mass balance non reactive 1

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Mass Balance



ECB 3013 MATERIAL AND ENERGY

BALANCES

The law of conservation of mass states that for any processunit (s), mass can neither be created nor destroyed but it

can change from one phase to another phase or converted

into other forms through reaction process.

The law concludes that:

Total mass INPUT = Total mass OUTPUT

FUNDAMENTAL OF MATERIAL BALANCES:

CONSERVATION OF MASS



PROCESS CLASIFICATION

Chemical process may be classified as:

1. Batch process: The feed is charged into a process unit at the

beginning of the process and the output is removed from the process

unit at the end of the process.

2. Continuous process: The inputs and outputs flow continuouslythroughout the duration of the process.

3. Semi batch: Any process that is neither batch nor continuous.

If the process variables such as flow rates, temperatures, pressures,

volume etc do not change with time, the process is said at STEADYSTATE. On the other hand, if the process variables do change with

time, the process is said at UNSTEDY STATE or TRANSIENT. By

nature, batch and semi-batch process are unsteady state operation,

while continuous process is steady state operation.



EXAMPLE 1

Process Classification Remark

A balloon is filled with air

at a steady state rate of 2

g/min

Water is boiled in an

open flaskWater is boiled in an

closed flask

Gasoline from car tank

Carbon dioxide andsteam are fed into

reactor to form carbon

dioxide and hydrogen



EXAMPLE 1

Process Classification Remark

A balloon is filled with air

at a steady state rate of 2

g/min

Semi-batch, transient Volume, T, P change

with time.

Water is boiled in an

open flask

Semi-batch, transient Volume, T, P change

with time.Water is boiled in an

closed flask

Batch, transient T & P change with

time.

Gasoline from car tank Semi-batch, transient Volume change with

time

Carbon dioxide and

steam are fed into

reactor to form carbon

dioxide and hydrogen

Continuous, steady state All process variables

do change with time.



MASS BALANCE EQUATION

Generally, the mass balance equation for any process unit is

given as:

Input + Generation - Output - Consumption = Accumulation

Enters

through

system

boundaries

Produced

within

system only

for reactive

system

Leaves

through

system

boundaries

Consumed

within

system only

for reactive

system

Buildup within

system only for

transient operation

[1]



Each year 50,000 people move into a city, 75,000 people move

out, 22,000 are born and 19,000 die. Write a balance on the

population of the city.

Solution:

Let P denote to people:

Input + generation – output – consumption = accumulation

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

A = -22,000 P/yr

Each year the city’s population decreases by 22,000 people.

EXAMPLE 2



1) If the balanced quantity is total mass, set generation = 0 and

consumption = 0. Except in nuclear reactions, mass can

neither be created nor destroyed.

2) If the balanced substance is a nonreactive species (neither a

reactant nor a product), set generation = 0 and consumption

= 0.

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

regardless of what is being balanced. By definition, in

steady-state system nothing can change with time, including

the amount of the balanced quantity.

Rules to simplify the material balance equation




Balance on Continuous Steady-State Processes

For continuous steady-state operation for reactive system, the

accumulation term is equal to zero. Hence, EQ [1] becomes,

Input + Generation - Output - Consumption = 0 [2]

For continuous steady-state operation for non reactive system,

EQ [2] is simplified as,

Input = Output [3]

See example 4.2-2 in Text book




Balances on Batch Processes

For batch processes, the input and output in Eq [1] is zero, and

the equation is simplified as,

For batch processes, accumulation is defined as,

Final Output - Initial Input = Accumulation [5]

[4]Generation - Consumption = Accumulation

Hence by equating Eq [4] and [5], for batch processes, the mass balance equation is given as,

Initial Input + Generation = Final Output + Consumption [6]

See example 4.2-3 in Text book




BALANCESSteps in Solving Material Balance

1. Read and try to understand on the process description. What type of processunit used and what type of process operation.

2. Draw a flowchart for the process description using boxes or other symbols to

represent process unit or unit operation (reactors, mixers, etc.) and lines with

arrows to represent inputs and outputs.

MASS BALANCE CALCULATION

DISTILLATION

COLUMN



3. Write all known stream variables i.e. inputs & outputs, on the flowchart.

4. Assign algebraic symbols to unknown stream variables.(e.g., m1 , n1 , yCO etc.)

5. If you are given mixed mass and mole units for a stream (such as a total

mass flow rate and component mole fractions or vice versa), convert all

quantities to one basis or the other.


