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Transcript of Lecture 3 – Materials Balances Introduction to Environmental Engineering Lecture3 Dr. Jawad...
![Page 1: Lecture 3 – Materials Balances Introduction to Environmental Engineering Lecture3 Dr. Jawad Al-rifai.](https://reader035.fdocuments.in/reader035/viewer/2022070411/56649f3c5503460f94c5b7e6/html5/thumbnails/1.jpg)
Lecture 3 – Materials Lecture 3 – Materials BalancesBalances
Introduction to Environmental EngineeringLecture3
Dr. Jawad Al-rifai
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The accounting of all mass in a chemical/Environmental process is referred to as a mass (or material) balance.
‘day to day’ operation of process for monitoring operating efficiency
Making calculations for design and development of a process i.e. quantities required, sizing equipment, number of items of equipment
Paul Ashall, 2008
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M.B. with a Single MaterialM.B. with a Single Material
Conservation of Mass – mass is neither created nor destroyed
Mass Flow – therefore mass flowing into a box will equal the flow coming out of a box
◦Black box – schematic representation
X0 X1
0
1
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M.B. with a Splitting Single M.B. with a Splitting Single MaterialMaterial
One or more effluent
One or more Feed Source / influent
X0X1
X2
01
2
X0
X1
X2
[Accumulation]= [In]– [Out] +[Produced] – [Consumed]
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State of Mixing-Steady StateState of Mixing-Steady State
1. Steady state The rate of input= rate of out put, mass
rate of accumulation is Zero
◦Conservation: In many problems conservation is assumed Material of concern is not consumed or produced
No chemical, biological or radioactive decay◦Ex. Salt in Sewer & stream
M.B. equation 0 = [In] – [Out] + 0 - 0
[In]= [Out]
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State of Mixing- State of Mixing- Reactions/ loss Reactions/ loss processprocess
2. Reactions/ loss process
dM/dt = [d(in)/dt – d(out)/dt]+ r
r=-KVCn
◦ K; reaction rate constant; S-1 or d-1
◦ C: Concentration of substances◦ n: reaction order◦ V: volume◦ - Indicate disappearance of substances
The reaction rate is often complex function of T, P
[Accumulation]= [In]– [Out] + [Produced] –
[Consumed]
[Accumulation]= [In]– [Out] + [Produced] –
[Consumed]
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Complex Processes with a Single Complex Processes with a Single MaterialMaterial
General Rules for solving M.B. Problems1. Draw the system as a diagram2. Add the available information3. Draw a dotted line around the component
being balanced4. Decide material to be balanced5. Write the basic M.B. equation6. If only one missing variable, solve7. If more that one unknown, repeat the
procedure
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ExampleExample
A completely mixed lake receives two inflows: natural stream flow 0.1 m3/s, wastewater discharge 0.054 m3/s and has a constant volume of 2 x 106 m3.Given:◦ 1) the wastewater has 20 mg/L NH3-N ◦ 2) stream has 1 mg/L NH3-Nbacteria in the lake convert NH3 to NO3
- by a process called nitrification.
-rN = k*CN
where k = a first-order rate constant = 0.03 day-1 and CN = concentration of ammonia-nitrogen mg/L
FIND: lake and outflow NH3-N Assume steady-state, non-conservative mass balance:
Ammonia is very toxic to fish, 1 mg/L NH4-N. Does the amount of natural nitrification in the lake allow wastewater discharge of 20 mg/L ammonia-N?
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Complex Processes with a Single Complex Processes with a Single MaterialMaterial
◦QW*CNW + QN*CNN - QTCN - V*k*CN = 0
◦where QW = wastewater flow, = 0.054 m3/s
◦CNW = wastewater ammonia-N = 20 mg/L
◦QN = stream flow = 0.1 m3/s
◦CNN =stream ammonia-N = 1 mg/L
◦QT = lake outflow = QW +QN = 0.154 m3/s
◦CN = lake and outflow ammonia-N = ?
◦V = lake volume = 2 x 106 m3 ◦ = 150 days
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Complex Processes with a Single Complex Processes with a Single MaterialMaterial
find CN: by rearranging mass balance:QTCN + V*k*CN = QW*CNW + QN*CNN
CN (QT + V*k) = QW*CNW + QN*CNN
Divide everything by QT;CN (1 + V/ QT *k) = (QW*CNW + QN*CNN)/ QT
CN =[ 1 / (1+ (V/QT)*k)]*[(QWCNW + QN*CNN)/QT] CN = [ 1/(1+ (t)*k)]*[(QW*CNW + QN*CNN)/QT]
CN = [1 /(l +(150d * 0.03d-1))]*[(0.054m3/s*20 mg/L + 0.1 m3/s*1 mg/l)/0.154m3/s]
CN = 1.4 mg/L ammonia-nitrogen1.4 mg/L ammonia-N > 1 mg/L standard.
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Complex Processes with a Single Complex Processes with a Single MaterialMaterial
Aside:What is the detention time of water in the lake (Hydraulic Residence Time)?
◦Define detention time, in the book:
= V/Q = volume/flow rate = time2 x 106 m3/(0.1 m3/s + 0.054 m3/s)*(1 day/86,400
s) = 150 days
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Batch cycleBatch cycle
Sequence of operations/steps repeated according to a cycle
Batch cycle timeBatch size
Paul Ashall, 2008
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Typical simple flowsheet Typical simple flowsheet arrangementarrangement
Paul Ashall, 2008
reactorSeparation & purification
Fresh feed(reactants, solvents,reagents, catalysts etc)
product
Recycle of unreacted material
Byproducts/coproductswaste
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Mass balance filtration/centrifugeMass balance filtration/centrifuge
Paul Ashall, 2008
feed suspension
wash water/solvent
solid
waste water
filtrate
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FiltrationFiltrationPaul Ashall, 2008
F1
5000 kg DM water
Impurity 55 kgWater 2600 kgAPI 450 kg
Water 7300 kgImpurity 50 kgAPI 2kg
Water 300 kgAPI 448 kgImpurity 5 kg
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Mass balance - drierMass balance - drierPaul Ashall, 2008
feed product
water/evaporated solvent
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Mass balance – extraction/phase Mass balance – extraction/phase splitsplit
Paul Ashall, 2008
A + B
S
A + B
S + B
A – feed solvent; B – solute; S – extracting solvent
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Example (single stage extraction; Example (single stage extraction; immiscible solventsimmiscible solvents))
Paul Ashall, 2008
E1feed
solvent
raffinate
extract
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Mass balance – absorption unitMass balance – absorption unit
Paul Ashall, 2008
feed gas stream
feed solvent
waste solvent stream
exit gas stream
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Multiple unitsMultiple units
E – evaporator; C – crystalliser; F – filter unitF1 – fresh feed; W2 – evaporated water; P3 – solid product;
R4 – recycle of saturated solution from filter unit
Paul Ashall, 2008
R4
E C FF1
W2
P3