h2 30.4 2
same configuration as the original. This allows
=
110 35.6
much greater flexibility in operations due to the
greater available pressure range for dredging.
∴ h2 = 80.2 kPa @ 1280 rpm (11.6 psi).
The next factor to be examined is the power
requirement. For this, the classic fluid power
equation is used:
Adjusting for speed using the head/speed rela-
tionship,
P = (0.9751)pQ
(A10)
h2 n2
2
=
where P = required power (kW)
(A9)
h1 n1
p = system pressure (MPa)
Q = fluid flow rate (L/s).
where n1 is the original impeller rotational rate,
The impeller drive motor requires about 0.065
and n2 is the rate of interest,
liters per revolution. Table A1 illustrates the vari-
=
1800 2
h2
ous power requirements for different system con-
80.2 1280
figurations.
∴ h2 = 159 kPa (16.4-m head)
Table A1. Slurry pump power
@ 1800 rpm (23.3 psi @ 53.8 ft).
requirements.
Impeller
The next series of calculations are for a 330-
speed
Pressure
Flow
Power
mm (13-in.) trim. The equations used are the
(rmp)
(MPa)
(L/s)
(kW)
same, giving us
1280
24.1
1.4
32.9
h2 33.0 2
1280
31.0
1.4
42.3
=
(Ref. A8)
110 35.6
1500
24.1
1.6
38.6
1800
24.1
1.9
46.3
1800
31.0
1.9
59.5
h2 = 94.5 kPa @ 1280 rpm (13.8 psi)
(Ref. A9)
A total of 74.5 kW of power is available at the
∴ h2 = 187 kPa (19.2-m head)
electric motor driving the hydraulic pumps. Two
@ 1800 rpm (27.3 psi/62.9 ft).
other systems are driven off the auxiliary hydrau-
lic pump. The auger motor requires about 8.9 kW
Finally, for the 35.6-cm trim, we need adjust for
at full power, 80% efficiency, and the traverse
impeller speed only:
drive requires about 0.4 kW at the same condi-
tions. Using these numbers, a total of 65.2 kW is
=
1800 2
h2
available to drive the slurry pump. Using an effi-
(Ref. A9)
94.5 1280
ciency factor of around 80%, the greatest power
utilization comes at an impeller speed of just be-
h2 = 218 kPa (22.3-m head)
low 1500 rpm with line pressure at 31 MPa (4500
@ 1800 rpm (31.6 psi/73 ft).
psi). These are the conditions for which the 35.6-
cm impeller works best.
Referring to Figure A1, we should be capable of
Finally, the net positive suction head available
around 45 m (150 ft) of head. Lack of line resis-
(NPSHA) needs to be examined to ensure cavita-
tance is the probable cause of this differential. Re-
tion does not occur. The equation normally used
ferring back to Figure A2, the outlet head is calcu-
is
lated for an impeller speed of 1500 rpm:
NPSHA = ha + hs hfs hvp
(A11)
h2 1500 2
=
where ha = atmospheric head
(Ref. A9)
110 1280
hs = static suction head
h2 = 151 kPa (15.6-m head)
hvp = vapor pressure head.
@ 1500 rpm (22 psi /51 ft).
Atmospheric pressure is taken as 10.3 m. For stat-
ic suction head, the intake for the pump is 0.48 m
If the 35.6-cm impeller option is chosen, the outlet
below water level. Friction head loss is calculated
pressure should be about 151 kPa (22 psi) for the
23
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