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MASS AND ENERGY BALANCE SEM 1 (2011/2012)

MASS AND ENERGY BALANCE SEM 1 (2011/2012). Chapter 3. MASS BALANCE. Process Classification. Semibatch. Batch. Feed is charge to the process and product is removed when the process is completed No mass is fed or removed from the process during the operation

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MASS AND ENERGY BALANCE SEM 1 (2011/2012)

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  1. MASS AND ENERGY BALANCESEM 1 (2011/2012) Chapter 3 MASS BALANCE

  2. Process Classification Semibatch Batch • Feed is charge to the process and product is removed when the process is completed • No mass is fed or removed from the process during the operation • Used for small scale production (pharmaceutical products) • Neither batch nor continuous • During the process a part of reactant can be fed or a part of product can be removed. Continuous • Input and output is continuously red and remove from the process • Used for large scale production

  3. Operation of Continuous Process Steady state Unsteady state • All the variables (i.e. temperatures, pressure, volume, flow rate, etc) do not change with time • Minor fluctuation can be acceptable • Process variable change with time, in particular mass flow rate.

  4. TEST yourself Define type and operation of process given below • A balloon is filled with air at steady rate of 2 g/min • A bottle of milk is taken from the refrigerator and left on the kitchen • Water is boiled in open flask Answer • Semibatch and unsteady state • Batch and unsteady state • Semibatch and unsteady state

  5. General balances equation General Balance • A balance on a conserved 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 way: INPUT + GENERATION– OUTPUT – CONSUMPTION = ACCUMULATION

  6. Types of balances Differential Integral • balances that indicate what is happening in a system at an instant time. • balance equation is a rate (rate of input, rate of generation, etc.) and has units of the balanced quantity unit divided by a time unit (people/yr, g SO2/s). • usually applied to a CONTINUOUS process. • Balances that describe what happens between two instants of time. • balance equation is an amount of the balanced quantity and has the corresponding unit (people, g SO2). • usually applied 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.

  7. Rules to simplify the material balance equations • If the balanced quantity is TOTAL MASS, set generation = 0 and consumption = 0. Mass can neither be created nor destroyed. • If the balanced substances is a NONREACTIVE SPECIES (neither a reactant nor a product), set generation = 0 and consumption = 0. INPUT = OUTPUT Generation and consumption = 0.

  8. Rules to simplify the material balance equations • If a system is at STEADY STATE, set accumulation = 0, regardless of what is being balanced. • Steady state: accumulation = 0 • INPUT + GENERATION = OUTPUT + CONSUMPTION

  9. Integral balances on batch process • Ammonia is produced from nitrogen and hydrogen in a batch reactor. At time t = 0 there are n0 mol of NH3 in the reactor, and at a later time tf the reaction terminates and the contents of the reactor, which include nf ammonia, are withdrawn. Between t0 and tf no ammonia enters or leaves through the reactor boundaries. From GMBE: (input=0; output=0) Generation – Consumption = Accumulation For batch reactor: Accumulation = Final output – Initial Input Final GMBE for batch process Initial input + Generation = Final output + Consumption

  10. flowcharts • When you are given process information and asked to determine something about the process, ORGANIZE the bydrawing a flowchart Represent INPUTS Represent PROCESS UNIT (Reactor, mixer, separation units, etc) Represent OUTPUTS

  11. Flowcharts LABELLING • Write the values and units of all known stream variables at the locations of the streams on the flowchart. • Example • A stream containing 21 mole% O2 and 79% N2 at 320˚C and 1.4 atm flowing at a rate of 400 mol/h might be labeled as: Usually we write of the stream! 400 mol/h 0.21 mol O2/mol 0.79 mol N2/mol T = 320˚C, P = 1.4 atm

  12. 100 kmol/min 0.6 kmol N2/kmol 0.4 kmol O2/kmol 60 kmol N2/min 40 kmol O2/min 10 lbm 0.3 lbm CH4/lbm 0.4 lbm C2H4/lbm 0.3 lbm C2H6/lbm 3.0 lbm CH4 4.0 lbm C2H4 3.0 lbm C2H6 Flowcharts LABELLING • Process stream can be given in two ways: • As the total amount or flow rate of the stream and the fractions of each component • Directly as the amount or flow rate of each component.

