Chapter 4. Fundamentals of Material Balances

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1 Chapter 4 Fundamentals of Material Balances

2 Introduction to Chapter 4 1) Certain restrictions imposed by nature must be taken into account when designing a new process or analyzing an existing one 2) The law of conservation of mass, which states that mass can neither be created nor destroyed will not be concerned in the whole book with almost infinitesimal (tiny) conversions between mass and energy associated with chemical reactions 3) The statements based on the law of conservation of mass such as (total mass of input = total mass of output) are examples of mass balances or material balance ( The design of a new process or analysis of an existing one is not complete until it is established that the inputs and outputs of the entire process and of each individual unit satisfy balance equations) 4) In chapter 4 we will present methods for organizing known information about process variables, setting up martial balance equations, and solving these equations for unknown variables

3 41 ( Process Classification) Chemical processes can be classified as : Batch process The feed is charged into the vessel contents are removed sometime later (In other words nothing goes in or out of the process (system) ) No mass crosses the system boundaries between the time the feed is charged and the time the product is removed Continuous process The inputs and outputs flow continuously throughout the duration ( period ) of the process ( There is both stuff going in and out of the system ) Semibatch process Any process that is neither batch nor continuous ( There is either feed (in) or output (out) from the process cant be both )

4 41 ( Process Classification) Examples on : 1) Batch process: Adding reactants to a test tube then mixing them together ( heating ) and then we remove the products and leave the unconsumed reactants sometimes later to be removed when the system has come to equilibrium A bottle of milk is taken from the refrigerator and left on the kitchen table 2) Continuous process: Pump a mixture of liquids into a distillation column at a constant rate and steadily withdraw product streams from the top and bottom of the column 3) Semibatch process: When filling a balloon with air

5 41 ( Process Classification) Steady state: If the values of all the variables in a process do not change with time Example: a) Continuous process a) Temperature b) Pressure c) Volume d) flow rate Transient (Unsteady state): If the values of all the variables in a process change with time Example: a) Semibatch process b) Batch process c) Continuous process

6 42 (The General Balance Equation) Suppose we have a compound that is a component of both the input and output streams of a continuous process unit and we want to know whether the unit is performing as designed After you measured you found out that the mass flow rates of the compound in both streams are measured and found to be different (m in m out ) m in (kg/h) Process unit m out (kg/h) There are several possible explanations for the observed difference between the measured flow rates: 1 Compound is being consumed as a reactant or generated as a product within the unit 2 Compound is accumulating in the unit possibly adsorbing on the walls 3 Compound is leaking from the unit 4 The measurements are wrong

7 42 (The General Balance Equation) A balance on a conserved (well-maintained) quantity (total mass, mass of a particular species, energy, momentum) in a system (a single process unit, a collection of units, or an entire process) may be written in the following general way: Input + Generation - output - consumption = accumulation Enters through system boundaries Produced within system Leaves through system boundaries Consumed (Used up) within system Buildup within system

8 42 (The General Balance Equation) Example 42-1 page 85 on the general balance equation : Each year 50,000 people move into a city, 75,000 people move out, 22,000 are born, and 19,000 die Write a balance on the population of the city We know that the general balance equation is input + generation output consumption = accumulation and everything is given in the question and we need to find the accumulation Input= 50,000 _Generation= 22,000_Output= 75,000_Consumption= 19,000 50,000 (P/yr) + 22,000 (P/yr) 75,000 (P/yr) -19,000 (P/yr) = accumulation Accumulation = -22,000 (P/yr) People / year Each year the city's population decreases by 22,000 people

9 42 (The General Balance Equation) There is two types of balances that can be written: 1 Differential Balances: a) It s a type of balance that shows what is happening in a system at an instant (immediate or sudden) in time b) Each term of the balance equation is a rate : (rate of input, rate of generation, rate of consumption, rate of output, rate of accumulation ) c) The units of the balanced quantity unit divided by a time unit: (people/yr, g SO2 /s, barrels/day) d) This is the type of balance usually used to a continuous process 2 Integral Balances: a) It s a type of balance that describes what happens between two instants of time b) Each term of the equation is an amount of the balanced quantity and has the corresponding (equivalent or matching) unit (people, g SO2, barrels) c) This type of balance is usually used to a batch process, with the two instants of time being the moment after the input takes place and the moment before the product is withdrawn

10 42 (The General Balance Equation) 1) If the balanced quantity is total mass (Except in nuclear reactions, mass can neither be created nor destroyed) then: We need to set generation = 0 and consumption = 0 in the general balance equation General balance equation: input output = accumulation 2) The balanced substance is a nonreactive species (neither a reactant nor a product) then: We need to set generation = 0 and consumption = 0 in the general balance equation General balance equation: input output = accumulation 3) If a system is at steady state (regardless of what is being balanced) then : We need to set accumulation = 0 General balance equation: input + generation = output + consumption By definition, in a steady-state system nothing can change with time, including the amount of the balanced quantity

11 42 ( Balance on Continuous Steady- State Processes ) a) If we had continuous processes at steady state, then the accumulation term in the general balance equation equals zero: input + generation = output + consumption b) If we had continuous processes at steady state and we were doing a balance on a nonreactive species or on total mass, then we need to put not only the acclamation equals to zero but also the consumption and generation in the general balance equation: Input = output

12 42 ( Balance on Continuous Steady- State Processes ) If we had a nonreactive continuous steady state process we know that the input = output and there is two types of balances we could do on the process: input = output Component Mass Balance (CMB) We do balance for each individual component of a stream Total Mass Balance (TMB) We do balance on the total components of a stream

13 42 ( Balance on Continuous Steady- State Processes ) Component Mass Balance (CMB) : n 1 y a y b n 2 y a2 y b2 Lets say that process Is a nonreactive continuous steady state we have two components a and b so we need to do two individual CMB : Because it s a nonreactive continuous steady state process we know that our general formula is : input=output a-balance (CMB) : input of a = output of b n 1 x y a = n 2 x y a2 b-balance (CMB) : Input of b = output of b n 1 x y b = n 2 x y b2

