A drinking water treatment plant is designed with three sedimentation basins in parallel. Each basin is...
A drinking water treatment plant is designed with three sedimentation basins in parallel. Each basin is designed with an overflow rate of 1,200 gpd/ft2 (gallons per day per square foot). (“In parallel” means that the flow is split evenly between the basins as shown below.) A. If the plant flow is 20 mgd (million gallons per day), find the plan area of each basin in square feet. B. Calculate the dimensions of the basins (length and width in feet) if each basin is 2.5 times as long as it is wide. C. How deep would each basin have to be to meet a design standard for the hydraulic retention time of 3 hours? Each basin is a rectangular box. Hint: A reasonable answer would be between 15 and 25 feet.
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A drinking water treatment plant is designed with three
sedimentation basins in parallel. Each basin is...
Water and Wastewater Treatment SP-WATER TREATMENT PLANT SEDIMENTATION BASIN Design a rectangular, horizontal flow sedimentation process...
Water and Wastewater Treatment
SP-WATER TREATMENT PLANT SEDIMENTATION BASIN Design a rectangular, horizontal flow sedimentation process for a water treatment plant for turbidity removal Design Maximum Day Flow-20,000 m2ld Two parallel sedimentation tank, each designed for half of the total flow Depth 4.5 m, Width 6 m, Surface Overflow Rate 50 m/day Chain and flight sludge collector will be used, hence the limiting width of 6 m. Minimum Hydraulic Retention Time (detention time) 2 hrs Maximum weir rate 250 m/day/m...
A water treatment plant is processing water at a flow rate of 1.5 m/s using three parallel treatm...
A water treatment plant is processing water at a flow rate of 1.5 m/s using three parallel treatment trains. The detention times for the rapid mix tank, flocculation basin, and rectangular sedimentation basin are 10 sec, 30 min and 60 min, respectively. The velocity gradients for the rapid mix and flocculation tanks are 900 s and 60:40:20 s1, respectively. The viscosity of water is 0.001 Pa-s. (a) Determine the size and power requirements for rapid mix tanks. (b) Determine the...
The total flow entering a wastewater treatment plant is constant at 9,000 m^3/day. The flow is...
The total flow entering a wastewater treatment plant is constant
at 9,000 m^3/day. The flow is evenly split between Three biological
reactors that are installed in parallel. Each of the reactors hasd
a volume of 750,000 L. What is the hydraulic retention time (in
hours) of each reactor?
1. (Ipts) The total flow entering a wastewater treatment plant is constant at 9,000 m/day. The flow is evenly split between three biological reactors that are installed in parallel. Each of the...
A wastewater treatment plant receives an average of 4.5 MGD of wastewater flow. Two rectangular primary...
A wastewater treatment plant receives an average of 4.5 MGD of wastewater flow. Two rectangular primary clarifiers operating in parallel will treat the flow. If the overflow rate is 700 gpd/ft2 and the depth of the clarifier is 10 ft, Determine: a) the dimensions of each clarifier assuming L:W ratio = 5:1. b) the detention time of each clarifier (in days).
4. (20 pts) A rapid mixing basin is designed to process 3.0 MGD of water. The...
4. (20 pts) A rapid mixing basin is designed to process 3.0 MGD of water. The basins will be square, and the depth is assumed to be 1.times of the width. Use a detention time of 30 s and velocity gradient of 750 s'. Two basins will be constructed each having a capacity to process exactly half the entire water flow. Determine: (a) Dimensions of the basin (L, W and H) (b) Power input (hp) (c) Rotational speed (rpm) of...
ShHT HMS BEHAN 1. Four million liter of water per day passes through a sedimentation basin...
ShHT HMS BEHAN 1. Four million liter of water per day passes through a sedimentation basin which is um wide, 15m long, and 3m deep. 9.) what is the average velocity of flow through the basin ? b.) what is the average retention time for this basin? c.) If the suspended solid content of water averages 40 mg 111 what is the weight of dry solids will be deposited every hour assuming 26% removal in the basin ?
