introduction to material balances

7/24/2019 Introduction to Material Balances

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Chapter 3:

Introduction to Material Balances

Learning Objectives

Upon completing this Chapter, you should be

able to understand what process flow sheet (PFS) or

process flow diagram (PFD) is

know a standard symbol for each of some

important process equipments

draw a simple process flow diagram (or a

block flow diagram or a flow chart) for the

given problem

make a necessary assumption/necessary

assumptions pertaining the given problem

set up an appropriate basis of calculation

(basis) of the given problem


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understand what steady-state process is

and how it affects the establishment/

set-up of material-balance equations understand what overall and species

balances are

establish overall- and species-balance

equations for the given problem

solve simple material-balance problems


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One of the mainresponsibilitiesof

chemical engineersis to create/construct/

analysechemical processes(or, at least,

to understandthe existing processes)

The layoutof a chemical processis

called process flow sheet (PFS) or

process flow diagram (PFD)

PFS or PFD can be for just a single

process unit of for the whole process,

either simple or complicated process


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Examples of PFS or PFD

PFD for an atmospheric

distillation of crude oil

(from Stoichiometry, 4thed; by Bhatt & Vora, 2004)


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PFD for a water-softeningby ion-exchange process

(from Stoichiometry, 4thed; by Bhatt & Vora, 2004)


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PFD for Ammonia Synthesis Plant

(from Introduction to Chemical Processes by Murphy, 2007)


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Normally, a PFS or a PFD comprises

all major process equipments/units

linesenteringor leavingthe

process/unit and/or linesconnecting

two or more process

equipments/units (these lines arecalled streams)

flow rateof each stream

compositionof each stream

operating conditionsof each stream

and/or unit/equipment (e.g., T, P)

energy/heat neededto be added to

and/or removed from any particularpart of the process or the entire

process


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Some important symbols of process

equipments are illustrated as follows

Mixer Packed-bed Batch Reactor

Reactor

Distillation Flash Drum Mixer-SettlerColumn

Pump Blower Turbine

Heat Exchanger Furnace

(from Introduction to Chemical Processes by Murphy, 2007)


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In order to be able to createor to

understandPFSor PFD, the knowledge

concerning MATERIAL & ENERGY

BALANCESis required

As a chemical engineer, you need to be

abletoperformmaterial and energy

balances for any particular process or for

the entire process efficiently/competently

We start our learning by doing MATERIAL

BALANCES, using an underlying knowledge

of law of conservation of mass


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Before starting learning material

balance, lets consider the following bank

account:

Date Description Deposit Withdrawal Balance

2/1/51

8/1/51

15/1/51

19/1/51

22/1/51

22/1/51

31/1/51

Beginning balance

Deposit from Mom

Deposit from Dad

ATM withdrawal

ATM withdrawal

Service charge

Closing balance

10,000

20,000

7,000

18,000

50

15,000

19,950

The data we obtain from the above

bank account are

The sum of deposits = 30,000

The sum of withdrawals = 25,050

The initialbalance = 15,000

Thefinalbalance = 19,950


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From the information in the previous

Page (Page 10), we can write the following

relationship:

The final balance The initial balance =

The sums of deposits The sums of withdrawals

(19,950) (15,000) =

(30,000) (25,050)

If we consider

the initial balance as the initial

mass of the process

the final balance as the final mass of

the process

the sums of deposits as the sums ofmass entering the process

the sums of withdrawals as the sums

of mass leaving the process


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Additionally, if we consider the

differencebetween thefinal balanceand

the initial balance, which is the amount

of money accumulatedin the bank

account as the amount of mass

accumulatedin the process, we can

establish the principle of material/mass

balanceas follows:


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Principles of MATERIAL BALANCE

In the case that there is NO Chemical

Rxn.

Total mass enteringa process/unit

Total mass leaving a process/unit

= Mass accumulation in a process/unit

However, the material balance

problemwill be more complicated(but

not too difficult believe me!) when thereis/are a Rxn./Rxns.in the process/unit,

as follows:

Total mass entering a process/unit

Total mass leaving a process/unit

+ Mass generating from a Rxn/Rxns

= Mass accumulation in a process/unit


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From the principles above, the following

equations can be written:

In the case that there is NO Rxn.

f isys sys sys

in out

m m m m m = =

(3.1)where

isysm = initialmass of a system

fsysm = final mass of a system

In the case that there is/are a

Rxn./Rxns.

.

sys

in out Rxn

m m m m + = (3.2)


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Normally, chemical processes are

continuous, the change in mass should,

therefore, be written in the rate form (i.e.

it changes with time)

Eqs. 3.1 & 3.2 can, then, be re-written,

as follows

sys

in out

dmdm dm

dt dt dt = (3.3)

.

sys

in out Rxn

dmdm dm dm

dt dt dt dt + =

(3.4)


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We can perform material balanceseither

for all species(called overall balance) or

for a specific/selected species(called

species balance)

In this level, we will carry out material

balances for steady-stateprocesses(do

you know what steady-state processes

are?)