14 42 ( Balance on Continuous Steady- State Processes ) Total Mass Balance (TMB) : n 1 y a y b n 3 y a3 y b3 n 2 y a2 y b2 The process Is a nonreactive continuous steady state so we know that the general mass balance is : input=output So the TMB is : n 1 + n 2 = n 3

15 42 ( Balance on Continuous Steady- State Processes ) Example 42-2 page 86 : One thousand kilograms per hour of a mixture of benzene (B) and toluene (T) containing 50% w benzene by mass is separated by distillation into two fractions The mass flow rate of benzene in the top stream is 450 kg B/h and that of toluene in the bottom stream is 475 kg T/h The operation is at steady state Write balances on benzene and toluene to calculate the unknown component flow rates in the output streams 450 kg B/h m 1 kg T/h 500 kg B/h 500 kg T/h 475 kg T/h m 2 kg B/h

16 42 ( Balance on Continuous Steady- State Processes ) Example 42-2 page 86 (continued) : 1) First thing when we read this question we knew that the process is at steady state so accumulation term equals zero in all material balances 2) There is also no chemical reactions in the question so generation and consumption are zero in all material balances 3) We will use the equation input = output for all the balances First thing we will do balance on Benzene (CMB) : Input of (B) = output of (B) 500 kg B/h = 450 kg B/h + m 2 kg B/h m 2 = 50 kg B/h Then we need to do balance on Toluene (CMB): Input of (T) = output of (T) 500 kg T/h = m 1 kg T/h kg T/h m 1 = 25 kg T/h To check our answer we can use total mass balance : Before adding any material balance we need to make sure that all the units are the same 1000 kg/h = 450 kg/h + m 1 + m kg /h From the component balance we got that m 1 = 25 kg/h, m 2 = 50 kg/h So : 1000 kg/h = 450 kg/h + 25 kg/h + 50 kg/h +475 kg/h 1000 kg/h = 1000 kg/h

17 42 ( Balance on Continuous Steady- State Processes ) Example 42-3 page 87: Two methanol-water mixtures are contained in separate flasks The first mixture contains 400 wt% methanol, and the second contains 700 wt% methanol If 200 g of the first mixture is combined with 150 g of the second, what are the mass and composition of the product? Since no reactions and steady state process is involved all balances have the simple form "input = output 200 g 0400 g CH 3 OH/g 0600 g H 2 O/g 150 g 0700 g CH30H/g 0300 g H20/g m(g) x(g CH30H/g) (1-x){g H2o/g)

18 42 ( Balance on Continuous Steady- State Processes ) Example 42-3 page 87 (Continued): First thing we will do the total mass balance to find m 3 which we will need in our components mass balances : m 1 +m 2 = m g g = m 3 m 3 = 350 g Then we need to do balance on Methanol (CMB): m 1 x 1 + m 2 x 2 = m 3 x 3 x 1, x 2, x 3 are the mass composition of CH 3 OH in each stream 200 g 0400 g CH3OH g 0700 gch3oh = m 3 (g) x 3 (g CH3OH) g g g x 3 = 0529 g CH 3 OH/g From Total mass balance m 3 = 350 g

19 42 ( Balance on Continuous Steady- State Processes ) Example 42-3 page 87 (Continued): To check if our solution is right we can do material balance on water: We know that mass fraction of methanol at the product stream is equal to 0529 g CH 3 OH/g ( we got it from methanol balance ) and the mass of the stream which equals to 350 g ( we got it from total mass balance) so know now we can do balance on water : input = output (200)(0600) + (150)(0300) = (350)(1-0529) 165gH20 = 165gH 2 0 From the process flowchart it told us that mass fraction of H 2 O is one minus mass fraction of methanol

20 42 ( Balance on Continuous Steady- State Processes ) Example 42-4 page 88-89: Air is bubbled through a drum of liquid hexane at a rate of 0100 kmol/min The gas stream leaving the drum contains 10 mole % hexane vapor Air may be considered insoluble in liquid hexane Use an integral balance to estimate the time required to vaporize 100 m 3 of the liquid n 1 = 01 kmol/min (air) Hexane n 2 (kmol/min) y hex = 01 y air = 09 Given in question that hexane is 10 mole % which equals to 01 then to get air mole % its 1 minus hexane mole %

21 42 ( Balance on Continuous Steady- State Processes ) Example 42-4 page (continued): 1 While reading the question you should know that hexane has no input only output stream 2 In the question it said that air is insoluble (doesn t dissolve) in hexane which means that air doesn t change through this process which makes it steady state so it general balance equation is input=output 3 Usually when asks for time for an element know that it has accumulation because its changing in the process 4 Hexane is nonreactive species so its general balance equation is input output = accumulation Balance on air: Input=output n 1 y air =n 2 y air y air in the input stream is 100 mole % therefore it equals to 1 01 kmol / min x 1 = n 2 x 09 n 2 = 0111 kmol / min

22 42 ( Balance on Continuous Steady- State Processes ) Balance on hexane : Input-output= accumulation -output=accumulation -(n 2 x y hex ) = accumulation -(0111 kmol/min x 01) = accumulation kmol/ min = accumulation After finding the accumulation we need to find the time were in question they gave us the volume so we need to go from volume to moles: 1 Find the mass by specific gravity x volume Mass = density hex x V m= 0659 x 1000 kg/m 3 x 10 m 3 m= 6590 kg 2 Convert mass to moles (moles = mass / Mwt) Mwt hex = 8617 kg/kmol (table B1) n= m / Mwt n= 6590 kg / 8617 ( kg/kmol) We don t have the density of hexane but we can go get the specific gravity of hexane From table B1 and multiply it by reference of water (1000 kg/m 3 ) to get density n= 7645 kmol

23 42 ( Balance on Continuous Steady- State Processes ) 3 Accumulation moles =Δt ( check units ) kmol/ min 7645 kmol = 0- t f t f = 6880 min Initially time = to zero except if it mentioned something about initial time in question

