Problem 1: A municipal waste water treatment plant discharges 1.2 m3/s of effluent having BOD of 60 mg/l that has a flow of 9.3 m3/s and BOD 6 mg/l. Estimate the mixing BOD of the river,
just downstream of the flow.
Solution:
We know that,
Xm= ( X1Q1 + X2Q2)/ (Q1 + Q2)
Here, Q1 = Sewage discharge
=1.2 m3/s
X1= Sewage effluent BOD
= 60 mg/l
Q2 = Flow of effluent
=9.3 m3/s
X2= Effluent BOD
= 6 mg/l
Xm= {(1.2*60)+(6*9.3)}/(1.2+9.3)
= 12.17 mg/l
Answer: 12.17 mg/l
Problem 2: The dilution factor =0.5, value of DO initial 7.4, after 5 days 4.4 and BOD reaction rate 0.2 per per day. Find out (1). BOD5 (2.) Ultimate BOD, (3). BOD remaining after 5 days.
Solution:
1. BOD5 =
(Initial DO - Final DO) / Dilution Factor
= (7.4 - 4.4 ) * 0.5
= 6
2. We know,
BOD5 = L (1 - 10-5R)
∴ 6 = L (1 - 10-5*.2)
So, L= 6.67
3. BOD remaining after 5 Days
= 6.67-6
= 0.67
Answer: 0.67
Problem 3: A Canal exerts waste water in a river at 55000 m3/day. BOD of waste water is given in 110 mg/l. What is the total amount of BOD exerted in the river?
If a treatment plant is treating the waste water to remove 70% BOD, what is the BOD exerted now?
Solution:
Total amount of BOD exerted in the river,
= (55000 * 103 * 110* 10-6) Kg
= 6050 Kg
Amount of BOD exerted now
= {6050 * (100 - 70)}*100
= 1815 Kg
Answer: 1815 Kg
Problem 4: If BOD5 at 200C of a sewage sample is 320 mg/l. Calculate it's 10 days BOD at 300C.
Solution:
We know that,
X5 = L (1 - 10-5R)
Here,
X5 = 320
Let, R= 0.18
∴320= L (1 - 10-5R)
320= L (1 - 10-5*18)
L= 366.09
Again,
R30=R20 (1.047)30-20
= 0.18 *
1.04710
= 0.285
∴ 10 days BOD = L (1 - 10-10R)
= 366.09 (1 - 10-10*0.285)
= 365.58 mg/l
Answer: 365.58 mg/l
Problem 5: If the BOD5 is 156 mg/l and K= 0.22 at 200C temperature. determine the ultimate BOD?
Solution:
We know that,
X5 = L (1 - 10-5R)
∴ 165 = L (1 - 10-5*0.22)
L= 179.24 mg/l
Ultimate BOD = 179.24 mg/l
Answer: Ultimate BOD 179.24 mg/l
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