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 FE Review for Environmental Engineering MATHEMATICAL/PHYSICAL FOUNDATIONS Problem Strategy Solution Complete the following chart: Total Solids (TS) ______ mg/L TDS TSS _____mg/L (_____%) _____mg/L (_____%) VDS FDS VSS FSS 70 mg/L (_____%) 270 mg/L (_____%) 180 mg/L (_____%) 180 mg/L (_____%) Problem Strategy Solution • Review the definitions of • • • • • • • TS TDS TSS VDS FDS VSS FSS • Assume a 1 liter sample • Divide 700 mg by the percentage shown or calculated Problem Strategy Solution Total Solids (TS) 700mg/L TDS TSS 340mg/L (49%) 360mg/L (51%) VDS FDS VSS FSS 70 mg/L (10%) 270 mg/L (39%) 180 mg/L (26%) 180 mg/L (26%) Complete a flow chart using the following information COM POUND C O N C E N T R A T IO N (m g/L ) D IS S O L V E S ? V O L A T IL IZ E S O R BU R N S A T 550 C S o d iu m C h lo rid e 45 Y es No C alciu m su lfate 30 Y es No C la y 100 No No C o p p er ch lo rid e 10 Y es No A cetic a cid 20 Y es Y es C o ffee gro u n d s 25 No Y es Problem Strategy Solution Total Solids (TS) ______ mg/L TDS TSS _____mg/L (_____%) _____mg/L (_____%) VDS FDS VSS FSS ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) Problem Strategy Solution • Review definitions • Fixed mean inorganic – it does not burn • Volatile means organic – it does burn Problem Strategy Solution Sodium chloride 45 mg/L Dissolves Doesn’t volatilize 45 Total Solids (TS) ______ mg/L 45 TDS TSS _____mg/L (_____%) _____mg/L (_____%) VDS FDS VSS FSS ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) 45 Problem Strategy Solution Calcium sulfate 30 mg/L Dissolves Doesn’t volatizes 45 30 Total Solids (TS) ______ mg/L 45 30 TDS TSS _____mg/L (_____%) _____mg/L (_____%) VDS FDS VSS FSS ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) 45 30 Problem Strategy Solution Clay 100 mg/L Doesn’t dissolve Doesn’t volatizes 45 30 100 Total Solids (TS) ______ mg/L 45 30 100 TDS TSS _____mg/L (_____%) _____mg/L (_____%) VDS FDS VSS FSS ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) 45 30 100 Problem Strategy Solution Copper chloride 10 mg/L Dissolves Doesn’t volatizes 45 30 10 100 Total Solids (TS) ______ mg/L 45 30 10 TDS TSS _____mg/L (_____%) _____mg/L (_____%) 100 VDS FDS VSS FSS ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) 45 30 10 100 Problem Strategy Acetic acid 20 mg/L Dissolves Volatizes Solution 45 20 30 10 100 Total Solids (TS) ______ mg/L 45 20 30 10 20 TDS TSS _____mg/L (_____%) _____mg/L (_____%) 100 VDS FDS VSS FSS ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) 45 30 10 100 Problem Strategy Coffee grounds 25 mg/L Doesn’t dissolves Volatizes Solution 45 20 30 10 100 25 Total Solids (TS) ______ mg/L 45 20 30 10 TDS TSS _____mg/L (_____%) _____mg/L (_____%) 100 25 VDS FDS VSS FSS ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) ____ mg/L (_____%) 20 45 30 10 25 100 Problem Strategy Solution Total Solids (TS) 230mg/L TDS TSS 105mg/L (46%) 125 mg/L (54%) VDS FDS VSS FSS 20 mg/L (9%) 85 mg/L (37%) 25 mg/L (11%) 100 mg/L (43%) Problem Strategy Solution • Water flows into a heated tank at a rate of 150 gal/min. • Evaporation losses are estimated to be 2000 lb/hr. • Assuming the tank volume to be constant, what is the flow rate out of the tank? Problem Strategy Solution • Draw a schematic (control volume) • Convert to like units (Weight of water 8.34 lb/gal) • Mass in = Mass out • concept of density (Volume in = Volume out) 150 gpm 2000 lb/hr ? 2000 lb 1 gal 1hr 4 gpm hr 8 .34 lb 60 m in dM 0 dt 150 4 x 0 F low rate out 146 gpm Example Solution Consider the following report from three supplies into a reservoir. Is it correct? Source A B C Flow 140 gpm 5 gpm 5 gpm Quality essentially clean 500 ppm toluene 500 ppm benzene Total 150 gpm 1000 ppm Example Solution Source A B C Flow 140 gpm 5 gpm 5 gpm Quality essentially clean 500 ppm toluene 500 ppm benzene Total 150 gpm 1000 ppm Do you see a problem here? Example Solution Important Rule: We can add mass (mass balance) but not concentrations Source A B C Flow 140 gpm 5 gpm 5 gpm Quality essentially clean 500 ppm toluene 500 ppm benzene Total 150 gpm 1000 ppm X Example Solution 1. What is the total volume of water per day? 140 gpm + 5 gpm + 5 gpm = 150 gpm Converting to liters/day (L/d) (150 gpm)(3.785 L/gal)(60 min/hr)(24 hr/day) = 817560 L/d Example Solution 2. What is the mass from source B? (5 gpm)(3.785 L/gal)(500 mg/L)(60 min/hr)(24 hr/d) = 1.36 x 107 mg/d Converting to ppm per day in