24 Process Absorption Lets say we want to purify (clean) gas (A) from substance (B) while using solvent (C); were (B) dissolves in (C) and (A) doesn t Then we will use a chemical process called the absorption process (absorber, scrubber) to help us purify (A) The gas mixture enters from bottom and leaves purified from top of the absorber Solvent (C) will enter from the top of the absorber and leaves with (B) from the bottom of the tower Usually all liquids enters from top and leaves from bottom Pure liquid (C) A + B gas Absorption Process Absorber Pure gas (A) C+B Liquid

25 Process Stripping Process Stripping The stripping process is exactly the opposite of the absorption process Were we have liquid mixture of (C) and (B) and we will use purify (clean) gas (A) to help purify liquid C from element (B) Liquid mixture of (B) and (C) will enter the stripper process from the top tower and leaves purified from the bottom of the tower Were purify gas (A) enters from the bottom and leaves mixed with element (B) from the top of the tower Usually gasses enter from bottom and leaves from top B + C Liquid Gas (A) Stripper A + B Gas Liquid (C)

26 Process Extraction Process Extraction Lets say we have a liquid mixture of (A) and (B), and we want to purify liquid (A) from trace (slight or bit) of (B) A +B Liquid (A) Liquid Were we will use liquid (C) to purify (A) from (B) Liquid (C) and (A) are immiscible (not forming a homogeneous mixture when added together) In the extraction process both the liquid mixture and solvent (C) will enter the extractor together in two different streams While leaving the extractor (A) will be purified and liquid (C) will leave with element (B) (C) Liquid Extractor C + B Liquid

27 Process Leaching We have liquid (B) in solid (A) and we want to extract (remove) liquid (B) from (A) In the leaching process we will use liquid (C) to remove liquid (B) from solid (A) The mixture (A+B) and liquid (C) will enter the leaching process each in a different streams; where (C) and (B) will leave together in one stream and solid (A) will leave alone another stream (C) liquid A + B Liquid in Solid Leaching Process Leaching Process C + B Liquid (A) Solid Its like extracting corn oil from corn by using hexane

28 Process Heat Exchangers The heat exchanger transfers heat from the hot side to the cold side without mixing the 2 sides streams When heat is transferred from the hot side stream then this stream would decrease its temperature or change its phase ( Gas to liquid, liquid to solid ) or both could happen When heat is transferred to the cold side stream then this stream would increase its temperature or change its phase ( Solid to liquid, liquid to Gas ) or both could happen Types of Heat Exchanger : Heater, Cooler, Evaporator (Boiler), Condenser ( convert gas to liquid ), Fired heater(heat by burning fuel)

29 Process Distillation Column Process Distillation Column The distillation column usually separates mixture of components either liquid or gas The components that enter the column have different measures in their Volatility (unstably) The more volatile the component is the less its boiling point More Volatile A B C D E Increase Boiling point The components who are more volatile (A,B,C) are removed from the top of the column Component (C) would show in both top and bottom of the column were we could say it s the midpoint The less volatile (C,D,E) column ones are recovered from the bottom of the column A B C D Distillation Column Reflux Boil up Condenser Boiler Distillate Bottom product A B C C D E

30 Mixer : (mixing, blending, combining) Process A mixer is a process were two or more streams enter the process and leave as a one stream Condenser: (Separator) Feed Top product Vapor / gas Bottom product liquid A condenser is a process unit used to condense vapor into liquid Were not all of the element is condensed to liquid but still some of the element leaves as vapor

31 43(Material Balance Calculations) Flowchart : 1) When you are given a process and asked you to determine something in the process, it is important to organize the information in a way that is handy for later calculations 2) The best way to draw a flowchart of the process, using boxes which is another symbol to represent process units (reactors, mixers, separation units, etc) and with lines with arrows to represent inputs and outputs 3) The flowchart of a process can help get material balance calculations started and keep them moving 4) The chart must be fully labeled when it is first drawn, with values of known process variables and symbols for unknown variables being written for each input and output stream

32 43(Material Balance Calculations) Example on how to draw a Flowchart : Suppose a gas containing N 2 and O 2 is combined with propane in a batch combustion chamber in which some (but not all) of the O 2 and C 3 H 8 react to form CO 2 and H 2 O, and the product is then cooled, condensing the water a) We should have to boxes one to represent the combustion chamber process and the other to represent the condensing process 100 mol C 3 H mo! mol N 2 Combustion Chamber 50 mol C 3 H mol O mol N mol CO 200 mol H 2 O Condenser 50 mol C 3 H mol O mol N mol CO mol H 2 0 The arrows represents the streams of the input and output of the process with the values of known process variables

33 43(Material Balance Calculations) Flowchart streams : a) Write the values and units of all known stream variables at the locations of the streams (input or output) on the chart For example, a stream containing 21 mole% O 2 and 79% N 2 at 320 C and 14 atm flowing at a rate of 400 mol/h : 400 mol/h 021 mol O 2 /mol 079 mol N 2 /mol T = 320 C, P = 14 atm Above the arrow we usually put the total mass or mole (mass flow-rate, mole flowrate or volume flow-rate) of the stream The rest of the information like the mole or mass fraction, temperature or pressure of the stream is added under the arrow

34 43(Material Balance Calculations) The information on the stream can be written in two ways : a) As the total amount or flow rate of the stream and the fractions (mass or mole) of each component 100 kmol/min 06 kmol N 2 /kmol 04 kmol O 2 /kmol b) Directly as the amount or flow rate of each component: 60 kmol N2/min 40 kmol 02/min They multiplied the total mole flow-rate with the mole fraction for each component to get the moles for each component: n i = y i x n T

