90229250 4 material balances

7/27/2019 90229250 4 Material Balances

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Fundamentals of

Material Balances

Ch E 201

Material and Energy Balances


2/214

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.


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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.


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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 developmentmathematical 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.


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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?


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The General Balance Equation

Consider the following continuous process unit forwhich 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).


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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


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Each year, 50,000 people move into a city; 75,000move out; 22,000 are born; 19,000 die. Perform a

balance on the population of the city (system).


input: 50,000 people/year

generation: 22,000 people/year

ouput: 75,000 people/year consumption: 19,000 people/year

accumulation: unknown


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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.


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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


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Continuous steady-state system

input + 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


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Benzene/Toluene distillation continuous process

steady-state operation

no reactions occurring

General species balance


input = output

0 0 0


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input = output Benzene balance

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

m2 = 50 kg B/h

Toluene balance

500 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


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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 nfmoles of NH3 in the reactor

between 0 and tf, no NH3 crosses system boundary

NH3 accumulation in system from 0 to tfis nf 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


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Batch Mixing Process Balance

Two methanol-watermixtures are contained

in flasks of amounts and

concentrations shown.

If the flasks are mixed, what is the mass andconcentration of the resulting product?

no reactions, generation = consumption = 0

input = output


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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?


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differential air balance

min

kmol111.0n

min

kmoln

kmol

iramolk900.0

min

iramolk100.0

outputinput


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integral hexane balance

min6880t

t111.0100.0nt100.0output

HCkmol45.76nonaccumulati

kg2.86

kmol1

m

L10

L

kg659.0m0.10nonaccumulati

outputonaccumulati

nconsumptiooutputgenerationinputonaccumulati

1

1minkmol

1

146

3

33


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Labeling Process Flowcharts

1. Write the values of all known stream variables onthe 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/h

0.21 mol O2/mol

0.79 mol N2/mol

320C, 1.4 atm

n (mol/h)

0.21 mol O2/mol

0.79 mol N2/mol

320C, 1.4 atm

. 400 mol/hy (mol O2/mol)

(1-y) (mol N2/mol)

320C, 1.4 atm


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Air Humidification and Oxygenation

An experiment on the growth rate of certainorganisms 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/min

B. 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.


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An experiment on the growth rate of certainorganisms 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/min

B. 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.


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Calculate n2 from volumetric flowrate and density:

minOHmol11.1

OgH02.18mol1

cmOgH00.1

minOHcm0.20n 2

2

322

3

2


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water balance:

min

mol1.74n

mol

OHmol015.0

min

moln

min

OmolH11.1

mol

OHmol015.0

min

moln

min

OmolHn

32

32

23

22


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Flowchart Scaling

Scaling theflowchart involves

changing all values

of stream flows by a

proportionalamount.

Note that mass (or

mole) fractions are

not scaled, but

remain unchanged.


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Flowchart Scaling

A 60/40 mixture (molar)of A and B is separated

batchwise into 2 fractions.

Scale the flowchart to a

continuous 1250 lbmol/hr feed rate.

bottom:h

lbmolA156

molA

hrlbmolA5.12molA12.5

mol

hrlbmol5.12

mol100

hlbmol1250=factorscale


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Flowchart Scaling





bottom:h

lbmolB469

molB

hrlbmolB5.12molB37.5

mol

hrlbmol5.12

mol100

hlbmol1250=factorscale


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Flowchart Scaling






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Basis of calculation

Since a flowchart can always be scaled, materialbalance 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 abasis of calculation; all unknown quantities are then

determined to be consistent with this basis.


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Balancing a Process

Suppose 3.0 kg/min of benzene and 1.0 kg/min oftoluene are mixed.

There are 2 unknown

quantities in this

process, mdot and x, thus 2 equations are needed tosolve 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).


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Balancing a Process

Balances:

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


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Balancing nonreactive processes

The maximum number of independent equationsthat 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 threebalance equations are independent, thus only two

unknowns can be found from these balances.

Write balances first that involve the fewest unknownvariables.


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Balances on a mixing unit

An aqueous solution of sodium hydroxide contains20.0% 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).


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


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

V1/100 (liters H2O/kg feed solution)

m2/100 (kg product solution/kg feed solution)


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3. Count unknown variables and equations. If thesequantities 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.


