4 material balances
TRANSCRIPT
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?
Integral Balances on Semibatch and Continous Processes
• differential air balance
minkmol
111.0n
minkmol
nkmol
ira molk 900.0min
ira molk100.0
outputinput
Integral Balances on Semibatch and Continous Processes
• 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
Air Humidification and Oxygenation
water balance:
minmol
1.74nmol
OH mol 015.0minmol
nmin
OmolH11.1
molOH mol 015.0
minmol
nmin
OmolHn
32
32
23
22
Air Humidification and Oxygenation
total mole balance:
minmol
1
minmol
minmol
1
3211
8.60n
1.7411.1n200.1
nnnn200.0
Air Humidification and Oxygenation
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
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.
• top:h
lbmol625
molhrlbmol
5.12mol 50.0
molhrlbmol
5.12mol 100
hlbmol1250=factor scale
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.
• bottom:h
lbmolA156
molAhrlbmolA
5.12molA 12.5
molhrlbmol
5.12mol 100
hlbmol1250=factor scale
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.
• bottom:h
lbmolB469
molBhrlbmolB
5.12molB 37.5
molhrlbmol
5.12mol 100
hlbmol1250=factor scale
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.
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.
Balances on a mixing unit
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)
Balances on a mixing unit
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.
Balances on a mixing unit
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
Balances on a mixing unit
5. NaOH balance (INPUT = OUTPUT):
(0.20 kg NaOH/kg)(100 kg) = (0.80 kg NaOH/kg)(m2)
m2 = 250 kg
= 250 kg NaOH
Balances on a mixing unit
6. Total mass balance (INPUT = OUTPUT):
100 kg + m1 = m2 =250 kg
m1 = 150 kg
= 250 kg
= 150 kg
Balances on a mixing unit
7. Diluent water volume:
V1 = m1 / ρH2O = 150 kg / (1 kg/L)
V1 = 150 L
= 250 kg
= 150 kg= 150 L
Balances on a mixing unit
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.
General Procedure – Single Unit Op
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
General Procedure – Single Unit Op
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
Distillation Column example• Ex. 4.3-5
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
Distillation Column example• Ex. 4.3-5
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
Extraction-Distillation Process
Simultaneously solve total mass and acetone balances to determine m1 and m3.
Solve MIBK balance to determine xM1.
Extraction-Distillation Process
Solve acetone, MIBK, and water balances to determine mA4, mM4, and mW4.
Extraction-Distillation Process
For either (just 1) extractor unit, solve acetone, MIBK, and water balances to determine mA2, mM2, and mW2.
Extraction-Distillation Process
ndf = 4 unknowns (mA6, mM6, mW6, and m5) – 3 balances = 1
underspecified
Extraction-Distillation Process
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
ndf = 2 variables (n2, n5) – 2 balances = 0
Balances on an Air Conditioner• Cooler
ndf = 2 variables (n2, n4) – 2 balances = 0
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
Evaporative Crystallization Process• Overall system:
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
Evaporative Crystallization Process• Overall system:
K2CrO4 balance water balance total mass balance specification
solve for m2 with knowns m4 and m5
hOH kg
222950m
hcrystals K kg
4 1470m
solu W/kg kg 0.636solu K/kg kg 364.0
5.77m hsolution kg
5
544
542hkg
52hK kg
54hK kg
mm95.0mmmm4500
m636.0m4500667.0
m364.0m4500333.0
Evaporative Crystallization Process• Overall system:
K2CrO4 balance water balance total mass balance specification
only 3 equations are independent
hOH kg
222950m
hcrystals K kg
4 1470m
solu W/kg kg 0.636solu K/kg kg 364.0
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
solu W/kg kg 0.636solu K/kg kg 364.0
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
solu W/kg kg 0.636solu K/kg kg 364.0
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
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
2.051
fA 25.082
fB
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
4.052
fA 5.084
fB
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
6.053
fA 75.086
fB
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
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
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
nB 8 2 8
nC 02 0
0
nA 5 5
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
1
nB 8 2 6
nC 02 2
nA 5 4
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
2
nB 8 2 4
nC 02 4
nA 5 3
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
3
nB 8 2 2
nC 02 6
nA 5 2
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
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
Reaction Stoichiometry• Acrylonitrile produced by reaction of ammonia,
propylene, and O2 at 30% conversion of 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
Reaction Stoichiometry• Acrylonitrile produced by reaction of ammonia,
propylene, and O2 at 30% conversion of limiting reactant:
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
Reaction Stoichiometry• Acrylonitrile produced by reaction of ammonia,
propylene, and O2 at 30% conversion of limiting reactant:
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
Reaction Stoichiometry• Acrylonitrile produced by reaction of ammonia,
propylene, and O2 at 30% conversion of limiting reactant:
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•
• Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00