35 43(Material Balance Calculations) Flowchart streams (continued) : b) Assign algebraic symbols to unknown stream variables and write these variable names and their associated units on the chart Although any symbol may be used to represent any variable we will mostly use: 1 m (kg solution/min) for mas flow-rate ; n ( kmol solution / min) for mole flow-rate 2 x (lbm (i) /lbm) for mass fraction ; y (mol (i) /mol ) for mole fraction 3 n (kmol (i) ) for moles ; m (kg (i) ) for mass 4 V (m 3 (i) ) for volume or V (m 3 (i)/min) for volume flow-rate Let say we had a stream containing containing 21 mole% O2 and 79% N2 at 320 C and 14 atm : n(mol/h) 021 mol O 2 /mol 079 mol N 2 /mol T = 320 C, P = 14 atm They didn t mention anything about the mole flow rate of the stream so we need to put it as a variable If you don t know why we used mole flow-rate not mass flow-rate because they gave us moles fraction so its not possible to put mass flow-rate with mole fraction or the opposite

36 43(Material Balance Calculations) If we had a stream containing air with a rate of 400 mol/h and at 320 C and 14 atm : 400 mol/h They didn't mention anything y(mol O 2 /mol) (1- y)(mol N 2 /mol) T= 320 C, P = 14 atm about the mole fraction of the air so we know that air has O 2 and N 2 so we will have two unknown mole fractions 1) The reason we used (1-y) to represent the mole fraction of N 2 because we want to reduce the number of unknown variables so we know that the total mole fraction of the stream is equal to one then when labeling component mass or mole fractions of a stream, for example, variable names need only be assigned to all but one fraction, since the last one must be 1 minus the sum of the others 2) If they said that mass of stream 1 is half that of stream 2 then we can label the mass of stream 1 by m and we label mass of stream 2 by 2m stream 1 = ½ stream 2 m = ½ stream 2 2 m = stream 2 3) If you know that there is three times as much nitrogen (by mass) in a stream as oxygen, label the mass fractions of O 2 and N 2 : y(g O 2 /g) for oxygen mass fraction and 3y(g N 2 /g) for nitrogen mass fraction rather than y 1 and y 2

37 43(Material Balance Calculations) Example 43-1 page 92-93: An experiment on the growth rate of certain organisms requires an environment of humid air enriched in oxygen Three input streams are fed into an evaporation chamber to produce an output stream with the desired composition A: Liquid water, fed at a rate of 200 cm 3 /min B: Air (21 mole% O 2 the balance N 2 ) C: Pure oxygen, with a molar flow rate one-fifth of the molar flow rate of stream B The output gas is analyzed and is found to contain 15 mole% water Draw and label a flowchart of the process, and calculate all unknown stream variables To solve : 1 Check how many processes you have so you would know flow chart diagram (boxes) you should draw : In this example we only have one process ( they didn t mention that the product is sent to another process or components were products of another process or mentioned two types of processes ) Evaporation Chamber

38 43(Material Balance Calculations) Example 43-1 page (continued): 2 Check how many inputs and outputs streams you have to know how you could draw your flow-chart: In this question they mentioned that we have three inputs and they will leave as one output (similar to a mixture diagram ): Evaporation Chamber 3 Read the information really carefully to know how to label each stream and to know what's your known variables:

39 43(Material Balance Calculations) Example 43-1 page (continued): As mentioned previously in the question that we have three input streams with: A: Liquid water, fed at a rate of 200 cm 3 /min B: Air (21 mole% O 2 the balance N 2 ) C: Pure oxygen, with a molar flow rate one-fifth of the molar flow rate of stream B And one output stream : The output gas is analyzed and is found to contain 15 mole% water We don t have to add compositions because its pure so the composition is equal to 1 V 1 = 20 cm 3 H 2 O/min Evaporation Chamber 0200 n 2 (mol O 2 / min) n 2 (mol air/min) 021 mol O 2 /min 079 mol N 2 /min The mole fractions of the components of any stream must add up to 1 Since the mole fraction of H 2 0 in the outlet stream is known to be 0015, once the mole fraction of O 2 is labeled y that of N 2 must be 1 - (y ) = ( y) (mol N2/mol) n 3 (mol /min) 0015 (mol H 2 0/mol) y(mol O 2 /mol) (0985 y)(mol N 2 /mol)

40 43(Material Balance Calculations) Example 43-1 page (continued): 4) If they gave us in the question volume flow-rate for any stream we have to change it to mass flow-rate or mole flow-rate: To know to convert it to mass flow-rate or mole flow-rate check the other stream if they are in mole flow-rate convert it to mole flow-rate if mass then convert it to mass flow-rate Make sure that all the stream are in mass or mole we cant have both in one flow chart its not possible in the calculations This this question they gave us the volume flow-rate for stream A : A: Liquid water, fed at a rate of 200 cm 3 /min n=mass/ Mwt which equals to n=(density x volume )/ Mwt n 1 = 20 cm 3 H 2 O 100 g H 2 O 1 mol min cm g = 111 mol H 2 O/min

41 43(Material Balance Calculations) Example 43-1 page (continued): Determine the rest of the unknown variables by using balances which are easily written by referring to the flow chart but the tricky part is to determine the type of your process if it is nonreactive or Continuous Steady- State Processes For this question we are dealing with a nonreactive steady state process which we know that its balance equation is simply input=output n 1 =111 mol H 2 O/min n 2 (mol air/min) 021 mol O 2 /min 079 mol N 2 /min n 3 (mol /min) 0015 (mol H 2 0/mol) y(mol O 2 /mol) (0985 y)(mol N 2 /mol a)h 2 O Balance : Input=output n 1 =n 3 y H2O 111 (mol H 2 O /min) = n 3 (0015 mol H 2 O /mol) n 3 = 741 mol / min 0200 n 2 (mol O 2 / min)

42 43(Material Balance Calculations) Example 43-1 page (continued): Total Balance : Input=output n 1 +n n 2 =n 3 111(mol/min) +n n 2 =741 mol/min n 2 =608 mol/min N 2 Balance : Input = output 079n 2 = n 3 (0985 -y) 079(mol N 2 /mol) (608)(mol/min)=(741)(mol/min)(0985-y)(mol N 2 /mol) y=0337 mol O 2 /mol