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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


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6. Total mass balance (INPUT = OUTPUT):

100 kg + m1 = m2 =250 kg

m1 = 150 kg

= 250 kg

= 150 kg


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7. Diluent water volume:

V1 = m1/ H2O = 150 kg / (1 kg/L)

V1 = 150 L

= 250 kg

= 150 kg

= 150 L


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Degree of Freedom Analysis

Process used to determine if a material balanceproblems has sufficient specifications to be solved.

a) draw and completely label the flowchart

b) count the unknown variables on the chart

c) count the independent equations relating these variablesd) calculate degrees of freedom by subtracting (b) from (c)

ndf= nunknowns nindep_eqns


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ndf= nunknowns nindep_eqns

If ndf = 0, problem can be solved (in principle).

if ndf > 0, problem is underspecified and at least ndfadditional 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.


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A stream of humid air enters a condenser in which95% 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.

f d l


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6 unknowns 3 material balances (1 each for O2, N2, H2O)

condensate volumetric to molar flow relation (MW and )

process specification: 95% of the water is condensed

ndf= 6 (3 + 1 + 1) = 1

Underspecified

cannot solve

f d l i


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5 unknowns 3 material balances (1 each for O2, N2, H2O)

condensate volumetric to molar flow relation (MW and )

process specification: 95% of the water is condensed

ndf= 5 (3 + 1 + 1) = 0

Solvable

f d l i


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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

kg100.18

OHolm1

L

OHkg

h

OHL

2

nnnn

nny;nny;nny

nn100.0n

n79.0900.0n

n21.0900.0n

n100.095.0n

00.1225n

222

3

2)l(2)l(2

G l P d Si l U i O


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General Procedure Single Unit Op

1. Choose as a basis of calculation an amount or flowrate 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 aknown composition.

G l P d Si l U i O


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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 components

b. total moles (or flow), and mole fractions of all stream components

c. 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

G l P d Si l U i O


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3. Express what the problem statement ask you to doin 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.


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Di till ti C l l


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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

Di till ti C l l


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Ex. 4.3-53. write expressions for quantities requested in problem statement

BT33BB x1x;mmx

3T3B3 mmm

312 mmm

Di till ti C l l


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Ex. 4.3-54. Convert mixed units in overhead product stream

kg

Tkg

2T

kg

Bkg

2B

Tkmol

Tkg

Bkmol

Bkg

058.0942.01y

942.0mixturekg7881Bkg7420y

mixturekg7881Tkg461Bkg7420

Tkg46113.92Tkmol0.5

Bkg742011.78Bkmol0.95

Di till ti C l l


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Ex. 4.3-55. Perform degree of freedom analysis

=0.942

=0.058

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

Di till ti C l l


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Ex. 4.3-56. Write system equations

7. Solve

h

kg

L

kg

hL

1 1744872.02000m i. volumetric flow conversion

h

Bkg

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

h

Bkg

2

3B2B21

766m

mymm45.0

iii. benzene balance

h

Tkg

3T

3T2B21

915m

my1mm55.0

iv. toluene balance

h

kg

h

kg

3T3B21

17441744

mmmm

iv. total mass balance (check)



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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

h

kg

h

Tkg

h

Bkg

3 9789158.62m

kg

Tkg

3T

kg

Bkg

h

kg

h

Bkg

33B3B

936.0064.01y

064.09788.62mmy

Balances on Multiple Unit Ops


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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.


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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)



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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


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Two-Unit Process Example

Variables for Streams 1, 2, and 3 are unknown



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Variables for Streams 1, 2, and 3 are unknown

Label unknown stream variables



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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)


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Extraction Distillation Process


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Extraction-Distillation Process

Simultaneously solve total

mass and acetone balances to

determine m1 and m3.

Solve MIBK balance to

determine xM1.

Extraction Distillation Process


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Solve acetone, MIBK, and

water balances to determine

mA4, mM4, and mW4.



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For either (just 1) extractorunit, solve acetone, MIBK, and

water balances to determine

mA2, mM2, and mW2.



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

underspecified



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

underspecified

Recycle


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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).


