Environmental treatment Solved problems

SP #1: Review What is the chemical formula for carbonate, bicarbonate, sulfate, calcium hydroxide, and potassium phosphate? Solution: Carbonate: CO32Bicarbonate: HCO3Sulfate: SO42Calcium hydroxide: Ca(OH)2 Potassium phosphate: K3(PO4)3 What are the radicals that cause alkalinity? (Problem 1.6 page 17 of Reynolds and Richards) Solution: Radicals that cause alkalinity are carbonate (CO32-), bicarbonate (HCO3-), and hydroxide (OH-) If a water has a pH= 8.0, what will be the concentration of H+ in mg/l Solution: pH= -log[H+], where [H+] is the concentration of H+ in mole/l Thus, [H+]= 10-8 mole/l Since the atomic weight of H+ =1.0 g/mol, then the concentration of H+ = 10-8 g/l = 10-5 mg/l What is the molarity of a solution that contains 400 mg/l calcium? Solution: # moles (mass / molecular weight ) (mass / l ) = = l l molecular weight 400 mg / l Thus, the molarity of the solution = = 10 −2 mol / l 40 × 10 mg / mol Note that the atomic weight of Ca is 40 g/mol Molarity = A water has a hardness of 185 mg/l as CaCO3 What is the hardness expressed in meq/l? (Problem 1.7 page 17 of Reynolds and Richards) Solution: Hardness (meq / l ) = 185 mg CaCO3 meq × = 3.7 meq / l l 50 mg CaCO3 Covert a flux of 15 gallon/min/ft2 to m3/hr/m2 Solution: 15 gallon 60 10.76 ft m3 Flux (m / hr / m ) = ( ⋅ )×( )( )( ) = 36.65 264.2 gallon hr ft m2 (Problem 4.5 page 92 of Reynolds and Richards) Note you need to solve problem 4.4 first to find the total solids concentration Solution From problem 4.4, the total solids concentration is determined as 44.6484 g − 44.6420 g 1000 mg 1000 ml TS (mg / l ) = ( )× × = 640 mg / l 10 ml g l a From the information given in problem 4.5, the total suspended solids concentration is 17.5504 g − 17.5482 g 1000 mg 1000 ml TSS (mg / l ) = ( )× × = 22 mg / l 100 ml g l b Total dissolved solids is now calculated as: TDS = TS- TSS = 640-22 = 618 mg/l The mass flow rate of a contaminant entering a treatment plant is calculated as: Mass flow rate = Volumetric flow rate × Concentration × conversion factor If the unit of the volumetric flow rate is million gallon per day, the unit of concentration is mg/l, and the unit of mass flow rate is lb/day, what is the value of the conversion factor? Solution:  lb million gallon mg  3785000l lb = × × ×  d day l  million gallon 453592.4mg  lb million gallon mg = × × [ 8.344] d day l Thus, the value of the conversion factor is 8.344 Obtain a snap shot from Goggle Earth for a water and a wastewater treatment plant in the UAE Solution: The top aerial photo below is for Al Fujairah Desalination Plant which is a water treatment plant) while the photo at the bottom is for Al Aweer Wastewater Treatment Plant in Dubai SP#2: Water Quality A solution has 46 mg/l of phosphate What is the concentration of phosphate in mg/l as P? Solution: Concentartion (mg / l as P) = concentration (mg / l as PO4 ) × = 46 × mass of P per mole of PO mass of PO per mole of PO 31 = 15 (31 + ×16) (Modified Problem 4.3 page 92 of Reynolds and Richards) A city has a present population of 53,678, and the average water consumption is 162 gal/cap/day A new water treatment plant is to be built to serve the city for the coming 10 years The expected population 10 years from now is 65,300 (a) What is the per capita water consumption 10 years from now? (b) What is the design capacity of the water treatment plant? Solution: (a) Percent increase in population= [(65300-53678)/53678]*100% =21.6% Per capita water consumption= 162 (100%+ 10%*21.6%) = 165.5 gal/cap/day (b) Capacity= (165.5 gal/cap/day)* 65,300 = 10807150 gal/day = 