i. e., multiply the pressure, in inches, of water, by the diameter of the pipe, also in inches; divide the product by the specific gravity of the gas, multiplied by the length of the pipe in yards; extract the square root of the quotient, which root, multiplied by the constant quantity, 1,350, and the square of the diameter of the pipe in inches, gives the number of cubic feet discharged in one hour. *

Example: It is required to find the number of cubic feet of gas, of the specific gravity of 57, which will be discharged in one hour, from a pipe six inches in diameter and three miles (5,280 yards) in length, under a pressure of ten pounds per square inch. This pressure equals, say, 276 inches, or the pressure given by a column of water 276 inches high.

(1

=

50 d2

1,350

.57

1.656

× 5,280)
)

hd

81

=

=

✓.5502; the square root is .74.

1,350 × 36 × .74 = 35,964 cubic feet per hour.

To make this as clear as possible to the non-technical reader, the

example will be still further explained.

Thus: h (the head or pressure) = 276 inches; this multiplied by 6 (the diameter, in inches, of the pipe) = 1,656; this divided by .57 (the specific gravity of the gas), multiplied by 5,250 (the number of yards in three miles), = .5502. The square root of the last sum equals .74 Χ, and this times 1,350 (constant quantity), and this again × 36 (the square of the diameter of the pipe) = 35,964, or the cubic feet of gas discharged per hour.

Second example: It is required to ascertain the quantity of gas, of the specific gravity of .6, which will be discharged through a pipe 8 inches in diameter and 10 miles in length, under the initial pressure of 75 pounds per square inch. Ten miles equals 17,600 yards, and 75 pounds equals 2,070 inches of water pressure, or head.

Thus: (hd) = 8 × 2,070 = 16,560.

= 1,350 × 64 × 1.252 = 108,172 cubic feet discharged Third example: Required to ascertain the quantity of gas, of the specific gravity of 5, that will be discharged through a pipe 10 inches in diameter and 30 miles long, under an initial pressure of 200 pounds per square-inch. Thirty miles equals 52,800 yards, and 200 pounds pressure equals 5,520 inches of water head.

Thus: (hd) = 10 × 5,520 = 55,200.

✓ 2.0909, the square root being 1.44.

1,350 × 100 × 1.44=194,400 cubic feet, or the

quantity that will be discharged in one hour.

Formula for ascertaining pressure required for the discharge of a given quantity of gas, through a given length and size of pipe:

Fourth example: Required to ascertain the pressure necessary to pass 100,000 cubic feet per hour, through eight miles (14,080) yards of 6-inch pipe; specific gravity of gas, .5.

Thus: 100,000 × 100,000 = 10,000,000,000 × .5 = 5,000,000,000 × 14,080 = 70,400,000,000,000 ÷ (1,350 × 1,350 × 7,776) = 4,968 (nearly), the pressure required (in inches of water), which equals 180 pounds per square inch.

This may be stated thus:

10.000.000 000 × 5 × 14.080
1,822,500 × 7,776

= 4,968 inches, or 180 pounds, in which

the ten billions is the square of the quantity, the decimal, five, the specific gravity, the fourteen thousand and eighty the length of pipe in yards, and the seven thousand seven hundred and seventy-six is the fifth power of the diameter of the pipe.

Fifth example: Required to ascertain the pressure necessary to discharge 20,000 feet of gas per hour; specific gravity, .58, through 3,000 feet (1,000 yards) of 3-inch pipe. Formula same as for preceding example.

Thus: 20,000 × 20,000 = 400,000,000 × .58 = 232,000,000 × 1,000= 232,000,000,000 ÷ (1,350 × 1,350 × 243) = 524 inches, or 18.9 pounds pressure.

To ascertain the diameter of a pipe, which will discharge a given quantity of gas, through a given length of pipe, under known pressure, we have the formula:

Sixth example: Required the diameter of a pipe which will discharge 50,000 feet of gas per hour, of the specific gravity of .7, through ten miles (17,600 yards) of pipe, with an initial pressure of eighty pounds (2,138 inches).

Thus:

2.500 000.000 ×.7 × 17.600

1,822,500 × 2,208

5

= 17,654 = 6-.

50,000 × 50,000 = 2,500,000,000 × .7 = 1 750,000,000 × 17,600 = 30,800,000,000,000 ÷ (1,350 × 1,350 × 2,136) = 7,654, which equals the fifth power of six inches, the required diameter of the pipe.

In gas engineers' hand-books are published tables, showing the discharge of gas through pipes ranging in size from a half-inch to thirtysix inches in diameter, in lengths of from ten yards to ten thousand yards, and under pressure from one-tenth of an inch to four inches. Believing that smaller pipe than three inches should not be used for mains, and that natural gas will seldom be used at a pressure below one inch, the discharge of gas through smaller sizes than three-inch, with less pressure than one inch, will not be given. The discharge through larger sizes of various lengths, and at different pressures up to 2.5 inches, are given.

These tables are calculated on the basis of .400 for the specific gravity of the gas, which is less than the specific gravity of any natural gas of which we have an analysis. The quantity of gas of any other specific gravity, which will be discharged in one hour, may be ascertained by multiplying the quantity indicated in the table by .6325 (the square root of .400), and dividing by square root of the specific gravity of other gas.

The quantity that will be discharged at any other pressure may be ascertained by multiplying the quantity indicated in the table by the square root of the new pressure, and dividing by the square root of the pressure given in the tables. These low pressures and short lengths will be applicable to the distribution of natural gas where municipal or other authorities prohibit high-pressure service in thickly populated portions of towns and cities. Even these low pressures are higher than should be used when consuming gas for illumination. Natural gas, as usually found, being of low illuminating power, should be consumed at a low pressure, and a pressure of 2.5 inches is probably as great as should be used when gas is consumed for heating purposes.

Quantity discharged in cubic feet, per hour, with 1.5 inches pressure

17,280

24,365

27,302