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Section 2. Machinery construction
Section 2. Machinery construction
Kaniev Djamshid, Head of Technopark Turin Polytechnic University in Tashkent, E-mail: j.kaniev@polito.uz
DETERMINATION OF THE SHOCK PULSE OF THE FLUID IN THE PRESSURE CHANNEL OF THE CONTROL VALVE
Abstract: In the article, the following issues will be discussed: increased temperature in impurity of bursting force pressure pipe hydraulic valve and mathematical forms for determination hydraulic stroke.
Keywords: pressure, control valve, hydraulic shock, hydraulic fluid flow, control valve, hydraulic impact.
The basic scheme of the physical process of the phenomenon of hydraulic impact according to the theory of N. E. Zhukovsky, is based on an instantaneous impact of the liquid on the membrane. In this case, the fluid is considered not viscous, but compressible and obeying Hooke's law, the pipeline is rigid, and the velocity of the pressure head is equal UH = -¡J2gH .
To obtain this velocity in the injection channel, it is necessary to create a pressure drop Ap, which is determined in the form of a pressure difference [7]:
U J
AP = Ph - pa =P-^d
Where u — the average flow velocity, a is the coefficient of kinetic energy, or the Coriolis coefficient, and is equal to the integral ratio:
a = — f (U fdm
Thus, we have the following equation for determining the pressure drop:
v2n 2gH Ap = ap= ap= ayH
Where y — is the specific gravity of the liquid. From equation (1), for a = 1,04 we obtain the following table for the relationship between the pressure drop and the head of the hydraulic fluid flow;
In the injection channel, due to the impact of the flow in the injection channel, flow through the throttle opening, a bypass valve with a diameter - d0.
Due to the sudden impact of two flows in the injection channel, a total hydraulic impact occurs in the bypass valve, which is perpendicular to the movement of the liquid. The impact arises from the collision of the flows in the injection channel, therefore, determine the dependence of the pressure drop on the momentum of the flow I* formed in the injection channel with a diameter of -D0. Is of great importance. For this purpose, we define the shock pulse in the flow I0, which is [2]:
1o = Po + YH
To determine the momentum of the flow, we use the Navier-Stokes equation for an incompressible viscous fluid, i. e. for hydraulic fluids. Then the equation of the number of motions of a viscous liquid has the form:
du du dp „2 pu--+ pup — = —— + vJV u
~ /-> ~ cp cp
dx dr dx
The equation of continuity of this liquid can be written in the form:
d(upr ) + d(vcppr) = o dx dr
Suppose that the pressure channel is round, axisymmetric. Then we have the following initial and boundary conditions:
uR) = 0, ddr
= 0,ucp (0) = 0,ucp (Ro) = 0
The equation of momentum with allowance for the continuity equation (3) can be reduced to the form [3]:
d( d(PUUcp) dp 2
— (pu ) +-— = —— + u V u
dx dr dx
To obtain an equation for the circular tube averaged over the thickness, we multiply the equation by rdrdd and integrate with respect to r and 0, and the inner integral of u before and then we have the following integral equation for the viscous fluid equation:
M rdr (pu 2) ^-^UA'
0 0 idx dr
2n «o ^ R0
= -J rdr + y\ de\rV2udr
0 0 dx o 0
From this, we obtain equality for the momentum of the flow:
d "f, 2 % , 1 du
2n—I (pu + p)rdr = y--
dx o r dr
Thus, we establish that the momentum of the flow is unchanged:
r=0
DETERMINATION OF THE SHOCK PULSE OFTHE FLU ID IN TH E PRESSU R E CHANN EL OF THE CONTROL VALVE
2n
s° R 2
J pu 2rdr + p—
0 2
= const
Suppose that in a shock pipeline the loss of pressure is invisible, ie:
\P - P I << 1
| eux ex.|
As is known, the first integral of the expression determines the momentum of the stream, then we will have an expression for the momentum of the stream in the following form.
I0 = 2nj pu2rdr = const.
0
According to Poiseuille's formula we have that the fluid flow is determined through the pressure drop in the form:
Q =
16^L
R - D0
(6)
Using the formula, we determine the momentum of the flow in the injection channel:
2n Ro Ro Ap 2
I0 = f def ru2pdr = 2nf p(——) [R04 - r2] rdr =
0 0 0 H
= 2np(-^)2 j0[RH -2RUr2 + r4]rdr =
= 2np(HE )2 f[RHr - 2 RUr3 + r3 ] dr =
H^L o
= 2np
TE
4^L
T}6 T}6 T}6
Ro - 2 RL + Ro 2 4 6
= 2np
Ap ) R
4^L
=p
ApR^ j
16^L I 6 2n R02
Thus, for a stream pulse, we have the expression: 16 1 ^
' •mQ
Where Q is the flow rate, i. e. the volume of the working fluid flowing through the cross section of the discharge channel, which is determined from the continuity equation, assuming that the cross section of the injection channel is a circle of radius R0, which is determined by the continuity formula:
Q = avcp = nR2„vrp
cp 0 cp
Here Ucp — is the average flow rate.
Then for the pulse, we have the following expression:
= —-pnu ] =
\2n R;
Or
T 2 2
10 =— PRUcp
Thus, the determination of the shock pulse of the liquid in the delivery channel of the control valve depends mainly on the diameter of the discharge channel and the speed of the working fluid.
References:
1. Argatov I. I. Asymptoticsolution of the problem about pressure of a solid body on a membrane // Applied Mathematics and Mechanics. - 2000. -T. 64. edition.4. -P.683-690 (InRussian).
2. Manual for usingprogramm"Calculation of hydraulic impact in closed irrigation networks on a computer" EC - 1022. - L.: Lenginro-vodkhoz, - 1980.(In Russian).
3. Sharypov K. A., Kaniev Dj., "Peak pressure of hydraulic shock during the directional control valve". Newsletter ACTA of Turin Polytechnic University in - Tashkent, - 2011. - No. 5. - P. 76-79.
4. Kaniev Dj, Sharypov K.A, Yusupov A. R., "Definition of the shock pulse of liquid in blowing channel of the hydraulic distributor". Newsletter of Kazakh National Technical University, - No. 1 (95). - 2013. (In Russian).
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