1. MATERIAL BALANCES
CHBI 201

2. CONSERVATION OF MASS
Mass is neither created nor destroyed
Distillation
Heat
Exchanger
Reactor
Seperator 1 2 6 7 8 13 14 12 11 10 4 5 3 9
{Input} + {Genn} - {Consumption} – {Output} = {Accumulation}
CHBI 201

3. SYSTEMS Systems Closed System Open System OPEN or CLOSED
Any arbitrary portion of or a whole process that you want to consider for analysis
Reactor, the cell, mitochondria, human body, section of a pipe
Closed System
Material neither enters nor leaves the system
Changes can take place inside the system
Open System
Material can enter through the boundaries
CHBI 201

4. Rate of addition = Rate of removal
STEADY-STATE
Steady-State
Nothing is changing with time
@ steady-state accumulation = 0
500 kg H2O
100 kg/min H2O
100 kg/min H2O
Rate of addition = Rate of removal
Unsteady-State (transient system)
{Input} ≠ {Output}
CHBI 201

5. PROCESSES Batch Process Continuous Process Semibatch Process
Feed is fed at the beginning of the process
Continuous Process
The input and outputs flow continuously throughout the duration of proces
Semibatch Process
Any process neither batch nor continuous
CHBI 201

6. Balances on Continuous Steady-state Processes
Input + Generation = Output + Consumption
If the balance is on a nonreactive species, the generation and consumption will be 0.
Thus, Input = Output
Example
Input of 1000 kg/h of benzene+toluene containing 50% B by mass is separated by distillation column into two fractions.
B: the mass flow rate of top stream=450 kg/h
T: the mass flow rate of bottom stream=475 kg/h
Distillation
1000 kg /h
Benzene + Toluene
%50 Benzene by mass
475 kg Toluene/h
M2 kg Benzene/h
m1 kg Toluene/h
450 kg Benzene/h
CHBI 201

7. Balances on Continuous Steady-state Processes
Solution of the example Input = Output
Benzene balance
1000 kg/h · 0.5 = 450 kg/h + m2
m2 = 50 kg/h Benzene
Toluene balance
1000 kg/h · 0.5 = 475 kg/h + m1
m1 = 25 kg/h Toluene . . . .
CHBI 201

8. BALANCES ON BATCH PROCESSES
Initial Input + Generation = Final Output + Consumption
Objective: generate as many independent equations as the number of unknowns in the problem
F = B + D
F.xF = D.xD + B.xB
F.yF = D.yD + B.xB
x: mole fraction of W
y: mole fraction of A F
(W+A) B D
CHBI 201

9. EXAMPLE (Batch Process)
Centrifuges are used to seperate particles in the range of 0.1 to 100 µm in diameter from a liquid using centrifugal force. Yeast cells are recovered from a broth ( a mix with cells) using tubular centrifuge. Determine the amount of the cell-free discharge per hour if 1000 L/hr is fed to the centrifuge, the feed contains 500 mg cells/L, and the product stream contains 50 wt% cells. Assume that the feed has a density of 1 g/cm3.
Feed (broth) 1000 L/hr
500 mg cells/L feed
( d= 1 g/cm3)
Centrifuge
Concantrated cells P(g/hr)
50 % by weight cells
Cell-free discahrge D(g/hr)
CHBI 201

10. EXAMPLE (Batch Process)
Centrifuges are used to seperate particles in the range of 0.1 to 100 µm in diameter from a liquid using centrifugal force. Yeast cells are recovered from a broth ( a mix with cells) using tubular centrifuge. Determine the amount of the cell-free discharge per hour if 1000 L/hr is fed to the centrifuge, the feed contains 500 mg cells/L, and the product stream contains 50 wt% cells. Assume that the feed has a density of 1 g/cm3.
Feed (broth) 1000 L/hr
500 mg cells/L feed
(d= 1 g/cm3)
Centrifuge
Concantrated cells P(g/hr)
50 % by weight cells
Cell-free discharge D(g/hr)
Cell balance
Fluid balance
Input: (106 – 500) g/h fluid
Output 1: 1000g/h = 500 g/h fluid
Output 2: D(g/h) = (106 – 500)g/h – 500 g/h = ( )g/h fluid
CHBI 201

