2DefinitionsProcess - an operation or series of operations that causes a physical or chemical change thereby converting raw materials into products. Chemical Engineering Examples: reactors, mixers, separators, biological systems, etc.Balance - an accounting or inventory of mass and changes.System – an arbitrary portion or whole of a process as specifiedby the engineer analyzing the problemCAMXRPPMixReactSeparateBW
3Accumulation = Input – Output + Generation - Consumption Steady State: Accumulation = 0 variables such as T, r, volume,flow rates etc. are not a function of time.Unsteady State (transient): At least one variable is a function of time.No reaction / conversion: Generation and consumption = 0
4Example - New York City People moving in People moving out Input = Output =BirthsGeneration =DeathsConsumption =Change in populationAccumulation =
5Batch Process – heating water in a beaker Fixed amount of materialIntegral balance – changes summed over timeContinuous Process – heating water flowing in pipesContinuous flowing streams in a pipeDifferential balance – analysis at an instant timeSemi-Batch Process – drain pot while heating
6Example– drug production process options BatchContinuousSemi-batchSteady stateTransient
7ExampleBioremediation is a method of cleaning up contaminated groundwater and soil. If a dilute solution of nutrients is pumped via a well into a closed soil layer underground at the rate of 1.5 kg/hr, and a recovery well removes 1.2 kg of depleted solution per hour, answer the following questions:a. What is the system (draw a picture)?b. What is the value of the input per hour?c. What is the value of the output per hour?d. What is the value of the accumulation per hour?e. What assumption has to be made to answer (d)?
8Reaction Balances CO2, CO, O2 CO2, CO, O2 2 CO + O2 = 2 CO2 In Total (kg/hr)kg CO /hrkg O2 /hrkg CO2 /hrkg C /hrkg O /hrOutTotal (kg/hr)kg CO /hrkg O2 /hrkg CO2 /hrkg C /hrkg O /hrTotal MassMolecular SpeciesAtomic Species
10Degree of Freedom Analysis A stream of humid air enters a condenser in which 95 % of the water vapor in the air is condensed. The flow rate of the condensate (the liquid leaving the condenser) is measured and found to be 225 L/hr. Dry air may be taken to contain 21 mol % O2 and 79 mol % N2. Calculate the flow rate of the gas stream leaving the condenser and the mole fractions of oxygen, nitrogen, and water in this stream.ProcessAFN1 (mole dry air / hr)0.21 mol O2 / mol0.79 mol N2 / molN2 (mol H2O / hr)N4 (mol O2 / hr)N5 (mol N2 / hr)N6 (mol H2O / hr)w225 L H2O/hrN3 (mol H2O / hr)95 % of water in feed
11Unknowns: N1, N2, N3, N4, N5, N6Givens: (1-3) Material balances(4) Volume to molar conversion for Stream W(5) 95% water specificationDegrees of freedom: 6-5 = 1 problem is underspecifiedAdd one specification: Entering stream is 10 mol % H2O
12Problem Solving Procedure READ and UNDERSTAND the problem statement.• Ask yourself• What information am I given?• What am I asked to do?• What other info might I need to solve the problem?2. Select a BASIS.• The first two questions in Step 1 should help you pickan amount or a flow rate to use as the BASIS.• or assume an amount or flow rate (typically a multiple of 10)3. DRAW and LABEL a process diagram• boxes (processes) and arrows (input & output)• Labels must include units.
134. ASSIGN ALGEBRAIC SYMBOLS to represent any unknowns using “Let x represent .....” statements.• Use as few unknowns / symbols as possible• Place symbols with units on process diagram.5. COLLECT and TABULATE any additional data thatmay be required.6. WRITE and BALANCE stoichiometric equations.7. Create a TABLE OF BALANCES and UNKNOWNS
148. WRITE MASS BALANCES• Always start with A = I + G – O – C andcancel unnecessary terms with justification.• Balances must be independent.• Write the balances in order, starting withthe balance with the fewest unknowns.• Number of independent balances should equalthe number of unknowns. If not, look forother relationships between unknowns.• for non-reactive systems, the max number ofindependent balances = number of molecular species
15What did I learn? 9. SOLVE BALANCES / EQUATIONS 10. CHECK ANSWER 11. ANSWER THE FOLLOWING QUESTION:What did I learn?
16ExampleA solution composed of 50 wt% ethanol (EtOH), 10 wt% methanol (MeOH), and 40 wt% water (H2O) is fed at the rate of 100 kg/hr into a separator that produces one stream at the rate of 60 kg/hrwith the composition of 80 wt% EtOH, 15 wt% MeOH, and 5 wt% H2O, and a second stream of unknown composition. Calculate the composition (in wt%) of the three compounds in the unknown stream and its flowrate in kg/hr.
