Brochure
3535
2.5 Dierential pressure across the pump - description of dierential pressure
2.5.1 Total pressure dierence
The total pressure dierence across the pump is calculated on the basis of
three contributions:
[ ]
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
Δp
tot
= Total pressure dierence across the pump [Pa]
Δp
stat
= Static pressure dierence across the pump [Pa]
Δp
dyn
= Dynamic pressure dierence across the pump [Pa]
Δp
geo
= Geodetic pressure dierence between the pressure sensors [Pa]
2.5.2 Static pressure dierence
The static pressure dierence can be measured directly with a dierential
pressure sensor, or a pressure sensor can be placed at the inlet and outlet
of the pump. In this case, the static pressure dierence can be found by the
following expression:
[ ]
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
2.5.3 Dynamic pressure dierence
The dynamic pressure dierence between the inlet and outlet of the pump
is found by the following formula:
[ ]
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