Multi-stage valve

The multi-stage closure is designed to limit cavitation. If the pressure at a certain point during the flow through the valve or pipe falls below the value of the saturated vapor pressure of the liquid, corresponding to its temperature, cavitation occurs. The cavitation bubbles suddenly disappear when they reach a higher-pressure area with the liquid flow, and cavitation wear of the material is caused.
For the formation of cavitation, it is decisive whether the liquid pressure falls below the critical value of cavitation pressure, which favorably corresponds to the saturated vapor pressure $P_T$ lies in the range of minimum pressure and pressure behind the valve, when cavitation occurs, and cavitation wear can be expected after a certain time. If the minimum pressure is greater than the saturated vapor pressure, steam cavitation will not occur.
The control valve, which must prevent cavitation, must be made for the given parameters and must be unique for the given situation (series-produced valves will not be functional for all situations). To prevent cavitation, it is necessary to use a multi-stage valve (there must be a gradual pressure drop in the valve by means of orifices), a single-stage valve is not suitable for larger pressure differences.

In the case of control valves, cavitation can develop if the condition is met:

$(P_1-P_2)\geq0,6\cdot(P_1-P_T)$
where:
$P_1$ inlet absolute static pressure $\mathrm{Pa}$
$P_2$ output absolute static pressure $\mathrm{Pa}$
$P_T$ saturated vapor pressure at a specific temperature $\mathrm{Pa}$

Pipeline speed:

$v=\cfrac{4Q}{\pi\cdot D^2}$
where:
$P_1$ inlet absolute static pressure $\mathrm{Pa}$
$P_2$ output absolute static pressure $\mathrm{Pa}$
$P_T$ saturated vapor pressure at a specific temperature $\mathrm{Pa}$
$v$ pipeline speed $\mathrm{m/s}$
$Q$ pipeline flow $\mathrm{m^3/s}$
$D$ internal pipe diameter $\mathrm{m}$

Pipeline flow:

$Q=\mu\cdot F\cdot\sqrt{2\cfrac{P_1-P_2}{ρ}}$
$\cfrac{4F}{\pi\cdot D^2}\le0,5$
where:
$Q$ pipeline flow $\mathrm{m^3/s}$
$\mu$ output coefficient $\mathrm{-}$
$F$ the flow area of the output hole $\mathrm{m^2}$
$P_1$ inlet absolute static pressure $\mathrm{Pa}$
$P_2$ output absolute static pressure $\mathrm{Pa}$
$ρ$ density $\mathrm{kg/m^3}$
$D$ internal pipe diameter $\mathrm{m}$

Output coefficient:
- Sharp-edged hole

$\mu=0,65$
$\cfrac{l}{d}=1,65$


sharp-edged-hole l d v
Fig. 1 - sharp-edged hole

- Beveled hole
$\mu=0,78$
$\cfrac{l}{d}=1,65$
$\cfrac{z}{d}=0,25$


beveled-hole z 90° v l d
Fig. 2 - beveled hole

- Rounded hole
$\mu=0,84$
$\cfrac{l}{d}=1,65$
$\cfrac{r}{d}=0,25$


rounded-hole r v l d
Fig. 3 - rounded hole

The flow area of the output hole:

$F=i\cdot\cfrac{\pi d^2}{4}$
$\cfrac{D}{50}\geq d$
where:
$F$ the flow area of the output hole $\mathrm{m^2}$
$i$ number of holes $\mathrm{-}$
$d$ diameter of the output hole $\mathrm{m}$
$D$ internal pipe diameter $\mathrm{m}$

Density:
Density $\rho\ \mathrm{[kg/m^3]}$ water depending on temperature and pressure