18Example1000 kg of FeCl 3 ·6 H2O are added to a mixture of crystals of FeCl3 ·H2O to produce a mixture of FeCl3 ·2.5 H2O crystals. How much FeCl3 ·H2O must be added to produce the most FeCl3 ·2.5 H2O ?
1910 Minute ProblemA laundry can purchase soap (the desired material) containing 30.0 wt % water at a price of $ 7.00 /kg. The same manufacturer offers a soap containing 5 wt % water. If the freight rate is $ 6.00 / 100 kg of soap solution, what is the maximum price the laundry should pay the manufacturer for the soap containing 5.0 wt % water ? Note that the buyer has to pay the freight cost.
20Extra Practice Problems Problem Set Handout: II-1 – II-53
22Example Write material balance equations around each block Calculate the degree of freedom analysis for each blockWrite the overall balance equations for the combined systemPerform a degree of freedom analysis for the combined systemSolve the equations using software on the Textbook CDwt %A 60B 20C 20wt %A 4B 36C 60P1Feed = 100 kg2P21W1wt %A 15B 30C 55wt %A 3.0B ??C ??W2 = 20 kg
2310 Minute ProblemA triple effect evaporator is designed to reduce water from an incoming brine (NaCl + H2O) stream from 25 wt % to 3 wt %. If the evaporator unit is to produce 14,670 lb/hr of NaCl (along with 3 wt % H2O), determine:a. the feed rate of brine in lb/hr.b. the water removed from the brine in each evaporator.Additional data are shown in the figure belowMass fractionH2OMass fractionH2OMass fractionH2OV2V3V1Mass fractionNaCl 0.25H2O3F12Mass fractionNaCl 0.33H2OP1P2Mass fractionNaCl 0.50H2OP314,670 lb/hrMass fractionNaCl 0.97H2O
24The Chemical Reaction Equation and Stoichiometry
25Stoichiometry Definitions • theory of the proportions in which chemical species combine with one another in a chemical reaction as represented by a reaction equation.Stoichiometric Equation:• statement of the relative number of molecules or moles of reactants and products that participate in the reaction.• relates molecules, atoms, or moles but not mass.Stoichiometric Coefficients: numbers preceding each species in the balanced reaction equation.Stoichiometric ratio: ratio of stoichiometric coefficients of any two species.
26ExampleGypsum (plaster of Paris : CaSO4 · 2H2O) is produced by the reaction of calcium carbonate and sulfuric acid (fed as a 98 wt% solution). A certain lime stone analyzes: CaCO %; MgCO %; inerts 1.70 %.For 5 metric tons (5000 kg) of limestone reacted completely, determine:The complete mass balance for all components entering and leaving the reactor.(MW : CaCO ; MgCO ; H2SO4 98; CaSO4 136; MgSO4 120;H2O 18; CO2 44)HINT: There are two reactions involved
27Important Definitions Limiting Reactant: RLReactant that would disappear first if a reaction proceeded to completion.A reactant is limiting if it is present in less than its stoichiometric proportion relative to all other reactants and therefore used up first. The excess reactant is “left over”.Fractional Excess:where i represents the excess reactant
28Important Definitions Let RL represent the limiting reactantLet R represent a specified reactant– if no reactant is specified, assume RL is to be usedLet PD represent desired productLet PunD represent undesired productsLet P represent productsWhere: x = extent of reactionni = moles after reactionnio = moles before reactionvi = stoichiometric coefficient
29Example Limiting Reactants / Selectivity 40 Guys (25 Nerds, 15 Jocks)30 Women (RL)Overall Reaction: Guys + Women → Couples (P)Desired Reaction: Nerds + Women → Couples (PD)Undesired Reaction: Jocks + Women → Couples (PunD)Prom
30ExampleIn a process for the manufacture of chlorine by direct oxidation of HCL with air over a catalyst to form CL2 and H2O (only), the exit product is composed of HCL (4.4 mol%), CL2 (19.8 mol %), H2O (19.8 mol%), O2 (4.0 mol %) and N2 (52.0 mol%)4 HCl + O2 → 2 Cl2 + 2 H2OWhat was:a) The limiting reactant (RL)b) The percent excess reactantc) The degree of completion of the reactiond) The extent of the reaction for HCl and O2e) The conversion of both HCL and O2
3110 Minute ProblemThe synthesis of ammonia proceeds according to the following reaction N2 + 3 H2 → 2 NH3 In a given plant, 4202 lb of nitrogen and 1046 lb of hydrogen are fed to the synthesis reactor per hour. Production of pure ammonia from this reactor is 3060 lb per hour.a. What is the limiting reactant.b. What is the percent excess reactant.c. What is the percent conversion obtained (based on the limiting reactant).