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Balances on an Air Conditioner


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Overall system

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



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Mixer




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Cooler




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Splitter


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


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overall mole balance

water balance

solved simultaneously:

mol290n;mol5.392n

n023.0n017.0n04.0

nnn

52

251

251

Reasons to recycle


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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


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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)



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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



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Feed/recycle mixer:

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

underspecified

544 mm95.0m



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Evaporator:

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

underspecified

544 mm95.0m



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apo at e C ysta at o ocess

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



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p y

Overall system:

K2CrO4 balance

water balance

total mass balance

specification

544

542h

kg

52h

Kkg

54h

Kkg

mm95.0m

mmm4500

m636.0m4500667.0

m364.0m4500333.0

solve simultaneously for m4 and m5

h

crystalsKkg

4 1470m

soluW/kgkg0.636soluK/kgkg364.0

5.77mh

solutionkg

5



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p y

Overall system:

K2CrO4 balance

water balance

total mass balance

specification

solve for m2 with knowns m4 and m5

hOHkg

222950m

h

crystalsKkg

4 1470m


5.77mh

solutionkg

5

544

542h

kg

52h

Kkg

54h

Kkg

mm95.0m

mmm4500

m636.0m4500667.0

m364.0m4500333.0



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p y

Overall system:

K2CrO4 balance

water balance

total mass balance

specification

only 3 equations are independent

hOHkg

222950m

h

crystalsKkg

4 1470m


5.77mh

solutionkg

5

544

542h

kg

52h

Kkg

54h

Kkg

mm95.0m

mmm4500

m636.0m4500667.0

m364.0m4500333.0



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p y

Crystallizer:

total mass balance

water balance

6h

kg

3

653

6h

kg

3

6543

m257.14.97m

m636.0m636.0m506.0

m5.771470m

mmmm

hOHkg

222950m

solve simultaneously for m3 and m6

h

kg

3 7200m hcrystalsKkg

4 1470m


5.77mh

solutionkg

5

h

kg

6 5650m



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p y

feed/recycle mixer:

total mass balance

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

h

kg

116h

kg 10150mmm4500

h

kg

1 10150m

hOHkg

222950m

h

kg

3 7200m

h

kg

6 5650m

h

crystalsKkg

4 1470m


5.77mh

solutionkg

5



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p y

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?


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Balances on Reactive Systems


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y

Material balance no longer takes the form

INPUT = OUTPUT

Must account for the disappearance of reactants and

appearance of products through stoichiometry.

Stoichiometric Equations


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q

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 numberof atoms of each atomic species is balanced.

2 S 2 S

4 O + 2 O 6 O



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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 speciesparticipating 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



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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?

2

2

22 Olbmol600

COlbmol4

Olbmol6COlbmol400

min

OHlbmol200

HClbmol1

OHlbmol450.0

min

HClbmol100 2

84

284


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Limiting and Excess Reactants


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Stoichiometric Proportion Reactants are present in

a ratio equivalent to the ratio of the stoichiometric

coefficients.

A + 2B 2C



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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.



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fractional excess (fXS) ratio of the excess to the

stoichiometric proportion.

A + 2B 2C

25.04

45

n

nnf

stoichA

stoichAfeedA

XS



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fractional conversion (f) ratio of the amount of a

reactant reacted, to the amount fed.

A + 2B 2C

fedA

reactedA

n

nf

0.05

0fA 0.0

8

0fB



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A + 2B 2C

fedA

reactedA

n

nf

2.05

1fA 25.0

8

2fB



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A + 2B 2C

fedA

reactedA

n

nf

4.05

2fA 5.0

8

4fB


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A + 2B 2C

fedA

reactedA

n

nf

8.05

4fA 0.1

8

8fB

Extent of Reaction


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extent of reaction () an extensive quantity

describing the progress of a chemical reaction .

stoichiometric coefficients: A = -1, B = -2, C = 2

A + 2B 2C

ni ni0 i

nA nA0 nB nB0 2 nC nC0 2

0

Extent of Reaction


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A + 2B 2C

ni ni0 i

nB 82 8 nC 02 0

0

nA 5 5

Extent of Reaction


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A + 2B 2C

ni ni0 i

1

nB 82 6 nC 02 2nA 5 4

Extent of Reaction


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A + 2B 2C

ni ni0 i

2

nB 82 4 nC 02 4nA 5 3


120/214

Extent of Reaction


121/214




A + 2B 2C

ni ni0 i

4

nB 82 0 nC 02 8nA 5 1

2C2H4 O2 2C2H4O


122/214

Assume an equimolar reactant feed of 100 kmol:

What is the limiting reactant?

2 4 2 2 4

ethylene

A reactant is limiting if it is present in less than stoichiometric proportion

relative to every other reactant.

1

1

n

n

1

2

n

n

feedO

HC

stoichO

HC

2

42

2

42

2C2H4 O2 2C2H4O


123/214


What is the percentage excess of each reactant?

2 4 2 2 4

%10000.1

50

50100

n

nn

fstoichO

stoichOfeedO

O,XS

2

22

2

2C2H4 O2 2C2H4O


124/214


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?

