1st. To find the pressure per square inch at which a given valve will open. Measure the following distances horizontally from the fulcrum to (1) the centre of the valve-stem; (2) the centre of the weight; (3) the centre of gravity of the lever, or the point on which it will balance if placed upon a knife-edge. Measure the diameter of the valve, and determine its area, either from a table or by multiplying the square of the diameter by 0.7854. Find the weight of (1) the valve; (2) the lever ; (3) the ball. .Multiply (i) the weight of the ball by its horizontal distance from the fulcrum ; the Weight of the lever by its horizontal distance from the fulcrum ; (3) the weight of the valve by its horizontal distance from the fulcrum ;(4) the area of the valve by its horizontal distance from the fulcrum. Add together the first three products and divide the sum by the fourth product.

Example. A given safety-valve has a weight of 50 lbs. 24 inches from the fulcrum, the lever weighs 6 lbs., and its centre of gravity is 15 inches from the- fulcrum; the weight of the valve is 2 Lbs., and its centre is 4 inches from the fulcrum. The diameter of the valve is 2 inches. At what pressure will the valve begin to rise?

(l) 50 times 24 is 1200; (2) 6 times15 is 90; (3) 2 times 4 is 8; (4) 4 times 3.1416 is 12.5664. The sum of (1), (2), and (3) is 1298, which divided by 12.5664 (the fourth product) is 103.03, the pressure in lbs. per square inch at which the valve will open.

2d. To find where to place the weight on a safety-valve so that it shall open at a given pressure of steam. Multiply (1) the weight of the lever by the horizontal distance of its centre of gravity from the fulcrum; (2) the weight of the valve by its horizontal distance from the fulcrum; (3) the area of the valve by the pressure of steam in lbs. per square inch, and by the horizontal distance of the valve from the fulcrum. Add together the first two products, subtract their sum from the third product, and divide the difference by the weight of the ball.

Example.-The ball of a safety-valve weighs 100 lbs., the lever weighs 10 lbs., the valve weighs 2 lbs., and has a diameter of 3 inches. The distance of the centre of gravity of the lever from the fulcrum is 25 inches, and the distance of the centre of the valve from the fulcrum is 5 inches. How far from the fulcrum must the valve be placed, in order that the lever may open at a pressure of 100 lbs.?

Area of valve, 7.07 square inches.

(1) 10 times 25 is 250; (2) 2 times 5 is 10; (3) the product of 7.07, 100, and 5, is 3535. Adding together products (1) and (2), we have as their sum 260; subtracting this from 3535, the third product, we have 3275. Dividing this difference by 100, the weight of the ball, we have 32.75, or 32f inches as the distance from fulcrum to ball.

To find what diameter a safety-valve must have, the other parts being known to open at a given steam-pressure. Multiply (1) the weight of the ball by its horizontal distance from the fulcrum; (2) the weight of the lever by the horizontal distance of its centre of gravity from the fulcrum; (3) the weight of the valve by the horizontal distance of its centre from the fulcrum; (4) the pressure of steam in pounds per square inch by the horizontal distance of the valve from the fulcrum, and by the number 0.7854. Add together the first three products, divide their sum by the fourth product, and take the square root of the quotient.

Example.-Weight of ball, 60 lbs.; lever, 7 lbs.; valve, 3 lbs. Distances from fulcrum: ball, 30 inches; centre of gravity of lever, 16 inches; centre of valve, 3 inches. Pressure of steam, 70 lbs. per square inch. What should be the diameter of the valve?

(1) The product of 60 and 30 is 1800; (2) the product of 7 and 16 is 112; (3) the product of 3 and 3 is 9; (4) the product of 70, 3, and 0.7854 is 161.934. The sum of the first three products, 1800, 112, and 9, is 1921. Dividing this sum by 164.934 (the fourth product), we have 11.647 inches. The square root of this number is 3.41+ inches, which by the rule is the required diameter of the valve. B.

Table Of Tangents From 20° To 50°

Example.-The greatest and least diameter of a valve are 4 6-10 and 4 inches, respectively, and the depth is 1/2 inch. What is the bevel?

From the table, it appears that the angle corresponding to this is nearly 31 . B.

Valve, To find the area of opening, in square inches, of a, due to a given lift-{a) When the lift of the valve is equal or to less than the depth of seat: Find the product of (1) the diameter of the valve, in inches; (2) the lift, in inches; (3) the sine of the angle of bevel of the valve, and (4) 3.1416. Add this to the product of (1) the square of the lift, in inches; (2) the square of the sine of angle of bevel of the valve; (3) the cosine of the angle of bevel of the valve, and (4) 3.1416.

(b) When the lift of the valve is greater than the depth of seat: Find the product of (1) the diameter of the valve, in inches; (2) the depth of seat, in inches; (3) the sine of the angle of bevel of the valve, and (4) 3.1416. Find the product of (1) the square of the depth of seat, in inches; (2) the square of the sine of the angle of bevel of valve; (3) the cosine of the angle of bevel of valve, and (4) 3.1416. Find the product of (1) the diameter of the valve, in inches; (2) the difference between the lift and the depth of seat, in inches, and (3) 3.1416. Take the sum of these three products.

Example.-The diameter of a valve is 4 inches, the bevel is 35°, and the depth of seat 1/4 of an inch. What is the area of opening for a lift of 3/8 of an inch?

The product of 4, 0.25, 0.574 (the sine of 35°), and 3 1416 is 1.8.

The product of the square of 0.25, the square of 0.574, 0.819, (the cosine of 35°), and 3.1416 is 1.85.

The product of 4, 0.125 (the difference between the lift and depth of seat), and 3.1416 is 1.57.

The sum of 1.8, 1.85, and 1.57 is 3.42 square inches, the area of opening required. B.