1. PTT108/108 MATERIAL AND ENERGY BALANCE
SEM 11 (2012/2013)
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

5. General balances equation
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with the two instants of time being the moment after the input takes place and the moment before the product is withdrawn.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. (Reactor, mixer, separation units, etc)
flowcharts
When you are given process information and asked to determine something about the process, ORGANIZE the by drawing a flowchart
Represent
INPUTS
Represent
PROCESS UNIT
(Reactor, mixer, separation units, etc)
Represent
OUTPUTS

10. Flowcharts LABELLING Example
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

11. 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.
0.6 kmol N2/kmol
0.4 kmol O2/kmol
100 kmol/min
60 kmol N2/min
40 kmol O2/min
0.3 lbm CH4/lbm
0.4 lbm C2H4/lbm
0.3 lbm C2H6/lbm
10 lbm
3.0 lbm CH4
4.0 lbm C2H4
3.0 lbm C2H6

12. 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.
mol/h
0.21 mol O2/mol
0.79 mol N2/mol
T = 320˚C, P = 1.4 atm
400 mol/h
y mol O2/mol
(1-y) mol N2/mol
T = 320˚C, P = 1.4 atm

13. 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

14. 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
y2 mol N2/mol

15. Consistent on notation

16. 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

17. 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 = kg T/h
(kg B/h)
475 kg T/h
= 25 kg T/h

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

19. 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.

20. solution 0.2 (mol O2/min) (mol/min) 0.015 mol H2O/mol y (mol O2/mol)
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.
(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)

21. solution H2O Balance 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)
20.0 cm3 H2O
1.00 g H2O
1 mol
min
cm3
18.02 g =
= mol/min
Nonreactive steady-state process input = output
H2O Balance
(mol/min) =
(mol)
0.015 mol H2O
(min)
mol
= 74.1 mol/min

22. solution Total Mole Balance 0.2 + + = N2 Balance 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 Mole Balance =
= 60.8 mol/min
N2 Balance
(mol)
0.79 mol N2 =
(0.985-y) (mol N2)
(min)
mol
y = mol O2/mol

23. Scaling down What? Scaling up Flowchart scaling
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 down
What?
Scaling up
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.

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

25. 3.0 kg/min of benzene and 1.0 kg/min of toluene are mixed
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 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
m (kg/min)
x (kg C6H6/kg)
(1-x) (kg C7H8/kg)
3 kg C6H6/min
1 kg C7H8/min

26. 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.

27. 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.

28. 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

29. 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).

30. 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)
0.2 kg NaOH/kg
0.8 kg H2O/kg
0.08 kg NaOH/kg
0.920 kg H2O/kg
kg H2O
liters H2O

31. solution NaOH Balance Total Mass Balance (kg) 100 (kg) 0.2 kg NaOH/kg
0.8 kg H2O/kg
0.08 kg NaOH/kg
0.920 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

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

33. Degree of freedom Procedure to perform a degree-of-freedom analysis:
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.

34. 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:

35. 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.

36. 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

37. 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
material balances ( since there are 3 molecular species in this nonreactive process)
density relationship (relating the mole flow rate to the given volumetric flow rate of
the condensate
the fractional condensation
0 degrees of freedom

38. 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

39. 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

40. 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.

41. 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

42. 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
Product
Recycle Stream
Fresh
Feed
Reactor
Separator

43. 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

44. EXERCISE

45. Forty-five hundred kilograms per hour of a solution that is one-third K2CrO4 by mass is joined by a recycled stream containing 36.4% K2CrO4 , and the combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains 49.4% K2CrO4 , this stream is fed into a crystallizer in which is cooled (causing crystals K2CrO4 to come out solution) and then filtered. The filter cake consist of K2CrO4 crystals and a solution that contains 36.4% K2CrO4 by mass; the crystal account for 95% of the total mass of the filter cake. The solution that passes through the filter, also 36.4% K2CrO4, is the recycle stream
1.Calculate the rate evaporation, the rate of production of crystalline K2CrO4, the feed rates that evaporator and the crystallizer must be designed to handle, and the recycle ratio (mass of recycle)/(mass of fresh feed)

46. 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.
Product
Bypass Stream
Fresh
Feed
Process Unit

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. C4H8 + 6 O2 --------> 4 CO2 + 4 H2O
Test Yourself
C4H8 + 6 O > 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

51. Fractional Conversion
Fractional Conversion (f)

52. Extent of Reaction Extent of Reaction, ξ ξ = extent of reaction
ni = moles of species i present in the system after the reaction occurred
nio = moles of species i in the system when the reaction starts
vi = stoichiometry coefficient for species i in the particular chemical reaction equation

53. EXERCISE

54. Acrylonitrile is produced in the reaction of propylene, ammonia, and oxygen
C3H6 + NH3 +3/2 O C3H3N + 3H2O
The feed contains 10.0 mole % propylene, 12.0% ammonia, and 78.0% air. A fractional conversion of 30.0 % of the limiting reactant is achieved. Taking 100 mol of feed as basis, determine which reactant is limiting, the percentage by which each of the other reactants is in excess, and the molar amounts of all product gas constituents for a 30% conversion of the limiting reactant (Assume basis 100 mol)

55. Chemical Equilibrium
For a given set reactive species and reaction condition, two fundamental question might be ask:
What will be the final (equilibrium) composition of the reaction mixture? – chemical engineering thermodynamics
How long will the system take to reach a specified state short of equilibrium? – chemical kinetics
Irreversible reaction
reaction proceeds only in a single direction (from reactants to products)
the concentration of the limiting reactant eventually approaches zero.
Reversible reaction
reactants form products for forward reaction and products undergo the reverse reactions to reform the reactants.
Equilibrium point is a rate of forward reaction and reverse reaction are equal
However the discussion to get the chemical equilibrium point is not covered in this text- learn in chemical engineering thermodynamic

56. Example 4.6-2

57. Multiples Reaction, Yield & Selectivity
Some of the chemical reaction has a side reaction which is formed undesired product- multiple reaction occurred.
Effects of this side reaction might be:
Economic loss
Less of desired product is obtained for a given quantity of raw materials
Greater quantity of raw materials must be fed to the reactor to obtain a specified product yield.
selectivity =
moles of desired product
moles of undesired product

58. Yield 3 definition of yield with different working definition Yield =
Moles of desired product formed
Moles that would have been formed if there were no side reaction and the limiting reactant had reacted completely
Yield =
Moles of desired product formed
Moles of reactant fed
Yield =
Moles of desired product formed
Moles of reactant consumed

59. Extent of Reaction for Multiple Reaction
Concept of extent of reaction can also be applied for multiple reaction
Only now each independent reaction has its own extent.

60. Example 4.6-3

61. Balance of Reactive Processes
Balance on reactive process can be solved based on three method:
Atomic Species Balance
Extent of Reaction
Molecular Species Balance

62. Atomic Species Balance
No. of unknowns variables
- No. of independent atomic species balance
- No. of molecular balance on indep.
nonreactive species
- No. of other equation relating the variable
=============================
No. of degree of freedom

63. Extent of Reaction No. of unknowns variables
+ No. of independent chemical reaction
- No. of independent reactive species
- No. of independent nonreactive species
- No. of other equation relating the variable
===================================
No. of degree of freedom

64. Molecular Species Balance
No. of unknowns variables
+ No. of independent chemical reaction
- No. of independent molecular species balance
- No. of other equation relating the variable
======================================
No. of degree of freedom

65. Example 4.7-1