Pressure $\mathrm{[MPa]}$ Temperature $\mathrm{[°C]}$
0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 100°
0,1 999,8 999,7 998,2 995,6 992,2 988,1 983,2 977,8 971,8 965,3 -
0,25 999,9 999,8 998,3 995,7 992,3 988,1 983,3 977,8 971,9 965,3 958,4
0,5 1000 999,9 998,4 995,8 992,4 988,2 983,4 978 972 965,5 958,5
1 1000,3 1000,1 998,6 996 992,7 988,4 983,6 978,2 972,2 965,7 958,8
1,5 1000,6 1000,4 998,8 996,3 992,9 988,6 983,9 978,4 972,4 966 959
2 1000,8 1000,6 999,1 996,5 993 988,8 984,1 978,6 972,7 966,2 959,2
2,5 1001,1 1000,8 999,3 996,7 993,3 989,1 984,3 978,9 972,9 966,4 959,5
3 1001,3 1001 999,5 996,9 993,4 989,2 984,5 979,1 973,1 966,6 959,7
3,5 1001,6 1001,3 999,8 997,2 993,7 989,5 984,6 979,2 973,3 966,8 960
4 1001,8 1001,6 1000 997,4 993,9 989,7 984,9 979,5 973,5 967,1 960,2
4,5 1002,1 1001,8 1000,2 997,6 994,1 989,9 985,1 979,7 973,8 967,3 960,4
5 1002,3 1002 1000,4 997,8 994,3 990,2 985,3 979,9 974 967,6 960,6
6 1002,8 1002,5 1000,9 998,3 994,8 990,6 985,8 980,4 974,5 968 961,1
7 1003,3 1003 1001,3 998,7 995,2 991 989,2 980,8 974,9 968,4 961,5
8 1003,8 1003,4 1001,8 999,1 995,6 991,5 986,6 981,3 975,3 968,9 962
9 1004,3 1003,9 1002,2 999,6 996,1 991,9 987,1 981,6 975,7 969,4 962,5
10 1004,8 1004,4 1002,7 1000 996,5 992,3 987,5 982,1 976,2 969,7 962,9
12,5 1006 1005,5 1003,8 1001,1 997,6 993,3 988,5 983,2 977,2 970,9 964
15 1007,3 1006,7 1004,9 1002,2 998,6 994,4 989,6 984,3 978,4 972 965,2
17,5 1008,5 1007,9 1006 1003,2 999,7 995,5 990,7 985,3 979,4 973,1 966,3
20 1009,7 1009 1007,2 1004,3 1000,8 996,5 991,7 986,4 980,5 974,2 967,4
25 1012,1 1011,3 1009,3 1006,5 1002,8 998,6 993,7 988,4 982,6 976,4 969,7
30 1014,5 1013,6 1011,4 1008,6 1004,9 1000,7 995,8 990,5 984,7 978,5 971,8
35 1016,9 1015,7 1013,6 1010,6 1007 1002,7 997,9 992,6 986,8 980,6 974
40 1019,3 1018 1015,7 1012,8 1009 1004,7 999,9 994,6 988,8 982,7 976,1
45 1021,6 1020,2 1017,8 1014,7 1011 1006,8 1001,9 996,6 990,9 984,7 978,2
50 1023,9 1022,3 1019,9 1016,8 1013 1008,7 1003,8 998,6 992,9 986,8 980,3
60 1028,3 1026,6 1024,1 1020,8 1017 1012,6 1007,8 1002,5 996,8 990,8 984,3
70 1032,7 1030,7 1028,1 1024,7 1020,8 1016,4 1011,5 1006,3 1000,7 994,6 988,3
80 1037 1034,9 1032 1028,5 1024,6 1020,2 1015,3 1010,1 1004,4 998,5 992,3

Saturated vapor pressure at a specific temperature:

Temperature Saturated vapor pressure
$\mathrm{°C}$ $\mathrm{Pa}$
0 611,3
10 1228,1
20 2338,8
30 4245,5
40 7381,4
50 12344
60 19932
70 31176
80 47373
90 70117
100 101320

Recommendation:
- The individual screens must be apart min. $5d$.
- The individual outlets located on one screen must be spaced apart min. $3d$.
- The individual outlet openings between the screens must not overlap.

where:
$d$ diameter of the output hole $\mathrm{m}$

Example 1:
We have to determine the output absolute static pressure at which cavitation does not occur with the following parameters: $P_1=65000000\ \mathrm{Pa}$, $P_T=2338,8\ \mathrm{Pa}$.

$P_2=P_1-0,6(P_1-P_T)=65000000-0,6(65000000-2338,8)=26001404,28\ \mathrm{Pa}$

Example 2:
We have to determine the input absolute static pressure at which cavitation does not occur with the following parameters: $P_2=101325\ \mathrm{Pa}$, $P_T=2338,8\ \mathrm{Pa}$.

$P_1=\cfrac{P_2-0,6P_T}{0,4}=\cfrac{101325-0,6\cdot2338,8}{0,4}=249804,3\ \mathrm{Pa}$

Literature:
- Prof. Ing. Jaromír Noskievič, DrSc a kolektiv: Kavitace v hydraulických strojích a zařízení.
- R. Mareš: Tabulky termodynamických vlastností vody a vodní páry.
- V. Kolář, S. Vinopal: Hydraulika průmyslových armatur. SNTL 1964.