32Extra Practice Problems Problem Set Handout: II-54 – II-63
33Material Balances for Processes Involving Chemical Reaction + Air (O2)H2O + N2 + CO2Carbon (Coal)ENERGY
34Combustion ProcessCombustion Process → burning or oxidation of fuel to release energy• fuel usually contains C, H, & S• examples: coal, fuel oil, natural gas (methane), liquefied petroleum gas (propane & butane)• two types of combustion reaction:1. “incomplete” – some of the C is converted to CO2. “complete” – no CO is formed• all C goes to CO2• all H goes to H2O• all S goes to SO2Consider the following equations:• incomplete combustion of C 2 C + O2 → 2 CO• complete combustion of C C + O2 → CO2• incomplete combustion of propane 2 C3H8 + 7 O2 → 6 CO + 8 H2O• complete combustion of propane C3H8 + 5 O2 → 3 CO2 + 4 H2O
35• even if both reactions occur in the same process, write the two reactions separately ... do not combine.• Usual source of O2 is air: 21 mole-% O2, 79 % N2• product gas is called “stack gas” or “flue gas”• stack gas is monitored in two ways:1. Volume of gas produced2. Chemical analysis of the stack gas – Orsat Analysis (dry basis)• gas leaving the reactor contains all products including H2O vapor• a sample is cooled to room temperature for chemical analysis H2O condenses• thus, chemical analysis gives analysis on DRY BASIS• to know real WET BASIS composition of the stack gas, add back in the H2O
36Definitions (combustion example) Example: A gas with 40% A, 40% B, and 20% H2O has a Orsat dry gas analysis of 50% A and 50% B.it is common to increase the amount of one reactant in order to (i) shift the equilibrium, and (ii) increase the conversion of the more expensive reactant• in combustion, the cheapest, but not free, component is air (O2)Definitions (combustion example)Theoretical Reactant (O2): moles of reactant (O2) required by stoichiometry for COMPLETE consumption of all the primary reactant to desired product (i.e. fuel to CO2 and H2O for combustion of a hydrocarbon).Theoretical Air: quantity of air that contains theoretical O2Excess Reactant (O2): amount of reactant (O2) in excess of that required for COMPLETE reaction (combustion).
37Combustion ExampleA furnace used to provide heat to anneal steel burns a fuel oil whose composition can be represented as (CH2)n. It is planned to burn this fuel with stoichiometric air.a. Assume complete combustion and calculate the Orsat analysis of the flue gas.b. Recalculate the Orsat analysis assuming that 5 % of the carbon in the fuel burns to CO only (the O2 feed rate is the same as in Part a)
38Element Balance Example Ethanol (CH3CH2OH) undergoes a oxidative (air is fed to the reactor) dehydrogenation reaction to produceacetaldehyde (CH3CHO). Multiple side reactions occur in addition to the primary reaction. The reactor productstream is separated into an acetaldehyde product stream and a gas stream which contains water and the followingcomponents obtained from an Orsat analysis.Determine the flow rate, in kg, of the product streams based on a feed basis of 100 kg of ethanol.Orsat Anaylsis (mole %)CO2 0.7O2 2.1CO 2.3H2 7.1CH4 2.6N2 85.2
40Bio-Example ProblemGlucose (C6H12O6) and ammonia (NH3) form a sterile solution (no live cells) fed continuously into a vessel containing a microorganism. Assume complete bio-reaction. One product formed from the reaction contains ethanol, cells (CH1.8O0.5N0.2 ) and water. (The gas produced is CO2 ). If the reaction occurs anerobically (without the presence of oxygen) what is the minimum amount of kg of feed (glucose and ammonia) required to produce 4.6 kg of ethanol ? Only 60 percent of the moles of glucose are converted to ethanol. The remainder is converted to cell mass, carbon dioxide, and water.
