Page 130 - Special Issue Modern Trends in Scientific Research and Development, Case of Asia

International Journal of Trend in Scientific Research and Development (IJTSRD) @ www. ijtsrd. com eISSN: 2456-6470

1. For experience II we have data:
2
V0 = 1,16 m/sec; Н0 = 40,0 m; Нтр = 55,0 м; Нвак.max. = 7,0 m; а V1, 5тр = 0, 66·0,08 = 0, 00
= 922 m/sec; tф = 2,50 sec; d = 50 mm.
V2 = V1, 5 - V* - V1, 5тр = 0, 08 – 0, 32 – 0, 00 = - 0, 24 m/sec;
According to the initial data of the second experiment, we
2
calculate: V2тр = 0, 66·0,24 = 0, 04 m/sec;
аV 0 1 l L V2, 5 = V2 - V* + V2тр = - 0,247 – 0,32 + 0,04 = 0,52 m/sec;
= 109,0 м; × = , 0 44 sec/м; V* = 0,50 m/sec; Vтр
g 2 d a 2
V2, 5тр = 0,66·0,52 = 0,18 m/sec;
2
= 0,44·0,50 = 0,11 m/sec;
V3 = V2, 5 - V* + V2,5тр = - 0,52 – 0,32 + 0,18 = - 0,66 m/sec.
*
V = V* - V 1тр = 0,50 – 0,11 = 39 m/sec; V1 = V0 – V* = 1,16 –
*
0,50 = 0,66 m/sec; Taking into account ∑ V1 = 0, we can write:
Q = 2(0, 78 + 0, 00 – 0, 04 – 0, 18) = 2·0, 56 = 1, 12 m/sec;
2
V1тр = 0,44·0,66 = 0,19 m/sec; V1,5 = V1 - V* - V1тр = 0,66 –
0,50 – 0,19 = 0,03 m/sec; R = 2V1тр + 3V1, 5тр - 4V2тр - 5V2, 5тр = 1, 56 – 0, 16 – 0, 90 = 0,
50 m/sec;
V1,5тр = 0,44·0,03 = 0,00; V2 = V1,5 - V* + V1,5тр = 0,03 – 0,50 +
2
0,00 = 0,53 m/sec; V0 – R = 1, 50 – 0, 50 = 1, 00 m/sec;

V2тр = 0,44·0,53 = 0,12 m/sec; V2,5 = V2 – V* + V2тр = - 0,53 – 2V0 + V* - Q = 3, 00 + 0, 32 – 1, 12 = 2, 20 m/sec; 2,20 = 4, 84;
2
2
0,50 + 0,12 = 0,91 m/sec.
4V* = 4·0, 32 = 1, 28 m/sec; 8V*(V0 – R) = 2, 56·1, 00 = 2, 56
Taking into account the condition (Vn) < V1 and taking into м /sec ;
2
2
account the friction forces, to determine n using formula
(18), we will first calculate: 2
(2V 0 +V * -Q ) 8- V * V (-R = ) 28,2 = , 1 51 m/sec
0
Q = 2(V1тр - V1, 5тр) = 2·0, 19 = 0, 38 m/sec;
, 2 20 + , 1 51 , 3 71
R = 2 V1тр - 3 V1, 5тр = 2·0, 19 = 0, 38 m/sec; n = = = , 2 90 =n , 2 90;
, 1 28 , 1 28
2V0 + V* - Q = 2, 32 + 0, 50 – 0, 38 = 2, 44 m/sec;

2
2
2
22 V0 + V* - Q = 2,44 = 5, 8536 м /sec ; tv = n·tф = 2,90·4,80 = 13,92 sec; по опыту tv = 11,10 sec;

4V* = 4·0, 50 = 2, 00 m/sec;
V2,90 = V2,5 – 0,40/0,50(V3 - V2,5) = - 0,52 – 0,11 = - 0,63
2
2
8V*(V0 – R) = 4, 00(1, 16 – 0, 38) = 4·0, 78 = 3, 12 m /sec ; m/sec;

2
(2V 0 +V * -Q ) 8- V * V (-R = ) 733,2 = , 1 65 m/sec. ΔН2 = 55, 56·0, 88 = 48, 9 м; по опыту ΔН2 = 49, 0 м.
0

4. CONCLUSIONS
Based on the above calculated calculations, we determine the
Taking into account the above, in conclusion, we can draw
n value:
conclusions.
, 2 44+ , 1 65 , 4 09 1. The dependence is obtained for calculating the
n = = = , 2 04 , n = 2, 04; maximum pressure during a hydraulic shock in the first
, 2 00 , 2 00
period of reduced pressure in the event of a rupture of
the continuity of the flow, taking into account the
tv = n·tф = 2,04·2,50 = 5,10 sec; from experience tv = 5,10 sec;
pressure loss due to friction in a long pressure pipeline.
V2,04 = - 0,56 m/sec; 2. The ratio for determining the number of phases during
the impact process is obtained.
ΔН2 = 93, 98(0, 39 + 0, 56) = 89, 28 м;
3. The obtained ratios are very important for the analysis
from experience ΔН2 = 86,0 м. of technical parameters of water hammer.

2. According to V experience, we have data: References
V0 = 1, 50 m/sec; Н0 = 10,0 м; Нтр = 82,5 м; Нвак.max. = 8, 0 м; [1] Rongsheng Zhu, Qiang Fu, Yong Liu, Bo He, Xiuli
а = 545 m/sec; tф = 4, 80 sec; d = 50 mm. Wang. The research and test of the cavitation
performance of first stage impeller of centrifugal
Next, we calculate: charging pump in nuclear power stations. Nuclear
1 l L a Engineering and Design. China. 2016. 74-84.
V* = 0, 32 m/sec; = , 0 66 m/sec; = 55 , 56 sec;
2 d a g [2] Bojan Ivljanin, Vladimir D. Stevanovic, Aleksandar
2
*
*
Vтр = 0, 66·0,32 = 0, 07 m/sec; V = V* - V тр = 0, 25 m/sec; Gajic. Water hammer with non-equilibrium gas
release. International Journal of Pressure Vessels and

Piping. Serbia. 2018. 229–240.
V1 = V0 - V* = 1, 50 – 0, 32 = 1, 18 m/sec;
[3] Arifjonov A. M., Djhonkobilov U. U. Water hammer in
V1тр = 0, 66·1,18 = 0, 78 m/sec; homogeneous and gas-liquid pressure pipelines.
2
Monograph. Toshkent, TIIIMSKh, 2018.- 142 p.
V1,5 = V1 - V* - V1тр = 1,18 – 0,32 – 0,78 = 0,08 m/sec;
ID: IJTSRD37970 | Special Issue on Modern Trends in Scientific Research and Development, Case of Asia Page 125