Lecture 7 Introduction to Environmental Engineering.

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Lecture 7 Introduction to Environmental Engineering

Transcript of Lecture 7 Introduction to Environmental Engineering.

Page 1: Lecture 7 Introduction to Environmental Engineering.

Lecture 7

Introduction to Environmental Engineering

Page 2: Lecture 7 Introduction to Environmental Engineering.

M.B. with Multiple Materials

Multiple compounds into and out of our “black box”Remember: you must have strong fundamentals and organizational skills to keep track of all components

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M.B. with Multiple Materials

Great Examples in this sectionAllegheny and Monongahela Rivers form the Ohio River in Pittsburgh, PAAllegheny, 340 cfs, 250 mg/L

Monongahela, 460 cfs, 1500 mg/L

Ohio River, ?, ?

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M.B. with Multiple Materials

Accum. = In – Out + Prod. – Consum.

0 = [(340 cfs)(250 mg/L) + (460 cfs)(1500 mg/L)] – (800 cfs)(C)

C = 969 mg/L

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M.B. with Multiple Materials

Sometimes the “black box”• Is an actual box, like a rectangular

reactor in WWTP’s or WTP• Is a theoretical representation of space,

like the “box model” in air pollution– Width, length, and a theoretical mixing depth

over a city– Wind blows through mixing with the pollutants– Example 3.7 is a good example of SO2

pollution of the air– Donora, PA, 27 deaths, “Donora Episode”

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Materials Separator

Goal• To split a mixed feed material into the

individual components by exploiting some difference in the material properties

• Example: clarifiers in wastewater treatment

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Materials Separator

Recovery and Purity

xo yo

x2 y2

x1 y1

01

2

yofovery Rec100

x ofovery Rec100

0

2

0

1

2

1

y

yR

x

xR

y

x

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Materials Separator

Recovery and Purity• If Rx and Ry equal 100%, then the

separator is perfect

Purity• We want to make sure only a small

portion of y1 is ending up in product stream 1

10011

11

yx

xPx

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Materials Separator

Example 3.9 – Sedimentation

Qi=40m3/hr, Ci = 5000 mg/LQo=30m3/hr, Co = 25 mg/L

Qu=?, Cu = ?

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Materials Separator• Qu = 40 m3/hr –30 m3/hr = 10 m3/hr

• 0 = CiQi – (CuQu + QoCo) + 0 – 0

0 = (5000 mg/L)(40 m3/hr) – [(Cu(10 m3/hr)+ (30 m3/hr)(25 mg/L)Cu = 19,900 mg/L

%5.99)/40)(/5000(

)/10)(/900,19(3

3

hrmLmg

hrmLmgRu

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Materials Separator

In some situations, purity and recovery will not give you the information you need to choose between two alternativesWorrell-Stessel EffectivenessRietema Effectiveness

100

100

0

1

0

1

2

1

0

2

0

1

y

y

x

xE

y

y

x

xE

R

WS

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Complex Processes with Multiple Materials

Most Env. Eng. Processes involve a series of unit operations, with different individual functions• M.B.’s can solve internal processes• M.B.’s can help determine the order of

unit processes

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Complex Processes with Multiple Materials

General Rules for process train Placement

1. Decide what material property to exploit, this becomes the code.

2. Decide how the code is to activate the switch.

3. Try to separate easiest materials first.4. Try to separate the highest quantity

material first.5. If possible do not add anything to

promote separation.

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Complex Processes

Activated Sludge System Waste Activated Sludge

Anaerobic DigestionVolatile SolidsDigested sludgeSecondary digesterSupernatantGases