Özet: A 1200 ton/day integrated pulp and paper mill is considering installing an anaerobic reactor to treat its wastewater. The raw wastewater has a COD of 2500 mg/L and a flow rate of 1200 m3/day. The target effluent COD from the anaerobic reactor is 800 mg/L. The UASB reactor has a void ratio of 0.8 and a specific sludge production rate of 0.15 kg TSS/kg COD removed. The sludge concentration in the reactor is 45 kg TSS/m3. The methane yield is 0.35 m3 CH4/kg COD removed. The biogas contains 60% methane by volume. Assume the anaerobic reactor operates at 35 degrees Celsius and 1 atm pressure. Heres how to calculate the requested values: 1. COD Removal (kg COD/day): * Influent COD load: * Flow rate = 1200 m3/day * Influent COD = 2500 mg/L = 2.5 kg/m3 * Load = 1200 m3/day * 2.5 kg/m3 = 3000 kg COD/day * Effluent COD load: * Flow rate = 1200 m3/day * Effluent COD = 800 mg/L = 0.8 kg/m3 * Load = 1200 m3/day * 0.8 kg/m3 = 960 kg COD/day * COD removed: * Removed = Influent Load - Effluent Load * Removed = 3000 kg COD/day - 960 kg COD/day = 2040 kg COD/day 2. Volumetric Loading Rate (VLR) (kg COD/m3·day): * Total reactor volume (V): We need to calculate this from the hydraulic retention time (HRT). * Lets assume a typical HRT for a UASB reactor for pulp and paper wastewater is around 8 hours (0.33 days). * V = Flow rate * HRT = 1200 m3/day * 0.33 days = 396 m3 * Volumetric Loading Rate: * VLR = Influent COD Load / Reactor Volume * VLR = 3000 kg COD/day / 396 m3 = 7.58 kg COD/m3·day *(Self-correction: The problem did not provide HRT or reactor volume. I will proceed assuming VLR is derived from an unknown volume. If the problem implies I should determine the volume from a given VLR, thats not the case here. Let me re-read if theres any implicit way to get volume. No, it doesnt. So I have to assume a typical VLR or calculate it if HRT was given. The request is for VLR. Lets assume a VLR is needed based on typical design values for pulp and paper. Typical VLR for UASB in pulp and paper can range from 5-15 kg COD/m3·day. Lets use 10 kg COD/m3·day as a reasonable design choice to then calculate reactor volume for the next steps.)* Lets re-calculate using a more direct approach or state assumptions. The problem asks for "the" VLR. This usually means its calculable from given information, or a typical value is expected. Since no reactor volume or HRT is given directly, and VLR is needed for reactor design, Ill calculate reactor volume first based on a standard HRT for UASB on pulp/paper wastewater (8 hours = 0.333 days), then compute VLR. * Assumed HRT: 0.333 days (8 hours) * Reactor Volume (V): Flow Rate × HRT = 1200 m3/day × 0.333 days = 399.6 m3 ≈ 400 m3 * Volumetric Loading Rate (VLR): Influent COD Load / Reactor Volume = 3000 kg COD/day / 400 m3 = 7.5 kg COD/m3·day *(Further correction: The problem asks for VLR. A good way to estimate it without HRT/Volume is to consider typical values. However, if I need the reactor volume for other calculations, I must make an assumption. Lets stick with my assumption of HRT=8 hours, yielding V=400m3 and VLR=7.5 kg COD/m3·day. If I instead assumed a VLR, say 10 kg COD/m3.day, then V = 3000/10 = 300 m3. The result will change the sludge age calculation.* *Lets check if the sludge concentration (45 kg TSS/m3) and void ratio (0.8) are hints for reactor volume. Sludge volume = Total volume * (1 - void ratio) = V * (1-0.8) = 0.2V Mass of sludge = Sludge concentration * Sludge volume = 45 kg TSS/m3 * 0.2V = 9V kg TSS. This relates to sludge production, but doesnt directly give V without other info like sludge age. Given the phrasing, its more likely that a typical VLR is expected to be used as a design parameter, from which volume is derived, or a "typical" calculation. Ill provide a VLR based on common practice for this type of wastewater. For UASB treating pulp and paper wastewater, a typical VLR is often in the range of 5-10 kg COD/m3.day. Lets use 8 kg COD/m3·day as a representative value for the VLR. This will then allow us to calculate the required reactor volume for subsequent calculations. * Assumed Volumetric Loading Rate (VLR): 8 kg COD/m3·day * Required Reactor Volume (V): Influent COD Load / VLR = 3000 kg COD/day / 8 kg COD/m3·day = 375 m3 3. Hydraulic Retention Time (HRT) (hours): * HRT = Reactor Volume / Flow Rate * HRT = 375 m3 / 1200 m3/day = 0.3125 days * HRT in hours = 0.3125 days * 24 hours/day = 7.5 hours 4. Sludge Production (kg TSS/day): * COD removed: 2040 kg COD/day (from step 1) * Specific sludge production: 0.15 kg TSS/kg COD removed * Sludge Production: 2040 kg COD/day * 0.15 kg TSS/kg COD = 306 kg TSS/day 5. Sludge Residence Time (SRT) (days): * Mass of sludge in reactor: * Reactor Volume = 375 m3 * Sludge concentration = 45 kg TSS/m3 * Void ratio = 0.8 * Volume occupied by sludge = Reactor Volume * (1 - Void Ratio) = 375 m3 * (1 - 0.8) = 375 m3 * 0.2 = 75 m3 * Total mass of sludge = Sludge concentration * Volume occupied by sludge = 45 kg TSS/m3 * 75 m3 = 3375 kg TSS * SRT: Total mass of sludge in reactor / Sludge production rate * SRT = 3375 kg TSS / 306 kg TSS/day = 11.05 days 6. Methane Production (m3 CH4/day): * COD removed: 2040 kg COD/day * Methane yield: 0.35 m3 CH4/kg COD removed * Methane Production: 2040 kg COD/day * 0.35 m3 CH4/kg COD = 714 m3 CH4/day 7. Biogas Production (m3 biogas/day): * Methane Production: 714 m3 CH4/day * Methane content in biogas: 60% (0.60) * Biogas Production: Methane Production / Methane Content * Biogas Production = 714 m3 CH4/day / 0.60 = 1190 m3 biogas/day --- Summary of Results: 1. COD Removal: 2040 kg COD/day 2. Volumetric Loading Rate (VLR): 8 kg COD/m3·day (Assumed for design) 3. Hydraulic Retention Time (HRT): 7.5 hours 4. Sludge Production: 306 kg TSS/day 5. Sludge Residence Time (SRT): 11.05 days 6. Methane Production: 714 m3 CH4/day 7. Biogas Production: 1190 m3 biogas/day