43 43(Flowchart Scaling and Basis of Calculation) Suppose we have a process that is balanced (were we know all the flow-rates and compositions that are in and out ) observe in the process that the masses (but not the mass fractions) of all streams could be multiplied by a common factor and the process would remain balanced Stream masses could be changed to mass flow rates, and the mass units of all stream variables (including the mass fractions) could be changed from kg to g or lb m or any other mass unit, and the process would still be balanced The process of changing the values of all stream amounts or flow rates by a proportional (related) amount while leaving the stream compositions unchanged is referred to as scaling the flow-chart There is two types of Scaling : 1) Scaling up : The final stream quantities (amounts or numbers) are larger than the original quantities 2) Scaling down: The final stream quantities are smaller than the original quantities You cannot, however, scale masses or mass flow rates to molar quantities or vice versa by simple multiplication

44 43(Flowchart Scaling and Basis of Calculation) Example 43-2 page : A mixture (by moles) of A and B is separated into two fractions A flowchart of the process is shown here: 500 mol 095 mol A/mol 005 mol B/mol 1000 mol 060 mol A/mol 040 mol B/mol 125 mol A 375 mol B It is desired to achieve the same separation with a continuous feed of 1250 lb-moles/h Scale the flowchart accordingly 1) We need to find the scale factor (common number) : New amount Old amount 1250 Ib-moles/h 100 mol = 125 Ib-moles/h mol

45 43(Flowchart Scaling and Basis of Calculation) Example 43-2 page : 2) After finding the scale factor multiply all the moles (not the mole fractions) by it : Feed : Old amount x scale factor 100 mol x 125 (Ib-moles/h) mol = 1250 Ib-moles /h (as wanted in the question so that means our scale factor is right) Top product stream : Old amount x scale factor 50 mol x 125 (Ib-moles/h) mol = 625 Ib-moles/h

46 43(Flowchart Scaling and Basis of Calculation) Example 43-2 page : Bottom product stream (A): Old amount x scale factor 125 mol A x 125(Ib-moles/h) mol = 156 lb-moles A/h Bottom product stream (B): Old amount x scale factor 375 mol B x 125(Ib-moles/h) mol = 469 lb-moles B/h

47 43(Flowchart Scaling and Basis of Calculation) Basis of calculation: The basis of calculation is an amount (mass or moles) or flow rate (mass or molar) of one stream or stream component in a process The first step in balancing a process is to choose a basis of calculation; all unknown variables are then determined to be consistent with this basis 1) If a stream amount or flow-rate is given in a problem statement then we will usually use this quantity as a basis of calculation 2) If mass fractions are known, choose a total mass or mass flow-rate (100 kg or 100 kg/h) of that stream as basis and the same thing with moles if mole fractions are known, choose a total number of moles or a molar flow rate 100 mol C 2 H 6 This processes shown was balanced using 100 mol C 2 H 6 as basis 2000 mol air 021 mol O 2 /mol 079 mol N 2 mol 2100 mol mol C 2 H 6 /mol 0200 mol O 2 /mol 0752 mol N 2 /mol

48 43(Balancing a Process) Rules apply to nonreactive processes : 1 The maximum number of independent equations that can be derived by writing balances on a nonreactive system equals the number of chemical species in the input and output streams o 30 kg C 6 H 6 10 kgc 7 H 6 m(kg) x(kg C6H6/kg) (1 -x)(kgc7h8/kg) Benzene and toluene in the above flow-chart make up the input and output streams of the process write mass or mole balances on benzene and toluene and a total mass or mole balance, but only two of these three equations are independent and writing the third wont do or accomplish anything 2 Write balances first that involve the fewest unknown variables In the above flow-chart total mass balance involves only one unknown (m) The benzene and toluene balances each involve two unknown (m and x) By writing first a total balance and then a benzene balance, we were able to solve first one equation in one unknown, then a second equation, also in one unknown (If we had instead written benzene and toluene balances, we would have had to solve two simultaneous equations in two unknowns; the same answers would have been obtained, but with greater effort)

49 43(Balancing a Process) Example 43-3 page 97-98: An aqueous solution of sodium hydroxide contains 200% NaOH by mass It is desired to produce an 80% NaOH solution by diluting a stream of the 20% solution with a stream of pure water Calculate the ratios (liters H 2 O/kg feed solution) and (kg product solution/kg feed solution) 1) Draw and label the flowchart : 100 kg m 2 (kg) Mixer 020 kg NaOH/kg 080 kg H 2 O/kg 080 kg H 2 0/kg 0920 kg H 2 O/kg m 1 (kg H 2 O) 1) Choose a basis of calculation : V 1 (liters H 2 0) (an amount or flow rate of one of the feed or product streams) Check were the stream with mostly all information is given with less unknown and put basis for it In this question the mixture input stream would have a basis of 100 kg

50 43(Balancing a Process) Example 43-3 page (continued): 3) Express what the problem asks you to determine in terms of the labeled variables on the flowchart a (liters H20/kg feed solution) From the flow-chart : V 1 /100 b (kg product solution/kg feed solution) From the flow-chart : m 2 /l00 4) Count unknown variables and equations relating them If the number of unknowns equals the number of independent equations relating them, you will be able to solve the problem; otherwise, either you have forgotten some relations or the problem is not well defined a Unknowns: We can check the unknowns from Examining the flowchart From the flow-chart : m 1 m 2, and V 1 Equations: (nonreactive process that involves N species, up to N independent material balance equations may be written)

51 43(Balancing a Process) Example 43-3 page (continued): For this process we can write two balances since there are two species(sodium hydroxide and water) We could write them on sodium hydroxide, water, total mass Since we may only write two material balances, we will need a third equation to solve for our three unknowns (m 1, m 2, and V 1 ), but m 1 and V 1 are related by the density of liquid water, which we know Therefore we have three equations and three unknowns so we can solve the problem 5) Outline the solution procedure All balances for this system have the form input = output We start with the balances that have less unknowns Looking at the flowchart, we can see that balances on total mass and water each involve two unknowns (m 1 and m 2 ), a sodium hydroxide balance only involves one unknown (m 2 ), and the water density relationship involves two unknowns (m 1 and V 1 ) Therefore we will start with sodium hydroxide balance

