The document contains multiple problems related to septic tank design and BOD calculations. The first problem provides a step-by-step calculation of the ultimate first stage BOD of a waste with a given 5-day BOD and reaction constant. The second problem asks the reader to calculate the reaction rate given ultimate and 5-day BOD values. The third problem involves calculating ultimate BOD and 8-day BOD values at a different temperature. The final problem provides a sample septic tank design calculation for a small colony.

1. WWE
Unit 1 Problems

2. Problems
1.Problem .
• A city has a projected population of 60,000 spread over area of 50
hectare. Find the design discharge for the separate sewer (out fall)
line by assuming rate of water supply of 250 LPCD and out of this
total supply only 75 % reaches in sewer as wastewater. Make
necessary assumption whenever necessary.
• Solution:
• Given data Q = 250 lit/capita/day
Sewage flow = 75% of water supply
= 0.75* 250
= 187.5 LPCD
Total sewage generated = 187.5*60000/(24*3600)
= 130.21 lit/sec
= 0.13 m3 /s
Assume peak factor = 2
Total design discharge = Total sewage generated * peak factor
Total design discharge = 0.26 m3 /s

3. 2.A main sewer is to be designed to receive a flow from
1 sq. km. area of a community where the population
density is 200 per hectare. The average sewage flow is
150 lpcd. What is the design flow for the main sewer?
• sol: DWF=150 lpcd
– Assumed peak factor=3
– Maximum flow , qm= 3 X150 = 450 lpcd
– Tributary area, A= 1 sq.km = 100 ha
– Population density, pd=200 persons/ha
– Therefore Q= 100 X 200 X 450 =90,00,000 lpd say
9 million litres per day

4. Problem:
An area consists of 20% roofs with runoff ratio 0.90, 25% of the
area consists of pavements with a runoff ratio 0.85, 50% of the area
consists of lawns and gardens with runoff ration 0.10, the remaining
area is wooded for which the runoff ratio is 0.05; determine the co-
efficent of runoff.
If the total area is 1.62 hectares. And the maximum intensity is
taken as 62.5 mm/hour, what is the total runoff for the district?
Problem contd…

5. An area consists of 20% roofs with runoff ratio 0.90, 25% of the area consists of
pavements with a runoff ratio 0.85, 50% of the area consists of lawns and gardens
with runoff ration 0.10, the remaining area is wooded for which the runoff ratio is
0.05; determine the co-efficent of runoff.
If the total area is 1.62 ha. And the maximum intensity is taken as 62.5 mm/hour,
what is the total runoff for the district?

6. problem
• Find the design discharge of a combined
sewer for a town having population of 25000
spreading in 75hectares. The average
coefficient of runoff is 0.60 and the time of
concentration is 30 minutes.
Q combined= Q dwf + Q wwf
Assumed per capita water supply = 150 lpcd and 80% of it reach to sewers
Q dwf- .8x150x25000 = 3MLD say 0.0347 cum/sec
Intensity of rain fall (R) =25.4 * a/tc+b
where tc = 30 min : a = 1020 b= 20
on substitution = 20.4 mm/hr
Q wwf = AIR/360=75X0.60X20.4/360= 2.54 cum/sec
:. Q = 0.0347 + 2.54 = 2.575 cum/sec

7. Problem

8. Q = 0.624 metre
cube/ sec

9. If it is not satisfactory either increase the slope/ increase
diameter of sewer

12. WWE
Unit 2 Problems

13. Formula to remember (BOD Problem)

14. BOD Problems
Problem 1:
Determine the ultimate first stage BOD of
a waste whose 5-day BOD at 20 o C is
200 mg/l. Assume the reaction constant
k1 = 0.2 days -1
Solution:
Given: y = 200 mg/l; k1 = 0.2 days -1
Estimate the ultimate BOD by equ
y = L (1 – 10-k1t)
L = y/ 1 – 10-k1t
= 200 / 1 – 10-0.2 x 5
= 200 / 0.9
= 222 mg/l

15. BOD problems
• Problem 2:
The 5 – day BOD of a waste is 280 mg/l. the ultimate
BOD is reported as 410 mg/l. At which rate is the
waste being oxidized?

16. Problem 3:
The 5 – day BOD at 20 o C of a waste is found to be
200 mg/l. Taking k1 = 0.15 days -1, estimate the
ultimate BOD. Also determine the 8 – day BOD
value at 15 o C.

19. Design criteria for septic tank
• Detention period = 24 hours
• Let depth of the liquid = 1.5 m (1.2 to 1.7 m usual)
• Length / breadth = 2 to 3 usual
• Allow free board = 0.5 m
CHECK FOR SPACES:
• Sedimentation volume for a clear space 37.5 cm. deep
• Sludge digestion @ 0.028 m3/cap.
• Sludge and scum storage @ 0.0002 m3/cap for 1 year

20. • Design a septic tank for a small colony of 100
persons with daily sewage flow of 135 litres per
head per day.
Septic tank design - problem
Solution:
Assume a detention period of 24 hours.
Daily sewage flow = 100 x 135
= 13,500 lit/day
= 13.5 m3 /day
There fore, tank capacity = 13.5 m3
Let depth of the liquid = 1.5 m (1.2 to 1.7 m usual)
Plan area of tank = 13.5/1.5
= 9 m2
Provide tank of size 4.5 m x 2.0 m
Length / breadth = 4.5 / 2.0 = 2.25 (2 to 3 usual)
Tank area provided = 9 m2

21. Septic tank design - problem
Allow free board = 0.5 m
Total depth of tank = 1.5 + 0.5 = 2.0 m
CHECK FOR SPACES:
i. Sedimentation volume for a clear space 37.5 cm. deep
= 9 x (37.5 / 100)
= 3.37 m3
ii. Sludge digestion @ 0.028 m3/cap.
therefore for 100 persons (given in question)
100 x 0.028 = 2.80 m3
iii. Sludge and scum storage @ 0.0002 m3/cap for 1 year
therefore for 100 persons (given in question)
=100 x 0.0002 x 365
= 7.30 m3
Total (i + ii+ iii)= 13.47m3
Against tank capacity of 13.5 m3 provided, hence design is
correct.