What is a horsepower?—How the Carthaginians manufactured horsepower—All that goes up must come down—How the sun lifts water up for us to use—Water the ideal power for generating electricity—The weir—Table for estimating flow of streams, with a weir—Another method of measuring—Figuring water horsepower—The size of the wheel—What head is required—Quantity of water necessary.
If a man were off in the woods and needed a horsepower of energy to work for him, he could generate it by lifting 550 pounds of stone or wood, or whatnot, one foot off the ground, and letting it fall back in the space of one second. As a man possesses capacity for work equal to one-fifth horsepower, it would take him five seconds to do the work of lifting the weight up that the weight itself accomplished in falling down. All that goes up must come down; and by a nice balance of physical laws, a falling body hits the ground with precisely the same force as is required to lift it to the height from which it falls.
The Carthaginians, and other ancients (who were deep in the woods as regards mechanical knowledge) had their slaves carry huge stones to the top of the city wall; and the stones were placed in convenient positions to be tipped over on the heads of any besieging army that happened along. Thus by concentrating the energy of many slaves in one batch of stones, the warriors of that day were enabled to deliver "horsepower" in one mass where it would do the most good. The farmer who makes use of the energy of falling water to generate electricity for light, heat, and power does the same thing—he makes use of the capacity for work stored in water in being lifted to a certain height. As in the case of the gasoline engine, which burns 14 pounds of air for every pound of gasoline, the engineer of the water-power plant does not have to concern himself with the question of how this natural source of energy happened to be in a handy place for him to make use of it.
The sun, shining on the ocean, and turning water into vapor by its heat has already lifted it up for him. This vapor floating in the air and blown about by winds, becomes chilled from one cause or another, gives up its heat, turns back into water, and falls as rain. This rain, falling on land five, ten, a hundred, a thousand, or ten thousand feet above the sea level, begins to run back to the sea, picking out the easiest road and cutting a channel that we call a brook, a stream, or a river. Our farm lands are covered to an average depth of about three feet a year with water, every gallon of which has stored in it the energy expended by the heat of the sun in lifting it to the height where it is found.
The farmer, prospecting on his land for water-power, locates a spot on a stream which he calls Supply; and another spot a few feet down hill near the same stream, which he calls Power. Every gallon of water that falls between these two points, and is made to escape through the revolving blades of a water wheel is capable of work in terms of foot-pounds—an amount of work that is directly proportional to the quantity of water, and to the distance in feet which it falls to reach the wheel—pounds and feet.
The Efficient Water Wheel
And it is a very efficient form of work, too. In fact it is one of the most efficient forms of mechanical energy known—and one of the easiest controlled. A modern water wheel uses 85 per cent of the total capacity for work imparted to falling water by gravity, and delivers it as rotary motion. Compare this water wheel efficiency with other forms of mechanical power in common use: Whereas a water wheel uses 85 per cent of the energy of its water supply, and wastes only 15 per cent, a gasoline engine reverses the table, and delivers only 15 per cent of the energy in gasoline and wastes 85 per cent—and it is rather a high-class gasoline engine that can deliver even 15 per cent; a steam engine, on the other hand, uses about 17 per cent of the energy in the coal under its boilers and passes the rest up the chimney as waste heat and smoke.
There is still another advantage possessed by water-power over its two rivals, steam and gas: It gives the most even flow of power. A gas engine "kicks" a wheel round in a circle, by means of successive explosions in its cylinders. A reciprocating steam engine "kicks" a wheel round in a circle by means of steam expanding first in one direction, then in another. A water wheel, on the other hand, is made to revolve by means of the pressure of water—by the constant force of gravity, itself—weight. Weight is something that does not vary from minute to minute, or from one fraction of a second to another. It is always the same. A square inch of water pressing on the blades of a water wheel weights ten, twenty, a hundred pounds, according to the height of the pipe conveying that water from the source of supply, to the wheel. So long as this column of water is maintained at a fixed height, the power it delivers to the wheel does not vary by so much as the weight of a feather.
This property of falling water makes it the ideal power for generating electricity. Electricity generated from mechanical power depends on constant speed for steady pressure—since the electric current, when analyzed, is merely a succession of pulsations through a wire, like waves beating against a sea wall. Water-power delivers these waves at a constant speed, so that electric lights made from water-power do not flicker and jump like the flame of a lantern in a gusty wind. On the other hand, to accomplish the same thing with steam or gasoline requires an especially constructed engine.
