# FE Review for Environmental Engineering Problems, problems, problems Presented by L.R. Chevalier, Ph.D., P.E. Department of Civil and Environmental Engineering.

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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

MATHEMATICAL/PHYSICAL FOUNDATIONS FE Review for Environmental Engineering

Complete the following chart:

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

Complete a flow chart using the following information

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

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

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

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

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

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

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

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?

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 ?

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

Do you see a problem here?

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

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

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

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

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 QuCuQuCu QdCdQdCd QeCeQeCe Industrial Complex

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.

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

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

....end of example

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

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

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

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?

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

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

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

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 )

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.

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

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:

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

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

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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