  13. mol/h 400 mol/h 0.21 mol O2/mol 0.79 mol N2/mol T = 320˚C, P = 1.4 atm y mol O2/mol (1-y) mol N2/mol T = 320˚C, P = 1.4 atm Flowcharts LABELLING • Assign algebraic symbols to unknown stream variables [such as m (kg solution/min), x (lbm N2/lbm), and n (kmol C3H8)] and write these variable names and their associated units on the flowchart.

  14. Flowcharts LABELLING If that the mass of stream 1 is half that of stream 2, label the masses of these streams as m and 2m rather than m1 and m2. m m1 m2 2m

  15. Flowcharts LABELLING If you know that the mass fraction of nitrogen is 3 times than oxygen, label the mass fraction as y g O2/g and 3y g N2/g rather than y1 and y2. y1 O2/g y2 g N2/g y g O2/g 3y g N2/g When labeling component mass fraction or mole fraction, the last one must be 1 minus the sum of the others y mol O2/mol (1-y) mol N2/mol y1 mol O2/mol y2mol N2/mol

  16. Consistent on notation

  17. exercise 1000 kg/h of a mixture of benzene (B) and toluene (T) containing 50 % 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 (kg T/h) 500 kg B/h 500 kg T/h (kg B/h) 475 kg T/h

  18. solution Steady state accumulation = 0 Since no chemical reactions occur generation & consumption = 0 INPUT = OUTPUT Benzene Balance 500 kg B/h = 450 kg B/h + 450 kg B/h (kg T/h) = 50 kg B/h 500 kg B/h 500 kg T/h Toluene Balance 500 kg T/h = + 475 kg T/h (kg B/h) 475 kg T/h = 25 kg T/h

  19. solution Benzene Balance 500 kg B/h = 450 kg B/h + 450 kg B/h (kg T/h) = 50 kg B/h 500 kg B/h 500 kg T/h Toluene Balance 500 kg T/h = + 475 kg T/h (kg B/h) 475 kg T/h = 25 kg T/h Check the calculation: Total Mass Balance 1000 kg/h = 450 + + + 475 (all kg/h) 1000 kg/h = 1000 kg/h

  20. exercise An experiment on the growth rate of certain organism 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 rate of 20 cm3/min B: Air (21% O2 and 79% N2) C: Pure O2 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 1.5 mole% water. Draw and label the flowchart of the process, and calculate all unknown stream variables.

  21. solution An experiment on the growth rate of certain organism 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 rate of 20 cm3/min B: Air (21% O2 and 79% N2) C: Pure O2 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 1.5 mole% water. Draw and label the flowchart of the process, and calculate all unknown stream variables. 0.2 (mol O2/min) (mol/min) 0.015 mol H2O/mol y (mol O2/mol) (0.985 – y)(mol N2/mol) (mol air/min) 0.21 mol O2/mol 0.79 mol N2/mol 20.0 cm3 H2O (l)/min (mol H2O/min)

  22. solution 0.2 (mol O2/min) (mol/min) 0.015 mol H2O/mol y (mol O2/mol) (0.985 – y)(mol N2/mol) (mol air/min) 0.21 mol O2/mol 0.79 mol N2/mol 20.0 cm3 H2O (l)/min (mol H2O/min) = = 50 kg B/h Nonreactive steady-state process input = output H2O Balance = 74.1 mol/min