The Simple Weir
Since a steady flow of water, and a constant head, bring about this ideal condition in the water wheel, the first problem that faces the farmer prospector is to determine the amount of water which his stream is capable of delivering. This is always measured, for convenience, in cubic feet per minute. (A cubic foot of water weighs 62.5 pounds, and contains 7½ gallons.) This measurement is obtained in several ways, among which probably the use of a weir is the simplest and most accurate, for small streams.
A weir is, in effect, merely a temporary dam set across the stream in such a manner as to form a small pond; and to enable one to measure the water escaping from this pond.
It may be likened to the overflow pipe of a horse trough which is being fed from a spring. To measure the flow of water from such a spring, all that is necessary is to measure the water escaping through the overflow when the water in the trough has attained a permanent level.
The diagrams show the cross-section and detail of a typical weir, which can be put together in a few minutes with the aid of a saw and hammer. The cross-section shows that the lower edge of the slot through which the water of the temporary pond is made to escape, is cut on a bevel, with its sharp edge upstream. The wing on each side of the opening is for the purpose of preventing the stream from narrowing as it flows through the opening, and thus upsetting the calculations. This weir should be set directly across the flow of the stream, perfectly level, and upright. It should be so imbedded in the banks, and in the bottom of the stream, that no water can escape, except through the opening cut for that purpose. It will require a little experimenting with a rough model to determine just how wide and how deep this opening should be. It should be large enough to prevent water flowing over the top of the board; and it should be small enough to cause a still-water pond to form for several feet behind the weir. Keep in mind the idea of the overflowing water trough when building your weir. The stream, running down from a higher level behind, should be emptying into a still-water pond, which in turn should be emptying itself through the aperture in the board at the same rate as the stream is keeping the pond full.
Your weir should be fashioned with the idea of some permanency so that a number of measurements may be taken, extending over a period of time—thus enabling the prospector to make a reliable estimate not only of the amount of water flowing at any one time, but of its fluctuations.
Under expert supervision, this simple weir is an exact contrivance—exact enough, in fact, for the finest calculations required in engineering work. To find out how many cubic feet of water the stream is delivering at any moment, all that is necessary is to measure its depth where it flows through the opening. There are instruments, like the hook-gauge, which are designed to measure this depth with accuracy up to one-thousandth of an inch. An ordinary foot rule, or a folding rule, will give results sufficiently accurate for the water prospector in this instance. The depth should be measured not at the opening itself, but a short distance back of the opening, where the water is setting at a dead level and is moving very slowly.
With this weir, every square inch of water flowing through the opening indicates roughly one cubic foot of water a minute. Thus if the opening is 10 inches wide and the water flowing through it is 5 inches deep, the number of cubic feet a minute the stream is delivering is 10 × 5 = 50 square inches = 50 cubic feet a minute. This is a very small stream; yet, if it could be made to fall through a water wheel 10 feet below a pond or reservoir, it would exert a continuous pressure of 30,000 pounds per minute on the blades of the wheel—nearly one theoretical horsepower.
This estimate of one cubic foot to each square inch is a very rough approximation. Engineers have developed many complicated formulas for determining the flow of water through weirs, taking into account fine variations that the farm prospector need not heed. The so-called Francis formula, developed by a long series of actual experiments at Lowell, Mass., in 1852 by Mr. James B. Francis, with weirs 10 feet long and 5 feet 2 inches high, is standard for these calculations and is expressed (for those who desire to use it for special purposes) as follows:
in which Q means quantity of water in cubic feet per second, L is length of opening, in feet; and H is height of opening in feet.
The following table is figured according to the Francis formula, and gives the discharge in cubic feet per minute, for openings one inch wide:
TABLE OF WEIRS
Inches | 0 | ¼ | ½ | ¾ |
---|---|---|---|---|
1 | 0.403 | 0.563 | 0.740 | 0.966 |
2 | 1.141 | 1.360 | 1.593 | 1.838 |
3 | 2.094 | 2.361 | 2.639 | 2.927 |
4 | 3.225 | 3.531 | 3.848 | 4.173 |
5 | 4.506 | 4.849 | 5.200 | 5.558 |
6 | 5.925 | 6.298 | 6.681 | 7.071 |
7 | 7.465 | 7.869 | 8.280 | 8.697 |
8 | 9.121 | 9.552 | 9.990 | 10.427 |
9 | 10.884 | 11.340 | 11.804 | 12.272 |
10 | 12.747 | 13.228 | 13.716 | 14.208 |
11 | 14.707 | 15.211 | 15.721 | 16.236 |
12 | 16.757 | 17.283 | 17.816 | 18.352 |
13 | 18.895 | 19.445 | 19.996 | 20.558 |
14 | 21.116 | 21.684 | 22.258 | 22.835 |
15 | 23.418 | 24.007 | 24.600 | 25.195 |
16 | 25.800 | 26.406 | 27.019 | 27.634 |
17 | 28.256 | 28.881 | 29.512 | 30.145 |
18 | 30.785 | 31.429 | 32.075 | 32.733 |
Thus, let us say, our weir has an opening 30 inches wide, and the water overflows through the opening at a uniform depth of 6¼ inches, when measured a few inches behind the board at a point before the overflow curve begins. Run down the first column on the left to "6", and cross over to the second column to the right, headed "¼". This gives the number of cubic feet per minute for this depth one inch wide, as 6.298. Since the weir is 30 inches wide, multiply 6.298 × 30 = 188.94—or, say, 189 cubic feet per minute.
