Question

Q Search The following information is given for an activated sludge system design: Flowrate Influent BOD Effluent BOD Unit m/d mg/L mg/L Value 10,000 150 SRT Synthesis yield, Y, g VSS/g bCOD Wastewater 1 Wastewater 2 Wastewater 3 Cell debris yield, f, 0.40 0.50 0.30 0.15 0.08 g VSS/g VSS Endogenous decay, b, gVSS/g VsSd mg/L nbVSS Temperature Note: Wastewater 1, 2, or 3 to be selected by instructor 10 Determine (a) the aeration tank oxygen requirements in kg/d, (b) the aeration tank oxygen uptake rate in mg/L h, (c) the aeration tank biomass concentration (mg/L), and (d) the portion of the total oxygen required that is needed for nitrification. Naranja
1 0
Add a comment Improve this question Transcribed image text
Answer #1

Sol O) demand biodijhadabe domital n 1.6 r 150 > 240 mg/L bio- ma» phoduetien tt k,.SRToo0o 05150-2 l4 (o.o8k6) 740 53.28 536 ka/d t.ur8 tont oxJgan んCute (OUR Ro OUR io - 142 536 1480 - T6l- 12- 718.88 ^/a- 1000(c) bio mob , bioSRT (536 kalas) (6J) 10000 mg ld 24 h 3 216 .929 kgL 1666.67 , 1929 mg IL

Add a comment
Know the answer?
Add Answer to:
Q Search The following information is given for an activated sludge system design: Flowrate Influent BOD...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • 2. A complete mix activated sludge process with recycle is being operated in the following conditions...

    2. A complete mix activated sludge process with recycle is being operated in the following conditions Unit Parameter Flow rate Influent BOD Value 12,500 mg/L 210 Efluent BOD HRT, λ SRT im hours days 0.50 Synthesis yieldVSS/gbCOD Cell debris yield, fa g VSS/g VSS Endogenous decay, ki nbVSS 0.13] g VSS/ g VSS.d 0.08 im 50 Temperature 10 Assume no nitrification occurs due to the SRT selected and low temperature. Find a) Oxygen requirement for the aeration tank, kg d...

  • 3. Determine the aeration tank volume (mgd), excess biomass in the waste sludge (kg/d), and the BOD f a completely mixed activated-sludge process by assuming the following: k 4.0 day r-0.55 lb vS...

    3. Determine the aeration tank volume (mgd), excess biomass in the waste sludge (kg/d), and the BOD f a completely mixed activated-sludge process by assuming the following: k 4.0 day r-0.55 lb vSS/1b BOD, K, 80 mg/l, ks-0.05 day', wastewater flow rate 2.5 mgd, soluble effluent BOD- 15 mgl, soluble influent BOD - 170 mgl, and MLVSS-2200 mg/l, and mean cell residence time -6 days. (15 pts) 3. Determine the aeration tank volume (mgd), excess biomass in the waste sludge...

  • A complete-mix activated sludge system is used to treat municipal wastewater after primary sedime...

    A complete-mix activated sludge system is used to treat municipal wastewater after primary sedimentation at 20 °C with MLVSS concentration of 2000 g/m3. The discharge standard for soluble BOD, concentration in the effluent is 0.5 g/m. The characteristics of the primary influent are: flow Q -2000 m'/d, soluble BOD -200 g/m3, Using these data and the kinetic coefficients in following, determine 1. Theoretical solids retention time (SRT) in unit of days 2. Pk vss in unit of kg VSS/d 3....

  • Secondary Treatment Based on a plant flow rate of 40 mgd, design an activated sludge, secondary t...

    Secondary Treatment Based on a plant flow rate of 40 mgd, design an activated sludge, secondary treatment system with recycle which maintains a MLVSS = 2,000 mg/L in the aeration reactor and has an average Solids Retention Time (SRT)-4 days. The biota kinetic constants are Kinetic Coefficients for the Activated-sludge Process for the Removal of Organic Matter from Domestic Wastewater Endogenous Decay Coefficient- k(d) 0.1 Half-velocity constant- K(s VSS/qVss-da 15 Mg/L bsCOD Max. Specific Substrate Utilization Rate-k 6 G bsCOD/gVSS-da...

  • A sewage treatment plant treats sewage at average flowrate of 10000m3/day. The plant applies activated sludge...

    A sewage treatment plant treats sewage at average flowrate of 10000m3/day. The plant applies activated sludge process for biological treatment. Find; 1. Volume of aeration tank. 2. Design air requirement. 3. Flowrate of waste sludge. Use the following data;  BOD of raw sewage=250mg/l.  Percent of BOD removal in primary sedimentation tank =32%.  MLVSS=2400 mg/l.  MLVSS in return sludge=10000mg/l.  Y=0.6 and kd=0.06/day.  Effluent BOD=20mg/l.  Effluent SS=30mg/l.

  • just part e please Activated sludge design A completely mixed activated sludge process is designed to...

    just part e please Activated sludge design A completely mixed activated sludge process is designed to treat 20,000 m/day of domestic wastewater having a BODs concentration of 250 mg/L following primary treatment. The NPDES permit requires that the effluent BOD, and TSS concentra- tions not exceed 20 mg/L on an annual basis. The following biokinetic coefficients are to be used in the design of the process: Y = 0.6 mg VSS/mg BODs, k = 5d"! Ks = 60 mg/L BODs,...

  • Environmental Engineering 3&4 So 3. An activated sludge plant treats 5 MGD and the average influent...

    Environmental Engineering 3&4 So 3. An activated sludge plant treats 5 MGD and the average influent BOD is 200 mg/L. The WWTP has 2 aeration basins, each 100 ft x 25 ft x 15 ft deep and 2 final clarifiers with diameters of 20 ft. About 35% of BOD is removed in primary clarifiers. Determine the BOD in the primary effluent, hydraulic retention time, F/M ratio, BOD exiting the aeration basin and mass of solids wasted (1b/d). The operation parameters...

  • 2. A completely mixed activated sludge process is designed to treat 20,000 m/d of domestic wastew...

    2. A completely mixed activated sludge process is designed to treat 20,000 m/d of domestic wastewater having a BODs concentration of 250 mg/L following primary treatment. The NPDES permit requirement for effluent BODs and TSS is 20 mg/L for each. The biokinetic parameters have beern established as: yield 0.6, k 5/d, K-60 mg/L, and kd 0.06/d. The MLVSS concentration in the aeration tank is to be maintained at 3000 mg/L and the ratio of VSS/TSS is 0.75. Assume that the...

  • This is all the information I was given For the completely mixed activated sludge process shown...

    This is all the information I was given For the completely mixed activated sludge process shown below, the system boundary is shown by the dash line. Q is the wastewater flow rate into the aeration tank; Qy is the flow rate of liquid containing microorganisms to be wasted; Xo is the microorganism concentration (VSS) entering the aeration tank; X is the microorganism concentration (MLVSS) in the aeration tank; He is the microorganism concentration (VSS) in the effluent from secondary settling...

  • Problem 2. An activated sludge unit is being designed to treat 375,000 gpd with an influent...

    Problem 2. An activated sludge unit is being designed to treat 375,000 gpd with an influent BOD of 250 mg/1. Parameters for the wastewater are: Y 0.6, m -2.0 days-1, kd0.15 days-1 and K, 40 mg/l. The recycle flow is 3750 gpd with a biomass concentration of 50 X, where X is the reactor biomass concentration. The waste sludge flow is 500 gpd. Calculate the mean cell residence time, e, and CSTR volume required for a 95% removal of influent...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT