1.a.b.State the differences and the significance of chemical oxygen demand(COD) and biological oxygen demand (BOD).Wastewater collected from a processing unit has a temperature of 20°C.About 25 mL of wastewater sample is added directly into a 300 mL BODincubation bottle. The estimated initial and final dissolved Oxygen (DO) ofthe diluted sample after 5 days are 9.5 mg/L and 2.5 mg/L, respectively. Thecorresponding initial and final DO of the seeded dilution water is 9.7 mg/Land 8.5 mg/L, respectively. Evaluate the effect of different key parameterson BOD values. Justify your answer with appropriate calculations.

Answered: Image /qna-images/answer/8f43026f-34ac-43f2-8df8-07669faead4e.jpg

Answered: 1. a. State the differences and the…

Engineering

Civil Engineering

1. a. State the differences and the significance of chemical oxygen demand (COD) and biological oxygen demand (BOD).

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A COD test is carried out on a batik factory wastewater. The relevant information is as follows. Volume of potassium dichromate standard solution 0.1N = 10.0mL Volume of wastewater sample = 20.0mL Wastewater sample dilution = 10X Fe^2+ standard solution = 0.1N Volume of Fe^2+ standard solution needed for wastewater sample = 13.55 mL Volume of Fe^2+ standard solution needed for distilled water = 34.70 mL. Calculate the COD value for the batik factory wastewater.
In a BOD test, the amount of organic matter remaining in the wastewater was measured at certain time intervals, instead of measuring the dissolved oxygen content. The amount of organic matterremaining after 4 and 9 d was measured as 52.86 mg/L and 8.11mg/L, respectively. Calculate the ultimate BOD and BOD rate constant k for the wastewater.
A 2% sewage sample (in distilled water) was incubated for 3 days at 27°C temperature. After incubation, dissolved oxygen depletion of 10 mg/Lwas recorded. The biochemical oxygen demand (BOD) rate constant at 27°C was found to be 0.23 day^-1 (at base e). What is the ultimate BOD (in mg/L) of the sewage?
a. The wastewater from a manufacturing treatment plant is discharged to a municipal `treatment plant for processing with domestic wastewater. The characteristics of the wastewater from the manufacturing treatment plant are 3500 mg/l BOD, 4500 mg/l SS, 70 mg/l nitrogen and 2 mg/l phosphorous. The characteristics of the domestic wastewater are 350 mg/l BOD, 285 mg/l SS, 50 mg/l N and 11 mg/l P. If the required BOD/N/P weight ratio is 100/5/1, what is the minimum quantity of domestic wastewater required per 1000 gallons of manufacturing wastewater to provide adequate nutrients for biological treatment and the concentration of SS in the combined wastewater? b. What additional of NH4OH and H3PO4 would be required in pounds per million gallons of wastewater to provide the necessary nutrient levels for the treatment of the manufacturing wastewater described above (a) if the domestic wastewater had not been combined with it.
It is desired to design an activated sludge process to treat 20,000 m3/d of settled wastewater with a BOD of 250 mg/L and nbVSS concentration of 10 mg/L. If the effluent BOD is not to exceed 20 mg/L, calculate the volume of the aeration tank required, sludge wastage rate, sludge return rate, oxygen requirements, F/M ratio, specific utilization rate, specific growth rate, and maximum specific growth rate. The following data and coefficients are applicable. i. Mean cell residence time = 5 daysii. Y = 0.5 gVSS/g BODiii. Kd = 0.10 /div. K = 5 /dv. Return sludge concentration = 8,000 mg/L VSSvi. Mixed liquor volatile suspended solids concentration = 2500 mg/Lvii. Effluent contains 20 mg/L suspended solids of which 65% are biodegradable.viii. Fraction of biomass remaining as cell debris = 0.15ix. BOD5/BODu = 0.68
