Environmental Engineering (Course Note 2_revised)

Joonhong ParkYonsei CEE Department

2015. 03. 09.

Extensive and Intensive Properties

Extensive properties: dependent upon the size of the system or the sample.

Intensive properties: independent of the size of the system or the sample.

Question: which are extensive properties among the followings?

Mass, Energy, Heat Capacity, volume, temperature heat capacity, and pressure.

Measure of Quantity and Concentration

The quantity of a substance: expressed in terms of mass or weight, volume, or number of moles.

Concentration: the quantity of a substrate (i) in a certain quantity of a particular phase (j)

quantity of a substrate (i)C(i,j) =

quantity of a particular phase (j)

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Concentrations and Other Units of Measure

How many molecules in a mole?

Concentration Units-Mass based, mg/L or g/L-Mole based (Molarity), mol/L or M-the concentration of charge associated with ions in water (Normality) eq/L = M X [charge/ion]

Example) 47.8 g of MgCl2 is dissolved in one liter of water. Calculate its mass-based concentration, molarity, and normality.

Ref) Atom mass for Mg = 24.3 g/mole; Atom mass for Cl = 35.5 g/mole

Concentration unitsConcentration units

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p-notation: negative logarithm of the molar concentration of a substrate, pC = -log [C]

e.g. pH = -log [H+]

Equivalent (eq): the quantity of a substrate equivalent to the quantity of the product in a particular reaction.

e.g. For the sulfuric acid reaction:

H2SO4 + 2H2O = 2H3O+ + SO42-

1 mol H2SO4 equivalent to 2 mol H+

Cf) centi-, milli-, micro-, nano-, pico-,…

Other expressions for concentrations

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Mass Fraction and Mole Fraction

Represent the ratio between the amount of the species and the total amount of the solution (fluid plus species)

% (percent): 1 part species per 100 parts solution%o (per mil): 1 part species per 1000 parts solutionppm (part per million): 1 part species per 106 parts solutionppb (part per billion): 1 part species per 109 parts solutionppt (part per trillion): 1 part species per 1012 parts solutioncf) molecular fraction, mass fraction, volumetric fraction

By convention (in general):-In air, fractions are generally expressed on a molar or volume basis (ex. 5ppb benzene = 5 x10-9 mole in a mole of air)-In water, fractions are generally expressed on a mass basis (ex. 5ppb benzene = 5 x10-9 gram in a gram of water)

FractionFraction

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Partial Pressure

Species amounts in air may be expressed in terms of partial pressure.

Idea gas law: PV = nRT (cf. M = n/V = P/RT) here P=pressure, V=volume, R=idea gas constant (82.05 *10-6 mol-1 m3 atm K-1) T=absolute temperature (oK)

pi =Pi/Pair = (ni/Vi)/(nair/Vair)

Example) 5ppb benzene in air at Pair = 5 x 10-9 atm

Ideal gas lawIdeal gas law

Partial pressurePartial pressure

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Unit Conversion

Mass Conc. = molecular weight of species i * molarity of i. Ex) Mass concentration of benzene (C6H6) = 5μg/L Molecular weight of benzene = 12 x 6 + 1 x 6 = 78g/mol Molarity of benzene = 5 (μg/L) / 78 (g/mol) = 0.0128 μmol/L or 0.0128 μM

Mass concentration = solution density * mass fraction cf) Solution of density in dilute solution in water = 1g/ml

The molecular fraction of a species: Yi = Ci / Cair

Cair = P/RT When P = 1atm and T=293 oK, Cair = 41.6 mol/m3

[R=82.05 x10-6 atm * m3/(mol * K)] Ci = 41.6 μmol/m3 corresponds to a mole fraction of 1 ppm.