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Example One thousand (1,000) kilograms

of a mixture of benzene (B) and toluene (T),

containing 40% by mass of B is to be

separated into two streams in a distillation

column. The top output stream of the

column contains 375 kg of B and the

bottom output stream contains 515 kg

of T.

(a)Perform the mass balance for B & T

(b)

Determine the composition of the

top and bottom streams


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Standard Procedure:

1)Make any necessary assumption(s)

For instance, in this case, we make an

assumptionthat the process is steady-

state

2)Draw a flow chart of the process/unit

Distilla tio n

co lumn

Input stream

1,000 kg mixture

40 wt% B

Top output stream

375 kg B

xkg T

Bottom output stream

ykg B

515 kg T


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3)Set a basis of calculations

In this case, we should set a basisas

1,000 kg of mixture

4)Determine the numbers of unknowns

In this case, there are 3 unknowns:

x = ?

y = ?

% of T of an input stream

5)

Establish material balance equations

In order to be able to solve for

unknowns, it is necessarythat the # of

Eqs.must be equalto the # of unknowns


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In this Example, since the input

stream consists of only 2 components, we

obtain the following equation:

wt% of B wt% of T 100+ =

Accordingly,

% T 100 % B

100 40

% T 60%

=

=

=

Note that one unknown is eliminated

(the # of unknowns are now only 2)

From the basis of calculationwe have set

(in Step 3) and from the percentages ofbenzene and toluene (the percentage of

toluene (%T) has been solved in Step 5), we

obtain the information that the input

streamcomprises:


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Benzene (B) = ( )40

1,000 kg 400 kg100

=

Toluene (T) = ( )60 1,000 kg 600 kg100

=

In a general form, let

Fm = mass of the input stream (feed)

Bm = mass of benzenein the input

stream

Tm = mass of toluenein the input

stream

By = mass fraction of benzenein the

input stream

Ty = mass fraction of toluenein the

input stream


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We can, then, write the following

equations

B B Fm y m=

T T Fm y m=

For the output streams, let

topm = mass of the topstream

bottomm = mass of the bottomstream

Since the process is steady-state(as we

made an assumption in Step 1) and has no

Rxn., the mass balance equation can be

written as follows

0sysin out

in out

m m m

m m

= =

=


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From the process flow chart (on Page

18), we obtain the facts that

F

in

m m=

and that

top bottom

out

m m m= +

Thus,

top bottomFm m m= + (3.5)

Eq. 3.5 is an example of an overall

mass balance equation

Substituting corresponding numerical

values into Eq. 3.5 yields

top bottom1,000 m m= + (3.6)


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In order to solve for 2 unknownsin Eq.

3.6 (i.e. topm & bottomm ), only 1 equationis

NOT enough

We needto have 2 equations; since we

have already got one, we, therefore, needanother equation

To obtain another equation, we need to

do a species balance

In this Example, we shall perform a

benzene balance, as follows

benzene benzenein out

m m

=


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Let

topBm = mass of benzenein the top

outputstream

bottomBm = mass of benzenein the

bottom outputstream

we can, then, write the following equation:

top bottomB B Bm m m= + (3.7)

ortop bottomB F B B

y m m m = + (3.8)

Eqs. 3.7 & 3.8 are the examples of

species balanceequations (in this case, it

is called benzene balance equations)


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Substituting numerical values into Eq.

3.8 gives

( ) ( ) top bottom0.4 1,000 400 B Bm m= = +

(3.9)

It is given that

topBm = 375 kg

Hence, from Eq. 3.9, we obtain

bottom

bottom

400 375

400 375

25 kg

B

B

m

m

y

= +

=

=


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We can also perform a toluene balance,

as follows

top bottomT T Tm m m= + (3.10)

top bottomT F T T x m m m = + (3.11)

It is given that

bottomTm = 515 kg

Thus, from Eq. 3.10, we obtain

top

top

600 515

600 515

85 kg

T

T

m

m

x

= +

=

=


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We can summarise our calculations as

illustrated in the following Table

SpeciesInput

[kg]

Output [kg]

Top Bottom

Benzene

Toluene

400

600

375

85

25

515

TOTAL 1,000 460 540


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Mole Balance

Performing mole balance is similartothat of the mass balance, but overall

mole balances are applicable ()

onlyfor theprocessesthat have no Rxns

Thus, in the case that there is NO Rxn.:

Overall balance

f isys sys sys

in out

n n n n n = =

(3.12)

Species balance

sys sys sys f ij j j j j

in out

n n n n n = =

(3.13)


