360

MISCELLANEOUS

TRACT 38.

PROBLEM XXVII.

T 5 determine the Time of . emptying a Vessel of Water by aSluice in the Bottom of it-, or in the Side near the Bottom,

. the Height of the Aperture being very small in respect ofthe Altitude of the Fluid.

Put a = the area of the aperture or sluice;d — the whole depth of water;

x — the variable alt. of the surface above the aperture;a — the area of the surface of the water.

Then V 16 : Vx : : 32 : 8 >/x the velocity with which thefluid will issue at the sluice and hence 8 Vx : Y --

a

the velocity with which the surface of the water will descendat the altitude x, or the space it would descend in 1 secondwith the velocity there, or it is only -"V— when reduced J- forthe .contraction of the stream. Now in descending the spacex, the velocity may be considered, as uniform ; and uniformdescents are as their times; therefore 6 " "f— : x :: l" : . a%

a 6 a ^/x

the time of descending Sc space, or the fluxion of the time of

. * — a'x

exhausting. That is, t = -—

Now, when the nature or figure of the vessel is given,there will be given a' in terms of x; which value of a’ beingsubstituted into this fluxion of the time, the fluent of the re-sult will be the time of exhausting sought.

So if, for example, the vessel be any prism, or everywhere of the same breadth; then a' is a constant quantity,and therefore the fluent is — ■^V'-r. But when x — d, this

becomes — £ a Vd, and should be 0; therefore the correct

fluent is t — — x ( \/d — V-r) for the time of the surfacedescending till the depth of the water be x. And whenX — 0, the whole time of exhausting is barely —

And hence if a' be 10000 square feet, a — 1 square foot,and d == 10 feet; the time is 10540 seconds, or 2 h 55' 40".