This document provides an overview of the activated sludge process for wastewater treatment. It describes that the activated sludge process uses microorganisms to consume organic matter in wastewater through biological oxidation, removing pollutants. The process involves wastewater flowing into an aeration tank where microorganisms are suspended and mixed with the water. The mixed liquor then flows to a secondary clarifier where the microorganisms are settled and separated from the treated water. Key factors like food to microorganism ratio, dissolved oxygen levels, hydraulic loading, and solids concentration must be controlled to maintain effective treatment. Calculations are provided to determine parameters like pounds of biomass and organic loading.
This document provides information about the activated sludge process for wastewater treatment. The activated sludge process uses microorganisms and oxygen to biologically treat wastewater. Microorganisms consume organic matter in the wastewater to grow, reproducing and removing pollutants through metabolic processes. Key components of an activated sludge system include the aeration tank where microorganisms and wastewater are mixed with air, and the secondary clarifier where microorganisms are separated from treated water. The food to microorganism ratio (F:M ratio) is important to balance to maintain effective treatment. Calculations are provided to determine pounds of biochemical oxygen demand (BOD), mixed liquor suspended solids (MLSS),
The document describes the principles and design of activated sludge wastewater treatment. Activated sludge processes use aerobic bacteria to consume organic matter in wastewater. Wastewater flows into an aeration tank where bacteria grow as they consume organic matter, then flows into a clarifier where bacteria settle as sludge. Sludge is partially recycled to the aeration tank to maintain bacteria populations. The design of activated sludge systems depends on parameters like hydraulic retention time, solids retention time, food to microorganism ratio, and dissolved oxygen levels. Typical designs include conventional, complete mix, extended aeration, and high rate aeration configurations.
This document contains solutions and additional questions related to water and wastewater treatment processes. It includes solutions to questions about batch reactor kinetics, anaerobic digester design, activated sludge process design, oxygen requirements, and sludge production calculations. Additional questions cover topics like sludge volume reduction from dewatering, volatile solids destruction during digestion, and methane production in anaerobic reactors.
This document provides information on operational mathematics for activated sludge systems. It defines key terms like F/M ratio, MCRT, and WAS. Formulas are given for calculating these values based on inputs like influent BOD, MLSS, aeration tank volume, and flows. The roles of the different zones in a BNR system are also summarized. Operational goals like maintaining the proper F/M ratio and sludge age are discussed.
The document analyzes the quality of greywater before and after treatment from four sites in Costa Rica. Samples were collected and tested for various water quality parameters like pH, conductivity, dissolved solids, nitrogen and more both before and after treatment. The results found that treatment improved water quality for most parameters, though some sites showed higher levels of certain contaminants after treatment likely due to design flaws or other site-specific factors.
The document discusses next generation wastewater treatment approaches at multiple scales from individual buildings to entire watersheds. At the building scale, technologies like greywater treatment and rainwater harvesting are discussed. At the cluster scale, technologies like extracting clean water from wastewater and energy extraction from organics are proposed. Finally, the catchment scale examines resource recovery opportunities at centralized wastewater treatment facilities through incremental process improvements.
This document presents a proposal for an anaerobic digestion system to process food waste from Clemson University's dining halls. It estimates that 262.5 tons of food waste is produced annually that could be used to produce biogas through anaerobic digestion. The goals of the project are to destroy 60% of volatile solids and produce 70% of the theoretical methane yield from the food waste. The document discusses governing equations, preliminary data collection, system design considerations, energy output estimates, and sustainability measures for the proposed anaerobic digestion system.
This document provides information about the activated sludge process for wastewater treatment. The activated sludge process uses microorganisms and oxygen to biologically treat wastewater. Microorganisms consume organic matter in the wastewater to grow, reproducing and removing pollutants through metabolic processes. Key components of an activated sludge system include the aeration tank where microorganisms and wastewater are mixed with air, and the secondary clarifier where microorganisms are separated from treated water. The food to microorganism ratio (F:M ratio) is important to balance to maintain effective treatment. Calculations are provided to determine pounds of biochemical oxygen demand (BOD), mixed liquor suspended solids (MLSS),
The document describes the principles and design of activated sludge wastewater treatment. Activated sludge processes use aerobic bacteria to consume organic matter in wastewater. Wastewater flows into an aeration tank where bacteria grow as they consume organic matter, then flows into a clarifier where bacteria settle as sludge. Sludge is partially recycled to the aeration tank to maintain bacteria populations. The design of activated sludge systems depends on parameters like hydraulic retention time, solids retention time, food to microorganism ratio, and dissolved oxygen levels. Typical designs include conventional, complete mix, extended aeration, and high rate aeration configurations.
This document contains solutions and additional questions related to water and wastewater treatment processes. It includes solutions to questions about batch reactor kinetics, anaerobic digester design, activated sludge process design, oxygen requirements, and sludge production calculations. Additional questions cover topics like sludge volume reduction from dewatering, volatile solids destruction during digestion, and methane production in anaerobic reactors.
This document provides information on operational mathematics for activated sludge systems. It defines key terms like F/M ratio, MCRT, and WAS. Formulas are given for calculating these values based on inputs like influent BOD, MLSS, aeration tank volume, and flows. The roles of the different zones in a BNR system are also summarized. Operational goals like maintaining the proper F/M ratio and sludge age are discussed.
The document analyzes the quality of greywater before and after treatment from four sites in Costa Rica. Samples were collected and tested for various water quality parameters like pH, conductivity, dissolved solids, nitrogen and more both before and after treatment. The results found that treatment improved water quality for most parameters, though some sites showed higher levels of certain contaminants after treatment likely due to design flaws or other site-specific factors.
The document discusses next generation wastewater treatment approaches at multiple scales from individual buildings to entire watersheds. At the building scale, technologies like greywater treatment and rainwater harvesting are discussed. At the cluster scale, technologies like extracting clean water from wastewater and energy extraction from organics are proposed. Finally, the catchment scale examines resource recovery opportunities at centralized wastewater treatment facilities through incremental process improvements.
This document presents a proposal for an anaerobic digestion system to process food waste from Clemson University's dining halls. It estimates that 262.5 tons of food waste is produced annually that could be used to produce biogas through anaerobic digestion. The goals of the project are to destroy 60% of volatile solids and produce 70% of the theoretical methane yield from the food waste. The document discusses governing equations, preliminary data collection, system design considerations, energy output estimates, and sustainability measures for the proposed anaerobic digestion system.
Irrigation of Controlled Environment Crops—Part 4: Balancing Light, Water, an...METER Group, Inc. USA
Are you unwittingly compromising your plants?
In a controlled environment many variables affect production. But if any one of those variables gets out of balance, it can undermine your whole operation. For example, if you apply enough nutrients for high production but only enough light for low production, you’ll increase costs and limit yield. To get the most out of your crop, you’ll need to measure and balance environmental inputs correctly to get the most efficient use out of them. If you’re not measuring the right variables, fixing problems that keep you from your goals will be a shot in the dark, because you won’t know what the real problem is.
Amplify your production and efficiency
In part 4 of our popular controlled-environment webinar series, world-renowned soil physicist, Dr. Gaylon Campbell, teaches what is required to ensure all environmental variables remain balanced for the highest possible efficiency and production. Discover:
- How to model biomass production from light, water, and nutrient resources
- Relationships between biomass production, light, and CO2
- Relationships between biomass production and water use
- Relationships between biomass production and nutrient uptake
- Limiting factors in the balance equations
- Examples and monitoring applications
The document discusses sludge treatment and disposal methods. It describes the processes of sludge digestion where sludge undergoes acid fermentation, acid regression and alkaline fermentation stages. Key factors like temperature, pH, seeding and mixing that affect digestion are also covered. Sludge digestion tanks are cylindrical with a conical bottom. Design considerations for sizing digestion tanks are provided. The document also discusses dewatering digested sludge using drying beds or mechanical methods like centrifuges.
This document provides an overview of anaerobic digestion and biogas production. It discusses the types and settings of digesters, as well as digester tank design, digestate management, gas upgrading and handling, gas use, project development, construction methodology, and examples of US projects. The document covers the chemistry and process of anaerobic digestion, which involves the breakdown of organic matter by microbes in oxygen-free conditions to produce biogas and digestate. It also addresses gas yield modeling and factors that influence biogas production rates.
