Page 881 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

836 Appendix F: Alum Data and Conversions

Example F.1 Determine Alum Feed Flow In designing space for alum storage, the feed rate can be
determined alum use on a monthly basis. Example F.2 illus-
Given trates the calculation, taking into account the monthly vari-
3
Q(plant) ¼ 1.00 m =s (22.83 mgd) ation over the annual cycle.
C(in-plant-flow) ¼ 20 mg Al (SO 4 ) 3 14H 2 O=L In terms of 100 lb sacks, the number from Example F.2
2
would be about 1,140 per month (i.e., 51,849 kg=month 2.2
3
¼ 0:020 kg=m (from jar tests or pilot testing) lb=kg=100 lb=sack). For a plant of this size, handling the
C(alumfeed) ¼ 100 g Al (SO 4 ) 3 14H 2 O=L sacks would average about 1.6 per hour. As a matter of
2
practice, plants using bagged alum will size a storage silo
3
¼ 100 kg Al (SO 4 ) 3 14H 2 O=m :
2 for less than a 10-day supply, e.g., normally 1–2 days. If, on
the other hand, the alum is purchased in bulk and stored in
Required
bulk, the silos will be significantly larger, e.g., 30-day supply
(a) Q(alum feed) each. To illustrate, the volume required for 1-month supply
(b) Mass feed rate of alum 3 3
in Example F.2 would be 47 m (51,848 kg=1,100 kg=m ).
As an example of dimensions, a tank 3.5 m diameter 4.0 m
Solution
(a) Calculate Q(alum feed) from (F.2), deep has such volume. Bulk storage should be designed for
continuous mechanical feed with conical bottom. Several
Q(plant) C(alum-in-plant-flow) such tanks, such as 2–6 would help to reduce the number
¼ Q(alum feed) C(alum feed) (F:2) of deliveries. The sizing would depend also on the capacity
of the trucks.
3
3
1:00 m =s 0:020 kg=m ¼ Q(alum feed) 100 kg=m 3
3
Q(alum feed) ¼ 0:00020 m =s ¼ 200 mL=s Example F.2 Determine Monthly Rate of Alum Use
(b) The mass feed rate of alum to the dissolution reactor Given
_
(for feeder selection) is Result from Example F.1, i.e., M ¼ 0.020 kg=s
Required
M(alum) ¼ Q(plant) C(alum plant) Monthly use of alum
3
¼ 1:00 m =s 0:020 kg=m 3 Solution
Monthly use of alum:
¼ 0:020 kg=s
¼ 1:2kgAl (SO 4 ) 3 14H 2 O=min _
2 M(alum-monthly) ¼ 0:020 kg=s 3600 24 s=day 30 day=month
¼ 1728 kg Al (SO 4 ) 3 14H 2 O=day ¼ 51,848 kg Al (SO 4 ) 3 14H 2 O=month
2 2
(c) Let the bulk density of granular alum be 1074 Comments
3
kg=m , i.e., from Table F.1, (1010 þ 1038)=2, to The bulk storage volume required is 51,848 kg
3
3
give, the volumetric feed rate, Al 2 (SO 4 ) 3 14H 2 O=month=1,100 kg=m ¼ 47 m . In other
words a storage volume of dimensions, 3.5 m 3.5 m 4.0 m
3
V(feedrate)¼ 1728kgAl (SO 4 ) 3 14H 2 O=day = 1074kg=m would store a 1-month supply of alum at the rate of use
2
3
1:6m =day calculated. Two or three such silos should be provided,
depending on preferences in operation. With three silos,
two could be active while the third would provide a supply
Comments
while the alum is ordered for the other two. Also, the first
To provide for say a 10-day supply of alum, about 16 m 3 two with metering could be cleaned and maintained while
would be required. This would be a silo that could have the third is placed on-line.
the dimensions, 2 m diameter 6 m high. Most probably
3–4 such silos would be installed, i.e., for a 30-day supply
plus one for standby. F.2.4 COST OF SOLID ALUM
The cost of alum depends on the amount ordered and
whether the form is bagged or bulk, and the shipping dis-
F.2.3 STORAGE OF ALUM
tance. All of these factors and others will be reflected in a
Alum crystals will store well and therefore a long-term supply bid price for alum (if the bid method of purchase is used).
can be provided, such as 1–2 years for smaller plants and Example F.3 illustrates the calculation of monthly cost.
shorter durations for large plants in which such an amount of To compare alum cost when there are different forms, such
storage could be excessive (Harringer, 1984). For the large as solid alum and liquid alum, the cost per unit of Al 3þ
plants, alum would be provided in bulk rather than in sacks. provides a means.