DISTILLATION

COLUMN

0.55 kg T/kg

0.05 mol T/mol

mT3 kg T/kg

2000 L/h

0.45 kg B/kg

0.95 mol B/mol

mB3 kg B/kg 8% of B in feed

m 2 kg/h

m 1 kg/h



6. Do the degree of freedom analysis. Count unknowns and identifyequations that relate them (zero degree of freedom). The equation:material balances, an energy balance, process specifications, physical

property relationships and laws, physical constraints and stoichiometricrelations.

7. Take basis of calculation. If no stream amount or flow rate is specified inthe problem statement, take as basis an arbitrary amount or flow rate ofthe stream with known composition (e.g., 100 kg or 100 kg/h or 100 molor 100 mol/h if all mass/mole fractions are known.)

8. Write mass balance equation for the overall system and for specificcomponent using selected Eq [1] to [6].

9. Perform mass balance for the process description. Always check theoverall mass balance for Total Inputs = Total Outputs




A liquid mixture of benzene (B) and toluene (T) containing

55% B by mass is fed continuously to a distillation column

with a feed rate of 100 kg/h. A product stream leaving the top

of the column (overhead product) contains 85% B and a

bottom product stream contains 10.6% B by mass. Determinethe mass flow rate of the overhead product stream and the

mass flow rate of the bottom product stream.

PROBLEM 4.3



DISTILLATION

COLUMN

0.450 kg T/kg

0.150 kg T/kg

0.894 kg T/kg

100 kg/h

0.550 kg B/kg

0.850 kg B/kg

0.106 kg B/kg

m V kg/h

m L kg/h

STEPS 2, 3 & 4

EXAMPLE 3



STEPS 5-DoF

DoF = unknown - independent equation

Unknown = 2

Independent equation = 2

Note: DoF must be zero to be solvable

mV

m L &

Material Balance for B and T

DoF = unknown - independent equation

= 2 - 2

= 0 (problem solvable)

EXAMPLE 3



Take the basis of calculation = 100 kg/h of feed

Since this operation is at steady state and non-reactive

system, hence

Input = output

DC

0.450 kg T/kg

0.150 kg T/kg

0.894 kg T/kg

100 kg/h

0.550 kg B/kg

0.850 kg B/kg

0.106 kg B/kg

m V kg/h

m L kg/h

Total balance:

100 kg/h = m V m

L + A

Benzene balance:

100 (0.550)

h

Bkg

= 0.850 m V

0.106 m L

B+

STEPS 6 & 7

EXAMPLE 3



Solve equation A and B simultaneously

The results are

m V

m L

= 59.7 kg/h

= 40.3 kg/h

EXAMPLE 3




BALANCES• Use to represent the process unit (reactors, mixers,

separation units)

• Lines with arrow - to represent inputs and outputs

Flowcharts

Flowchart scaling and basis of calculation

• Scaling up – final stream quantities are larger than the

original quantities

• Scaling down – final stream quantities are smaller than

the original quantities

• Basis of calculation – an amount (mass or moles) or

flow rate (mass or moles) of one stream



Balancing a process

For non reactive process:

1. Maximum no of independent equations equals the number of

chemical species in the input and output

Ex: input has B and T,

• Independent equations involve mass or mole balance on Band T

• Balance on total mass or moles

2. Write balances first that involve the fewest unknown variables



21

A stream containing four components (14 % w

/w A,36.1% w/w B, 23.6% w/w C and 26.3% w/w D) flows

at a rate of 984.0 kg/hr into a separator. The

separator produces two streams of differing

compositions. The upper product stream has a

composition of 16.5 % w/w A and 40.9 % w/w B with

C and D making up the remainder. The lower

product stream contains all four components but

only the weight percentages of B (20.9% w/w) and

C (37.6% w/w) are known. Draw and label theprocess and calculate the compositions and flow

rates of the two product streams.

PROBLEMS



Degree of freedom

•

A way to identify that enough information isavailable for material balance calculation

• ndf = nunknowns – nindep eqns

– ndf =0, can be solved

– ndf >0, relations has been overlooked, problem is

underspecified

– ndf <0, flowchart is incompletely labeled, problem

is overspecified with redundant and inconsistentrelations



Degree of freedom

•

Sources of relating unknown variables:1. Material balance

2. Energy balance

3. Process specification –

related4. Physical properties and laws

5. Physical constraint

6. Stoichiometric relations



PROBLEMS

See problems 4.6, 4.7, 4.9, 4.10




BALANCESPROBLEM 4.20 OF TEXT BOOK

Wet air containing 4.0 mole% water vapor is passedthrough an adsorption column containing calcium chloride

pellets. The pellets adsorb 97.0% of the water and none

of the other constituents of the air. The column packing

was initially dry and had a mass of 3.40 kg. Following 5.0

hours of operation, the pellets are reweigh and found to

have a mass of 3.54 kg. Calculate the molar flow rate

(mol/h) of the feed gas and mole fraction of water vapor in

the product gas.



