CTB3365x Introduction to Water Treatment

Transcription

1 CTB3365x Introduction to Water Treatment W4a Aeration tank loading and dimensioning Merle de Kreuk Today we start with designing the biological part of the sewage treatment plant. After this module, you will be able to make rough size and flow calculations of the aeration tanks, the final clarifier and the return sludge, 1

2 based on average influent characteristics. We will first start with the loading rate of the aeration tank and its dimensioning. This lecture consists of dimensioning via rules of thumb instead of using biomass kinetics. This is because we expect that you are not too familiar with biomass kinetics and it will go beyond this MOOC to introduce this more complex, but more accurate, design method. I rather redirect you to our online master Water management to learn more about design via biomass kinetics. One of the important parameters to base your simple design on, as how we explain the design in this MOOC, is the sludge loading rate. This determines the treatment efficiency of your system and the occurrence of nitrogen removal. First we will look into this parameter, what is it exactly? Well, the sludge loading rate is also known as the Food to Microorganism, or food to mass, ratio applied to a system. In other words: how much food is fed to the bugs during a day. In sewage treatment, we often express food in terms of a substrate concentration, or S, in kilograms BOD per cubic meter. If we want to know how much food we feed per day, we need to multiply the influent substrate concentration with the dry weather flow Q, expressed as cubic meters per day. We also need to know how much biomass is present in the system. Therefore we take the biomass concentration in the aeration tank and multiply it with the volume of this tank. The biomass loading rate is now easily calculated by dividing the substrate fed per day by the total biomass weight in the aeration tank. As said, one of the main parameters set by the sludge loading rate is the treatment efficiency of the sewage treatment plant. With the effluent demand, a sludge loading rate can be chosen. The higher the sludge loading, the more BOD will end up in the effluent, mainly because the organisms present won t have enough time to convert all the substrate. 2

3 Besides the treatment efficiency, the sludge loading rate will influence some other crucial parameters. Growth of the bacteria is determined by the Yield of the biomass, which is defined as the amount of cell mass produced per unit of substrate consumed. The more substrate that is introduced to the system, the more the biomass will grow. As mentioned, the total amount of biomass in the system is the product of the concentration in the aeration tank and the volume of this tank. If the BOD load to the tanks increases so will the biomass growth, thus more sludge will need to be wasted to maintain a stable biomass concentration which will lead to a lower solid retention time or sludge age. The sludge age becomes an important parameter when trying to cultivate a specific type of biomass. In the lectures about nitrogen removal, you will learn that nitrogen removal requires a typical sludge age of at least 3 days due to the low growth rate of the nitrifying organisms; therefore low sludge loading rates, typically below 0.15 kg/kg/day are required. Another parameter that is determined by the loading rate is the degree of stabilization of the sludge. Stabilization is the process that decreases the organic fraction of the sludge and therefore reduces the natural fermentation capacity. Sludge that was lightly loaded during aeration will have been starved and thus already consumed all available organics within the sludge itself, increasing its inorganic fraction. For example, waste sludge from ultralow loaded systems such as oxidation ditches will produce less gas and fewer odors upon storage due than high loaded sludge because there is less organic sludge to ferment. Last but not least, the oxygen demand of the system is determined by the sludge loading rate as well: the more food, the more oxygen is needed to convert it. 3

4 Here you will see some typical design parameters for activated sludge plants. Sludge concentration in the aeration tank is in general controlled between 3 and 5 gram per liter. Designs are often made at 4 g/l. It is advisable to increase the activated sludge concentration as much as possible, since that determines the conversion rate of the total system. Of course within the limits allowed by the sludge settleability and clarifier design and care must be taken to assure not negatively influencing aeration capacity. This will be further explained later in this module. The growth yield of heterotrophic biomass is in principle between 0.3 and 0.5 kg VSS/kg BOD. Low loaded systems with relatively high endogenous respiration will show lower yields than high loaded systems, where part of the organics is stored within the activated sludge flocs. Also the type of BOD plays a role in the overall yield. Influent consists of particulate, colloidal and dissolved substrate. The first two can be incorporated in the sludge floc, but will not necessarily degrade completely. This will increase the measured Yield. Finally, the sludge age of activated sludge systems with nitrification is typically between 5 and 25 days. The sludge age or solid retention time is determined by the mass of sludge in the system, divided by the waste sludge or excess sludge production per day. The calculation of the aerated tank volume, including the anoxic zones, is very straightforward when you understand the different parameters discussed. First you should select the sludge loading rate to apply, guided by your effluent demands. Secondly, you decide upon a sludge concentration you can maintain in your system. Often 4 g/l is chosen. With these two values and given influent characteristics, you are able to calculate the volume of the aeration tank. Would you like to give it a try? 4

5 Of course you like that, since now you can practice with what you just have learned. If you need more time than you will see in the timer at the bottom of the slide. Simply pause the movie until you found the answer by yourself. This time you have 10 seconds per sub-question. Influent characteristics and biomass concentration in the aeration tank are given for this sewage treatment plant. Try to calculate the size of the aeration tank if only BOD removal is required. Let me help you a little; you should first choose a biomass concentration in your aeration tank. Let s say 4 g/l as I suggested earlier to be a very general value for activated sludge systems. Next you question yourself what is the sludge loading rate if you just want BOD removal? It can be quite high, so choose for example the conventional loading rate of 0.4 kg BOD per kg biomass per day. Now it is a simple exercise solving the sludge loading rate equation. Did you find a volume of 1563 m3? Very good. Now you could do the same for a system with nitrification. Remember the loading rate that is allowed for nitrifying systems? This is much lower than the first example. Take 0.15 kilo per kilo per day as rule of thumb. Now you can solve the equation again. Did you find the same answer? No you did not; you can see that low loaded systems will require much more space. You can practice a little bit more with the exercises given at the website. Have fun! 5