Absolutely, let’s walk through the problem together step by step, just like we would if we were at a whiteboard! We’re dealing with a **Sequencing Batch Reactor (SBR)** system that treats 900 m³ of wastewater per day. There are three reactors, each one handles part of the flow, and each reactor goes through 3 cycles per day. --- ### **a) Calculate the cycle time, \( T_C \), of the SBR** First, since there are 3 reactors and 3 cycles per reactor per day, that means each reactor must complete 3 cycles in 24 hours. To find the cycle time (\( T_C \)), we just divide the total hours in a day by the number of cycles per day per reactor: \[ T_C = \frac{24 \text{ hours}}{3 \text{ cycles}} = 8 \text{ hours per cycle} \] --- ### **b) What shall be the fill volume (\( V_F \)) for each SBR?** The total flow treated per day is 900 m³. Since there are 3 reactors and 3 cycles per reactor, the total number of cycles per day is \( 3 \times 3 = 9 \). So, the fill volume per cycle is: \[ V_F = \frac{900 \text{ m}^3/\text{day}}{9 \text{ cycles/day}} = 100 \text{ m}^3 \text{ per cycle} \] --- ### **c) Calculate the anoxic fill time (\( T_F, T_{anox} \)) of SBR** We’re told that the anoxic fill time to effective process time (\( T_E \)) ratio is 44%. The effective process time (\( T_E \)) is the total cycle time minus the time for settling and decantation. First, let’s find out how much time is for settling and decantation. That’s given as 1.5 hours per cycle. So, \[ T_E = T_C - (T_S + T_D) = 8 \text{ h} - 1.5 \text{ h} = 6.5 \text{ h} \] Now, the anoxic fill time: \[ T_{F, anox} = .44 \times T_E = .44 \times 6.5 \text{ h} = 2.86 \text{ h} \] --- ### **d) Write the operation steps of SBR in one cycle** Here’s how a typical SBR cycle would go, based on the info: - **Fill/Anoxic Phase:** Wastewater is added to the reactor (anoxic fill takes about 2.86 h). - **React/Aeration Phase:** The reactor contents are mixed/aerated after the fill phase for the remaining effective process time. - **Settle:** The contents are allowed to settle (1.5 h for settling and decantation together). - **Decant:** The treated supernatant is removed from the top (this is included in the 1.5 h settle + decantation time). - **Idle (if any):** Sometimes there's a small idle time, but here, the steps fill the cycle. --- ### **e) Calculate the initial volume (\( V_ \)) of each reactor** The volumetric exchange ratio is given as \( \frac{V_F}{V_T} = .3 \) (or 30%). This means that with every cycle, 30% of the reactor volume is exchanged with new wastewater. We know \( V_F = 100 \) m³ from earlier, so: \[ .3 = \frac{100}{V_T} \implies V_T = \frac{100}{.3} = 3333.33 \text{ m}^3 \] Now, usually, after decantation, what remains is the initial volume (\( V_ \)), which is the reactor volume minus the fill volume: \[ V_ = V_T - V_F = 3333.33 - 100 = 2333.33 \text{ m}^3 \] --- **In summary:** - Each cycle is 8 hours long. - Each SBR handles 100 m³ per cycle. - Anoxic fill lasts about 2.86 hours. - The cycle steps are: Fill (Anoxic) → React → Settle → Decant. - The initial volume in each reactor is about 2333 m³. Let me know if you want to dig into any of these steps further, or if you want a diagram to help visualize the process!