Chat. XVII.

THEORY OF THE MOTION OF FLUIDS.

11 29

by m, and t the time of flowing, we shall have M. m l aa — abx. TV , dd — bdxtu,since the expenditures are in the compound ratio of the orifices, the times, and thevelocities ; whence we have aa — ab x T V m = dd—bd x faM, which formula will give thatof the four required dimensions d,m,t,u.

If we wish to ascertain the diameter of the orifice of a reservoir whose height or velocityis given, in order that it may effectually expend in a given time a determined quantity ofwater, designated by m, taking the mean of all the other quantities comprised in the formula,we shall substitute a for d, to have aa — ab x T Y m — xx — bx x tu M, which will be reduced

to A b +

bb TVm ,

+ -^ ir x (aa-

ab) = x.

• ni TVM_dd—bd f M X '

tu M

If the diameter of the orifice, the height of the reservoir, or the velocity of the water,and the time are given, and we require the effective expenditure, we must substitute x in

place of m , to have aa — ab x TYx = dd — bd xtu'M; whence we have x = ——x

1 aa—abiy

If the diameter of the orifice, the effective expenditure, and the times are given, and wewish to know the velocity of the water per second, in order to deduce the height of the

* , , „ , (Zfl -

reservoir, we must put x in place of u, to have •

When similar quantities have the same value, we must efface them from the formula,which will render the calculation more easy. For example, it is required what diametershould be given to the orifice of a reservoir, 13 feet high, in order that its effectual expendi-

M 4

ture per minute may be quadruple that of M. Mariotte; as we haveT = t, Y = u, and —

1 m 1

4 a a—4 ab bb b . 9 ,

--+ — + - = x, having a = 3 lines, b = —, and

-= — • making the calculation we shall find that the diameter required should be nearly

2 20 °

5$ lines instead of 6, which appeared to be that required.

To show that the orifice just found will effectively expend quadruple that of the ex-periment ; as 9 (the square of the diameter of 3 lines), is to 30$ (the square of the diameterjust found, 5$ lines), so 4 lbs. of water (the natural expenditure of the first orifice), is to thenatural expenditure of the second, which will be 134$ lbs. To subtract the waste, multiplythe denominator fg by the diameter 5$, we shall have for the ratio of the waste to the na-tural expenditure ; then, as 55 is to 46, so 134$, the natural expenditure of the second orifice,is to its effective expenditure, which will be 112$ lbs., a number quadruple 28; that is to say,the effective expenditure of the first orifice, neglecting the four ounces arising from the dia-meter having been estimated at a little more than it ought, by supposing =

We shall have with geometrical precision therequired diameter by constructing the equation

the equation will be changed into

y

4 aa

M , bb b

-4 a 5 4--f - = r.

4 2

For this purpose we must

Fig. 1713.

draw the line AB equal to 2 a; prolong it from B toC, so that B C may be equal to 2 6; describe on A C as adiameter the semicircle ADC; elevate the perpendicularB D, whose square will be 4a6; raise also on the ex-tremity A the perpendicular AF equal to draw the

line FB, whose square will be 4aa +^ ; describe onthis line the demicirele BIIF; make BH equal to

BD; then draw the line FII, which must be prolonged from F to G by the lengthFA; then the line GH will be exactly the diameter required, since we shall have

^/fB s ^aa + ^'j - BH ! (4o6) + GF Q j = G H (x).

Of whatever figure the orifices may be, whether similar or not, their natural expenditurebeing always in the ratio of their superficies, and the wastes in the ratio of their circuit, itfollows that when the superficies are in the ratio of the circuits, the natural expenditurewill be as the decreases, and this happens when one of two orifices is a circle, and the other

c d

the square of its diameter; for calling d the diameter, and c the circumference, — will be the