Environmental Engineering (Course Note 2_revised) Joonhong Park Yonsei 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)

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 units CEE3330 Y2008 WEEK1

Other expressions for concentrations 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-,… CEE3330 Y2008 WEEK1

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 solution ppm (part per million): 1 part species per 106 parts solution ppb (part per billion): 1 part species per 109 parts solution ppt (part per trillion): 1 part species per 1012 parts solution cf) 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) CEE3330 Y2008 WEEK1

Partial Pressure Ideal gas law 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 law Partial pressure CEE3330 Y2008 WEEK1

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. CEE3330 Y2008 WEEK1

Some Basic Units and Conversion Factors Quantity SI units SI symbol Conversion Factor USCS units Length Mass Temperature Area Volume Power Velocity Flow rate Density Meter Kilogram Celsius Square meter Cubic meter Kilojoule Watt Meter/sec Cubic meter/sec Kilogram/cubic meter M Kg oC M2 M3 kJ W m/s m3/s Kg/m3 3.2808 2.2046 1.8 (oC) + 32 10.7639 35.3147 0.9478 3.4121 2.2369 0.06243 ft Lb oF ft2 ft3 Btu Btu/hr Mi/hr ft3/s Lb/ft3

Common Prefixes Quantity Prefix Symbol 10^-15 0.000000000001 0.000000001 0.000001 0.001 0.01 0.1 10 100 1000 1000000 1000000000 1000000000000 10^15 10^18 10^21 10^24 femto pico nano micro milli centi deci deka hecto kilo mega giga tera peta exa zetta yotta f p n μ m c d da h k M G T P E Z Y

Precision and Accuracy Precise Imprecise Accurate Inaccurate

Material Balance Concepts Characteristic times Contents: Material Balance Concepts Characteristic times CEE3330 Y2013 WEEK3

Components in Mass/Material Conservation Law (Mass/Material Balance) Accumulation rate = Input rate – Output rate + Reaction rate CEE3330 Y2013 WEEK3

Material Balance Model containers and flow paths in environmental system closed open Converging flow paths Diverging flow paths CEE3330 Y2013 WEEK3

Material Balances (MBs) Different types of MBs 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. CEE3330 Y2013 WEEK3

Batch (Closed) Systems with Conservative Pollutants (MB1) V: Control Volume (Reactor Volume) C: Cons. Pollutant Conc. 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) CEE3330 Y2013 WEEK3

In the reactor, what would be concentrations at different locations? Example for MB1 Reactor Observation Situation: At t = 0, add C pollutant at Co concentration Co Cons. Poll. Conc. Completely Mixed Batch Reactor (CMBR) Volume (V) Time (t) Question: In the reactor, what would be concentrations at different locations? CEE3330 Y2013 WEEK3

Batch (Closed) Systems with Nonconservative Pollutants (MB2) V: Control Volume (Reactor Volume) C: NC Pollutant Conc. 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) CEE3330 Y2013 WEEK3

Question: What if t = infinitely long? Example for MB2 Reactor Observation Situation: At t = 0, add NC pollutant at Co concentration Co NC Poll. Conc. Completely Mixed Batch Reactor (CMBR) Volume (V) Time (t) 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 Question: What if t = infinitely long? CEE3330 Y2013 WEEK3

Conservative Pollutants at SS (MB3) Open Systems with Conservative Pollutants at SS (MB3) Completely Mixed Continuous Flow (V, C) Cout =C, Qout = Q Cin=C Qin =Q Terminology: CSTR: Completely Stirred Tank Reactor CMCFR: Completely Mixed Continuous Flow Reactor V: Control Volume (Reactor Volume) C: Cons. Pollutant Conc. Accumulation rate = Input rate – Output rate + Reaction rate At Steady-State (Water & Pollutant), Accumulation = 0 Since the pollutant is a conservative material, intrinsic reaction rate = 0 Input rate = Output rate Question: By the way, what is steady state?? CEE3330 Y2013 WEEK3

Steady-State Conservative Systems at SS Steady State: Accumulation rate = 0 Conservative Pollutant: No transformation rate CEE3330 Y2013 WEEK3

Example: Steady-State Conservative Systems CEE3330 Y2013 WEEK3

Steady-State Conservative Systems Step 1: Make simplified the situations (Sketch) Qs Assumptions: conserved properties converging flow paths steady state Cs Qm Cm What materials? What Processes? (Transformation? Transport?) Qw Cw Cw: contaminant concentration in wastewater Qw: volumetric flow rate of wastewater Cs: contaminant concentration in upstream river Qs: volumetric flow rate in upstream river Cm: contaminant concentration in downstream river Qm: volumetric flow rate in downstream river CEE3330 Y2013 WEEK3

Steady-State Conservative Systems Step 2: Set the material balances Accumulation Rate = Input Rate – Out Rate + Reaction Rate Steady-State means Accumulation Rate = 0 Conservative Materials means Reaction Rate = 0 Therefore, Input Rate = Out Rate Qs Cs Qm 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 Cm Qw Cw CEE3330 Y2013 WEEK3

Steady-State Systems with Nonconservative Pollutants (MB4) Completely Mixed Continuous Flow (V, C) Cout =C, Qout = Qin =Q Cin Qin =Q V: Control Volume (Reactor Volume) C: NC Pollutant Conc. 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 CEE3330 Y2013 WEEK3

Characteristic times Characteristic time (τ) 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 residence time (τr) 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. S (stock): the amount of the materials in a system Fin (flow in) Assume Fin = Fout= F τr = S/F Fout (flow out) CEE3330 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 CEE3330 Y2013 WEEK3

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 CEE3330 Y2013 WEEK3