The goal of the study was to design a continuous stirred-tank reactor (CSTR) to capture 10 mg/L-day of atmospheric carbon using algal cultures. A model of the CSTR was developed using the Monod equations to describe microbial growth. The model found that a hydraulic retention time of 70 hours would allow the reactor to capture carbon at a rate of 10.55 mg/L-day. This meets the design goal of 10 mg/L-day. The optimal conditions include a biomass concentration of 12 mg/L, substrate concentration of 1 mg/L, and specific growth rate of 0.1023 hr-1.
Question 1. Two parallel flocculation basins are to be used to tre.docxIRESH3
Question 1. Two parallel flocculation basins are to be used to treat water flow of 150m3/s. If the design detention time is 20minute, what is the volume of each tank? If the average velocity gradient in these two tanks is 124/s, calculate the velocity gradient is each basin if the gradient in second basin is half of the first one.
Question 2. Determine the volume of the aeration tank for the following operating conditions:
Influent BOD5 concentration after the primary is = 150mg/L
Wastewater flow rate = 10MGD
F/M ratio = 0.2/d
Mixed Liquor volatile suspended solid concentration = 2200mg/L
Question 3. Given below is the wastewater characteristics, determine the F/M ratio? (10 Points)
Influent BOD5 concentration = 84mg/L
Wastewater flow rate = 0.150m3/s
Volume of the aeration tanks = 970m3
Mixed Liquor volatile suspended solid concentration = 2000mg/L
Question 4. What is the terminal settling velocity of a particle with a specific gravity of 1.4 and a diameter of 0.010mm in 20oC water? Would this particle be completely removed in a settling basin with a width of 10.0m, depth of 3.0m, a length of 30.0m, and a flow rate of 7500m3/d? What is the smallest diameter particle of specific gravity 1.4 that would be removed in the sedimentation basin described above?
Question 5. Will grit particle with a radius of 0.04mm and a specific gravity of 2.65 be collected in a horizontal grit chamber that is 13.5m in length if the average grit-chamber flow is 0.15m3/s, the width of the chamber is 0.56m, and the horizontal velocity is 0.25m/s? The wastewater temperature is 22oC.
Question 6. Wastewater treatment plant flow rate is 20MGD. Chlorine dosage is 10mg/L. Determine chlorine requirement (lb/day)
Question 7. If a particle having a 0.0170-cm radius and density of 1.95g/cm3 is allowed to fall into quiescent water having a temperature of 4oC, what will be the terminal settling velocity? Assume the density of water = 1000kg/m3. Assume Stoke’s law applies?
Question 8. If the terminal settling velocity of a particle falling in quiescent water having a temperature of 15oC is 0.0950cm/s, what is its diameter? Assume a particle density of 2.05g/cm3 and density of water equal to 1000kg/m3.
µ@15oC = 1.139 mPa-s
ρ @15oC = 999.103kg/m3
Question 9. Determine the diameter of a single-stage rock media filter to reduce an applied BOD5 of 125mg/L to 25mg/L. Use a flow rate of 0.14m3/s, a recirculation ratio of 12.0 and a filter depth of 1.83m. Assume the NRC equations apply and that the wastewater temperature is 20oC?
Question 10. Bacterial kill rate typically follows Chick’s law. If the first-order kill rate for a certain weak disinfectant is 0.067/h. Determine the time it will take to reduce the bacterial population to half of its original concentration?
Question 11. A town discharges 17,360 m3/d of treated wastewater into the Creek. The Creek has a flow rate of 0.43m3/s and the DO of the creek is 6.5 mg/L and DO of the wastewater is 1.0 mg/L. Compute the DO?
Questio ...
This document discusses alternatives for offsetting nitrogen loads from a wastewater treatment plant expansion in Virginia. It analyzes potential offsets from agricultural best management practices, urban stormwater controls, and nutrient assimilation projects. For agriculture, options like cover crops and reduced nitrogen application could offset the load but feasibility is uncertain. Urban options like wet ponds and septic retirements may be feasible but are costly at $600/lb and $30/lb respectively. Nutrient assimilation through oyster aquaculture, algal harvesting, or wetland restoration could also work but costs per pound removed are unknown. The analysis finds no single clear lowest cost alternative and that further study is needed to determine feasibility and costs.
The document discusses developing and implementing biodigesters to reduce greenhouse gas emissions from organic waste. It proposes designing a small-scale biodigester at a local high school to process food waste from the cafeteria, which could generate renewable energy. Key questions are posed about how to eliminate food waste, engage students, and use the waste to produce clean energy instead of sending it to landfills where it emits methane.
Adding Wellness to Diet-based Chesapeake Bay Best Management Practices Michael Collins
1) Wellness-based best management practices (BMPs) aim to add "exercise" to traditional diet-based BMPs by promoting healthy soil biology and native plant biodiversity.
2) Examples include using compost extract and avoiding tilling to protect soil biology, as well as naturalizing landscapes instead of using turfgrass.
3) Case studies found that biologically healthy soil can naturally provide all the nitrogen needs of plants, eliminating the ongoing costs of synthetic fertilizers. Native plantings also significantly reduced phosphorus pollution loads.
Tom Newmark - Field Trials in Costa Rica bio4climate
Tom Newmark - Field Trials in Costa Rica
From Biodiversity for a Livable Climate conference: "Restoring Ecosystems to Reverse Global Warming"
Saturday November 22nd, 2014
www.bio4climate.org
Tom Newmark - Field Trials in Costa Rica
From Biodiversity for a Livable Climate conference: "Restoring Ecosystems to Reverse Global Warming"
Saturday November 22nd, 2014
biological treatment i activated sludge processManish Goyal
The activated sludge process uses microorganisms to biologically treat wastewater. Wastewater enters an aeration tank where microbes consume organic matter, producing new cells and reducing biochemical oxygen demand. Mixed liquor flows to a clarifier where microbes are separated from treated water; some microbes are recycled to the aeration tank while excess sludge is wasted. The process reduces organic matter through microbial growth and substrate utilization under aerobic conditions.
The wastewater treatment plant is experiencing issues like poor sludge settling and high pH. Recommendations include decreasing soda ash to lower alkalinity and pH, reducing dissolved oxygen levels, cleaning one of the treatment trains, installing pumps to better control return activated sludge between aerobic and anaerobic zones, using flocculants to aid settling, and adding a coagulant to remove emulsions. The goal is to improve treatment and meet discharge standards.
Evaluation of Anaerobic digestion by Influent of Ammonia Nitrogen Concentrati...simrc
This study evaluated anaerobic digestion using food waste water by investigating the effect of nitrogen loading rate. Without feeding, ammonia nitrogen concentration increased from 451 mg/L to 1,060 mg/L over 30 days as nitrogen was converted to ammonia. With feeding, ammonia nitrogen, pH, and alkalinity increased proportionally with nitrogen loading rate. Higher loading rates led to accumulation of volatile fatty acids and soluble COD as acidogenesis outpaced methanogenesis, inhibiting the digestion process. Maintaining appropriate ammonia nitrogen concentration through water addition and sludge recirculation was necessary for stable operation of anaerobic digestion using high-strength food waste water.
Economics of biogas plants and their role in saving the environmentDhananjay Rao
This document discusses biogas production through anaerobic digestion. It begins by outlining the composition of biogas and common organic materials used for production. These include cattle dung, kitchen waste, and crop residue. The document then describes the 3-phase digestion process and provides breakdowns of gas yield from different feedstocks. Key uses of biogas are also listed, such as cooking, lighting, and power generation. Installation rates of biogas plants in India are then assessed, followed by economics of family-sized plants and their role in environmental protection.