26

)/(1 hmol n

)/(2 hmol n

x mol H2O / mol(1 - x ) mol DA / mol

)/( 23 hadsorbed O H mol n

0.040 mol H2O / mol

0.960 mol DA / mol

97.0% of H2O in feed

Adsorption Unit



27

Need H2O

Adsorption Rate

Problem Statement: xnnn ,,, 321

4 unknowns :3 independent equations

Degree of Freedom:

DoF = Unks. - IE = 4 – 3

DoF = + 1



28

Overall mass balance

INPUT=OUTPUT + ACCUMULATION

321 nnn 1

Component mass balances

321040.0 nn xn H2O :

21 )1(960.0 n xn -DA :

2

3

Water in CaCl2 pellets



29

Take the basis of calculation= 100 mol/hr of feed stream

mol/h12.96

[3]insubsituteand0.12/x,

/88.3497.0

ascalculated becanthatknow

)1(96

4

100

2

2

3

3

2

32

2

32

321

-

n

n Hence

hmol n

nwe

n x

DA

nn x

O H

nn

nnn

[1]

[2]

[3]



30

Stream INPUT OUTPUT

H2O

(mol/h)

DA

(mol/h)

H2O

(mol/h)

DA

(mol/h)

n1 4 96 - -

n2 - - 0.12 96

n3 - - 3.88 -

TOTAL 4 96 4 96

Take the basis of calculation= 100 mol/hr of feed stream



31

Stream INPUT OUTPUT TOTALH2O

(mol/h)

DA

(mol/h)

H2O

(mol/h)

DA

(mol/h)

n1 1.6 38.4 - - 40

n2 - - 0.048 38.4 38.448

n3 - - 1.552 - 1.552

TOTAL 1.6 38.4 1.6 38.4

AdsorptionRatehO H mol O H kg

O H mol

h

kg n /56.10180.0

1

5

)40.354.3(2

2

23

-

Based on the given problem,

Scale down factor = 1.56/3.88 = 0.40. hence the new INPUT/OUTPUT values

will be multiplied by a factor of 0.4 to meet the process description.



32

Adsorption

RatehO H mol

O H kg

O H mol

h

kg n /56.1

0180.0

1

5

)40.354.3(2

2

23

-

hmol nnnn /54.3856.11.402321 - Total balance :

Given: 97% H2O (of the input stream) is adsorbed. Therefore

hmol nn /1.4056.1)040.0(97.0 11

H2O balance : x = 1.25×10-3

Note : DA balance may also be used to calculate x.

Take basis calculation of 0.14 kg of water vapor adsorbed in 5 hour operation



33

Gas absorption of gas scrubbing is a commonly usedmethod for removing environmentally undesired species

from waste gases in chemical manufacturing and

combustion processes. A waste gas containing 10 mol%

SO2 (a precursor of acid rain) and several other species

(collectively as A) is fed to scrubbing tower where it

contacts a solvent (B) that absorbs SO2. The ratio of

liquid stream to gas stream fed to the column is 5. The

solvent feed rate to the tower is 1000 mol/min. The

absorption unit has an efficiency of 98%. Assume that theabsorption of A and evaporation of B in the scrubbing can

be neglected, calculate the mole fraction of SO2 in the

liquid and gas streams.

LOOK LIKE PROBLEM 4.26 OF TEXT BOOK



34

10 mol% SO2

90 mol% A

G1=200 mol/min

y1 mol% SO2y2 mol% A

G2

1000 mol/min B

L1=1000 mol/min

L2

x1 mol% SO2

x2 mol% S

Gasabsorption

unit



35

Degree of Freedom

Problem Statement:G2, L2, x1, x2, y1 and y2 = 6 unknown

4 independent equations, 1 overall, 3 specific equations

Degree of Freedom:

DoF = Unks – IE= 6-4 =+2

The remaining 2 independent equations can be obtained from the column

efficiency.



36

Overall Balance

G1 + L1 = G2 + L2

200 + 1000 = G2 +

L2

1200= G2 + L2

Solvent Balance

1000 = (1-x1)L2

L2=1000 +

(0.98)20

L2=1019.6mol/min

Hence, x1=0.019

Gas A Balance

0.9(200) = (1-y1)G2

180 = (1-y1)G2

G2=Gas A +

Remaining SO2 ingas A

G2=180 + (0.02)20

=180.4 mol/min

Hence, y1=2.21x10-3

Take basis of calculation L1= 1000 mol/min solvent

SO2 Balance

0.1(200) =

y1G2+x1L2

20 = y1G2 + x1L2

mass balance non reactive 1

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