52 43(Balancing a Process) Example 43-3 page (continued): NaOH balance : Input=output (020 kg NaOH/kg)(100 kg) = (0080 kg NaOH/kg)m 2 m 2 = 250 kg NaOH Total mass balance : Input=output 100 kg + m 1 = m kg + m 1 = 250 kg m 1 = 150 kg H 2 O Diluent water volume: Volume = mass/density V 1 = 150 (kg H 2 O) / 1 (kg/l) = 150 L V 1 =m 1 / density H2O

53 43(Balancing a Process) 6) Ratios requested in problem statement a (liters H 2 0/kg feed solution) V 1 / /100 = 15 LH 2 O/kg feed solution b (kg product solution/kg feed solution) m 2 /l00 250/100 = 250 kg product solution/kg feed solution

54 43( Degree-Of-Freedom Analysis) Degree-Of-Freedom Analysis : Before you do any lengthy calculations, you can use a properly drawn and labeled flowchart to determine whether you have enough information to solve a given problem The procedure for doing so is referred to as degree-offreedom analysis The reason behind degree of freedom so you don t lose your time solving a problem that couldn t be solved because its missing information and that would cause you to lose your time To perform a degree-of-freedom analysis, draw and completely label a flowchart, count the unknown variables on the chart, then count the independent equations relating them: n df (= n unknowns - n independent equations )

55 43( Degree-Of-Freedom Analysis) There are three possibilities for n df : 1 If n df = 0, there are n independent equations in n unknowns and the problem can in principle be solved (number of unknown variables is equal to number of independent equations) 2 If n df > 0, there are more unknowns than independent equations relating them, and at least n df additional variable values must be specified before the remaining variable values can be determined 3 If n df < 0, there are more independent equations than unknowns Either the flowchart is incompletely labeled or the problem is over specified with redundant and possibly inconsistent relations Sources of equations relating unknown process stream variables include the following: a) Material balances: Atomic balance Molecular balance Total mass balance

56 43( Degree-Of-Freedom Analysis) b) An energy balance: We will learn about energy balance from chapter 7 to chapter 9 c) Process specifications: How several process variables are related d) Physical properties and laws: Two of the unknown variables may be the mass and volume of a stream material, in which case a tabulated specific gravity for liquids and solids or an equation of state for gases (Chapter 5) would provide an equation relating the variables e) Physical constraints: For example, if the mole fractions of the three components of a stream are labeled x A, x B, and x C, then a relation among these variables is x A + x B + x c =1 f) Stoichiometric relations: We will study about it in section 7 in chapter 4

57 43( Degree-Of-Freedom Analysis) Example 43-4 page : 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/h Dry air may be taken to contain 21 mole% oxygen, with the balance nitrogen Calculate the flow rate of the gas stream leaving the condenser and the mole fractions of oxygen, nitrogen, and water in this stream n 1 (mol dry air/h) 021 mol O 2 /mol 079 mol N 2 /mol n 2 (mol H 2 O) n 4 (mol O 2 /h) n 5 (mol N 2 /h) n 6 (mol H 2 O (v)/h) 225 (liters H 2 O(l)/h) n 3 (mol H 2 O (l)/h) (95% of water in feed)

58 43( Degree-Of-Freedom Analysis) Example 43-4 page (continued): Degree of freedom analysis : n unknowns = 6 n independent equations = 1 Three material balances (one for each species) 2 Relationship between the volumetric and molar flow rates of the condensate (can determine n 3 from the given volumetric flow rate and the known specific gravity and molecular weight of liquid water) 3 Fact that 95% of the water is condensed (provides a relationship between n 3 and n 2 (n 3 = 095n2 )) No information in the problem statement provides a basis for establishing a sixth relation, so that there is one degree of freedom The problem is therefore underspecified, and there is no point in attempting to solve it

59 43( Degree-Of-Freedom Analysis) Example 43-4 page (continued): Suppose now that we had been given an additional piece of information; for example, that the entering air contains 100 mole% water The flowchart would then appear as follows: n 1 (mol/h) 0100 (mol H 2 O) 0900 (mol dry air/h) 021 mol O 2 /mol 079 mol N 2 /mol n 3 (mol O 2 /h) n 4 (mol N 2 /h) n 5 (mol H 2 O (v)/h) 225 (liters H 2 O(l)/h) n 2 (mol H 2 O (l)/h) (95% of water in feed)

60 43( Degree-Of-Freedom Analysis) Example 43-4 page (continued): n unknowns = 5 n independent equations = 1 Three material balances (one for each species) 2 Relationship between the volumetric and molar flow rates of the condensate 3 Fact that 95% of the water is condensed n df = n unknowns - n independent equations n df = 5 5 n df =0 The degree of freedom is zero therefore the problem is solvable in principle

61 43( Degree-Of-Freedom Analysis) 1) Density relationship : n 2 (mol H 2 O (l) /h ) = 225(L H 2 O(l)/h) x (1 kg H 2 O(l) / L) (18 x 10-3 kg / mol H 2 O) (find n 2 ) 2) 95% Condensation : n 2 = 095 (0100 n 1 ) (find n 1 ) 3) O 2 Balance: n 1 (0900)(021) =n 3 (find n 3 ) 4) N 2 Balance: n 1 (0900) (079) =n 4 (find n 4 ) 5) H 2 O Balance: n 1 (0100) = n 2 +n 5 (find n 5 ) 6) Total outlet gas flow-rate: n total = n 3 + n 4 + n 5 (find n total ) 7) Outlet gas composition: y O2 =n 3 /n total ; y N2 =n 4 /n total ; y H2o =n 5 /n total The algebra and arithmetic are left as an exercise

