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

FE Review for Environmental EngineeringProblems, problems, problemsPresented by L.R. Chevalier, Ph.D., P.E.Department of Civil and Environmental EngineeringSouthern Illinois University Carbondale

MATHEMATICAL/PHYSICAL FOUNDATIONS

FE Review for Environmental Engineering

Complete the following chart:

Problem Strategy Solution

Total Solids (TS)______ mg/L

TDS_____mg/L (_____%)

VDS70 mg/L (_____

%)

FDS270 mg/L (_____%)

TSS_____mg/L (_____%)

VSS180 mg/L (_____%)

FSS180 mg/L (_____%)

• 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

Problem Strategy Solution

Total Solids (TS)700mg/L

TDS340mg/L (49%)

VDS70 mg/L (10%)

FDS270 mg/L (39%)

TSS360mg/L (51%)

VSS180 mg/L (26%)

FSS180 mg/L (26%)

Complete a flow chart using the following information

COMPOUND CONCENTRATION(mg/L)

DISSOLVES? VOLATILIZES ORBURNS AT 550 C

Sodium Chloride 45 Yes No

Calcium sulfate 30 Yes No

Clay 100 No No

Copper chloride 10 Yes No

Acetic acid 20 Yes Yes

Coffee grounds 25 No Yes

Problem Strategy Solution

Problem Strategy Solution

Total Solids (TS)______ mg/L

TDS_____mg/L (_____%)

VDS____ mg/L (_____%)

FDS____ mg/L (_____%)

TSS_____mg/L (_____%)

VSS____ mg/L (_____%)

FSS____ mg/L (_____%)

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

Problem Strategy Solution

45

45

Sodium chloride 45 mg/L Dissolves Doesn’t volatilize

45

Problem Strategy Solution

Total Solids (TS)______ mg/L

TDS_____mg/L (_____%)

VDS____ mg/L (_____%)

FDS____ mg/L (_____%)

TSS_____mg/L (_____%)

VSS____ mg/L (_____%)

FSS____ mg/L (_____%)

4530

4530

Calcium sulfate 30 mg/L Dissolves Doesn’t volatizes

4530

Problem Strategy Solution

Total Solids (TS)______ mg/L

TDS_____mg/L (_____%)

VDS____ mg/L (_____%)

FDS____ mg/L (_____%)

TSS_____mg/L (_____%)

VSS____ mg/L (_____%)

FSS____ mg/L (_____%)

4530

4530

4530

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

100

100

Problem Strategy Solution

Total Solids (TS)______ mg/L

TDS_____mg/L (_____%)

VDS____ mg/L (_____%)

FDS____ mg/L (_____%)

TSS_____mg/L (_____%)

VSS____ mg/L (_____%)

FSS____ mg/L (_____%)

453010

453010

453010

100

100

100

Copper chloride 10 mg/L Dissolves Doesn’t volatizes

Problem Strategy Solution

Total Solids (TS)______ mg/L

TDS_____mg/L (_____%)

VDS____ mg/L (_____%)

FDS____ mg/L (_____%)

TSS_____mg/L (_____%)

VSS____ mg/L (_____%)

FSS____ mg/L (_____%)

45 203010

45 203010

453010

100

100

100

Acetic acid 20 mg/L Dissolves Volatizes

20

Problem Strategy Solution

Total Solids (TS)______ mg/L

TDS_____mg/L (_____%)

VDS____ mg/L (_____%)

FDS____ mg/L (_____%)

TSS_____mg/L (_____%)

VSS____ mg/L (_____%)

FSS____ mg/L (_____%)

45 203010

45 203010

453010

10025

10025

100

Coffee grounds 25 mg/L Doesn’t dissolves Volatizes

2025

Problem Strategy Solution

Total Solids (TS)______ mg/L

TDS_____mg/L (_____%)

VDS____ mg/L (_____%)

FDS____ mg/L (_____%)

TSS_____mg/L (_____%)

VSS____ mg/L (_____%)

FSS____ mg/L (_____%)

Problem Strategy Solution

Total Solids (TS)230mg/L

TDS105mg/L (46%)

VDS20 mg/L (9%)

FDS85 mg/L (37%)

TSS125 mg/L (54%)

VSS25 mg/L (11%)

FSS100 mg/L (43%)

• 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)

150gpm

2000 lb/hr

?

Problem Strategy Solution

20001

8 34

1

604

0

150 4 0

146

lb

hr

gal

lb

hrgpm

dM

dtx

Flow rate out gpm

. min

Problem Strategy Solution

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

Source Flow Quality

A 140 gpm essentially cleanB 5 gpm 500 ppm tolueneC 5 gpm 500 ppm benzene

Total 150 gpm 1000 ppm

Example Solution

Source Flow Quality

A 140 gpm essentially cleanB 5 gpm 500 ppm tolueneC 5 gpm 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 Flow Quality

A 140 gpm essentially cleanB 5 gpm 500 ppm tolueneC 5 gpm 500 ppm benzene

Total 150 gpm 1000 ppmX

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?

Example Solution

Simple Model Of Stream Pollution Based On Mass Balance

Qu

Cu

Qd

Cd

Qe

Ce

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 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

Qd = ?Cd = 0.005 mg/L

Qe =0.1 m3/sCe = ?

Industrial Complex

Qu= 0.5 m3/sCu = 0 mg/L

Problem Strategy Solution

Lmg

sm

Lmg

sm

Lmg

sm

Q

CQCQC

CQCQCQdt

dm

e

uudde

ddeeuu

03.0

1.0

05.0005.06.0

0

3

33

....end of example

Problem Strategy Solution

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 m3/s P=0.01 mg/L

Waste water treatment plant Q = 0.2 m3/s P=5.0 mg/L

Surface area oflake 80 x 106 m2

Settling ratevs = 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?

Problem Strategy Solution

• Evaluate all inputs and outputs to the control volume• Qin = Qout

• QCin = QCout

Problem Strategy Solution

Using mass balance approach:Rate of addition of P = Rate of removal of P

Sources

Problem Strategy Solution

QwwtPwwt

QstreamPstream

Settling rate, AvsPlake

Area, A

Outflow rate, QtPlake

Concentration, Plake

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)

Problem Strategy Solution

AvQ

SP

sTlake

which results in a steady-state concentration of

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

Problem Strategy Solution

sm

s shr

hrd

dyr

yrm

v

71017.3

36002436510

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:

Problem Strategy Solution

Lmgmg

m

AvQ

SP

sm

sm

sg

sTLake

028.0

028.0

10801017.32.15

15.1

3

2673

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

Problem Strategy Solution

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

Problem Strategy Solution

sg

msm

sm

mg

AvQPS stlake

41.0

10801017.32.1501.0 2673

3

Summary of Problem

• We basically did a mass balance for• the water supply• the contaminant

Q1

Q2

Q3=Q1+Q2

M1/T

M2/T

M4/T=M1/T+M2/T-M3/T

M3/T(settling)

where M4 is the mass in both the lake and the outgoing stream