Introduction to Primary Clarifier and Activated Sludge Process

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[00:00:01]Hello everyone. Uh myself Dr. G. Wang Nalendra, assistant professor at Assad University uh working in the department of civil engineering. Today we will discuss about the primary sequence tank and activity slots and trickling filters.

[00:00:17]So in the previous uh lecture means we discussed about the um what is the purpose of wastewater treatment and uh we discuss the screens, grid chambers and equilis tank. In this lecture we will discuss about the primary segmentation tank and followed by ASP and filters. Now means primary segmentation tank. So it is located after the grid chamber. It removes lightweight organic suspended solids.

[00:00:52]Settling type will be the fl settling.

[00:00:55]Then uh so here means we need to provide particular time that is called as detection time. Then only the suspended solids get settled bottom of the tank.

[00:01:10]So here detention time will be the uh major factor uh to remove the suspended solids.

[00:01:20]So see here from the uh primary cmentation tank the b removal is around 30 to 40% and suspended solids remove around 50 to 70%. So this is the design criteria. The velocity should be around 1 cm/s and detection time will be providing around 90 to 150 minutes. This uh ding time will be provided based on the uh concentration of suspended solids as well as the p concentration. The effective particles if it is greater than 0.1 mm then this primary sedimement tank will be effective. Then what are the tank features? Means uh here uh design purpose it should be in a rectangular shape maybe in a circular tanks. So in a rectangular tanks the flow along the length and circular tanks the flow will be radial flow. So what kind of effent weights? See rectangular then another circular rectangular here means outlet end the circular will be the per very. So where loading rate is uh means less than 185 m cube per meter per t.

[00:02:39]So what are the factors or affect uh factors affecting the efficiency? So one will be the ed currents wind turbulence then thermal convention currents density currents and statisfication.

[00:02:55]So these are the factors it influencing the efficiency. So this is the efficiency relationship.

[00:03:03]So R will be the removal efficiency equal to time divided by A + B into time. Here A B are the constants and T will be the extinction time or we already discussed it is the removal efficiency.

[00:03:24]So design corrections is required for the field uh conditions. Then overflow rate should be multiply with 65 to 085.

[00:03:36]Detection time also it needs to be multiply with 1.25 to 1.

[00:03:43]It accounts for the non ideal conditions.

[00:03:47]These are the design conditions for the primary sedimentation tank.

[00:03:53]So we uh we will discuss about the one numerical problem about the primary sedimentation tank. See here a primary segment tank treats waste water with the following data. What are the data? Flow rate is given. Initial detection time is given as 1.8 hours. Then these are the empirical constants. We discussed a b values. The surface overflow road overflow rate theoretically it is given as 35 m cube per meter squared per day.

[00:04:27]So due to field conditions the correction factors also given the correction factors in previously we discussed it is between this range to this range. So here fixed values given for the detention time it is around 1.4 four and then overflow rate is around cure factor will be 0.75.

[00:04:49]So in this question they ask to calculate the theoretical removal efficiency that is the R1 then corrected detection time. So detection time theoretical they provided for a practical application it may vary the factor also is given that actual remove efficiency if the detection time will be varied the removal part removal efficiency also will be varied how much that also needs to be calculate then required surface area of tank also needs to be calculated before and after correction so correction needs to be done here for warflow rate here also is given 0.75 What will be the before correction? What will be the after correction? And last one of the comment and how correction affect the design. Is it this correction will affect the design conditions or not? We need to check. These are the five means uh uh problems we need to solve in this design problem. So as for our discussion uh the efficiency means theoretical uh removal efficiency how to calculate R1 equal to T by A + B into T.

[00:06:04]See here so T is given as 1.8 hours then A also is given 4 then B also is given 6 then time will be the 1.8 it is coming around 121.6%.

[00:06:22]So since efficiency cannot be more than 100%.

[00:06:27]So here efficiency we are considering as 100%.

[00:06:32]So why we are means from 121 to 100 means the maximum is 100% only for practically. So we are limited to 100%.

[00:06:43]Next step will be the we need to go for the detection time correction. So here the correction factor is given as 1.4 in the given question. So 1.8 is the theoretically detection time. Then for practical applications we are multiplying with the 1.4 then it is coming is around 2.52 hours. So next step is the actual removal efficiency.

[00:07:12]we need to calculate. So when we need to calculate after the correction of detection time so time is a new time will be the 2.52 instead of 1.8 hours.

