Pollution Control Engineering Numerical.

1. Problems
2021
Mujeeb UR Rahman
17CH106
Chemical Engineering Department
Assigned by: Prof.Dr. Abdul Rehman
Chemical Dept: Mehran UET
Pollution Control Engineering
Mehran University of Engineering & Technology
Jamshoro, Pk
Date: 10Feb2021

2. Problems
1. A storm sewer is carrying snow melt containing 1.3 g/L of NaCl enters
into a small stream. The stream has a naturally occurring NaCl
concentration of 20mg/L. If the storm sewer flow rate is 2000L/min and
the stream flow rate is 2m3
/s, what is the concentration of salt in the stream
after the discharge point.
Assume that the sewer flow and the stream flow are completely mixed,
that the salt is a conservative substance and hence it does not react, and
that the system is at steady state.
Given Data:
Concentration of sewer Csewer = 1.3 g/L
Converting into mg/L = Csewer = 1.3 g/L x 1000 mg/g = 1300 mg/L
Flow rate of sewer Qsewer = 2000 L/min
Concentration of sewer Cstream = 20 mg/L
Flow rate of sewer Qstream = 2m3
/s
Converting Q into L/min = 2m3
/s x 1000 x 60 = 120000 L/min
Mass diagram
Cstr = 20mg/L
Qstr = 120000mg/L Cmix
Qsew = 2000mg/L
Csew = 1300mg/L
Sewer
Control Volume
Stream

3. Problems
Explanation:
In the steady state total volume going out will be equal to total volume
coming in, and
total salt coming in = total salt going out
Total volume coming in per minute = 2000 + 2 x 10³ x 60
= 122000 L.
Total salt coming in per minute = 1300 x 2000 + 20 x 2 x 10³ x 60
= 2600000 + 2400000 mg
= 5000000 mg
volume of water going out per minute = 122000 L
Total salt going out per minute = 5000000 mg
concentration of water going out = 5000000 / 122000
= 40.983 mg / L
Rate of accumulation of salt is given by the formula
= [Cstream Qstream + Csewer Qsewer] – Cmix Qmix
In the steady state rate of accumulation of salt is zero,
Cmix Qmix = Cstream Qstream + Csewer Qsewer
Concentration of salt in the stream is given by the formula
𝐶𝑚𝑖𝑥 =
𝐶𝑠𝑡𝑟𝑒𝑎𝑚𝑄𝑠𝑡𝑟𝑒𝑎𝑚 + 𝐶𝑠𝑒𝑤𝑒𝑟𝑄𝑠𝑒𝑤𝑒𝑟
𝑄𝑚𝑖𝑥
𝐵𝑢𝑡 𝑄𝑚𝑖𝑥 = 𝑄𝑠𝑡𝑟𝑒𝑎𝑚 + 𝑄𝑠𝑒𝑤𝑒𝑟
𝐶𝑚𝑖𝑥 =
(20 × 120000) + (1300 × 2000)
120000 + 2000
𝐶𝑚𝑖𝑥 = 40.983𝑚𝑔/𝐿

4. Problems
2. If 100 mg of H2SO4 is added to 1L water, what is the final pH?
Given:
H2SO4 = 100mg
H2O = 1liter
pH = ?
Solution:
H2SO4 ↔ 2𝐻+
+ 𝑆𝑂4
2−
So molar ratio between H2SO4 and 2H+
is 1:2, or one mole of H2SO4
produces 2 moles of H+
. By converting 100mg of H2SO4 to molar amount
[H+
] can be calculated, then pH can be calculated by -log[H+
].
H+
= 1.02 × 10−3
𝑀
100𝑚𝑔
𝐿
×
𝑚𝑜𝑙𝑒
98𝑔
×
𝑔
1000𝑚𝑔
= 1.02 × 10−3
𝑀
𝑝𝐻 = −𝑙𝑜𝑔(2 × 1.02 × 10−3) = 2.69