Some Basic Units and Conversion Factors

Quantity SI units SI symbol ConversionFactor

USCS units

LengthMassTemperatureAreaVolumePowerVelocityFlow rateDensity

MeterKilogramCelsiusSquare meterCubic meterKilojouleWattMeter/secCubic meter/secKilogram/cubic meter

MKgoCM2M3kJWm/sm3/sKg/m3

3.28082.20461.8 (oC) + 3210.763935.31470.94783.41212.236935.31470.06243

ftLboFft2ft3BtuBtu/hrMi/hrft3/sLb/ft3

Common Prefixes

Quantity Prefix Symbol

10^-150.0000000000010.0000000010.0000010.0010.010.110100100010000001000000000100000000000010^1510^1810^2110^24

femtopiconanomicromillicentidecidekahectokilo

megagigaterapetaexa

zettayotta

fpnμmcddahkMGTPEZY

Precision and Accuracy

PrecisePrecise ImpreciseImprecise

AccurateAccurate

InaccurateInaccurate

Contents:

1. Material Balance Concepts

2. Characteristic times

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Components in Mass/Material Conservation Law

(Mass/Material Balance)

Accumulation rate = Input rate – Output rate + Reaction rate

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Material Balance

Model containers and flow paths Model containers and flow paths in environmental systemin environmental system

closedopen

Converging flow paths Diverging flow paths

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MB1: the amount of a conserved property in a closed container does not change

MB2: the rate of change in the amount of a nonconserved property within a closed container is equal to the net rate of production of that property within the container.

MB3: the rate of change in the amount of a conserved property in an open container is equal to the difference between the rate of flow of that property into the container and the rate of flow out of the container.

MB4: the rate of change in the amount of a nonconserved property in an open container is equal to the rate of flow in, minus the rate of flow out, plus the net rate of production within the container.

Material Balances (MBs)

Different types of MBsDifferent types of MBs

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Batch (Closed) Systems with Conservative Pollutants (MB1)

Accumulation rate = Input rate – Output rate + Reaction rate

Input rate = 0, Out rate = 0 since closed system

Accumulate rate = Reaction rate = 0 since cons. material

(assumption: completely mixed; no sink or no source)

Closed

V: Control Volume (Reactor Volume)C: Cons. Pollutant Conc.

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CompletelyMixedBatch

Reactor(CMBR)

Reactor Volume (V)

Situation: At t = 0, add C pollutant

at Co concentration

Example for MB1

Co

Time (t)

Con

s. P

oll.

Con

c.

0

Reactor Observation

Question:

In the reactor, what would be concentrations at different locations?

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Batch (Closed) Systems with Nonconservative Pollutants (MB2)

Accumulation rate = Input rate – Output rate + Reaction rate

In this case, Input rate = Output rate = 0 (since closed system)

Accumulation rate = Reaction rate for NC Pollutant(assumption: completely mixed; no sink or no source)

Closed

V: Control Volume (Reactor Volume)C: NC Pollutant Conc.

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CompletelyMixedBatch

Reactor(CMBR)

Reactor Volume (V)

Situation: At t = 0, add NC pollutant

at Co concentration

Example for MB2

Co

Time (t)

NC

Pol

l. C

onc.

0

Accumulation rate at t = V * (dC/dt) observed in the reactor at t(function of reactor configuration and other system factors)

Reaction rate = V * (dC/dt) intrinsic reaction rate(function of chemical property and transformation process)

At CMBR, V * (dC/dt)reactor = V * (dC/dt)reaction

Reactor Observation

Question: What if t = infinitely long?

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Open Systems with Conservative Pollutants at SS

(MB3)

Accumulation rate = Input rate – Output rate + Reaction rate

At Steady-State (Water & Pollutant), Accumulation = 0Since the pollutant is a conservative material, intrinsic reaction rate = 0 Input rate = Output rate

CompletelyMixed

ContinuousFlow (V, C)

V: Control Volume (Reactor Volume)C: Cons. Pollutant Conc.