10.81 million gal/day The US Primary Drinking Water Standard requires that nitrate in drinking water should not exceed 10 mg/l as N A groundwater has a nitrate (NO3-) concentration of 15.5 mg/l If this water is to be used for drinking purposes, is there a need to reduce the nitrate concentration to meet the US Primary Drinking Water Standard? Explain Solution: Convert the concentration of nitrate in groundwater from mg/l to mg/l as N concentartion (mg / l as N ) = concentration (mg / l as NO3 ) × = 15.5 × mass of N per mole of NO3 mass of NO3 per mole of NO3 14 = 3.5 < 10 (14 + × 16) Thus, there is no need to reduce the nitrate concentration as it is below the maximum contaminant level Water samples from the effluent of a water treatment plant of a small town (18,000 people) were analyzed at regular intervals over a month period The numbers of coliform per 100 ml sample were as shown in the table below: Sample Coliform per 100ml 0 Sample 10 Coliform per 100ml Sample 17 18 Coliform per 100ml 1 0 11 12 13 14 15 16 19 20 21 22 23 24 0 1 According to the EPA regulations, are the number of samples and effluent bacterial quality acceptable? Explain Solution The population is 18,000, so the minimum number of samples should be 20 Since we have 24 samples then the number of samples taken is OK Since the average number of coliforms is 0.875 which is not more than 1, then this condition is OK Since the number of samples is 24>20, then 5% of the samples are allowed to have more than coliform/100 ml Number of samples allowed to have more than coliform/100 ml= (5/100) (24)= 1.2 Which means only sample is allowed to have more than coliform/100 ml Inspection of the table above shows that there is one sample (sample # 14) that has more than coliform/100ml, which is Ok Treated wastewater samples from the effluent of a wastewater treatment plant were analyzed at regular intervals The influent of the plant receives waste that has a BOD of 300 mg/l and suspended solids of 220 mg/l The characteristics of the effluent are shown in the table below for 30 successive days From the data in the table, determine if the wastewater treatment plant violates the NPDES discharge requirements Sample pH 10 11 12 13 7.8 7.4 7.6 7.4 7.7 7.8 7.9 7.8 7.7 7.8 7.4 7.5 7.4 Oil and grease (mg/l) 12 14 11 18 21 7 11 13 12 14 BOD (mg/l) SS (mg/l) Sample pH 33 36 24 48 51 25 24 18 37 23 26 21 24 19 24 20 24 18 27 23 26 21 24 28 27 33 16 17 18 19 20 21 22 23 24 25 26 27 28 7.9 7.6 7.7 7.7 8.1 8.2 7.9 7.7 7.7 7.8 7.9 7.8 7.7 Oil and grease (mg/l) 17 7 11 13 12 14 11 12 21 BOD (mg/l) SS (mg/l) 23 26 25 21 24 28 37 33 26 31 24 18 37 24 17 15 15 22 23 26 21 24 18 37 23 26 14 15 7.7 7.8 11 14 18 27 26 31 29 30 7.8 7.3 15 23 36 21 24 Solution A spreadsheet is prepared to calculate average of and 30-consuctive days for BOD, SS, and oil and grease (O&G) Also, the percent removal of BOD and SS is shown in the spreadsheet below Sample pH BOD (mg/l) SS (mg/l) O&G (mg/l) 7.8 33 19 7.4 36 24 7.6 24 7.4 7.7 BOD(7) BOD( 30) SS(7) SS(30) O&G(7) O&G(30 ) BOD removal (%) SS removal (%) 12 89.0 91.4 14 88.0 89.1 20 11 92.0 90.9 48 24 18 84.0 89.1 51 18 21 83.0 91.8 7.8 25 27 91.7 87.7 7.9 24 23 34.4 22.1 12.6 92.0 89.5 7.8 18 26 32.3 23.1 11.9 94.0 88.2 7.7 37 21 32.4 22.7 11.0 87.7 90.5 10 7.8 23 24 11 32.3 23.3 11.0 92.3 89.1 11 7.4 26 28 13 29.1 23.9 10.3 91.3 87.3 12 7.5 21 27 12 24.9 25.1 9.0 93.0 87.7 13 7.4 24 33 14 24.7 26.0 10.3 92.0 85.0 14 7.7 18 26 11 23.9 26.4 10.9 94.0 88.2 15 7.8 27 31 14 25.1 27.1 11.9 91.0 85.9 16 7.9 23 24 17 23.1 27.6 13.1 92.3 89.1 17 7.6 26 17 23.6 26.6 12.3 91.3 92.3 18 7.7 25 15 23.4 24.7 