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1.85 WATER AND WASTEWATER TREATMENT ENGINEERING
HOMEWORK 3
Question 1 (6 points)
A wastewater was tested in a settling column test with the following results in terms of
suspended sediment concentration:
At t = 30 minutes
Depth below
surface, h
(cm)
Concentration
remaining,
c/c0
At t = 60 minutes
Depth below
surface, h
(cm)
Concentration
remaining,
c/c0
At t = 90 minutes
Depth below
surface, h
(cm)
Concentration
remaining,
c/c0
38
0.23
34
0.05
32
0.03
118
0.81
114
0.46
112
0.23
198
0.94
194
0.73
192
0.52
278
0.97
274
0.86
272
0.70
358
0.98
354
0.92
352
0.83
a. What type of settling is indicated by these data? (2 points)
b. This particular wastewater is proposed to be treated in a rectangular sedimentation
tank having a detention time of 2 hours and a depth of 4 meters. Estimate the
percent of suspended sediments that will be removed. (2 points)
c. If the wastewater flow rate is 7,500 m3/day, what needs to be the area and volume of
the sedimentation tank? (2 points)
Question 2 (2 points)
The city of Hong Kong has a dual water-supply system that provides freshwater for drinking
and bathing, and salt water for flushing toilets. At the Hong Kong wastewater treatment
plant, as in most other wastewater treatment plants, wastewater is passed through a primary
clarifier to settle out suspended solids. The Hong Kong plant achieves greater removal in
their primary clarifier than most plants elsewhere in the world. Why?
Question 3 (2 points)
A rectangular sedimentation basin is to be designed for a flow of 1.0 mgd (million gallons
per day) using a 2:1 length:width ratio, an overflow rate of 0.00077 fps (feet per second),
and a detention time of 3.0 hr. What are the dimensions of the basin?
1
Solution to Homework 3, Question 1
a.
0
40
Depth
below
surface
(cm)
97
95
77
90
80
70
54
19
120
77
50
160
200
27
6
40
240
48
30
280
30
14
3
20
10
320
360
17
8
2
400
15
30
45
60
75
Detention time (minutes)
Straight removal efficiency lines indicate discrete particle settling.
Homework 3.xls 10/24/2005
90
105
120
Solution to Homework 3, Question 1
Detention
time
(min)
t
30
30
30
30
30
60
60
60
60
60
90
90
90
90
90
Depth
Conc.
remaining
(cm)
c/c0
0.23
0.81
0.94
0.97
0.98
0.05
0.46
0.73
0.86
0.92
0.03
0.23
0.52
0.70
0.83
z
38
118
198
278
358
34
114
194
274
354
32
112
192
272
352
Settling
velocity
(cm/min)
v
1.3
3.9
6.6
9.3
11.9
0.6
1.9
3.2
4.6
5.9
0.4
1.2
2.1
3.0
3.9
F, Fraction of particles with less
than settling velocity v
Particle velocity distribution
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Curve drawn by "eyeball"
0
1
2
3
4
5
6
7
8
v, settling velocity
Homework 3.xls 10/24/2005
9
10
11
12
13
Solution to Homework 3, Question 1
Fraction removed from Equation 9.30, pg. 230, of Reynolds and Richards
1 - F0 =
1 - 0.75
0.25
v
F
3.3
3.0
2.5
2.0
1.5
1.0
0.5
0.0
Fraction removed =
Fraction removed =
v - ave
0.76
0.70
0.64
0.49
0.34
0.18
0.04
0.00
∆F
3.2
2.8
2.3
1.8
1.3
0.8
0.3
v ∆F
0.06
0.06
0.15
0.15
0.16
0.14
0.04
0.19
0.17
0.34
0.26
0.20
0.11
0.01
Sum
1.27
Sum/v0
0.38
0.25 + 0.38
0.63
Alternative calculation using graph and formula from Lecture 6, page 18
R
∆h
h
100
90
70
50
40
30
26
0
100
190
230
325
370
400
100
90
40
95
45
30
95
80
60
45
35
28
Sum
Percent removed =
Homework 3.xls 10/24/2005
64.5 %
∆h/hn * Rave
Rave
23.8
18.0
6.0
10.7
3.9
2.1
64.5
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