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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

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MATHEMATICAL/PHYSICAL FOUNDATIONS FE Review for Environmental Engineering

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Complete the following chart:

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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

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Complete a flow chart using the following information

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Review definitions Fixed mean inorganic – it does not burn Volatile means organic – it does burn

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45 Sodium chloride 45 mg/L Dissolves Doesn’t volatilize 45

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30 45 30 Calcium sulfate 30 mg/L Dissolves Doesn’t volatizes 45 30

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45 30 45 30 45 30 Clay 100 mg/L Doesn’t dissolve Doesn’t volatizes 100

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45 30 10 45 30 10 45 30 10 100 Copper chloride 10 mg/L Dissolves Doesn’t volatizes

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45 20 30 10 45 20 30 10 45 30 10 100 Acetic acid 20 mg/L Dissolves Volatizes 20

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45 20 30 10 45 20 30 10 45 30 10 100 25 100 25 100 Coffee grounds 25 mg/L Doesn’t dissolves Volatizes 20 25

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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?

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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 ?

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Consider the following report from three supplies into a reservoir. Is it correct?

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Do you see a problem here?

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Important Rule: We can add mass (mass balance) but not concentrations X

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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

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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 10 7 mg/d Converting to ppm per day in total water 1.36 x 10 7 mg/ 817560 L = 16.67 mg/L = 16.67 ppm

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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 10 7 mg/d Converting to ppm per day in total water 1.36 x 10 7 mg/ 817560 L = 16.67 mg/L = 16.67 ppm

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4. Therefore, we have 16.67 ppm benzene, and 16.67 ppm toluene! Not 1000 ppm! 5. Can we add these concentrations?

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Simple Model Of Stream Pollution Based On Mass Balance QuCuQuCu QdCdQdCd QeCeQeCe Industrial Complex

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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 m 3 /s. Concentration of copper in upstream flow is below detection limits Effluent flow from plating factory 0.1 m 3 /s Determine the maximum concentration allowable in the effluent from the factory’s wastewater.

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Draw a control volume diagram Determine Q total = Q stream + Q effleuent Convert concentrations to mass (mass flux) Use mass balance to determine the allowable concentration (based on mass) of effluent

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Q d = ? C d = 0.005 mg/L Q e =0.1 m 3 /s C e = ? Industrial Complex Q u = 0.5 m 3 /s C u = 0 mg/L

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....end of example

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SIMPLE PHOSPHOROUS MODEL Want to estimate the amount of phosphorous control needed to prevent eutrophication due to the over- production of algae

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Simple Phosphorous Model Completely mixed lake Steady state Constant settling rate Phosphorous is the controlling nutrient Assumptions:

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Stream Q = 15.0 m 3 /s P=0.01 mg/L Waste water treatment plant Q = 0.2 m 3 /s P=5.0 mg/L Surface area of lake 80 x 10 6 m 2 Settling rate v s = 10 m/yr Schematic Of System

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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?

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Evaluate all inputs and outputs to the control volume Q in = Q out QC in = QC out

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Using mass balance approach: Rate of addition of P = Rate of removal of P Sources Q wwt P wwt Q stream P stream

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Settling rate, Av s P lake Area, A Outflow rate, Q t P lake Concentration, P lake Using mass balance approach: Rate of addition of P = Rate of removal of P

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Rate of addition of P = Rate of removal of P S = Q T P lake + v s AP lake where: S = rate of addition of phosphorus from all sources (g/s) P = concentration of phosphorus (g/m 3 ) Q T = stream outflow rate (m 3 /s) v s = the phosphorus settling rate (m/s) A = surface area of the lake (m 2 )

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which results in a steady-state concentration of Of note, v s is empirically derived and difficult to predict with any confidence. Suggest a settling rate of 3-30 m/year.

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1. Determine the mass loading from both sources Phosphorous loading from incoming stream: S s = (15.0 m 3 /s)(0.01 mg/L)(g/1000mg)/(1000 L/m 3 ) = 0.15 g/s From the wastewater treatment plant: S w = (0.2 m 3 /s)(5.0 mg/L)(1 g/m 3 )/(mg/L) = 1 g/s For a total loading of S = 0.15 g/s + 1.0 g/s = 1.15 g/s

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2. Determine the volume (mass) of water entering over time: Neglecting evaporation Q T = 15 m 3 /s + 0.2 m 3 /s = 15.2 m 3 /s 3. Estimate the settling rate:

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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 4. Apply model

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The amount that the wastewater treatment plant could contribute would be: S w = 0.41g/s – 0.15 g/s = 0.26 g/s Since S w is now at 1.0 g/s, there is a need for 74% phosphorous removal

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Summary of Problem We basically did a mass balance for the water supply the contaminant Q1Q1 Q2Q2 Q 3 =Q 1 +Q 2 M 1 /T M 2 /T M 4 /T=M 1 /T+M 2 /T-M 3 /T M 3 /T (settling) where M 4 is the mass in both the lake and the outgoing stream

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