CHEE 221: Chemical Processes and Systems What is a process?

CHEE 221: Chemical Processes and Systems What is a process? An operation (or series of operations) by which a particular objective is accomplished Chem Eng: operations that cause a physical or chemical change in material, or changes the condition of energy These changes are accomplished in a series of process units, linked together with a set of input and output process streams What is a system? A region of space defined by a real or imaginary closed envelope (envelop=system boundary) CHEE 221 1

Accumulation = In Out + (Generation Consumption) Inputs to system Accum. Gen. Cons. System boundary Outputs from system Input: Amt of the quantity entering the system through the system boundary Output: Amt of the quantity exiting the system through the system boundary Generation: Amt of the quantity formed inside the system boundary Consumption: Amt of the quantity consumed inside the system boundary Accumulation: (final initial) amount of the quantity inside the system boundary To apply M&E balances, need to define the system and the quantity (quantities) of interest CHEE 221 2

Gasoline + Air Combustion Engine Exhaust Gas C 8 H 18 + 25/2 O 2 8 CO 2 + 9 H 2 O Conserved Quantities Cannot be generated or consumed Total mass* Total energy* Atoms* Accumulation = In Out * Non nuclear systems Material (substance) can be generated or consumed by chemical rxn Accumulation = In Out + (Generation Consumption) Volume, enthalpy, entropy are not necessarily conserved CHEE 221 3

Forms of the General Balance Equation Acc = In Out + Gen Con Integral balance: written in terms of integrated (total) amounts of the specified quantity over a period of time E.g.; At the start of the year, my bank balance was $4328. Now it is $2555. Accumulation = (final initial) Often applied to a batch chemical process Differential balance: written in terms of rates of change of the specified quantity with respect to time E.g.; I get paid $10.50/hr. Accumulation is a differential term Usually applied to a continuous process In M&E balances: A quantity is meaningless without units Any equation that is not dimensionally consistent is incorrect CHEE 221 4

Assumed knowledge Know how to: Perform conversions (always show units and conversion factors) Convert volumetric quantities to mass or molar quantities: Liquid or solid: tabulated specific gravities (Appendix B1) Vapour/gas: equation of state ideal gas law (PV=nRT) Know difference between P abs and P gauge ; different P units; T conversions (K, C, F ) Report appropriate number of sig figs with your answer Convert between mass and molar flows and compositions 1. Week 1 tutorial examples 2. F&R textbook Ch 2: 2.1 2.5b Ch 3 (all) Ch 5.2 (PV=nRT) CHEE 221 5

Measurable Units Measurable units are specific values of dimensions that have been defined by convention, such as grams for mass, seconds for time and centimeters for length. Quantity Unit (Base Unit) Symbol Length Mass Time Temperature meter centimeter kilogram gram second day Celsius Kelvin m cm kg g s day C K Units can be treated like algebraic variables when quantities are added, subtracted, multiplied or divided. CHEE 221 6

Derived Units Derived units can be obtained by multiplying or dividing base units (i.e. units for length, mass, time, etc.) Quantity Unit Symbol In Terms of Base Units Volume litre L 0.001 m 3 Force Newton N 1 kg m/s 2 Energy Joule J 1 N m = 1 kg m 2 /s 2 Pressure Pascal Pa N/m 2 Density g/cm 3 Molecular Weight Dalton MW g/mol CHEE 221 7

Systems of Units 1. Système Internationale d Unités or SI Length = meter (m) Mass = kilogram (kg) Temperature = Kelvin (K) or Celsius (C) Time = second (s) 2. CGS System Length = centimeter (cm) Mass = gram (g) Temperature = Kelvin (K) or Celsius (C) Time = second (s) In this course we will use primarily SI units, but be prepared for American (English) units. 3. American (English) Engineering System Length = foot (ft) Mass = pound mass (lb m ) Temperature = Fahrenheit (F) or Rankine ( R) Time = second (s) CHEE 221 8

