4 Material Balances

Fundamentals of Material Balances

Ch E 201Material and Energy Balances

Objectives• Classify processes as batch, semibatch, continuous,

transient, and steady-state.• Define: recycle, purge, degrees of freedom, fractional

conversion of a limiting reactant, percentage excess of a reactant, yield, selectivity.

• Draw and label process flowcharts.• Select a calculation basis.• Perform a degree of freedom analysis.• Define/solve equations to calculate process variables.• Perform combustion calculations.

Process Classification• Batch process

No mass crosses system boundaries between the time feed is charged and the time product is removed.

Typically used for making small quantities, particularly those products of sparatic manufacture.

• Continuous process Feeds and effluents continously flow across the system

boundary through the duration of the process. Suited for the production of large quantities.

• Semibatch process Any process that is neither batch nor continuous.

Process Operation• Steady state

There is no change in the value of all process variables (temperature, pressure, flowrates, heat-transfer rates) except for minor flucctuations about the mean value.

Continuous processes may be steady-state.• Transient (Unsteady-State)

The values of process variables change with time. Batch and semibatch process are transient by nature. Continuous processes may be transient.

Process Operation• Pseudo Steady State

A transient process for which the rate of change of particular process variables is small may be considered to operate at close to a steady state over short time periods.

Classification is assumed in the development mathematical relationships that permit the design or describe the performance of unit operations that involve fundamental phenomena such as heat or mass transfer, heterogeneous catalysis, phase equilibrium, etc.

Classify this!• Filling an empty swimming pool with water.• Leave the milk on the counter.• Cook pasta in boiling water in a dutch oven.• Filling a full swimming pool with water.• Drinking beer at a party all night on a Friday.• Baking a pizza.• Operation of a photovoltaic cell.• Other examples?

The General Balance Equation• Consider the following continuous process unit for

which methane is a component of both the input and output, but the measured methane inlet and outlet mass flowrates are not the same.

• Maybe methane is… consumed as a reactant, or generated as a product within

the process unit; or accumulating within or leaking from the unit; or the measurements are wrong (though we will assume they

are correct).

The General Balance Equation• A balance of a conserved quantity (mass, energy,

momentum) in a system may be written generally as:

input + generation – output – consumption = accumulation

input: enters through system boundaries generation: produced within the system ouput: leaves through system boundaries consumption: consumed within the system accumulation: buildup within the system

The General Balance Equation• Each year, 50,000 people move into a city; 75,000

move out; 22,000 are born; 19,000 die. Perform a balance on the population of the city (system).input + generation – output – consumption = accumulation

input: 50,000 people/year generation: 22,000 people/year ouput: 75,000 people/year consumption: 19,000 people/year accumulation: unknown

The General Balance Equation• Each year, 50,000 people move into a city; 75,000

move out; 22,000 are born; 19,000 die. Perform a balance on the population of the city (system).input + generation – output – consumption = accumulation

50,000 P/yr + 22,000 P/yr - 75,000 P/yr - 19,000 P/yr = A

A = -22,000 P/yr

∴ the city’s (system) population is decreasing by 22,000 people each year.

Balance Types• Differential balances

Indicate state of various rates occuring in a system at an instant in time. Typically applied to a continuous process.

• Integral balances indicate total amounts of a balanced quantity between

two instants of time. Typical applied to a batch process.

Rules of MB simplification• If the balanced quantity is total mass,

set generation =0 and consumption = 0

• If the balance substance is a nonreactive species, set generation =0 and consumption = 0

• If a system is at steady state, set accumulation = 0

Continuous steady-state systeminput + generation – output – consumption = accumulation

input + generation = output + consumption

• If the balance is for a nonreactive species or on total mass, the generation and consumption terms equal zero, and the balance reduces as

input = output

0

Continuous steady-state system• Benzene/Toluene distillation

continuous process steady-state operation no reactions occurring

• General species balance

input + generation – output – consumption = accumulation

input = output

0 0 0

Continuous steady-state system

input = output• Benzene balance

500 kg B/h = 450 kg B/h + m2

m2 = 50 kg B/h

• Toluene balance500 kg T/h = m1 + 475 kg T/h

m1 = 25 kg T/h

• Total mass balance1000 = 450 + m1 + m2 + 475 (all with units of kg/h)

1000 kg/h = 1000 kg/h ✓

Integral Balances on Batch Processes

consider the reaction N2 + H2 → NH3 in a batch reactor at t=0, there is n0 moles of NH3 in the reactor at t=tf, there is nf moles of NH3 in the reactor between 0 and tf, no NH3 crosses system boundary NH3 accumulation in system from 0 to tf is n∴ f – n0.

• therefore, for a batch process,accumulation = final output – initial input

= generation – consumption ⇒ initial input + generation = final output + consumption

Identical to continuous steady-state balance except in/out terms denote discrete amounts instead of flow rates

Batch Mixing Process Balance• Two methanol-water

mixtures are contained in flasks of amounts and concentrations shown.

• If the flasks are mixed, what is the mass and concentration of the resulting product? no reactions, generation = consumption = 0∴ input = output

Batch Mixing Process Balance• Total Mass Balance

200 g + 150 g = m = 350 g• Methanol balance

• Water balance200(0.6) + 150(0.3) = 350(1-0.529)165 g H2O = 165 g H2O ✓

gOHCH g

529.0x

gOHCH gx

gmg

OHCH g 700.0g150

gOHCH g 400.0

g200

3

333

Integral Balances on Semibatch and Continous Processes

• Air is bubbled through a drum of liquid hexane. • Gas stream leaving contains air and hexane.• How long does it take to vaporize 10.0 m3 of liquid?


• differential air balance

minkmol

111.0n

minkmol

nkmol

ira molk 900.0min

ira molk100.0

outputinput


• integral hexane balance

min6880t

t111.0100.0nt100.0output

HC kmol 45.76nonaccumulati

kg2.86kmol1

mL10

Lkg659.0

m0.10nonaccumulati

outputonaccumulati

nconsumptiooutputgenerationinputonaccumulati

1

1minkmol

1

146

3

33

Process Flowcharts• A process flowchart is a method for organizing

information about a process in a format that permits convenient and easy to understand.

• A process flowchart uses boxes and lines with arrows to represent inputs and outputs of a process.

Labeling Process Flowcharts

1. Write the values of all known stream variables on the locations of the streams on the chart.

1. Assign algebraic symbols to unknown stream variables and write these variable names and their associated units on the chart.

400 mol/h0.21 mol O2/mol0.79 mol N2/mol320°C, 1.4 atm

n (mol/h)0.21 mol O2/mol0.79 mol N2/mol320°C, 1.4 atm

. 400 mol/hy (mol O2/mol)(1-y) (mol N2/mol)320°C, 1.4 atm

Air Humidification and Oxygenation• 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 20.0 cm3/minB. air (21 mol% O2, 79 mol% N2)

C. pure O2 with a molar flow rate 1/5 of that of Stream B

Output gas is found to contain 1.5 mol% water.