11.3 91.7 93.2 19 7.7 21 15 23.4 23.0 10.6 93.0 93.2 20 8.1 24 22 23.4 21.4 9.6 92.0 90.0 21 8.2 28 23 24.9 21.0 9.1 90.7 89.5 22 7.9 37 26 11 26.3 20.3 8.7 87.7 88.2 23 7.7 33 21 13 27.7 19.9 8.1 89.0 90.5 24 7.7 26 24 12 27.7 20.9 9.1 91.3 89.1 25 7.8 31 18 14 28.6 21.3 10.3 89.7 91.8 26 7.9 24 37 11 29.0 24.4 10.9 92.0 83.2 27 7.8 18 23 12 28.1 24.6 11.6 94.0 89.5 28 7.7 37 26 21 29.4 25.0 13.4 87.7 88.2 29 7.8 23 21 27.4 24.3 12.6 92.3 90.5 30 7.3 36 24 15 27.9 28.2 24.7 23.6 12.9 11.4 88.0 89.1 Regulation 6-9 45 30 45 30 20 10 Status OK OK OK OK OK OK Not OK >85% times violatio n >85% time violatio n The table below summarizes the regulations and the existing violations 7-consuctive days 30-consuctive days Oil and grease BOD Suspended solids pH BOD removal Suspended solids removal 20 (OK) 10 (Exceeded) 45 (OK) 30 (OK) 45 (OK) 30 (OK) 6-9 (OK all values within this range) >85% (violated once) >85% (violated twice) SP # 3: Reactors A completely mixed reactor has a volume of 40 m3 and receives a pollutant at a concentration of 40 mg/l with a flow rate of 10 m3/d The pollutant leaves the reactor at a concentration of mg/l Answer the following questions: (i) What is the influent concentration? (ii) What is the effluent concentration? (iii) What is the concentration inside the reactor? (iv) Is the pollutant conservative? (v) What is the incoming mass flow rate of the pollutant Solution: (i) 40 mg/l (ii) mg/l (iii) mg/l because the reactor is completely mixed (iv) No, because the concentration dropped which means the pollutant decayed (v) The incoming mass flow rate = (40 mg/l) (10 m3/d) (1000 l/m3) = 400,000 mg/d = 0.4 kg/day In the schematic shown below two streams (1 and 2) are joined to form a single stream (3) with a suspended solids concentration (SS3) of 2500 mg/l What is the value of Q2 in million gallons per day (mgd)? Assume complete mixing at the joint point Q1=2 mgd SS1= 200 mg/l SS3= 2500 mg/l Q2=? mgd SS2= 12000 mg/l Solution: Apply mass balance at the joint point, assuming no decay: Mass input rate = Mass output rate Q1*SS1 + Q2*SS2 = (Q1 + Q2)*SS3 (2)(200) + Q2(12000) = (2 + Q2)*2500 Q2= 0.484 mgd A treatment unit is to be designed to remove 90% of a contaminant by a reaction with a rate constant of 3.0 per day Calculate the detention times required in completely-mixed and in a plug-flow reactor Solution: For a completely mixed reactor: Co + k ( V / Q) Since (Co-C)/Co= 0.9, then C/Co=0.1 Now solve for the detention time (V/Q) using k =3 V/Q= days C= For a plug-flow reactor C = e − k (V / Q ) Co Using C/Co=0.10 and k = per day, V/Q will be 0.77 day The data presented in the following table show the effluent concentration (C) of a dispersed plug flow reactor resulting from injecting a pulse of an ideal tracer into the influent and measured effluent concentrations versus time after injection Determine the actual retention time (min) and the dispersion number of the reactor t, 15 30 45 60 75 90 105 120 135 150 180 210 240 270 300 330 360 C, ug/L 0 3.5 16.5 46.5 72 89 95 88 78.2 55.2 33.5 20 12.1 7.5 4.6 2.6 Solution t, 15 30 45 60 C, ug/L 0 3.5 16.5 tC 0 157.5 990 t2C 0 7087.5 59400 75 90 105 120 135 150 180 210 240 270 300 330 360 Sum 46.5 72 89 95 88 78.2 55.2 33.5 20 12.1 7.5 4.6 2.6 624.2 3487.5 6480 9345 11400 11880 11730 9936 7035 4800 3267 2250 1518 936 85212 261562.5 583200 981225 1368000 1603800 1759500 1788480 1477350 1152000 882090 675000 500940 336960 13436595 The actual retention time: t= ∑t c ∑c i i 85212 = 136.5 624.2 = i The dispersion number (D/vxL) can be found as follows: σ2 = ∑t C ∑C i i − t = 2890.05 i σ θ2 = σ2 = 0.15508 t2 