The Importance of Proper Units Media Relations Office Jet Propulsion Laboratory California Institute Of Technology National Aeronautics And Space Administration Pasadena, Calif. 91109 Telephone (818) 354 5011 http://www.jpl.nasa.gov FOR IMMEDIATE RELEASE September 30, 1999 A failure to recognize and correct an error in a transfer of information between the Mars Climate Orbiter spacecraft team in Colorado and the mission navigation team in California led to the loss of the spacecraft last week, preliminary findings by NASA's Jet Propulsion Laboratory internal peer review indicate. "People sometimes make errors," said Dr. Edward Weiler, NASA's Associate Administrator for Space Science. "The problem here was not the error, it was the failure of NASA's systems engineering, and the checks and balances in our processes to detect the error. That's why we lost the spacecraft." The peer review preliminary findings indicate that one team used English units (e.g., inches, feet and pounds) while the other used metric units for a key spacecraft operation. This information was critical to the maneuvers required to place the spacecraft in the proper Mars orbit. "Our inability to recognize and correct this simple error has had major implications," said Dr. Edward Stone, director of the Jet Propulsion Laboratory. "We have underway a thorough investigation to understand this issue." CHEE 221 9

Dimensional Homogeneity Every equation must be dimensionally homogeneous (or dimensionally consistent), that is, every term on both sides of the equation must have the same units. From now on, this is a given. If not, you will get incorrect results, and you will be penalized. Writing out the units for every term in your equation will ensure errors are not made in your calculations. For example, in a problem you may be given information about Stream A flowing at 266 kg/h, Stream B at 90 moles/s, and Stream C at 0.885 m 3 /day. You cannot do the mass balance unless you choose a Basis (e.g. kg/h) and convert all streams to the same units. This is usually your choice, but you may be asked to report your answer using some specified units. Accumulation = In Out + (Generation Consumption) CHEE 221 10

Conversion of Units A measured quantity can be expressed in terms of units having the appropriate dimension. For example, velocity may be expressed in terms of ft/s, miles/h, cm/yr or any other ratio of length/time. The numerical value of velocity depends on the units chosen. Example: 20.0 m/s = (20.0 m/s ) (3.28 ft/m)= 65.6 ft/s To convert a quantity expressed in terms of one unit to its equivalent in terms of another unit, multiply the given quantity by the conversion factor (new unit/old unit). Example: Car fuel economy is generally expressed in two ways; find the equivalent expression (in L/100km) for a fuel economy of 30 miles/gallon (Imperial). [Answer 9.4 L/100 km. Try to get this answer.] *Useful conversion factors can be found inside the front cover of F&R.* CHEE 221 11

Significant Figures and Scientific Notation The significant figures of a number are the digits from the first nonzero digit on the left to either: (a) the last nonzero digit of the number if there is no decimal point, or (b) the last digit (zero or non zero) on the right if there is a decimal point. The number of significant figures is readily seen if scientific notation, generally a more convenient way of representing large numbers, is used. Conventional number: 123,000,000 Scientific notation: 1.23 x 10 8 3 significant figures CHEE 221 12

Significant Figures Arithmetic Multiplication and Division: The number of significant figures in your answer should equal the lowest number of significant figures of any of the numbers being multiplied or divided Addition and Subtraction: Compare the positions of the last significant figure of each number; the position farthest to the left is the position of the last significant figure in your answer Note: F&R often does not include a decimal place after a number. Assume that the decimal place is there. Thus, 100 L/s in a problem statement means the flow is between 99.5 and 100.4 L/s, with 3 sig. figs. CHEE 221 13

Force According to Newton s 2 nd Law of Motion, force is proportional to the product of mass and acceleration (length/time 2 ), F = m a and has units of kg m/s 2 (SI), g cm/s 2 (CGS) and lb m ft/s 2 (American engineering). For the metric systems, the following derived units are commonly used: 1 newton (N) 1 kg m/s 2 1 dyne 1 g cm/s 2 In the American engineering system, the derived force unit is called a poundforce (lb f ) and is defined as the product of a unit mass (1 lb m ) and the acceleration of gravity, which is 32.174 ft/s 2. 1 lb f 32.174 lb m ft/s 2 F&R Ch 2.4 CHEE 221 14

Pressure Pressure ratio of a force to the area on which the force acts; pressure units are force units divided by area units (e.g., N/m 2 or Pascal (Pa), dynes/cm 2,lb f /in 2 or psi) The pressure at the base of a vertical column of fluid of density and height h is called the hydrostatic pressure, and is given by: h (m) P 0 (N/m 2 ) P (N/m 2 ) A (m 2 ) P P 0 Fluid density (kg/m 3 ) gh where, P 0 is the pressure exerted on the top of the column and g is the acceleration of gravity. F&R Ch 3.4 CHEE 221 15

