5.The information for an activated-sludge system design is given in TABLE Q5. Th
ID: 300496 • Letter: 5
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5.The information for an activated-sludge system design is given in TABLE Q5. There is no nitrification at the selected SRT and temperature TABLE Q5: Activated-sludge System Design Data arameter nit alue Flow m3ld 10 000 Influent BOD g/m3 150 Effluent BOD g/m3 4 SRT Wastewater Synthesis yield, Y Cell debris yield, fa Endogenous decay, ka g VSSlg bCOD g VSSIg VSS g VSSIg VSS.d g/m3 oC 0.40 0.15 0.08 nbVSS 40 Temperature Assess the biodegradable COD (g/m3) and aeration tank oxygen requirement (kg/d) a. 8 marks] b. Estimate the oxygen demand and aeration tank uptake rate (in mg/L.hr) and the aeration tank biomass concentration in mg/L. [12 marks]Explanation / Answer
The expression "secondary treatment" refers to all biological treatment processes of both aerobic and anaerobic wastewater. The active sludge process has been used for the treatment of both industrial and urban wastewater. This process was born from the observation made a long time ago that if any residual water, urban or industrial, is subjected to aeration for a period of time, its content of organic matter is reduced, forming at the same time a flocculent mud. The active sludge process has been developed as a continuous operation through the recycling of biological sludge. For the active sludge process to function properly, the MLVSS of the reactor effluent must be quickly separated in the secondary clarifier. The condition that sometimes occurs where the mud is light and dispersed (inflated mud) and therefore difficult to settle is called "bulking". The mud of these characteristics passes over the separation dumps and escapes with the effluent from the secondary clarifier. Since the concentration of substrate in the effluent is small, there is not enough food to maintain the growth of the microorganisms that make up the sludge. Therefore, microorganisms are forced to function in an endogenous respiration regime. Due to the oxygen consumption of the endogenous respiration, this effluent would have a relatively high BOD, which is not desirable.
(a) Biodegradable COD:
Oxygen requirement in the tank to aeration: The following relationship should be followed:
Req. O2 = [Q * (So - S) / f] * (1 Kg / 10000 g) - 1.42 * Px
Where:
Q = Flow
So = Soluble inlet substrate
S = Effluent soluble substrate
Px = Sludge production
f = fraction of biomass 0.1 gr VSS / gr VSS
Req. O2 = [10000 m3 / day * (150-2) g / m3 / 0.15 gr vSS / gr vSS] - 1.42 (Px)
To calculate the sludge production we follow the following relationship:
Px = [Q * y * (So-S) / (1 + Kd * T)] + fd * Kd * X * V + Q * Xo
Where:
y = shield
Kd = endogenous decay
T = cell retention time
X = Substrate in aeration tank
V = Reactor volume
Xo = Non-biodegradable suspended solids
Px = [10000 m3 / day * 0.4 * (150-2) / (1 + 0.08 * 4)] + 0.15 * 0.08 * X * V + 10000 m3 / day * 40 g / m3
Next we calculate the Substrate in aeration tank and the volume of the reactor:
V = SRT * Q = 6 d * 10000 m3 / day = 60000 m3
X = (T / SRT) * [y * (So-S) / (1 + Kd * T)] = (4/6 * 24) * [0.4 * (150-2) / (1 + 0.08 * 4) ] = 1.25
We return to calculate
Px = 448484.84 + 900 + 400000 = 849384.84 g / d = 850 KgvSS / d
Finally we calculate the oxygen requirement:
Req. O2 = 9866.67 + 1.42 * 850 = 11073.67 Kg O2 / day.
Biodegradable COD = Total COD - Non-biodegradable COD = 150-2-40 = 108 g / m3
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