11. FLOW CHARTS
Boxes and other symbols are used to represent process units.
Write the values and units of all known streams
Assign algebraic symbols to unknown stream variables
100 mol C3H8
Combustion
Chamber
Condenser
50 mol C3H8
750 mol O2
3760 mol N2
150 mol CO2
1000 mol O2
3760 mol N2
200 mol H2O
CHBI 201

12. EXAMPLE (Flow charts)
Humidification and Oxygenation Process in the Body: An exp. on the growth rate of certain organisms requires an environemnt of humid air enriched in oxygen. Three input streams are fed into an evaporator to produce an output stream with the desired composition. A: liquid water, fed at a rate of 20 cm3/min, B: Air, C: Pure oxygen (with a molar flow rate one-fifth of the molar flow rate of stream B) . .
n3 mol/min
0.015 mol H2O/mol
y mol O2/mol
(0.985 – y ) mol N2/mol
0.2 n1 mol O2/min C . B A
n1 mol air/min
0.21 mol O2/mol
0.79 mol N2/mol .
20 cm3 H2O /min
n2 mol H2O/min
CHBI 201

13. EXAMPLE (Flow chart) EXAMPLE
n2 = 20 cm3 H2O/min . 1 g H2O/cm3 . 1 mol/18.02 g
n2 = 1.11 mol H2O/min
H2O Balance
n2 mol H2O/min = n3 mol/min mol H2O/mol
n3 = 74.1 mol/min
Total Mole Balance
0.2 n1 + n1 + n2 = n3
n1 = 60.8 mol/min
N2 Balance
n1 mol/min mol N2/mol = n3 mol/min . (0.985-y) mol N2/mol
y = mol O2/mol
CHBI 201

14. FLOWCHART SCALING A A Scale factor: 100 n1 n3 n2 100 n1 100 n3 100 n2
CHBI 201

15. DEGREE OF FREEDOM ANALYSIS (df)
ndf = nunknowns – nindep.eqn’s
If ndf = 0
Problem can be solved (determined)
If ndf > 0
Unknowns > knowns (underspecified)
If ndf < 0
overspecified (no solution)
Material balances,
Energy balances,
Process specificaitons,
Physical props&laws,
Physical constraints
CHBI 201

16. CHBI 201

17. CHBI 201

18. EXAMPLE 1 Example ρH20 is given In the condenser, Condenser (n4) O2
95% of H2O in the inlet air is condensed.
Humid air
(n0) O2
(n1) N2
(n2) H2O
(n3) H2O
225 L/h
Condenser
Dry air
(n4) O2
(n5) N2
(n6) H2O
7 unknowns (n0 -> n6) 7 equations needed
3 independent material balance
n3 = ρ.V
n0/n1 = 21/79
n3 = 0.95 n2
One more equation is needed
Volume is not conserved!
Use consistent units (mole, kg)
Do not make mole balances in reactive processes.
CHBI 201

19. EXAMPLE 2
EXAMPLE
A continuous mixer mixes NaOH with H2O to produce an aqueous solution of NaOH. Determine the composition and flow rate of the product, if the flow rate of NaOH is 1000 kg/hr and the ratio of the flow rate of H2O to the product solution is 0.9.
Nsp = number of species
Ns = number of streams
Nu = total number of variables
System boundary
H2O
NaOH M
Product
CHBI 201

20. EXAMPLE 2 - continue Nu = 3(2+1) = 9
Streams
Species
FEED
WATER
PRODUCT
NaOH
FNaOH
WNaOH
PNaOH
H2O
FH2O
WH2O
PH2O
Total F W P
Nu = 3(2+1) = 9
Last row in the table
Specifications: ratio of two streams
the % conversion in a reaction
the value of each concentration, flow rate, T, P, ρ, V, etc.
a variable is not present in a stream, hence ,it is 0
CHBI 201

21. EXAMPLE 3
EXAMPLE
A cylinder containing CH4, C2H6, and N2 has to be prepared containing a CH4 to C2H6 mole ratio of 1.5 to 1. Avaliable to prepare the mixture are
1) a cylinder containing a mixture of 80% N2 and 20% CH4
2) a cylinder containing a mixture of 90% N2 and 10% C2H6
3) a cylinder containing a mixture of pure N2
What is the number of degrees of freedom?
CHBI 201

