24 1.46 1.64 1.80 2.08 2.32 2 1.21 1.41 1.59 1.75 2.02 2.26 3 1.17 1.37 1.55 1.70 1.97 2.20 4 1.14 1.34 1.51 1.66 1.92 2.14 5 1.12 1.31 1.47 1.62 1.87 2.09

then look in table XI for the factor corresponding with the assumed drop and the higher initial pressure to be used. The quantity given in table X multiplied by this factor will give the actual capacity of the pipe under the given conditions.

Example - What weight of steam will be discharged through a 3" pipe, 100 feet long, with an initial pressure of 60 pounds and a drop of 2 pounds?

Looking in table X we find that a 3" pipe will discharge 25.4 pounds of steam per minute with a 2-pound drop. Then looking in table XI we find the factor corresponding to 60 pounds initial pressure and a drop of 2 pounds to be 2.02. Then according to the rule given, 25.4 X 2.02 = 51.3 pounds which is the capacity of a 3" pipe under the assumed conditions.

Sometimes the problem will be presented in the following way: What size of pipe will be required to deliver 80 pounds of steam a distance of 100 feet with an initial pressure of 40 pounds and a drop of 3 pounds?

. TABLE XII.

 Feet. Factor. 10 3.16 20 2.24 30 1.82 40 1.58 50 1.41 60 1.29 70 1.20 80 1.12 90 1.05 100 1.00 110 .95
 Feet. Factor. 120 .91 130 .87 140 .84 150 .81 160 .79 170 .76 180 .74 19.0 .72 200 .70 225 .66 250 .63
 Feet. Factor. 275 .60 300 .57 325 .55 350 .53 375 .51 400 .50 425 .48 450 .47 475 .46 500 .45 550 .42
 Feet Factor. 600 .40 650 .39 700 .37 750 .36 800 .35 850 .34 900 .33 950 .32 1,000 .31

We have seen that the higher the initial pressure with a given drop, the greater will be the quantity of steam discharged; therefore a smaller pipe will be required to deliver 80 pounds of steam at 40 pounds than at 3 pounds initial pressure. From table XI we find that a given pipe will discharge 1.7 times as much steam per minute with a pressure of 40 pounds, and a drop of 3 pounds, as it would with a pressure of 3 pounds, dropping to zero. From this it is evident that if we divide 80 by 1.7 and look in table X under "3 pounds drop" for the result thus obtained, the size of pipe corresponding will be that required.

80/ 1.7 = 47.

The nearest number in the table marked "3 pounds drop" is 47.8 which corresponds to a 3 1/2 pipe and is the size required.

These conditions will seldom be met with in low-pressure heating, but apply more particularly to combination power and heating plants, and will be taken up more fully under that head. For lengths of pipe other than 100 feet, multiply the quantities given in table X by the factors found in table XII.

Example - What weight of steam will be discharged per minute through a 3 1/2 pipe, 450 feet long with a pressure of 5 pounds and a drop of 1/2 pound ?

Table X, which may be used for all pressures below 10 pounds, gives for a 3 1/2" pipe, 100 feet long, a capacity of 18.3 pounds for the above conditions. Looking in table XII, we find the correction factor for 450 feet to be .47. Then 18.3 X .47 = 8.6 pounds, the quantity of steam which will be discharged if the pipe is 450 feet long.

Examples involving the use of tables X, XI and XII in combination are quite common in practice. The following shows the method of calculation:

What size of pipe will be required to deliver 90 pounds of steam per minute a distance of 800 feet, with an initial pressure of 80 pounds and a drop of 5 pounds? Table XII gives the factor for 800 feet as .35 and table XI that for 80 pounds pressure and 5 pounds drop as 2.09. Then _______90______ = 123; which

.35 X 2.09 is the equivalent quantity, we must look for in table X. We find that a 4" pipe will discharge 91.9 pounds, and a 5" pipe 163 pounds. A 4 1/2" pipe is not commonly carried in stock and we should probably use a 5" in this case, unless it was decided to use a 4" and allow a slightly greater drop in pressure. In ordinary heating work with pressures varying from 2 to 5 pounds, a drop of 1/4 pound in 100 feet has been found to give satisfactory results.

In computing the pipe sizes for a heating system by the above methods it would be a long process to work out the size of each branch separately so the following table has been prepared for ready use in low-pressure work.