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8/10/2019 Cofee Maker Pp t

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Material Balance Calculations

Department of Chemical Engineering

Based on Notes from Prof. M. Ioannidis


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Outline

Single Unit in the Absence of

Chemical Reactions

Multiple Units in the Absence of

Chemical Reactions

Multiple Units in the Presence ofChemical Reactions


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1. Single Unit Analysis

An everyday example

The Process: Brewing Coffee (technical

term: leaching)

The Machinery: Coffeemaker (technical

term: solid-liquid contactor)


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Process Description*

W (water)

C (Brewed Coffee)

G (Dry

Grounds)

F (Filtered

Grounds)

(*) Batch process


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Stream Description

One knows everything about a stream if one

knows: its mass(or mass flow rate forcontinuous processes) andits composition.

n variables needed to describe a stream withn components. What may these variables

be?


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Stream C (two components: WC, GC) needs

2 variables to describe it.

The 2 mass fractions: xWCand xGCare a

poor choice, because xGC+ xWC= 1 (i.e.,

they are notindependent). Thus, at leastonevariable must be a mass (total or component

mass).

Stream Description (contd)


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To describe stream D, one could use either:

the total mass Dand any two mass fractions

(say, xWC

and xGC

), or

the mass of each component: DWC, DGC, or

any other combination, exceptxWCand xGC!

For example, upon findingDWC, DGC, the

mass fractions are also known:

xWC

=DWC

/(DWC

+ DGC

), etc.

Stream Description (contd)


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WF, GF

W

WC, GC

G

Process Description (contd)

Verify that if you knew G, W, E, WC,GC, WF, and GF one knows everything there

is to know about the mass and composition of all four streams.

E


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To determine unknowns one must solve an

equal number of independent equations that

relate them. Around a process unitinvolving ncomponents one can write at

most nindependentmaterial balance

equations:Total balanceplus n-1component balances, or

ncomponent balances.

Material Balance Equations


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Suppose one writes:Water balance, W = WF + WC +E

Coffee Grounds balance, G = GC + GF

then,Total balance (not independent): G + W = E + C + F

To solve this problem, one must have no more than

7 unknowns (because one can write at most 2

independent equations). The remaining 5 variables

must be obtained from data. Data are frequently

interpreted as extra (auxiliary) equations.

Material Balance Equations (contd)


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The following equations are not material balances, they are

auxiliary equations coming from data (as they may be given ina problem statement):

10% of H20 was lost to evaporation:

E = 0.1W,

Coffee contains 1.25% G:

GC/(GC+WC) = 0.0125,

Water to Ground ratio:

100/40 = WF/GF,

20% of the dry coffee is in the cup of coffee:

0.2*G = GC,

Total mass of C

C = GC + WC

Data and Auxiliary Equations


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Thus far, there are 7 independent simultaneous linear

equations in 7 unknowns (AWis directly available):

(1)G+ (0)W+ (0)E

+ (-1)GF

+ (0)WF

+ (-1)GC

+ (0)WC

= 0

(0)G+ (1)W+ (-1)E+ (0)GF+ (-1)WF+ (0)GC+ (-1)WC = 0

(0)G+ (-.01)W+ (1)E+ (0)GF+ (0)WF+ (0)GC+ (0)WC = 0

(0)G+ (0)W+ (0)E+ (0)GF+ (0)WF+ (1)GC+ (0)WC = 2.5

(0)G+ (0)W+ (0)E+ (100)GF+ (-40)WF+ (0)GC+ (0)WC = 0

(0.2)G+ (0)W+ (0)E+ (0)GF+ (0)WF+ (-1)GC+ (0)WC = 0

(0)G+ (0)W+ (0)E+ (0)GF+ (0)WF+ (1)GC+ (1)WC = 200

Linear_Algebra@Work


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Alternatively one writes , [M]{X} = {b}, where

Solution: {X} = M-1{b}

Excel and Mathcad can both solve linear systems easily...

Linear_Algebra@Work (contd)

M

0

1

0

0.01

0

0

0

0

0

1

0.99

0

0

0

0

0

1

0

0.996

0

0

1

0

0

0

0.004

0

0

0

1

0

0

0

0.2

0.196

0

0

1

0

0

0.8

0.196

1

0

0

0

0

0.8

0.804

b

1

0

0

0

0

0

0

X

BCG

BCS

CCS

CW

DCG

DCS

DW


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Analytical clarity

Ability to investigate what-if scenarios

Convenient treatment of processes

involving many streams and many

components

Advantages of Matrix Solution


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2. Multiple Unit Analysis

1 2

Two-distillation columnprocess to separate benzene

(B), toluene (T) and xylene

(X).

First column producesoverhead product containing

mostlyB.

Second column produces

overhead product containingmostlyT and bottom product

containing mostlyX.

All chemicals (B,T,X) are

present in all streams.


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Pretend that you have

no data.

One Methodology

1 2 Then give a unique

name to each stream.

F

A

C

D

E


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Recall that each stream

has 3 components

each stream is fully

described by 3 variables. Lets use component

mass flowsto describe

them.