22. EXAMPLE 3 - continue Unknowns: 3 xi and 4 Fi F4 F1 CH4 xCH4 CH4 0.2
N2 xN2
C2H6 xC2H6 F3
N2 1
Unknowns: 3 xi and 4 Fi
CHBI 201

23. EXAMPLE 3 - continue Equations: Material balance (CH4, C2H6, N2)
One specified ratio xCH4/xC2H6 = 1.5
One summation of mole fractions
5 independent equations
Ndf = 7 – 5 = 2
If you pick a basis as F4=1, one other value has to be specified in order to solve the problem.
CHBI 201

24. Balances on Multiple-unit Processes
30 kg/hr
0.6 kg A/kg
0.4 kg B/kg
40 kg/hr
0.9 kg A/kg
0.1 kg B/kg 1 Q1 x1 Q2 x2 3 Q3 x3
100 kg/hr
0.5 kg A/kg
0.5 kg B/kg 2
30 kg/hr
0.3 kg A/kg
0.7 kg B/kg 4
CHBI 201

25. Balances on Multiple-unit Processes
Q : mass flow rate
xA : mass fraction of A
1-xA : mass fraction of B
Number of unknowns = 6
Number of equations = = 6
Therefore, solution exists
100 = 40 + Q Q1 = 60 kg/hr
100.(0.5) = 40.(0.9) + 60.(x1) x1 = 0.233
30 + Q1 = Q2 Q2 = 90 kg/hr
x2 = 0.256
30 + Q3 = Q2 Q3 = 60 kg/hr x3 = 0.083
You should treat any junction as a process unit! 1 2 3
CHBI 201

26. RECYCLE & BYPASS STREAM
It is rare that a chemical reaction A  B proceeds to completion in a reactor. Its efficiency is never 100. Some A in the product !
To find a way to send the “A” back to feed you need a seperation and recycle equipment, this would decrease the cost of purchasing more A.
If a fraction of the feed to a process unit is diverted around the unit and combined with the output stream, this process is called bypass.
rxn
Sep.
Feed
Product
Recycle
Process
Unit
Feed
Bypass stream
CHBI 201

27. Fresh air is combined with a recycle stream of dehumidified air.
EXAMPLE (pg 110)
Feed: Fresh air with 4 mole% H2O(v) is “cooled” and “dehumidified” to a water content of 1.7 mole% H2O.
Fresh air is combined with a recycle stream of dehumidified air.
The blended stream entering unit contains 2.3 mole% H2O. In the air conditioner some of the water is removed as liquid.
Take 100 mole of dehumidified air delivered to the room, calculate moles of feed, water condensed, dehumidified air recycled.
CHBI 201

28. EXAMPLE - continue n5 (mol) 0.983 DA, 0.017 W n1 (mol) 0.04 W 0.96 DA
AIR
CONDITIONER
n4 (mol)
0.017 W
0.983 DA
100 mol
0.983 DA
0.017 W(v)
n3 mole W(ℓ)
n2 (mol)
0.977 DA
0.023 W(v)
CHBI 201

29. EXAMPLE - continue Overall system: 2 variables (n1, n3)
2 balance equations (two species)
Degree of freedom = 0
 (n1, n3) are determined!!!
Mixing point: variables (n2, n5)
Cooler: variables (n2, n4)
Splitting point: 2 variables (n4, n5) Donot use SP in the solution
1 balance equation
Degree of freedom = 1
CHBI 201

30. EXAMPLE - continue Overall DA balance:
0.96 n1 = (100)  n1 = mol fresh feed
Overall mole balance:
n1 = n  n3 = 2.4 mol H2O condensed
Mole balance on Mixing point:
n1 + n5 = n2
Water blance on Mixing point:
0.04n n5 = 0.023n2
n2 = mol
n5 = 290 mol recycled
CHBI 201

31. CHBI 201

32. CHBI 201

33. CHBI 201

34. CHBI 201

35. CHEMICAL REACTION STOICHIOMETRY
If there is a chemical reaction in a process
 More complications
The stoichiometric ratios of the chemical reactions
 Constraints
The stoichiometric equation 2SO2 + O2  2SO3
2 molecules of SO2 reacts with 1 molecule of O2 and yields 2 molecules of SO3
2, 1 and 2 are stoichiometric coefficients of a reaction
CHBI 201