Beverage industry wastewater treatment with two–stage MBBR plantIJRES Journal
Beverage industry wastewater treatment was studied, first on pilot-scale and then on full-scale, with
two-stage moving bed biofilm reactor (MBBR). The pilot plant was made of a pure biofilm aerated MBBR (210
L) with filling degree 60%, a hybrid aerated MBBR (370 L) with filling degree 60%, and a lamellar settler (350
L) with plates sloped 60° from horizontal; carriers specific surface was 500 m2/m3. The pilot plant treated 12 L/h
wastewater with 5000-10000 mg/L COD and removed COD with average efficiency 73%. The full-scale plant
was made of two parallel pure biofilm aerated MBBR (18 m3 each) with filling degree 60%, two parallel hybrid
aerated MBBR (32 m3 each) with filling degree 60%, two parallel lamellar settlers (7 m3 each) with plates
sloped 60° from horizontal, and a final quarzite filter. The full scale plant treated 39-175 m3/d (average 70 m3/d)
wastewater with 490-4900 mg/L COD (average 1793 mg/L) and removed COD with average efficiency 97%;
the final effluent respected always emission limits.
This document discusses organic fertilizers for avocado cultivation. It describes different types of organic fertilizers commonly used, including compost, vermicompost, bokashi, animal bedding, manure, pruning waste, green manures, crop residues, compost tea, and leachate. It focuses on compost, explaining what it is, its components, the composting process, and benefits of using compost in soils. It also discusses calculating the appropriate amount of compost to apply based on desired increases in organic matter or nutrient levels like nitrogen. Companion crops that fix nitrogen like alfalfa are also covered.
The document discusses the carbon cycle and the role of carbon in the environment. It notes that nature's carbon cycle is self-regulating, with plants absorbing CO2 through photosynthesis and microbes releasing carbon back into the atmosphere through decomposition. However, anthropogenic carbon emissions are excess and not part of the natural cycle. It then discusses how human carbon emissions amplify greenhouse gas concentrations and the greenhouse effect, threatening the environment by causing global warming. The concepts of carbon neutrality, carbon footprint, carbon sequestration, carbon offsets, and carbon credits in addressing this issue are also summarized.
This lab aims to study aerobic respiration and alcoholic fermentation through experiments with seeds and yeast. Students will measure oxygen consumption of germinating and ungerminated pea seeds over 60 minutes using a volumeter. They will also observe gas production from yeast fermenting glucose, starch, and starch treated with amylase. The document provides background on metabolism, respiration, ATP, aerobic respiration, and fermentation. It outlines the objectives, materials, procedures, and data table for the oxygen consumption experiment and fermentation demonstration.
Irrigation of Controlled Environment Crops—Part 4: Balancing Light, Water, an...METER Group, Inc. USA
Are you unwittingly compromising your plants?
In a controlled environment many variables affect production. But if any one of those variables gets out of balance, it can undermine your whole operation. For example, if you apply enough nutrients for high production but only enough light for low production, you’ll increase costs and limit yield. To get the most out of your crop, you’ll need to measure and balance environmental inputs correctly to get the most efficient use out of them. If you’re not measuring the right variables, fixing problems that keep you from your goals will be a shot in the dark, because you won’t know what the real problem is.
Amplify your production and efficiency
In part 4 of our popular controlled-environment webinar series, world-renowned soil physicist, Dr. Gaylon Campbell, teaches what is required to ensure all environmental variables remain balanced for the highest possible efficiency and production. Discover:
- How to model biomass production from light, water, and nutrient resources
- Relationships between biomass production, light, and CO2
- Relationships between biomass production and water use
- Relationships between biomass production and nutrient uptake
- Limiting factors in the balance equations
- Examples and monitoring applications
The document discusses sludge treatment and disposal methods. It describes the processes of sludge digestion where sludge undergoes acid fermentation, acid regression and alkaline fermentation stages. Key factors like temperature, pH, seeding and mixing that affect digestion are also covered. Sludge digestion tanks are cylindrical with a conical bottom. Design considerations for sizing digestion tanks are provided. The document also discusses dewatering digested sludge using drying beds or mechanical methods like centrifuges.
This document provides an overview of anaerobic digestion and biogas production. It discusses the types and settings of digesters, as well as digester tank design, digestate management, gas upgrading and handling, gas use, project development, construction methodology, and examples of US projects. The document covers the chemistry and process of anaerobic digestion, which involves the breakdown of organic matter by microbes in oxygen-free conditions to produce biogas and digestate. It also addresses gas yield modeling and factors that influence biogas production rates.
The goal of the study was to design a continuous stirred-tank reactor (CSTR) to capture 10 mg/L-day of atmospheric carbon using algal cultures. A model of the CSTR was developed using the Monod equations to describe microbial growth. The model found that a hydraulic retention time of 70 hours would allow the reactor to capture carbon at a rate of 10.55 mg/L-day. This meets the design goal of 10 mg/L-day. The optimal conditions include a biomass concentration of 12 mg/L, substrate concentration of 1 mg/L, and specific growth rate of 0.1023 hr-1.
Question 1. Two parallel flocculation basins are to be used to tre.docxIRESH3
Question 1. Two parallel flocculation basins are to be used to treat water flow of 150m3/s. If the design detention time is 20minute, what is the volume of each tank? If the average velocity gradient in these two tanks is 124/s, calculate the velocity gradient is each basin if the gradient in second basin is half of the first one.
Question 2. Determine the volume of the aeration tank for the following operating conditions:
Influent BOD5 concentration after the primary is = 150mg/L
Wastewater flow rate = 10MGD
F/M ratio = 0.2/d
Mixed Liquor volatile suspended solid concentration = 2200mg/L
Question 3. Given below is the wastewater characteristics, determine the F/M ratio? (10 Points)
Influent BOD5 concentration = 84mg/L
Wastewater flow rate = 0.150m3/s
Volume of the aeration tanks = 970m3
Mixed Liquor volatile suspended solid concentration = 2000mg/L
Question 4. What is the terminal settling velocity of a particle with a specific gravity of 1.4 and a diameter of 0.010mm in 20oC water? Would this particle be completely removed in a settling basin with a width of 10.0m, depth of 3.0m, a length of 30.0m, and a flow rate of 7500m3/d? What is the smallest diameter particle of specific gravity 1.4 that would be removed in the sedimentation basin described above?
Question 5. Will grit particle with a radius of 0.04mm and a specific gravity of 2.65 be collected in a horizontal grit chamber that is 13.5m in length if the average grit-chamber flow is 0.15m3/s, the width of the chamber is 0.56m, and the horizontal velocity is 0.25m/s? The wastewater temperature is 22oC.
Question 6. Wastewater treatment plant flow rate is 20MGD. Chlorine dosage is 10mg/L. Determine chlorine requirement (lb/day)
Question 7. If a particle having a 0.0170-cm radius and density of 1.95g/cm3 is allowed to fall into quiescent water having a temperature of 4oC, what will be the terminal settling velocity? Assume the density of water = 1000kg/m3. Assume Stoke’s law applies?
Question 8. If the terminal settling velocity of a particle falling in quiescent water having a temperature of 15oC is 0.0950cm/s, what is its diameter? Assume a particle density of 2.05g/cm3 and density of water equal to 1000kg/m3.
µ@15oC = 1.139 mPa-s
ρ @15oC = 999.103kg/m3
Question 9. Determine the diameter of a single-stage rock media filter to reduce an applied BOD5 of 125mg/L to 25mg/L. Use a flow rate of 0.14m3/s, a recirculation ratio of 12.0 and a filter depth of 1.83m. Assume the NRC equations apply and that the wastewater temperature is 20oC?
Question 10. Bacterial kill rate typically follows Chick’s law. If the first-order kill rate for a certain weak disinfectant is 0.067/h. Determine the time it will take to reduce the bacterial population to half of its original concentration?
Question 11. A town discharges 17,360 m3/d of treated wastewater into the Creek. The Creek has a flow rate of 0.43m3/s and the DO of the creek is 6.5 mg/L and DO of the wastewater is 1.0 mg/L. Compute the DO?
Questio ...
This document discusses alternatives for offsetting nitrogen loads from a wastewater treatment plant expansion in Virginia. It analyzes potential offsets from agricultural best management practices, urban stormwater controls, and nutrient assimilation projects. For agriculture, options like cover crops and reduced nitrogen application could offset the load but feasibility is uncertain. Urban options like wet ponds and septic retirements may be feasible but are costly at $600/lb and $30/lb respectively. Nutrient assimilation through oyster aquaculture, algal harvesting, or wetland restoration could also work but costs per pound removed are unknown. The analysis finds no single clear lowest cost alternative and that further study is needed to determine feasibility and costs.
