Presentation on theme: "FE Review for Environmental Engineering Problems, problems, problems Presented by L.R. Chevalier, Ph.D., P.E. Department of Civil and Environmental Engineering."— Presentation transcript:
FE Review for Environmental Engineering Problems, problems, problems Presented by L.R. Chevalier, Ph.D., P.E. Department of Civil and Environmental Engineering Southern Illinois University Carbondale
WATER TREATMENT FE Review for Environmental Engineering
Calculate the carbonate hardness (CH) and the noncarbonate hardness (NCH) for two water samples listed below. Reported concentrations are as mg/L CaCO 3 Sample 1Sample 2 Alkalinity327498 Total Hardness498327 CH NCH
Sample 1Sample 2 Alkalinity327498 Total Hardness498327 CH327 NCH1710
Compute the TH, CH, NCH of the following water at a pH of 7.2. CONSTITUENTmg/L as CaCO 3 Magnesium107.7 Potassium3.2 Phosphate12.2 Calcium296.3 Fluoride0.8 Bicarbonate136.5 Carbon dioxide19.8
Estimate alkalinity from bicarbonate and carbonate Sum the multivalent metallic cations to get TH Determine which is larger Determine CH Determine NCH
Alkalinity = 136.5 mg/L as CaCO 3 TH = 404.0 mg/L as CaCO 3 CH = 136.5 mg/L as CaCO 3 NCH = 404.0 - 136.5 = 267.5 as CaCO 3
Consider a groundwater source that contains 2x10 -4 moles of H 2 CO 3 (carbonic acid). The rate of pumping from the aquifer is 1,000 m 3 /day. Determine the amount of hydrated lime (Ca(OH) 2 ) needed each day for neutralizing the carbonic acid and the amount of calcium carbonate (CaCO 3 ) produced as a result. Report your answer in kg/day.
Determine the molecular weight of Ca(OH) 2 and CaCO 3 Review the governing chemical reaction to determine the ratio between the moles of H 2 CO 3, Ca(OH) 2 and CaCO 3 Use this ratio, the pumping rate and the MW to determine the mass per day (in kg/day)
Determine the molecular weight of Ca(OH) 2 and CaCO 3 Ca(OH) 2 = 74.1 g/mol CaCO 3 = 100 g/mol
Review the governing chemical reaction to determine the ratio between the moles of H 2 CO 3, Ca(OH) 2 and CaCO 3 1 mole H 2 CO 3 : 1mole Ca(OH) 2 : 1 mole CaCO 3 2x10 -4 mole H 2 CO 3 : 2x10 -4 mole Ca(OH) 2 : 2x10 -4 mole CaCO 3
Use this ratio, the pumping rate and the MW to determine the mass per day (in kg/day)
WASTEWATER TREATMENT FE Review for Environmental Engineering
One wastewater treatment process, activated sludge, which will be discussed later, requires either a detention time of 4 hrs, or the ability to treat approximately 20 gal/capita-day. If a city has a population of 10,000, and an average flow of 1.2 MGD, what approximate tank volume is required?
The tank size can be estimated based on flow and typical detention times or on population and the size per capita. SAME ANSWER EITHER WAY! a) The typical detention time, is 4 hours. Thus the tank volume is:
The tank size can be estimated based on flow and typical detention times or on population and the size per capita. b) Based on population requirements, the volume is:
Estimate the area needed for bar racks given a city population of 150,000. Clearly state all assumptions.
(Answers May Vary Depending On Assumptions) Assume a peaking factor of 2.8 ( range 2-5) Assume 150 gal/capita-day Q peak = (2.8)(150 gal/capita-day)(150,000) Q peak = 63.0 MGD = 238,140 m 3 /day Limit approach velocity to 0.8 m/s (acceptable range 0.6 - 1.0 m/s) A = Q/v
0.8 m/s = 69120 m/day A = (238140 m 3 /day) / (69120 m/day) A = 3.45 m 2 = 3.5 m 2 Want to order 2 in case one is off line for maintenance or repair
Estimate the settling velocity of sand (density = 2650 kg/m 3 ) with a mean diameter of 0.21 mm. Assume the sand is approximately spherical. Using a safety factor of 1.4 to account for inlet and outlet losses, estimate the area required for a grit chamber to remove sand if the flow rate is 0.25 m 3 /s. The density and viscosity of water at 20°C is 998 kg/m 3 and 1.01 x 10 -3 N·s/m 2, respectively.
Review the governing equations Note the given parameters d = 0.21 mm g =9.8 m 2 /s p = 2650 kg/m 3 At 20°C, w = 1.01 x 10 -3 N·s/m 2 w = 998 kg/m 3 Q = 0.25 m 3 /s SF = 1.4
The Stokes’ settling velocity can thus be calculated:
Knowing the overflow rate, we can calculate the area required for the grit chamber. Note, the safety factor 1.4 So the area of the grit chamber must be 9 m 2 to remove 0.21mm grit.
Sizing a Primary Clarifier for WWT Use the typical design values to estimate the size for two circular clarifiers used to treat wastewater at a design flow of 20 MGD. Each clarifier is to treat half the flow. Report the diameter and depth. diameter depth
Determine the Design Data needed for your solution surface over flow rate of 1,000 gal/ft 2 /day average detention time, , of 2.0 hr Each clarifier should receive half the flow, Q/2 = 10 MGD Estimate the area (A=Q/v) Estimate the diameter (d) assuming a circular clarifier Clarifiers diameters are generally available in multiples of 5 ft in the US, or in multiples of 2 m outside the US Estimate the volume using the detention time (V=Q Estimate the depth of the tank (V/A = d)
Each clarifier should receive half the flow, or 10 MGD. Using the typical design value for the surface over flow rate of 1,000 gal/ft 2 /day, we can compute the area of each clarifier
From this area, we can now calculate the diameter of the clarifier. Clarifiers are generally available in multiples of 5 ft in the US, or in multiples of 2 m outside the US.
To determine volume, we will need a detention time. Again, we will use a design value. In this case, consider the typical design value for the average detention time, , of 2.0 hr
Clarifiers should be as shallow as possible, but not less that 7 ft deep. The approximate depth, h, can now be calculated.
A township has been directed to upgrade its primary WWTP to a secondary plant that can meet an effluent standard of 30 mg/L BOD 5 and 30 mg/L SS. Assuming that the BOD 5 of the SS may be estimated as equal to 63% of the SS concentration, estimate the required volume of the aeration tank. The following data are available from the existing primary plant.
Existing Plant Effluent Characteristics Assume that the secondary clarifier can produce an effluent with only 30 mg/L SS Assume MLVSS = 2000 mg/L Want to meet an effluent standard SS = 30 mg/L BOD 5 = 30 mg/L BOD 5 = 84 mg/LFlow = 0.150 m 3 /s