36. LIMITING & EXCESS REACTANTS
If the reactants are not in stoichiometric proportion
 one of them will be excess, the other will be limiting
CHBI 201

37. EXAMPLE (pg 120) C3H6 + NH3 + 3/2 O2  C3H3N + 3 H2O
Feed: 10 mol % of C3H6, 12 mole % NH3 and 78 mole % air
A fractional converison of limiting reactant = 30%
Taking 100 mol of feed as a basis, determine which reactant
is limiting, and molar amounts of all product gas constituents
for a 30% conversion of the limiting reactant.
REACTOR
100 mol
0.1 mol C3H6/mol
0.12 mol NH3/mol
0.78 mol air/mol
0.21 mol O2/mol air
0.79 mol N2/mol air
nC3H6
nNH3
nO2
nN2
nC3H3N
nH2O
CHBI 201

38. EXAMPLE – continue
Feed: nC3H6= 10 mole nNH3=12 mole nO2= 78.(0.21) =16.4 mole
nNH3/nC3H6= 12/10 =  NH3 is excess (stoich. 1)
nO2/nC3H6= 16.4/10 =  O2 is excess (stoich. 1.5)
(nNH3)stoich.= 10 mole (nO2)stoich.= 15 mole
(% excess)NH3 = (12-10) /10 x 100 = 20% excess NH3
(% excess)O2 = ( ) /15 x 100 = 9.3% excess O2
(nC3H6)out=0.7 x (nC3H6)0= 7 mole C3H6 (since the fractional conversion of C3H6 is 30%)
Extent of reaction = ζ = 3 mole (since ni = ni0 + ni ξ => 7= ξ)
nNH3 = 12- ζ =9 mole nO2=16.4 – 1.5.(ζ)= 11.9
nC3H3N= ζ = 3 mole nH2O=3.(ζ) = 9 mole
nN2= (nN2)0=61.6 mole
Moles reacted
Moles fed
CHBI 201

39. CHEMICAL EQUILIBRIUM
If you are given a set of reactive species and reaction conditions;
What will be the final (equilibrium) composition of the reaction mixture?
How long will the system take to reach a specified state short of equilibrium?
Chemical equilibrium thermodynamics & Chemical Kinetics
A reaction can be
Reversible
Irreversible
CHBI 201

40. EXAMPLE CO (g) + H2O (g) CO2 (g) + H2 (g) Given @ T=1105 K, K=1
nCO= 1 mol, nH2O= 2mol, initially no CO2 and H2
Calculate the equilibrium composition and the fractional converison of the limiting reactant.
Equilibrium constant;
K(T) =
CHBI 201

41. EXAMPLE – continue nCO = 1-ζe , nH2O = 2-ζe , nCO2 = ζe , nH2 = ζe
yCO = (1-ζe)/3 yH2O = (2-ζe)/3
yCO2 = ζe / yH2 = ζe /3
K(T) = (ζe)2 / (1-ζe) (2-ζe) = 1
ζe = mole
yCO = yH2O = 0.444
yCO2 = yH2 = 0.222
Limiting reactant is CO nCO = = 0.333
Fractional conversion = ( ) / 1 mol feed = 0.667
CHBI 201

42. CHBI 201

43. CHBI 201

44. CHBI 201

45. Step 1. Draw a block diagram of the process.
Procedure
Step 1. Draw a block diagram of the process.
Show each significant step as a block, linked by lines and arrows to show the stream connections and flow direction.
Step 2. List all the available data.
Show on the block diagram the known flows (or quantities) and stream compositions.
Step 3. List all the information required from the balance.
Step 4. Decide the system boundaries.
Step 5. Write out all the chemical reactions involved for the main products and byproducts.
Step 6. Note any other constraints, such as: specified stream compositions, azeotropes, phase equilibria, tie substances.
The use of phase equilibrium relationships and other constraints in determining stream compositions and flows is discussed in more detail in.
Step 7. Note any stream compositions and flows that can be approximated.
Step 8. Check the number of conservation (and other) equations that can be written, and compare with the number of unknowns. Decide which variables are to be design variables;
This step would be used only for complex problems.
Step 9. Decide the basis of the calculation;
The order in which the steps are taken may be varied to suit the problem.
CHBI 201