Using σ θ2 = D D − 2( ) (1 − e −v x L / D ) vx L vx L Solve for D/vxL using the graph of σ θ2 versus D/vxL Thus, D/vxL is about 0.0847 What is the effluent concentration of a contaminant with a retention time of hrs injected into a dispersed plug-flow reactor at a concentration of 75 mg/l The axial velocity and length of the reactor are cm/hr and 80 cm, respectively The dispersion coefficient is 140 cm2/hr Assume a first-order reaction with a rate constant of 0.35 hr-1 Solution: C 4ae ( Pe / 2) = Co (1 + a ) e ( aPe / ) − (1 − a ) e ( − aPe / 2) and The Peclet number = Pe= vxL/D= 1.142  4k t R  a = 1 +  Pe   0.5 14.2 28.1 37.0 46.7 a a a a a= data showed an increase in solids concentration 48.7 a Plot lines of equal percent removal as shown in the curve below Depth 100% 34 41 59 69.5 75.2 22.9 32 43.6 50.6 63.8 14.2 30 % 28.1 10 20 37 40 % 46.7 50 % 60 % 48.7 30 40 50 60 70 Time Find the percent removal at four different times, say 25, 37, 50, and 60 minutes 100% Depth 30 % 60 % 50 % 40 % 10 20 30 40 50 60 70 Time 5.3 2.4 1.8 0.8 ) + (50 − 40)( ) + (60 − 50)( ) + (100 − 60)( ) = 45.9% 8 8 5.6 2.5 1.0 R37 = 40 + (50 − 40)( ) + (60 − 50)( ) + (100 − 60)( ) = 55.1% 8 R25 = 30 + (40 − 30)( 6.2 3.8 0.9 ) + (60 − 50)( ) + (100 − 60)( ) = 65.7% 8 7.0 4.8 1.7 R60 = 46 + (50 − 46)( ) + (60 − 50)( ) + (100 − 60)( ) = 69.2% 8 R50 = 44 + (50 − 44)( Plot the removal efficiency versus time and from the curve determine the time to achieve 60% removal The time will be 43 For design multiply the time by 1.75 Thus, the design detention time will be 75 Now, find the surface area using: Q× t ( 2.5 × 10 × 0.134 /(24 × 60)(75) As = = = 2181 ft Depth Note: 0.134 is a conversion factor from gallon to ft3 SP # 6: Desalination A reverse osmosis unit is to demineralize 1750 m3/d of tertiary treated effluent Pertinent data are as follows: mass transfer coefficient = 0.22 liter/{(d-m2) (kPa)}, pressure difference between the feed and the product water = 2100 kPa, osmotic pressure difference between the feed and the product water = 300 kPa, and membrane area per unit volume of equipment = 2000 m2/m3 Determine a The membrane area required b The space required for the equipment, m3 Solution: a 0.22l (2100 − 300)kPa = 396 l / d m 2 d m kPa 1750 m / d Q A= = = 4419 m Fw 396 × 10 −3 m / d m Fw = K (∆p − ∆ψ ) = b Space needed = 4419 m = 2.2 m 3 2000 m / m An electrodialysis stack is to be used to partially demineralize 500 m3/d of brackish water so that it can be used by an industry The raw water has a TDS of 4000 mg/l and the product water must not have more than 1000 mg/l Given the followings: The normality of the raw water is 0.075 eq/l The membrane dimensions are 75×75 cm Stack resistance= 4.5 ohms Current efficiency = 84% Ratio of current density to normality= 400 (with current density as milli ampere/cm 2) Power cost= 2.5 cents/ kwh Brine TDS concentration= 70 g/l Determine a The salt removal efficiency b Product water flow c Number of membranes d Power cost per m3 Solution: a Salt removal efficiency = 4000 − 1000 × 100% = 75% 4000 b 500 × 4000 = Q(1000) + (500 − Q)(70000) ⇒ Q = 478.26 m / d c current density = 400 ⇒ current density = 400 × 0.075 = 30 N ⇒ current = (30) × (75 × 75) = 168750 milli ampere I= FQNE r nE c 168.75 amp = 96,500 amp sec/ eq × 500 m / d × 0.075 eq / l × 0.75 d 10 l × × n × 0.84 24 × 60 × 60 sec m ⇒ n ≅ 222 ⇒ number of membranes = n − = 221 d Power = RI = (4.5)(168.75) = 128145 watts = 128.1 kw 2.5 cents day 24h Cost = × (128.1kw) × × = 15.4 cents / m 3 kwh day 500m SP #7: Precipitation Lime-soda ash is to be used to soften a water with the following composition: