A Table of Angles for Windmill Sails.

The radius is supposed to be divided into six equal parts, and from the centre is called 1, the extremity being denoted by 6.

The first column contains the angles according to Smeaton, but experience has taught us that the angles in the second column are preferable.

OF PUMPS AND PUMPING ENGINES.

Pumps are chiefly designated by the names of lifting and force pumps: lifting pumps are applied to wells, &c., where the height of the bucket, from the surface of the water, must not exceed 33 feet; this being nearly equal to the pressure of the atmosphere, or the height to which water would be forced up into a vacuum by the pressure of the atmosphere. Force pumps are applicable on all other occasions, as raising water to any required height, supplying boilers against the force of the steam, hydrostatic presses, &c.

NOTE. Hot liquor pumps, or pumps to be employed in raising any fluid where steam is generated, requires to be placed in the fluid, or as low as the bottom of it, on account of the steam filling the pipes, and acting as a counterpoise to the atmosphere.

The power required to raise water to any height is as the weight and velocity of the water, with an addition of about of the whole power for friction; hence the rule.-Multiply the perpendicular height of the water, in feet, by the velocity, also in feet, and by the square of the pump's diameter in inches, and again by .341; (this being the weight of a column of water 1 inch diameter, and 12 inches high, in libs. avoirdupois ;) divide the product by 33,000 and of the quotient, added to the whole quotient, will be the number of horses' power required.

EXAMPLE.-Required the power necessary to overcome the resistance and friction of a column of water

4 inches diameter, 60 feet high, and flowing with a velocity of 130 feet per minute.

60 x 130 x 42 x .341

33000

horses' power nearly.

1.3.26+1.3 = 1.56

5

The diameter of a Pump and velocity of the Water given to find the quantity discharged in gallons, or cubic feet, in any given time.

RULE.-Multiply the velocity of the water, in feet, per minute, by the square of the pump's diameter in inches, and by .034 for imperial gallons; or, .0005454 for cubic feet, and the product will be the number of gallons, or cubic feet, discharged in the given time nearly.

EXAMPLE. What is the number of imperial gallons of water discharged per hour by a pump 4 inches diameter, the water flowing at the rate of 130 feet per minute?

130 × 60 7800 feet per hour.

And, 7800 × 42 × .034 = 4243.2 gallons nearly.

The length of stroke, and number of strokes given, to find the diameter of a pump, and number of horses' power that will discharge a given quantity of water in a given time.

RULE 1.-Multiply the number of imperial gallons required, in the given time, by 353, or the number of cubic feet by 2201, and divide the product by the velocity of the water, in inches, and the square root of the quotient will be the pump's diameter, in inches, nearly.

2.-Multiply the number of gallons, per minute, by 10, or the number of cubic feet by 62.5, and by the perpendicular height of the water in feet, divide the product by 33,000, then will of the quotient, added to the whole quotient, be the number of horses' power required.

EXAMPLE.-Required the diameter of a pump, and number of horses' power, capable of filling a cistern 20 feet long, 12 feet wide, and 6 feet deep, in 45 minutes, whose perpendicular height is 53 feet; the pump to have an effective stroke of 26 inches, and make 30 strokes per minute.

20 × 12 × 6.5 = 1560 cubic feet, and

To find the time a Cistern will take in filling, when a known quantity of water is going in, and a known portion of that water is going out, in a given time.

RULE. Divide the content of the cistern, in gallons, by the difference of the quantity going in, and the quantity going out, and the quotient is the time in hours and parts that the cistern will take in filling.

EXAMPLE.-If 30 gallons per hour run in and 221⁄2 gallons per hour run out of a cistern capable of containing 200 gallons, in what time will the cistern be filled?

30 22.5 7.5. and 2007.5 26.666, or 26 hours and 40 minutes nearly.

To find the number of Imperial Gallons contained in a Yard of Pipe of any given diameter.

RULE.-Square the diameter of the pipe in inches, cut off one integer for a decimal; again, multiply the square by 2, the product is hundredths, &c. of a gallon, which add to the former product, and the sum will be the content of the pipe in imperial gallons nearly.

EXAMPLE 1.-Required the number of imperial gallons contained in each yard of a 64 inch pipe.

6.25239.0625 and 390625 × 2 = 78125,

= 3.984375 gallons nearly.

EXAMPLE 2.-Required the content of a yard of 4 inch pipe in imperial gallons.

4216, and 16 x 232, then 1.6

+

32

1.632 gallons nearly.

To find the Weight that a given power can raise by one of Bramah's Pumps or Hydrostatic Presses.

RULE.-Multiply the square of the diameter of the ram in inches by the power applied in lbs., and by the effective leverage of the pump handle; divide the product by the square of the pump's diameter also in inches, and the quotient is the weight that the power is equal to.

EXAMPLE.-What weight will a power of 50 lbs. raise by means of an hydrostatic press, whose ram is 7 inches diameter, pump, and the effective leverage of the pump handle being as 6 to 1?

72 × 50 × 6

.8752

= 19200 lbs., or 8 tons, 11 cwt.

In the following rules for pumping engines the boiler is supposed to be loaded with about 24 lbs. per square inch, and the barometer attached to the condenser indicating 26 inches on an average, or 13 lbs. = 15 lbs., from which deduct for friction, leaves a pressure of 10 lbs. nearly, upon each square inch of the piston.