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where

isysn = initialtotal # of moles of

all speciesin the system

fsysn = finaltotal # of moles of

all speciesin the system

sysijn = # of moles of species jin the

system at the initialstate

sysfjn = # of moles of species jin the

system at thefinalstate

Eqs. 3.12 & 3.13 can also be written in

the rate form (i.e.with respect to time),

as follows

Overall balance

sys

in out

dndn dn

dt dt dt

= (3.14)


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Species balance

sysjj j

in out

dndn dn

dt dt dt =

(3.15)

In the case that there is a Rxn/are

Rxns, mole balanceis still applicable,

but only for species balances(NOT for

an overall balance), as follows

.sysj j j j

in out Rxn

n n n n + = (3.16)

or, in the rate form

.

sysjj j j

in out Rxn

dndn dn dn

dt dt dt dt + =

(3.17)


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Example An experiment on the growth rate of

organisms requires an environment of humid

air enriched in oxygen

Three input streams are fed into an

evaporation chamber:

liquid water, fed at the volumetric flow

rate of 20 cm3/min

air(21.0 vol% O2& 79 vol% N2)

pure O2, with the molar flow rate of

one-fifth (1/5) of the stream of air

to produce an output stream with the desired

composition

The output gas is analysed and is found to

contain 1.5 mol% of water

Calculate all unknowns

(Data: Density of water is 1.0 g/cm3; and

MW of water = 18.02, of O2= 32.00, & of

N2= 28.02)


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Assumption: Steady-state process

Flow chart:

Since the process is steady-state, (as we

made an assumption),

0sysdn

dt =

Eq. 3.14:

sys

in out

dndn dn

dt dt dt = (3.14)

Evaporation

chamber

Air

21 vol% O2

79 % N2

Pure O2

1

3

2

4

H2O (l)

20 cm3/min

Output stream

1.5 mol% H2O

x % N2

y % O2


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is then reduced to

0in out

dn dn

dt dt

=

orin out

dn dn

dt dt = (3.18)

From the flow chart (see Page 33), we

can write an overallmole balance equation,

as follows

2 31 4dn dndn dn

dt dt dt dt

+ + =

(3.19)

Let i idn

n

dt

=

Thus, Eq. 3.19 can be re-written as

follows

[ ]1 2 3 4n n n n + + = (3.20)


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It is given that

2 1 1

10.2

5

n n n= = (3.21)

Combining Eq. 3.21 with Eq. 3.20 gives

( )[ ]1 1 3 40.2n n n n + + =

and

1 3 41.2n n n+ = (3.22)

It is also given that the volumetricflow

rate of stream 3 = 20 cm3/min

Thus, the massflow rate of stream 3 can

be calculated as follows

3

3

cm g g20 1.0 20

min cm min

=


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which can be converted to the molarflow

rate as follows

20 g 1 g-mol= 1.11 g-mol/min

min 18.02 g

Thus,

3n = 1.11 g-mol/min

Performing species balances:

H2O balance:

( )23 H O 44

n x n= (3.23)

where

( )2H O 4x = mole fraction of H2O in

stream 4


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Substituting corresponding numerical

values into Eq. 3.23:

( )23 H O 44n x n= (3.23)

results in

4

4

1.51.11

10074.0 g-mol/min

n

n

=

=

Substituting 3n & 4n into Eq. 3.22:

1 3 41.2n n n+ = (3.22)

gives

1 3 4

1

1.2

1.2 1.11 74.0

n n n

n

+ =

+ =

1 60.7 g-mol/minn =


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Thus, the flow rate of stream 2 can be

computed, using Eq. 3.21, as follows

( )

2 115

160.7

5

n n=

=

2 12.1 g-mol/minn =

N2balance:

( ) ( )

( ) ( )

2 2N 1 N 41 4

79 60.7 g-mol/min 74.0 g-mol/min100 100

64.8 mol%

x n x n

x

x

=

=

=

(note that, for gases, vol% = mol%)

In other words, mol% of N2in the output

stream (stream 4) is 64.8%


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Thus, mol% of O2in the output stream

(stream 4) or the value of ycan be

computed as follows

2 2 2mol% O 100 mol% H O mol% N

100 1.5 64.8

33.7 %mol

y

=

=

=


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Finally, we can summarise our

calculations in the following Table:

INPUT OUTPUT

Stream 1*

1 60.7n =

Stream 2*

2 12.1n =

Stream 3*

3 1.11n =

Stream 4*

4 74.0n =

Composition

(mol% orvol%)

Composition

(mol% orvol%)

Composition

(mol% orvol%)

Composition

(mol% orvol%)

O2 21O2 100 H2O 100

H2O 1.5

N2 79 O2 33.7

N2 64.8

* Flow rateof each stream is in the

unitof g-mol/min

introduction to material balances

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