The document discusses developing and implementing biodigesters to reduce greenhouse gas emissions from organic waste. It proposes designing a small-scale biodigester at a local high school to process food waste from the cafeteria, which could generate renewable energy. Key questions are posed about how to eliminate food waste, engage students, and use the waste to produce clean energy instead of sending it to landfills where it emits methane.
Adding Wellness to Diet-based Chesapeake Bay Best Management Practices Michael Collins
1) Wellness-based best management practices (BMPs) aim to add "exercise" to traditional diet-based BMPs by promoting healthy soil biology and native plant biodiversity.
2) Examples include using compost extract and avoiding tilling to protect soil biology, as well as naturalizing landscapes instead of using turfgrass.
3) Case studies found that biologically healthy soil can naturally provide all the nitrogen needs of plants, eliminating the ongoing costs of synthetic fertilizers. Native plantings also significantly reduced phosphorus pollution loads.
Tom Newmark - Field Trials in Costa Rica bio4climate
Tom Newmark - Field Trials in Costa Rica
From Biodiversity for a Livable Climate conference: "Restoring Ecosystems to Reverse Global Warming"
Saturday November 22nd, 2014
www.bio4climate.org
Tom Newmark - Field Trials in Costa Rica
From Biodiversity for a Livable Climate conference: "Restoring Ecosystems to Reverse Global Warming"
Saturday November 22nd, 2014
biological treatment i activated sludge processManish Goyal
The activated sludge process uses microorganisms to biologically treat wastewater. Wastewater enters an aeration tank where microbes consume organic matter, producing new cells and reducing biochemical oxygen demand. Mixed liquor flows to a clarifier where microbes are separated from treated water; some microbes are recycled to the aeration tank while excess sludge is wasted. The process reduces organic matter through microbial growth and substrate utilization under aerobic conditions.
The wastewater treatment plant is experiencing issues like poor sludge settling and high pH. Recommendations include decreasing soda ash to lower alkalinity and pH, reducing dissolved oxygen levels, cleaning one of the treatment trains, installing pumps to better control return activated sludge between aerobic and anaerobic zones, using flocculants to aid settling, and adding a coagulant to remove emulsions. The goal is to improve treatment and meet discharge standards.
Evaluation of Anaerobic digestion by Influent of Ammonia Nitrogen Concentrati...simrc
This study evaluated anaerobic digestion using food waste water by investigating the effect of nitrogen loading rate. Without feeding, ammonia nitrogen concentration increased from 451 mg/L to 1,060 mg/L over 30 days as nitrogen was converted to ammonia. With feeding, ammonia nitrogen, pH, and alkalinity increased proportionally with nitrogen loading rate. Higher loading rates led to accumulation of volatile fatty acids and soluble COD as acidogenesis outpaced methanogenesis, inhibiting the digestion process. Maintaining appropriate ammonia nitrogen concentration through water addition and sludge recirculation was necessary for stable operation of anaerobic digestion using high-strength food waste water.
Economics of biogas plants and their role in saving the environmentDhananjay Rao
This document discusses biogas production through anaerobic digestion. It begins by outlining the composition of biogas and common organic materials used for production. These include cattle dung, kitchen waste, and crop residue. The document then describes the 3-phase digestion process and provides breakdowns of gas yield from different feedstocks. Key uses of biogas are also listed, such as cooking, lighting, and power generation. Installation rates of biogas plants in India are then assessed, followed by economics of family-sized plants and their role in environmental protection.
Beverage industry wastewater treatment with two–stage MBBR plantIJRES Journal
Beverage industry wastewater treatment was studied, first on pilot-scale and then on full-scale, with
two-stage moving bed biofilm reactor (MBBR). The pilot plant was made of a pure biofilm aerated MBBR (210
L) with filling degree 60%, a hybrid aerated MBBR (370 L) with filling degree 60%, and a lamellar settler (350
L) with plates sloped 60° from horizontal; carriers specific surface was 500 m2/m3. The pilot plant treated 12 L/h
wastewater with 5000-10000 mg/L COD and removed COD with average efficiency 73%. The full-scale plant
was made of two parallel pure biofilm aerated MBBR (18 m3 each) with filling degree 60%, two parallel hybrid
aerated MBBR (32 m3 each) with filling degree 60%, two parallel lamellar settlers (7 m3 each) with plates
sloped 60° from horizontal, and a final quarzite filter. The full scale plant treated 39-175 m3/d (average 70 m3/d)
wastewater with 490-4900 mg/L COD (average 1793 mg/L) and removed COD with average efficiency 97%;
the final effluent respected always emission limits.
This document discusses organic fertilizers for avocado cultivation. It describes different types of organic fertilizers commonly used, including compost, vermicompost, bokashi, animal bedding, manure, pruning waste, green manures, crop residues, compost tea, and leachate. It focuses on compost, explaining what it is, its components, the composting process, and benefits of using compost in soils. It also discusses calculating the appropriate amount of compost to apply based on desired increases in organic matter or nutrient levels like nitrogen. Companion crops that fix nitrogen like alfalfa are also covered.
The document discusses the carbon cycle and the role of carbon in the environment. It notes that nature's carbon cycle is self-regulating, with plants absorbing CO2 through photosynthesis and microbes releasing carbon back into the atmosphere through decomposition. However, anthropogenic carbon emissions are excess and not part of the natural cycle. It then discusses how human carbon emissions amplify greenhouse gas concentrations and the greenhouse effect, threatening the environment by causing global warming. The concepts of carbon neutrality, carbon footprint, carbon sequestration, carbon offsets, and carbon credits in addressing this issue are also summarized.
This lab aims to study aerobic respiration and alcoholic fermentation through experiments with seeds and yeast. Students will measure oxygen consumption of germinating and ungerminated pea seeds over 60 minutes using a volumeter. They will also observe gas production from yeast fermenting glucose, starch, and starch treated with amylase. The document provides background on metabolism, respiration, ATP, aerobic respiration, and fermentation. It outlines the objectives, materials, procedures, and data table for the oxygen consumption experiment and fermentation demonstration.
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precisely delineate tumor boundaries from magnetic resonance imaging (MRI)
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the state-of-the-art Deeplabv3+ architecture with the ResNet18 backbone. The
model is rigorously trained and evaluated, exhibiting remarkable performance
metrics, including an impressive global accuracy of 99.286%, a high-class accuracy of 82.191%, a mean intersection over union (IoU) of 79.900%, a weighted
IoU of 98.620%, and a Boundary F1 (BF) score of 83.303%. Notably, a detailed comparative analysis with existing methods showcases the superiority of
our proposed model. These findings underscore the model’s competence in precise brain tumor localization, underscoring its potential to revolutionize medical
image analysis and enhance healthcare outcomes. This research paves the way
for future exploration and optimization of advanced CNN models in medical
imaging, emphasizing addressing false positives and resource efficiency.
Optimizing Gradle Builds - Gradle DPE Tour Berlin 2024Sinan KOZAK
Sinan from the Delivery Hero mobile infrastructure engineering team shares a deep dive into performance acceleration with Gradle build cache optimizations. Sinan shares their journey into solving complex build-cache problems that affect Gradle builds. By understanding the challenges and solutions found in our journey, we aim to demonstrate the possibilities for faster builds. The case study reveals how overlapping outputs and cache misconfigurations led to significant increases in build times, especially as the project scaled up with numerous modules using Paparazzi tests. The journey from diagnosing to defeating cache issues offers invaluable lessons on maintaining cache integrity without sacrificing functionality.
Software Engineering and Project Management - Introduction, Modeling Concepts...Prakhyath Rai
Introduction, Modeling Concepts and Class Modeling: What is Object orientation? What is OO development? OO Themes; Evidence for usefulness of OO development; OO modeling history. Modeling
as Design technique: Modeling, abstraction, The Three models. Class Modeling: Object and Class Concept, Link and associations concepts, Generalization and Inheritance, A sample class model, Navigation of class models, and UML diagrams
Building the Analysis Models: Requirement Analysis, Analysis Model Approaches, Data modeling Concepts, Object Oriented Analysis, Scenario-Based Modeling, Flow-Oriented Modeling, class Based Modeling, Creating a Behavioral Model.