[00:07:25]So 2.52 divided by a b parameters we know the same t value. So we we got around 1 131.8% 8 percentage here. Here also the efficiency percentage will be more than 100%.

[00:07:46]So we are considering upper limit as 100%. So efficiency here also is getting around 100%.

[00:07:55]Next one is the surface area calculations before correction.

[00:08:01]So area how to correct the area? area equal to what are the parameters existed based on the given data how we need to calculate the area. So Q discharge divided by overflow rate. So here area equal to overflow rate will be done. So Q value will be 36,000 m cube per day. So here also is given 35 m cube per me square per day. So based on these parameters the area could be around,28.6 m².

[00:08:38]This is the before correction surface area. So next we need to go for the area after correction. The character overflow rate will be the means theoretically it is around 35. The corrector factor will be 0.75 given in the question. Then correct workflow rate will be 26.25 m cube per day per me.

[00:09:06]So here area means modified area correction after area will be 36,000 divided by 26.25.

[00:09:16]It is coming around 1371.4 male.

[00:09:24]So these are the means area is increased then efficiency will be the same after correction. So the area increased from,28 to 1371 then deduction time increased from 1.8 to 2.52 hours ensues. So better settling under real conditions.

[00:09:48]So what we means discussed here in the primary sedimentation tank we need to calculate the removal efficiency. So here removal efficiencies basically depend upon the detection time as well as the surface area. If you are means changing the these two parameters how the efficiency will be varied. Basically this question will be understood that the these parameters are very important uh to settling of suspended maybe to degrade of organic matter that is BOD.

[00:10:23]So it's very important. So in the means uh uh in this problem we calculated these all the parameters what we discussed.

[00:10:36]So next uh process will be secondary treatment that is activated sludge process. So activated sludge process is used during secondary treatment of waste water. Activate sludge is a mixture of bacteria fungi protojava and rotifiers and maintain in suspension by aation and mixing. So this basically means uh in aation tank we need to means provide a kind of bacteria as well as the organic matter present. So uh that's why means it is kind of mixture of bacteria fungi proto and rotifiers maintained in suspension by aation and mixing so that whatever food is coming uh in the influen can degrade by the microorganisms.

[00:11:29]So in this process a biomass of aerobic organisms is grown in a large erated basins. These organisms break down the waste and use it as their food to grow themselves. So activated sludge process retain sludge uh settled sludge to the aation patient in order to maintain the right amount of organisms to handle the incoming food. activated sludge process have been removal efficiency in the in the range of 95.95 to 98 percentage. Uh then the trickling filters see in the primary clarifier or what will be the removal of B it is around 30 to 35%.

[00:12:17]So while means after next to primary clarifier if you are means uh choosing the means right secondary uh treatment unit as ASP then it should be removal of B is around 95 95 to 80 98% but here in the two separate units is there one is the aation tank and followed by secondary clarify in the primary Means uh uh after 5 minutes segment is tank there will be aation tank. Here we need to provide the aation so that whatever the microorganisms present in the tank should be in a suspended state.

[00:13:00]Then whatever the organic matter present in the water it gets eaten by the microorgans and microorganisms population will be grown and the next to the aation tank there will be secondary clarifier. Here we need to provide some kind of uh residency time so that means whatever the sludge it is there it gets settled at the bottom of the tank.

[00:13:25]Then to maintain the microorganisms to organic matter. So some of the return sludge we need to send back to back again to the aeration tank. So means it is the recycling ratio. Recycling means what are the means after treating sludge needs to be sent again to the main tank.

[00:13:48]So that ration needs to be calculated and also what will be the and means uh means uh recycle biomass concentration and what what kind of means concentration concentration is coming from the uh secondary clarifier we need to calculate.

[00:14:11]See next slide uh working of a activated sludge process. So here so we already discussed the primary clarifier that is over the removes the art of means part of suspended solids reduce the organic load and the ASP. So that's why primary clarifier is there before of ASP. So next after picop we explain aation tank.