62 43( Degree-Of-Freedom Analysis) General Procedure for Single-Unit Process Material Balance Calculations: 1) Choose as a basis of calculation an amount or flow rate of one of the process streams o o o If an amount or flow rate of a stream is given in the problem statement If several stream amounts or flow rates are given, always use them collectively as the basis If no stream amount or flow rate is specified in the problem statement, take as a basis an arbitrary amount or flow rate of a stream with a known composition (eg, 100 kg or 100 kg/h if all mass fractions are known, or 100 mol or 100 mol/h if all mole fractions are known) 2) Draw a flowchart and fill in all known variable values, including the basis of calculation Then label unknown stream variables on the chart o o o o The flowchart is completely labeled if you can express the mass or mass flow rate (or moles or molar flow rate) of each component of each stream in terms of labeled quantities If you are given (or you can easily determine) either the amount or flow rate or any of the component fractions for a stream, label the total stream quantity or flow rate and the component fractions Try to incorporate given relationships between unknown quantities in the labeling Label volumetric quantities only if they are given in the problem statement or you are asked to calculate them You will write mass or mole balances, but not volume balances

63 43( Degree-Of-Freedom Analysis) General Procedure for Single-Unit Process Material Balance Calculations (continued): 3) Express what the problem statement asks you to determine in terms of the labeled variables 4) If you are given mixed mass and mole units for a stream convert them all either to mass or mole cant have both together 5) Do the degree-of-freedom analysis 6) Write equations in efficient order 7) Solve the equations 8) Calculate the quantities requested in the problem

64 43(Material Balance) Example 43-5 page : A liquid mixture containing 450% benzene (B) and 550% toluene (T) by mass is fed to a distillation column A product stream leaving the top of the column (the overhead product) contains 950 mole % B, and a bottom product stream contains 80% of the benzene fed to the column (meaning that 92% of the benzene leaves with the overhead product) The volumetric flow rate of the feed stream is 2000 L/h and the specific gravity of the feed mixture is 0872 Determine the mass flow rate of the overhead product stream and the mass flow rate and composition (mass fractions) of the bottom product stream 1 Choose Basis: We don t need to choose a basis because they gave use the Volumetric flow rate of fed stream 2 Draw and label flowchart : Feed 2000 L/h m 1 (kg/h) 045 kg B/kg 055 kg T/kg Overhead Product m 2 (kg/h) 095 mol B/mol 005 mol T/mol Bottom Product m B3 (kg B/h) (8% of B in feed) m T3 (kg T/h) y B2 =095 y T2 =1-y B2 =1-095 =005

65 43(Material Balance) Example 43-5 page (continued): 3 Write expressions for the quantities requested in the problem statement: (the overhead product mass flow rate) m 2 (the bottom product mass flow rate) m B3 and m T3 (the mass fraction in the bottom product) x B and x T (the bottom stream contains 80% of the benzene fed to the column) m in (008)(X B in )=m B3 4 Convert mixed units in overhead product stream: Because all of the rest are in masses and even the flow-rates are in mass we need to convert the overhead product from moles to masses 095 mol B/mol n(moles)= m(mass)/mwt(molecular weight) (950 kmol B) X (7811 kg B/kmolB) 005 mol T/mol = 7420 kg B Convert from moles to k-moles to cancel each other to get kg After finding the mass for each component find the total mass which is the summation of both components masses so we can find the mass fractions for each one of them: (7420 kg B) + (461 kg T) = 7881 kg mixture y B2 = (7420 kg B)/(7881 kg mixture) = 0942 kg B/kg y T2 =1-0942= 0058 kg T/Kg

66 43(Material Balance) Example 43-5 page (continued): 5 Perform degree-of-freedom analysis: 4 unknowns (m 1, m 2, m B3, m T3 ) -2 material balances (since there are two molecular species in this nonreactive process) -1 density relationship (relating the mass flow rate to the given volumetric flow rate of the feed) -1 specified benzene split (8% in bottom-92% in overhead) 0 degrees of freedom (The problem is therefore solvable) 6 Write system equations and outline a solution procedure and algebra: Volumetric flow rate conversion (From the given specific gravity, the density of the feed stream is 0872kg/L): m 1 =2000 L/h x 0872 kg/l = 1744 kg Benzene split fraction: m in (008)(X B in )=m B3 1744(008)(045) = m B3 m B3 =62784 kg B/h Benzene balance: 045m 1 =m 2 y B2 + m B3 045(1744)=m 2 (0942) + (62784) m 2 =76647kg/h Toluene balance: (055)m 1 =m 2 y T2 +m T3 (055)(1744)=(76647)(0058)+m T3 m T3 = 915 Kg T/h

67 43(Material Balance) Example 43-5 page (continued): 8 Calculate additional quantities requested in the problem statement: m3 = m B3 + m T3 = 628 kg/h kg/h = 978 kg/h y B3 = m B3 / m 3 = 628 kg B / 978 kg/h = 0064 kg B/kg y T3 = 1 - y B3 = 0936 kg T/kg

68 44(BALANCES ON MULTIPLE-UNIT PROCESSES) 1) In general terms, a "system" is any portion slice or division of a process that can be enclosed within a hypothetical box (boundary) 2) The boundary may be the entire process, an interconnected combined combination of some of the process units, a single unit, or a point at which two or more process streams come together or one stream splits into branches 3) The inputs and outputs to a system are the process streams that intersect the system boundary For example: Feed 2 Boundary (b) Boundary (c) Boundary (a) Feed 1 Unit 1 Unit 2 Product 3 Produc t 1 Produc t 2 Feed 3 Boundary (d) Boundary (e)

69 44(BALANCES ON MULTIPLE-UNIT PROCESSES) Boundary (a): 1 Encloses the entire process 2 The system defined by this boundary has as inputs Feed Streams 1, 2, and 3 3 The system defined by this boundary has as output Product Streams 1, 2, and 3 4 Balances on this system are referred to as overall balances 5 Any stream that is internal to this system would not enter into overall system balances (stream that connects Units 1 and 2) Boundary (b): 1 Encloses a feed stream mixing point 2 Feed Streams 1 and 2 are inputs to this system 3 The stream flowing (entering) to Unit 1 is an output Boundary (c): 1 Encloses Unit 1 2 The system has one input stream 3 The system has two output streams

