Volume 6, Issue 3, March – 2021 International Journal of Innovative Science and Research Technology

ISSN No:-2456-2165

Determination of the Magneto Hydraulic Pusher’s

Electromagnet Armature-Piston Attraction Time to
the Core Using Mathematical Modeling
Rusudan Bitsadze Simon Bitsadze
Department of Mathematics Department of Engineering Graphics and
Georgian Technical University Engineering Mechanics
Tbilisi, Georgia Georgian Technical University
Tbilisi, Georgia

Abstract:- Small lifting time of pusher’s rod used as a When two flat surfaces approach each other in the di-
drive gear in vehicles and machinery is one of the key rection of their normal, a thin liquid layer located between
factors of their reliable operation, and increasing the these surfaces acts as a shock-absorber, that’s why a certain
productivity and durability. In its turn, the push rod time is necessary for liquid extrusion. In case of heavy thick-
lifting time depends on both electromagnetic characteris- ness of a liquid layer located between these two surfaces,
tics of pusher’s electromagnet and on those mechanical when these two surfaces approach each other, a resistance to
or hydraulic resistances that are predetermined by their motion is sufficiently small and these two surfaces
design features of the pusher. One of the reasons of push quickly approach each other, while reduction of liquid layer
rod lifting time increment is the resisting force of work thickness causes increase in the liquid extrusion resistance
fluid extrusion situated between the armature-piston and force and the approach velocity of these two surfaces re-
the electromagnet core. In order to make the magneto duces.
hydraulic pusher design even more perfect it is desirable
to determine armature-piston attraction time taking hyd- In case of membrane magneto hydraulic pusher elabo-
raulic resistance into account. In the work there is rated by us [2], when the armature of pusher’s electromagnet
established time distribution of pressure in work fluid attracts to the core, causing the extrusion of a working fluid
area, is calculated the work fluid resistance when arma- located between ring-shaped faces of armature and core, if
ture-piston attracts to the core and there is determined the liquid layer thickness is small, a liquid pressure on the
attraction time with consideration to hydraulic ring surface circles of maximum and minimum radius is
resistance. equal to the magnitude of work chamber pressure. In all
other MHPs elaborated by us [3-7], when armature-piston
Keywords:- Magneto Hydraulic Pusher; Armature-Piston; attracts to the core a liquid is extruded through holes of their
Core; Boundary Conditions; Integration. ring-shape faces. That’s why, in case of liquid extrusion the
fluid pressure at the minimum-radius circles of the ring-
I. INTRODUCTION shape faces equals to the pressure value in the work cham-
ber, i.e. to the minimum pressure in the liquid to be ex-
Different kinds of pushers find manifold use in many truded. In its turn, a liquid pressure at the maximum-radius
industry branches, where there is a necessity of electric pro- circle of the ring-shape face equal to the maximum pressure
cesses transformation into mechanic ones, in particular into in the liquid to be extruded.
rectilinear translation movement [1]. Push rod movement ti-
me when lifting is the important technical characteristic of We set a goal to determine the movement time of ar-
the pusher. mature-piston during attraction, taking into account liquid
extrusion resistance force.
In magneto hydraulic pushers (MHPs) of any design a
MHP electromagnet switches on right after the electric ener- II. MAIN RESULTS
gy supply to pushers, and when electromagnet armature (in
membrane MHP) or armature-piston (in membraneless When armature-piston attracts to the core, the pressure
MHP) completely attracts to the electromagnet core, it squ- P in the liquid located between them can be described by
eezes out the work fluid (oil) situated between them, forces the equation [8]
it into the push rod sub-piston area and lifts push rod by
working stroke value.  2 P  2 P 12 dh
  3  , (1)
x 2 y 2 h dt
In case of complete attraction of armature or armature-
piston they fit tightly by upper flat ring-shape butt-end
surface to flat ring-shape butt end of the core and stay in this where  – is an process fluid viscosity; h – armature
position unless electricity supply to MHP cuts off. stroke value; t – movement time during armature attraction.

