Page 128 - Special Issue Modern Trends in Scientific Research and Development, Case of Asia

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the check valve moves in the pipeline towards the pool. This The duration of the first period of reduced pressure without
speed decreases from phase to phase (friction forces are not regard to friction forces is
taken into account here) until it reaches zero. tv = n t ф. (11)

The check valve has a vacuum space filled with air bubbles It is possible to determine over the geodetic pressure and
and steam, the pressure in it will be equal to, i.e. H = - Hвак. the duration of the first period of reduced pressure, taking
max., the velocity of the separated water mass of the flow in into account the head loss due to hydraulic resistance along
the subsequent phases according to (2) will accordingly be the length of the discharge pipeline. We use the Darcy-
equal to: Weisbach formula to determine the head loss
V 1 =V 0 -V *  L V 2
 H тр = l , (12)
V 2 =V 0 -3V *  d 2 g

V 3 =V 0 -5V *  . (4) where λ – is the coefficient of hydraulic friction resistance,
.......... .........  d – is the discharge line diameter.


V =V - ( -12i V )
i 0 * Then the loss of kinetic energy will be equal to:
We determine the total values of the traversed paths g 1 L l
2
of the water mass for each phase, taking into account that V = H = × × V . (13)
тр тр
this path first increases to a certain value S, and then a 2 a d
decreases to zero. Then, equating to zero the indicated sum,
we find the number of vacuum spaces n, i.e. At the moment of rupture of the continuity of the flow
through the pipeline from the place of the rupture of
n
t V =0 propagation, the wave of reduced pressure (Н0 + Нвак.max),
ф ∑ 1
=1 which corresponds to the instantaneous velocity V*.
i

or This speed, formed due to the deformation of the walls of the
pipeline and changes in the density of water in it, propagates
n
∑ 1 =n ( -nVV 0 * ) =0 , (5) through the pipeline.
V
=1
i
Calculating the loss in speed by formula (13), through each
From it half-phase we will have
*
*
V = V* - V тр. (14)
V
n = 0 . (6)
V * where
1 L
* 2
Substituting the value of n in (4), we obtain the velocity of V тр = × V .
*
the water mass at the moment of its impact on the check 2 a
valve, equal to
Vn = -(V0 – V *) = -V1 . (7) Obviously, for the entire time that there is a vacuum space
(in the case under consideration, at the beginning of the

pressure pipeline, at the check valve), there will be Н= -
In this case, the additional pressure of the water hammer in
excess of the geodesic pressure will be equal to Нвак.max., and in the last section of the pressure pipeline
adjacent to the pressure basin of sufficiently large
aV
D H = D H + D H доб. = 0 , (8) dimensions positive pressure equal to atmospheric pressure
2
1
g plus the immersion of the last section under the pressure
basin horizon. Based on this, in the initial section of the
pressure pipeline (at the check valve), the instantaneous
where ΔH1 = H0 + Hвак.max. the largest drop in shock pressure
compared to geodesic; velocity during the rupture of the pressure flow changes by
*
the value V*, and in the last section by the value V = V* - V тр
*
a a
D H = V = (V - V ). (9) (taking into account the speed loss for half phase of water
доб . 1 0 *
g g hammer).

The largest value of the shock pressure will be equal to: Therefore, at the moment of restoring the continuity of the
flow, the super geodetic pressure of the hydraulic shock will
aV 0
H макс. = H + D H = H + . (10) be equal to
0
2
0
g a * a a *
2
D Н = V + V = (V + V n ) , (15)
n
When the water supply through the pump is instantaneously g g g
interrupted with the formation of a discontinuity in the flow
at the check valve in the pipeline, an increased pressure a *
above the geodetic pressure arises, equal to [8]: where V the value of the pressure increase, equal to the
g
aV
D H = 0 . previous pressure decrease relative to the geodetic, taking
2
g
ID: IJTSRD37970 | Special Issue on Modern Trends in Scientific Research and Development, Case of Asia Page 123