Lecture 7 Introduction to Environmental Engineering.
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Transcript of Lecture 7 Introduction to Environmental Engineering.
![Page 1: Lecture 7 Introduction to Environmental Engineering.](https://reader038.fdocuments.us/reader038/viewer/2022102906/56649d175503460f949ed7bd/html5/thumbnails/1.jpg)
Lecture 7
Introduction to Environmental Engineering
![Page 2: Lecture 7 Introduction to Environmental Engineering.](https://reader038.fdocuments.us/reader038/viewer/2022102906/56649d175503460f949ed7bd/html5/thumbnails/2.jpg)
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