CIVE 3610 – Water Supply and Treatment
Fall Semester 2004
Homework #5 – Solution
1) A flocculation basin is to be designed and the design flow for the basin is 13.0
MGD. The basin is to be a cross–flow horizontal- shaft, paddle-wheel type with a
mean velocity gradient of 26.7 sec-1 (at 50 F), a detention time of 45 min, and a
GT value from 50,000 to 100,000. Tapered flocculation is to be provided in three
compartments of equal depth. The G values are to be 50, 20, and 10 sec -1. The
basin is to have a width of 90 ft to adjoin an existing basin. The paddle wheels
are to have blades with a 6-in width and length of 10 ft. The outside blades
should clear the floor by 1 ft and be 1 ft below the water surface. There are to be
six blades per paddle wheel, and the blades should have a clear spacing of 12 in.
Adjacent paddle wheels should have a clear spacing of 30 to 36 in. between
blades. The wall clearance is 12-18 in. Determine:
a. The basin dimensions (1 in. increments).
b. The paddle–wheel design.
c. The power to be imparted to the water in each compartment.
Solution
a. The basin dimensions.
V=
13000000 gal
ft 3
= 54,311 ft3
 45 min 
1440 min
7.48 gal
Area of the profile section = 54311 ft3/ 90 ft = 603.46 ft2
Let X = Compartment width and depth
(3X)*X = 603.46 ft2
X =14.183 ft = 14 ft- 3 in
3X = (3) (14 ft-3 in) = 42 ft-9 in
V= (14.25) (42.75) (90.0) = 54, 827 ft3
b. Assume 7 wheels with 6 in  10 ft blades.
Let the spacing between wheels be s and between wheels and wall be ½ s
7s+ 7(10 ft) = 90 ft
s = 2.86 ft or 34.3 in which is between 30 to 36 in.
Try 6 blades per wheel as shown in the drawing
6” by 10’
Wall
12”
12”
12”
14.25’
D3
12”
12”
12”
Floor
D1 = 14.25’ – (2)(1’) – 2(3/12’) = 11.75’
D2
D1
D2 = 11.75’-(2)(1’)- 2(3/12’) –2(3/12’) = 8.75’
D3 = 8.75’ – (2)(1’)- 2(3/12’) –2(3/12’) = 6.25’
Area of blades per shaft = (0.5’*10’)(6)(7) = 210 ft3
Cross section of basin = (14.25’) (90’) = 1282.5 ft3
% Area = (210/1282.5)*100% = 16.4% (between 15 to 20 %)
c. The power to be imparted to the water in each compartment
1st compartment
P = μVG2
P = (2.73*10-5 lb-sec/ft2) * (50/sec)2 * (54827 ft3/ 3)
= 1.2473 ft-lb/sec = 2.27 HP
2nd compartment
P = μVG2
P = (2.73*10-5 lb-sec/ft2) * (20/sec)2 * (54827 ft3/ 3)
= 199.57 ft-lb/sec = 0.363 HP
3rd compartment
P = μVG2
P = (2.73*10-5 lb-sec/ft2) * (10/sec)2 * (54827 ft3/ 3)
= 49.89 ft-lb/sec = 0.091 HP
Total HP = 2.72 HP
2) A circular primary clarifier is to be designed for a municipal wastewater
treatment plant having an average flow of 1.5 MGD. The overflow rate based on
average daily flow = 800 gal/ft2day. The minimum depth =10 ft. Determine:
a. The diameter of the tank assuming sludge rake equipment is only
available in %ft increments.
b. The design depth of the tank.
c. The overflow rate during the peak hourly flow (2.6X average flow rate)
d. Will the design be adequate during peak hourly flow?
a. The diameter assuming sludge rake equipment is only available in %ft
increments.
Overflow rate VO = Q/ As
where, As = Surface Area
As = Q/ VO = (1.5*106 gal/day) / (800 gal/ft2day)
= 1875 ft2
As = 1875 = (π/4)*d2
D = 48.86 ft = 50ft
b. The design depth of the tank.
Assume design depth = minimum depth = 10 ft.
We don’t have any θ (detention time). Usually the detention time is between 2-4 hrs.
Checking if θ lies between 2-4 hrs for the depth assumed
θ = V/Q =
 / 4(50) 2 ft 2
24h 7.48 gal

1.5  10 gal / day 1day
ft 3
6

= 2.34 h
Therefore, design depth = minimum depth assumption ok.
c. The overflow rate during the peak hourly flow (2.6X average flow rate).
VO = Q/ As = (1.5*106 gal/day*2.6) / [π/4 * (50)2] ft
= 1986 gal/ft2 day
d. Will the design be adequate during peak hourly flow?
The overflow rate is too high. It should usually be between 500-1000. Therefore, the
design is not adequate for peak hourly flow.
3. Sedimentation Basin theory and design questions.
a. What is the difference between Type I and Type II settling?
b. What environmental factors may affect the sedimentation efficiency?
c. List one advantage and one disadvantage of rectangular clarifiers.
a. What is the difference between Type I and Type II settling?
Type I –
1) Discrete particle settling
2) Vs stays same with depth
Type II –
1) Flocculent particle settling
2) Vs increases with the depth as particles grow
b. What environmental factors may affect the sedimentation efficiency?
Wind and Temperature
c. List one advantage and one disadvantage of rectangular clarifiers.
Advantage:
More efficient – closer to PFR
Disadvantage:
Higher Operational and Maintenance costs
More concrete required
Siting more difficult