[00:14:38]It here contains influent and activated sludge. Activated sludge means here mixer liquid. It contains different kinds of pathogens. May bacteria means bacter contain me within bacteria kind of sulfate reducing bacteria. These kind of different kinds of bacteria existed on the uh this clutch. So MLSS control here maintain for effective treatment efficiency we need to maintain the particular microbial growth. Then next one the ariation supplies oxygen and keep sludge in the suspension. So generally means we are taking SL it contains different bacteria. If it is settled at bottom it won't take the organic matter present in the waste water. So it should been always in the suspended state. So for that we are providing the irriation to the tank. So next one is the followed by secondary setting separates the treated water and sludge. Here if you are providing some kind of detection time what will happen the means sludge will be settled then remaining treated water will be sent to the further treatment if it is required otherwise they will dispose on the water bodies. Then efficient clear water dised after treatment for suppose municipal waste water it is not required tiary treatment. If it is industrial waste water after secondary treatment also needs to go for the ter treatment that is special treatment is required then return sl. So here the portion of recycle sl back to the aeration tank.

[00:16:23]Why we need to provide again the settled salts to the aeration tank means so it should be balanced between the organic matter as well as the MLSS concentration that is the mixed liquid suspended solids that is the bacteria growth. If it is not maintained what will happen the organic matter present in the waste water will not be degraded. So effent concentration of B will be higher. So to maintain uh this uh eine concentration low we need to maintain the MLSS concentration and organic matter should be in the balance that that is called as food to microorganisms needs to be calculated then excess sl.

[00:17:05]So whatever the means some portion of sl we are sending back to the aeration tank then remaining sl we need to go for the further treatment before disposing on the environment. So here we need to go for the anorobic digestion. Maybe aerobic digestion of is there after kind of this treatment for excessive sl it is called as nutrientrich uh compost that we can use for another purposes. This is the basically working of activation process. So main main mechanism here is the so to maintain bacteria in a suspended state. So water is coming in the reactor in horizontal for direction. So that it is in suspended state back means organic matter is break down by the microorgan.

[00:18:01]So what are the means b present in the water get reduced? It is eaten by the microorgans. If they are taking this food then it it goes by the so yield also will be increases means growth of bacteria will be increases. So micro growth will be increases means b concentration also will be reduced. Why we need to send again to the cetus also to the aeros tank means here to maintain the microorgan.

[00:18:43]So this is the typical diagram of ASB. C is the influent. Q will be the influent flow. It is coming from the primary clarif concentration. Initial BOD concentration that is called as food. Food d will be the bod xnaut. Xnaut means here the microorganisms concentration that is called as xnot. That influent is going to the aiation tank. This is the aiation tank. uh so here we are providing the some kind of mechanical aeration maybe we are providing aeration to the reactor so that what will happen the sludge won't be settled it will be always in the means suspended state so whatever the bacteria is there in the mean whatever the organic matter present in the waste water it can be eaten by the microorganisms then next unit is followed by aation will be the clarifier Here we are providing some retention time. So here the sludge will be settled at the bottom of the tank.

[00:19:59]Some portion of sludge we are sending back to the aeration tank. Those concentration see QR will be the uh written flow rate.

[00:20:12]So XR will be the return or under uh underwater biomass concentration. SR will be the underwater uh effilient V concentration. Then SL here the amount of sludge wasted here means flow will be wasted flow will be QW XR will be the underwater biomass concentration. Here efflant will be Q minus QW. Here Qw we are wasting.

[00:20:46]So remaining water will be in a circular. So here the effluent flow rate will be Q minus QW and XC will be the effluent biomass concentration. S E will be the effilient B concentrations. The concentration if you want to dispose on any water body the concentration B concentration should be below 30 mg per liter. So this is the one of the efficiency uh efficient process to remove the BOD it is around 85 to 90 90 percentage.

[00:21:27]So these are the abbreviations I have explained in the previous slide and the units also given it is very important.

[00:21:37]You need to see volume in a matrix cube.

[00:21:39]Sn what is the S not initial B concentration substrate concentration. S will be the aient. X not will be the concentration of biomass. XR will be the concentration of biomass in the return line. XR is the con bomass sludge drain.

[00:21:56]Here XR and XR these both are same only the XC value will be the concentration of biomass in the effluent.

[00:22:09]So design considerations.

[00:22:14]So here for a design of activated sludge process here is the two equations already derived uh from it is equations as taken from the metaf and ad. So next class I will show the uh means uh derived equations.

[00:22:36]C. Now equations used for design of erration tank. Theta C will be volume multiply with the biomass concentration in the aeration tank divided by QW into XR.

[00:22:56]QW that is the wasted means water. The XR will be the underground micron concentration. XC will be the airing tank biomass concentration. Here also one more equation 1 by theta C equal to Q will be the flow rate. Y will be the yield. S will be the initial B concentration. S will be the effluent and V X KD will be the decay constant. M during the growth of microns.