2 4 2 2 4

50

21000

nn424242 HC

o

HCHC

50n

501100n

nn

2

2

222

O

O

O

o

OO

100n

5020n

nn

OHC

OHC

OHC

o

OHCOHC

42

42

424242

2C2H4 O2 2C2H4O


125/214


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 ?

2 4 2 2 4

5.0nn

ffedHC

reactedHC

42

42

25

210050100

nn424242 HC

o

HCHC

75n

251100n

nn

2

2

222

O

O

O

o

OO

50n

2520n

nn

OHC

OHC

OHC

o

OHCOHC

42

42

424242

5.0n

nnf

o

HC

HC

o

HC

42

4242

50n

5.0100

n100

42

42

HC

HC

2C2H4 O2 2C2H4O


126/214


If the reaction proceeds to a point where 60 mol of O2 is left,

what is the fractional conversion of C2H4? What is ?

2 4 2 2 4

8.010020100

n

n

ffedHC

reactedHC

42

42

20n

402100n

nn

42

42

424242

HC

HC

HC

o

HCHC

40

110060

nn222 O

o

OO

Reaction Stoichiometry


127/214

Acrylonitrile produced by reaction of ammonia,

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

C3H

6NH

3 3

2O

3 C

3H

3N 3H

2O

nNH3

nC3H

6

0 0.120100 0.100100 1.20

nNH3

nC3H

6

stoich

1 1 1

nO2 nC3H6 0 0.780 0.21100 0.100100 1.64nO

2nC

3H

6

stoich

1.5 1 1.5

limiting

determine limiting reactant



128/214



C3H

6NH

3 3

2O

3 C

3H

3N 3H

2O

nNH3 stoich 10.0 mol C3H61 mol NH3

1 mol C3H6

10.0 mol NH3

fXS

NH3

NH3

0 NH3

stoich

NH3 stoich

12.010.010.0

0.20

nO2

stoich

10.0 mol C3H

6

1.5 mol O2

1 mol C3H6

15.0 mol O

2

fXS O2

O2

0 O2

stoich

O2 stoich

16.415.015.0

0.093

fXS 0.20

fXS 0.093

determine fractional excesses

limiting



129/214



C3H

6NH

3 3

2O

3 C

3H

3N 3H

2O

n

C3H6 1

f

n

C3H6

0

1

0.30

10.0 mol C

3

H6

7.0 mol C

3

H6

fXS 0.093

fXS 0.20

use fractional conversion to

determine amount of

propylene that leaves the

reactor

limiting



130/214

ni ni0 i



C3H

6NH

3 3

2O

3 C

3H

3N 3H

2O

nC3H6

7.0 mol C3

H6

fXS 0.093

fXS 0.20

nC

3H

6 n

C3H

6

0

1 7.0 mol 10.0 mol

3 mol

3 moldetermine extent of

reaction by applying molebalance to propylene

limiting



131/214



OHNHCONHHC23322

3

3633

90.330

6.613010078.079.0

30.310

9.11310078.021.0

93110012.0

2

2

33

2

3

2

3

OH

N

NHC

O

NH

n

n

n

n

n

fXS 0.093

fXS 0.20limiting

nC3H6

7.0 mol C3

H6

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


132/214

Given

a set of reactive species, and

reaction conditions

Determine

1. the final (equilibrium) composition of the reactionmixture

2. how long the system takes to reach a specified state

short of equilibrium

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



133/214

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.



134/214

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.