  23. solution 0.2 (mol O2/min) (mol/min) 0.015 mol H2O/mol y (mol O2/mol) (0.985 – y)(mol N2/mol) (mol air/min) 0.21 mol O2/mol 0.79 mol N2/mol 20.0 cm3 H2O (l)/min (mol H2O/min) Total Mass Balance 0.2 + + = = 60.8 mol/min N2 Balance y = 0.337 mol O2/mol

  24. Procedure of changing the values of all stream amounts or flow rates by a proportional amount while leaving the stream compositions unchanged. if final stream quantities are smaller than the original quantities. If final stream quantities are larger than the original quantities. Scaling up Scaling down What? Flowchart scaling • Suppose you have balanced a process and the amount or flow rate of one of the process streams is n1.You can scale the flow chart to make the amount or flow rate of this stream n2 by multiplying all stream amounts or flow rate by the ratio n2/n1. You cannot, however, scale masses or mass flow rates to molar quantities or vice versa by simple multiplication; conversions of this type must be carried out using the methods as discussed in mass fraction and mol fraction section.

  25. 1 kg C6H6 2 kg 0.5 kg C6H6/kg 0.5 kg C7H8/kg 1 kg C7H8 x 300 300 kg C6H6 600 kg 0.5 kg C6H6/kg 0.5 kg C7H8/kg 300 kg C7H8 kg kg/h Replace kg with lbm 300 lbm/h 600 lbm/h 0.5 lbm C6H6/lbm 0.5 lbm C7H8/lbm 300 lbm/h Flowcharts scalING

  26. 3 kg C6H6/min m (kg/min) x (kg C6H6/kg) (1-x) (kg C7H8/kg) 1 kg C7H8/min Balancing a process 3.0 kg/min of benzene and 1.0 kg/min of toluene are mixed • Two unknown quantities – m and x, associated with process, so two equations are needed to calculate them. • For NONREACTIVE STEADY STATE process input = output. • 3 possible balance can be written – Balance on total mass, benzene, and toluene – any two of which provide the equations needed to determine m and x. For example, Total Mass Balance: 3.0 kg/min + 1.0 kg/min = m kg/min = 4.0 kg/min Benzene Balance: 3.0 kg C6H6/min = 4.0 kg/min (x kg C6H6/kg) x = 0.75 kg C6H6/kg

  27. Balancing a process • Rules of thumb for NONREACTIVE process • 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. • Write balances first that involve the fewest unknown variables.

  28. General Procedure for Single Unit Process Material Balance Calculation • Choose as basis of calculation an amount or flow rate of one of the process streams. • Draw a flowchart and fill in all unknown variables values, including the basis of calculation. Then label unknown stream variables on the chart. • Express what the problem statement asks you to determine in terms of the labeled variables. • If you are given mixed mass and mole units for a stream (such as a total mass flow rate and component mole fractions or vice versa), convert all quantities to one basis. • Do the degree-of-freedom analysis. • If the number of unknowns equals the number of equations relating them (i.e., if the system has zero degree of freedom), write the equations in an efficient order (minimizing simultaneous equations) and circle the variables for which you will solve. • Solve the equations. • Calculate the quantities requested in the problem statement if they have not already been calculated. • If a stream quantity or flow rate ng was given in the problem statement and another value nc was either chosen as a basis or calculated for this stream, scale the balanced process by the ratio ng/nc to obtain the final result.

  29. Basis of calculation A basis of calculation is an amount (mass or moles) of flow rate (mass or molar) of one stream or stream component in a process. All unknown variables are determined to be consistent with the basis. light If a stream amount or flow rate is given in problem, choose this quantity as a basis Basis of calculation If no stream amount or flow rate are known, assume one stream with known composition. If mass fraction is known, choose total mass or mass flow rate as basis. If mole fraction is known, choose a total moles or molar flow rate as basis

  30. exercise An aqueous solution of NaOH contains 20% NaOH by mass. It is desired to to produce an 8 % NaOH solution by diluting a stream of the 20 % solution with a stream of pure water. Calculate the ratios (liters H2O/kg feed solution) and (kg product solution/ kg feed solution).