Once the weir is set, it is the work of but a moment to find out the quantity of water a stream is delivering, simply by referring to the above table.
Another Method of Measuring a Stream
Weirs are for use in small streams. For larger streams, where the construction of a weir would be difficult, the U. S. Geological Survey engineers recommend the following simple method:
Choose a place where the channel is straight for 100 or 200 feet, and has a nearly constant depth and width; lay off on the bank a line 50 or 100 feet in length. Throw small chips into the stream, and measure the time in seconds they take to travel the distance laid off on the bank. This gives the surface velocity of the water. Multiply the average of several such tests by 0.80, which will give very nearly the mean velocity. Then it is necessary to find the cross-section of the flowing water (its average depth multiplied by width), and this number, in square feet, multiplied by the velocity in feet per second, will give the number of cubic feet the stream is delivering each second. Multiplied by 60 gives cubic feet a minute.
Figuring a Stream's Horsepower
By one of the above simple methods, the problem of Quantity can easily be determined. The next problem is to determine what Head can be obtained. Head is the distance in feet the water may be made to fall, from the Source of Supply, to the water wheel itself. The power of water is directly proportional to head, just as it is directly proportional to quantity. Thus the typical weir measured above was 30 inches wide and 6¼ deep, giving 189 cubic feet of water a minute—Quantity. Since such a stream is of common occurrence on thousands of farms, let us analyze briefly its possibilities for power: One hundred and eighty-nine cubic feet of water weighs 189 × 62.5 pounds = 11,812.5 pounds. Drop this weight one foot, and we have 11,812.5 foot-pounds. Drop it 3 feet and we have 11,812 × 3 = 35,437.5 foot-pounds. Since 33,000 foot-pounds exerted in one minute is one horsepower, we have here a little more than one horsepower. For simplicity let us call it a horsepower.
As an example, let us say that we have a stream whose weir measurement shows it capable of delivering 376 cubic feet a minute, with a head (determined by survey) of 13 feet 6 inches. What is the horsepower of this stream?
This is theoretical horsepower. To determine the actual horsepower that can be counted on, in practice, it is customary, with small water wheels, to figure 25 per cent loss through friction, etc. In this instance, the actual horsepower would then be 7.2.
The Size of the Wheel
Water wheels are not rated by horsepower by manufacturers, because the same wheel might develop one horsepower or one hundred horsepower, or even a thousand horsepower, according to the conditions under which it is used. With a given supply of water, the head, in feet, determines the size of wheel necessary. The farther a stream of water falls, the smaller the pipe necessary to carry a given number of gallons past a given point in a given time.
A small wheel, under 10 × 13.5 ft. head, would give the same power with the above 376 cubic feet of water a minute, as a large wheel would with 10 × 376 cubic feet, under a 13.5 foot head.
This is due to the acceleration of gravity on falling bodies. A rifle bullet shot into the air with a muzzle velocity of 3,000 feet a second begins to diminish its speed instantly on leaving the muzzle, and continues to diminish in speed at the fixed rate of 32.16 feet a second, until it finally comes to a stop, and starts to descend. Then, again, its speed accelerates at the rate of 32.16 feet a second, until on striking the earth it has attained the velocity at which it left the muzzle of the rifle, less loss due to friction.
The acceleration of gravity affects falling water in the same manner as it affects a falling bullet. At any one second, during its course of fall, it is traveling at a rate 32.16 feet a second in excess of its speed the previous second.
In figuring the size wheel necessary under given conditions or to determine the power of water with a given nozzle opening, it is necessary to take this into account. The table on page 51 gives velocity per second of falling water, ignoring the friction of the pipe, in heads from 5 to 1000 feet.
The scientific formula from which the table is computed is expressed as follows, for those of a mathematical turn of mind:
Velocity (ft. per sec.) = sqrt(2gh); or, velocity is equal to the square root of the product (g = 32.16,—times head in feet, multiplied by 2).