4. Microorganisms consume organic matter
from the wastewater, using oxygen for
respiration
Food
Food
Food
O2
O2
O2
O2
Millions of aerobic and facultative micro-organisms
remove pollutants thru living and growing process
SECONDARY TREATMENT
Biological Wastewater Treatment
7. Growth rate produces about
0.7 lbs of biological solids per
lb BOD removed
Activated Sludge
Need favorable conditions for growth and
for separation from the water
Biological solids are used
over and over
Suspended Growth,
Biological Treatment
15. 2. Conversion of Food to New Cells
and Byproducts.
Acclimated Biomass
Useable Food Supply
Adequate D.O.
Proper Nutrient Balance
100 : 5 : 1
C : N : P
Biological Wastewater
Treatment
17. 3. Flocculation and Solids Removal
Biological Wastewater
Treatment
Proper Growth Environment
Filamentous Bacteria – Form Strings
Mixed Liquor Does Not Compact - Bulking
Must Have Controls
20. Pounds =
CALCULATION OF POUNDS
Concentration
Of STUFF
In the
Water
Conc. x Flow (or Volume) x 8.34 Lbs/gallon
X
Quantity
Of Water
The STUFF
Is In
Weight
Of The
Water
X
21. Pounds =
CALCULATION OF POUNDS
Conc. x Flow (or Volume) x 8.34 Lbs/gallon
Flow (Volume) and Concentration must be
expressed in specific units.
22. Concentration must be expressed
as parts per million parts.
Concentration is usually reported as
milligrams per liter.
This unit is equivalent to ppm.
1 mg = 1 mg = 1 mg = ppm
liter 1000 grams 1,000,000 mg
ppm = Parts
Mil Parts
= Lbs.
Mil Lbs.
23. Flow or Volume must be expressed
as millions of gallons:
= MG
i.e.) A tank contains 1,125,000 gallons of water.
How many million gallons are there?
1,125,000 gal = 1.125 MG
1,000,000 gal/MG
1,000,000 gal/MG
gallons
24. Lbs. =
When Volume is expressed as MG
and concentration is in ppm,
the units cancel to leave only Pounds.
Concentration x Volume x 8.34 Lbs/gallon
Lbs.
M Lbs. X X
M gal Lbs.
gal
= Lbs
25. Lbs./Day =
When Flow is expressed as MGD
and concentration is in ppm,
the units cancel to leave Pounds/Day.
Concentration x Flow x 8.34 Lbs/gallon
Lbs.
M Lbs. X X
M gal Lbs.
gal
Day
= Lbs/Day
26. EXAMPLE:
How many pounds of suspended solids leave
a facility each day if the flow rate is
150,000 gal/day and the concentration of
suspended solids is 25 mg/L?
Lbs/day = Conc. (mg/L) x Flow (MGD) x 8.34 Lbs
gal
Lbs/day = 25 mg/L x 150,000 gal/day x 8.34 Lbs
1,000,000 gal/MG gal
= 25 x 0.15 x 8.34
= 31 Lbs/day
28. Example Problem
BOD Loading
An activated sludge plant receives 2.0 MGD from the primary
clarifiers at 120 mg/L BOD. Calculate the organic loading
(Lbs/D BOD) on the activated sludge process.
Work Calculation on Separate Paper
Answer Given on Next Slide
29. Example Problem
BOD Loading
An activated sludge plant receives 2.0 MGD from the primary
clarifiers at 120 mg/L BOD. Calculate the organic loading
(Lbs/D BOD) on the activated sludge process.
Lbs
Day
= 120 mg/L X 2.0 MGD X 8.34 Lbs
Gal
= 2001.6 Lbs BOD
Day
Lbs/day = Conc. (mg/L) x Flow (MGD) x 8.34 Lbs
gal
30. OXYGEN DEMAND
Biochemical Oxygen Demand
B.O.D.
The Quantity of Oxygen Used in
the Biochemical Oxidation of
Organic Material.
5 Day Test
31. OXYGEN DEMAND
Biochemical Oxygen Demand
B.O.D.
Best to Use a “Moving Average”
to Determine the Average Impact
on a Treatment System.
5 Day Test
32. BOD Moving Average
Date Pounds of BOD
9/29 2281
9/30 2777
10/1 1374
10/2 2459
10/3 960
10/4 1598
10/5 2076
10/6 1577
10/7 2351
10/5
2281
2777
1374
2459
960
1598
2076
13,525
13,525
7
= 1932
10/6
13,525
- 2281
+ 1577
12,821
12,821
7
= 1832
Calculate the 7 day moving average of pounds of BOD
for 10/5 and 10/6.
33. Need to Balance Organic Load (lbs BOD)
With Number of Active Organisms in
Treatment System
Food to Microorganism
Ratio
F:M or
F
M
34. How Much Food ?
How is M (Microorganisms) measured?
Mixed Liquor Volatile Suspended Solids
(MLVSS)
Lbs/D BOD = FLOW (MGD) X 8.34 Lbs/Gal X P.E. BOD (mg/L)
M = Pounds MLVSS
(In Aeration Tank)
F = Pounds BOD
(Coming into Aeration Tank)
Primary Effluent BOD
37. Wt. of Solids + Paper, mg
Wt. of Paper, mg
Wt. of Solids, mg
Determining MLSS
Solids
Wt. of Solids, mg
Volume of Sample, L
MLSS, mg/L
38. Determining MLVSS
Wt. of Dish + Solids, mg
Wt. of Dish + Ash, mg
Wt. of Volatile Solids, mg
Solids
Volatile
Solids
550 oC
Wt. of Volatile Solids, mg
Volume of Sample, L MLVSS, mg/L
39. How Much Food ?
How is M (Microorganisms) measured?
Mixed Liquor Volatile Suspended Solids
(MLVSS)
Lbs/D BOD = FLOW (MGD) X 8.34 Lbs/Gal X P.E. BOD (mg/L)
M = Pounds MLVSS
(In Aeration Tank)
F = Pounds BOD
(Coming into Aeration Tank)
Primary Effluent BOD
40. Analysis Gave Us M (MLVSS)
In mg/L
How Do We Get Volume ?
Lbs/D BOD =
Volume (MG) X 8.34 Lbs/Gal X MLVSS (mg/L)
Volume Of What ?
Where Microorganisms Are
Aeration Tank
How Do We Get To Pounds?
41. Aeration Tank Volume
(MG)
L (ft) X W (ft) X SWD (ft) = Volume (ft3)
ft3 X 7.48 gal/ft3 = gallons
gallons / 1,000,000 = million gallons (MG)
42. Aeration Tank Volume
(MG)
Example Calculation:
Calculate the volume in million gallons of an aeration
tank that is 120 ft long, 35 ft wide, with a SWD of 15 ft.
A.
V = L X W X D
V = 120 ft X 35 ft X 15 ft = 63,000 ft3
63,000 ft3 X 7.48 gal
ft3
= 471,240 gallons
= 0.471 MG
471,240 gallons / 1,000,000
43. Aeration Tank Volume
(MG)
Example Calculation:
The average BOD load on this aeration tank is 1954 lbs/day.
Calculate the organic loading in lbs/day/1000ft3.
B.
1954 lbs/day
63,000 ft3
X 1,000 = 31.0 lbs/day/1000ft3
44. Need to Balance Organic Load (lbs BOD)
With Number of Active Organisms in
Treatment System
Food to Microorganism Ratio
F:M or
F
M
45. How Much Food (F) ?
Pounds BOD
How is M (Microorganisms) measured?