A wastewater treatment plant that treats wastewater to secondary treatment standards uses a primary sedimentation process...
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...
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...
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...
Hydrology ENV 321 Homework #1 Due: Friday, September 6 at 1:00pm 1. Convert (show all conversion...
Hydrology ENV 321 Homework #1 Due: Friday, September 6 at 1:00pm 1. Convert (show all conversion steps for full credit; convert only one unit for cach step as demonstrated in class): a) acre-feet/day to mgd (mgd is million gallons per day] b) cfs to cms c) inches/hr rainfall to mm/day rainfall d) short tons/acre-day to Mg/hectare-year [Mg is megagrams e) inches"F to cm/ C [be careful, this one may trick you] 2. Define the hydrologic cycle component and indicate how...
Water and Wastewater Treatment
SP-WATER TREATMENT PLANT SEDIMENTATION BASIN Design a rectangular, horizontal flow sedimentation process for a water treatment plant for turbidity removal Design Maximum Day Flow-20,000 m2ld Two parallel sedimentation tank, each designed for half of the total flow Depth 4.5 m, Width 6 m, Surface Overflow Rate 50 m/day Chain and flight sludge collector will be used, hence the limiting width of 6 m. Minimum Hydraulic Retention Time (detention time) 2 hrs Maximum weir rate 250 m/day/m...
A water treatment plant is processing water at a flow rate of 1.5 m/s using three parallel treatment trains. The detention times for the rapid mix tank, flocculation basin, and rectangular sedimentation basin are 10 sec, 30 min and 60 min, respectively. The velocity gradients for the rapid mix and flocculation tanks are 900 s and 60:40:20 s1, respectively. The viscosity of water is 0.001 Pa-s. (a) Determine the size and power requirements for rapid mix tanks. (b) Determine the...
The total flow entering a wastewater treatment plant is constant
at 9,000 m^3/day. The flow is evenly split between Three biological
reactors that are installed in parallel. Each of the reactors hasd
a volume of 750,000 L. What is the hydraulic retention time (in
hours) of each reactor?
1. (Ipts) The total flow entering a wastewater treatment plant is constant at 9,000 m/day. The flow is evenly split between three biological reactors that are installed in parallel. Each of the...
A wastewater treatment plant receives an average of 4.5 MGD of wastewater flow. Two rectangular primary clarifiers operating in parallel will treat the flow. If the overflow rate is 700 gpd/ft2 and the depth of the clarifier is 10 ft, Determine: a) the dimensions of each clarifier assuming L:W ratio = 5:1. b) the detention time of each clarifier (in days).
4. (20 pts) A rapid mixing basin is designed to process 3.0 MGD of water. The basins will be square, and the depth is assumed to be 1.times of the width. Use a detention time of 30 s and velocity gradient of 750 s'. Two basins will be constructed each having a capacity to process exactly half the entire water flow. Determine: (a) Dimensions of the basin (L, W and H) (b) Power input (hp) (c) Rotational speed (rpm) of...
ShHT HMS BEHAN 1. Four million liter of water per day passes through a sedimentation basin which is um wide, 15m long, and 3m deep. 9.) what is the average velocity of flow through the basin ? b.) what is the average retention time for this basin? c.) If the suspended solid content of water averages 40 mg 111 what is the weight of dry solids will be deposited every hour assuming 26% removal in the basin ?
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...
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...
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...
Hydrology ENV 321 Homework #1 Due: Friday, September 6 at 1:00pm 1. Convert (show all conversion steps for full credit; convert only one unit for cach step as demonstrated in class): a) acre-feet/day to mgd (mgd is million gallons per day] b) cfs to cms c) inches/hr rainfall to mm/day rainfall d) short tons/acre-day to Mg/hectare-year [Mg is megagrams e) inches"F to cm/ C [be careful, this one may trick you] 2. Define the hydrologic cycle component and indicate how...
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