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Volume 6, Issue 3, March – 2021 International Journal of Innovative Science and Research Technology
ISSN No:-2456-2165
The following boundary conditions will take place 
A0    An cos n  Bn sin n 
n 1
P   r  P1 , (2)  n
r
     Cn cos n  Dn sin n 
n 1  R 
P   R  P2 , (3) 3 h
 C0 ln R  3 R 2  P2 .
h
where the following notation is used
P1  Pmin , A0 , C0 coefficients are determined from the system:
P2  Pmax , 3 h 2 1
2
A0  C0 ln r 
h3
r 
2  Pd  P ,
0
1 1
(7)
 is a polar radius, r and R are the small and big radiuses
2
of the ring. If we use polar coordinates: 3 h 2 1
 x   cos  , A0  C0 ln R 
h 3
R 
2  P d  P .
2 2 (8)
 0
 y   sin  ,
If we subtract (7) from (8) we obtain that
Equation (1) is written in the following form:
 2 P 1 P 1  2 P 12 dh 3 h 2
   3  . (4) C0  ln R  ln r    R  r 2   P2  P1 ,
 2    2  2 h dt h3

In order to solve the Dirichlet boundary problem (2), from where

3 h 2 2
(3), (4) we note that the general solution of equation is a sum
of general P0 solution of Laplace equation, i.e.
P2  P1 
h3
R  r 
C0  , (9)
corresponding homogenous equation and particular solution ln R  ln r
P of Reynolds non-homogenous equation. As is known the
solution of Laplace equation in the collar neighborhood has and taking (7) into account
to be sought in the following form:
3 h 2 2
P2  P1 
h3
R  r  3 h

n

A0  P1   ln r  3 r 2 . (10)
P0  A0      An cos n  Bn sin n  ln R  ln r h
n 1  R 

n
An , Bn , Cn and Dn coefficients are determined by the equ-

r
    Cn cos n  Dn sin n   Co ln  . (5) ations
n 1     r  n 2
1
  An  Cn   P1 cos n d  0,
 R   0
Using a direct verification we will make sure that we can  n 2
12 dh  2  r 1
 0
take 3  in the role of P .  A n    C n  P2 cos n d  0
h dt 4  R

Thus, the general solution of (4) equation is sought in and

the following form  r  n 2
1
 0 1


n
  n B  D  P sin n d  0,
P0  A0      An cos n  Bn sin n 
n
 R 
n 1  R   n 2
 r 1
 n  R  Dn    P2 sin n d  0.
B 

r
n
   0
    Cn cos n  Dn sin n 
n 1   
From these four latter ratios we obtain that

3 h 2 An  Bn  Cn  Dn  0 .
 C0 ln   . (6)
h3
Thus, the solution of Dirichlet problem for Reynolds
In order that the pressure value in the hydrosystem equation will be of the following form
would satisfy boundary conditions (2), (3), the following
ratios have to be observed based on (4)

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Volume 6, Issue 3, March – 2021 International Journal of Innovative Science and Research Technology
ISSN No:-2456-2165
3 h 2 2
R  r 
Note that when using the partial integration formula
P2  P1 
h3 3 h R
 R
  2 
P  P1  ln r  3 r 2
ln R  ln r h r r
 ln d   r r d  2 
ln
3 h 2 2
P2  P1  3  R  r   2
R R
2 r 1
h 3 h  ln     d
 ln   3  2 , 2  r
ln R  ln r h r 2 r r

which we re-write as follows R2 R R2  r 2

3 h  ln  . (13)
P2  P1  3  R 2  r 2  2 r 4
P  P1  h  ln   ln r 
ln R  ln r Taking (13) into account, from (12) we obtain
F   P1  R 2  r 2 
3 h  2

h3
   r2  . (11)   P2  P1   2 R R 2  r 2 
  R ln  
ln R r  r 2 
3 h  R 2  r 2   2 R R 2  r 2 
(11) shows the time distribution of pressure along the radius
(from r to R) in the process fluid area to be extruded.   R ln  
h3 ln R r  r 2 
In the dimensionless form the (11) equation will be
written as follows
3 h  R  r 
2 2 2

h3 h3  3 
P  P1 , (14)
 hr 2  hr 2 h 2
 
 3   R   1 ln 
2
i.e. in dimensionless form
h3  2 1  r
P  P ln  r   r
    h3 P1   R  
2
 h3
F  
2    1
h r 2
R R  hr 4
 hr   r  
ln ln 
r r
   2 
 h  P2  P1    R 2 R  1  R 2  
3

  ln      1 
3     1 .
 r    hr 2 ln R r   r  r  2  r  

 