Air Humidification and Oxygenation• 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 20.0 cm3/minB. air (21 mol% O2, 79 mol% N2)

C. pure O2 with a molar flow rate 1/5 of that of Stream B

Output gas is found to contain 1.5 mol% water.

Air Humidification and Oxygenation

Calculate n2 from volumetric flowrate and density:

minOH mol

11.1OgH02.18

mol1cm

OgH00.1min

OHcm 0.20n 2

23

223

2


water balance:

minmol

1.74nmol

OH mol 015.0minmol

nmin

OmolH11.1

molOH mol 015.0

minmol

nmin

OmolHn

32

32

23

22


total mole balance:

minmol

1

minmol

minmol

1

3211

8.60n

1.7411.1n200.1

nnnn200.0


nitrogen balance:

molO mol

molN mol

minmol

molN mol

minmol

molN mol

3molN mol

minmol

1

2

22

22

337.0y

y985.01.7479.08.60

y985.0n79.0n

Flowchart Scaling• A kilogram of benzene is mixed with a kilogram of

toluene. The output of this process is 2 kilograms of a mixture that is 50% mass of each component.

• The flowchart is balanced because material balances on both species are satisfied. 1 kg × 1.0 kg CnHm/kg = 2 kg × 0.5 kg CnHm/kg

Flowchart Scaling• Scaling the

flowchart involves changing all values of stream flows by a proportional amount.

• Note that mass (or mole) fractions are not scaled, but remain unchanged.

Flowchart Scaling• A 60/40 mixture (molar)

of A and B is separatedbatchwise into 2 fractions.

• Scale the flowchart to acontinuous 1250 lbmol/hr feed rate.

• feed:

molhrlbmol

5.12mol 100

hlbmol1250=factor scale

hlbmol

1250mol

hrlbmol5.12mol 100




• top:h

lbmol625

molhrlbmol

5.12mol 50.0

molhrlbmol

5.12mol 100





• bottom:h

lbmolA156

molAhrlbmolA

5.12molA 12.5

molhrlbmol

5.12mol 100





• bottom:h

lbmolB469

molBhrlbmolB

5.12molB 37.5

molhrlbmol

5.12mol 100





Basis of calculation• Since a flowchart can always be scaled, material

balance calculations can be performed on the basis of any convenient set of stream amounts or flow rates and the results can subsequently be scaled to any desired extent.

• A basis of calculation is an amount or flow (mass or molar) of one stream or component in a process.

• The first step in balancing a process is to chose a basis of calculation; all unknown quantities are then determined to be consistent with this basis.

Basis of calculation• If a stream amount or flow is given in a problem

statement, it is usually the most convenient basis to use.

• If no stream amounts or flows are known, assume a value of 1, preferrably for a stream of known composition. If mass fractions are known, set a total mass or flow of that

stream (i.e., 100 kg or 100 kg/h) as the basis. If mole fractions are known, chose a total number of moles

or molar flow rate.

Balancing a Process• Suppose 3.0 kg/min of benzene and 1.0 kg/min of

toluene are mixed.• There are 2 unknown

quantities in this process, mdot and x, thus 2 equations are needed to solve for these unknowns.

• For non-reacting processes, the material balance takes the form: INPUT = OUTPUT.

• 3 balances can be written: one for total mass, and one for each component (benzene and toluene).

Balancing a Process

• Balances:

total mass: 3.0 kg/min + 1.0 kg/min = mdot

mdot = 4.0 kg/min

benzene: 3.0 kg C6H6/min = mdot (kg/min) + x (kg C6H6/kg) 3.0 kg C6H6/min = 4.0 kg/min + x (kg C6H6/kg) x = 0.75 kg C6H6/kg

Balancing nonreactive processes• 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. In the benzene/toluene example, only two of the three

balance equations are independent, thus only two unknowns can be found from these balances.

• Write balances first that involve the fewest unknown variables.

Balances on a mixing unit• An aqueous solution of sodium hydroxide contains

20.0% mass NaOH. It is desired to produce an 8.0% mass NaOH solution by diluting with pure water.

• Calculate the ratios (liters H2O /kg feed solution) and (kg product solution/kg feed solution).

Balances on a mixing unit

1. Chose basis of calculation and draw/label flowchart.


2. Express what the problem asks you to determine in terms of the labeled variables on the flowchart.

V1/100 (liters H2O/kg feed solution) m2/100 (kg product solution/kg feed solution)


3. Count unknown variables and equations. If these quantities are not equal, problem cannot be solved.

3 unknowns: m1, m2, V1 (need 3 equations) equations:

• 2 species → 2 independent material balances• density relates V1 to m1.


4. Outline solution procedure:balances have the form INPUT = OUTPUT

1. NaOH balance contains 1 unknown: m2

2. total mass balance contains 2 unknowns: m1 and m2

3. water balance contains 2 unknowns: m1 and m2

4. density relation contains 2 unknowns: V1 and m1

only need 1 of Equations 2 and 3 above


5. NaOH balance (INPUT = OUTPUT):

(0.20 kg NaOH/kg)(100 kg) = (0.80 kg NaOH/kg)(m2)

m2 = 250 kg

= 250 kg NaOH


6. Total mass balance (INPUT = OUTPUT):

100 kg + m1 = m2 =250 kg

m1 = 150 kg

= 250 kg

= 150 kg


7. Diluent water volume:

V1 = m1 / ρH2O = 150 kg / (1 kg/L)

V1 = 150 L

= 250 kg

= 150 kg= 150 L


7. Ratios:

V1 /100 kg = 150 L / (100 kg) = 1.50 L H2O / kg feed solution

m2 /100 kg = 250 kg / 100 kg = 2.50 kg product solution/ kg feed solution

= 250 kg

= 150 kg= 150 L

Degree of Freedom Analysis• Process used to determine if a material balance

problems has sufficient specifications to be solved.a) draw and completely label the flowchartb) count the unknown variables on the chartc) count the independent equations relating these variablesd) calculate degrees of freedom by subtracting (b) from (c)

ndf = nunknowns – nindep_eqns

Degree of Freedom Analysis

ndf = nunknowns – nindep_eqns

• If ndf = 0, problem can be solved (in principle).• if ndf > 0, problem is underspecified and at least ndf

additional variables must be specified before the remaining variable values can be determined.

• if ndf < 0, the problem is overspecified with redundant and possibly inconsistent relations.

Degree of Freedom Analysis• Sources of equations relating unknown process

stream variables include: Material balances. For a nonreactive process, no more

than nms (number of molecular species) independent material balances may be written.

Energy balance. An energy balance provides a relationship between inlet and outlet material flows and temperatures.

Process specifications. Physical properties and laws. Physical constraints. Stoichiometric relations. (for reacting systems)

Degree of Freedom Analysis• 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 (liquid leaving the

condenser) is measured and found to be 225 L/h. • Calculate the flow rate of the gas stream leaving the

condenser and the mole fractions of O2, N2, and H2O.

