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Material Balances. Process - an operation or series of operations that causes a physical or chemical change thereby converting raw materials into products.

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Presentation on theme: "Material Balances. Process - an operation or series of operations that causes a physical or chemical change thereby converting raw materials into products."— Presentation transcript:

1 Material Balances

2 Process - 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 specified by the engineer analyzing the problem Definitions Mix React Separate A B MX C RP P W

3 Accumulation = Input – Output + Generation - Consumption Steady State: Accumulation = 0 variables such as T, , 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

4 Input = People moving in Output = People moving out Generation = Births Consumption = Deaths Accumulation = Change in population Example - New York City

5 Batch Process – heating water in a beaker Fixed amount of material Integral balance – changes summed over time Continuous Process – heating water flowing in pipes Continuous flowing streams in a pipe Differential balance – analysis at an instant time Semi-Batch Process – drain pot while heating

6 Example– drug production process options Batch Continuous Semi-batch Steady state Transient

7 Example Bioremediation 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)?

8 Reaction Balances 2 CO + O 2 = 2 CO 2 CO 2, CO, O 2 In Total (kg/hr) kg CO /hr kg O 2 /hr kg CO 2 /hr kg C /hr kg O /hr Out Total (kg/hr) kg CO /hr kg O 2 /hr kg CO 2 /hr kg C /hr kg O /hr Molecular Species Atomic Species Total Mass

9 Example Steam Cracker C2= Reactions C3= Reactions C4= Reactions Aromatics Reactions Naphtha 250,000 ton /yr CH 4, H 2 (39,500 T/yr) Polyethylene (30,000 T/yr) Polystyrene (5,000 T/yr) PVC (40,000 T/yr) Ethylene (55,000 T/yr) Fuel Gas Propylene (45,000 T/yr) Acryonitrile (20,000 T/yr) Butene (30,700 T/yr) Synthetic rubber (10,000 T/yr) Butenes (24,000 T/yr) Aromatics (48,000 T/yr) Fuel Oil (3,800 T/yr) Dodecylbenzene (8,000 T/yr) Phenol / acetone (15,750 T/yr) Aromatics (68,000 T/yr) Heavy Oil

10 Degree of Freedom Analysis Process 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 % O 2 and 79 mol % N 2. Calculate the flow rate of the gas stream leaving the condenser and the mole fractions of oxygen, nitrogen, and water in this stream. F A w N 1 (mole dry air / hr) 0.21 mol O 2 / mol 0.79 mol N 2 / mol N 2 (mol H 2 O / hr) N 4 (mol O 2 / hr) N 5 (mol N 2 / hr) N 6 (mol H 2 O / hr) 225 L H 2 O/hr N 3 (mol H 2 O / hr) 95 % of water in feed

11 Unknowns: N1, N2, N3, N4, N5, N6 Givens: (1-3) Material balances (4) Volume to molar conversion for Stream W (5) 95% water specification Degrees of freedom: 6-5 = 1 problem is underspecified Add one specification: Entering stream is 10 mol % H 2 O

12 Problem Solving Procedure 1.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 pick an 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.

13 4. 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 that may be required. 6. WRITE and BALANCE stoichiometric equations. 7. Create a TABLE OF BALANCES and UNKNOWNS

14 8. WRITE MASS BALANCES Always start with A = I + G – O – C and cancel unnecessary terms with justification. Balances must be independent. Write the balances in order, starting with the balance with the fewest unknowns. Number of independent balances should equal the number of unknowns. If not, look for other relationships between unknowns. for non-reactive systems, the max number of independent balances = number of molecular species

15 9. SOLVE BALANCES / EQUATIONS 10. CHECK ANSWER 11. ANSWER THE FOLLOWING QUESTION: What did I learn?

16 Example A solution composed of 50 wt% ethanol (EtOH), 10 wt% methanol (MeOH), and 40 wt% water (H 2 O) is fed at the rate of 100 kg/hr into a separator that produces one stream at the rate of 60 kg/hr with the composition of 80 wt% EtOH, 15 wt% MeOH, and 5 wt% H 2 O, 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.

17 Material Balances Without Reactions

18 Example 1000 kg of FeCl 3 ·6 H 2 O are added to a mixture of crystals of FeCl 3 ·H 2 O to produce a mixture of FeCl 3 ·2.5 H 2 O crystals. How much FeCl 3 ·H 2 O must be added to produce the most FeCl 3 ·2.5 H 2 O ?

19 10 Minute Problem A 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.

