- Questions & Answers
- + Post
- Get Expert Help
- Get Coins
Solution:- the values given in the question are as follows:
flow rate(Q)=44 L/s
BOD concentration in raw sewage=440 mg/L
BOD removed by primary sedimentation tank=30%
filter volume in first stage=826000 L
filter volume in second stage=2563000 L
recirculation rate(second tank)=0.5
the effluent concentration of BOD in primary tank=30% of BOD concentration in raw sewage
the effluent concentration of BOD in primary tank=(30/100)*440
the effluent concentration of BOD in primary tank=132 mg/L
the effluent concentration of BOD in secondary tank=BOD concentration in raw sewage-the effluent concentration of BOD in primary tank
the effluent concentration of BOD in secondary tank=440-132
the effluent concentration of BOD in secondary tank=308 mg/L
treatment efficiency of primary tank(p)=[volume filter by primary tank/total volume]*100
treatment efficiency of primary tank(p)=[826000/(826000+2563000)]*100
treatment efficiency of primary tank(p)=24.3729%
treatment efficiency of secondary tank(s)=[volume filter by secondary tank/total volume]*100
treatment efficiency of secondary tank(s)=[2563000/(826000+2563000)]*100
treatment efficiency of secondary tank(s)=75.627%
treatment efficiency of plant()=[total filtered volume/total volume]*100
where, total filtered water=volume filter by primary tank+volume filter by secondary tank
total filtered water=826000+2563000
total filtered water=3389000 L
total volume=volume filter by primary tank+volume filter by secondary tank+recirculation volume
total volume=4670500 L
treatment efficiency of plant()=[3389000/4670500]*100
treatment efficiency of plant()=72.5618%
depth of each tank=5 m
let diameter of primary tank is D1 and secondary tank is D2
volue(V)=area of tank*depth of tank
volue of primary tank(V)=area of primary tank*depth of primary tank
primary tank filter volume=826000 L
where, 1000 L=1 m^3
diameter of primary tank(D1)=14.5030 m
volue of secondary tank(V)=area of secondary tank*depth of secondary tank
secondary tank filter volume=2563000 L, or 2563 m^3
diameter of secondary tank(D2)=25.54726 m
:A wastewater treatment plant consisted of a primary sedimentation tank and two stages of trickling filter...
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process followed by conventional activated sludge process composed of covered aeration tanks and secondary clarifiers. You are asked to calculate the different operation parameters to make sure that the system is operated within normal process ranges listed in the reference tables below. Table 5-20 Typical design information for primary sedimentation tanks U.S. customary units Item Unit Range Typical Primary sedimentation tanks followed by secondary treatment...
10) Determine the size of two stage trickling filter using the NRC equations. Assume both filters have the same efficiency BOD removal and the same recirculation ratio. Other conditions are as follows o Wastewater temperature 20 C o Wastewater flow, Q 3785 n/d o Influent BODs 195 mg/l Design effluent BOD = 20mg/ o Depth of each filter 2m o Recirculation for filters 1 and 2, r 218 10) Determine the size of two stage trickling filter using the NRC...
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...
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.
Problem 5 A primary effluent with Q- 1.0 MGD, BODs 138 mg/, and SS - 104 mg/l is treated by a single stage tricking filter with the following characteristics: 6 feet deep rock trickling filter media; design BOD loading of 50 lb/day/1000 ft3; design hydraulic loading of 0.30 gpm/ft2; recirculation ratio of 1.5. Determine: 1) the volume (ft3) and diameter (ft) of the trickling filter; 2) the effluent BOD concentration. Problem 6
2.  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%...
Q4:Design rectangular primary sedimentation tank to remove 30% of BOD from the total an equivalent load of influent BOD and SS=700kg/day when the maximum expected flow is 8000m2/day and the detention time is 2.5hr .(take L:W=3:1). Then estimate: 1-The quantity of the sludge that is removed in an efficiency of 60% (assume moisture content of sludge =90%) 2-The effluent concentration of BOD and SS.
Two trickling filters operating in parallel are being designed to treat the domestic wastewater from a town of 5500 people where the average per-capita wastewater generation is 450 liters per person per day. A plastic trickling filter media and filter height of 3.5 m have been selected. The influent BOD5 is 120 mg/L and the facility must meet an NPDES permit limiting effluent BOD5 to 35 mg/L. The filter media constant (n) is 0.67 and the BOD decay constant for...
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....
Describe in detail the function of the wastewater treatment plant components below and what stage of treatment they are (primary, secondary, tertiary) use the lines provided for the second part of this question. For the descriptions use any space on the next page. In the descriptions you must state what is specifically being treated, and if compounds are generated in a particular treatment stage you must describe that as well. In addition, if a particular process has two stages -...