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and the nature of the parabola is given by its equation

y = p.cť.

727.

If x denote the distance of any element of the orifice ydx (parallel to the horizon) from the surface of the fluid, and Q the quantity of fluid issuing through the whole orifice in a second, we have

dQ = 2ydx x N 2g«

:: Q = 2fydx 2gx integrated between x = 0 and x = x, the altitude of the fluid.

Let now the given area of the orifice be A, and that of the descending surface, which is also given, be M; then the velocity of the descending surface is

A

x Qaz" by the question
M

M
:: Sydx » 202 = x сxx"

2A

nMC
and

Х

2A / 29 or the required curve is a parabola.

The time of emptying any depth x of the fluid is

728. (696)

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.

But here

62 X = m =

be 을 (2ax - 2) a and 6 being the semiaxes of the generating ellipse.

62

x f(2a xdx xdx)
ao AJ 25

2762
a? AN 29

5

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Make rus a, and then x = ., and the difference of the results gives the time required, viz.,

292 - 17 X

60

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But since, as it may be easily shewn, the section of an oblate spheroid made by a plane I to its major axis at the distance : from the vertex, is an ellipse, whose semiaxes are

b

2ax x, and 2ax x

e 음

a

a and 8 being the semiaxes of the spheroid, we have

b X = X

(2ax xa)

.

a

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a AW 2g
16 ba?

15 AN 9 for the whole spheroid.

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HYDRODYNAMICS IN GENERAL.

731. LET « be the angular velocity round the axis ; then when the water and cylinder are relatively at rest, the surface of the water will be that of a certain solid of revolution, whose ordinate, normal, and subnormal, referred to the axis of rotation, will represent the centrifugal force, the re-action of the fluid, and the force of gravity, respectively. Hence if y denote the ordinate, and v the linear velocity of any particle of the fluid in the surface, the centrifugal force is

a?y

= a'y

y and the subnormal is 1

9 a'y or the generating curve of the volume of the part evacuated by the fluid is the common parabola, whose parameter is

29

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X gy

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Hence, if r be the radius of the base of the cylinder, the quantity of water thrown out by the revolution of the cylinder is equal in volume to the paraboloid, whose base is ars, and altitude 2ge?

i. e. to 29

na? X ar x x

704.

29 49 and this taken from the given volume of the whole cylinder, will leave the quantity of water required.

732. Let o be the angle required; then (Vince's Hydrostatics, Prop. XXVIII.) the effect cc sin. 0 cos. 6 = max. and by the rule for maxima and minima, we get

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733. If p be the parameter of the given paraboloid, andr the radius of its base; then the angular velocity necessary to empty the vessel is (731)

2g

P and therefore the time of revolution is

360

р. 29

734. Let a be the required angular velocity; then the volume of fluid evacuated is that of a paraboloid, whose base is an (r being the radius of the base of the cylinder,) and altitude (see 731,)

Hence, and by the question, 29

Tr2 e22 Tr? x h

2 h being the height of the cylinder.

2gh.

Х

2g

2

.

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735. The descending force is w, and the resistance is

s x v* (Vince's Hydrostatics, Prop. XXV.) s being the specific gravity of the fluid, and v the velocity. i the force which accelerates the whole apparatus downwards is

nd'sva F =

4

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