Degree of Freedom Analysis• 6 unknowns

3 material balances (1 each for O2, N2, H2O) condensate volumetric to molar flow relation (MW and ρ) process specification: 95% of the water is condensed

• ndf = 6 – (3 + 1 + 1) = 1

Underspecifiedcannot solve

Degree of Freedom Analysis• 5 unknowns

3 material balances (1 each for O2, N2, H2O) condensate volumetric to molar flow relation (MW and ρ) process specification: 95% of the water is condensed

• ndf = 5 – (3 + 1 + 1) = 0

Solvable

Degree of Freedom Analysis• Density relationship• 95% condensation specification• O2 Balance• N2 Balance• H2O Balance• outlet gas composition• total outlet gas flow rate

543total

total5OHtotal4Ntotal3O

521

41

31

12

kg100.18OH olm 1

LOH kg

hOH L

2

nnnn

nny ;nny ;nnynn100.0n

n79.0900.0nn21.0900.0n

n100.095.0n

00.1225n

222

32)l(2)l(2

General Procedure – Single Unit Op

1. Choose as a basis of calculation an amount or flow rate of one of the process streams.

If an amount or flow of a stream is given, it is usually convenient to use it as the basis of calculation. Subsequently calculated quantities will be correctly scaled.

If several stream amounts or flows are given, always use them collectively as the basis.

If no stream amount or flow rate is specified, take as a basis an arbitrariy amount or flow rate of a stream with a known composition.


2. Draw flowchart and fill in all variable values, including the basis. Label unknown stream variables.

Flowchart is completely labeled if you can express the mass / mass flow rate (moles / molar flow rate) of each component of each stream in terms of labeled quantities.

Labeled variables for each stream should include 1 of:a. total mass (or flow), and mass fractions of all stream componentsb. total moles (or flow), and mole fractions of all stream componentsc. mass, moles (or flow) of each component in each stream• use (c) if no steam information is known

incorporate given relationships into flowchart label volumetric quantities only if necessary


3. Express what the problem statement ask you to do in terms of the labeled variables.

4. If given mixed mass and mole units, convert.5. Do a degree-of-freedom analysis.6. If ndf = 0, write equations relating unknowns.

7. Solve the equations in (6).8. Calculate requested quantities.9. Scale results if necessary.

Distillation Column example• Ex. 4.3-5

1. basis is given as a volumetric quantity

2a. Flowchart drawn from description

2b. Convert mole to mass fractions

2c. no stream information knownwrite in terms of species flows


2d. confirm every component mass flow in every process stream can be expressed in terms of labeled quantities and variables.

2e. process specification

Distillation Column example• Ex. 4.3-5 3. write expressions for quantities requested in problem statement

BT33BB x1x ;mmx

3T3B3 mmm

312 mmm

Distillation Column example• Ex. 4.3-5 4. Convert mixed units in overhead product stream

kgT kg

2T

kgB kg

2B

T kmolT kg

B kmolB kg

058.0942.01y

942.0mixture kg 7881B kg 7420y

mixture kg 7881T kg 461B kg 7420T kg 46113.92T kmol 0.5

B kg 742011.78B kmol 0.95

Distillation Column example• Ex. 4.3-5 5. Perform degree of freedom analysis

=0.942=0.058

4 unknowns-2 material balances-1 density relationship-1 process specification0 degrees of freedom


6. Write system equations7. Solve

hkg

Lkg

hL

1 1744872.02000m i. volumetric flow conversion

hB kg

13B 8.62m45.008.0m ii. benzene split fraction

h

B kg2

3B2B21

766m

mymm45.0

iii. benzene balance

h

T kg3T

3T2B21

915m

my1mm55.0

iv. toluene balance

hkg

hkg

3T3B21

17441744

mmmm

iv. total mass balance (check)

Distillation Column example• Ex. 4.3-5 8. Calculate additional quantities

=0.942=0.058

=1744 kg/h

=915 kg T/h=62.8 kg B/h

=766 kg/h

hkg

hT kg

hB kg

3 9789158.62m

kgT kg

3T

kgB kg

hkg

hB kg

33B3B

936.0064.01y

064.09788.62mmy

Balances on Multiple Unit Ops• A system is any portion of a process that can be

enclosed within a hypothetical box (boundary). It may be the entire process, a single unit, or a point where streams converge or combine.

Balances on Multiple Unit Ops• Boundary A encloses the entire process.

inputs: Streams 1, 2, and 3 products: 1, 2, and 3 Balances on A would be considered overall balances internal streams would not be included in balances

Balances on Multiple Unit Ops• B: an internal mixing point (2 inputs, 1 product)• C: Unit 1 (1 input, 2 products)• D: an internal splitting point (1 input, 2 products)• E: Unit 2 (2 inputs, 1 product)

Balances on Multiple Unit Ops• The procedure for solving material balances on multi-

unit processes is the same as for a single unit; though, it may be necessary to perform balances on several process subsystems to get enough equations to determine all unknown stream variables.

Two-Unit Process Example• Variables for Streams 1, 2, and 3 are unknown

Two-Unit Process Example• Variables for Streams 1, 2, and 3 are unknown• Label unknown stream variables

Two-Unit Process Example• Degree-of-freedom analysis

overall system: 2 unknowns – 2 balances = 0 (find m3, x3) mixer: 4 unknowns – 2 balances = 2 Unit 1: 2 unknowns – 2 balances = 0 (find m1, x1) mixer: 2 unknowns – 2 balances = 0 (find m2, x2)

Extraction-Distillation Process


Simultaneously solve total mass and acetone balances to determine m1 and m3.

Solve MIBK balance to determine xM1.


Solve acetone, MIBK, and water balances to determine mA4, mM4, and mW4.


For either (just 1) extractor unit, solve acetone, MIBK, and water balances to determine mA2, mM2, and mW2.


ndf = 4 unknowns (mA6, mM6, mW6, and m5) – 3 balances = 1

underspecified


ndf = 4 unknowns (mA6, mM6, mW6, and m5) – 3 balances = 1

underspecified

Recycle• It is seldom cost effective to waste reactant fed that

does not react to product. More often, this material is separated (recovered), and recycled (returned to its point of origin for reuse).

Balances on an Air Conditioner• process cools and dehumidifies feed air• unknowns: n1, n2, n3, n4, n5 (requested by problem)• degree-of-freedom analysis critical to solution

basis

Balances on an Air Conditioner• Overall system

ndf = 2 variables (n1, n3) – 2 balances = 0

Balances on an Air Conditioner• Mixer


Balances on an Air Conditioner• Cooler


Balances on an Air Conditioner• Splitter

ndf = 2 variables (n4, n5) – 1 balances = 1• only 1 independent balance can be written on the splitter because

the streams entering/leaving have the same composition.

100017.0n017.0n017.0

100983.0n983.0n983.0

54

54

Balances on an Air Conditioner Overall: ndf = 2 variables (n1, n3) – 2 balances = 0 Mixer: ndf = 2 variables (n2, n5) – 2 balances = 0 Cooler: ndf = 2 variables (n2, n4) – 2 balances = 0 Splitter: ndf = 2 variables (n4, n5) – 1 balances = 1

To find requested unknowns, solve overall balances followed by mixing balances.There is no need to solve the cooler or splitter balances.