135/214

Equilibrium Composition


136/214

Water-gas shift reaction: Assume 1 mole CO and 2 mole H2O

K(1105 K) = 1.00

OHCO

HCO

2

22

yy

yy

TK

3nnnnn1nn

1nn

21nn

11nn

222

22

22

22

HCOOHCOtotal

0HH

0COCO

0OHOH

0COCO

i0ii nntotal

ii

n

ny

g2g2g2g HCOOHCO

Equilibrium Composition

1


137/214


K(1105 K) = 1.00

OHCO

HCO

2

22

yy

yy

TK

i0ii nntotal

ii

n

ny

3n

n

n

2n

1n

total

H

CO

OH

CO

2

2

2

121

mol667.0

22

21

22

2

667.0

667.0333.1

333.0

g2g2g2g HCOOHCO

Equilibrium Composition11103/3330


138/214


K(1105 K) = 1.00

3n

222.03/667.0y

222.03/667.0y444.03/333.1y

111.03/333.0y

total

H

CO

OH

CO

2

2

2

121

mol667.0

22

21

22

2

OHCO

HCO

2

22

yy

yy

TK

i0ii nntotal

ii

n

ny

g2g2g2g HCOOHCO

Equilibrium Composition11103/3330y


139/214


K(1105 K) = 1.00

limiting reactant is CO

i0ii nn

0.667mol

mol333.0

667.011nCO

at equilibrium,

fractional conversion at equilibrium

667.0f00.1

333.000.1

g2g2g2g HCOOHCO

3n

222.03/667.0y

222.03/667.0y444.03/333.1y

111.03/333.0y

total

H

CO

OH

CO

2

2

2

Multiple Reactions

1


140/214

C2H

4 1

2O

2 C

2H

4O

C2H4 3O2 2CO2 2H2O

j

jij0iinnforjreactions ofi species,

mole balance becomes

10OHCOHC

2121

0OO

210HCHC

1nn

3nn

11nn

4242

22

4242

20OHOH20COCO

2nn

2nn

22

22

Multiple Reactions

1


141/214

C2H

4 1

2O

2 C

2H

4O

C2H4 3O2 2CO2 2H2O

reactionssidenowithconversion100%atformedmoles

formedproductdesiredmolesyield

formedproductundesiredmoles

formedproductdesiredmolesyselectivit

j

jij0iinnforjreactions ofi species,

mole balance becomes

Multiple Reactions


142/214

100 moles A fed to a batch reactor

product composition: 10 mol A, 160 B, 10 C

What is

1. fA?

2. YB?

3. SB/C?

4. 1, 2

CA

B2A

9.0

100

10100fA

Multiple Reactions


143/214



What is

1. fA?

2. YB?

3. SB/C?

4. 1, 2

889.0

10100

160Y

1

2B

CA

B2A

Multiple Reactions


144/214



What is

1. fA?

2. YB?

3. SB/C?

4. 1, 2

16

10

160S C/B

CA

B2A

Multiple Reactions


145/214



What is

1. fA?

2. YB?

3. SB/C?

4. 1, 2 1090

10010

nn

12

21

22A11AAoA

80

20160

nn

1

1

11BBoB

CA

B2A

Balances on Reactive Processes


146/214

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



147/214


First consider molecular balances:

Molecular H2 balance: generation = output

generation H2 = 40 kmol H2/min

24262 HHCHC



148/214



C2H6 balance: input = output + consumption

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

24262 HHCHC



149/214



C2H4 balance: generation = output

(C2H4 generated) = n2

24262 HHCHC



150/214


Atomic C balance: input = output

Atomic H balance: input = output

24262 HHCHC

426262 HCmol1Cmol2

2HCmol1Cmol2

1HCmol1Cmol2

62 nnHCmol100

4262262 HCmol1Hmol4

2HCmol1Hmol6

1Hmol1Hmol2

HCmol1Hmol6

62 nn40HCmol100

Independent Equations


151/214

To understand the number ofindependent species

balances in a reacting system requires anunderstanding ofindependent algebraic equations.

Algebraic equations are independent if you cannot

obtain any of them by adding/subtracting multiplesof the others.

2][12y6x3

[1]4y2x

[5]6zy4

[4]2zx2

[3]4y2x

3[1] = [2] 2[3] [4] = [5]

Independent Equations


152/214

To understand the number ofindependent species

balances in a reacting system requires anunderstanding ofindependent algebraic equations.

Algebraic equations are independent if you cannot

obtain any of them by adding/subtracting multiplesof the others.

2][12y6x3

[1]4y2x

1212

12y6y612

12y6y243

Independent Species


153/214

If two molecular or atomic species are in the same

ratio to each other where ever they appear in aprocess and this ratio is incorporated in the

flowchart labeling, balances on those species will not

be independent equations.

31

31

n76.3n76.3

nn

Independent Chemical Reactions


154/214

When using molecular species balances or extents of

reaction to analyze a reactive system, the degree offreedom analysis must account for the number of

independent chemical reactions among the species

entering and leaving the system.

Independent Chemical Reactions


155/214

Chemical reactions are independent if the

stoichiometric equation of any one of them cannotbe 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


156/214

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 chemicalequilibrium problems.

