ME 309 Fall 2008
Section 2
Homework # 29
Due Friday 7 November 2008
Problem 291
.
A centrifugal pump is used to pump water from a reservoir.
The
centerline of the pump is 2.2
m
below the water level in the reservoir.
The pipe is
PVC with and ID of 24
mm
and negligible average roughness.
The pipe length
from the submerged inlet to the pump inlet is 2.8
m
. There are two minor losses in
the piping system from the pipe inlet to the pump inlet:
a sharpedged inlet with
K
= 0.6 and a flanged elbow with
K
= 0.3.
The pump’s NPSHR is provided by the
manufacturer as
NPSHR = 2.2
+0.0013 Q
2
with
Q
in
m
3/
s
and NPSHR in
m
.
Estimate the maximum volume flow rate that may be pumped without cavitation
when the reservoir is at a temperature of 300
K
..
Given
:
Pump 2.2 m below reservoir
Smooth PVC Pipe
D = 0.24 m
L = 2.8 m
Inlet K = 0.6, elbow K = 0.3
NPSHR = 2.2
+0.0013 Q
2
Find
:
Max flow rate without cavitation
Assumptions
:
Steady, incompressible
Properties
:
rho
997,
mu 8.91e4,
p
v
= 3169 Pa
p
atm
= 101325 Pa
Solution
:
Mechanical energy equation:
(
)
[
]
tot
gh
z
z
g
u
u
p
p
W
Q
ρ
ρ
+

+

+

=

1
2
2
1
2
2
1
2
2
1
&
&
No heat and no work up to pump.
u
1
= 0,
p
1
= 101325, z
2
– z
1
= 2
m
Stagnation pressure at pump inlet:
[
]
tot
gh
z
z
g
p
u
p
ρ
ρ



=
+
1
2
1
2
2
2
2
1
[
]
tot
v
v
h
z
z
g
p
p
g
p
g
u
g
p
NPSHA




=

+
=
1
2
1
2
2
2
2
2
ρ
ρ
ρ
Know everything but h_tot—evaluate:
+
=
∑
s
K
D
L
f
g
u
h
D
tot
'
2
2
2
Minor losses:
9
.
0
3
.
0
6
.
0
=
+
=
∑
K
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Use Colebrook equation for friction factor:
D
2
10
0.9
0.25
,
5.74
log
3.7
D
f
D
Re
ε
=
+
Smooth pipe—use with eps = 0
Need Reynolds number to compute friction factor
Solve for a sequence of flow rates and compare with NPSHR equation:
Set u
2
= 2 and increment by two:
See Table below:
Table of NPSHR vs NPSHA
Problem 291 NSPH
L
2.8
2.8
2.8
2.8
2.8
2.8
2.8
2.8
ID
0.24
0.24
0.24
0.24
0.24
0.24
0.24
0.24
K_elb
0.3
0.3
0.3
0.3
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 Spring '08
 MERKLE
 Fluid Dynamics, pump inlet, Orifice plate

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