Balances on an Air Conditioner• overall dry air balance• overall mole balance

0.960 n1 0.983 100 n1 102.4mol

n1 n3 100 n3 2.4 mol H2O condensed

Balances on an Air Conditioner• overall mole balance• water balance

solved simultaneously:

mol 290n ;mol 5.392n

n023.0n017.0n04.0nnn

52

251

251

Reasons to recycle• recover catalyst

typically most expensive chemical constituent• dilute a process stream

reduce slurry concentration• control a process variable

control heat produced by highly exothermic reaction• circulation of a working fluid

refrigerant

Evaporative Crystallization Process• Calculate:

rate of evaporation rate of production of crystalline K2CrO4

feed rates to evaporator and crystallizer recycle ratio (mass or recycle/mass of fresh feed)

Evaporative Crystallization Process• Overall system:

ndf = 3 unknowns (m2, m4, m5) – 2 balances – 1 spec = 0 specification: m4 is 95% of total filter cake mass

544 mm95.0m

Evaporative Crystallization Process• Feed/recycle mixer:

ndf = 3 unknowns (m6, m1, x1) – 2 balances = 1 underspecified

544 mm95.0m

Evaporative Crystallization Process• Evaporator:

ndf = 3 unknowns (m3, m1, x1) – 2 balances = 1 underspecified

544 mm95.0m

Evaporative Crystallization Process• Crystallizer:

ndf = 2 unknowns (m3, m6) – 2 balances = 0 solvable Once m3, m6 are known, mixer or evaporator balances can

be solved.

544 mm95.0m


K2CrO4 balance water balance total mass balance specification

544

542hkg

52hK kg

54hK kg

mm95.0mmmm4500

m636.0m4500667.0

m364.0m4500333.0

solve simultaneously for m4 and m5

hcrystals K kg

4 1470m

solu W/kg kg 0.636solu K/kg kg 364.0

5.77m hsolution kg

5



solve for m2 with knowns m4 and m5

hOH kg

222950m

hcrystals K kg

4 1470m


5.77m hsolution kg

5

544

542hkg

52hK kg

54hK kg

mm95.0mmmm4500

m636.0m4500667.0

m364.0m4500333.0



only 3 equations are independent

hOH kg

222950m

hcrystals K kg

4 1470m


5.77m hsolution kg

5

544

542hkg

52hK kg

54hK kg

mm95.0mmmm4500

m636.0m4500667.0

m364.0m4500333.0

Evaporative Crystallization Process• Crystallizer:

total mass balance

water balance

6h

kg3

653

6hkg

3

6543

m257.14.97m

m636.0m636.0m506.0m5.771470m

mmmm

hOH kg

222950m

solve simultaneously for m3 and m6

hkg

3 7200m hcrystals K kg

4 1470m


5.77m hsolution kg

5

hkg

6 5650m

Evaporative Crystallization Process• feed/recycle mixer:

total mass balance water or K2CRO4 balance could be used tp find x1 if desired

hkg

116hkg 10150mmm4500

h kg

1 10150m

hOH kg

222950m

hkg

3 7200m

hkg

6 5650m

hcrystals K kg

4 1470m


5.77m hsolution kg

5

Evaporative Crystallization Process• If recycle is not used,

crystal production is 622 kg/h vs 1470 kg/h (w/ recycle) discarded filtrate (m4) is 2380 kg/h, representing 866 kg/h

of potassium chromate• What are cost consequences of using recycle vs not?

Bypass Stream• Similar to a recycle, but a fraction of a stream is

diverted around a process unit, rather than being returned to it.

• Calculation approach is identical.

Balances on Reactive Systems• Material balance no longer takes the form

INPUT = OUTPUT

• Must account for the disappearance of reactants and appearance of products through stoichiometry.

Stoichiometric Equations• The stoichiometric equation of a chemical reaction is

a statement of the relative amounts of reactants and products that participate in the reaction.

2 SO2 + O2 → 2 SO3

• A stoichiometric equation is valid only if the number of atoms of each atomic species is balanced.

2 S → 2 S4 O + 2 O → 6 O

Stoichiometric Equations• The stoichiometric equation of a chemical reaction is

a statement of the relative amounts of reactants and products that participate in the reaction.

2 SO2 + O2 → 2 SO3

• A stoichiometric rato of two molecular species participating in a reaction is the ratio of their stoichiometric coefficients:

2 mol SO3 generated / 1 mol O2 consumed

2 mol SO3 generated / 2 mol SO2 consumed

Stoichiometric Equations

C4H8 + 6 O2 → 4 CO2 + 4 H2O• Is this stoichiometric equation balanced?• What is the stoichiometric coefficient of CO2?• What is the stoichiometric ratio of H2O to O2?• How many lb-mol O2 react to form 400 lb-mol CO2?

• 100 lbmol/min C4H8 is fed and 50% reacts. At what rate is water formed?

22

22 O lbmol 600

CO lbmol 4O lbmol 6CO lbmol 400

minOH lbmol 200

HC lbmol 1OH lbmol 450.0

minHC lbmol 100 2

84

284

Limiting and Excess Reactants• Two reactants are said to be in stoichiometric proportion

if the ratio (moles A present/moles B present) equals the stoichiometric ratio from the balanced reaction equation.

2 SO2 + O2 → 2 SO3

• the feed ratio that would represent stoichiometric proportion is nSO2/nO2 = 2:1

• If reactants are fed in stoichometric proportion, and the reaction proceeds to completion, all reactants are consumed.

Limiting and Excess Reactants• Stoichiometric Proportion – Reactants are present in

a ratio equivalent to the ratio of the stoichiometric coefficients.

A + 2B → 2C

Limiting and Excess Reactants• Limiting reactant – A reactant is limiting if it is present

in less than stoichiometric proportion relative to every other reactant.

A + 2B → 2C

• Excess reactant – All other reactants besides the limiting reactant.

Limiting and Excess Reactants

• fractional excess (fXS) – ratio of the excess to the stoichiometric proportion.

A + 2B → 2C

25.04

45

n

nnf

stoichA

stoichAfeedAXS

Limiting and Excess Reactants• fractional conversion (f) – ratio of the amount of a

reactant reacted, to the amount fed.