Molecular Species Balances


157/214

To use molecular species balances to analyze a

reactive system, the balances must containgeneration 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

24262 HHCHC


158/214

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 state

Molecular Species Balances24262 HHCHC


159/214

H2 Balance: generation = output

2H Hkmol40gen 2

at steady state



160/214

C2H6 Balance: input = output + consumption

min

HCkmol

1

Hkmol1

HCkmol1

min

Hkmol

1min

HCkmol

62

2

62262

60n

40n1000

at steady state



161/214

C2H4 Balance: generation = output

min

HCkmol

2

Hkmol1

HCkmol1

min

Hkmol

2

42

2

422

40n

40n

at steady state


162/214


163/214

Atomic Species Balance24262 HHCHC


164/214

C Balance: input = output

21

HCkmol1Ckmol2

2HCkmol1Ckmol2

1HCkmol1Ckmol2

min

HCkmol

nnmolk100

nn100426262

42



165/214

H Balance: input = output

21

HCkmol1Hkmol4

2HCkmol1Hkmol6

1

Hkmol1Hkmol2

min

Hkmol

HCkmol1Hkmol6

min

HCkmol

n4n6+molk80molk600

nn

40100

4262

2

2

62

42



166/214

Solve simultaneously

min/HCkmol40nmin/HCkmol60n

n4n6+molk80molk600:H

nnmolk100:C

422

621

21

21

Extent of Reaction

d


167/214

The 3rd method by which to determine molar flows

in a reactive system is using expressions for eachspecies flow rate in terms ofextents of reaction().

Degree-of-freedom analysis for such an approach:

ndf = # of unknown labeled variables

+ # independent reactions

- # independent nonreactive species

- # other relationships or specifications

j

jij0iinn

Incomplete Combustion of CH4

h b d h d


168/214

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 3

2

O2 CO2 H2O

CH4 2 O2 CO2 2 H2O



169/214

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 3

2

O2 CO2 H2O

CH4 2 O2 CO2 2 H2O

fCH4

= 0.9



170/214

ndf= 5 unknowns+ 2 independent reactions

- 5 expressions for (CH4, O2, CO, CO2, H2O)

- 1 nonreactive species balance (N2)

- 1 specified methane conversion

=0

CH4 3

2O2 CO2 H2O

CH4 2 O2 CO2 2 H2O

fCH4

= 0.9



171/214

N2 balance: nonreactive species, INPUT = OUTPUT

CH4 conversion specification:

CH4 3

2O2 CO2 H2O

CH4 2 O2 CO2 2 H2O

2molNmol

N Nmol8.72mol100728.0n2

2

4molCHmol

CH CHmol780.0mol1000780.0900.01n4

4

fCH4

= 0.9



172/214

Extent of reaction mole balances:

CH4 3

2O2 CO2 H2O

CH4 2 O2 CO2 2 H2O

21230OO

210OHOH

20COCO

10COCO

210CHCH

2nn

22nn

1nn

1nn

11nn

22

22

22

44

0.78 7.80 1 2nCO 1nCO2 8nCO 2nH

2O 21 22

nO2 19.4 3

21 22

fCH4

= 0.9

Product Separation and Recycle

T d fi iti f t t i d i


173/214

Two definitions ofreactant conversion are used in

the analysis of chemical reactors with productseparation and recycle of unconsumed reactants.

reactortoinput

reactorfromoutput-reactortoinputreactant

conversion

passglesin

processtoinput

processfromoutput-processtoinputreactantconversion

overall



174/214

reactortoinput

reactorfromoutput-reactortoinputreactant

conversion

passglesin

processtoinput

processfromoutput-processtoinputreactantconversion

overall



175/214

%75%100A/minmol100

A/minmol25-A/minmol100

conversion

passglesin

%100%100A/minmol75

0-A/minmol75

conversion

overall

Catalytic Propane Dehydrogenation95% overall

conversion


176/214

C3H8 C3H6 H2

conversion


conversion


177/214

C3H8 C3H6 H2

conversion

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


178/214


conversionn

df= 3


179/214

C3H8 C3H6 H2

conversion

reactor

ndf= 5 unknowns (n3, n4, n5, n1, n2)

2 balances (C and H)

