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less than 0,5 mm in size, and we are rather doing more of
a dredging operation where the gravel at the suction inlet is
much larger and could be termed a ‘heterogeneous slurry’.
In this case, it is probably more practical to choose the
size of suction and discharge lines, assume their lengths
from the probable location on the stream bank, and using
the principal of a minimum velocity V
L
in the suction line
to maintain the gravel in suspension, use these numbers
to determine the production rate of gravel and the pump
duty. We can then vary the suction line size to obtain the
optimum velocity.
There are a number of historical formulae, deduced
mainly from empirical testing, that allow us to determine
the minimum velocity, V
L
, required in the suction hose to
prevent settling.
Depending on the assumed average gravel size (d
50
) the
formulae used could vary from the pumping of a slurry where
the average size is small, right up to a fairly large gravel
size which then almost constitutes a dredging operation.
An initial assumption of a 6” (15,41 cm) diameter suction
hose produced an excess of material to sort; thus the suc-
tion hose has been reduced to 2” (5,25 cm) diameter. For
the purposes of example, assume we install a 2” diameter
plastic suction hose of about 50 m in length.
Then assume we are pumping a gravel/water mixture of
about a C
V
or volume concentration of 15 %, the gravel and
mud average size of about 3 mm, or d
50
= 3 mm.
Assume the SG of the wet gravel
S=2,65
With an average particle size of
d
50
=3 mm
Concentration of solids
Cw=20 % by mass
Discharge head, static
Zb=2 m
Suction head
Za= -5 m (negative)
Length of hoses
Suction hose 50 m
of 0,0525 m dia,
Discharge hose 3 m
Number of valves and fittings
Foot strainer on
the suction hose
From Durand’s settling velocity equation V
L
[2]:
V
L
= F
L
√(2gD(S-1)) where
V
L
= minimum velocity in the line to maintain gravel flow
F
L
= is the Modified Froude number =1,34 an empirical
coefficient [3] for this slurry size
2g = g being gravitational acceleration or 9,81 m/sec
2
D = diameter of pipe in metres for 2” line = 0,0525 m
S = solids specific gravity = 2,65
V
L
=1,4√(2 x 9,81 x 0,0525 x (2,65-1)) =1,825 say 1,8m/sec
With a safety factor of about 20 % we will use 1,8 x 1,2 =
2,2 m/sec as the velocity in the line.
There are a number of other correlations available to
determine the critical velocity, or V
C,
which is similar to the
V
L
or minimum velocity, to maintain gravel suspension.
For example, the K C Wilson [4] equations; however, the
Durand values are more conservative and so are used here
as an example.
V= 4Q/πD
2
so Q in m
3
/sec = 3,142 x 0,00275 x 2,2/4 =
0,00475 m
3
/sec or 17,1 m
3
/hr (2” dia = 0,0525 m)
The suction hose will be in the stream and the discharge will
be from the pump to the ‘trommel’. The illustration above
describes the system.
What would be the pump type, materials of construction,
speed, required power, Net Positive Suction head require-
ments, etc? As many of the variables in the pumping system
will probably be confirmed on the site, v-belts from the motor
to the pump will be installed to allow changes in the pump
speed to ensure optimum operation.
We will refer to the ‘mixture’ of gravel, sand and mud as
a ‘slurry’ for the purposes of the calculation.
Depending on the coarseness of the gravel, the slurry can
be categorised and, for example, a Category ‘C’ slurry is the
suction dredging of gravel and coarse sand [5]. Cw is usually
less than 20 % as it is difficult to manually ensure a higher
concentration by dragging the suction hose through the silt.
Assume the Cw (or Concentration of solids in the slurry, by
mass %) = 20 % = x t/hr
Mass of water will be 80 % of slurry or x/20 % x 80 % =
4 x t/hr
Mass of water + mass of gravel = (4
x
+ x) = 5
x
t/hr
SG of gravel is 2,65
So slurry mass will have a mass of water equivalent to
x/2,65 t/hr
Total mass of water would be 4
x
+
x
/2,65 = 4,38
x
t/hr
The SG of slurry mix would be = 5
x
/4,38
x
=1,14
The water has a density of 1 000 kg/m
3
, slurry 1 140 kg/m
3
,
1 000 kg/tonne
Slurry flow of 17,1 m
3
/hr converted to tonnes/hr = 17,1
x1 140/1 000 = 19,494 t/hr
5
x
= 19,494 hence
x
= 3,9 t/hr of gravel.
Assuming 50 kg sorting bags, production would be about
450 bags per 6-hour day.
Friction head Hf for the pipeline to
select the pump and power
The friction losses in the lines can be calculated by a num-
ber of means, the most common being to determine the
equivalent lengths of pipe for the hoses and fittings and so
obtain the friction loss.
The Reynold’s number for the system can be calculated
Re
and using the Moody ‘Friction factors for pipe flow’,
graph, we can obtain the fanning friction factor
f .
Using Darcy’s formula for friction head loss h
L
h
L
= (f
L
/D +
K
ent
+ K
exit
) V
2
/2g
The Dynamic head loss is now calculated as the friction
PUMPS & VALVES
Chemical Technology • April 2016