Brochure

p
stat
p
tot
p
dyn
p
tot
p
stat
p
stat
p
tot
Q
p
dd
3232
2. Performance curves
2.2 Pressure
Pressure (p) is an expression of force per unit area and is split into static and
dynamic pressure. The sum of the two pressures is the total pressure:
[ ]
Pa
ppp
dyn
(2.1)
(2.2)
(2.5)
(2.6)
(2.7)
stattot
+ =
[ ]
PaV
2
1
2
1
2
1
p
2
dyn
⋅ ⋅ = ρ
[ ]
Papppp
geodynstattot
∆ + ∆ + ∆
p∆
p∆
∆ =
[ ]
Papp
stat, instat, outstat
− =
[ ]
PaVV
2
in
2
outdyn
⋅⋅−⋅ ⋅ = ρ ρ
(2.8)
2
1
[
]
Pa
D
1
D
1
4
Q
p
4
in
4
out
2
dyn








− ⋅








⋅ ⋅ =
π
ρΔ
(2.9)
[ ]
Pagzp
geo
⋅ ⋅ ∆ = ∆ ρ
(2.10)
(2.3)
(2.4)
(2.11)
(2.13)
(2.14)
(2.12)






= ⋅ + +
2
22
s
m
Constantzg
2
V
p
ρ
[ ]
Pappp
barrelabs
+ =
[ ]
m
g
p
H
tot
⋅
=
ρ
Δ
[ ]
WQpQgHP
tothyd
⋅ ∆ = ⋅⋅ ⋅ = ρ
[ ]
⋅ 100 %
[ ]
⋅ 100 %
[ ]
⋅ 100 %
=
2
hyd
hyd
P
P
η
=
1
hyd
tot
P
P
η
[ ]
WP
2
P
1
P
hyd
> >
(2.15)
(2.16)
(2.17)
(2.17a)
(2.18)
(2.19)
⋅⋅=
hydmotorcontroltot
ηηηη
( )
[ ]
m
g
pp
NPSH
vapourabs,tot,in
A
⋅
−
=
ρ
[ ]
mNPSH = NPSH
3%
NPSH
RA
0.5 +>
NPSH
A
>
[ ]
mNPSH = NPSH
3%
or
R
S
A
.
[ ]
m
g
p
H
g
p
NPSH
p
vapour
suction pipe
,loss
geo
bar
A
⋅
∆
− −
+
⋅ ⋅
=
ρ
ρ
9.81m
23
A
Pa 7375 3500 Pa
m 3
sm992.2kg
101300 Pa
NPSH −−−
⋅ 9.81m
23
sm992.2kg ⋅ 9.81m
23
sm992.2kg ⋅
=
9.81m
23
A
47400 Pa
1 m 3 m
sm973 kg
-27900 Pa + 101000 Pa
+ 500 Pa
NPSH − −+
⋅ 9.81m
23
sm973 kg ⋅
=
6.3mNPSH
A
=
4.7mNPSH
A
=
[ ]
m
g
p
HH
g
pp
NPSH
vapour
loss, pipegeo
barstat,in
A
⋅
−−+
⋅
+ +
=
ρ
ρ
[
( )
0.5
.
ρ
.
V 1
2
where
p
tot
= Total pressure [Pa]
p
stat
= Static pressure [Pa]
p
dyn
= Dynamic pressure [Pa]
Static pressure is measured with a pressure gauge, and the measurement of
static pressure must always be done in static fluid or through a pressure tap
mounted perpendicular to the flow direction, see figure 2.3.
Total pressure can be measured through a pressure tap with the opening
facing the flow direction, see figure 2.3. The dynamic pressure can be found
measuring the pressure dierence between total pressure and static pressure.
Such a combined pressure measurement can be performed using a pitot tube.
Dynamic pressure is a function of the fluid velocity. The dynamic pressure can
be calculated with the following formula,where the velocity (V) is measured
and the fluid density ( ρ ) is know:
[ ]
Pa
ppp
dyn
(2.1)
(2.2)
(2.5)
(2.6)
(2.7)
stattot
+ =
[ ]
PaV
2
1
2
1
2
1
p
2
dyn
⋅ ⋅ = ρ
[ ]
Papppp
geodynstattot
∆ + ∆ + ∆
p∆
p∆
∆ =
[ ]
Papp
stat, instat, outstat
− =
[ ]
PaVV
2
in
2
outdyn
⋅⋅−⋅ ⋅ = ρ ρ
(2.8)
2
1
[
]
Pa
D
1
D
1
4
Q
p
4
in
4
out
2
dyn








− ⋅








⋅ ⋅ =
π
ρΔ
(2.9)
[ ]
Pagzp
geo
⋅ ⋅ ∆ = ∆ ρ
(2.10)
(2.3)
(2.4)
(2.11)
(2.13)
(2.14)
(2.12)






= ⋅ + +
2
22
s
m
Constantzg
2
V
p
ρ
[ ]
Pappp
barrelabs
+ =
[ ]
m
g
p
H
tot
⋅
=
ρ
Δ
[ ]
WQpQgHP
tothyd
⋅ ∆ = ⋅⋅ ⋅ = ρ
[ ]
⋅ 100 %
[ ]
⋅ 100 %
[ ]
⋅ 100 %
=
2
hyd
hyd
P
P
η
=
1
hyd
tot
P
P
η
[ ]
WP
2
P
1
P
hyd
> >
(2.15)
(2.16)
(2.17)
(2.17a)
(2.18)
(2.19)
⋅⋅=
hydmotorcontroltot
ηηηη
( )
[ ]
m
g
pp
NPSH
vapourabs,tot,in
A
⋅
−
=
ρ
[ ]
mNPSH = NPSH
3%
NPSH
RA
0.5 +>
NPSH
A
>
[ ]
mNPSH = NPSH
3%
or
R
S
A
.
[ ]
m
g
p
H
g
p
NPSH
p
vapour
suction pipe
,loss
geo
bar
A
⋅
∆
− −
+
⋅ ⋅
=
ρ
ρ
9.81m
23
A
Pa 7375 3500 Pa
m 3
sm992.2kg
101300 Pa
NPSH −−−
⋅ 9.81m
23
sm992.2kg ⋅ 9.81m
23
sm992.2kg ⋅
=
9.81m
23
A
47400 Pa
1 m 3 m
sm973 kg
-27900 Pa + 101000 Pa
+ 500 Pa
NPSH − −+
⋅ 9.81m
23
sm973 kg ⋅
=
6.3mNPSH
A
=
4.7mNPSH
A
=
[ ]
m
g
p
HH
g
pp
NPSH
vapour
loss, pipegeo
barstat,in
A
⋅
−−+
⋅
+ +
=
ρ
ρ
[
( )
0.5
.
ρ
.
V 1
2
where
V = Velocity [m/s]
ρ = Density [kg/m
3
]
Dynamic pressure can be transformed to static pressure and vice versa. Flow
through a pipe where the pipe diameter is increased converts dynamic pressure
to static pressure, see figure 2.4. The flow through a pipe is called a pipe flow, and
the part of the pipe where the diameter is increasing is called a diusor.
Figure 2.4: Example of conversion of
dynamic pressure to static pressure in
a diusor.
Figure 2.3: This is how static pressure p
stat
,
total pressure p
tot
and dynamic pressure
p
dyn
are measured.