Pressure cont d The earth s atmosphere can be considered a column of fluid with zero pressure at the top. The fluid pressure at the base of this column (e.g., at sea level) is atmospheric pressure, P atm. P atm = 760 mm Hg = 1.01325 x 10 5 N/m 2 (Pa)= 1 atm = 14.7 psi Theabsolutepressure,P abs, of a fluid is the pressure relative to a perfect vacuum (P =0). Many pressure measurement devices (e.g. a tire gauge) report the gauge pressure, which is the pressure relative to atmospheric pressure. A gauge pressure of 0 indicates that the absolute pressure of the fluid is equal to atmosphericpressure. Atiregaugereadingof32psig is 46.7 psia. The relationship for converting between absolute and gauge pressure is: P absolute = P gauge + P atmospheric This is important because the Ideal Gas Law considers Absolute Pressure. CHEE 221 16

Temperature The most commonly used temperature scales are Celsius (C), Fahrenheit (F), and their corresponding absolute temperature scales Kelvin (K) and Rankine (R), respectively. The Kelvin and Rankine scales are defined such that absolute zero has a value of 0 (lowest temperature attainable in nature: 273.15C and 459.67F), and the size of the degree is the same as the Celsius degree (Kelvin scale) or a Fahrenheit degree (Rankine scale). Actual temperatures expressed in one of these scales may be converted to equivalent temperatures in another scale by using the following relationships: T ( K) T ( C) 273.15 T ( R) T ( F) 459.67 T ( R) 1.8T ( K) T ( F) 1.8T ( C) 32 F&R Ch 3.5 CHEE 221 17

Temperature cont d Using previous formulae it is possible to calculate actual temperature values in different scales. For example: For T = 20 C, T = (20 x 1.8) + 32 = 68 F and for T = 200 C, T = (273 + 200) = 473 K = (1.8 x 473) = 851.4 R For temperature differences, recall that there are 100 Celsius degrees between water freezing and water boiling, and 180 Fahrenheit degrees between water freezing and water boiling. Thus there are 1.8 Fahrenheit degrees for each Celsius degree (i.e. one Celsius degree represents a larger amount of temperature). Use this information when converting units containing temperature scales (e.g. the Ideal Gas Law Constant, R. For a temperature difference of 20 Celsius degrees (i.e. between 20 and 40 C), there are (1.8 x 20) = 36 Fahrenheit degrees. Thus for T = 20 C, T = 36 F. CHEE 221 18

Flow Rate Flow Rate the rate at which a material (gas, liquid or solid) is transported in a process stream mass flow rate volume flow rate mass m ; molar flow rate n time volume V time where, the dot above the m, n and V refers to a flow rate. mol time m (kg fluid/s) or V (m 3 fluid/s) To convert a given volumetric flow rate to the mass flow rate of that stream, the density (or specific volume or SG) of a stream can be used. Density CHEE 221 19 m V F&R Ch 3.2 m V

Mass and Volume Volume balances don t work; if you are given information as volumes, change volume to mass or moles. Gases (vapours) are compressible and density changes with pressure and temperature. Since volumes of gases are (almost) never conserved, volume balances generally do not work (i.e. Vol in Vol out ). For low pressures, convert volumes of gases to moles/mass via Ideal Gas Law (F&R Ch 5.2). Solids and liquids are incompressible; density is constant with pressure and varies slightly with temperature. However, volumes should be changed to masses using densities, specific volumes, or specific gravities (e.g. F&R Table B.1, F&R Ch 3.1) Volume balances don t work; if you are given information as volumes, change volume to mass or moles. Always. Always. Always! CHEE 221 20

Converting Solid and Liquid Volumes to Mass (i.e., if you are given information as volumetric flows). Density () mass (m) per unit volume (V) of a substance; units are mass/length 3 (kg/m 3, g/cm 3, lb m /ft 3 ) Specific volume (1/) volume occupied by a unit mass of a substance; units are length 3 /mass (reciprocal of density) Specific gravity (SG) ratio of the density () of a substance to the density ( ref ) of a reference substance at a specific condition; dimensionless SG ref The reference commonly used for solids and liquids is water at 4C: ref = H2O(l) (4C) = 1.000 g/cm 3 = 1000 kg/m 3 = 62.43 lb m /ft 3 The following notation signifies that the specific gravity of a substance at 20C with reference to water at 4C is 0.6. F&R Ch 3.1 SG 20 0.6 4 Appendix B.1 F&R CHEE 221 21

