Читать книгу A Catechism of the Steam Engine - C.E. John Bourne - Страница 6

15. Q.--How do you determine the velocity of falling bodies of different kinds?

A.--All bodies fall with the same velocity, when there is no resistance from the atmosphere, as is shown by the experiment of letting fall, from the top of a tall exhausted receiver, a feather and a guinea, which reach the bottom at the same time. The velocity of falling bodies is one that is accelerated uniformly, according to a known law. When the height from which a body falls is given, the velocity acquired at the end of the descent can be easily computed. It has been found by experiment that the square root of the height in feet multiplied by 8.021 will give the velocity.

16. Q.--But the velocity in what terms?

A.--In feet per second. The distance through which a body falls by gravity in one second is 16–1/12 feet; in two seconds, 64–4/12 feet; in three seconds, 144–9/12 feet; in four seconds, 257–4/12 feet, and so on. If the number of feet fallen through in one second be taken as unity, then the relation of the times to the spaces will be as follows:--

Number of seconds	1	2	3	4	5	6
Units of space passed through	1	4	9	16	25	36	&c.

so that it appears that the spaces passed through by a falling body are as the squares of the times of falling.

17. Q.--Is not the urging force which causes bodies to fall the force of gravity?

A.--Yes; the force of gravity or the attraction of the earth.

18. Q.--And is not that a uniform force, or a force acting with a uniform pressure?

A.--It is.

19. Q.--Therefore during the first second of falling as much impelling power will be given by the force of gravity as during every succeeding second?

A.--Undoubtedly.

20. Q.--How comes it, then, that while the body falls 64–4/12 feet in two seconds, it falls only 16–1/12 feet in one second; or why, since it falls only 16–1/12 feet in one second, should it fall more than twice 16–1/12 feet in two?

A.--Because 16–1/12 feet is the average and not the maximum velocity during the first second. The velocity acquired at the end of the 1st second is not 16–1/12, but 32–⅙ feet per second, and at the end of the 2d second a velocity of 32–⅙ feet has to be added; so that the total velocity at the end of the 2d second becomes 64–2/6 feet; at the end of the 3d, the velocity becomes 96–3/6 feet, at the end of the 4th, 128–4/6 feet, and so on. These numbers proceed in the progression 1, 2, 3, 4, &c., so that it appears that the velocities acquired by a falling body at different points, are simply as the times of falling. But if the velocities be as the times, and the total space passed through be as the squares of the times, then the total space passed through must be as the squares of the velocity; and as the vis viva or mechanical power inherent in a falling body, of any given weight, is measurable by the height through which it descends, it follows that the vis viva is proportionate to the square of the velocity. Of two balls therefore, of equal weight, but one moving twice as fast as the other, the faster ball has four times the energy or mechanical force accumulated in it that the slower ball has. If the speed of a fly-wheel be doubled, it has four times the vis viva it possessed before--vis viva being measurable by a reference to the height through which a body must have fallen, to acquire the velocity given.

21. Q.--By what considerations is the vis viva or mechanical energy proper for the fly-wheel of an engine determined?

A.--By a reference to the power produced every half-stroke of the engine, joined to the consideration of what relation the energy of the fly-wheel rim must have thereto, to keep the irregularities of motion within the limits which are admissible. It is found in practice, that when the power resident in the fly-wheel rim, when the engine moves at its average speed, is from two and a half to four times greater than the power generated by the engine in one half-stroke--the variation, depending on the energy inherent in the machinery the engine has to drive and the equability of motion required--the engine will work with sufficient regularity for most ordinary purposes, but where great equability of motion is required, it will be advisable to make the power resident in the fly-wheel equal to six times the power generated by the engine in one half-stroke.

22. Q.---Can you give a practical rule for determining the proper quantity of cast iron for the rim of a fly-wheel in ordinary land engines?

A.--One rule frequently adopted is as follows:--Multiply the mean diameter of the rim by the number of its revolutions per minute, and square the product for a divisor; divide the number of actual horse power of the engine by the number of strokes the piston makes per minute, multiply the quotient by the constant number 2,760,000, and divide the product by the divisor found as above; the quotient is the requisite quantity of cast iron in cubic feet to form the fly-wheel rim.

23. Q.--What is Boulton and Watt's rule for finding the dimensions of the fly-wheel?

A.--Boulton and Watt's rule for finding the dimensions of the fly-wheel is as follows:--Multiply 44,000 times the length of the stroke in feet by the square of the diameter of the cylinder in inches, and divide the product by the square of the number of revolutions per minute multiplied by the cube of the diameter of the fly-wheel in feet. The resulting number will be the sectional area of the rim of the fly-wheel in square inches.