CO2= mg/l, Ca= 110 mg/l, Mg= 24.4 mg/l, Na= 11.5 mg/l, K= 19.6 mg/l, HCO3= 200 mg/l as CaCO3, SO4= 86.4 mg/l, and Cl= 45.5 mg/l a Draw a meq/l bar graph for the raw water b Determine the amount of lime and soda ash needed for excess lime softening Assume the practical limits of hardness removal for Ca is 30 mg/l and for Mg is 10 mg/l as CaCO3 c Determine the pH of the water after softening but before recarbonation d Determine the concentration of sodium in the softened water Solution: a Constituent CO2 Ca Mg Na K HCO3 SO4 Cl 0.36 Conc mg/l 110 mg/l 24.4 mg/l 11.5 mg/l 19.6 mg/l 200 mg/l as CaCO3 86.4 mg/l 45.5 mg/l CO2 M.Wt 44 40 24.4 23 39 100 96 35.5 5.5 7.5 Ca Mg SO4 HCO3 4.0 Eq # 2 1 2 Na Cl 5.8 … … 8.0 Eq Wt 22 20 12.2 23 39 50 48 35.5 meq/L 0.36 5.5 2.0 0.5 0.5 4.0 1.8 1.28 8.5 K … 7.08 Note that there are some negative ions that not appear in the bar graph due to incomplete water analysis b meq/l of CO2 Ca(HCO3)2 Mg(HCO3)2 Ca-noncarbonate Mg-noncarbonate Excess Total meq/l needed from lime 1 1.25 Soda ash 0 1 Actual conc present 0.36 4.0 1.5* 2** Actual meq/l needed from lime 0.36 4.0 0 1.25 7.61 Soda ash 0 1.5 3.5 * 1.5=5.5-4.0 (This is the concentration of Ca that does not have alkalinity) 2=7.5-5.5 (This is the concentration of Mg that does not have alkalinity) ** Amount of lime needed in mg/l = (7.61 meq/l)× (37mg/meq) = 281.6 Amount of soda ash needed in mg/l = (3.5 meq/l)× (53 mg/meq) = 185.5 c The concentration of OH after treatment includes that from excess lime (1.25 eq/l) and that from the equilibrium dissolution of Mg(OH)2 which is 0.2 meq/l So the total OH is 1.45 meq/l which equals to 1.45 mmol/ = 1.45*10-3 mol/l 10 −14 [H + ] = = 6.9 × 10 −12 −3 1.45 × 10 pH= -log[H+]= 11.16 d The concentration of Na= original concentration in water plus the contribution from the addition of soda ash Na= 0.5 + 3.5= meq/l = 4meq/l × 23mg/meq= 92 mg/l SP #8: Sorption A wastewater that has an organic compound at 10 mg/l is to be treated by granular activated carbon A batch isotherm test has been performed in the laboratory and the following results were obtained: Bottle Carbon mass Solution volume C equilibrium (g) (ml) (mg/l) 0.12 100 5.5 0.22 100 2.7 0.33 100 1.25 0.38 100 0.98 0.43 100 0.56 0.54 100 0.3 0.65 100 0.15 0.82 100 0.11 Determine (a) The most appropriate sorption isotherm model that describe these data (b) Use the model selected in part (a) to find the carbon mass required to treat a batch of 200 m3 of this wastewater in the field such that the treated water has a phenol concentration of no more than mg/l Solution: (a) Calculate S for each bottle using CoV=CV+mS and then plot graphs of S versus C (linear model), logS versus logC (Freundlich model), and 1/S versus 1/C (Langmuir model) Find the best fit line for each case along with the coefficient of determination (R2) Note that for the case of the linear model, the data were fit to a line with zero-intercept since the linear model takes the form S=KC From the above, the Freundlich model (logS vrs logC) has the highest R2 value But since the model contains two parameters and the linear model contains one parameter, then to judge which model is more suitable we need to use the corrected Akaike Information Criteria (AICc) The AICc is given by: 2( P + 1)( P + 2)  SSR  AICc = N ln  + 2( P + 1) + N−P  N  Find first the SSR values associated with the linear and Freundlich models Linear Bottle C 5.5 2.7 1.25 0.98 0.56 0.3 0.15 0.11 S S 3.75 3.32 2.65 2.37 2.20 1.80 1.52 1.21 Res 5.07 2.49 1.15 0.90 0.52 0.28 0.14 0.10 SSR Freundlich S Res2 1.75 0.69 2.25 2.16 2.82 2.31 1.90 1.22 15.09 3.94 3.23 2.61 