A + 2B → 2C

fedA

reactedA

n

nf

0.050

fA 0.080

fB



A + 2B → 2C

fedA

reactedA

n

nf

2.051

fA 25.082

fB



A + 2B → 2C

fedA

reactedA

n

nf

4.052

fA 5.084

fB



A + 2B → 2C

fedA

reactedA

n

nf

6.053

fA 75.086

fB



A + 2B → 2C

fedA

reactedA

n

nf

8.054

fA 0.188

fB

Extent of Reaction• extent of reaction (ξ) – an extensive quantity

describing the progress of a chemical reaction . ν – stoichiometric coefficients: νA = -1, νB = -2, νC = 2

A + 2B → 2C

ni ni0 i

nA nA0

nB nB0 2

nC nC0 2

0



A + 2B → 2C

ni ni0 i

nB 8 2 8

nC 02 0

0

nA 5 5



A + 2B → 2C

ni ni0 i

1

nB 8 2 6

nC 02 2

nA 5 4



A + 2B → 2C

ni ni0 i

2

nB 8 2 4

nC 02 4

nA 5 3



A + 2B → 2C

ni ni0 i

3

nB 8 2 2

nC 02 6

nA 5 2



A + 2B → 2C

ni ni0 i

4

nB 8 2 0

nC 02 8

nA 5 1

• Assume an equimolar reactant feed of 100 kmol: What is the limiting reactant?

2C2H4 O2 2C2H4O

ethylene

A reactant is limiting if it is present in less than stoichiometric proportion relative to every other reactant.

11

n

n

12

n

n

feedO

HC

stoichO

HC

2

42

2

42

• Assume an equimolar reactant feed of 100 kmol: What is the percentage excess of each reactant?

2C2H4 O2 2C2H4O

%10000.150

50100

n

nnf

stoichO

stoichOfeedOO,XS

2

22

2

• Assume an equimolar reactant feed of 100 kmol: If the reaction proceeds to completion: (a) how much of the excess

reactant will be left?

(b) How much C2H4O will be formed?

(c) What is the extent of reaction?

2C2H4 O2 2C2H4O

50

21000

nn424242 HC

oHCHC

50n

501100n

nn

2

2

222

O

O

OoOO

100n

5020n

nn

OHC

OHC

OHCo

OHCOHC

42

42

424242

• Assume an equimolar reactant feed of 100 kmol: If the reaction proceeds to a point where the fractional

conversion of the limiting reactant is 50%, how much of each reactant and product is present at the end? What is ξ?

2C2H4 O2 2C2H4O

5.0n

nf

fedHC

reactedHC

42

42 25

210050100

nn424242 HC

oHCHC

75n

251100n

nn

2

2

222

O

O

OoOO

50n

2520n

nn

OHC

OHC

OHCo

OHCOHC

42

42

424242

5.0n

nnf o

HC

HCo

HC

42

4242

50n

5.0100

n100

42

42

HC

HC

• Assume an equimolar reactant feed of 100 kmol: If the reaction proceeds to a point where 60 mol of O2 is left,

what is the fractional conversion of C2H4? What is ξ?

2C2H4 O2 2C2H4O

8.0

10020100

n

nf

fedHC

reactedHC

42

42

20n

402100n

nn

42

42

424242

HC

HC

HCo

HCHC

40

110060

nn222 O

oOO

Reaction Stoichiometry• Acrylonitrile produced by reaction of ammonia,

propylene, and O2 at 30% conversion of limiting reactant:

C3H6 NH3 32

O3 C3H3N 3H2O

nNH3nC3H6 0 0.120100 0.100100 1.20

nNH3nC3H6 stoich

1 1 1

nO2nC3H6 0 0.7800.21100 0.100100 1.64

nO2nC3H6 stoich

1.5 1 1.5

limiting

determine limiting reactant



C3H6 NH3 32

O3 C3H3N 3H2O

nNH3 stoich10.0 mol C3H6

1 mol NH3

1 mol C3H6

10.0 mol NH3

fXS NH3

NH3 0 NH3 stoich

NH3 stoich

12.0 10.010.0

0.20

nO2 stoich10.0 mol C3H6

1.5 mol O2

1 mol C3H6

15.0 mol O2

fXS O2

O2 0 O2 stoich

O2 stoich

16.4 15.015.0

0.093

fXS 0.20

fXS 0.093

determine fractional excesses

limiting



C3H6 NH3 32

O3 C3H3N 3H2O

nC3H6 1 f nC3H6 0 1 0.30 10.0 mol C3H6 7.0 mol C3H6

fXS 0.093

fXS 0.20

use fractional conversion to determine amount of propylene that leaves the reactor

limiting

ni ni0 i



C3H6 NH3 32

O3 C3H3N 3H2O

nC3H67.0 mol C3H6

fXS 0.093

fXS 0.20

nC3H6 nC3H6 0 1

7.0 mol 10.0 mol 3 mol

3 mol determine extent of reaction by applying mole balance to propylene

limiting



OHNHCONHHC 233223

363 3

90.330

6.613010078.079.0

30.310

9.11310078.021.0

93110012.0

2

2

33

2

3

23

OH

N

NHC

O

NH

n

n

n

n

n

fXS 0.093

fXS 0.20limiting

nC3H67.0 mol C3H6

3 mol

ni ni0 i

units not included and sig fig rules not followed to permit fit of all calculations

apply mole balance to all remaining species

Chemical Equilibrium• Given

a set of reactive species, and reaction conditions

• Determine1. the final (equilibrium) composition of the reaction

mixture2. how long the system takes to reach a specified state short

of equilibrium

• This course will cover #1 (Ch E 441 will cover #2)

Chemical Equilibrium• Irreversible reaction

reaction proceeds only in a single direction A → B concentration of the limiting reactant eventually

approaches zero (time duration can vary widely)

Equilibrium composition of an irreversible reaction is that which corresponds to complete conversion.

Chemical Equilibrium• Reversible reaction

reaction proceeds in both directions A ↔ B net rate (forward – backward) eventually approaches zero

(again, time can vary widely)

Equilibrium composition of a reversible reaction is that which corresponds to the equilibrium conversion.

Equilibrium Composition• An equilibrium reaction proceeds

to an extent at temperature T based on the equilibrium constant, K(T). where yi is the mole fraction of species i

• Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00

g2g2g2g HCOOHCO

OHCO

HCO

2

22

yyyy

TK

i0ii nntotal

ii n

ny

Equilibrium Composition•


OHCO

HCO

2

22

yyyy

TK

3nnnnn

1nn

1nn

21nn11nn

222

22

22

22

HCOOHCOtotal

0HH

0COCO

0OHOH

0COCO

i0ii nntotal

ii n

ny

g2g2g2g HCOOHCO

Equilibrium Composition


OHCO

HCO

2

22

yyyy

TK

i0ii nntotal

ii n

ny

3n

n

n

2n1n

total

H

CO

OH

CO

2

2

2

121

mol667.022

2122

2

667.0667.0333.1333.0

g2g2g2g HCOOHCO



3n

222.03/667.0y

222.03/667.0y

444.03/333.1y111.03/333.0y

total

H

CO

OH

CO

2

2

2

121

mol667.022

2122

2

OHCO

HCO

2

22

yyyy

TK

i0ii nntotal

ii n

ny

g2g2g2g HCOOHCO



limiting reactant is CO

i0ii nn

0.667mol mol 333.0

667.011nCO

at equilibrium,

fractional conversion at equilibrium

667.0f 00.1333.000.1

g2g2g2g HCOOHCO

3n

222.03/667.0y

222.03/667.0y

444.03/333.1y111.03/333.0y

total

H

CO

OH

CO

2

2

2

Multiple Reactions

C2H4 12

O2 C2H4OC2H4 3O2 2CO2 2H2O

j

jij0ii nnfor j reactions of i species,mole balance becomes

10OHCOHC

2121

0OO

210HCHC

1nn

3nn

11nn

4242

22

4242

20OHOH

20COCO

2nn

2nn

22

22

Multiple Reactions

C2H4 12

O2 C2H4OC2H4 3O2 2CO2 2H2O

reactions side no with conversion 100% at formed molesformed product desired molesyield

formed product undesired molesformed product desired molesyselectivit

j

jij0ii nnfor j reactions of i species,mole balance becomes

Multiple Reactions• 100 moles A fed to a batch reactor• product composition: 10 mol A, 160 B, 10 C

What is1. fA?