Mixed Liquor Volatile Suspended Solids
(MLVSS)
Lbs/D BOD =
FLOW (MGD) X 8.34 Lbs/Gal X Pri. Eff. BOD (mg/L)
M = Pounds MLVSS
46. Pounds =
CALCULATION OF POUNDS
Concentration
Of STUFF
In the
Water
Conc. x Flow (or Volume) x 8.34 Lbs/gallon
X
Quantity
Of Water
The STUFF
Is In
Weight
Of The
Water
X
47. Pounds of Volatile Solids
in the Aeration Tank
Volume Aeration Tank, MG X MLVSS, mg/L X 8.34 Lbs/gal
Example Problem:
Calculate the pounds of volatile solids in an aeration tank
that has a volume of 0.471 MG and the concentration of
volatile suspended solids is 1700 mg/L.
Lbs MLVSS =
Lbs = 0.471 MG X 1700 mg/L X 8.34 lbs/gal
= 6678 lbs MLVSS
48. Food to Microorganism Ratio
Lbs of BOD
Lbs of MLVSS
Example Problem:
The 7-day moving average BOD is 2002 lbs and the mixed
liquor volatile suspended solids is 6681 pounds.
Calculate the F/M ratio of the process.
F
M
= 2002 lbs BOD
6681 lbs MLVSS
= 0.30
=
49. Conventional Activated Sludge
F:M
F:M
Extended Aeration Activated Sludge
0.25 - 0.45
0.05 - 0.15
Typical Range:
The F/M Ratio for Best Treatment Will Vary for
Different Facilities
Determined by Regular Monitoring and
Comparing to Effluent Quality
Often Will Vary Seasonally
Food to Microorganism Ratio
50. Food to Microorganism Ratio
Lbs of BOD
Lbs of MLVSS
Calculate Often to Monitor/Control
=
Monthly (Minimum)
Weekly (Better)
Use Moving Average
F
M
=
51. Food to Microorganism Ratio Calculations
F/M Ratio is Used to Determine the Lbs of MLVSS
Needed at a Particular Loading Rate
suppose F/M of 0.30 is desired
and BOD loading is 1200 lbs/day
F
M
= 0.30
1200 lbs
0.30
= 4000 lbs MLVSS
F
F/M
= M (Lbs MLVSS)
Lbs BOD
lbs MLVSS
=
F/M
F
0.30
= M
FOR DAILY USE
52. If we Know the Pounds of MLVSS Needed and the Volume
of the Aeration Tank We Can Calculate MLVSS, mg/L.
Calculate the MLVSS, mg/L given
an Aeration Tank Volume of 0.20 MG.
4000 lbs = 0.20 MG X 8.34 lbs X ? mg/L
gal
2398 mg/L
4000 lbs
0.20 MG X 8.34 lbs/gal
=
Food to Microorganism Ratio Calculations
53. Problem A:
How many pounds of MLVSS should be maintained in an
aeration tank with a volume of 0.105 MG receiving primary
effluent BOD of 630 lbs/d ? The desired F:M is 0.3.
= 630 lbs/d
0.3
= 2100 lbs MLVSS
F
F/M
= M
F:M Calculations
54. Problem B:
What will be the MLVSS concentration in mg/L ?
2100 lbs = Conc X 0.105 MG X 8.34 lbs/gal
2100 lbs
0.105 MG X 8.34 lbs/gal
= 2398 mg/L
F:M Calculations
55. Food to Microorganism Ratio Calculations
F/M Ratio is Used to Determine the Lbs of MLVSS
Needed at a Particular Loading Rate
Can you Calculate the Pounds of MLVSS Needed for a
Specific F/M
and
What Concentration That Would Be in an Aeration Tank?
Prove It !
F
F/M
= M (Lbs MLVSS)
Lbs BOD
lbs MLVSS
=
F/M
56. Problem C:
How many pounds of MLVSS should be maintained in an aeration
tank with a volume of 0.471 MG receiving primary effluent BOD of
2502 lbs/d ? The desired F:M is 0.3.
F:M Calculations II
Problem D:
What will be the MLVSS concentration in mg/L ?
Work Calculations on Separate Paper
Answers Given on Next Slides
57. Problem C:
How many pounds of MLVSS should be maintained in an
aeration tank with a volume of 0.471 MG receiving primary
effluent BOD of 2502 lbs/d ? The desired F:M is 0.3.
F .
F/M
= M = 2502 lbs/d
0.3
= 8340 lbs MLVSS
F:M Calculations II
58. Problem D:
What will be the MLVSS concentration in mg/L ?
8340 lbs = Conc X 0.471 MG X 8.34 lbs/gal
8340 lbs .
0.471 MG X 8.34 lbs/gal
= 2123 mg/L
F:M Calculations II
67. This graph illustrates that
the activities of
Microorganisms in a
biological treatment
system is related to the
Average Age of the
Organisms in the System
or the “CRT” of the System
Note: The CRT is
Controlled in an
Activated Sludge System
by Wasting which will be
discussed later.
Graph Showing Growth Phases in a
Biological System
68. Cell Residence Time, CRT
Mean Cell Residence Time, MCRT
Sludge Age, SA
MCRT =
Total MLVSS, lbs (Aerator + Clarifier)
Total MLVSS Wasted + Effluent TSS, lbs/d
SA, days =
Suspended Solids in Aerator, lbs
Suspended Solids in PE, lbs/day
The Average Length of Time in Days
that an Organism Remains in the
Secondary Treatment System
Biomass Age
The SA and MCRT Calculations are Seldom Used
The Most Common (and Best for Most Processes) Is the
Cell Residence Time
69. Cell Residence Time
The Average Length of Time in Days that an
Organism Remains in the Secondary
Treatment System
CRT, days = Total MLVSS, lbs
Total MLVSS Wasted, lbs/d
Cell Residence Time, CRT
70. The Average Length of Time in Days that an
Organism Remains in the Secondary
Treatment System
CRT, days = Total MLVSS, lbs
Total MLVSS Wasted, lbs/d
Example:
MLVSS = 6681 lbs
MLVSS Wasted = 835 lbs/d
Calculate the CRT.
6681 lbs
835 lbs/d
CRT =
CRT = 8.0 Days
Cell Residence Time, CRT
Cell Residence Time
71. Like The F/M Ratio
The CRT for Best
Treatment
Will Vary for Different
Facilities
Determined by Regular
Monitoring and
Comparing to Effluent
Quality
Often Will Vary
Seasonally
Cell Residence Time
72. Conventional Activated Sludge
Aerator Detention Time
F:M
CRT
Aerator Detention Time
F:M
CRT
Extended Aeration Activated Sludge
4 - 8 Hrs.
0.25 - 0.45
4 - 6 Days
16 - 24 Hrs.
0.05 - 0.15
15 - 25 Days
79. Cell Residence Time
The Average Length of Time in Days that an
Organism Remains in the Secondary
Treatment System
CRT, days = Total MLVSS, lbs
Total MLVSS Wasted, lbs/d
Waste Activated Sludge (WAS)
The CRT for Facility is Controlled/Maintained
by Wasting the Appropriate Amount of
Excess Biomass
81. Sludge Wasting Rates
CRT(days) = Lbs of MLVSS in aerators
Lbs/day WAS VSS
Therefore:
Lbs WAS VSS =
day
Lbs of MLVSS in aerators
CRT (days)
82. Sludge Wasting Rates
With a known RAS VSS concentration, the
WAS Flow in MGD can be calculated:
= WAS (MGD)
lbs/day WAS VSS
RAS VSS (mg/L) x 8.34 lbs
gal
MGD x 1,000,000 = gallons per day
Lbs/ day = mg/L x 8.34 lbs
gal
x ? MGD
83. Sludge Wasting Rates
If wasting is to be done over a 24 hr. period:
WAS (gpm) = gallons/day .
1440 minutes/day
If wasting is to be done over a
shorter period:
WAS (gpm) =
gallons/day .
min wasting to be done/day
84. Sludge Wasting Rates
Example Calculations
Problem #1:
A cell residence time of 5.8 days is desired. With 5800
pounds of MLVSS in the aeration tanks, calculate the
pounds of VSS that must be wasted per day.
lbs/day = 5800 lbs
5.8 days
= 1000 lbs/day
Need to Waste 5800 lbs in 5.8 Days
85. Sludge Wasting Rates
Problem #2:
Calculate the flow rate in MGD that must be pumped in order to waste
the number of pounds calculated in Problem #1 given a Return
Sludge concentration of 9000 mg/L VSS.
lbs/day = conc. x 8.34 lbs/gal x MGD
1000 lbs = 9000 mg/L x 8.34 lbs/gal x MGD
day
= 0.0133 MGD = 13,300 gal/day
1000 lbs/day
9000 mg/L x 8.34 lbs/gal
= MGD Wasted
86. Problem #3:
Calculate the wasting rate in gallons per minute if the
wasting was done in 24 hours.
