Hydropneumatic Tank design for surge protection
1/ Hydropneumatic tanks
It is a closed cylindrical tank containing compressed air in its upper part and a certain volume of
water in the lower part, commonly referred to as an air tank or hydropneumatic tank. These are tanks
in which the volume of water is balanced by the operating pressure in steady state conditions. This
tank is located at the base of the discharge pipe, containing a determined quantity of water. In the
event of a vacuum in the piping, the air above the water level in the tank exerts pressure on it,
forcing it to be evacuated into the pipe. Conversely, in case of overpressure in the discharge pipe,
the tank plays an opposite role: excess water in the pipe seeks to occupy more space, and the
connection between the tank and the discharge pipe acts as a discharge path. The water then exits
through a throttling device to dampen oscillations, thus entering the tank with less energy, while the
air is compressed to make room for the water.
2/ Water hammer analysis
The analysis of this phenomenon involves calculating the surges and depressions to ensure that the
values obtained do not exceed the intended operating pressure. In the context of discharge pipes,
water hammer occurs during sudden start-ups or shutdowns of the pump system. In gravity flow
pipes, water hammer happens when valves are closed or opened abruptly.
The wave speed is given by the expression:
1
a ＝ ―――――――
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
⎛ 1
DN ⎞
+ ―――⎟
ρ ⋅ ⎜――
⎝ Ew Th ⋅ Epipe ⎠
a : wave speed (m/s)
Ew : elasticity modulus of water (Pascal) = 2.188 ⋅ 10 9 Pa
Epipe : elasticity modulus of the discharge pipe(Pascal)
Th : pipe wall thickness (m)
DN : nominal diameter of the discharge pipe
ρ : water density(kg/ m 3 )
1
If the closure is considered sudden:
If the closure is considered sudden:
a ⋅ V0
2⋅L
T < ―― ---> b ＝ ――
a
g
If the closure is considered gradual:
2 ⋅ L ⋅ V0
2⋅L
T > ―― ---> b ＝ ―――
a
g⋅T
T : time required for the pressure wave to travel (sec)
L : length of the discharge pipe(m)
V0 : flow velocity in normal condition (m/s)
b : maximum/minimum pressure head(m)
Head : total head (m)
Z0 : normal pressure (m)
ΔP : pressure head change (m)
In case of maximum pressure created by the water hammer:
ΔP ＝ Z0 + b
Z0 = Hgéo +10
In case of minimum pressure:
ΔP ＝ Z0 - b
3/ Machinery and accessories
2
Nozzle:
This is a constriction device placed at the outlet of the air reservoir. The role of the nozzle consists
of:
Damping oscillations: When water hammer occurs, it generates pressure waves within the pipeline.
Nozzle:
This is a constriction device placed at the outlet of the air reservoir. The role of the nozzle consists
of:
Damping oscillations: When water hammer occurs, it generates pressure waves within the pipeline.
These waves can bounce back and cause significant oscillations in the water flow, which can damage
the pipelines and associated equipment. The constriction device acts as a damping mechanism by
either reducing or increasing the water flow velocity, thereby decreasing or increasing the kinetic
energy and consequently mitigating the oscillations.
Pressure reduction: By decreasing the water flow velocity at the outlet of the air reservoir, the
constriction device also helps to reduce the pressure in the system. This aids in protecting the
pipelines and equipment from excessive pressures that could potentially damage them.
4/ Principle of hydropneumatic tank calculation
The Bergeron diagram will be used to determine the maximum depression and overpressure values in
the pipeline after establishing beforehand the characteristics of the air reservoir (normal regime air
volume U0) and its constriction device (nozzle). The Bergeron diagram method can be applied
regardless of the flow rate, pipe section, and discharge length.
1- The intervals follow according to the values of
2⋅L
T ＝ ――
a
2- Variation in the volume of air within the hydropneumatic tank
ΔUi ⎛⎝vi⎞⎠ ＝ S ⋅ T ⋅ Vmi ⎛⎝vi⎞⎠
with :
vi_1 + vi
Vmi ⎛⎝vi⎞⎠ ＝ ―――
2
Vmi : average velocity within interval i (m/s)
vi_1 : flow velocity within interval i-1 (m/s)
vi : flow velocity within interval i (m/s)
S : discharge pipeline section ( m 2 )
ΔUi : Variation in air volume within interval i ( m 3 )
Example :
i=1
vi_1 ＝ v0
vi ＝ v1
v0 + v1
Vm1 ＝ ―――
2
ΔU1 ＝ S ⋅ T ⋅ Vm1
3
3- The air volume U of the tank.
