Brochure

H
geo
>0
p
stat, in
Reference plane
System
4343
Example 2.2 Pump in a closed system
In a closed system, there is no free water surface to refer to. This example
shows how the pressure sensor’s placement above the reference plane can
be used to find the absolute pressure in the suction line, see figure 2.13.
The relative static pressure on the suction side is measured to be p
stat,in
=
-27.9 kPa
2
. Hence, there is negative pressure in the system at the pressure
gauge. The pressure gauge is placed above the pump. The dierence in
height between the pressure gauge and the impeller eye H
geo
is therefore a
positive value of +3m. The velocity in the tube where the measurement of
pressure is made results in a dynamic pressure contribution of 500 Pa.
Barometric pressure = 101 kPa
Pipe loss between measurement point (p
stat,in
) and pump is calculated to
H
loss,pipe
= 1m.
System temperature = 80°C
Vapour pressure p
vapour
= 47.4 kPa and density is ρ = 973 kg/m
3
, values are
found in the table ”Physical properties of water”.
For this system, formula 2.16 expresses the NPSH
A
as follows:
[ ]
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
Inserting the values gives:
[ ]
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
Despite the negative system pressure, a NPSH
A
value of more than 4m is
available at the present flow.
Figure 2.13: Sketch of a closed system.