Sludge Wasting Rates
13,300 gal
day
1 day
24 hrs
x x 1 hour
60 min
= 9.24 gals
min
87. Sludge Wasting Rates
Problem #4:
Calculate the wasting rate in gallons per minute if the
wasting was done in 4 hours.
13,300 gal
4 hr
x 1 hr
60 min
= 55.4 gal/min
88. Sludge Wasting
Excess Biological Solids eliminated from the
secondary treatment system to control the
cell residence time of the biomass
When to Waste:
Continuous (Whenever Possible)
Or If Necessary (Piping, Pumping or Valve Limitations)
Intermittent - During Low Load Conditions
90. Sludge Wasting
Where to:
Solids Handling
Disadvantage – Thinner Solids to Solids Process
Advantage – Know Solids are Out of the System
Sludge Processing (Thickening, Stabilization, etc.)
RAS
WAS
91. Sludge Wasting
How Much:
Secondary Sludge Wasting One of the Most
Important Controls
Wasting Controls the Most Important Aspect
of Treatment, the Biomass Population
95. Growth rate produces about
0.7 lbs of biological solids per
lb BOD removed
Activated Sludge
Need favorable conditions for growth and
for separation from the water
Biological solids are used
over and over
Suspended Growth,
Biological Treatment
97. Yield Coefficient (Y)
Growth Rate
Pounds of Biological Solids Produced
Per Pounds of BOD Removed
Example
Average Concentration Of BOD Entering Aeration
125 mg/L
Average Concentration of BOD from Secondary System
5 mg/L
Average Plant Flow
2.0 MGD
Average RAS Concentration (Wasting from Return)
8000 mg/L
98. Yield Coefficient (Y)
Growth Rate
Pounds of Biological Solids Produced
Per Pounds of BOD Removed
Example
BOD Removed = 125 mg/L – 5 mg/L = 120 mg/L
At 2.0 MGD
Lbs BOD Removed = 2 MGD X 8.34 X 120 mg/L
= 2002 Lbs/Day
At Y= 0.7
Biomass Produced = 2002 Lbs/Day X 0.7
= 1401 Lbs/Day
99. Yield Coefficient (Y)
At 2.0 MGD
Lbs BOD Removed = 2 MGD X 8.34 X 120 mg/L
= 2002 Lbs/Day
At Y= 0.7
Biomass Produced = 2002 Lbs/Day X 0.7
= 1401 Lbs/Day
100. Yield Coefficient (Y)
At 2.0 MGD
Lbs BOD Removed = 2 MGD X 8.34 X 120 mg/L
= 2002 Lbs/Day
At Y= 0.7
Biomass Produced = 2002 Lbs/Day X 0.7
= 1401 Lbs/Day
RAS at 8000 mg/L
1401 lbs/day .
8000 mg/L X 8.34 lbs/gal
= 20,998 gallons WAS
(To Balance Solids Produced)
101. Yield Coefficient (Y)
At 2.0 MGD
Lbs BOD Removed = 2 MGD X 8.34 X 120 mg/L
= 2002 Lbs/Day
At Y= 0.5
Biomass Produced = 2002 Lbs/Day X 0.5
= 1001 Lbs/Day
RAS at 8000 mg/L
1001 lbs/day .
8000 mg/L X 8.34 lbs/gal
= 15,002 gallons WAS
(To Balance Solids Produced)
102. Yield Coefficient (Y)
The Difference:
20,998 gallons
15,002 gallons
5,996 gallons
6000 gal/day X 365 day/year = 2,190,000 gallons per year
Per Day
103. 0 0.20 0.40 0.60 0.80 1.00 1.20
F:M Ratio
Conventional
Extended
Air
High Rate
300
200
100
Sludge
Volume
Index
A Major Advantage of Extended Aeration
(Old Sludge Age)
Less Solids Produced
104. Yield Coefficient (Y)
Growth Rate
Y =
Pounds of Biological Solids Produced
Per Pound of BOD Removed
How to Determine Y for a Facility?
Use Monthly Average of Pounds of Solids Wasted
Divided by
the Monthly Average of Pounds of BOD Removed
Should be Monitored Regularly (Monthly)
105. Returned from Secondary
Clarifier
Activated Sludge
Need favorable conditions for growth and
for separation from the water
Biological solids are used
over and over
Suspended Growth,
Biological Treatment
108. Return Activated Sludge
Biological Solids (Mixed Liquor Solids) which
have settled in the secondary clarifier,
continuously returned to the aeration system.
Why:
• Control sludge blanket in clarifier
• Maintain a sufficient population of
active organisms in service
It’s Not the Food
It’s the Bugs
109. Return Activated Sludge
Biological Solids (Mixed Liquor Solids) which
have settled in the secondary clarifier,
continuously returned to the aeration system.
Why:
• Control sludge blanket in clarifier
• Maintain a sufficient population of
active organisms in service
Not a Means of
Controlling MLSS
110. Return Activated Sludge
Biological Solids (Mixed Liquor Solids) which
have settled in the secondary clarifier,
continuously returned to the aeration system.
Why:
• Control sludge blanket in clarifier
• Maintain a sufficient population of
active organisms in service
Controls Solids
Depth in
Seconday Clarifier
111. • 1 – 3 Feet Depth
• Too Much – Solids Over Weir
• Too Little – Thin RAS Concentration
(More Volume When Wasting)
RAS Control:
Return Activated Sludge
112. • Consistent Flow Rate
• % Influent Flow
• RAS Metering
RAS Control:
Setting the RAS Rate
How Much is Enough
(or Too Much)
Sludge Blanket Depth
Sludge Judging
Return Activated Sludge
113. Q
RQ
Q + RQ
RAS
MLSS
(RQ + Q) X 8.34 lbs/gal X MLSS (mg/L)
Lbs of Material Into Clarifier
RQ X 8.34 lbs/gal X RAS (mg/L)
Lbs of Material Out of Clarifier
RAS Mass Balance
114. Lbs Into Clarifier = Lbs Out of Clarifier
Q
RQ
Q + RQ
RAS SS
MLSS
(Q + RQ) X 8.34X MLSS = RQ X 8.34 X RAS
(Q + RQ) X MLSS = RQ X RAS SS
(RQ X MLSS) + (Q X MLSS) = RQ X RAS SS
Q X MLSS = RQ X RAS SS - RQ X MLSS
Q X MLSS = RQ X (RAS - MLSS)
Q X MLSS
RAS SS - MLSS
= RQ
RAS Mass Balance
115. Q X MLSS
RAS SS - MLSS
RQ =
Most People Forget the Derivation of the
Formula and Just Memorize the Formula
Return Activated Sludge
116. Q X MLSS
RAS SS - MLSS
RQ =
Units for RQ will Match Units for Q
100 X MLSS
RAS - MLSS
% RQ =
Return Activated Sludge
To Express RQ as % of Influent Flow:
117. Return Rates - Example Calculations
Given: MLSS = 2400 mg/L
RAS SS = 6500 mg/L
Flow = 2.0 MGD
1. Calculate the Return Sludge Rate in MGD
needed to keep the solids in the process in
balance.
RAS, MGD = 2.0 MGD X 2400 mg/L
6500 mg/L - 2400 mg/L
= 4800
4100
= 1.17 MGD
118. Given:
MLSS = 2400 mg/L
RAS SS = 6500 mg/L
2. Calculate the Return Sludge Rate in
% of plant influent flow needed to
keep the solids in the process in balance.
% RAS = 100 X 2400 mg/L
6500 mg/L - 2400 mg/L
% RAS = 240000
4100
= 58.5 %
Return Rates - Example Calculations
119. Return Rates - Practice Calculations
Given: MLSS = 2700 mg/L
RAS SS = 8200 mg/L
Flow = 2.5 MGD
1. Calculate the Return Sludge Rate in MGD needed
to keep the solids in the process in balance.
