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A domestic wastewater source has 1560 mg L-1 BOD, 82 mg L-1 total nitrogen (comprising Total Kjeldahl Nitrogen (TKN), nitrate and nitrite, and ammonia) and 15 mg L-1 soluble phosphorus (SP). These...

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A domestic wastewater source has 1560 mg L-1 BOD, 82 mg L-1 total nitrogen (comprising Total Kjeldahl Nitrogen (TKN), nitrate and nitrite, and ammonia) and 15 mg L-1 soluble phosphorus (SP). These parameters need to be brought to < 30 mg L-1 BOD, < 15 mg L-1 for total nitrogen and < 2.5 mg L-1 for SP before being discharged into a local river. Investigate possible wastewater treatment methods that could be applied to remove these pollutants to the levels required. Calculate the percentage removal required for each pollutant. If the wastewater is to be used for a recycled water supply to domestic households for non-potable use, such as watering gardens, toilet flushing and washing of pavements, investigate what further treatments should be applied.
Answered Same DayNov 01, 2021

Solution

Yasodharan answered on Nov 01 2021
52 Votes
Domestic Waste Water Source
In domestic waste water contains 99.9% water and 0.1% waste which we need to remove comprised of organic matter, microorganisms and inorganic compounds. There are major measures needs to be taken care of what is to be in treated water based on oxygen demand (BOD, COD), Indicator organisms (F.coli, E.coli etc.,) and Solid contents (TSS, TDS), in addition to this other measures to consider are ammonia & nitrate, total & reactive P, pH, alkalinity, odour and dissolved gases [1]. For the given parameter condition dissolved oxygen needs to be used after pre-treatment of wastewater to reduce down the organic ca
ons to energy, ca
on dioxide, water and residue as expressed below,
Organic ca
on + O2 Energy + CO2 + H2O + Residue
Nitrogen will be eliminated using nitrogen cycle scheme with the influence of microorganisms as most of the nitrogen present in water are released by protein degradation.
Organic – N + microorganisms NH3/NH4+ [2]
And the phosphorous content is removed using chemical treatment at later stage using cations which gets easily bind to phosphate (Ca2+, Al3+, Fe3+)
The % removal based on given data is,
BOD % removal = (1560 – 30) / 1560 *100% = 98.07 %
TKN % removal = (82 – 15) / 82 *100% = 81.71%
SP % removal = (15 – 2.5) / 15 *100% = 83.33...
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