= 3

ndf

= 2


conversionn

df= 3 ndf= 0


180/214

C3H8 C3H6 H2

conversion

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


conversion


181/214

C3H8 C3H6 H2

conversion

overall

836 HCmol5mol10095.01n

conversionrelationship


conversion


182/214

C3H8 C3H6 H2

conversion

overall

637

HCmol1Cmol3

7HCmol1Cmol3

83HCmol1Cmol3

HCmol95n

nHCmol5mol100638383

C atomic balance

836 HCmol5n


183/214


conversion


184/214

C3H8 C3H6 H2

conversion

separator

6310710

83336

HCmol75.4nn0500.0n

HCmol900nn00555.0n

given relations

637 HCmol95n 836 HCmol5n

28 Hmol95n


conversion


185/214

C3H8 C3H6 H2

conversion

separator

n3 n6 n9 n9 895 mol C3H8

propane balance

n10 4.75 mol C3H6

833 HCmol900n


28 Hmol95n


conversion


186/214

C3H8 C3H6 H2

conversion

mixer

100n9 n1 n1 995 mol C3H8

propane balance

n10 4.75 mol C3H6

n9 895 mol C3H8

833 HCmol900n


28 Hmol95n


conversion


187/214

C3H8 C3H6 H2

conversion

mixer

n10 n2 n2 4.75 mol C3H6

propylene balance

n10 4.75 mol C3H6

n9 895 mol C3H8

n1 995 mol C3H8 833HCmol900n


28 Hmol95n


conversion


188/214

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

HCmol1 Cmol34HCmol1 Cmol383

HCmol1Cmol3

63HCmol1Cmol3

83

HCmol75.99n

nHCmol009

HCmol.754HCmol959

6383

6383

833 HCmol900n


28 Hmol95n


conversion


189/214

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

Hmol1 Hmol25HCmol1 Hmol663HCmol1 Hmol883

HCmol1Hmol6

63HCmol1Hmol8

83

Hmol95n

nHCmol75.99HCmol009

HCmol.754HCmol959

26383

6383

833 HCmol900n

634 HCmol75.99n 637 HCmol95n 836 HCmol5n

28 Hmol95n


conversion


190/214

C3H8 C3H6 H2

single-pass

conversion

n10 4.75 mol C3H6

n9 895 mol C3H8

n1 995 mol C3H8

n2 4.75 mol C3H6

%55.9%100HCmol959HCmol009HCmol959

f 83

8383

passsingle

833 HCmol900n

634 HCmol75.99n

25 Hmol95n


28 Hmol95n


conversionfsinglepass 9.55%


191/214

C3H8 C3H6 H2

recycle

ratio


28 Hmol95n

833 HCmol900n

n10 4.75 mol C3H6

n9 895 mol C3H8

n1 995 mol C3H8

n2 4.75 mol C3H6

feedfreshmolrecyclemol109

0.9mol001

mol.754mol958

feedmol001

nn

R

634 HCmol75.99n

25 Hmol95n

Purging

Necessary with recycle to prevent accumulation of a


192/214

Necessary with recycle to prevent accumulation of a

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

CO2 3H2 CH3OHH2O

mixed fresh feedand recycle is a

convenient

basis selection

fsinglepass 60%

Methanol Synthesis

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


193/214

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

- 5 independent species balances = 3

CO2 3H2 CH3OHH2O

fsinglepass 60%

Methanol Synthesis

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


194/214

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

4 independent species balances

1 single pass conversion = 0

CO2 3H2 CH3OHH2O

fsinglepass 60%

Methanol Synthesis

ndf = 3 unknowns (n5, x5C, X5H)


195/214

ndf 3 unknowns (n5, x5C, X5H)


= 0

CO2 3H2 CH3OHH2O

fsinglepass 60%

Methanol Synthesis

ndf = 3 unknowns (n0, x0C, nr)


196/214

df ( 0, 0C, r)


= 0

CO2 3H2 CH3OHH2O

fsinglepass 60%

Methanol Synthesis

ndf = 1 unknowns (np)investigate

l b l d


197/214

df ( p)

1 independent species balance

= 0

CO2 3H2 CH3OHH2O

fsinglepass 60%

mole balances and

their solution inthe text

Combustion Reactions

Combustion - rapid reaction of a fuel with oxygen.


198/214

p yg

Valuable class of reactions due to the tremendousamount of heat liberated, subsequently used to

produce steam used to drive turbines which

generates most of the worlds 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


199/214

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

S forms SO2

N forms NO2 (above 1800C)

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


Stack (flue) gas product gas that leaves a furnace.


200/214

(f ) g p g

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


Stack gas contains (mol) on a dry basis:


201/214

g ( ) y

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

gaswetlbmol

gasrydlbmol

gaswetlbmol

OHlbmol9300.00700.0 2

gasdrylbmol

OHlbmol 20753.0

xH2O 7.53

107.5 0.0700

xN2 65

107.5 0.605

xCO 2 14

107.5 0.130

xO2 10107.5 0.0930

xCO 11

107.5 0.102

Theoretical and Excess Air

The less expensive reactant is commonly fed in


202/214

p y

excess of stoichiometric ratio relative to the morevaluable 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 oxygen is the exact amount of O2 needed


203/214

2

to completely combust the fuel to CO2 and H2O. Theoretical airis that amount of air that contains the

amount of theoretical oxygen.