2. Calculate the Return Sludge Rate in
% of plant influent flow needed to
keep the solids in the process in balance.
Work Calculations on Separate Paper
Answers Given on Next Slides
120. Return Rates - Practice Calculations
Given: MLSS = 2700 mg/L
RAS SS = 8200 mg/L
Flow = 2.5 MGD
Calculate the Return Sludge Rate in MGD needed
to keep the solids in the process in balance.
RAS, MGD = 2.5 MGD X 2700 mg/L
8200 mg/L - 2700 mg/L
= 6750
5500
= 1.23 MGD
1.
121. Given:
MLSS = 2700 mg/L
RAS SS = 8200 mg/L
2. Calculate the Return Sludge Rate in
% of plant influent flow needed to
keep the solids in the process in balance.
% RAS = 100 X 2700 mg/L
8200 mg/L - 2700 mg/L
% RAS = 270,000
5500
= 49.1 %
Return Rates - Example Calculations
122. Return Activated Sludge
Q X MLSS
RAS SS - MLSS
RQ =
Units for RQ will Match Units for Q
100 X MLSS
RAS - MLSS
% RQ =
To Express RQ as % of Influent Flow:
In Summary
125. Biological Wastewater
Treatment
Three Steps
3. Flocculation and Solids Removal
Even if the First Two Steps are Effective,
If Settling and Separation is Poor
RAS Will be Thin and/or Solids May Be Lost in the Effluent
128. Although a 1000mL
Graduated Cylinder
May be Used
A Settleometer
Designed for this
Test is Best
The Wider Container More Approximates a Clarifier
Settleometer Test
129. Although a 1000mL
Graduated Cylinder
May be Used
A Settleometer
Designed for this
Test is Best
Settleometer Test
A Settleometer has a
Capacity of 2000 mL
Graduated in
mL/Liter
132. Settleometer Test
While Settling Observe:
Color of ML and Supernatant
Supernatant Turbidity
Straggler Floc
Record
Settled
Sludge
Volume
Every 5
Minutes for
30 Minutes
139. Watch for Indications of Denitrification
Gas Bubbles in Settled Sludge
Rising Sludge
Solids Separation
Rate
Characteristics
140. Sludge Volume Index (SVI)
The volume in milliliters occupied by one gram
of activated sludge which has settled for 30 min.
mLs Settled in 30 min
SVI = MLSS Conc, grams/L
=
mLs Settled
MLSS, mg/L
1000
The volume compared to weight.
(Weight [in grams] of the solids that occupy the Volume.)
141. The volume in milliliters occupied by one gram of
activated sludge which has settled for 30 min.
SVI Practice Problem:
30 minute settling 260 mL
MLSS Conc. 2400 mg/L
mLs Settled
MLSS, mg/L
1000
mLs Settled in 30 min
SVI = MLSS Conc, grams/L
=
Work
Calculations on
Separate Paper
Answer Given
on Next Slide
Sludge Volume Index (SVI)
142. The volume in milliliters occupied by one gram of
activated sludge which has settled for 30 min.
SVI Practice Problem:
30 minute settling 260 mL
MLSS Conc. 2400 mg/L
260
2.4
= 108
SVI =
260 mL
2400 mg/L
1000
SVI =
mLs Settled
MLSS, mg/L
1000
mLs Settled in 30 min
SVI = MLSS Conc, grams/L
=
Sludge Volume Index (SVI)
143. The volume in milliliters occupied by one gram of
activated sludge which has settled for 30 min.
mLs Settled
MLSS, mg/L
1000
mLs Settled in 30 min
SVI = MLSS Conc, grams/L
=
Typical Range for Good Settling 80 - 120
The higher the number, the less compact the sludge
Sludge Volume Index (SVI)
144. The grams of activated sludge which occupies a
volume of 100 mL after 30 min. of settling.
SDI =
grams/L of MLSS
mLs settled in 30 min.
100
The weight compared to volume.
Sludge Density Index (SDI)
145. The grams of activated sludge which occupies a
volume of 100 mL after 30 min. of settling
30 min. Settling / 100
SDI =
MLSS / 1000
The weight compared to volume.
Sludge Density Index (SDI)
146. The grams of activated sludge which occupies a
volume of 100 ml after 30 min. of settling
SDI Practice Problem:
30 minute settling 260 mL
MLSS Conc. 2400 mg/L
mLs settled in 30 min.
SDI =
grams/L of MLSS
100
Work
Calculations on
Separate Paper
Answer Given
on Next Slide
Sludge Density Index (SDI)
147. The grams of activated sludge which occupies a
volume of 100 ml after 30 min. of settling
2.4
2.6
SDI = = 0.92
SDI Practice Problem:
30 minute settling 260 mL
MLSS Conc. 2400 mg/L
mLs settled in 30 min.
SDI =
grams/L of MLSS
100
2400 mg/L / 1000
260 mL / 100
SDI =
Sludge Density Index (SDI)
148. The grams of activated sludge which occupies a
volume of 100 ml after 30 min. of settling
mLs settled in 30 min.
SDI =
grams/L of MLSS
100
Typical Range for Good Settling 0.8 - 1.2
The lower the number, the less compact the sludge
Sludge Density Index (SDI)
149. SVI - SDI Relationship
SDI =
100
SVI
SVI =
100
SDI
150. SVI - SDI Relationship
SVI =
100
SDI
SDI =
100
SVI
Practice Problems:
a) What is the SDI if the SVI is 133?
b) What is the SVI if the SDI is 0.6?
Work Calculations on Separate Paper
Answers Given on Next Slide
151. SVI - SDI Relationship
SVI =
100
SDI
SDI =
100
SVI
Practice Problems:
a) What is the SDI if the SVI is 133?
b) What is the SVI if the SDI is 0.6?
100/133 = 0.75
100/0.6 = 167
152. Return Sludge Concentration
and SDI
With the Clarifier Solids in Balance, the Settled
Sludge Concentration in the Settleometer
Will Approximate the RAS SS Concentration
153. Return Sludge Concentration and SDI
SDI = mLs settled in 30 minutes
MLSS, G/L
100
SDI 1.0 =
100 mLs settled
1.0 G
1 G
100 mL
1 G
100 G
= 1 %
=
1 G
100 mL
= 1000 mg
100 mL
=
10,000 mg
1,000 mL
10,000 mg
L
=
154. With Clarifier Solids in Balance :
SDI = RAS SS Conc. in Percent
SDI of 0.8
RAS SS = 0.8 % Solids
SDI X 10,000 = RAS SS in mg/L
SDI = 0.8
RAS SS = 8,000 mg/L
Return Sludge Concentration and SDI
155. Sludge Volume Index
The volume in milliliters occupied by one gram
of activated sludge which has settled for 30 min.
The volume compared to weight.
Sludge Density Index
The grams of activated sludge which occupies a
volume of 100 mL after 30 min. of settling.
The weight compared to volume.
In Summary
156. Sludge Volume Index
Sludge Density Index
mLs settled in 30 min.
SDI =
grams/L of MLSS
100
mLs Settled
MLSS, mg/L
1000
SVI =
157. SVI - SDI Relationship
SDI =
100
SVI
SVI =
100
SDI
158. SVI - SDI
Typical SDI Range for Good Settling 0.8 - 1.2
Typical SVI Range for Good Settling 80 - 120
159. This graph illustrates the
Relationship Between The
F:M of a System to the
Ability of the Biomass to
Settle in Clarifier
Relationship of F:M to Settleability
System
It Shows that there are
Three Areas of Operation
where the Biomass
Normally Settles Well
160. These Areas as Defined by
F:M Ratio Are
Note: The High rate
Mode is Seldom Used
Except when Followed
by Additional Treatment
Relationship of F:M to Settleability
System
High Rate
F:M 0.9 to 1.2
Conventional
F:M 0.25 to 0.45
Extended Air
F:M Less than 0.2
161. The Graph Also Shows the
Potential Consequences of
Operation with an F:M Out
Of these Ranges
Relationship of F:M to Settleability
System