The hydrodynamic mass of ship and its determination by electrical methods of analogy

(1)

nu, A /s edt 1-.-"Lcs

3,

171; eic

The hydrodynamic mass

of

ship and its determination

by

electrical methods of analogy

by

Dipl.-Ing. Martin Lemke and Prof.Dr.-Ing. Erwin Metzmeier

I.S.S.C,

1967

(2)

The following analysis shows the determination of ship's hydrodynami mass. Especially the analysis is composed to show not only the motic of the whole hull but also include vibrations of parts. hence it is neccessary to know the distribution of the hydrodynamic mass along

a construction unit.

The solution of this problem - only the plain problem shall be

considered here - is already difficult with infinitely large hydro-dynamic mass, it hardly can be conducted with a hydrohydro-dynamic mass

limited by the cross section of a canal. The described electrical method of analogy was used to get also a sufficiently exact solution

for the mentioned cases.

The knowledge of the hydrodynamic equations is supposed. To relate to the analogy at the corresponding points the neccessary references

will be written once more.

In the main the potential equation must be represented:

In this equation p means the pressure, means the density, v means

the velocity of the water particles. Because the squared speed can

be neglected compared with the component F)/ the potential equatio

with = const. takes from like:

er" (2)

Hereby a fault results which will not be considered in the following

Some authors fixed it about 0,1 %. In addition to the equations (1) and(2) Euler's dynamicalequation is neccessary:

t

rod

p

(3)

With the mentioned neglect it becomes:

'CT

ot, t

t

The following relations result from equation (3):

--c) V -4) p) V:(1,

p

----(5a ,b , c)

t

x

They are neccessary for representin the boundary conditions.

Because potential equation (2) is true for the pressure it is

obvious to find an electrical analogy, because e.g. the electrical

field in conductors satisfies the potential equation.

Then it is only neccessary to realize the boundary conditions.

If voltage is placed instead of pressure in equation (2) it can be

written:

- (6)

To reduce the theoretical consideration and to ease the execution

the following analysis only attends to the plane case. Therefore

we suppose a cylinder which longitudinally extends to infinity.

We consider a conductive foil of graphite paper according to

figure 1. This paper simultaneously can be considered as a cross

section of a water filled canal. If the problem is symmetrical to

the y-axis representing one half of the cross section is sufficient.

The left upper edge is cut out and represents the cross section of

a ship below water line.

The electrical and hydrodynamical boundary conditions are represented

in figure 1. At the water surface the pressure is constant which

is unimportend for the dynamical case, so that p = 0 can be placed.

Because the pressure corresponds to the electrical voltage u=const.,

respectively u = 0 must be demanded as electrical boundary condition

along the surface. The condition is satisfied if the graphite paper

is connected with a very good conductor at that boundary which shall

represent the water surface.Where the fluid is limited by the side,

respectively by the bottom of the canal the velocity vertical to

the side ever must be zero. Thus at this point we get:

2

(4)

-On the other hand we get by Ohm's law

Thus we get with the current densities i, i , i and the resistance

x y

W per square centimeter:

4 "()

S k

(11) 51v1

x

( 9 ) -Dp

By equation (5) the result for it is ---= 0, respectively = 0

2)y

of the bottom as boundary condition for pressure. Analogously comes

true for the electrical field intensity E 0, respectively

K -ax

E = 0. These conditions are represented by an insolated

Y

157

boundary. At the line of symmetry of the canal also are

that means _{Ex =} 0.

The boundary condition at the hull result because the velocity of the water particles vertical to the side of the ship must be as

great as the velocity va of the side of the ship. The result is:

V

= V S

fe7 oc, 4-

V Go S

( 7 )

differentiated with respect to t:

Vx

6

c.c) S _{( 8 )}

t

t-As boundary condition for the pressure along the hull we get with

equation (5):

i)p

C)

vi 0c, bcr--Q Do

=7 q

? x

y

t

To get the electrical boundary conditions consider figure 2.