2. YB?

3. SB/C?

4. ξ1, ξ2

CAB2A

9.0100

10100fA


What is1. fA?

2. YB?

3. SB/C?

4. ξ1, ξ2

889.010100

160Y12B

CAB2A


What is1. fA?

2. YB?

3. SB/C?

4. ξ1, ξ2

1610

160S C/B

CAB2A


What is1. fA?

2. YB?

3. SB/C?

4. ξ1, ξ21090

10010nn

12

21

22A11AAoA

8020160

nn

1

1

11BBoB

CAB2A

Balances on Reactive Processes• Continuous, steady-state dehydrogenation of ethane• Total mass balance still has INPUT = OUTPUT form• Molecular balances contain consumption/generation• Atomic balances (H and C) also have simple form

24262 HHCHC

Balances on Reactive Processes• Continuous, steady-state dehydrogenation of ethane• First consider molecular balances:

Molecular H2 balance: generation = output

generation H2 = 40 kmol H2/min

24262 HHCHC


C2H6 balance: input = output + consumption

100 kmol C2H6/min = n1 + (C2H6 consumed)

24262 HHCHC


C2H4 balance: generation = output

(C2H4 generated) = n2

24262 HHCHC

Balances on Reactive Processes• Continuous, steady-state dehydrogenation of ethaneAtomic C balance: input = output

Atomic H balance: input = output

24262 HHCHC

426262 HC mol 1

C mol 22HC mol 1

C mol 21HC mol 1

C mol 262 nnHC mol 100

4262262 HC mol 1

H mol 42HC mol 1

H mol 61H mol 1

H mol 2HC mol 1H mol 6

62 nn40HC mol 100

Independent Equations• To understand the number of independent species

balances in a reacting system requires an understanding of independent algebraic equations.

• Algebraic equations are independent if you cannot obtain any of them by adding/subtracting multiples of the others.

2][ 12y6x3[1] 4y2x

[5] 6zy4[4] 2zx2[3] 4y2x

3×[1] = [2] 2×[3] – [4] = [5]

Independent Equations• To understand the number of independent species

balances in a reacting system requires an understanding of independent algebraic equations.

• Algebraic equations are independent if you cannot obtain any of them by adding/subtracting multiples of the others.

2][ 12y6x3[1] 4y2x

121212y6y61212y6y243

Independent Species• If two molecular or atomic species are in the same

ratio to each other where ever they appear in a process and this ratio is incorporated in the flowchart labeling, balances on those species will not be independent equations.

31

31

n76.3n76.3nn

Independent Chemical Reactions • When using molecular species balances or extents of

reaction to analyze a reactive system, the degree of freedom analysis must account for the number of independent chemical reactions among the species entering and leaving the system.

Independent Chemical Reactions• Chemical reactions are independent if the

stoichiometric equation of any one of them cannot be obtained by adding and subtracting multiples of the stoichiometric equations of the others.

[3] 2CA[2] CB[1] 2BA

2×[2] + [1] = [3]

Solving Reactive Systems• There are 3 possible methods for solving balances

around a reactive system:1. Molecular species balances require more complex

calculations than the other methods and should be used only for simple (single reaction) systems.

2. Atomic species balances generally lead to the most straightforward solution procedure, especially when more than one reaction is involved

3. Extents of reaction are convenient for chemical equilibrium problems.

Molecular Species Balances• To use molecular species balances to analyze a

reactive system, the balances must contain generation and/or consumption terms.

• The degree-of-freedom analysis is as follows: # unknown labeled variables+ # independent chemical reactions- # independent molecular species balances- # other equations relating unknown variables # of degrees of freedom

Molecular Species Balances

2 unknown labeled variables + 1 independent chemical reactions - 3 independent molecular species balances - 0 other equations relating unknown variables 0 degrees of freedom

at steady state24262 HHCHC


H2 Balance: generation = output

2H H kmol 40gen2



C2H6 Balance: input = output + consumption

min

HC kmol1

H kmol 1HC kmol 1

minH kmol

1minHC kmol

62

2

62262

60n

40n 1000


Molecular Species Balances24262 HHCHC

C2H4 Balance: generation = output

min

HC kmol2

H kmol 1HC kmol 1

minH kmol

2

42

2

422

40n

40n

at steady state

Atomic Species Balance

• All atomic species balances take the form INPUT = OUTPUT

• Degree-of-freedom analysis, ndf = # unknown labeled variables - # independent atomic species balances - # molecular balances on independent nonreactive species- # other equations relating unknown variables

24262 HHCHC


• All atomic species balances take the form INPUT = OUTPUT

• Degree-of-freedom analysis, ndf = 0 = 2 unknown labeled variables - 2 independent atomic species balances - 0 molecular balances on independent nonreactive species- 0 other equations relating unknown variables

24262 HHCHC


C Balance: input = output

21

HC kmol 1C kmol 2

2HC kmol 1C kmol 2

1HC kmol 1C kmol 2

minHC kmol

nnmolk 100

nn 100426262

42

24262 HHCHC


H Balance: input = output

21

HC kmol 1H kmol 4

2HC kmol 1H kmol 6

1

H kmol 1H kmol 2

minH kmol

HC kmol 1H kmol 6

minHC kmol

n4n6+molk 80molk 600

nn

40 100

4262

2

2

62

42

24262 HHCHC


Solve simultaneously

min/HC kmol 40n min/HC kmol 60n

n4n6+molk 80molk 600 :Hnnmolk 100 :C

422

621

21

21

24262 HHCHC

Extent of Reaction• The 3rd method by which to determine molar flows in

a reactive system is using expressions for each species flow rate in terms of extents of reaction (ξ).

• Degree-of-freedom analysis for such an approach: ndf = # of unknown labeled variables

+ # independent reactions - # independent nonreactive species - # other relationships or specifications

j

jij0ii nn

Incomplete Combustion of CH4

• Methane is burned with air in a continuous steady-state reactor to yield a mixture of carbon monoxide, carbon dioxide, and water.

• The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO2 per mole CO.

CH4 32

O2 CO2 H2O

CH4 2 O2 CO2 2 H2O


• The feed to the reactor contains 7.80 mol% CH4, 19.4 mol% O2, 72.8 mol% N2. Methane undergoes 90.0% conversion, and the effluent gas contains 8 mol CO2 per mole CO.

