Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Rule.-Multiply the radius in the table, by the pitch given; and the product will be the radius of the wheel required.

Or,- Divide the radius of the wheel, by the radius in the table, and the quotient will be the pitch of the wheel required.

Example 1.-Required the radius of a wheel to contain 64 teeth, of 3 inch pitch.

10.19 x 3 = 30.57 inches nearly, ExamPLE 2.-What is the pitch of a wheel to contain 80 teeth, when the radius is 25.47 inches.

25.47 = 12.735 = 2 inch pitch. Or, set off upon a straight line seven times the pitch given, divide that, or another exactly the same length, into eleven equal parts; call each of those divisions four, or each of those divisions will be equal to four teeth upon the radius.

EXAMPLE.—Were it required to find the diameter of a wheel to contain 21 teeth, the construction would be as follows:

il 21 31 41 51 61 il 21 31 41 51 61 71 81 91 10 4...4...8... 12...16...

Thus, 5 divisions and of another equal the radius of the wheel.

And multiply the pitch by .8, the product will be the length of the teeth. --Also, multiply the pitch by .46, and the product will be the thickness.

7

11

20.>

K

ON THE MAXIMUM VELOCITY AND

POWER OF WATER WHEELS.

Since publishing the first edition of this work, I have endeavoured, as far as possible, to acquire the most improved practical principles of water wheels as a moving power; and

[blocks in formation]

The term undershot is applied to a wheel when the water strikes at, or below, the centre. And the greatest effect is produced when the periphery of the wheel moves with a velocity of .57 that of the water; - hence, to find the velocity of the water, multiply the square root of the perpendicular height of the fall in feet by 8, and the product is the volocity in feet per second.

EXAMPLE.—Required the maximum velocity of an undershot wheel, when propelled by a fall of water 6 feet in height.

V 6 = 2.45 x 8 = 19.6 feet velocity of water, And 19.6 x .57 = 11.17 feet per second for the wheel.

2. Of Breast and Overshot Wheels.

Wheels that have the water applied between the centre and the vertex are styled breast wheels, and overshot when the water is brought over the wheel and laid on the opposite side; however, in either case, the maximum velocity is įrds that of the water; hence, to find the head of water proper for a wheel at any velocity, say,

As the square of 16.083, or 258.67, is to 4, so is the square of the velocity of the wheel in feet per second to the head* of water required.

EXAMPLE.--Required the head of water necessary for a wheel of 24 feet diameter, moving with a velocity of 5 feet

per

second. 5 x 3

= 7.5 feet velocity of the water,

2 And 258.67 : 4 :: 7.52 : .87 feet, head of water

required. But one-tenth of a foot of head must be added for every foot of increase in the diameter of the wheel, from 15 to 20 feet, and .05 more for every foot of increase from 20 to 30 feet, commencing with five. tenths for a 15 feet wheel.

This additional head is intended to compensate for the friction of water in the aperture of the sluice to keep the velocity as 3 to 2 of the wheel; thus, in place of .87 feet head for a 24 feet wheel, it will be .87 + 1.2 = 2.07 feet head of water.

If the water flow from under the sluice, multiply the square root of the depth in feet by 5.4, and by the area of the orifice also in feet, and the product is the quantity discharged in cubic feet per second.

Again, if the water flows over the sluice, multiply the square root of the depth in feet by 5.4; and į of the product multiplied by the length and depth, also in feet, gives the number of eubic feet discharged per second nearly.

EXAMPLE 1.-Required the number of cubic feet per second that will issue from the orifice of a sluice 5 feet long, 9 inches wide, and 4 feet from the surface of the water.

* By head is understood the distance between the aperture of the sluice and where the water strikes upon the wheel.

V4 = 2 x 5.4 = 10.8 feet velocity, And 5 x .75 X 10.8 = 40.5 cubic feet per second.

EXAMPLE 2.-What quantity of water per second will be expended over a wear, dam, or sluice, whose length is 10 feet, and depth 6 inches?

1.20744 x 2 7.5 = .2236 x 5.4 =

= .80496

3 feet velocity ; then, 10 x .5 = 5 feet, and .80496

x 5 = 4.0248 cubic feet per second nearly. In estimating the power of water wheels, half the head must be added to the whole fall, because 1 foot of fall is equal to 2 feet of head; call this the effective perpendicular descent; multiply the weight of the water per second by the effective perpendicular descent and by 60; divide the product by 33,000, and the quotient is the effect expressed in horses' power.

EXAMPLE 1.–Given 16 cubic feet of water per second to be applied to an undershot wheel, the head being 12 feet, required the power produced.

6 x 16 x 62.5 X 60 .12 = 2 = 6 and

= 10.9

33000 horses' power nearly. EXAMPLE 2.Given 16 cubic feet of water per second, to be applied to a high breast, or an overshot wheel, with 2 feet head, and 10 feet fall; required the power.

1 + 10 x 16 x 62.5 x 60 2:2= 1 and

33000

= 20

horses' power.

N.B.Only about two-thirds of the above results can be taken as real communicative power to machinery.

OF THE CIRCLE OF GYRATION IN WATER WHEELS.

The centre or circle of gyration is that point in a revolving body into which, if the whole quantity of matter were collected, the same moving force would generate the same angular velocity, which renders it of the utmost importance in the erection of water wheels, and the motion ought always to be communicated from that point when it is possible.

To find the Circle of Gyration.

Rule.- Add into one sum twice the weight of the shrouding, buckets, &c., multiplied by the square of the radius, į of the weight of the arms multiplied by the square of the radius, and the weight of the water multiplied by the square of the radius also; divide the sum by twice the weight of the shrouding, arms, &c. added to the weight of the water, and the square root of the quotient is the distance of the circle of gyration from the centre of suspension nearly.

ExamPLE.- Required the distance of the centre of gyration from the centre of suspension in a water wheel 22 feet diameter, shrouding, buckets, &c. = 18 tons, arms = 12 tons, and water = 10 tons.

22 ; 2= 11 and 112 = 121
Then, 18 x 2 = 36 x 121 = 4356

f of 12 = 8 x 121 = 968
water = 10 x 121 = 1210

6534

And, 18 + 12 x 2 = 60 + 10= 70 hence
6534

= 9.6 feet from the centre of suspen

70 sion nearly

« ΠροηγούμενηΣυνέχεια »