3- The air volume U of the tank.
Ui ⎛⎝vi⎞⎠ ＝ Ui_1 + ΔUi ⎛⎝vi⎞⎠
Example :
i=1
U1 ＝ U0 + ΔU1
U0 : Pre-selected normal condition tank volume( m 3 )
Comment :
The chosen value for U0 should be proportional to the flow rate; the higher the flow rate, the
higher the value of U0 should be as well
It's better if the value of U0 is not below 5 m 3 .
4- The pressure Z in the tank:
It is determined by Poisson's law:
⎛⎝Z0 + δ0⎞⎠ ⋅ U0 1.4
⎛
⎞
Zi ⎝vi⎠ ＝ ―――――
1.4
Ui ⎛⎝vi⎞⎠
Warning !
⎛⎝Z0 + δ0⎞⎠ ⋅ U0 1.4 Remains constant regardless of interval i.
Example :
i=1
⎛⎝Z0 + δ0⎞⎠ ⋅ U0 1.4
Z1 ＝ ―――――
U1 1.4
i=2
⎛⎝Z0 + δ0⎞⎠ ⋅ U0 1.4
Z2 ＝ ―――――
U2 1.4
Z0 ＝ Head + 10 m
δ0 ＝ R ⋅ Q 2
8⋅λ⋅L
R ＝ ――――
g ⋅ π2 ⋅ D5
δ0 : Linear head loss under normal (steady-state) conditions (m)
R : Resistance of the discharge pipeline.
λ : Roughness of the material
4
5- Calculation of velocity and head losses in the nozzle :
a) Water descent (from the reservoir to the discharge pipeline)
By convention, we assume that velocity is positive (+v) for water descent and negative otherwise
(water ascent), thus for v>0:
5- Calculation of velocity and head losses in the nozzle :
a) Water descent (from the reservoir to the discharge pipeline)
By convention, we assume that velocity is positive (+v) for water descent and negative otherwise
(water ascent), thus for v>0:
D2
K1 ＝ ―――――
2
⎛⎝Cdischarge ⋅ d⎞⎠
Comment :
- d will be chosen so that ultimately it falls between 15 and 20.
- Cdischarge : Nozzle discharge coefficient = 0.92, the formula can then be written as:
D2
K1 ＝ ――――
2
((0.92 ⋅ d))
Vnozzle_i ⎛⎝vi⎞⎠ ＝ K1 ⋅ vi
Vnozzle_i : flow velocity in the nozzle within interval i(m/s)
2
1
⋅ Vnozzle_i ⎛⎝vi⎞⎠
Δh1 ⎛⎝vi⎞⎠ ＝ C1 ⋅ ――
2⋅g
Δh1 : Head loss due to constriction within interval i if v>0 (m)
C1 : Singular head loss coefficient if v>0
C1 is determined from a graph (chart) providing the values of based on the coefficient .
2
((0.92 d))
m ＝ ――――
dnozzle 2
D
dnozzle ＝ ―
2
b) Case of water ascent (from the discharge pipeline to the tank).
v < 0:
2 ⋅ D2
K2 ＝ ――
d2
Vnozzle_i ⎛⎝vi⎞⎠ ＝ K2 ⋅ vi
2
1
⋅ Vnozzle_i ⎛⎝vi⎞⎠
Δh2 ⎛⎝vi⎞⎠ ＝ C2 ⋅ ――
2⋅g
d2
1
⋅―
m ＝ ―――
2
2
dnozzle
5
C ＝ f ((m))
6- Total Pressure Ptot :
δ ⎛⎝vi⎞⎠ ＝ R ⋅ S 2 ⋅ vi 2
δ : Linear head loss under transient flow conditions within interval i (m)
If v > 0:
Ptot ⎛⎝vi⎞⎠ ＝ Z ⎛⎝vi⎞⎠ - δ ⎛⎝vi⎞⎠ - Δh1 ⎛⎝vi⎞⎠
If v < 0:
Ptot ⎛⎝vi⎞⎠ ＝ Z ⎛⎝vi⎞⎠ + δ ⎛⎝vi⎞⎠ + Δh2 ⎛⎝vi⎞⎠
6
7- Construction principle of the BERGERON diagram:
a) Point 1R :
7- Construction principle of the BERGERON diagram:
a) Point 1R :
We first start by representing the horizontal line Z0, H = Z0. Point 1R is given by the intersection
of the normal condition velocity v0 and the horizontal line passing through the value of Z0.
