Question

A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process fol

You have gathered data below from the plant flowmeters, and the construction drawings. Primary Sedimentation Process The foll

16. What is the secondary clarifiers detention time? (answer will be given in the unit of (hr)) 17. Are the processes operat

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Let's Proceed

16, 17

Page 16 GivenamG Detention time is given by De= Volume of one clasifier one Clarifies # total flowetz GOMGD Flow over #flow o

18,19

18 Flm ratio is Food/micoo organisms It is amount of sludge recieing into the aeration Colo activated sludge unit. = when the

Hope this helps

All the best :)

Add a comment
Know the answer?
Add Answer to:
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process...
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
  • 1. A circular secondary clarifier (a.k.a. a final clarifier) is to be designed for an activated...

    1. A circular secondary clarifier (a.k.a. a final clarifier) is to be designed for an activated sludge treatment plant serving a municipality. The state regulatory agency criteria for final clarifiers used for activated sludge are: Peak SOR = 1,200 gal/day-ft2 Average SOR = 500 gal/day-ft2 Peak solids loading 50 lb/day-ft Peak weir loading-30,000 gal/day.ft The average flow to the aeration basin prior to junction with the recycle line is 3.5 MGD ·The recycled sludge flow is 30 percent of the...

  • 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.

  • Given: There is a wastewater treatment plant that has a Primary Clarifier, Activated Sludge, Secondary clarifier...

    Given: There is a wastewater treatment plant that has a Primary Clarifier, Activated Sludge, Secondary clarifier and the average daily flow was 1 mgd. Calculate the volume needed if the activated sludge is 24 hours 10 Mgal 100 Mgal 1 Mgal 1000 Mgal

  • :A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter...

    :A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter received flow 44L/Sec , the BOD concentration in raw sewage =440 mg/L , 30% BOD removed in the primary sedimentation tank, volume of filter (first stage )=826000L, volume of filter (second stage)=2563000L, recirculation rate =0.5 in second tank while no recirculation in first tank. Determine the following: 1-The effluent concentration of BOD in each unit. 2-Treatment efficiency of each unit. 3-Treatment efficiency of plant....

  • A wastewater treatment plant, which serves a population of 300,000 people, receives an average daily volume...

    A wastewater treatment plant, which serves a population of 300,000 people, receives an average daily volume of 24 million gallons per day (MGD) at an average influent 5-day biochemical oxygen demand concentration of 200 mgBODs/L and an average influent total suspended solids concentration of 220 mgTSS/L. The plant operates a primary sedimentation process that remove 65% of the incoming TSS and 35% of the incoming BOD %. This process is followed by a secondary treatment process before discharge to the...

  • 2. [20] A small town has been directed to upgrade its primary wastewater treatment plant to...

    2. [20] A small town has been directed to upgrade its primary wastewater treatment plant to a secondary plant that can meet an effluent standard of 25.0 mg/L BODs and 30.0 mg/L suspended solids. They have selected a completely mixed activated sludge system. The following data are available from the existing primary plant. Existing plant effluent characteristics Flow = 0.2000 m 3/s soluble BODs-80 mg/L Assume that the BOD5 of the suspended solids may be estimated as equal to 50%...

  • 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....

  • 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...

  • Question 1 The town of Lawrence has been directed to upgrade its primary WWTP to a...

    Question 1 The town of Lawrence has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 11 mg/L BODS. They have selected a completely mixed activated sludge system with an aeration tank (in the secondary plant) that can produce an effluent with 2,000 mg MLSS/L. a) Estimate the required volume of the aeration tank needed in the secondary plant The following data are available from the existing primary plant. Existing primary...

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