On the problem of horizontal hydrodynamic impact of a sphere

ON THE PROBLEM OF HORIZONTAL HYDRODYNAMIC IMPACT OF A SPHERE (K ZADACHE

PMM

Vo1.22,

I.

v.

0 GOBIZONTAL'NOM GIDRODINAWICHESKOM UDARE SPERY) No.6,

of

sphere

on a free

series

containing

method

by which

pp.847-849

YOSSAKOVSKII and V.L. (Dnepropetrovsk-Berdiansk) (Received

The solution

1958,

29

the problem fluid

of

surface

spherical a solution

1957)

October

the horizontal was found

(harmonic) of

this

RVACHEV

hydrodynamic

by Blokh [ 1

functions.

problem

1

This

impact

in the note

of

outlines

may be determined

a

form of

a a

in closed

form. 1. Let

a spherical

bowl

be immersed

space z 2 0. so that its wetted surface that the radius Z2 = 1. (The assumption obviously

does

Now suppose

not

affect

that

the

axis Ox. Then, allowing of the perturbed fluid domain,

connected

where p is

the

with

density

the generality sphere

suddenly

the of

impulsive the

of

fills

the half-

r2 = L2 + y2 + equal to unity

the

argument.)

acquires

a velocity

pressure we arrive

fluid,

by the

pt

at the

when

x2 + Y2> i

u,x

when

x2 + y2 + 22 = I

grad ‘p + 0

when

a’p a’p z = z

may easily

which

has the equation of the sphere is

U0 along

for the fact that the velocity potential motion is a harmonic function within the

‘p(z, y, 0) = 0

It

in a fluid

=

be shown that

the

relation

following

the

(G(z, y, z) fluid pt=-p$h

conditions: (1.1) (1.3)

lz > 0)

12 _1-Y2 + 22- co

(1.3)

function

(f .4)

like 4(x, y, z), is (1.2) we find that

a harmonic

function.

1217

From the

conditions

(1.1)

and

V.I.

1218

Mossakovskii

and

V.L.

Rvachev

when x~+Y~>I

J,65 YI 0) = 0

(f-5)

whenx2-kY2-#-Z2=f

4J (I, Y, 4 = u,x

U.fi)

(2 > 0)

Let

y (2, Y, I) = 0 (JJ,Y, 4

Then the

\Y (x,

Z) = 0 when

y,

It

r = 1.

-u, -g

follows

(1.7)

from

Kelvin’s

theorem

that

function

will be harmonic in r = 1 and, moreover,

the domain we have

It

r < 1.

is

obvious

that

y = y* = 0 when

(i 3) Thus of

the

function

the

function

Let

F(x,

P(r, Y, half-space

Z)

Y

\II*(x, (t,

y,

Z)

y, z) proves across the

y, Z) be the function when r < 1. Then F(x, z > 0.

satisfying

the

F (I, Y, 0) =

to be the analytical sphere r = 1. equals Y (x, y, Z) will be a harmonic

which y,

Z)

following

-

boundary

2.

Let

F(x,

y,

z) = F1(z,

functions

y,

satisfying

z)

the

when 2’ i- Y2 < 1

UC9

when ~2 + IJ*> 1

+ F2 (x,

boundary

“a

0 Fz (2, ~1 0) =

1102 -

It

so

that

may easily

be

shown

that

(“2

on

z = 0:

(1.2)

I,

z),

where

Fi

and

F2

are

conditions

when~2+ Fi (2, Y, 0) =

when r > 1 and function in the

conditions

UC? (22 + y’)“’

harmonic

continuation

~2 < I

(2.1)

when X2 -k Y2> i whenza+y2 i

Horizontal

impact

hydrodynamic

1

F (5, y, z) = F, (x, y, z)

We determine for

the

the

function

FI(z,

--r-

of

x y - -i r2 ’ r!2

PI -.-

J 9

(2.3)

-p>

Z) by solving

y,

1219

a sphere

the Dirichlet

problem

half-space: 6dSh ss E,+‘l’
F1 (2, y, z) = - -$

Using

the

formulas

(1.41,

(1.71,

(2.3)

(2.4)

to spherical

and transforming

coordinates z=rsinOcoso, we

y=rsinOsino,

z=rcosO

get U, sin 0 co8 0

r

QJ

-

it

In particular,

sI

dr ’ -r @ o [I -2t

P dt sin 0 cos (o-a)

can be shown that

(2.5)

+ t2]‘Iz

when r = 1, rw

~p(i, e,0)=

--!$sinec0s~--~c0secos~

s

P (sin 0 cos a) cos a da

(2.6)

0

where

l-332 I f3z -;1n p @) = (1 + ;) v 2 (i _ z) + 2 (1 --z”) Similarly,

it

is

problem ” internal” filled with fluid.

possible

to obtain

of the horizontal In this problem,

the

(

i+.&

solution

(2.7)

I1

of

the

impact of a spherical for instance,

so called bowl

half

2x ‘p*(1, 8, 0) = u, sin 8 cos 0 - 2!- cos e cos o

s

P’ (sin 8 cos a) cos a da

(2.8)

0

where r’(3)--

3. virtual

For comparison mass coefficient

--(I

with

1 + 3: +z)I+q_-l)+

the of

solution

the

sphere

2s y+;In 1-z

(

1 +e;

1/z

VI-Z

)

given by Blokh, we calculate A L, given by the formula

(2.9)

the

1220

V.I.

Mossakovskii

and

V.L.

Rvachcv

(3.1)

where P, wetted

is

the

surface

resultant of

of

the

impulsive

pressure

forces

acting

on the

the sphere:

sspt

P( = -

cos (n,

(ds = sin 6 d6 do)

OS)ds

(3.2)

(8)

Remembering

that

Pt = - p $,

we thus

have

‘IG

2s

cos o do

sin* 69 (I, 0, 0) de

(3.3)

0

Hence,

using

the

relation

(2.6)

for

the

internal

the

$-

1 = 0.27323954

problem,

we find

h, =

Similarly

and evaluating

integrals,

we get

(3.4)

that

h; = 4x-i - ‘Is = 0.39823954 The following of

Blokh:

values

AZ = 0.2’7322.

of

the same coefficients

are given

in the

work

AZ* = 0.39622.

BIBLIOGRAPHY 1.

Blokh, E. L. , Gorizontal’nyi svobodnoi poverkhnosti pact

of

a sphere

gidrodinamicheskii zhidkosti

on a free

fluid

udar sfery

(The horizontal surface).

pri

nalichii

hydrodynamic

PMM Vol.

17.

No. 5,

im1953.