42Extra Practice Problems Problem Set Handout: II-69 – II-105
43Material Balances Involving Chemical Reaction and Multiple Units
44ExamplePlants in Europe sometimes use the mineral pyrites (the desired compound in the pyrites isFeS2) as a source of SO2 for the production of sulfite pulping liquor. Pyrite rock containing 48.0 wt % sulfur, 43.0 wt % iron, and 9.0 wt % inerts is burned completely by flash combustion. All of the iron forms Fe3O4 in the cinder (the solids) and a negligible amount of SO3 occurs in either the cinder or the product gas.The exit gas from the absorber analyzes: SO2 0.7 mol %, O2 2.9 mol % and N mol %.Calculate the kg of air supplied to the burner per kg of the pyrites burned.(MW : S 32; Fe 56; O 16; N 14)GasSep 2PyritesBurnerSep 1SolidsAirSO2
45Extra Practice Problems Problem Set Handout: II-106 – II-109
46Recycle, Bypass, and Purge AspenPlus Cyclohexane Process Simulation
47Review Cyclohexane process diagram Recycle is an example of a multi-unit system• Most often used with reactive system but our first example will be non-reactive• used in reactive systems to feed un-reacted reactants back into a reactor thereby achieving a higher conversion of expensive reactants.Bypass is also an example of a multi-unit system• used in both reactive and non-reactive systemsPurge is also an example of a multi-unit system• used to eliminate the buildup of undesirable material in processReview Cyclohexane process diagram
48Guidelines / Definitions OVERALL & REACTOR Balances involve reaction.All others do not involve reactions.• always identify process, species, and type of balance• Reactor C2H4 Mole Balance: A = I – O + G – CAdd the following to existing terminology:
49Non-Reactive ExampleThe flow chart of a process to recover crystalline potassium chromate (K2CrO4) from an aqueous solutionof this salt is shown below. Forty-five hundred kilograms per hour of a feed solution that is 33.3 wt %K2CrO4 is joined by a recycle stream containing wt % K2CrO4, and the combined stream is fed into anevaporator. The concentrated stream that leaves the evaporator contains 49.4 wt % K2CrO4. This stream is fed into a crystallizer in which it is cooled (causing crystals of K2CrO4 to come out of solution) and then filtered. The filter cake consists of K2CrO4 crystals and a solution that contains wt% K2CrO4. The crystals account for 95 % of the total mass of the filter cake. The solution that passes through the filter, also36.36 wt % K2CrO4, is the recycle stream.Calculate the kg/hr of water removed in the evaporator, the rate of production of crystalline K2CrO4 in kg/hr,the ratio (kg recycle) / (kg fresh feed) ,and the feed rates that the evaporator and crystallizer must bedesigned to handle in kg/hr.Water (Qw)Filter Cake (Pc)Fresh Feed (F)FeConcentrate (Fc)Crystallizer /filterEvaporatorSolution (Ps)Recycle (R)
50Reactive ExampleHydrogen is used to reduce Fe2O3 to metallic iron (Fe) as shown belowThe reaction is: Fe2O3 + 3 H2 → 2Fe + 3 H2OSingle pass conversion of Fe2O3 is 60 %.Determine:a) The overall conversion of Fe2O3b) R in moles / hrc) Moles / hr of each component in the purge stream.:Purge – 0 % waterGasesCondenserFF mol/hr H22 mol/hr CO2ReactorW -100 % waterGFF - 14 mol/hr Fe2O3SeparatorR 80 mol % Fe2O320 mol % FeProduct – mol/hr Fe2 mol/hr Fe2O3
51Reactive Example Consider the simple recycle operation shown below. Ethylene oxide is produced by the reaction 2CH2 =CH2 + O2 → 2CH2 CH2 O. Assume the separator is perfect; that is, it separates all unreacted ethylene and oxygen from the productethylene oxide. The once-through or single-pass conversion is 50 %. The reactants in the fresh feed (FF) stream are in a 2:1 ratio. All unreacted reactants are recycled in the RC stream to join with the FF stream and form the mixed feed (MF) stream. Under these conditions, what is the recycle ratio required to produce and overall conversion of 100 %? Recycle ratio is defined as total moles of recycle (RC) per mole of P.Reconsider the above problem but this time assume the separator has been taken out of service; that is, the reactor product (RP) stream is simply split so that some of it is recycled (RC) and some of it is collected as product stream (P). Under these conditions, what recycle ratio is required to achieve an overall conversion of 75 percent?FFMFRPSeparatorPReactorRC
5210 Minute ProblemConsider a more realistic process for the production of ethylene oxide. In this case air rather than pureoxygen is mixed with the ethylene gas to form the FF stream. The ethylene/oxygen ratio that results isnot the stoichiometric 2:1. The separator (an absorber) is used as shown to the right. The separator product stream (SP) is split: some of it becomes the RC stream and some of it becomes the waste stream (W). The ethylene oxide product is removed from the bottom of the absorber as the P stream. Assume the separator is ideal as in the example problem.In this example the ethylene/air ratio being fed to the process is 1:10. The conversion of ethylene to ethylene oxide on a once-through basis is 55 %. What will be the overall conversion if 65 % of the gases leaving the absorber as SP are recycled?Reconsider the problem above from another point of view. Assume the separator is ideal. In this case, we analyze the W stream and find that it is 81.5 mole % N2 , 16.5 % O2 , and 2 % ethylene. We know that the recycle ratio RC/W is 3.0 in this problem. Calculate (a.) the ethylene/air ratio in the FF stream, and (b.) the conversion on a once-through basis.PFFMFRPSPAbsorberWReactorRC
53Extra Practice Problems Problem Set Handout: II-110 – II-135