CH4 32

O2 CO2 H2O

CH4 2 O2 CO2 2 H2O

fCH4 = 0.9


ndf = 5 unknowns

+ 2 independent reactions- 5 expressions for ξ (CH4, O2, CO, CO2, H2O)

- 1 nonreactive species balance (N2)

- 1 specified methane conversion =0

CH4 32

O2 CO2 H2O

CH4 2 O2 CO2 2 H2O

fCH4 = 0.9


N2 balance: nonreactive species, INPUT = OUTPUT

CH4 conversion specification:

CH4 32

O2 CO2 H2O

CH4 2 O2 CO2 2 H2O

2molN mol

N N mol 8.72mol 100728.0n 2

2

4molCH mol

CH CH mol 780.0mol 1000780.0900.01n 4

4

fCH4 = 0.9


Extent of reaction mole balances:

CH4 32

O2 CO2 H2O

CH4 2 O2 CO2 2 H2O

212

30OO

210OHOH

20COCO

10COCO

210CHCH

2nn

22nn

1nn1nn

11nn

22

22

22

44

0.78 7.80 1 2

nCO 1

nCO28nCO 2

nH2O 21 22

nO219.4 3

21 22

fCH4 = 0.9

Product Separation and Recycle• Two definitions of reactant conversion are used in

the analysis of chemical reactors with product separation and recycle of unconsumed reactants.

reactor to inputreactor from output - reactor to inputreactant

conversionpass glesin

process to inputprocess from output - process to inputreactant

conversionoverall

Product Separation and Recycle

reactor to inputreactor from output - reactor to inputreactant

conversionpass glesin

process to inputprocess from output - process to inputreactant

conversionoverall

Product Separation and Recycle

%75%100A/min mol 100

A/min mol 25 - A/min mol 100conversion

pass glesin

%100%100A/min mol 75

0 - A/min mol 75conversion

overall

Catalytic Propane Dehydrogenation

C3H8 C3H6 H2

95% overallconversion


C3H8 C3H6 H2


Overall Process

ndf = 3 unknowns (n6, n7, n8) – 2 independent atomic balances (C and H) – 1 relation (overall conversion) = 0

consider n6, n7, n8 known for further DOF analyses


C3H8 C3H6 H2


Mixingpoint

ndf = 4 unknowns (n9, n10, n1, n2) – 2 balances (C3H8 and C3H6) = 2

ndf = 2


C3H8 C3H6 H2


reactor

ndf = 5 unknowns (n3, n4, n5, n1, n2) – 2 balances (C and H) = 3

ndf = 2 ndf = 3


C3H8 C3H6 H2


separator

ndf = 5 unknowns (n3, n4, n5, n9, n10) – 3 balances (C3H8, C3H6, and H2) – 2 relations (reactant and product recovery fractions) = 0

ndf = 2 ndf = 3 ndf = 0


C3H8 C3H6 H2


overall

836 HC mol 5mol 10095.01n

conversionrelationship


C3H8 C3H6 H2


overall

637

HC mol 1C mol3

7HC mol 1C mol3

83HC mol 1C mol3

HC mol 95n

nHC mol 5mol 100638383

C atomic balance

836 HC mol 5n


C3H8 C3H6 H2


overall

263

8383

H mol 1H mol 2

8HC mol 1H mol 6

63

HC mol 1H mol8

83HC mol 1H mol 8

nHC mol 95

HC mol 5mol 100

H atomic balance

28 H mol 95n

637 HC mol 95n 836 HC mol 5n


C3H8 C3H6 H2


separator

6310710

83336

HC mol 75.4nn0500.0nHC mol 900nn00555.0n

given relations


28 H mol 95n


C3H8 C3H6 H2


separator

n3 n6 n9 n9 895 mol C3H8

propane balance

n10 4.75 mol C3H6

833 HC mol 900n


28 H mol 95n


C3H8 C3H6 H2


mixer

100n9 n1 n1 995 mol C3H8

propane balance

n10 4.75 mol C3H6

n9 895 mol C3H8

833 HC mol 900n


28 H mol 95n


C3H8 C3H6 H2


mixer

n10 n2 n2 4.75 mol C3H6

propylene balance

n10 4.75 mol C3H6

n9 895 mol C3H8

n1 995 mol C3H8 833 HC mol 900n


28 H mol 95n


C3H8 C3H6 H2


reactorC atomic balance

n10 4.75 mol C3H6

n9 895 mol C3H8

n1 995 mol C3H8

n2 4.75 mol C3H6

634

HC mol 1C mol 3

4HC mol 1C mol3

83

HC mol 1C mol 3

63HC mol 1C mol3

83

HC mol 75.99n

nHC mol 009

HC mol .754HC mol 959

6383

6383

833 HC mol 900n


28 H mol 95n


C3H8 C3H6 H2


reactorH atomic balance

n10 4.75 mol C3H6

n9 895 mol C3H8

n1 995 mol C3H8

n2 4.75 mol C3H6

25

H mol 1H mol 2

5HC mol 1H mol 6

63HC mol 1H mol 8

83

HC mol 1H mol 6

63HC mol 1H mol 8

83

H mol 95n

nHC mol 75.99HC mol 009

HC mol .754HC mol 959

26383

6383

833 HC mol 900n

634 HC mol 75.99n 637 HC mol 95n 836 HC mol 5n

28 H mol 95n


C3H8 C3H6 H2


single-passconversion

n10 4.75 mol C3H6

n9 895 mol C3H8

n1 995 mol C3H8

n2 4.75 mol C3H6

%55.9%100

HC mol 959HC mol 009HC mol 959f

83

8383passsingle

833 HC mol 900n

634 HC mol 75.99n

25 H mol 95n 637 HC mol 95n

836 HC mol 5n

28 H mol 95n


C3H8 C3H6 H2


recycleratio


28 H mol 95n

833 HC mol 900n

n10 4.75 mol C3H6

n9 895 mol C3H8

n1 995 mol C3H8

n2 4.75 mol C3H6

feed fresh molrecycle mol109 0.9

mol 001mol .754mol 958

feed mol 001nnR

634 HC mol 75.99n

25 H mol 95n

fsingle pass 9.55%

Purging• Necessary with recycle to prevent accumulation of a

species that is both present in the fresh feed and is recycled rather than separated with the product.

