In the above tables we find the relation between the diameter of pipe, cubic feet per hour, fall of pressure per
mile, and the gauge pressure of the flowing gas in the pipe at the lower end of that mile. But it is to be observed
that the fall of pressure per mile is not the same for each mile of a long pipe-line of uniform size carrying gas in
continuity of flow. Thus, if the drop in pressure in one mile, in such a pipe, is one pound, it will not be fifty pounds
in the same pipe line fifty miles long. To determine the fall of pressure in a long line of many miles for a
given diameter and flow, is the object of Table X, as more fully indicated below.

To explain Table X more fully, it should be noted that this, as well as Tables VII and VIII, are only applicable to a pipe-line of uniform size for the length considered, and for the suppositions of uniformity or continuity of flow.

In working out a long pipe-line of several sizes of pipe, each size must be taken by itself, and the several results of fall of pressure for each added.

In the first column of the Tables VII and X the same figures are intentionally given for the purpose of facilitating passing from one table to the other, thus connecting more directly the diameter of pipe and cubic feet of discharge, with the length and total fall of pressure.

To show the application of the tables the following examples are

given:

1st. Example: Required the cubic feet of gas discharged by a pipe-line twenty miles long and six inches in diameter throughout, where the fall of gauge-pressure is from 200 pounds per square-inch to 40 pounds per square-inch; the forty pounds being, of course, marked by the gauge at the down stream end of the pipe-line, and the temperature of flowing gas being taken at 50° Fahr.

160 54.6

=

2.936, which figure is to be looked

The total fall of pressure is 20040 160 pounds; and this, divided by 14.6 40 54.6, is for in Table X in the column under the stated length twenty miles. It lies between the figures 2.786 and 4.012, and about one-eighth the way from the first toward the second. Hence, the figure in the first column will be one-eighth the way from .291 toward .485. Passing to Table VII we find the same figures in the first column. Hence, the required cubic feet is at the one-eighth point between the two lower values of the table, and under the stated diameter six inches. Hence, the cubic feet discharged by the pipe per hour is 35,257.

But it is to be observed that this volume is at the temperature of 50° Fahr., and at the gauge-pressure of forty pounds per square-inch, as the volume given by Table VII is at the pressure of the flowing gas at the down stream end of pipe, forty pounds by gauge in this case.

To change this to the volume at atmospheric pressure, or "storage pressure." Tables IX gives the multiplier, 3.740, which, multiplied by the above volume 35,257, gives 131,861 cubic feet per hour, as the volume of gas that would be discharged by the pipe-line into a gasholder at atmospheric pressure and at a temperature of 50° Fahr.

2d. Example: Required the fall of pressure in a pipe-line eighty miles long and eighteen inches in diameter that will deliver 6,564,285

cubic feet per hour of gas at atmospheric pressure, and 50° temperature, the gauge-pressure at the down stream end of pipe being taken at 200 pounds per square-inch.

To change this atmospheric pressure-volume to that of 200 pounds by gauge, Table IX gives the multiplier 14.7, which is now to be used as a divisor. Dividing and we obtain 446,550 cubic feet for the volume per hour at the 200-pound gauge-pressure.

Table VII, in the column under eighteen inches diameter, gives the figure 446,550 in the third line from the bottom, and opposite the figure .272 in the first column.

Then, looking in Table X, opposite the first column figure.272, and in the column for the stated length of line, eighty miles, we find the value 6.094, which is to be multiplied by the 200 pounds gauge pressure plus 14.6 pounds = 214.6 pounds or 214.6 × 6.094 = 1308. pounds, as the fall of pressure for the eighty miles of eighteen-inch pipe-line. Adding the 200 pounds and we obtain the 1,508 pounds gauge pressure for the gas at entrance into the eighty mile-line.

3d example: Pipe-line in two sizes, viz., forty miles of four-inch pipe discharging into twenty miles of six-inch pipe.

Gauge pressure at delivery end of six-inch pipe, thirty pounds; cubic feet discharged 40,000 per hour at atmospheric pressure and temperature of 50° Fahr.

Temperature of flowing gas in the pipes 65°, and specific gravity of gas, 0.8.

Table IX, multiplier = 3.055. Dividing 40,000, gives 13,098 cubic feet at thirty pounds pressure and 50° temperature.

Table VIII multiplier for 65° temperature of flow and 0.8 specific gravity is .115, and the correction 13,098 × .115 = 1,470, which we now add, because working backwards relative to application of Table VIII, giving 14,568 cubic feet to be discharged at the lower end of the six-inch pipe at the gauge pressure thirty pounds and temperature 65° Fahr.

Table VII, under six-inch pipe, the figure 14,568 comes opposite the value.031 in first column.

Table X, under twenty miles, we find just above top figure opposite .031 the value .176. Fall of pressure in the twenty mile six-inch pipe = (30 + 14.6) × .176 = 7.85 pounds.

Gauge pressure at upper end of six-inch pipe or lower end of fourinch pipe = 37.85 pounds, and the cubic feet, by aid of Table IX,

=

3.055

14,568

= 11,760.

Table VII, under four-inch pipe, the figure 11,760 comes opposite the value.222 in the first column.

Table X, opposite .222 and under forty miles, we get 3,560 to multiply into 37.85 to give the fall of pressure in the four-inch pipe, which = 135. pounds.

Hence, gauge pressure at entrance into the four-inch pipe = 135 + 37.85 = 172.85 pounds. Total fall of pressure in both pipes = 135. +7.85 = 142.85 pounds.

Acknowledgments are due to B. G. Lamme, a student in the Department of Mechanical Engineering in the State University, for efficient aid in working out most of the tables of this chapter.

CHAPTER X.

THE PITTSBURGH COAL SEAM IN JEFFERSON, BELMONT AND GUERNSEY COUNTIES.

BY PROFESSOR C. NEWTON BROWN, OHIO STATE UNIVERSITY.

The Pittsburgh coal seam (No. 8 coal of Newberry's scale) is the most important found in the Upper Coal Measures of the state. It lies from 190 feet to 210 feet above the Ames or Crinoidal limestone, and from 80 feet to 100 feet below the Meigs Creek coal, and forms the base of the Upper Coal Measures.

There are two areas in the state where it is mined in a large way, both for local use and shipment by railway and river. The largest and most important of these cover parts of Jefferson, Harrison, Belmont and Guernsey counties. The other covers parts of Athens, Meigs and Gallia counties, where it is known as the Pomeroy coal. Only that portion in Jefferson, Belmont and Guernsey counties is touched in this chapter.

Through this area the coal is remarkable for its regularity in quality, thickness, structure and freedom from wants, clay-veins and horse-backs.

There are usually two partings that divide the seam into three benches. The lower parting-from twelve to fourteen inches from the bottom-is a thin, black slate, sometimes carrying balls of pyrites in it.

The other parting is at, or a little above, the center. It is usually made up of two bands of clay with two to four inches of coal between them, the entire parting being from five to six inches thick. The bearing-in is frequently made at this place.

There is a coal from one to three feet thick, of poor quality, above the main coal and separated from it by about twelve inches of clay. The clay comes down as soon as the main coal is mined out and the thin coal above left for a roof. Above the roof coal is a bed of clay shale several feet thick, and under the main coal is a thin bed of clay underlaid by a bed of non-fossiliferous limestone.

This coal-field is crossed by five lines of railroad and bounded on