1CIEG -125 Introduction to Civil EngineeringFall 2005Lecture 6Career Services and Fundamental Concepts inEnvironmental EngineeringToday• Career Services• Ethics – Groups 10, 11 and 12• Fundamental Concepts in Environmental Engineering• Scheduling the Final ExamFundamental Concepts in Environmental Engineering Outline• Important Dimensions and Units• density• concentration• flow rate• residence time• Mass balance w/one materialDensity• Mass divided by unit volumeρ = M/Vρ = density (kg/m3, lbM/ft3)M = massV = volume• ρwater= 1 x 103kg/m3 = 1 g/cm3 = 62.4 lbM/ft3Concentration• Gravimetric definition: mass of material A in a unit volume of material consisting of material A and other materials B.CA= MA/ (VA+ VB)CA= concentration of A ( kg/m3, mg/L used in EE)MA= mass of material AVA= volume of material AVB= volume of material BExamplePlastic beads with a volume of 0.04 m3and a mass of 0.48 kg are placed in a container and 100 L of water are poured into the container.What is the concentration of plastic beads in mg/L?CA= MA/ (VA+ VB)= (0.48 kg) / (0.04 m3+ 100L(1 m3/ 1000L))= 3.43 kg/ m3= 3.43 kg/ m3(106 mg/kg) / (103L/m3)= 3420 mg/L2Example 2Plastic beads with a volume of 0.04 m3and a mass of 0.48 kg are placed in a 100 L container into which water is poured filling the container to the brim..What is the concentration of plastic beads in mg/L?VA+ VB= 100 LCA= MA/ (VA+ VB)= 0.48 kg / 100L= 0.0048 kg/ L= 0.0048 kg/ L (106 mg/kg)= 4800 mg/LConcentration: ppm• Another measure of concentration is ppm or parts per million• If the fluid is water (ρ = 1 g/cm3) :1 mg/L = (0.001 g)/1000 mL= (0.001 g)/1000 cm3= (0.001 g)/1000 g= 1 g / 106g = 1 ppmConcentration: percentage• Some material concentrations are expressed as percentages, by mass:ΦA= 100 * MA/ (MA+ MB) • The percentage can also be expressed by volume:ΦA= 100 * VA/ (VA+ VB) ExampleA wastewater sludge has a solids concentration of 10,000 ppm. Express this concentration in percent solids (mass basis), assuming that the density of the solids is 1 g/cm3.10,000 ppm = 104g / 106g = 1/100 = 1 %Pollution Concentrations in Air• In air pollution, concentrations are usually expressed in µg/m3of air • 1 microgram (µg) = 10-6g• Sometimes, concentrations are expressed in terms of ppm, by volume• Conversion of µg/m3to ppm (Volume/Volume) requires knowledge of the gram molecular weight of the gasMoles and GMW of Gases• 1 mole is made up of 6.02 x 1023molecules • 1 mole is the amount of a gas in grams numerically equivalent to its molecular weight• 1 mole of CO2weighs:• 12 + 16 + 16 = 44 g CO2• A.k.a. gram molecular weight (GMW)• At standard condition (0°C and 1 atm of pressure), 1 mole of any gas occupies 22.4 L3Converting µg/m3 to ppmppm (V/V) = (1 m3of pollutant / 106m3of air) X µg/m3= (X µg of pollutant / 1 m3of air)= [X µg of pollutant * (1 g / 106µg) * (1 mole / GMW g) * (22.4 L/mole)] / [1 m3of air * (106/ 106) ] = [X µg of pollutant * (1 g / 106µg) * 106* (1 mole / GMW g) * (22.4 x 10-3m3/mole)] / [106m3of air ]= [(X * 22.4 x 10-3/ GMW) m3of pollutant] / [106m3of air ]= [ X * 22.4 x 10-3/ GMW] ppm= [ X * 22.4 / (GMW * 1000)] ppmExample: Gaseous Pollution• If a concentration of 600 µg/m3of sulfer dioxide is in air at standard conditions, what is the ppm of this pollutant?Sulfer dioxide (SO2) has one sulfer atom and two oxygen atomsGMW = (32 + 16 + 16) g/mole = 64 g/moleppm = [(600 * 22.4 ) / (64 * 1000) ] = m3/mol)= 0.21 ppmFlow Rate• Flow Rate can either be:• gravimetric (mass) flow rate (kg/s or lbM/s); or• volumetric (volume) flow rate (m3/s or ft3/s)• dependent quantities because:[mass] = [density] x [volume]• Mass flow rate, QM, is the amount of mass passing a point during a unit of time• Volume flow rate, QV, would be the volume of that same mass of material passingFlow rate, con’t• QM= QV*ρ• Suppose that we have a volumetric flow of materials A and B, QV(A+B), and we know the concentration of A, CA.• The mass flow rate of A is:QMA= CA* QV(A+B)• This is different from first equation, which is only applicable for one material.ExampleA wastewater treatment plant discharges a flow of 1.5 m3/s (water plus solids) at a solids concentration of 20 mg/L (20 mg of solids per liter of flow, solids plus water). How much solid material is the plant discharging per day?QM-SOLIDS= CSOLIDS* QV(WATER + SOLIDS)= [20 mg/L * (10-6kg/mg) ] * [(1.5 m3/s) * (103L/ m3) * (86,400 s/day) ]= 2592 kg/dayExampleA drinking water treatment plant adds fluorine at a concentration of 1 mg/L and the average daily demand is 18 million gallons. How many pounds of fluorine must the community purchase?QM-FLOURINE = CFLOURINE* QWATER= (1 mg/L) * (3.79 L/gal) * (2.2 x 10-6lb/mg) * (18 x 106gal/day)= 150 lb/day4Residence TimeWhen a fluid flows through a container,how long does the average particle of fluid stay in the container?Called residence time, detention time or retention time.QinQoutResidence Time• If the volume of the container is V (m3) • the flow rates Qout= Qin= Q (m3/s) , then the average residence time is:tavg= V/Q• Note that residence time can be increased by increasing the volume of the tank or reducing the flow rate.ExampleA lagoon has a volume of 1500 m3and the flow into the lagoon is 3 m3/hour. What is the residence time in the lagoon?T = V/Q = (1500 m3)/ (3 m3/hour) = 500 hoursWhy do you think we might be concerned about residence time?Mass Balances w/ Single Material• Except for processes involving nuclear reactions, a pound of any material in the beginning of a process will yield a pound of that material in the end, albeit in a different form.• In other words, mass is conserved.• Because mass is conserved, we can do a mass balance before/after a processMass Balance using Black Boxes• We do mass balances by:• drawing a boundary around a process (or set of processes), called a black box• identifying all input mass flows to and output mass flows from this processQ0Q1Conservative, Steady-State Systems• If the process inside the black box does not create or destroy material; and• The flows in/out do not change (i.e., the system is at steady-state),• The system is said to be a steady-state conservative system.• For such systems, Q0= Q1• Thus, mass in = mass out.5Splitting Single Material Flows• Suppose we have two output mass flows instead of one?• If the system is still