In the example of the graph above, where, v0 = 1.4 m/s and Z0 = 96m (Head = 86m) we have:
1R (v=v0=v1 , H= Z0)
b) Point 2P:
Starting from point 1R, we draw a line with a slope of + a/g , Point 2P is determined by selecting a
velocity v2 such that, upon calculation, we arrive at a total pressure Ptot .
The intersection of Ptot and v2 should meet the line with a slope of(+a/g) at a point; otherwise, we
adjust the velocity and iterate the process.
c) Point 3R:
From point 2P, we draw a line with a slope of - a/g , Point 3R is the intersection of the line Z0 and
the line -a/g.
Comment : all points P and R will be plotted following the same procedure
7
Point 2P lies outside the line +a/g for a velocity v2 =0.8m/s --> Another value of v2 needs to be
chosen.
Point 2P indeed corresponds to the intersection between velocity v2 =1.03m/s and the line with a
slope of +a/g
Comment:
- By convention, we assume that the portion of the diagram located below the line Z0 corresponds to
the depression zone, whereas the portion located above Z0 corresponds to the maximum pressure
zone.
8
Comment:
- We will stop the diagram when the maximum pressure wave begins to attenuate.
Example:
In this example, the pressure wave starts to attenuate at a total pressure P total = 104.57 m, just after
reaching its maximum value, P total max = 110.63 m.
8- Sizing of the tank:
The sizing of the hydropneumatic tank is done through the following method:
From the corresponding table in the Bergeron diagram, we extract Umax ( the maximum air
volume)
Example:
9
The volume of the tank URA is given by:
The volume of the tank URA is given by:
URA ＝ 1.2 ⋅ Umax
HRA ＝ 2 ⋅ DRA
HRA : Height of the tank(m)
DRA : Diameter of the tank (m)
π ⋅ DRA 2
URA ＝ SRA ⋅ HRA ＝ ―――⋅ 2 ⋅ DRA
4
DRA ＝
3
‾‾‾‾‾‾
2 ⋅ URA
―――
π
5/ Example of a practical calculation case
Network Characteristics :
- Material: Ductile iron
Th ≔ 0.020 m
Ewater ≔ 2.188 ⋅ 10 9 Pa
kg
ρ ≔ 1000 ――
m3
D ≔ 0.8 m
Epipe ≔ 1.72 ⋅ 10 11 Pa
1
m
a ≔ ――――――――= ⎛⎝1.204 ⋅ 10 3 ⎞⎠ ―
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
s
⎛ 1
⎞
D
ρ ⋅ ⎜―― + ―――⎟
⎝ Ewater Th ⋅ Epipe ⎠
π ⋅ D2
= 0.503 m 2
S ≔ ―――
4
L ≔ 1200 m
U0 ≔ 5 m 3
Head ≔ 50 m
8⋅λ⋅L
s2
=
4.238
R ≔ ――――
――
g ⋅ π2 ⋅ D5
m5
a
―= 122.795 s
g
m3
Q ≔ 0.58 ――
s
Q
m
v0 ≔ ―= 1.154 ―
S
s
λ ≔ 0.014
Z0 ≔ Head + 10 m = 60 m
2⋅L
= 1.993 s
T ≔ ――
a
δ0 ≔ R ⋅ Q 2 = 1.426 m
i = 1:
vi_1 + vi
Vmi ⎛⎝vi⎞⎠ ＝ ―――
2
vi_1 ＝ v0
vi ＝ v1 ＝ v0
v1 ≔ v0
v0 + v1
m
= 1.154 ―
Vm1 ≔ ―――
2
s
10
ΔU1 ≔ 0 m 3
Ui ⎛⎝vi⎞⎠ ＝ Ui_1 + ΔUi ⎛⎝vi⎞⎠
U1 ≔ U0 + ΔU1 = 5 m 3
⎛⎝Z0 + δ0⎞⎠ ⋅ U0 1.4
Z1 ≔ ―――――
= 61.426 m
U1 1.4
δ1 ≔ δ0 = 1.426 m
m
Vnozzle1 ≔ 0 ―
s
Δh1 ≔ 0 m
v1 > 0
Ptot1 ≔ Z1 - δ1 - Δh1 = 60 m
i = 2:
m
v2 ≔ 0.95 ― (from the diagram)
s
m
v1 = 1.154 ―
s
U1 = 5 m 3
v1 + v2
m
= 1.052 ―
Vm2 ≔ ―――
2
s
ΔU2 ≔ S ⋅ T ⋅ Vm2 = 1.054 m 3
U2 ≔ U1 + ΔU2 = 6.054 m 3
⎛⎝Z0 + δ0⎞⎠ ⋅ U0 1.4
Z2 ≔ ―――――
= 46.996 m
U2 1.4
v2 > 0 :
d ≔ 0.2 m
D2
= 18.904
K1 ≔ ――――
2
((0.92 ⋅ d))
15 < K1 < 20
Trick !