Converting Gaseous Volumes to Moles/Mass: Ideal Gas Law The ideal gas law relates the molar quantity and volume of a gas to temperature and pressure. The law is derived from the kinetic theory of gases which assumes that gas molecules have negligible volume, exert no forces on one another, and collide elastically with the walls of their container. The equation usually appears in the form: PV nrt or PV nrt Valid only at low P P absolute pressureofagas V( V ) volume (volumetric flow rate) of the gas n( n) number of moles (molar flow rate) of the gas R gas constant, value depends on units of P,V, n and T (values found inside back cover of F&R) T absolute temperature of the gas F&R Ch 5.2 CHEE 221 22

Standard Temperature and Pressure (STP) Standard temperature and pressure (STP) is used widely as a standard reference point for expression of the properties and processes of ideal gases. standard temperature T s 0C 273 K standard pressure P s 1atm 760 mm Hg 101.3 kpa absolute standard specific molar volume Vˆ s 22.4 m 3 (STP) kmol L(STP) 22.4 mol 359 ft 3 (STP) lb mole Canada is the world s third largest producer of natural gas with average annual production of 6.4 trillion cubic feet how much is this??? 1 g mole of an ideal gas at 0C and 1 atm occupies 22.4 liters 1 kg mole of an ideal gas at 0C and 1 atm occupies 22.4 m 3 1 lb mole of an ideal gas at 0C and 1 atm occupies 359 ft 3 CHEE 221 23

Concentration The concentration of a component of a mixture or solution is the quantity of this component per unit volume of the mixture. Mass Concentration mass of a component per unit volume of the mixture (g/cm 3, lb m /ft 3, kg/in 3, ) Molar Concentration number of moles of the component per unit volume of the mixture (kmol/m 3, lb moles/ft 3, ) Molarity molar concentration expressed in moles solute/litre solution Trace species (species present in minute amounts) in mixtures of gases or liquids are typically expressed in units of parts per million (ppm) or parts per billion (ppb). If y i is the fraction of component i,then ppm i = y i x 10 6 ppb i = y i x 10 9 F&R Ch 3.3c CHEE 221 24

Moles and Molecular Weight Atomic Weight the mass of an atom of an element based on 12 Chavinga mass of exactly 12; see back cover of F&R Mole the amount of a species whose mass in grams is numerically equal to its molecular weight; one mole of any species contains 6.02 x 10 23 (Avogadro s number) molecules of that species. Moles can also be expressed as kg moles or lb moles, e.g. 27.0 kg of aluminum is equivalent to 1 kg mole, and 1 lb mole of magnesium sulfate has a mass of 120.4 lb. Molecular Weight the sum of the atomic weights of the atoms that constitute a molecule of the compound (same as molar mass); units of the form kg/kmol, g/mol or lb m /lb mole; see Appendix B.1 in F&R Molecular weight is the conversion factor that relates the mass and the number of moles (of any kind,e.g.kg moles) of a quantity of a substance. F&R Ch 3.3a CHEE 221 25

Mass and Mole Fractions Process streams occasionally contain one substance, but more often they consist of mixtures of liquids or gases, or solutions of one or more solutes in a liquid solvent. The following terms may be used to define the composition of a mixture of substances, including a species A: Mass Fraction : x A mass of A kg A total mass kg total or g A g total lb or lb m A total m Mole Fraction : y A moles of A total moles kmol A kmol or mol A mol or lb -moles A lb -mole The mass percent of A is 100x A, and the mole percent of A is 100y A. Mass fractions can be converted to mole fractions or vice versa by assuming a basis of calculation see F&R examples. Remember, mass/mole fractions are unitless. CHEE 221 F&R Ch 3.3b 26

Assumed knowledge Know how to: Perform conversions (always show units and conversion factors) Convert volumetric quantities to mass or molar quantities: Liquid or solid: tabulated specific gravities (Appendix B1), densities or specific volumes Vapour/gas: equation of state ideal gas law (PV=nRT) Know difference between P abs and P gauge ; T and P conversions Report appropriate number of sig figs with your answer Convert between mass and molar flows and compositions Week 1 tutorial (problems posted) F&R textbook Ch 2: 2.1 2.5b Ch 3 (all) Ch 5.2 (PV=nRT) We are moving forward from here CHEE 221 27