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Design Example

Worked-out Examples:
Population Forecast by Different Methods
Sedimentation Tank Design
Rapid Sand Filter Design
Flow in Pipes of a Distribution Network by Hardy Cross Method
Trickling Filter Design
Population Forecast by Different Methods
Problem: Predict the population for the years 1981, 1991, 1994, and 2001 from
the following census figures of a town by different methods.
Year
1901
Population: 60
(thousands)

1911
65

1921
63

1931
72

1941
79

1951
89

1961
97

1971
120

Solution:
Year
1901
1911
1921
1931
1941
1951
1961
1971
Net values
Averages

Population: Increment per Incremental Percentage Increment
(thousands)
Decade
Increase
per Decade
60
65
+5
(5+60) x100=+8.33
63
-2
-3
(2+65) x100=-3.07
72
+9
+7
(9+63) x100=+14.28
79
+7
-2
(7+72) x100=+9.72
89
+10
+3
(10+79) x100=+12.66
97
+8
-2
(8+89) x100=8.98
120
+23
+15
(23+97) x100=+23.71
1
+60
+18
+74.61
8.57
3.0
10.66

+=increase; - = decrease
Arithmetical Progression Method:
Pn = P + ni
Average increases per decade = i = 8.57
Population for the years,
1981= population 1971 + ni, here n=1 decade
= 120 + 8.57 = 128.57
1991= population 1971 + ni, here n=2 decade
= 120 + 2 x 8.57 = 137.14
2001= population 1971 + ni, here n=3 decade
= 120 + 3 x 8.57 = 145.71
1994= population 1991 + (population 2001 - 1991) x 3/10
= 137.14 + (8.57) x 3/10 = 139.71
Incremental Increase Method:
Population for the years,
1981= population 1971 + average increase per decade + average incremental
increase
= 120 + 8.57 + 3.0 = 131.57

1991= population 1981 + 11.57
= 131.57 + 11.57 = 143.14
2001= population 1991 + 11.57
= 143.14 + 11.57 = 154.71
1994= population 1991 + 11.57 x 3/10
= 143.14 + 3.47 = 146.61
Geometric Progression Method:
Average percentage increase per decade = 10.66
P n = P (1+i/100) n
Population for 1981 = Population 1971 x (1+i/100) n
= 120 x (1+10.66/100), i = 10.66, n = 1
= 120 x 110.66/100 = 132.8
Population for 1991 = Population 1971 x (1+i/100) n
= 120 x (1+10.66/100) 2 , i = 10.66, n = 2
= 120 x 1.2245 = 146.95
Population for 2001 = Population 1971 x (1+i/100) n
= 120 x (1+10.66/100) 3 , i = 10.66, n = 3
= 120 x 1.355 = 162.60
Population for 1994 = 146.95 + (15.84 x 3/10) = 151.70
Sedimentation Tank Design
Problem: Design a rectangular sedimentation tank to treat 2.4 million litres of raw
water per day. The detention period may be assumed to be 3 hours.
Solution: Raw water flow per day is 2.4 x 106 l. Detention period is 3h.
Volume of tank = Flow x Detention period = 2.4 x 103 x 3/24 = 300 m3
Assume depth of tank = 3.0 m.
Surface area = 300/3 = 100 m2
L/B = 3 (assumed). L = 3B.
3B2 = 100 m2 i.e. B = 5.8 m
L = 3B = 5.8 X 3 = 17.4 m
Hence surface loading (Overflow rate) = 2.4 x 106 = 24,000 l/d/m2 < 40,000 l/d/m2
(OK)
100
Rapid Sand Filter Design
Problem: Design a rapid sand filter to treat 10 million litres of raw water per day
allowing 0.5% of filtered water for backwashing. Half hour per day is used for
bakwashing. Assume necessary data.
Solution: Total filtered water = 10.05 x 24 x 106 = 0.42766 Ml / h
24 x 23.5
Let the rate of filtration be 5000 l / h / m2 of bed.
Area of filter = 10.05 x 106 x 1
23.5
5000