[00:23:28]So decay rate also will be there. So we need to consider decay decay rate also.

[00:23:34]Then next the mass balance around clarify. So is the final equation here will be how to calculate the QR. QR = Q into X minus QW into XR divided by XR - X. This is the way we need to calculate the recycle flow rate. Then recycle ratio will be QR divided by Q.

[00:24:03]So now we will solve one numerical problem. How to calculate the recycle ratio? How to calculate the means wasted volume? How to calculate the uh kind of volume of the aeration tank? So you see read the question properly. So an activated sludge process is to be used for secondary treatment of 15,000 m cube per of municipal waste water. So water is coming from the municipal waste water. So after primary clarification the B is 170 mg per liter. So what is the initial purity concentration 170 and it is desired to have not more than 25 mg per liter. It is a given target the effent concentration won't be more than 25 means that is s e equal to 25 mg per liter is given. So a completely mixed reactor is to be used and pilot plant analysis has established the following values. See here hydraulic detection time that is theta C is given as 10 days. E coefficient Y = 0.5 kg per kg.

[00:25:26]Then KD value decay constant is given as 0.05 inverse. So next will be the assuming an MLSS concentration of 4,500 mg per liter and and underflow concentration of 12,000 means here MLS concentration means X = to 4,500 mg per liter is given underflow concentration means XR will be 12,000 mg per liter from the secondary Clarify determine the volume of the reactor. The first question the second one will be the mass and volume of solids that must be wasted each day. So mass and volume means QW need to be calculated. Then recycle ratio also needs to be calculated.

[00:26:24]Recycle ratio means here QR by Q. These are the three questions we need to address.

[00:26:32]First in the question is given as Q = 10,000 m cube per day. Theta C also is given 10 days. So we know the formula 1 by theta C equal to this is the equation. If we are substituting these equations. So s not we know s we know y also know q also know kd also given x also given theta c also is given only unknown parameter will be the volume. So after substituting these all values the final volume will be 1611 m cube. That means so volume of the aiation tank will be 1,61 m cube. So there will be another equation theta c = v into x divided by qw into x. Theta C we know volume also we know X also we know XR also known only QW value needs to be calculated here qw XR equal to so vx divided by 10 that will gives the 724 95 kg per day so this is the mass is wasted so along the mass we need to calculate the volume also what would be the volume if the concentration of XR will be 12,000 mg per liter. QW you know QW we need to calculate QW equal to 72495 divided by XR concentration will be 12,000 mg per liter that is 12 kg per me cube. It is getting around 60.41 m cube per day.

[00:28:23]So here we need to calculate the QR by Q. So we know the formula of QR qx - qw xr / xr - x q we know x value also know qw xr calculated xr no x also given so as for this equation the qr is getting around 8,934 m cube per day this much of water is circulating out of 15,000 m cube per day. So in that we need to calculate the recycle ratio 8,00 93.34 divided by 15,000 that is the actual inflow rate then it is around 0.59.

[00:29:13]So as for the design conditions the means recycle ratio should be around 0.6.

[00:29:22]So I will briefly means what we discuss in this lecture especially ASP here means main mechanism B mechanism involved in the actual sludge process will be the micro needs to be introduced in the aation tank that also should be in the suspended state for that we are providing the aiation means here B will be the organic matter that will be there in the waste water so if we micro will be there in suspended state it will be eaten organic matter by microorgans. So that in this process what will happen? B concentration will be reduced. Next will be the secondary clarifier. Here the sludge will be set in the uh some portion of uh sludge needs to be introduced into the uh aeration tank.

[00:30:17]Why we are doing again back to the means eras tank will reason behind this we need to maintain the ratio between microance as well as the organic matter then only b removal will be happen particularly in this uh design problem we discussed how to calculate the volume of aeration tank and how to calculate the amount of mass wasted that is qw into xr Next will be the what will be the volume to be wasted that is the QW also calculated and also last one the needs to be calculate the recycle ratio so that is QR by Q that is also calculated is getting around 6. So these are the important parameters when someone is designing a ASP. These are the some of the design parameters needs to be considered. Then only efficiency will achieved around 85 to 90%.

[00:31:19]In uh upcoming class uh we will discuss about the remaining secondary treatment units. Thank you very much.