total water 1.36 x 107 mg/ 817560 L = 16.67 mg/L = 16.67 ppm Example Solution 3. What is the mass from source C? (5 gpm)(3.785 L/gal)(500 mg/L)(60 min/hr)(24 hr/d) = 1.36 x 107 mg/d Converting to ppm per day in total water 1.36 x 107 mg/ 817560 L = 16.67 mg/L = 16.67 ppm Example Solution 4. Therefore, we have 16.67 ppm benzene, and 16.67 ppm toluene! Not 1000 ppm! 5. Can we add these concentrations? Simple Model Of Stream Pollution Based On Mass Balance Industrial Complex Qu Cu Qe Ce Qd Cd Problem Strategy Solution • A factory for copper and brass plating is dumping its wastewater effluent into a near-by stream. • Local regulations limit the copper concentration in the stream to 0.005 mg/L. • Upstream flow in stream, 0.5 m3/s. Concentration of copper in upstream flow is below detection limits • Effluent flow from plating factory 0.1 m3/s • Determine the maximum concentration allowable in the effluent from the factory’s wastewater. Problem • • • • Strategy Solution Draw a control volume diagram Determine Qtotal = Qstream + Qeffleuent Convert concentrations to mass (mass flux) Use mass balance to determine the allowable concentration (based on mass) of effluent Problem Strategy Solution Industrial Complex Qu= 0.5 m3/s Cu = 0 mg/L Qe =0.1 m3/s Ce = ? Qd = ? Cd = 0.005 mg/L dm dt Qu C u Q eC e Q d C d 0 Ce Qd C d QuC u Qe 0 .6m s 0 .005 mg L 0 .5 m s 0 mg L 3 3 0 . 1m 0 . 03 3 s mg L ....end of example SIMPLE PHOSPHOROUS MODEL Want to estimate the amount of phosphorous control needed to prevent eutrophication due to the overproduction of algae Simple Phosphorous Model Assumptions: • • • • Completely mixed lake Steady state Constant settling rate Phosphorous is the controlling nutrient Schematic Of System Waste water treatment plant Q = 0.2 m3/s P=5.0 mg/L Stream Q = 15.0 m3/s P=0.01 mg/L Settling rate vs = 10 m/yr Surface area of lake 80 x 106 m2 Problem Strategy Solution • Estimate P • What rate of phosphorous removal at the wastewater treatment plant would be required to keep the concentration of phosphorous in the lake at an acceptable level of 0.01 mg/L? Problem Strategy Solution • Evaluate all inputs and outputs to the control volume • Qin = Qout • QCin = QCout Problem Strategy Sources QwwtPwwt QstreamPstream Using mass balance approach: Rate of addition of P = Rate of removal of P Solution Problem Strategy Solution Outflow rate, QtPlake Area, A Concentration, Plake Settling rate, AvsPlake Using mass balance approach: Rate of addition of P = Rate of removal of P Problem Strategy Solution Rate of addition of P = Rate of removal of P S = QTPlake + vsAPlake where: S = rate of addition of phosphorus from all sources (g/s) P = concentration of phosphorus (g/m3) QT = stream outflow rate (m3/s) vs = the phosphorus settling rate (m/s) A = surface area of the lake (m2) which results in a steady-state concentration of Plake S QT v s A Of note, vs is empirically derived and difficult to predict with any confidence. Suggest a settling rate of 3-30 m/year. Problem Strategy Solution 1. Determine the mass loading from both sources Phosphorous loading from incoming stream: Ss = (15.0 m3/s)(0.01 mg/L)(g/1000mg)/(1000 L/m3) = 0.15 g/s From the wastewater treatment plant: Sw = (0.2 m3/s)(5.0 mg/L)(1 g/m3)/(mg/L) = 1 g/s For a total loading of S = 0.15 g/s + 1.0 g/s = 1.15 g/s 2. Determine the volume (mass) of water entering over time: Neglecting evaporation QT = 15 m3/s + 0.2 m3/s = 15.2 m3/s 3. Estimate the settling rate: 10 m yr d hr vs yr 365 d 24 hr 3600 s 3 . 17 10 7 m s 4. Apply model PLake S QT v s A 1 . 15 15 . 2 m 3 . 17 10 3 s 0 . 028 g 0 . 028 mg m g s 7 m s 80 10 6 m 2 3 L This is above the 0.01 mg/L suggested for acceptable concentration. We cannot reduce background levels in the stream. Therefore, we need to determine the reduction at the plant. To start with, solve for S with a known value of P = 0.01 mg/L S Plake Q t v s A g 6 2 15 . 2 m 3 3 . 17 10 7 m 0 . 01 80 10 m 3 s s m 0 . 41 g s The amount that the wastewater treatment plant could contribute would be: Sw = 0.41g/s – 0.15 g/s = 0.26 g/s Since Sw is now at 1.0 g/s, there is a need for 74% phosphorous removal Summary of Problem • We basically did a mass balance for • the water supply • the contaminant Q1 where M4 is the mass in both the lake and the outgoing stream M1/T M4/T=M1/T+M2/T-M3/T Q3=Q1+Q2 M2/T Q2 M3/T (settling)

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# Math Foundations - Civil and Environmental Engineering | SIU