Assume that DLSUD lake is 100m2 and its average height is 5ft. Wastewater that flows into the lake are the following: Food square kitchen waste 7 m3/sec; Total Dissolved Solids (TDS) is 25,000 mg/L which operates 8hrs per day CIH 5m3 /hr kitchen waste; TDS is 20mg/L which operates 4hrs per day PCH 20m3 /day diluted laboratory waste ; TDS = 40mg/L (1)Determine the volumetric flow rate of the lake water can still accommodate prior to rainfall.
Process Selection, Reactor Sizing, and Biogas Yield Estimation The average wastewater characteristics produced by a food processing plant are shown in Table A.2. If anaerobic treatment is proposed as a treatment or pretreatment option, provide initial working plans and estimates for the following: Pretreatment requirements b. Reactor size Biogas production d. Sludge (or biosolids) production TABLE A.2 Average Wastewater Characteristics Parameter Value Flow rate, m3/day 500 COD, mg/L 11,000 BODU, mg/L 10,000 TKN, mg N/L 200 TSS, mg/L 2,000 VSS, mg/L 1,800 SO4=, mg/L 200 Alkalinity, mg/L CaCO3 500 Temperature, °C 30 pH 4.8
The specific gravity of particles in wastewater is 2.7 and the kinematic viscosity of water is 1.02x10-2 cm2/s. Compute the required area of tank for removing particles of size 0.07 mm if wastewater is flowing at the rate of 0.2 m3/s in a settling tank.
Question 4. An industrial process produces a by-product compound ‘X’ which exceeds the sewer use limit. Therefore, before the wastewater can be discharged to the sewer, the concentration of ‘X’ must be reduced to below 30 mg C/L. The industrial wastewater, with a concentration of 1780 mg C/L, flows through a long series of channels before reaching the treatment reactor. The long series of channels can be modelled as a PFR with a volume of 23 m3 and the treatment reactor can be modelled as a CSTR with a volume of 15 m3. The compound ‘X’ naturally degrades in the series of channels before the treatment reactor with a reaction rate constant of 0.75 mg/(L·min). In the treatment reactor the addition of chemicals and catalysts cause compound ‘X’ to degrade according to second order kinetics described with a reaction rate constant of 0.015 L/(mg·min). a) Derive the appropriate HRT equations for the PFR channels and the CSTR treatment reactor (state all assumptions and show all work). b) Does…
A wastewater treatment plant discharges 2.16x107 L/hr of treated effluent with an ultimate BOD of 22.0 mg/L and dissolved oxygen concentration of 5.0 mg/L to a river. Upstream of the waste discharge, the river has a flow of 9.0 m3/s, an ultimate BOD of 10.0 mg/L, and dissolved oxygen concentration of 7.0 mg/L. Both the river and effluent are at 20°C. After the river and effluent mix, the river flows at a velocity of 0.13 m/s and has a depth of 1.8 m. The deoxygenation rate constant for the waste is 0.45 day-1. A fish hatchery is located 9.6 km downstream of the waste discharge. The hatchery requires 5.0 mg/L DO for healthy fish growth and reproduction. Is the hatchery requirement being met? show all work
A primary clarifer removes 35% of BOD5 and 55% of suspendedsolids from the incoming wastewater. Calculate the mass of solidsand mass of BOD5 removed in kg/d for a plant processing 4500 m3/dof wastewater with 275 mg/L BOD5 and 400 mg/L SS.
Answer and show complete solution. Pls don't rejectA factory discharges waste water to a nearby river. The concentration of dissolved oxygen (DO) a few meters downstream of the factory’s discharge point was recorded for 10 consecutive days as shown in the table below. The lower the DO, the more “polluted” the river is and as the DO reaches zero, the river becomes unable to sustain marine life. Day DO (milligram/liter) 1 1.8 2 2 3 2.1 4 1.7 5 1.2 6 2.3 7 2.5 8 2.9 9 1.9 10 2.2 Determine the 95% confidence interval for the true mean DO concentration.