By Kirchhoff the total of the being added currents is equal to the total of the draining currents. For the triangular element with the -ides dx, dy, ds we get with the current densities i, ix

i , the relation _c,

s

01V +

dX

wi th ClY

_-.7.Sitlkand 2'

x

C.4.5

x y

S

that changes over into: 5

=

1/. ki C-0 Oc, ₍₁₀₎

The voltage is connected with field intensity by -

graat

(A,

(5)

+-Comparing of the conditions (9) and (11) gives

=

(12)

t

GI/

where i is the current density at the line of contact between water

and hull. Equation (12) shows that a constant current must be

supplied along the ship line (frame line). That means that the

current lines resulting from the vibration of the side of the ship

must be vertical to the side of the ship. Simultanously that is the

direction of va . The electrical current lines act the same way.

JP..

f

As known the hydrdynamic mass is defined by 11/ - where is

't6112, t

u cti.3.te

the acceleration of the frame line

PA, = the pressure at the frame line

Corresponding to equation (13) the definition of the hydrodynamic

P Of S

mass is with equation (12) of 1,14

- .4-

or by using voltage ua

rA(

instead of pressure pa at the hull:( ___

of, frvi

s

(14)

Because it is difficult according to equation (12) to inject a

constant current it is better to supply the current point-for-point

to the frame curve and to rectify the single currents. That satisfie

the changing over into the differential equation:

/\

frvi

A s

k,

A s

kqq.

h(AP

* L"

/AS

(15)

By introducind a dimensionless factor of the following form

k =6114 ,where length

I. 't+L

corresponding to figure 7,

equation (15) and .A.Sz-

!with

n as total number of the single

currents become:

- ₍₁₆₎

Partitioning into twenty fields was chosen in the present case.

Figure 5 shows the formation of the current supplies at the dotted

shown curve. The longer the supplies are improved the more the

garantee is given that with equal currents i1 = i2 = i3 = = i8

(6)

the electrical current lines are square with the frame curve.

The graphite paper has a resistance value of 1903 per square

centimeter. The electrodes for the current supply (power supply)

directly will be plotted with silver contact ink to the graphite

layer.

This layer naturally is very thin, but even if the transfer

resistance is valued some Ohms it still is very small opposite the

resistance of the graphite layer and therefore the result must be

correct.

The specific resistance of the foil was measured at four stips

which were removed at different points and in different directions

in order to control the homogenity of the foil. The latitude of th

strips is about 10 mm, the length about 350 mm.

The amplitude of the foil's resistance requires a corresponding

amplitude of the balancing resistances which are neccessary for

representing the boundary conditions along the frame curve. To

avoid unneccessary long adjusting trouble any one of these

resistances must be large enough against the specific foil resistance

Than unimportant changing of the current's adjustment in a

neigboring resistance will not change the current in the considered

resistance. The supply voltage was chosen 50 to 100 volts. Direct

voltage could be used because thermoelectric voltages are unimportant

and because the working resistance are about 10 volts. Potentiometers

with a resistance value of 100 kOhms were used as series resistance.

Figure (3) and (4) show (an impression of) the assembly of the

arrangement.

The current directly will be measured. After current measuring the

measuring points will be connected by a short circuit plug.

Figure (5) shows the circuit.

For current in 0,5, respectively 1 mA was chosen in order to avoid

overloading of the graphite layer. Number n = 20 in this case and

W = 1903 Ohms, the resistance of one square centimeter of the

graphite paper.

With the so attained kn the mass distribution can be determined

(7)

The kn-values are to take from figure 6, while figure 7 shows

the mass distribution.

Finally some should be said about the precision of the method.

For that it es neccessary to represent a case with an exactly

known solution.

The resulting fault is composed of

Inaccuracy of instrument indication Inaccuracy of reading indication Inhomogenity of graphite foil

Inaccuracy when taking up the dimensions of the foil

Using differential method instead of continuous current

injection.

Inaccuracy resulting by neglecting v2/2 against p/q here is not

considered.

At first we will consider the theoretical case, that a cylinder

with the radius a is in a second cylinder (as shown in figure 8)

with the radius b, whereby the value a may change periodically

e.g. by the relation a Le, (4+ .5 /./r2 ijand

6

In this central symmetrical case we get with v=grad Cr

Then the pressure is

p = --qg

r

(17)

(18)

Using the continuity equation we can write 2_.

4-- 61,v ,Lrit-v(r)

a,

or

V

r

= - V

(19)

M,

where

va is the velocity of the cylinder barrel and therefore

of the water particles at this point too.

Equation (17) offers the velocity potential as:

_---. V

r

C(4.- I₍ _{2o )} 6 Further on cf

=

-5

+

(8)

The unknown f1(t) can be determined by the boundary conditions.