SPOUTING VELOCITY OF WATER, IN FEET PER SECOND, IN HEADS OF FROM 5 TO 1,000 FEET
Head | Velocity |
---|---|
5 | 17.9 |
6 | 19.7 |
7 | 21.2 |
8 | 22.7 |
9 | 24.1 |
10 | 25.4 |
11 | 26.6 |
11.5 | 27.2 |
12 | 27.8 |
12.5 | 28.4 |
13 | 28.9 |
13.5 | 29.5 |
14 | 30.0 |
14.5 | 30.5 |
15 | 31.3 |
15.5 | 31.6 |
16 | 32.1 |
16.5 | 32.6 |
17 | 33.1 |
17.5 | 33.6 |
18 | 34.0 |
18.5 | 34.5 |
19 | 35.0 |
19.5 | 35.4 |
20 | 35.9 |
20.5 | 36.3 |
21 | 36.8 |
21.5 | 37.2 |
22 | 37.6 |
22.5 | 38.1 |
23 | 38.5 |
23.5 | 38.9 |
24 | 39.3 |
24.5 | 39.7 |
25 | 40.1 |
26 | 40.9 |
27 | 41.7 |
28 | 42.5 |
29 | 43.2 |
30 | 43.9 |
31 | 44.7 |
32 | 45.4 |
33 | 46.1 |
34 | 46.7 |
35 | 47.4 |
36 | 48.1 |
37 | 48.8 |
38 | 49.5 |
39 | 50.1 |
40 | 50.7 |
41 | 51.3 |
42 | 52.0 |
43 | 52.6 |
44 | 53.2 |
45 | 53.8 |
46 | 54.4 |
47 | 55.0 |
48 | 55.6 |
49 | 56.2 |
50 | 56.7 |
55 | 59.5 |
60 | 62.1 |
65 | 64.7 |
70 | 67.1 |
75 | 69.5 |
80 | 71.8 |
85 | 74.0 |
90 | 76.1 |
95 | 78.2 |
100 | 80.3 |
200 | 114.0 |
300 | 139.0 |
400 | 160.0 |
500 | 179.0 |
1000 | 254.0 |
In the above example, we found that 376 cubic feet of water a minute, under 13.5 feet head, would deliver 7.2 actual horsepower. Question: What size wheel would it be necessary to install under such conditions?
By referring to the table of velocity above, (or by using the formula), we find that water under a head of 13.5 feet, has a spouting velocity of 29.5 feet a second. This means that a solid stream of water 29.5 feet long would pass through the wheel in one second. What should be the diameter of such a stream, to make its cubical contents 376 cubic feet a minute or 376/60 = 6.27 cubic feet a second? The following formula should be used to determine this:
Substituting values, in the above instance, we have:
That is, a wheel capable of using 30.6 square inches of water would meet these conditions.
What Head is Required
Let us attack the problem of water-power in another way. A farmer wishes to install a water wheel that will deliver 10 horsepower on the shaft, and he finds his stream delivers 400 cubic feet of water a minute. How many feet fall is required? Formula:
Since a theoretical horsepower is only 75 per cent efficient, he would require 10 × 4/3 = 13.33 theoretical horsepower of water, in this instance. Substituting the values of the problem in the formula, we have:
What capacity of wheel would this prospect (400 cubic feet of water a minute falling 17.6 feet, and developing 13.33 horsepower) require?
By referring to the table of velocities, we find that the velocity for 17.5 feet head (nearly) is 33.6 feet a second. Four hundred feet of water a minute is 400/60 = 6.67 cu. ft. a second. Substituting these values, in formula (B) then, we have:
Quantity of Water
Let us take still another problem which the prospector may be called on to solve: A man finds that he can conveniently get a fall of 27 feet. He desires 20 actual horsepower. What quantity of water will be necessary, and what capacity wheel?
Twenty actual horsepower will be 20 × 4/3 = 26.67 theoretical horsepower. Formula:
Substituting values, then, we have:
A head of 27 feet would give this stream a velocity of 41.7 feet a second, and, from formula (B) we find that the capacity of the wheel should be 30 square inches.
It is well to remember that the square inches of wheel capacity does not refer to the size of pipe conveying water from the head to the wheel, but merely to the actual nozzle capacity provided by the wheel itself. In small installations of low head, such as above a penstock at least six times the nozzle capacity should be used, to avoid losing effective head from friction. Thus, with a nozzle of 30 square inches, the penstock or pipe should be 180 square inches, or nearly 14 inches square inside measurement. A larger penstock would be still better.