Brochure

3636
2. Performance curves
In practise, the dynamic pressure and the flow velocity before and after the
pump are not measured during test of pumps. Instead, the dynamic pressure
dierence can be calculated if the flow and pipe diameter of the inlet and
outlet of the pump are known:
[ ]
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
The formula shows that the dynamic pressure dierence is zero if the pipe
diameters are identical before and after the pump.
2.5.4 Geodetic pressure dierence
The geodetic pressure dierence between inlet and outlet can be measured
in the following way:
[ ]
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
Δz is the dierence in vertical position between the gauge connected to the
outlet pipe and the gauge connected to the inlet pipe.
The geodetic pressure dierence is only relevant if Δz is not zero. Hence,
the position of the measuring taps on the pipe is of no importance for the
calculation of the geodetic pressure dierence.
The geodetic pressure dierence is zero when a dierential pressure gauge
is used for measuring the static pressure dierence.