20 Extra Practice Problems Problem Set Handout: II-1 – II-53

21 Material Balances Involving Multiple Units

22 1 2 P1 W1 W2 = 20 kg Feed = 100 kg P2 wt % A 60 B 20 C 20 wt % A 15 B 30 C 55 wt % A 4 B 36 C 60 wt % A 3.0 B ?? C ?? A)Write material balance equations around each block B)Calculate the degree of freedom analysis for each block C)Write the overall balance equations for the combined system D)Perform a degree of freedom analysis for the combined system E)Solve the equations using software on the Textbook CD Example

23 10 Minute Problem A triple effect evaporator is designed to reduce water from an incoming brine (NaCl + H 2 O) stream from 25 wt % to 3 wt %. If the evaporator unit is to produce 14,670 lb/hr of NaCl (along with 3 wt % H 2 O), 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 below F V1 V2V3 P1 P2P3 Mass fraction H 2 O 1.00 Mass fraction NaCl 0.25 H 2 O 0.75 Mass fraction H 2 O 1.00 Mass fraction H 2 O 1.00 Mass fraction NaCl 0.33 H 2 O 0.67 Mass fraction NaCl 0.50 H 2 O ,670 lb/hr Mass fraction NaCl 0.97 H2O 0.03

24 The Chemical Reaction Equation and Stoichiometry

25 Stoichiometry Definitions Stoichiometry: 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.

26 Example Gypsum (plaster of Paris : CaSO 4 · 2H 2 O) 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 ; H 2 SO4 98; CaSO 4 136; MgSO 4 120; H 2 O 18; CO 2 44) HINT: There are two reactions involved

27 Limiting Reactant: R L Reactant 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 Important Definitions

28 Let R L represent the limiting reactant Let R represent a specified reactant – if no reactant is specified, assume R L is to be used Let P D represent desired product Let P unD represent undesired products Let P represent products Important Definitions Where:  extent of reaction n i = moles after reaction n io = moles before reaction vi = stoichiometric coefficient

29 Example Limiting Reactants / Selectivity 40 Guys (25 Nerds, 15 Jocks) 30 Women (R L ) Prom Overall Reaction: Guys + Women → Couples (P) Desired Reaction: Nerds + Women → Couples (P D ) Undesired Reaction: Jocks + Women → Couples (P unD)

30 Example In a process for the manufacture of chlorine by direct oxidation of HCL with air over a catalyst to form CL 2 and H 2 O (only), the exit product is composed of HCL (4.4 mol%), CL 2 (19.8 mol %), H 2 O (19.8 mol%), O 2 (4.0 mol %) and N 2 (52.0 mol%) 4 HCl + O 2 → 2 Cl H 2 O What was: a) The limiting reactant (R L ) b) The percent excess reactant c) The degree of completion of the reaction d) The extent of the reaction for HCl and O 2 e) The conversion of both HCL and O 2

31 10 Minute Problem The synthesis of ammonia proceeds according to the following reaction N H 2 → 2 NH 3 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).

32 Extra Practice Problems Problem Set Handout: II-54 – II-63

33 Material Balances for Processes Involving Chemical Reaction + Air (O 2 ) H 2 O + N 2 + CO 2 Carbon (Coal) ENERGY

34 Combustion 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 CO 2. “complete” – no CO is formed all C goes to CO 2 all H goes to H 2 O all S goes to SO 2 Consider the following equations: incomplete combustion of C 2 C + O 2 → 2 CO complete combustion of C C + O 2 → CO 2 incomplete combustion of propane2 C 3 H O 2 → 6 CO + 8 H 2 O complete combustion of propaneC 3 H O 2 → 3 CO H 2 O Combustion Process

35 even if both reactions occur in the same process, write the two reactions separately... do not combine. Usual source of O 2 is air: 21 mole-% O 2, 79 % N 2 product gas is called “stack gas” or “flue gas” stack gas is monitored in two ways: 1. Volume of gas produced 2. Chemical analysis of the stack gas – Orsat Analysis (dry basis) gas leaving the reactor contains all products including H 2 O vapor a sample is cooled to room temperature for chemical analysis..... H 2 O condenses thus, chemical analysis gives analysis on DRY BASIS to know real WET BASIS composition of the stack gas, add back in the H2O

36 Example: A gas with 40% A, 40% B, and 20% H 2 O 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 (O 2 ) Definitions (combustion example) Theoretical Reactant (O 2 ): moles of reactant (O 2 ) required by stoichiometry for COMPLETE consumption of all the primary reactant to desired product (i.e. fuel to CO 2 and H 2 O for combustion of a hydrocarbon). Theoretical Air: quantity of air that contains theoretical O 2 Excess Reactant (O 2 ): amount of reactant (O 2 ) in excess of that required for COMPLETE reaction (combustion).