There is a total of 15

unknowns(remember,

one pretends one has no

data!)

One Methodology (contd)

1 2

FB, FT, FX

AB, AT, AX

CB, CT, CX

DB, DT, DX

EB, ET, EX


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1 2

FB, FT, FX

AB, AT, AX

CB, CT, CX

DB, DT, DX

EB, ET, EX

Recall that around each

unitone can write as many

independent mass balances

as the number ofcomponents involved, that

is 3 balances.

For unit 1:

[1]: FB= AB+ CB(B-balance)

[2]: FT= AT+ CT(T-balance)

[3]: FX= AX+ CX(X-balance)


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1 2

FB, FT, FX

AB, AT, AX

CB, CT, CX

DB, DT, DX

EB, ET, EX

For unit 2:[4]: CB= DB+ EB(B-balance)

[5]: CT= DT+ ET(T-balance)

[6]: CX= DX+ EX(X-balance)

Total of 6 independent

mass balances

Anything more (e.g.,

overall balance for B):FB= AB+ DB+ EB

is redundant (to see this

add equations [1] and [4]!)


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1 2

FB, FT, FX

AB, AT, AX

CB, CT, CX

DB, DT, DX

EB, ET, EX

Observe that without dataone cannot proceed,

because there is 6

equations and 15

unknowns!

Data can be translated into

auxiliary equations;one

needs 9 such equations andone wants them to be

independent!

Suppose they give only the

composition of stream A...


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1 2

FB, FT, FX

AB, AT, AX

(4%B, 91%T, 5%X)

CB, CT, CX

DB, DT, DX

EB, ET, EX

Knowledge of stream Acomposition allows us to write

no more than 2 auxiliary

equations, e.g.,

[7] AB/(AB+AT+AX) = 0.04[8] AT/(AB+AT+AX) = 0.91

The following would not be

independent (why?)

AX/(AB+AT+AX) = 0.05 Generalize: knowing the

composition of a stream of n

components affords us n-1

auxiliary equations.


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1 2

FB

, FT

, FX

(35%B, 50%T, 15%X)

AB, AT, AX

(4%B, 91%T, 5%X)

CB, CT, CX

DB, DT, DX

(4.3%B, 91.2%T, 4.5%X)

EB, ET, EX

Knowledge of the composition ofstreams Fand D would give 4

additional auxiliary equations:

[9] DB/(DB+DT+DX) = 0.043

[10] DT/(DB+DT+DX) = 0.912

[11] FB/(FB+FT+FX) = 0.35

[12] FT/(FB+FT+FX) = 0.50


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1 2

FB

, FT

, FX

(35%B, 50%T, 15%X)

AB, AT, AX

(4%B, 91%T, 5%X)

CB, CT, CX

DB, DT, DX

(4.3%B, 91.2%T, 4.5%X)

EB, ET, EX

A basisprovides one more

auxiliary equation, e.g.:

[13] FB+FT+FX = 100

The last two auxiliary equations

may come from knowing that

stream E contains 10% of B in

the feed and 93.3% of X in thefeed:

[14] EB= 0.1FB

[15] EX = 0.933FX


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Think: Why do one prefer to work with component

mass flows as our stream variables, e.g., CB, CTandCXfor stream C, and not with total mass flow and

mass fractions (e.g., C, xCBand xCT) ?


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1 2

FB

, FT

, FX

(35%B, 50%T, 15%X)

AB, AT, AX

(4%B, 91%T, 5%X)

C, xCT, xCB

DB, DT, DX

(4.3%B, 91.2%T, 4.5%X)

EB, ET, EX

Suppose we used C, xCBand xCTto describe stream C. Then, the

mass balances for, say, unit 2

would be:

CxCB= DB+ EB(for B)

CxCT= DT+ ET(for T)

C(1-xCT-xCB)= DX+ EX(for X)

This formulation would result

in non-linearequations

which are more difficult to

solve!


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3. Single Unit Balance with

Reaction

4NH3+ 5O24NO + 6H2O

A (NH3)

B (Air: O2, N2) C (O2, N2, NO,

NH3, H2O)

NOTE: Since no data are available at this stage, one must

assume that allreactants and products are present in the

effluent stream (8 stream variables)


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If Xis the number of ammonia moles that reacted:

Ammonia: CNH3 = ANH3- X

Nitrogen monoxide: CNO= X

Oxygen: CO2= BO2- (5/4)X

Nitrogen: CN2= BN2

Water: CH2O= (6/4)XTo the 8 stream variables one must add the extent of

reaction, i.e., one has 9 unknowns and 5 equations

from mass balances (one for each chemical species)

Mass balances using the extent of

reaction...


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In addition to the 8 stream variables, the extent of reaction is

a also an unknown, i.e., one has 9 unknowns and only 5

equations from mass balances (one for each of the 5

chemical species). One needs 4 auxiliary equations

before the problem can be solved. These may be:

Onefrom the basis: e.g.,100 mol of NH3 feed Onefrom the composition of stream B (why not two?)

Onefrom knowledge of NH3 fractional conversion

Onefrom knowledge of the percent excess air

Auxiliary equations from data...

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