  31. solution An aqueous solution of NaOH contains 20% NaOH by mass. It is desired to to produce an 8 % NaOH solution by diluting a stream of the 20 % solution with a stream of pure water. Calculate the ratios (liters H2O/kg feed solution) and (kg product solution/ kg feed solution). (kg) 100 (kg) 0.08 kg NaOH/kg 0.920 kg H2O/kg 0.2 kg NaOH/kg 0.8 kg H2O/kg kg H2O liters H2O

  32. solution (kg) 100 (kg) 0.08 kg NaOH/kg 0.920 kg H2O/kg 0.2 kg NaOH/kg 0.8 kg H2O/kg kg H2O liters H2O Nonreactive steady-state process input = output NaOH Balance (0.2 kg NaOH/kg)(100 kg)=(0.08 kg NaOH/kg) = 250 kg NaOH Total Mass Balance 100 kg + = = 150 kg NaOH

  33. Diluted water volume = = 150 liters Ratios requested in problem statement = 1.50 liters H2O/kg feed solution =2.50 kg product solution/kg feed solution

  34. Degree of freedom A degree of freedom analysis (DFA) is simply an accounting of the number of unknowns in a problem and the number of independent equations that can be written. light • Procedure to perform a degree-of-freedom analysis: • draw and completely label a flowchart • count the unknown variables on the chart (n unknowns) • count the independent equations (n indep. eq.) • find number of degree-of-freedom (ndf) ndf= n unknowns- n indep. eq. Degree of freedom Independent equations Equations are independent if none of them can be derived from the others. For example, not one of the set of equations can be obtained by adding or subtracting multiples of the others.

  35. Number of Degree of freedom light – n df = 0, there are n independent equations and n unknowns. The problem can be solved. – n df > 0, there are more unknowns that independent equations. The problem is underspecified. n df more independent equations or specifications are needed to solve the problem. – n df < 0, there are more independent equations than unknowns. The problem is overspecified with redundant and possibly inconsistent relations. Possible outcomes of a DFA:

  36. 6 Sources of Equation for Balance • Material balances. • For a nonreactive process, number of independent equation can be written is not more than number of molecules species (n ms) of the process • If benzene and toluene is involve in stream, we can write balance on benzene, toluene, total mass, atomic carbon and etc., but only TWO INDEPENDENT balance equation exist • An energy balance. • If the amount of energy exchanged between the system and its surroundings is specified or if it is one of the unknown process variables, an energy balance provides a relationship between inlet and outlet material flows and temperatures. • To be discussed in later chapters • Process specifications • The problem statement may specify how several process are related. • i.e: Outlet flow rate is two times than flow rate stream 1 or etc.

  37. 6 Sources of Equation for Balance • 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 would provide an equation relating the variables. • Physical constraints • For example, if the mole fractions of the three components of a stream labeled xA, xB, and xC, then the relation among these variables is xA + xB + xC = 1. • Instead label as xc, the las fraction should be 1-xA-xB • Stoichiometric relations • If chemical reactions occur in a system, stoichiometric equation provide a relationship between the quantities of reactant and the product • To be discussed later

  38. exercise 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. The entering air contains 10.0 mole % water. Calculate the flow rate of the gas stream leaving the condenser and the mole fractions of oxygen, nitrogen, and water in this stream. Degree of freedom analysis: 5 unknowns - 3 material balances ( since there are 3 molecular species in this nonreactive process) - 1 density relationship (relating the mole flow rate to the given volumetric flow rate of the condensate - 1 the fractional condensation 0 degrees of freedom

  39. Balances on Multiple Unit Processes • In real chemical industries, more than one unit processes exist such as a separation unit after reactor and so on. • Need to know term called SYSTEM in order to solve material problem • SYSTEM: • Any portion of process that can be enclosed within a hypothetical box (or boundary) • It can be the entire process, an interconnected of process unit, a single unit, a point which two or more stream come together into one stream or etc. • The inputs and outputs to a system are the process streams that are intersect to the system boundary

  40. System of Multiple Unit Processes FEED 2 A E C PRODUCT 3 B D FEED 1 UNIT 2 UNIT 1 PRODUCT 1 PRODUCT 2 FEED 3

  41. Balances on Multiple Unit Processes • Solving material balances in multiple unit process is basically the same as single unit processes • In multiple unit, must isolate and write balance on several subsystems to obtain enough equation to determine all unknowns stream variables • Always perform degree-of-freedom analysis before solving a material balance of system.