Excess airis the amount by which the air fed to the

reactor exceeds the theoretical air.

% excess air =moles air fed - moles air theoretical

moles air theoretical

100%


C4H10 13

2O2 4CO2 5H2O


204/214

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

4 10 2 2 2 2

hr

Omol650

HCmol

Omol5.6

hr

HCmol100n 2

104

2104

ltheoreticaO

2

hr

airmol3094

Omol

airmol.764

hr

Omol506n

2

2

ltheoreticaair

% excess air

50003094

3094100% 61.6%

Combustion Reactors

Procedure for writing/solving material balances for a


205/214

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

outlet stream (the stack gas) includes

a. 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 air3. If multiple reactions, use atomic balances

Combustion of EthaneC2H6

7

2O2 2CO2 3H2O

C2H6 5

2O2 2CO3H2O

degree-of-freedom


206/214

fC2H6 = 0.9

25% of the ethane burned forms CO

degree-of-freedom

analysis

ndf= 7 unknowns

- 3 atomic balances

- 1 nitrogen balance

- 1 excess air specification

- 1 ethane conversion specification- 1 CO/CO2 ratio specification

= 0


207/214


7

2O2 2CO2 3H2O

C2H6 5

2O2 2CO3H2O

ethane


208/214

fC2H6 = 0.9


ethane

conversionspecification

62621 HCmol0.10HCmol10090.01n

n0 2500 mol air


7

2O2 2CO2 3H2O

C2H6 5

2O2 2CO3H2O

CO/CO2 ration1 10.0 mol C2H6


209/214

fC2H6 = 0.9


CO/CO2 ratio

specification

OCmol0.45reactHCmol1

genCOmol2HCmol1009.025.0n

62

624

n0 2500 mol air


7

2O2 2CO2 3H2O

C2H6 5

2O2 2CO3H2O

nitrogenn1 10.0 mol C2H6


210/214

fC2H6 = 0.9


nitrogen

balance

23 Nmol1975iramol250079.0n

n0 2500 mol air

n4 45.0 mol CO


7

2O2 2CO2 3H2O

C2H6 5

2O2 2CO3H2O

atomic Cn1 10.0 mol C2H6


211/214

fC2H6 = 0.9


atomic C

balance

25

COmol1Cmol1

5COmol1Cmol1

4HCmol1Cmol2

1HCmol1Cmol2

62

COmol135n

nnnHCmol10026262

n0 2500 mol air

n4 45.0 mol CO

n3 1975 mol N2


7

2O2 2CO2 3H2O

C2H6 5

2O2 2CO3H2O

atomic Hn1 10.0 mol C2H6


212/214

fC2H6 = 0.9


atomic H

balance

OHmol270n

nHCmol10HCmol100

26

OHmol1Hmol2

6HCmol1Hmol6

62HCmol1Hmol6

62 26262

n0 2500 mol air

n4 45.0 mol CO

n3 1975 mol N2

n5 135 mol CO2


7

2O2 2CO2 3H2O

C2H6 5

2O2 2CO3H2O

atomic On1 10.0 mol C2H6


213/214

fC2H6 = 0.9


atomic O

balance

22

OHmol1Omol1

2COmol1Omol2

2

COmol1Omol1

Omol1Omol2

2Omol1Omol2

2

Omol232n

OHmol702COmol135

COmol54nOmol255

22

22

n0 2500 mol air

n4 45.0 mol CO

n3 1975 mol N2

n5 135 mol CO2

n6 270 mol H2O


7

2O2 2CO2 3H2O

C2H6 5

2O2 2CO3H2O

stack gasn1 10.0 mol C2H6

n2 232 mol O2


214/214

fC2H6 = 0.9


stack gas

composition(dry basis)

sum10232197445135 2396

y1 10 2396 0.00417 mol C2H6 mol

y2 232 2396 0.0970 mol O2 mol

1974 2396 0 824 l N l

n0 2500 mol air

n4 45.0 mol CO

n3 1975 mol N2

n5 135 mol CO2

n6 270 mol H2O

270 mol H2O

2396 mol dry stack gas

0 113mol H2O

90229250 4 material balances

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