Burst Steam Pipes

Steam-pipes which have cracks in them from having burst, maybe repaired by heating and then soldering them.

Condensation In Subterranean Steam Pipes

To prevent this, inclose the pipe in another larger pipe, and fill the space between the two with plaster-of-Paris or charcoal. The outside pipe should be water-tight.

Isolating Material For Steam Pipes

Take 132 lbs. limestone, 385 lbs. coal, 275 lbs. clay, and 330 lbs. sifted coal-ashes. This is finely pulverized, and mixed with 660 lbs. water, 11 lbs. sulphuric acid at 50° B., and about 1G0 lbs. calves' hair or bog-bristles. The compound is applied to the pipes in coats of 0.4 inch thickness, repeated until a thickness of an inch and a half is obtained, when a light covering of oil is given.

How To Prevent Cracking Steam Pipes From Freezing

Steam-pipes apt to fill with condensed water and burst from freezing should have small holes with plugs to them, the plugs to be taken out at night.

Safe Pressure Stayed Surface, In Lbs. Per Square Inch, For

Divide the square of the thickness of the plate in inches, by the square of the distance between stays, in inches, and multiply the quotient by L6, 875 for a copper plate, by 27, 000 for a wrought iron plate, and by 45, 000 for a steel plate

Example.- What is the safe pressure for a plate of wrought iron, 1/4 of an inch thick, secured by stays 6 inches from centre to centre?

The quotient arising from dividing 0.0625 (the square of 1/4) by 36, is 0.00174, Multiplying 0.00174 by 27, 000, the product is the required pressure, about 47 lbs. per square inch. B.

Thickness Of Stayed Surface, In Inches

Multiply the square root of the pressure, in Lbs. per square inch, by the dis-tance between centres of stays in inches, and multiply this product by 0.007698 for a copper plate, by 0.0060858 for a wrought iron plate, by 0.0047141 for a steel plate.

For a copper fire-box, in which the stays are 10 inches apart from centre to centre, and the pressure of steam is 60 lbs.: The thickness of plate is the product of 7.746 (the square root of 60), 10 and 0.007698; which is equal to 0.596, or about19/32 of an Inch. B.

Proper Diameter For A Stay, In Inches

Multiply the distance between stays, in inches, by the square root of the pressure, in pounds per square inch, and multiply this product by 0.0206 for a copper stay, by 0.01784 for a wrought iron stay.

Example.- What is the proper diameter for wrought-iron stays, 6 inches between centres, the pressure of steam being 75 pounds per square inch?

This is the product of 6, 8.66 (the square root of 75), and 0.01784; which is equal to 0.92697, or about 15/16 of an inch. B.

How To Find The Angle Of Bevel Of Valve

This is the angle of inclination f e a, ore d b, to a vertical line. Make the following measurements: (1) greatest diameter, g h, of valve, in inches; (2) least diameter, a b, of valve, in inches; (3) depth, a k, of valve, in inches. Divide the difference of the greatest and least diameters by the depth of seat. Find the angle whose tangent is nearest this quotient, in the accompanying table of tangents.