  42. Discussion in class 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. 3 streams whose flow rates and/or compositions are not known are labeled 1, 2 and 3. Calculate the unknown flow rates and compositions of stream 1, 2 and 3. 40.0 kg/h 0.900 kg A/kg 0.100 kg B/kg 30.0 kg/h 0.600 kg A/kg 0.400 kg B/kg 100.0 kg/h 0.500 kg A/kg 0.500 kg B/kg 3 1 2 Solution: X1=0.233 kg A/kg X2=0.255 kgA/kg m1= 60.0 kg/h m2= 90.0 kg/h m3= 60.0 kg/h 30.0 kg/h 0.300 kg A/kg 0.700 kg B/kg

  43. Fresh Feed Reactor Separator Product Recycle Stream Recycle • Normally in chemical reaction, some of unreacted reactant also found in the product. • This unreacted reactant can be separated and recycle back to the reactor

  44. Purpose of Recycle • Recovery of catalyst – catalyst is very expensive • Dilution of process stream – typically for slurry solution • Control of process variables – especially for the reaction that release heat, heat can be reduce by lowering the feed concentration • Circulation of working fluid such as in refrigerator system

  45. Class Discussion Example 4.5-2

  46. Fresh Feed Process Unit Product Bypass Stream Bypass • Fraction of the feed to a process unit is diverted around the unit and combined with the output stream from the unit • Used to control the composition of a final exit stream from a unit by mixing the bypass stream & the unit exit stream in suitable proportions to obtain desired final composition.

  47. Stoichiometry • Stoichiometry – theory of proportions in which chemical species combine with one another. • Stoichiometric equation of chemical reaction – statement of the relative number of molecules or moles of reactants and products that participate in the reaction. 2 SO2 + O2 ---> 2 SO3 • Stoichiometric ratio • ratio of species stoichiometry coefficients in the balanced reaction equation • can be used as a conversion factor to calculate the amount of particular reactant (or product) that was consumed (produced). 2 mol SO3 generated 2 mol SO2 consumed 2 mol SO2 consumed 1 mol O2 consumed

  48. Test Yourself C4H8 + 6 O2 --------> 4 CO2 + 4 H2O • Is the stochiometric equation balance? • Yes • What is stochiometric coefficient for CO2 • 4 • What is stochiometric ratio of H2O to O2 including it unit • 4 mol H2O generated/ 6 mol O2 consumed • How many lb-moles of O2 reacted to form 400lb-moles CO2 • 600 lb-moles O2 reacted • 100 mol/min C4H8 fed into reactor and 50% is reacted. At what rate water is formed? • 200 mol/min water generated

  49. Limiting Reactant & Excess Reactant The reactant that would run out if a reaction proceeded to completion is called the limiting reactant, and the other reactants are termed excess reactants. A reactant is limiting if it is present in less than its stoichiometric proportion relative to every other reactant. If all reactants are present in stoichiometric proportion, then no reactant is limiting.

  50. Example C2H2 + 2H2 ------> C2H6 Inlet condition: 20 kmol/h C2H2 and 50 kmol/h H2 What is limiting reactant and fractional excess? (H2:C2H2) o = 2.5 : 1 (H2:C2H2) stoich = 2 : 1 H2 is excess reactant and C2H2 is limiting reactant Fractional excess of H2 = (50-40)/40 = 0.25

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