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CS5308 WATER AND WASTEWATER ENGINEERING

   

Added on  2022-09-09

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CS5308 WATER AND WASTEWATER ENGINEERING
Tutorial 5
Question 1: Wastewater discharge calculations
a) The town of Viracocha discharges 17,360m3 d-1 of treated wastewater into the
Pachacamac Creek. The treated wastewater has a BOD5 of 12mgL-1 and a BOD decay
constant, k, of 0.12d-1 at 20ºC. Pachacamac Creek has a flow rate of 0.43m3s-1 and an
ultimate (BODL) of 5.0mg/L. The DO of the stream is 6.5mgL-1, and DO of the
wastewater is 1.0mgL-1. The stream temperature is 10ºC, while the wastewater
temperature is also 10ºC. Compute the dissolved oxygen value and initial ultimate BOD
(of the mixture) immediately after mixing.
Solution
Parameter Stream Wastewater
Discharge 0.43m3s-1 17,360 m3d-1
Dissolved oxygen DO 6.5 mgL-1 1.0 mgL-1
BOD5 12 mgL-1
Temperature 10ºC 10ºC
Reaction rate constant 0.12d-1
BODL 5.0 mgL-1
To compute the DO and initial ultimate BOD of the mixture, we first calculate the
discharge of the mixture in m3/S
Waste water discharge = 17360
24∗60∗60 = 0.201m3/S (conversion of units from m3d-1 to m3/S)
Stream discharge = 0.43 m3/S
Total discharge = 0.43 + 0.201=0.631 m3/S

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From the basic theory of conservation of mass,
Cm=𝐶𝑠𝑄𝑠+𝐶𝑤𝑄𝑤
𝑄𝑠+𝑄𝑤 where Cs= concentration and Qs= Discharge
DO of the mixture Cm=6.5∗0.43+1.0∗0.201
0.631 = 𝟒. 𝟕𝟒𝟖mg/l
Initial ultimate BOD
Given the ultimate BOD as 5mg/l, we work backwards by converting it to BOD5 so as to
compute the ultimate BOD of the mixture.
BOD ultimate= 𝐵𝑂𝐷5
1−𝑒(−𝑘𝑡)
Thus, rearranging the above equation and substituting values gives,
BOD5 =5*(1-е-0.12*5)
BOD5=5*0.4512
BOD5=2.256mg/l
Ultimate BOD of the mixture is then given by;
BOD mix Cm=𝐶𝑠𝑄𝑠∗𝐶𝑤𝑄𝑤
𝑄𝑠+𝑄𝑤 , where the parameters remain as described earlier
Cm=2.256∗0.43+12∗0.201
0.631 = 5.360mg/l
BODL= 𝐵𝑂𝐷5
1−𝑒(−𝑘𝑡)
BODL=5.3597 0.4512⁄
BODL=11.879mg/l

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b) Calculate the initial DO deficit of the Pachacamac Creek after mixing with the
wastewater from the town of Viracocha.
Initial DO deficit=DO sat – DO mix
But as calculated previously, DO mix=4.748mg/l
Calculating DO sat
DO sat = 𝐷.𝑂 𝑚𝑖𝑥
100% 𝑠𝑜𝑙𝑢𝑏𝑖𝑙𝑖𝑡𝑦 𝑎𝑡 10º𝐶 * 100
Where solubility is the amount of DO that distilled water can hold at a given temperature.
Temperature of the mix T mix=10∗0.43+10∗0.201
0.631 = 10ºC
DO sat = 4.748
11.3 * 100 =42.018 mg/l
Alternatively, a saturation monogram can be used comparing the two variables (water
temperature and oxygen solubility) to determine the percentage saturation, in this case
solubility at 10ºC and DO of 4.7 gives a percentage saturation of 43mg/l
DO deficit=42.018-4.748 = 37.27mg/l

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c) Given average speed μ=0.03ms-1, depth h=5.0m and bed activity coefficient ɳ=0.35; the
rate of deoxygenation is calculated as follows;
Kd=k + μ
ℎ * ɳ
Kd= 0.12 + 0.03
5 * 0.35
Kd=0.122d-1
d) Determine the DO concentration in the Pachacamac Creek at a point 5km downstream
from the Viracocha discharge point. Also, determine the critical DO and the distance
downstream at which the DO is reached.
DO concentration= DO sat – deficit at that point, but DO sat=42.018mg/l
Time to reach 5km t=500
0.03 = 166666.67
24∗60∗60 = 1.929 days
Deficit= 𝑘1𝑐𝐿˳
𝐾2𝑐−𝐾1𝑐{е-k1ct – e-k2ct} + DOe-k2ct
Given DO=37.27mg/l, L˳=11.879mg/l, t=1.929 days;
K1c= k1θ(Tmix-20) and θ=1.047, while for k2c, θ=0.40
K1c=0.12 * 1.047(10-20) = 0.0758d-1 and k2c=0.134d-1
Deficit=0.0758∗11.879
0.134−0.0758 {e-0.0758*1.929 – e-0.134*1.929}+37.27e-0.134*1.929

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