2.44 2.08 1.75 1.44 1.32 SSR 0.036 0.008 0.002 0.004 0.012 0.002 0.005 0.014 0.083 Now calculate the AICc value for the two models Model N P P+1 SSR AICc Linear Freundlich 8 2 15.09 0.083 11.47 -24.55 Since the Freundlich model has a lower AICc value than the linear model, it is more suitable to use Therefore, logS= 0.2787logC + 0.3892 b Find S for a concentration of mg/l using the above Freundlich equation: logS= 0.2787log1 + 0.3892= 0.3892 Thus, S=2.45 mg/g Apply CoV=CV+MS (10mg/L) (200m3*1000 L/m3)= (1 mg/L) (200m3*1000 L/m3) + M(2.45 mg/g) Thus, M=734,636.5 g = 734.6 kg A wastewater having an organic contaminant with a concentration of 250 mg/l is to be treated by a fixed bed activated carbon at a flow rate of 120 m3/d The allowable effluent concentration is not to exceed 30 mg/l Breakthrough data shown in the table below have been obtained from a laboratory column V (liters) 1300 2300 2600 2900 3200 3600 4100 C (mg/l) 0.015 10 38 90 150 220 247 The laboratory column has the following characteristics: diameter= 10 cm, length= 1.20 m, weight of activated carbon = 3.0 and an applied flow rate = 15 liter/hr a What is the volume and mass of carbon that can be used in a field column? b Use the scale-up approach to determine the volume of wastewater that can be treated in the field with the mass of activated carbon determined in part a Solution: a Since the hydraulic detention time in the lab and filed, the volume of the column in the field is calculated from: V V ( column ) lab = ( column ) field Q Q The volume of the column in the laboratory is 9420 cm3= 9.42 liters Thus, Vcolumn 9.42 l ( ) lab = ( ) field 15 l / hr (120 m / d )(d / 24 hr ) So, Vcolumn in the field= 3.14 m3 Since the bulk density in the laboratory and the field are the same, the mass of activated carbon in the field is calculated from: M carbon M ) lab = ( carbon ) field ρ= ( Vcolumn Vcolumn M 3kg ( ) = ( carbon3 ) field −3 lab 9.42 × 10 m 3.14m M in the field is 1000 kg b Plot effluent concentration (C) versus volume of water out of the laboratory column as shown in the figure below From the figure, the volume of water treated in the laboratory until the point of an allowable concentration of 30 mg/l is about 2500 Liters Vtreated V ) lab = ( treated ) field M M Given that M lab=3 kg, Vtreated in the lab = 2500 Liters (from graph), and M filed=1000 kg, then Vtreated in the field will be 833,333 liters or 833 m3 Using ( SP #9: Disinfection Results of a chlorine demand test on a raw water are as follows: Chlorine dosage (mg/l) NH3, mg/l 0.2 0.4 0.6 0.8 1.2 1.4 1.6 1.8 1.72 1.48 1.15 0.75 0.41 0.05 - Residual chlorine after 10-min contact (mg/l) 0.16 0.32 0.42 0.52 0.36 0.15 0.35 0.55 a Sketch a chlorine demand curve showing also the changes in NH3 level b What is the breakpoint chlorine dosage? c What is the chlorine demand at a dosage of 1.5 mg/l? d What chlorine dose is necessary to provide free available residual chlorine of 0.6 mg/l? Solution: a The chlorine demand curve is shown below NH3 depletion curve c b d b Breakpoint chlorine dosage = 1.2 mg/l (see figure above) c Chlorine demand at an applied dose of 1.5 is = 1.5-0.45=1.05 mg/l (see figure above) d Chlorine dose to provide a free residual of 0.6 mg/l is 1.7 mg/l (see figure above) Inactivation of coliforms using chlorination is affected by chlorine dose and contact time as shown in the following equation N = (1 + 0.23C ⋅ t ) −3 No where N and No are the concentration of coliforms after and before chlorination, respectively The chlorine residual concentration maintained in the reactor is 1.2 mg/l and the initial concentration of