 Hello friends. My name is Mahirubali. I am an assistant professor in WIT, Singapore. Today we are going to learn about activated sludge process and its analysis. So what are the learning outcomes from this session? Basically we are going to learn about what is the analysis of activated sludge process in wastewater treatment. So let's study that how we are going to do whole this process just to create the activated sludge process with a proper quantity. How much we have to add the BOD? How much we have to fix the detention time of aeration? Then how much we are going to fix the nutrient content? For these whole questions, we are going to study about the four analysis and the four criteria for activated sludge process. Let's study first one is hydraulic retention time that is HRD which is mostly focused on the detention time of wastewater in the aeration tank. The second one is volumetric BOD loading. How much BOD we are adding in the aeration time? Organic loading which is F by M ratio. What is F and what is M? We are going to study further. Fourth one is solids retention time that is SRD. Let's study the first one hydraulic retention time. It is mostly the volume of aeration time with respect to the detention time on which we are putting the wastewater into the aeration time. How we can calculate it? We are calculating by using the formula V by Q into 24 where V stands for volume of aeration tank in meter cube and Q is the flow rate in meter cube per day. So understand the concept of formula. Here it is a volume and it is a discharge. So by dividing it, we are getting the time parameter and what it is representing? It is representing the time of wastewater in the aeration time. The next one is volumetric BOD loading. So how it is different than the HRD? BOD loading means how much carbonaceous matter which is present in the wastewater is going to applied in the aeration time. Now we have to understand see the volume of wastewater is different and the BOD loading is different. It may happen that a required volume of water can be increased BOD and it can be with the reduced BOD. It means the typical volume of wastewater can have higher rate of carbonaceous matter present in it or lower rate of carbonaceous matter present in it. So while adding the wastewater into the aeration time, we have to calculate how much carbonaceous matter we are adding into the aeration time. So for that we have to calculate a BOD applied per unit volume. How much carbonaceous matter we are going to add? Because the carbonaceous matter and the quantity of carbonaceous matter will give us a proper formation of microorganisms and secondarily a activated microorganisms. That's why we have to calculate how much BOD we have to add. So how we can calculate it? First we have to calculate by experimentation how much BOD is present in the wastewater and it is termed as LA and it is measured in MG per liter. So it is multiplied by the flow rate of wastewater and divided by volume of aeration tank. This formula will give us a volumetric load in terms of Kg BOD5 per meter cube. Now what is F-bomb ratio? Here F means food. What is our food? It is carbonaceous matter which we are going to degrade. It is the food which is measured in terms of BOD5 which is coming into the aeration tank and M stands for microorganisms which are present in the aeration tank. They can be coming up from the activated sludge or they can be coming up with the food also, with the influent wastewater also because the influent wastewater also contains some amount of microorganisms. So addition of activated sludge microorganisms plus the influent microorganisms totally it will give us the microorganisms present in the aeration tank. Don't confuse which microorganisms basically we are going to take in it. For the particular proportion which we want to get in the aeration tank with respect to the nutrients which we had seen in the earlier, the C into N raised to P we maintain the nutrient proportion in the aeration tank for the proper degradation of carbonaceous matter. For proper proportion we have to fix the F-bomb ratio. So how much we keep it? Typically it is kept by 0.25 to 0.6. If you see here this is our food and these are our microorganisms in the aeration tank. So for the proper decomposition of proper degradation of carbonaceous matter by microorganisms we have to provide a proper food and proper microorganisms. If anyone of these fails to maintain this ratio whole activated sludge process is going to give us the poor efficiency. For a greater efficiency we have to fix our food also and our microorganisms also because these both points are very much important for us. There must be one producer and there must be one decomposer. Then and only then the food chain will properly grows the food chain is properly maintained and the same logic has been applied in the activated sludge process also. So how the F-bomb ratio is important for us and where we are focusing on. In the earlier session we had studied about the declination of food in the tank and the increase of microorganisms to the peak and declination and going towards the regeneration process of microorganisms we have studied. In this growth phase where our food to microorganisms ratio is going to be focused is more important for us. If you see if we are putting the higher F-bomb ratio it means the food is present too much at a higher than the microorganisms at that time the dispersed growth will be form of microorganisms it means a very less quantity of microorganisms are going to form. So these stage will not give us a proper activated microorganisms which we are required as the activated microorganisms. So these phase will going to fail us. Now the second one is ideal F-bomb ratio. In the ideal F-bomb ratio the microorganisms are very higher rate than the food. Here the food is in the minimum condition and the microorganisms are relatively higher. What they will do they will take all this food from here and they will decrease the food and they will increase themselves. Fine in the aeration tank and then a large quantity of a flocked formation region will be there in the aeration tank. So this condition is our ideal condition. This condition will maintain the food to microorganisms ratio because food is continuously coming up. It is not a batch process it is a continuous process. So food is continuously coming up and it is going down continuously coming up and going down. That is what we wanted. These food we have to decrease it and we have to maintain the microorganisms to the higher level. If we are putting the microorganisms in the higher level then and only then the activated microorganisms are going to increase. So this condition is our favorable condition. Fine then if we are putting the low food to mass ratio. Now you have to understand here the microorganisms are going to decrease and relatively the food is very much lower. That's why the food to microorganisms ratio is what we call a lower ratio. Now you have to understand here also a very dispersed flock generation will be there. Therefore these microorganisms will giving out our dispersed flock. That's why we don't use these kind of food to microorganisms ratio. Now how we can calculate the food to mass ratio. Here we use F-bomb ratio where the food is calculated in terms of BOD in mg-polluter and microorganisms is calculated in terms of PPM or mg-polluter of MLSS. That is our small XT. Where F-bomb ratio is calculated as Q into La divided by V divided by 1000 into Qt. What does it mean? Q is a flow rate, La is our BOD and V is the volume and XT is our microorganisms in MLSS which we calculate experimentally. The fourth one is solids retention time. Here as we are adding food as we are calculating the microorganisms and simultaneously we are calculating the carbonaceous matter. After eating there is some degradation and there is some formation of sludge. Whatever the sludge and the MLSS combined together and the presence of such kind of a combination in the aeration tank gives out SRT. It is a time in which this sludge is present in the aeration tank so SRT is highly different than HRT. HRT is the volume of wastewater and SRT is the volume of sludge or MLSS which is present in the aeration tank. So both are different. SRT is donated as Qc and it is formulated as X divided by delta X divided by delta T. Where X is a total microbial mass present in the reactor which is our aeration tank and delta X divided by delta T is the total quantity of solids withdrawn daily including solids deliberately wasted and those in the effluent. So let's have some review questions. First one is increase in HRT reduces efficiency of ASP. True or false? F by M ratio reduces by increase in and reduced by CRT. Third one in ASP increase in return activated sludge increases. So answers for the review questions are increase in HRT reduces the efficiency of ASP it's true. F by M ratio reduces by increasing the COD and reduced BOD. In ASP increase in return activated sludge increases F by M ratio SRT, BOD file loading rate and all of the above. These are the references I have used for the making of this PPPT. Thank you.