5. Problems
3. A stream feeding a lake has an average flow of 1ft3
/sec and a phosphate
concentration of 10mg/L. The water leaving the lake has a phosphate
concentration of 5mg/L.
• How much phosphate in metric tons is deposited in the lake each year
(1000 kg = 1 metric ton)?
• Where does this phosphorus go since the outflowing concentration is
less than the inflowing concentration?
• Would the average phosphate concentration be higher near the surface
of the lake or near the bottom?
• Would you expect eutrophication of the lake to be accelerated? Why?
(1) 1ft3
= 28.317 liters
Deposition in 1liter = 10mg – 5mg = 5mg
Deposition in 1sec = 5mg x 28.317 = 141.585mg
Deposition in 1yr = 141.585 x 3600 x 24 x 365.25
= 4465.4004 kg
Deposition in 1yr = 4.4654 metric ton
(2) Here the outflowing conc. is less than the inflowing conc. so the
phosphorus stays in lake and as it is heavier than the water and
residence time of phosphorus is also high, so they get deposited
in the lake.
(3) Average phosphate concentration will be higher near bottom
near the sulphate as the weight of phosphate is higher.
(4) Yes, the eutrophication of the lake will accelerate at phosphate
is the main factor of eutrophication due to phosphate excessive
growth of plant in lake will occur and organism lives in lake
will get less oxygen to survive.

6. Problems
4. MUET discharges 18000 m3
/d of untreated wastewater into the Jamshoro
canal. The untreated wastewater has BOD5 of 812mg/L and a k of 0.62d-1
at
20o
C. The canal has a flow rate of 0.43m3
/s and an ultimate BOD of
2005mg/L. The DO of the canal water is 6.5mg/L and the DO of wastewater
is 1.0mg/L. Compute the DO and initial ultimate of BOD of the canal after
mixing.
Given:
Qw = 18000 m3
/day =
18000
24 ×60×60
= 0.208 m3
/s
Qc = 0.43m3
/s
BOD5 = 812 mg/L
k@20o
C = 0.62.d-1
Sol:
Ultimate BOD of wastewater
BOD5 = Lo(1-𝑒−𝑘𝑡
)
k@20o
C = k(0)T-20
= 0.62(1.135)20-20
k@20o
C = 0.62.d-1
812mg/L = Lo (1-𝑒0.62 (2005)
)
812mg/L = Lo (1-7.438)
812mg/L = Lo (-6.438)
Lo =
812
−6.438
𝑚𝑔/𝐿
Lo = -126.126 mg/L

7. Problems
DOmix =
6.5(0.43) + 1(0.208)
0.43+0.208
= 4.706𝑚𝑔/𝐿
After mixing:
Ultimate BOD mix =
𝑄𝑐𝐶𝑐 + 𝑄𝑊𝐶𝑊
𝑄𝑐 + 𝑄𝑊
0.43(5) + 0.208(−126.126)
0.43 + 0.208
−24.0842
0.638
Initial ultimate BODmix = -37.74 mg/L

8. Problems
5. Keenjhar lake has water carrying capacity of 11 million m3
with a reaction
rate constant (k) of 0.21/day. The lake water is polluted when it receives
sewage outfall with concentration (CW) of 105mg/L at a discharge rate (QW)
0.6m3
/s as well as an incoming stream with a concentration (CS) of 15mg/L
at a discharge rate (QS) of 6m3
/s. Draw the relevant image of the lake
showing the given data and determine the concentration and discharge rate
of the mixed contents of outgoing stream from lake.
Given:
Total capacity = 11 million m3
Reaction rate constant = 0.21/day
Sewage concentration = CW = 105 mg/L
Sewage discharge = QW = 0.6 m3
/s
Incoming Stream conc. = 15 mg/L
Stream discharge = Qs = 6 m3
/s
Sol:
➢ Swage moles = nw =
𝑚𝑎𝑠𝑠
𝑀𝑀
Converting mg to g =
105
1000
= 0.105g
Now,
𝑚𝑎𝑠𝑠
𝑀𝑀
=
0.105𝑔
18 𝑔
𝑚𝑜𝑙
nw = 5.83x10-3
moles
Lake
6m3
/s
Imcoming Stream
Sewage outfall
0.6m3
/s

9. Problems
➢ Steam moles = ns =
𝑚𝑎𝑠𝑠
𝑀𝑀
Converting mg to g =
15
1000
= 0.015g
Now,
𝑚𝑎𝑠𝑠
𝑀𝑀
=
0.015𝑔
18
𝑚𝑜𝑙
ns = 8.33x10-4
➢ Total number of moles = nw + ns = 5.83x10-3
+ 8.33x10-4
nt = 6.663x10-3
mol
➢ Molarity =
𝑚𝑜𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
6.663×10−3
1𝐿
= 6.663x10-3
M
➢ Concentration =
𝑚𝑜𝑙𝑒𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑚𝑜𝑙
𝑀𝑀
𝑣𝑜𝑙𝑢𝑚𝑒
=
6.66310−3×18
𝐼𝐿
Concentration = 0.1199 g/L or 119.934 mg/L
➢ Total discharge = 0.6m3
/s + 6m3
/s
QT = 6.6 m3
/s