= 85.5 m2

Provide two units. Each bed area 85.5/2 = 42.77. L/B = 1.3; 1.3B2 = 42.77
B = 5.75 m ; L = 5.75 x 1.3 = 7.5 m
Assume depth of sand = 50 to 75 cm.
Underdrainage system:
Total area of holes = 0.2 to 0.5% of bed area.
Assume 0.2% of bed area = 0.2 x 42.77 = 0.086 m2
100
Area of lateral = 2 (Area of holes of lateral)
Area of manifold = 2 (Area of laterals)
So, area of manifold = 4 x area of holes = 4 x 0.086 = 0.344 = 0.35 m2 .
\ Diameter of manifold = (4 x 0.35 /p)1/2 = 66 cm
Assume c/c of lateral = 30 cm. Total numbers = 7.5/ 0.3 = 25 on either side.
Length of lateral = 5.75/2 - 0.66/2 = 2.545 m.
C.S. area of lateral = 2 x area of perforations per lateral. Take dia of holes = 13 mm
Number of holes: n p (1.3)2 = 0.086 x 104 = 860 cm2
4
\ n = 4 x 860 = 648, say 650
p (1.3)2
Number of holes per lateral = 650/50 = 13
Area of perforations per lateral = 13 x p (1.3)2 /4 = 17.24 cm2
Spacing of holes = 2.545/13 = 19.5 cm.
C.S. area of lateral = 2 x area of perforations per lateral = 2 x 17.24 = 34.5 cm2.
\ Diameter of lateral = (4 x 34.5/p)1/2 = 6.63 cm
Check: Length of lateral < 60 d = 60 x 6.63 = 3.98 m. l = 2.545 m (Hence
acceptable).
Rising washwater velocity in bed = 50 cm/min.
Washwater discharge per bed = (0.5/60) x 5.75 x 7.5 = 0.36 m3/s.
Velocity of flow through lateral =
0.36
= 0.36 x 10 4 = 2.08 m/s (ok)
Total lateral area
50 x 34.5
Manifold velocity = 0.36
0.345

= 1.04 m/s < 2.25 m/s (ok)

Washwater gutter
Discharge of washwater per bed = 0.36 m3/s. Size of bed = 7.5 x 5.75 m.
Assume 3 troughs running lengthwise at 5.75/3 = 1.9 m c/c.
Discharge of each trough = Q/3 = 0.36/3 = 0.12 m3/s.
Q =1.71 x b x h3/2
Assume b =0.3 m
h3/2 = 0.12
= 0.234
1.71 x 0.3
\ h = 0.378 m = 37.8 cm = 40 cm
= 40 + (free board) 5 cm = 45 cm; slope 1 in 40

Clear water reservoir for backwashing
For 4 h filter capacity, Capacity of tank = 4 x 5000 x 7.5 x 5.75 x 2 = 1725 m3
1000
Assume depth d = 5 m. Surface area = 1725/5 = 345 m2
L/B = 2; 2B2 = 345; B = 13 m & L = 26 m.
Dia of inlet pipe coming from two filter = 50 cm.
Velocity <0.6 m/s. Diameter of washwater pipe to overhead tank = 67.5 cm.
Air compressor unit = 1000 l of air/ min/ m2 bed area.
For 5 min, air required = 1000 x 5 x 7.5 x 5.77 x 2 = 4.32 m3 of air.
Flow in Pipes of a Distribution Network by Hardy Cross Method
Problem: Calculate the head losses and the corrected flows in the various pipes of
a distribution network as shown in figure. The diameters and the lengths of the
pipes used are given against each pipe. Compute corrected flows after one
corrections.

Solution: First of all, the magnitudes as well as the directions of the possible flows
in each pipe are assumed keeping in consideration the law of continuity at each
junction. The two closed loops, ABCD and CDEF are then analyzed by Hardy Cross
method as per tables 1 & 2 respectively, and the corrected flows are computed.

