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(2)

Fig. 2 Schematic of hydrodynamic system in the meridian plan

The mathematical formulation of a torque con- verter has been developed by Ishihara and Emori Kotwicki, and Hrovat and Tobler [28-30].

The kinetic energy of the total mass of the fluid is:

2

c V2

E =ρ AdL, (1)

where is integral for full closed fluid flow; A is merid- ian section of the fluid flow, which is perpendicular to the meridian plane.

So the variation of the kinetic energy according to the time is:

2 2

2 2

1 ;

2 2

1 ,

2

c

c

dE d V dV

AdL AdL

dt dt dt

dE V V z

dt t z t AdL

ρ ρ

ρ

= =

= +

∫ ∫

(2)

where z is length of the average current line.

As dt

V =dz then the relation (2) becomes:

1 2

2

dEc d V d z

dt =ρd z d tAdL (3)

The absolute velocity V is expressed as follows:

2 2 2

V =U +C ; U r= ω.

The integral Eq. (3) becomes:

( )

( ) ( )

2 2

2

1 ;

2

;

,

c

c

c

d U C

dE AdL

dt d t

dE d r dC

r AdL C AdL

dt d t dt

dE d r dC

r AdL Q

dt d t dt

ρ ρ ω ω

ρ ω ω Φ

+

=

⎞ ⎪

= + ⎟ ⎬

⎠ ⎪

= +

(4)

with Q Ac= and A2 dL Φ A

= ⎜ ; A2 is meridian section of the fluid flow; C2 is meridian speed at the entering of turbine.

The moment of the pump is expressed as follows:

( )

2

( )

12 2 2

1 2 1 1 2

11

M r A dL Q r r

t

ω

ρ ρ ω ω

= + . (5)

The moment of the turbine will be:

( )

2

( )

22 2 2

2 1 2 2 1

21

M r A dL Q r r

t

ω

ρ ρ ω ω

= + . (6)

The hydraulic power of the pump and the turbine will be respectively:

( )

( )

12 2 2

1 1 2 1 1 2

11

P r r AdL Q r r

t

ρω ∂ ω ω ω

= +

; (7)

( )

( )

22 2 2

2 2 2 1 1 2

21

P r r AdL Q r r

t

ρω ∂ ω ω ω

=

. (8)

During the fluid circulation in the coupling, there are energy losses EL due to the frictions in the blades, and to the shocks during sudden change of the direction of inlet velocity at the pump and particularly at inlet of the turbine.

Energy losses EL are expressed per unit of time as:

(

12 22

)

( 1 2)2 22

1 2 dEL

Q r r LC

dt = ρ + ω ω + , (9)

where L is total friction coefficient by full circulation.

The equation of non-steady motion of fluid is ob- tained by substituting the Eqs. (4), (7)-(9) in the following relationship:

1 2

c L

dE dE

P P

dt = + − dt . (10)

It follows:

(

2 2

)(

2 2

)

2

2 1 2 1 1 2 2

2

dc r r LC

Φ dt = ω ω . (11)

To obtain the motion equation of the pump, the inertial moment of the pump and of all rotating mechanical parts without the presence of oil, which is denoted J1.

The fluid act on the pump with a reaction time is expressed by Eq. (5).

Equation of motion of the pump:

(

2 2

)

1' 1 2 2 2 1 1 2

P

J d M A c r r

dt

ω = ρ ω ω . (12)

Equation of motion of the turbine:

(

2 2

)

2' 2 2 2 2 1 1 2

T

J d A c r r M

dt

ω =ρ ω ω , (13)

J′1 is inertia moment of the rotating parts of the pump side without the presence of oil; J′2 is inertia moment of the rotating parts of the turbine side without the presence of oil; ω1 - angular velocity of the pump wheel; ω2 is angu- r1

r2

21

11 22

߱ ߱

22

T P

(3)

397 lar velocity of the turbine wheel; MP is output torque on the pump; MT is output torque on the turbine.

By neglecting the inertia moment of the fluid act- ing on the pump and the turbine, we obtain:

1' 1

J =J and J2' =J2,

where J1 is inertia moment of the pump side with oil, J2 is inertia moment of the turbine side with oil.

