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                                                               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



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