CO2 3H2 CH3OHH2O

mixed fresh feedand recycle is a

convenientbasis selection

fsingle pass 60%

Methanol Synthesis

• ndf = 7 unknowns (n0, x0C, np, x5C, x5H, n3, n4) + 1 rxn

- 5 independent species balances = 3

CO2 3H2 CH3OHH2O

fsingle pass 60%

Methanol Synthesis

• ndf = 4 unknowns (n1, n2, n3, n4) + 1 rxn

– 4 independent species balances – 1 single pass conversion = 0

CO2 3H2 CH3OHH2O

fsingle pass 60%

Methanol Synthesis

• ndf = 3 unknowns (n5, x5C, X5H)

– 3 independent species balances = 0

CO2 3H2 CH3OHH2O

fsingle pass 60%

Methanol Synthesis

• ndf = 3 unknowns (n0, x0C, nr)

– 3 independent species balances = 0

CO2 3H2 CH3OHH2O

fsingle pass 60%

Methanol Synthesis

• ndf = 1 unknowns (np)

– 1 independent species balance = 0

CO2 3H2 CH3OHH2O

fsingle pass 60%

investigatemole balances andtheir solution in the text

Combustion Reactions• Combustion - rapid reaction of a fuel with oxygen.• Valuable class of reactions due to the tremendous

amount of heat liberated, subsequently used to produce steam used to drive turbines which generates most of the world’s electrical power.

• Common fuels used in power plants: coal fuel oil (high MW hydrocarbons) gaseous fuel (natural gas) liquified petroleum gas (propane and/or butane)

Combustion Chemistry• When a fuel is burned

C forms CO2 (complete) or CO (partial combustion) H forms H2O S forms SO2

N forms NO2 (above 1800°C)

• Air is used as the source of oxygen. DRY air analysis: 78.03 mol% N2

20.99 mol% O2

0.94 mol% Ar 0.03 mol% CO2

0.01 mol% H2, He, Ne, Kr, Xe

usually safe to assume:79 mol% N2

21 mol% O2

Combustion Chemistry• Stack (flue) gas – product gas that leaves a furnace.• Composition analysis:

wet basis – water is included in mole fractions dry basis – does not include water in mole fractions

• Stack gas contains (mol) on a wet basis: 60.0% N2, 15.0% CO2, 10.0% O2, 15.0% H2O Dry basis analysis:

• 60/(60+15+10) = 0.706 mol N2/mol• 15/(60+15+10) = 0.176 mol CO2/mol• 10/(60+15+10) = 0.118 mol O2/mol

Combustion Chemistry• Stack gas contains (mol) on a dry basis:

65% N2, 14% CO2, 10% O2, 11% CO xH2O = 0.0700 (humidity measurement)

Wet basis analysis:

• assume 100 mole dry gas basis– 7.53 mole H2O– 65 mole N2

– 14 mole CO2

– 10 mole O2

– 11 mole CO– total = 107.5 mole

gas wet lbmolgas ryd lbmol

gas wet lbmolOH lbmol 9300.00700.0 2

gas dry lbmolOH lbmol 20753.0

xH2O 7.53107.5

0.0700

xN2 65

107.50.605

xCO2 14

107.50.130

xO2 10

107.50.0930

xCO 11107.5

0.102

Theoretical and Excess Air• The less expensive reactant is commonly fed in

excess of stoichiometric ratio relative to the more valuable reactant, thereby increasing conversion of the more expensive reactant at the expense of increased use of excess reactant.

• In a combustion reaction, the less expensive reactant is oxygen, obtained from the air. Conseqently, air is fed in excess to the fuel.

Theoretical and Excess Air

• Theoretical oxygen is the exact amount of O2 needed to completely combust the fuel to CO2 and H2O.

• Theoretical air is that amount of air that contains the amount of theoretical oxygen.

• Excess air is the amount by which the air fed to the reactor exceeds the theoretical air.

% excess air = moles air fed - moles air theoreticalmoles air theoretical

100%

Theoretical and Excess Air

nC4H10 = 100 mol/hr; nair = 5000 mol/hr

C4H10 132

O2 4CO2 5H2O

hr

O mol650HC molO mol 5.6

hrHC mol 100n 2

104

2104ltheoreticaO2

hr

air mol3094O mol

air mol .764hr

O mol 506n2

2ltheoreticaair

% excess air 5000 3094

3094100% 61.6%

Combustion Reactors• Procedure for writing/solving material balances for a

combustion reactor1. When you draw and label the flowchart, be sure the

outlet stream (the stack gas) includesa. unreacted fuel (unless the fuel is completely consumed)b. unreacted oxygenc. water and carbon dioxide (and CO if combustion is incomplete)d. nitrogen (if air is used as the oxygen source)

2. Calculate the O2 feed rate from the specifed percent excess oxygen or air

3. If multiple reactions, use atomic balances

Combustion of Ethane

C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9

25% of the ethane burned forms CO

degree-of-freedomanalysis

ndf = 7 unknowns- 3 atomic balances - 1 nitrogen balance- 1 excess air specification- 1 ethane conversion specification- 1 CO/CO2 ratio specification

= 0


C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9


excess airspecification

262

262ltheoreticaO O mol 350

HC molO mol 5.3HC mol 100

n2

air mol 2500n

O mol 3505.1n21.0

0

20


C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9


ethaneconversion

specification

62621 HC mol 0.10HC mol 10090.01n

n0 2500 mol air


C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9


CO/CO2 ratiospecification

OC mol 0.45react HC mol 1gen CO mol 2HC mol 1009.025.0n

62624

n0 2500 mol air

n1 10.0 mol C2H6


C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9


nitrogenbalance

23 N mol 1975ira mol 250079.0n

n0 2500 mol air

n1 10.0 mol C2H6

n4 45.0 mol CO


C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9


atomic Cbalance

25

CO mol 1C mol 1

5CO mol 1C mol 1

4HC mol 1C mol 2

1HC mol 1C mol 2

62

CO mol 135n

nnnHC mol 10026262

n0 2500 mol air

n1 10.0 mol C2H6

n4 45.0 mol CO

n3 1975 mol N2


C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9


atomic Hbalance

OH mol 270n

nHC mol 10HC mol 100

26

OH mol 1H mol 2

6HC mol 1H mol 6

62HC mol 1H mol 6

62 26262

n0 2500 mol air

n1 10.0 mol C2H6

n4 45.0 mol CO

n3 1975 mol N2

n5 135 mol CO2


C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9


atomic Obalance

22

OH mol 1O mol 1

2CO mol 1O mol 2

2

CO mol 1O mol 1

O mol 1O mol 2

2O mol 1O mol 2

2

O mol 232n

OH mol 702CO mol 135

CO mol 54nO mol 255

22

22

n0 2500 mol air

n1 10.0 mol C2H6

n4 45.0 mol CO

n3 1975 mol N2

n5 135 mol CO2

n6 270 mol H2O


C2H6 72

O2 2CO2 3H2O

C2H6 52

O2 2CO3H2O

fC2H6 = 0.9


stack gascomposition(dry basis)

sum 102321974 45135 2396y1 10 2396 0.00417 mol C2H6 moly2 232 2396 0.0970 mol O2 moly3 1974 2396 0.824 mol N2 moly 4 45 2396 0.019 mol CO moly 5 135 2396 0.0563 mol CO2 mol

n0 2500 mol air

n1 10.0 mol C2H6

n4 45.0 mol CO

n3 1975 mol N2

n5 135 mol CO2

n6 270 mol H2O

n2 232 mol O2

270 mol H2O2396 mol dry stack gas

0.113 mol H2Omol dry stack gas

Documents

4 Material Balances