The resistance F of the displaced process fluid, when


3  R r   1
2

the armature approaches the core, is calculated via the ring ln R r

D D pressure integration
F   P dx dy .
  R 2 R 1   R 2   3
2
D   R 2 
    ln      1       1 . (15)
 r 
 r 2   r   2
  r  
In polar coordinates we will have
F   P  d  d .
D
Now let us re-write (15) in the following form
h F  h3 P1   R  
3 2

 
2    1
If we move to multiple integrals, we obtain  hr  hr   r 
4 
2 R 
F   P  d  d 
 h  P2  P1    R 2 R  1  R 2  
3

 ln      1 
 hr 2 ln R r   r  r  2  r  
0 r


 3 h
 P2  P1  3  R 2  r 2 
R
   R 2 
2
 2   P1   h  ln 3
R      1
r
r


ln
r
2  r  

   R 2 
3 h 3 2  3     1
 r    R 2 R 1  R 2 
 3   r   d  .     ln     1 
(12)     
h  ln R r   r  r 2   r  
  

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Volume 6, Issue 3, March – 2021 International Journal of Innovative Science and Research Technology
ISSN No:-2456-2165
from where As h0 we have to take that thickness, in case of which
dt  3   R 
2
2
 the resistance to movement becomes so sufficient that
     1 armature’s retarded motion starts.
dh  2   r  

 
III. CONCLUSION
3  R r   1   R 2 R 1   R  2   
2

    ln      1  
ln R r  r 
 r 2   r    

In magneto hydraulic pushers [3-7], for the case of
direct current electromagnet armature-piston attraction to the
core, a resistance force of working fluid extrusion located
 F between them is established. Taking this force into account,
P   R 2 
  4  12     1 the armature-piston movement time when pusher’s
 r r
  r   electromagnet armature-piston attracts to the core is
determined. This fact can be used afterwards, when
1 projecting magneto hydraulic pushers design in order to
  P  P    R 2 R 1   R 2    1 improve their characteristics, in particular, for push rod
 2 2 1    ln      1    3 . (16)
 r ln R r   r  r 2   r   h

lifting time reduction.

REFERENCES
Via integration of (16) equation we obtain a ratio
between time and armature stroke. If armature-piston will [1]. M. P. Aleksandrov, Brakes of carrying and lifting
move from ho to h1, when time changes from to to t1 then machines. Mashinostroenie, 1976.
h1  2 [2]. О. S. Ezikashvili and S. G. Bitsadze, Magneto
 3   R  
t1 2

 dt  h  2   r   1 hydraulic pusher. Authorship certificate № 582188,

o 
to
   USSR, 1977.
[3]. S. Bitsadze and R. Bitsadze, Magneto hydraulic


3  R r   1   R  ln R  1   R 
2
2


2
  
  
pusher. Patent of invention P5536, Georgian Patent,
  r    1
2   r 
ln R r r  2012.
    [4]. S. Bitsadze and R. Bitsadze, Magneto hydraulic
pusher. Patent of invention P5869, Georgian Patent,
2013.
 F P   R 2 
  4  12     1 [5]. S. Bitsadze and R. Bitsadze, Magneto hydraulic
 r r
  r   pusher. Patent of invention P6572, Georgian Patent,
2016.
1
[6]. S. Bitsadze and R. Bitsadze, Magneto hydraulic
 (P  P )   R  R 1   R 
2
 1 2
pusher. Patent of invention P6744, Georgian Patent,
 2 2 1    ln      1    3 dh . (17)
r 2   r   h 2017.
 r ln R r   r   [7]. S. Bitsadze and R. Bitsadze, Patent of invention P2019
6975B, Georgian Patent, 2019.
Note that the armature attraction force F with an [8]. N. P. Petrov et al., Hydrodynamic theory of
adequate accuracy can be considered as a constant one for lubrication. State Technical and Creative Publisher,
sufficient small h. Based on this fact. Moscow, 1934.

1  1 1   3
2
  R 2 
t  t1  t0   2  2       1
2  ho h1   2
  r  


3  R r   1   R  ln R  1   R 
2
2 2


  
  
  r    1
ln R r  r 2   r  
  
 F  P  R 
2

  4  12     1
 r
  r   r  

1
  P  P    R 2 R 1   R 2   
 2 2 1    ln      1   .
 r ln R r   r  r 2   r  


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