Geoscience Reference
In-Depth Information
Compute the head loss passing through the sand. Use the same data but insert:
k
= 5
d
= 0.0007 m
L
= 0.6 m
2
−
6
2
1 131 10
981
.
×
(. )
.
058
042
8
0 0007
×
h
=×
5
×
×
0 00272
.
×
06
.
=
0.5579 m
3
.
.
Compute total head loss (
h
):
h
= 0.0410 m + 0.5579 m = 0.599 m
23.6.16 h
ead
l
oss
through
a
F
fluidized
b
ed
If the upward water flow rate through a filter bed is very large the bed mobilizes pneumatically and
may be swept out of the process vessel. At an intermediate flow rate the bed expands and is in what
we call an
expanded
state. In the fixed bed the particles are in direct contact with each other, sup-
porting each other's weight. In the expanded bed, the particles have a mean free distance between
particles and the drag force of the water supports the particles. The expanded bed has some of the
properties of the water (i.e., of a fluid) and is called a
fluidized bed
(Chase, 2002). Simply,
fluidi-
zation
is defined as upward flow through a granular filter bed at sufficient velocity to suspend the
grains in the water. Minimum fluidizing velocity (
U
mf
) is the superficial fluid velocity required to
start fluidization; it is important in determining the required minimum backwashing flow rate. Wen
and Yu (1966) proposed that the
U
mf
equation include the constants (over a wide range of particles)
33.7 and 0.0408 but exclude porosity of fluidization and shape factor (Wen and Yu, 1966):
µ
33 7
.
µ
(
)
−
05
.
U
=
×
1135 69
.
+
0 0408
.
G
(23.83)
mf
n
pd
pd
eq
eq
where
µ = Absolute viscosity of water (N·s/m
2
, lb·s/ft
2
).
p
= Density of water (kg/m
3
, lb/ft
3
).
d
eq
= d
90
sieve size is used instead of d
eq
.
G
n
= Galileo number, which is equal to
(
)
/ µ
3
2
dpppg
−
(23.84)
eq
s
Note:
Based on the studies of Cleasby and Fan (1981), we use a safety factor of 1.3 to ensure
adequate movement of the grains.
■
EXAMPLE 23.100
Problem:
Estimate the minimum fluidized velocity and backwash rate for a sand filter. The
d
90
size
of the sand is 0.90 mm. The density of the sand is 2.68 g/cm
3
.
Solution:
Compute the Galileo number. From the data given and the applicable table at 15°C:
p
= 0.999 g/cm
3
µ = 0.0113 N s/m
2
= 0.00113 kg/ms = 0.0113 g/cm s
µ
p
= 0.0113 cm
2
/s
g
= 981 cm/s
2
d
= 0.90 cm
p
s
= 2.68 g/cm
3
Search WWH ::
Custom Search