It is highly likely that if you choose an arbitrary value for d, it will not meet the condition 15
< K1 < 20 right away. Therefore, it is advisable to choose one-quarter of D, as in all
scenarios, K1 would be equal to 18.904.
m
Vnozzle2 ≔ K1 ⋅ v2 = 17.958 ―
s
2
((0.92 d))
m1 ≔ ――――
= 0.212
dnozzle 2
D
dnozzle ≔ ―= 0.4 m
2
C1 ≔ 0.63 (refer to the chart on page 12)
1
⋅ Vnozzle2 2 = 10.359 m
Δh1 ≔ C1 ⋅ ――
2⋅g
1
⋅ Vnozzle2 2 = 10.359 m
Δh1 ≔ C1 ⋅ ――
2⋅g
δ2 ≔ R ⋅ S 2 ⋅ v2 2 = 0.966 m
Ptot2 ≔ Z2 - δ2 - Δh1 = 35.671 m
After all calculations are done, we obtain the following results :
Ptot_max ≔ 71.58 m
Ptot_min ≔ 35.55 m
Umax ≔ 7.18 m 3
Umin ≔ 4.48 m 3
The diagram concludes at interval i = 13 with a value of Ptot ≔ 69.15 m
12
Sizing of the tank:
URA ≔ 1.2 ⋅ Umax = 8.616 m 3
DRA ≔
3
‾‾‾‾‾‾
2 ⋅ URA
= 1.764 m
―――
π
HRA ≔ 2 ⋅ DRA = 3.527 m
13
6/ Introduction of the "Hydropneumatic V1.0" application
6/ Introduction of the "Hydropneumatic V1.0" application
What is the purpose of this application?
As observed in the preceding paragraphs, the calculations are laborious and require testing and
iterations. It is with this idea in mind that we (my colleague and I) developed an application using
a Python program, which significantly simplifies the calculations for the user.
1- Introduction to the interface:
The application is available in two languages: French and English.
2- Inputs
14
Inputs:
- D (m)
- d (m)
- L(m)
- U0( m 3 )
Inputs:
- D (m)
- d (m)
- L(m)
- U0( m 3 )
- Total Head(m)
-λ
3
- Q( m /s)
- Pipe thickness (m)
- Pipe material (Cast iron , Reinforced concrete, Ductile iron , Steel , PVC, etc...)
3- Outputs
15
Manually tracing the diagram can be quite tedious, especially when multiple iterations need to be
managed (multiple velocities). A simple click on the "Validate" button allows you to avoid such a
hassle.
16
The Hydropneumatic V1.0 application even automates the calculation of parameters m1, m2, C1,
and C2, which were previously determined through a chart.
By pressing "Show results" in the "Options" tab, a new window opens. This window displays the
results of v , ΔU , U, Z et P total.
17
18
The values of Umax , D tank and H tank are automatically calculated..
Outputs :
- m1
- m2
- C1
- C2
- S( m 2 )
- E pipe (Pascals)
- K1
- a(m/s)
- a/g
- T(sec)
- The diagram
- The dimensions of the hydropneumatic tank
4- Error messages
19
Before pressing "Validate," ensure that K1 falls between 15 and 20. Opt for a value of d equal
to a quarter of the pipe diamete D thus, you ensure that K1 will be equal to 18.90.
Before pressing "Validate," ensure that K1 falls between 15 and 20. Opt for a value of d equal
to a quarter of the pipe diamete D thus, you ensure that K1 will be equal to 18.90.
5- Saving of inputs / diagram
When you press "Save data" in the "File" tab, the inputs will be saved in a .txt file.
You can import the inputs by pressing "Open" in the "File" tab from a previously saved .txt file.
20
You can even save the diagram in .png format by simply pressing "Save Graph" in the "File" tab.
Made by :
Oussama Ben Rejeb
21