Table 1
Consider loop ABCD
Dia of pipe Length of
K
pipe (m) = L
470
in
in
d in
d4.87
d4.87
l/sec cumecs m
(1) (2)
(3)
(4)
(5)
(6)
(7)

Qa1.85

AB

Pipe

Assumed
flow

(+)
43

BC

(8)

(9)

(10)

500

373

3 X10-3

+1.12

26

+0.023 0.20

300

1615

9.4 X10-4

+1.52

66

500

2690

-1.94

97

300

1615

-3.23

92

-2.53

281

CD

-0.020

0.20

DA

(-) 20

-0.035

0.20

S

lHL/Qal

K.Qa1.85

+0.043 0.30 2.85 X10-3

(+)
23

(-) 35

HL=

3.95 X10-4
3.95 X10-4
3.95 X10-4

7.2 X10-4
2 X10-3

* HL= (Qa1.85L)/(0.094 x 100 1.85 X d4.87)
or K.Qa1.85= (Qa1.85L)/(470 X d4.87)
or K =(L)/(470 X d4.87)

For loop ABCD, we have d =-SHL / x.S lHL/Q al
=(-) -2.53/(1.85 X 281) cumecs
=(-) (-2.53 X 1000)/(1.85 X 281) l/s
=4.86 l/s =5 l/s (say)
Hence, corrected flows after first correction are:
Pipe
AB
Corrected flows
after first correction + 48
in l/s

BC

CD

DA

+ 28

- 15

- 30

Table 2
Consider loop DCFE
Assumed
Dia of pipe Length K = L
HL= lHL/Qal
Qa1.85
flow
of pipe
470 d4.87
K.Qa1.85
(m)
in
in
d in
4.87
l/sec cumecs m d
(1) (2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)

Pipe

DC
CF

(+) 20
(+) 28

+0.020
+0.028

0.20
0.15

FE

(-) 8

-0.008

0.15

ED

(-) 5

-0.005

0.15

3.95
X10-4
9.7
X10-5
9.7
X10-5

500

2690

300

7.2 X104

6580

500

10940

300

6580

9.7
X10-5

1.34
X10-3
1.34
X10-4

+1.94

97

+8.80

314

-1.47

184

-0.37

74

+8.9

669

5.6 X105

S
For loop ABCD, we have d =-SHL / x.S lHL/Q al

=(-) +8.9/(1.85 X 669) cumecs
=(-) (+8.9 X 1000)/(1.85 X 669)) l/s
= -7.2 l/s
Hence, corrected flows after first correction are:
Pipe
Corrected flows after
first correction in l/s

DC

CF

FE

ED

+ 12.8 + 20.8 - 15.2 - 12.2

Trickling Filter Design
Problem: Design a low rate filter to treat 6.0 Mld of sewage of BOD of 210 mg/l. The
final effluent should be 30 mg/l and organic loading rate is 320 g/m3/d.
Solution: Assume 30% of BOD load removed in primary sedimentation i.e., = 210 x
0.30 = 63 mg/l. Remaining BOD = 210 - 63 = 147 mg/l.
Percent of BOD removal required = (147-30) x 100/147 = 80%
BOD load applied to the filter = flow x conc. of sewage (kg/d) = 6 x 106 x 147/106 =
882 kg/d
To find out filter volume, using NRC equation

E2=

100
1+0.44(F1.BOD/V1.Rf1)1/2

80 =

100

Rf1= 1, because no circulation.

1+0.44(882/V1)1/2
V1= 2704 m3
Depth of filter = 1.5 m, Fiter area = 2704/1.5 = 1802.66 m2, and Diameter = 48 m <
60 m
Hydraulic loading rate = 6 x 106/103 x 1/1802.66 = 3.33m3/d/m2 < 4 hence o.k.
Organic loading rate = 882 x 1000 / 2704 = 326.18 g/d/m3 which is approx. equal to
320.