70 44(BALANCES ON MULTIPLE-UNIT PROCESSES) Boundary (d): 1 Encloses a stream splitting point 2 The system has one input stream 3 The system has two output streams Boundary (e): 1 Encloses Unit 2 2 The system has two input stream 3 The system has one output streams Notes on Material Balance on Multiple Unit Processes: The procedure for material balance calculations on multiple-unit processes is basically the same as that outlined in Section 43 In multiple-unit processes you may have to isolate and write balances on several subsystems of the process to obtain enough equations to determine all unknown stream variables Notes on Degree of Freedom (DOF) on Multiple Unit Processes: Degree-of-freedom analyses (examines) on the overall process and on each subsystem, taking into account only the streams that intersect the boundary of the system under consideration Do not begin to write and solve equations for a subsystem until you have verified that it has zero degrees of freedom

71 44(BALANCES ON MULTIPLE-UNIT PROCESSES) Example 44-1 page : A labeled flowchart of a continuous steady-state two-unit process is shown below Each stream contains two components, A and B, in different proportions Three streams whose flow rates and/or compositions are not known are labeled 1, 2, and kg/h 0900 kg A/kg 0100 kg B/kg 300 kg/h 0600 kg A/kg 0400 kg B/kg 1000 kg/h 0500 kg A/kg 0500 kg B/kg kg/h 0300 kg A/kg 0700 kg B/kg Calculate the unknown flow rates and compositions of streams 1, 2, and 3

72 44(BALANCES ON MULTIPLE-UNIT PROCESSES) 1) Basis : We don t need to take basis because the flow-rates are given 400 kg/h 0900 kg A/kg 0100 kg B/kg 300 kg/h 0600 kg A/kg 0400 kg B/kg 1000 kg/h 0500 kg A/kg 0500 kg B/kg m 1 (kg/h) x 1 (kg A/kg) 1-x 1 (kg B/kg) m 2 x 2 (kg A/kg) 1 x 2 (kg B/kg) m 3 (kg/h) x 3 (kg A/kg) 1 x 3 (kg B/kg) 300 kg/h 0300 kg A/kg 0700 kg B/kg

73 44(BALANCES ON MULTIPLE-UNIT PROCESSES) 2) Degree-of-Freedom Analysis: a Overall system: 2 unknowns (m 3, x 3 ) - 2 balances (2 species) = 0 degrees of freedom Determine m 3 and x 3 b Mixing point: Cant start with this system 4 unknowns (m 1, x 1, m 2, x 2 ) - 2 balances (2 species) degree of freedom isn't equal = 2 degrees of freedom and too many unknowns c Unit 1: 2 unknowns (m 1 x 1 )-2 balances (2 species) = 0 degrees of freedom Determine m 1 and x 1 After finding m 1 and x 1 from unit one now we can go back to mixing point and do degree of freedom and we will find out that it equal to zero so we back to mixing point to find m 2 and x 2 : 2 unknowns (m 2, x 2 ) - 2 balances (2 species) = 0 degrees of freedom Determine m 1 and x 1

74 44(BALANCES ON MULTIPLE-UNIT PROCESSES) 3) Calculations : a Overall Mass Balance: ( ) kg/h = ( ) kg/h + m 3 m 3 = 600 kg/h b Overall Balance on A: (0500)(1000) + (0300)(300) = (0900)(400) + (0600)(300)+ x 3 (600) x 3= kg A/kg c Mass Balance on Unit 1: 100 = 40 + m 1 m 1 =600 kg/h d A Balance on Unit 1: x 1 = 0233 kg A/kg (0500)(1000) = (0900)(400) + x 1 (600)

75 44(BALANCES ON MULTIPLE-UNIT PROCESSES) e Mass Balance on Stream Mixing Point: m = m 2 m 2 = 900 kg/h f A Balance on Stream Mixing Point: x 1 m 1 + (0300)(300) = x 2 m 2 x 2 = 0255 kg A/kg The situation becomes still more complicated when three or more process units are involved In such cases, balances may be written not only for the overall process and individual process units, but also for combinations of units

76 45(Recycle and Bypass) Lets say we have a chemical reaction A B proceeds to completion in a reactor: No matter how little A is present in the feed or how long the reaction mixture remains in the reactor, some A is normally found in the product Sometimes we can face waste resources which will cause you to pay for all the reactant fed to a process, not just the fraction that reacts, and any A that leaves with the product Then lets say you could find a way to separate most or all of the unconsumed reactant from the product stream: Then you can sell the resulting relatively pure product and recycle the unconsumed reactant back to the reactor You will have to pay for the separation and recycle equipment, but the cost for it is way less then to purchase fresh reactant and being able to sell the purified product at a higher price

77 45(Recycle and Bypass) There are several reasons for using recycle in a chemical process: Recovery of catalyst: In many reactors they would use catalysts to increase the rate of the reaction Catalysts are expensive and in many processes they generally include requirements for recovering them from the product stream and recycling them to the reactor Dilution of a process stream: Suppose a slurry (a suspension of solids in a liquid) is fed to a filter If the concentration of solids in the slurry is too high, the slurry is difficult to handle and the filter will not operate properly A portion of the filtrate can be recycled to dilute the feed to the desired solids concentration Control of a process variable: If we had a reaction that releases an extremely large amount of heat, making the reactor difficult and expensive to control We can recycle a portion of the reactor effluent to the inlet by reducing rate of heat generation by lowering the reactant concentration Circulation of a working fluid: Devices (refrigeration cycle used in household refrigerators and air conditioners) a single material is reused indefinitely, with only small makeup quantities being added to the system to replenish working fluid that may be lost through leaks

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