Finally, the motion equations of the pump and the turbine will be respectively:

(

2 2

)

1d 1 P 2 2 2 1 1 2

J M A c r r

dt

ω = ρ ω ω (14)

(

2 2

)

2d 2 2 2 2 1 1 2 T

J A c r r M

dt

ω =ρ ω ω (15)

2.3. Resolution of the equation system

The system of fundamental equations of non- steady movement of the hydrodynamic coupling is written as follows:

( )( )

( )

( )

2 2 2 2 2

2 2 1 1 2 2

2 2

1 1 2 2 2 1 1 2

2 2

2 2 2 2 2 1 1 2

1 ;

2

; .

P

T

dc r r L c

dt

J d M A c r r

dt

J d A c r r M

dt

Φ ω ω

ω ρ ω ω

ω ρ ω ω

=

=

=

⎪⎩

(16)

We have a system of Eq. (3) to Eq. (5) unknowns, namely: ω1 is angular velocity of the wheel pumps, rad/s;

ω2is angular velocity of the wheel turbines, rad/s;

c2 is meridian velocity of the fluid, m/s; MP is output torque of the pump, daNm; MT is output torque of the tur- bine, daNm.

It is more practical to make the system of equa- tions in dimensionless form. The system (16) becomes then:

( )( )

( )

( )

2 2 2 2

11

1 1 2

2 2 2

1 1 ;

2

;

dx a y z L x

dt p

Q dy T x y a z dt

Q dz x y a z T . dt

=

= −

=

(17)

2 2 0

x c rω

= is dimensionless speed; ω0 is engine angular speed; 1 3 2

2 2 0

MP

T = ρA r ω is dimensionless moment of the pump wheel; 2 3 2

2 2 0

MT

T = ρA rω is dimensionless moment on the turbine wheel, 1 1 3

2 2

Q J

A r

= ρ is dimensionless moment of inertia of the pump side; 2 2 3

2 2

Q j

A r

= ρ is dimensionless

moment of inertia of the turbine side;

2

p r

=Φ is dimen- sionless parameter length.

Fig. 3 Flowchart to solving dimensionless equation system The system (17) becomes then:

( )( )

( )

( )

2 2 2 2

1 11 1 11

2

1 1 1 1 1 1 1

11 1 11

2

2 1 2 1 1 1 1

11 1 11

1 1 ;

2

;

.

i i

i i

i i

i i i i

i i

i i

i i i i

i i

x x a y z L x

t t p

y y

T Q x y a z

t t

z z

T Q x y a z

t t

+ +

+ + + + +

+

+ + + + +

+

=

= +

= − +

⎪⎩

( )

( )( )

( )

2 2 2 2

1 11 1 11

1 1 1 2 1

1 1 1 1 1

11 1 11

2 1 2 1 1

11 1 11

1 1 ;

2

;

;

i i i i i i i

i i i i

i i

i i

i i

i i

i i

i i

x x t t a y z L x

p

F x y a z

y y

T Q F

t t

z z

T Q F .

t t

+ +

+ + + +

+ + +

+

+ + +

+

=

=

⎪⎪ = +

= − +

⎪⎩

(18)

11 0

t =ωt is dimensionless time.

Start

ܫ=ܫ+ 1

I = m

ݐଵଵ,ܻ,ܼ,ܺ,ܶ,ܶ

RETURN

Yes No

௜ାଵ=+ 1

2 ∗ଵଵ௜ାଵଵଵ௜1 −

௜ାଵ=௜ାଵ ௜ାଵ௜ାଵ ଵ௜ାଵ=௜ାଵ

ଵଵ௜ାଵଵଵ௜ +௜ାଵ ଶ௜ାଵ=௜ାଵ

ଵଵ௜ାଵଵଵ௜ +௜ାଵ

ݐଵଵ,ܻ,ܼ

ܲ,ܽ,ܳ,ܳ,ܮ ݔሺ1ሻ,ܶሺ1ሻ,ܶ(1)

ܫ= 1

Figure

Fig. 2 Schematic of hydrodynamic system in the meridian  plan
Fig. 3 Flowchart to solving dimensionless equation system  The system (17) becomes then:
Fig. 9 Test of braking - modus operandi-  5. Analysis and discussion of results

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