37 Combustion Example A furnace used to provide heat to anneal steel burns a fuel oil whose composition can be represented as (CH 2 ) 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 O 2 feed rate is the same as in Part a)

38 Element Balance Example Ethanol (CH 3 CH 2 OH) undergoes a oxidative (air is fed to the reactor) dehydrogenation reaction to produce acetaldehyde (CH 3 CHO). Multiple side reactions occur in addition to the primary reaction. The reactor product stream is separated into an acetaldehyde product stream and a gas stream which contains water and the following components 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 %) CO O CO2.3 H CH N

39 Polymath Solution POLYMATH Results , Rev LEQ SOLUTION [1] x1 = 46.9 (F) [2] x2 = 44.1 (B) [3] x3 = 40.2 (W) LEQ REPORT Coefficients matrix and beta matrix x1 x2 x | | | The equations [1] 2·x1 - 2·x2 = 5.6 [2] 3·x1 - 2·x2 - x3 = 12.3 [3] 0.5·x ·x ·x3 = General Number of equations : 3

40 Bio-Example Problem Glucose (C 6 H 12 O 6 ) and ammonia (NH 3 ) 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 (CH 1.8 O 0.5 N 0.2 ) and water. (The gas produced is CO 2 ). 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.

41 Polymath Solution POLYMATH Results , Rev LEQ SOLUTION [1] x1 = 0.8 [2] x2 = 4 [3] x3 = 0.8 [4] x4 = 1.8 LEQ REPORT Coefficients matrix and beta matrix x1 x2 x3 x | | | | 0 The equations [1] x2 + x3 = 4.8 [2] -3·x ·x2 + 2·x4 = 8.4 [3] 0.5·x2 + 2·x3 + x4 = 5.4 [4] -x ·x2 = 0 General Number of equations : 4

42 Extra Practice Problems Problem Set Handout: II-69 – II-105

43 Material Balances Involving Chemical Reaction and Multiple Units

44 Example Plants in Europe sometimes use the mineral pyrites (the desired compound in the pyrites is FeS 2 ) as a source of SO 2 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 Fe 3 O 4 in the cinder (the solids) and a negligible amount of SO 3 occurs in either the cinder or the product gas. The exit gas from the absorber analyzes: SO mol %, O 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) Burner Sep 2 Sep 1 Pyrites Air Solids Gas SO 2

45 Extra Practice Problems Problem Set Handout: II-106 – II-109

46 Recycle, Bypass, and Purge AspenPlus Cyclohexane Process Simulation

47 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 systems Purge is also an example of a multi-unit system used to eliminate the buildup of undesirable material in process Review Cyclohexane process diagram

48 OVERALL & REACTOR Balances involve reaction. All others do not involve reactions. always identify process, species, and type of balance Reactor C 2 H 4 Mole Balance: A = I – O + G – C Add the following to existing terminology: Guidelines / Definitions

49 Non-Reactive Example Evaporator Crystallizer / filter Recycle (R) Concentrate (Fc)Fresh Feed (F) Filter Cake (Pc) Solution (Ps) Fe The flow chart of a process to recover crystalline potassium chromate (K 2 CrO 4 ) from an aqueous solution of this salt is shown below. Forty-five hundred kilograms per hour of a feed solution that is 33.3 wt % K 2 CrO 4 is joined by a recycle stream containing wt % K 2 CrO 4, and the combined stream is fed into an evaporator. The concentrated stream that leaves the evaporator contains 49.4 wt % K 2 CrO 4. This stream is fed into a crystallizer in which it is cooled (causing crystals of K 2 CrO 4 to come out of solution) and then filtered. The filter cake consists of K 2 CrO 4 crystals and a solution that contains wt% K 2 CrO 4. The crystals account for 95 % of the total mass of the filter cake. The solution that passes through the filter, also wt % K 2 CrO 4, is the recycle stream. Calculate the kg/hr of water removed in the evaporator, the rate of production of crystalline K 2 CrO 4 in kg/hr, the ratio (kg recycle) / (kg fresh feed),and the feed rates that the evaporator and crystallizer must be designed to handle in kg/hr. Water (Qw)

50 Reactive Example F - 14 mol/hr Fe 2 O 3 R 80 mol % Fe 2 O 3 20 mol % Fe FF - 60 mol/hr H 2 2 mol/hr CO 2 Gases Condenser W -100 % water Purge – 0 % water Product – 24 mol/hr Fe 2 mol/hr Fe 2 O 3 Separator Reactor GF Hydrogen is used to reduce Fe 2 O 3 to metallic iron (Fe) as shown below The reaction is: Fe­ 2 O H 2 → 2Fe + 3 H 2 O Single pass conversion of Fe 2 O 3 is 60 %. Determine: a) The overall conversion of Fe 2 O 3 b) R in moles / hr c) Moles / hr of each component in the purge stream. :

51 Reactive Example Consider the simple recycle operation shown below. Ethylene oxide is produced by the reaction 2CH 2 =CH 2 + O 2 → 2CH 2 CH 2 O. Assume the separator is perfect; that is, it separates all unreacted ethylene and oxygen from the product ethylene 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? Reactor Separator RC RP FF PMF

52 10 Minute Problem Reactor Absorber RC RPFFSP MF Consider a more realistic process for the production of ethylene oxide. In this case air rather than pure oxygen is mixed with the ethylene gas to form the FF stream. The ethylene/oxygen ratio that results is not 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 % N 2, 16.5 % O 2, 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. W P

53 Extra Practice Problems Problem Set Handout: II-110 – II-135


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