Workedout 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 
1911 
1921 
1931 
1941 
1951 
1961 
1971 
Population: (thousands) 
60 
65 
63 
72 
79 
89 
97 
120 
Solution:
Year 
Population: (thousands) 
Increment per Decade 
Incremental Increase 
Percentage Increment per Decade 
1901 
60 
 
 
 
1911 
65 
+5 
 
(5+60) x100=+8.33 
1921 
63 
2 
3 
(2+65) x100=3.07 
1931 
72 
+9 
+7 
(9+63) x100=+14.28 
1941 
79 
+7 
2 
(7+72) x100=+9.72 
1951 
89 
+10 
+3 
(10+79) x100=+12.66 
1961 
97 
+8 
2 
(8+89) x100=8.98 
1971 
120 
+23 
+15 
(23+97) x100=+23.71 
Net values 
1 
+60 
+18 
+74.61 
Averages 
 
8.57 
3.0 
10.66 
+=increase;  = decrease
Arithmetical Progression Method:
P_{n} = 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 10^{6} l. Detention period is 3h.
Volume of tank = Flow x Detention period = 2.4 x 10^{3} x 3/24 = 300 m^{3}
Assume depth of tank = 3.0 m.
Surface area = 300/3 = 100 m^{2}
L/B = 3 (assumed). L = 3B.
3B^{2 } = 100 m^{2} i.e. B = 5.8 m
L = 3B = 5.8 X 3 = 17.4 m
Hence surface loading (Overflow rate) = 2.4 x 10^{6} = 24,000 ^{}l/d/m^{2 }< 40,000 l/d/m^{2} (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 10^{6} = 0.42766 Ml / h
24 x 23.5
Let the rate of filtration be 5000 l / h / m^{2 } of bed.
Area of filter = 10.05 x 10^{6} x 1 = 85.5 m^{2}
23.5 5000
Provide two units. Each bed area 85.5/2 = 42.77. L/B = 1.3; 1.3B^{2 }= 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 m^{2}
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 m^{2} .
\ 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 10^{4 } = 860 cm^{2}
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 cm^{2}
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 cm^{2}.
\ 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 m^{3}/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 = 1.04 m/s < 2.25 m/s (ok)
0.345
Washwater gutter
Discharge of washwater per bed = 0.36 m^{3}/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 m^{3}/s.
Q =1.71 x b x h^{3/2}
Assume b =0.3 m
h^{3/2} = 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 m^{3}
1000
Assume depth d = 5 m. Surface area = 1725/5 = 345 m^{2}
L/B = 2; 2B^{2 }= 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/ m^{2} bed area.
For 5 min, air required = 1000 x 5 x 7.5 x 5.77 x 2 = 4.32 m^{3} 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
Pipe 
Assumed flow 
Dia of pipe 
Length of pipe (m) 
K = L
470 d^{4.87} 
Q_{a}^{1.85} 
H_{L}= K.Q_{a}^{1.85} 
lH_{L}/Q_{a}l 

in l/sec 
in cumecs 
d in m 
d^{4.87} 
(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(8) 
(9) 
(10) 
AB
BC
CD
DA 
(+) 43
(+) 23
() 20
() 35 
+0.043
+0.023
0.020
0.035 
0.30
0.20
0.20
0.20 
2.85 X10^{3}
3.95 X10^{4}
3.95 X10^{4}
3.95 X10^{4} 
500
300
500
300 
373
1615
2690
1615 
3 X10^{3}
9.4 X10^{4}
7.2 X10^{4}
2 X10^{3} 
+1.12
+1.52
1.94
3.23 
26
66
97
92 
S 







2.53 
281 
* H_{L}= (Q_{a}^{1.85}L)/(0.094 x 100 ^{1.85} X d^{4.87})
or K.Q_{a}^{1.85}= (Q_{a}^{1.85}L)/(470 X d^{4.87})
or K =(L)/(470 X d^{4.87})
For loop ABCD, we have d =SH_{L} / x.S lH_{L}/Q_{a}l
=() 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 
BC 
CD 
DA 
Corrected flows after first correction in l/s 
+ 48 
+ 28 
 15 
 30 
Table 2
Consider loop DCFE
Pipe 
Assumed flow 
Dia of pipe 
Length of pipe (m) 
K = L
470 d^{4.87} 
Q_{a}^{1.85} 
H_{L}= K.Q_{a}^{1.85} 
lH_{L}/Q_{a}l 

in l/sec 
in cumecs 
d in m 
d^{4.87} 
(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(8) 
(9) 
(10) 
DC
CF
FE
ED 
(+) 20
(+) 28
() 8
() 5 
+0.020
+0.028
0.008
0.005 
0.20
0.15
0.15
0.15 
3.95 X10^{4}
9.7 X10^{5}
9.7 X10^{5}
9.7 X10^{5} 
500
300
500
300 
2690
6580
10940
6580 
7.2 X10^{4}
1.34 X10^{3}
1.34 X10^{4}
5.6 X10^{5} 
+1.94
+8.80
1.47
0.37 
97
314
184
74 
S 







+8.9 
669 
For loop ABCD, we have d =SH_{L} / x.S lH_{L}/Q_{a}l
=() +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 
DC 
CF 
FE 
ED 
Corrected flows after first correction in l/s 
+ 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/m^{3}/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 = (14730) x 100/147 = 80%
BOD load applied to the filter = flow x conc. of sewage (kg/d) = 6 x 10^{6} x 147/10^{6} = 882 kg/d
To find out filter volume, using NRC equation
E_{2}= 100
1+0.44(F_{1.BOD}/V_{1}.Rf_{1})^{1/2}
80 = 100 Rf_{1}= 1, because no circulation.
1+0.44(882_{}/V_{1})^{1/2}
V_{1}= 2704 m^{3}
Depth of filter = 1.5 m, Fiter area = 2704/1.5 = 1802.66 m^{2}, and Diameter = 48 m < 60 m
Hydraulic loading rate = 6 x 10^{6}/10^{3} x 1/1802.66 = 3.33m^{3}/d/m^{2} < 4 hence o.k.
Organic loading rate = 882 x 1000 / 2704 = 326.18 g/d/m^{3} which is approx. equal to 320. 