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International Journal of Trend in Scientific Research and Development (IJTSRD) @ www. ijtsrd. com eISSN: 2456-6470
here k – the number of the phase from which the mass of
* a
into account the loss in speed V for half the phase; V is water begins its reverse movement, and the speed calculated
n
g
by formula Vi тр (i = 1; 1,5; …; n – 0,5) (13).
the value of the additional shock pressure obtained as a
result of a water hammer a mass of water at a speed of Similarly to formula (5), the algebraic sum of all velocities
V about a check valve. (17) for the first period of reduced pressure is equal to zero:
n
(2n – 1)V0 – n(2n - 1)V* - (2n - 2)V1тр – (2n - 3)V1,5тр - (2n -
4)V2тр –
In this case, the maximum pressure of the water hammer
will be equal …. – [2n – 2(k – 0,5)]V(k-0,5)тр + (2n – 2k)Vk тр + [2n – 2(k +
0,5)]V(k + 0,5)тр + … = 0
a *
Н = Н + D Н = Н + ( V + V ) . (16)
макс 0 2 0 n
g or
2
2V* n – (2V0 + V* - Q)n + (V0 - R) = 0, (18)
To determine the duration of the first period of reduced
where
pressure and the velocity Vn of the water mass at the
Q = 2(V1тр + V1,5тр + … + V(k-0,5)тр – Vkтр – V(k+0,5)тр – …)
moment it hits the check valve, we will write expressions for
the column velocity for each subsequent half of the water R = 2V1тр + 3V1,5тр + 4 V2тр + … + 2(k – 0,5)V(k-0,5)тр – 2k Vkтр –
hammer phase. - 2(k + 0,5)V(k+0,5)тр - …
Residual speed value, i.e. the velocity of the mass of water at Solving the quadratic equation (18) with respect to n, we
the beginning of the first phase is equal to: find the number of phases of a water hammer (during which
there was a cavitation-vacuum space) equal to
V1 = V0 – V*.
= 1 (V2 ) + 2 ( +V 2 - ( )
)V8
For the next half phase, this speed will be: n 4 V 0 +V * -Q V 0 * -Q * V -R
V1,3 = V1 – V* - V1 тр = V0 - 2V* - V1 тр, *
, (19)
and at the beginning of the second phase, the speed value
here only a positive value of the root is taken by equation
will be:
V2 = V1,5 - V* - V1,5 тр = V0 - 3V* - (V1 тр + V1,5 тр) (18), since the plus sign satisfies the condition of the
problem being solved.
and so on.
Thus, the velocities of the mass in the next half of the phase Having found n, from (17) we can determine the water mass
velocity at the moment of its impact on the check valve.
will respectively be equal:
-V , 3. STUDIES RESULTS
V 1 =V 0 * Below are calculations to determine the magnitude of the
V 5 , 1 =V 0 - V2 * -V 1 тр , shock pressure and the duration of the first period of
reduced pressure, taking into account the hydraulic
V 2 =V 0 - V3 * - ( V 1 тр +V 5 , 1 тр ) resistance along the length of the pipeline in the event of a
, (17) rupture of the flow continuity. The calculation results are
.......... .......... .......... .......... .. compared with the results of the experiments of D.N.
V =V - ( - Vn 12 ) - V( +V +....... + Smirnov [6].
n 0 * 1 тр 5 , 1 тр
-.....
+V ( -k 0 5 , ) тр -V k тр V n ( - 5,0 тр ) As can be seen from Table 1, comparative calculations by the
proposed method with the experimental data of D.N.
Smirnov give satisfactory results.
Table 1
ΔН2 tv
по
аV
№ of Нвак. 0 according to according to
V0, Н0 Нтр L а, ΔН1 по опытам расче-
Smirnovs max. g tф Smirnov's the author
м/с m/sec Смирного там
experiments experiments calculations
автора
М М Sec
1 0,42 36,5 7,0 - 1148 39,4 922 35,0 33,0 33,8 2,49 2,85 2,69
2 1,16 40,0 55,0 7 1148 109,0 922 47,0 86,0 89,3 2,50 5,10 5,10
3 1,32 21,0 70,5 7 1180 105,6 787 28,0 82,0 74,6 3,00 8,10 8,22
4 1,44 17,0 75,0 7 1176 91,2 622 24,0 56,0 61,5 3,80 9,00 9,72
5 1,50 10,0 82,5 8 1170 83,5 545 18,0 49,0 48,9 4,80 11,10 13,92
6 1,69 5,0 87,0 8 1165 71,7 416 13,0 32,0 36,0 5,60 14,70 13,10
7 1,22 36,5 56,0 7 1148 114,7 922 43,5 88,5 89,3 2,49 5,40 5,55
8 0,93 36,5 36,5 6,5 1148 87,8 922 43,0 64,5 71,4 2,49 4,50 4,68
9 0,76 37,0 22,0 6 1148 71,3 922 43,0 61,0 62,0 2,49 3,90 3,96
10 0,58 37,0 14,0 5 1148 54,3 922 42,0 46,0 46,0 2,49 3,45 3,19
11 0,42 37,0 6,5 - 1148 39,4 922 35,0 33,8 2,49 2,49 2,85 2,69
ID: IJTSRD37970 | Special Issue on Modern Trends in Scientific Research and Development, Case of Asia Page 124