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
• Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00
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
Equilibrium Composition
• Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00
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
Equilibrium Composition
• Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O K(1105 K) = 1.00
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
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
889.010100
160Y12B
CAB2A
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
1610
160S C/B
CAB2A
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, 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
Balances on Reactive Processes• Continuous, steady-state dehydrogenation of ethane• First consider molecular balances:
C2H6 balance: input = output + consumption
100 kmol C2H6/min = n1 + (C2H6 consumed)
24262 HHCHC
Balances on Reactive Processes• Continuous, steady-state dehydrogenation of ethane• First consider molecular balances:
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
Molecular Species Balances
H2 Balance: generation = output
2H H kmol 40gen2
at steady state24262 HHCHC
Molecular Species Balances
C2H6 Balance: input = output + consumption
min
HC kmol1
H kmol 1HC kmol 1
minH kmol
1minHC kmol
62
2
62262
60n
40n 1000
at steady state24262 HHCHC
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
Atomic Species Balance
• 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
Atomic Species Balance
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
Atomic Species Balance
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
Atomic Species Balance
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
Incomplete Combustion of CH4
• 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
Incomplete Combustion of CH4
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
Incomplete Combustion of CH4
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
Incomplete Combustion of CH4
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
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
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
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
Mixingpoint
ndf = 4 unknowns (n9, n10, n1, n2) – 2 balances (C3H8 and C3H6) = 2
ndf = 2
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
reactor
ndf = 5 unknowns (n3, n4, n5, n1, n2) – 2 balances (C and H) = 3
ndf = 2 ndf = 3
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
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
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
overall
836 HC mol 5mol 10095.01n
conversionrelationship
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
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
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
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
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
separator
6310710
83336
HC mol 75.4nn0500.0nHC mol 900nn00555.0n
given relations
637 HC mol 95n 836 HC mol 5n
28 H mol 95n
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
separator
n3 n6 n9 n9 895 mol C3H8
propane balance
n10 4.75 mol C3H6
833 HC mol 900n
637 HC mol 95n 836 HC mol 5n
28 H mol 95n
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
mixer
100n9 n1 n1 995 mol C3H8
propane balance
n10 4.75 mol C3H6
n9 895 mol C3H8
833 HC mol 900n
637 HC mol 95n 836 HC mol 5n
28 H mol 95n
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
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
637 HC mol 95n 836 HC mol 5n
28 H mol 95n
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
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
637 HC mol 95n 836 HC mol 5n
28 H mol 95n
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
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
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
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
Catalytic Propane Dehydrogenation
C3H8 C3H6 H2
95% overallconversion
recycleratio
637 HC mol 95n 836 HC mol 5n
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
Combustion of Ethane
C2H6 72
O2 2CO2 3H2O
C2H6 52
O2 2CO3H2O
fC2H6 = 0.9
25% of the ethane burned forms CO
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
Combustion of Ethane
C2H6 72
O2 2CO2 3H2O
C2H6 52
O2 2CO3H2O
fC2H6 = 0.9
25% of the ethane burned forms CO
ethaneconversion
specification
62621 HC mol 0.10HC mol 10090.01n
n0 2500 mol air
Combustion of Ethane
C2H6 72
O2 2CO2 3H2O
C2H6 52
O2 2CO3H2O
fC2H6 = 0.9
25% of the ethane burned forms CO
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
Combustion of Ethane
C2H6 72
O2 2CO2 3H2O
C2H6 52
O2 2CO3H2O
fC2H6 = 0.9
25% of the ethane burned forms CO
nitrogenbalance
23 N mol 1975ira mol 250079.0n
n0 2500 mol air
n1 10.0 mol C2H6
n4 45.0 mol CO
Combustion of Ethane
C2H6 72
O2 2CO2 3H2O
C2H6 52
O2 2CO3H2O
fC2H6 = 0.9
25% of the ethane burned forms CO
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
Combustion of Ethane
C2H6 72
O2 2CO2 3H2O
C2H6 52
O2 2CO3H2O
fC2H6 = 0.9
25% of the ethane burned forms CO
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
Combustion of Ethane
C2H6 72
O2 2CO2 3H2O
C2H6 52
O2 2CO3H2O
fC2H6 = 0.9
25% of the ethane burned forms CO
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
Combustion of Ethane
C2H6 72
O2 2CO2 3H2O
C2H6 52
O2 2CO3H2O
fC2H6 = 0.9
25% of the ethane burned forms CO
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