Supposing the pressure shall be zero at the point r=b, we get

Ct, LGI respectively for the pressure:

V-P

=

a'

t_av

Equation (22) then gives the pressure at the cylinder:

Lvi

b

Equation (23), integrated over the cylinder surface becomes:

s/10,t,

r(-- 01,2-

tfri

va/

Zy

pier

Equation (24) is related from the unit of length of the cylinder.

With that help the hydrodynamic mass can be defined as:

Lt/1

r a,

(2-1raij52;

4

.2-r

The dimensionless factor k of equation (25) _{becomes with the}

later on considered relation b/a=16,56

=

4 1,171

4 2,2°2 _0,17,-44-6S

2_ ir

-

(26)

The case here considered simply is to _{represent electrically.}

Figure 9 shows the position of the electrodes _{on the graphite}

foil and the circuit of the connected instruments. A quater circle

was represented for total utilization of the foil. Then _{it is}

neccessary to attend that when calculating a fourfold of _{the measuref}

values must be used for the currents. To avoid the fault becoming

too great, conditioned by the inhomogenity of _{the foil, the}

resistance of the foil was measured _{at a test piece close at the}

used piece. The resulded resistance _{was W = 1827J2.. Voltage of}

50 V and current of 15,6 mA give the _{factor k as:}

-(22)

(23)

(,-t1

(9)

11 IA/ It- I / v

-

3

4-3g6

4

. 4 6, ,(27)

The agreement of (26) and (27) shows a derivation of 1,8 %. A further case Is worked by other authors and represented in

figure 10.

A rectangle cross section carries out horizontal vibrations at infinitely enlarged water surface and limited depth of water.

For the case h/b =0,0 which will be reached for h 7,7°40 (unlimited

depth 'Of water) and for b = 0 too, a factor k of

At first 20 currents

in

_{(n=20) were injected at shown in figure 11.}

Each lug hat

a

width of 5 mm.. For

in

_{= lmA and the foil resistance}

W =

182712:un

= 479,12Vwere measured. That gives the Value:

1,1

4- 7-9, 4k

IA,

0 6

(30>

.A 2_ 7- 2-0

- 40

and a deviation of 10,6 % against (28).

A, further Measurement was tade with 10 lugs of a width of 2 mm.

At a current

in, = 1 mA

Zun

= 121,1

V

and therefore:

4"z4I1

4S7 27- 401

_{/10--C =}

_{6 3 3}

(31)

That gives a deviation of 11,8 % against (28). But we always must

consider _{that the theoretical values of (28Y are valid for}

unfinitely enlarged water, 'while _{30) and (31) were measured at}

limited canal width, For the value of (30) the relation

canal

(29)

8

11

7s-sir

₃

₍₂₈₎

will be stated. The unfinitely enlarged water surface can't be

realized in the

test!

The section of the graphite foil is shown

in figure 11. The representation of the diagram of connection hat been neglected..

(10)

width-plate width is 7,2 , for the value of (31) the relation

is 36.

At last 2 lugs of 2 mm width were used. With _{un =} 4,23 V and a

current of 1 mA the factor k is:

11-,

5 7-6i

lt- . A 0 3

and the fault 2,4 % .

Recapitulating we can say, that the cases, theoretically and practically representable without any limitation and directly

coparable, show faults of 1,8% respectively 0,85%. Therefore we

can expect to get the values of the test with an accuracy of

-7- 2%, if the theoretical result can't be determined.

(32)

(11)

y,

_Fig.

P Ex-r. = ax

by symmetry

VG, Vx sinoc Vy COS cC aVg du sin cc+ du at ax P =0'

Ea I ds

dVy aP

-o

dt ay

Electrical and hydrodynamic boundary conditions

(12)

Fig. 2

Fig. 3

Fig. 4

(13)

cutoff

_connect/6 Potentiometer

fff

fffL4

Fig. 5

circuit diagram

F'ower supply

graphite paper

(14)

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- Pei . -0'14' !I . iniii ' It sem:

(15)

2

water line

profile of ship

Fig. 7

Distribution

of mass

(16)

111 vII

Fig. 9

I it 1

-Fig. 8

(17)

Fig. 10

Fig. 11

n y 720,.,, a

-// -//-//-//-//-//-///

/////////

= n