Cin=CQin =Q

Cout =C, Qout = Q

Terminology:•CSTR: Completely Stirred Tank Reactor•CMCFR: Completely Mixed Continuous Flow Reactor

Question: By the way, what is steady state??

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Steady-State Conservative Systems at SS

Steady State: Accumulation rate = 0Conservative Pollutant: No transformation rate

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Example: Steady-State Conservative Systems

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Steady-State Conservative Systems

Step 1: Make simplified the situations (Sketch)Step 1: Make simplified the situations (Sketch)

Qs

Cs

Qw

Cw

Assumptions: conserved properties converging flow paths steady state

Qm

Cm

What materials?What Processes? (Transformation? Transport?)

Cw: contaminant concentration in wastewaterQw: volumetric flow rate of wastewaterCs: contaminant concentration in upstream riverQs: volumetric flow rate in upstream riverCm: contaminant concentration in downstream riverQm: volumetric flow rate in downstream river

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Steady-State Conservative Systems

Qs

Cs

Qw

Cw

Water Balance: Water Density *Qm = Water Density * (QS + QW)

Contaminant Balance: CmQm = QSCS +QWCW

Qm = Qs + Qw = 10.0 m3/s + 5.0 m3/s = 15.0 m3/s

Cm = (CsQs +CwQw)/Qm = (20.0*10.0 + 40.0*5.0) (mg/L * m3/s) / (15.0 m3/s) = 26.7 mg/L

Accumulation Rate = Input Rate – Out Rate + Reaction RateSteady-State means Accumulation Rate = 0Conservative Materials means Reaction Rate = 0Therefore, Input Rate = Out Rate

Qm

Cm

Step 2: Set the material balancesStep 2: Set the material balances

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Steady-State Systems with Nonconservative Pollutants (MB4)

Accumulation rate = Input rate – Output rate + Reaction rate

At Steady-State (Water & Pollutant), Accumulation = 0 Reaction rate = - V * (dC/dt) 0 = Input rate - Output rate + Reaction rate 0 = Cin * Q - C * Q - V * (dC/dt) => C*Q = Cin*Q - V*(dC/dt) => C = Cin – V/Q * (dC/dt) here V/Q = hydraulic retention time

CompletelyMixed

ContinuousFlow (V, C)

V: Control Volume (Reactor Volume)C: NC Pollutant Conc.

CinQin =Q

Cout =C, Qout = Qin =Q

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Characteristic times

Def: magnitude estimates of the time required for a process to approach completion.

When t >> τ (fast process); When t << τ (slow process)

When t ~ τ (intermediate cases)

Characteristic time (Characteristic time (ττ))

Def: a magnitude estimate of the average time that a molecule of a specific matter(or material) will remain in the system before being removed by flow out of the system.

Characteristic residence time (Characteristic residence time (ττrr))

S (stock): the amount of the materials in a system

Fin (flow in) Fout (flow out) Assume FAssume Finin = = FFoutout= F= F

ττr r = S/FCEE3330 Y2013 WEEK3

Example: Characteristic residence time

Given) In Nanji Wastewater Treatment Plant (Seoul, Korea), wastewater flows into its treatment reactor at a steady flow rate (Q) of 1,000 m3/day, and flows out of the treatment reactor at the sample steady flow rate. The volume of treatment reactor (V) was 50 m3.

Mission) Calculate the characteristic residence time of a water molecule

Step 1: Identify S

☞ NOTE: What assumptions may be needed?

Step 2: Identify F

☞ NOTE: Any assumptions to be made?

Step 3: Calculate residence time by S/F in hours

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Reading Assignment and HW#2

Reading Assignment: EES (Environmental Engineering Science, Nazaroff & Alvarez-Cohen) pp. 1-25

H.W.#2

In EES, solve some problems in Chapter 1 (p.26-29)

-1.1, 1.2, 1.3, 1.8, 1.9, 1.12, 1.14

-Due( 제출마감일시 ): March 23 (6pm) at C218