coliforms is 10000 per 100 ml Sketch a curve showing the effluent number of coliform/100 ml (N) versus contact time Assume the contact time ranges from to 30 minutes Solution: Solve the above equations (using No= 10000 and C= 1.2 mg/l) for different values of t to find N as tabulated below t (min) 10 15 20 25 30 N 742 188 74 36 20 13 Now plot N versus t as shown in the curve below Note how the number of coliform bacteria drops dramatically as the contact time increases SP #10: Biological Treatment A wastewater treatment plant consists of extended aeration tanks without primary treatment The extended aeration tanks have a volume of 4.0 million gallon The incoming flow rate is 4.5 mgd with a BOD of 220 mg/l The MLSS in the aeration tanks is maintained at 2400 mg/l by circulating settled sludge in the secondary clarifiers at a rate of mgd Calculate the aeration period, BOD loading per 1000 ft3, sludge age and the F/M ratio Q=4.5mgd BOD=220mg/l SS=210 mg/l MLSS= 2400 mg/l V= million gallon SST BOD=20 mg/l SS=20 mg/l QR=1 mgd SS=12250mg/l Qw= 30000gpd Solution: Aeration period = V/Q= 4/(4.5+1)= 0.66 day= 17.4 hrs Volume of the tanks = 4,000,000 gal/(7.48 gal/ft3)= 534,759 ft3 Thus the BOD loading is: BOD loading = 220 × 4.5 × 8.34 = 15.4 lb / d / 1000 ft 534.76 Sludge age is given by: 2400 × × 8.34 t sludge = = 21 day 4.5 × 20 × 8.34 + 0.03 × 12250 × 8.34 The food-to-microorganisms ratio is given by: 220 × 4.5 × 8.34 F /M = = 0.103 lb BOD per day / lb MLSS 2400 × × 8.34 A trickling filter plant consists of four 180-ft diameter by 7-ft-deep filters The design flow rate is 13.5 mgd with a BOD of 195 mg/l The primary clarifier removes 25% of the BOD Calculate the effluent BOD at a temperature of 15.5 oC QR=5mgd TF Q=13.5 mgd BOD=195 mg/l PST TF Q=18.5mgd Q=13.5 mgd TF TF Solution: π (180) × Each TF volume = = 178 thousand ft 1000 BOD concentration after PST= 0.75×195= 146.2 mg/l BOD remaining after PST = (146.2 mg / l )(13.5 mgd )(8.34) = 16,466 lb / day BOD loading on each TF = 16,466lb / day = 23.1 lb / 1000 ft 3 × 178 thousand ft QR (18.5 − 13.5) = = 0.37 or 37% Q 13.5 From the graph of percent removal and BOD loading find the percent removal with R=0.37 (between R=0 and R=0.5) and BOD loading =23.1 lb/1000 ft3 The removal efficiency (E 20oC) is about 80% R= Now find the removal at 15.5oC using E15.5= E20 (1.035)(15.5-20) =68.5% Effluent BOD after TF= (1-0.685)×146.2= 46 mg/l [...]... plot N versus t as shown in the curve below Note how the number of coliform bacteria drops dramatically as the contact time increases SP #10: Biological Treatment 1 A wastewater treatment plant consists of extended aeration tanks without primary treatment The extended aeration tanks have a volume of 4.0 million gallon The incoming flow rate is 4.5 mgd with a BOD of 220 mg/l The MLSS in the aeration...a = (1 + 4 × 0.35 × 6 0.5 ) = 2.89 1.142 C = 0.26 Co ⇒ C = 19.7 mg/l SP #4: Equalization and Filtration 1 The flow (every two hours) at an industrial wastewater treatment plant is as given in the following table The wastewater treatment plant does not have an equalization tank You are asked to design an equalization tank for the plant What is the design capacity of the tank? Solve the problem... that does not have alkalinity) ** Amount of lime needed in mg/l = (7.61 meq/l)× (37mg/meq) = 281.6 Amount of soda ash needed in mg/l = (3.5 meq/l)× (53 mg/meq) = 185.5 c The concentration of OH after treatment includes that from excess lime (1.25 eq/l) and that from the equilibrium dissolution of Mg(OH)2 which is 0.2 meq/l So the total OH is 1.45 meq/l which equals to 1.45 mmol/ = 1.45*10-3 mol/l 10