Material Balances for Non-reactive Process

Introduction

Elementary Principles of Chemical Processes,

by Richard M. Felder and Ronald W. Rousseau

by

MOHD FADHIL MAJNIS

Department of Chemical

& Petroleum Engineering

CHAPTER 3

Learning Outcome

Define the law of material balance and the general equation

of material balance.

Briefly and clearly explain in your own words the meaning

of the following terms:

Batch, semi-batch, continuous, transient and steady state

processes.

By the given process description, able to:

Draw and label the full flowchart

Choose a convenient basis of calculation

Definition “Materials Balance analysis (MB)

is a systematic reconstruction of the way in which a chemical

element, a compound or material passes through a natural

cycle and/or its economical benefits. An analysis of the

material flow, usually is based on the origin of a physical

balance.”

Open vs. closed systems System:

Arbitrary portion/whole process that is considered for analysis

Limits of the system can be drawn as the system boundary

Boiling system

Overall system

Reactor system

Distillation system

Distilla

tion

A, B

C

Re

acto

r

A

B

A + B C

Chemical process system

System boundary

Open vs. closed systems

Closed system Material neither enters nor leaves i.e. no material crosses the system boundary Changes take place inside the system No mass exchange occurs with the surroundings

Open system Material crosses the system boundary

1000 kg H2O

System boundary

Closed system

1000 kg H2O

Open system

100 kg/min 100 kg/min

Process classification

• Feed is charge to the process and product is removed when the process is completed

• No mass is fed or removed from the process during the operation

• Used for small scale production

• Operate in unsteady state

• Example: rice cooking, polymerization

Batch

• Input and output flow continuously through the duration of the process

• Operate in steady state

• Used for large scale production

• Example: hydro power generation, distillation

Continuous

• Neither batch nor continuous

• During the process a part of reactant can be fed or a part of product can be removed.

Semibatch

Distillation

Batch

Continuous

Semibatch

Some daily operations

Processes Batch Continuous Semi-batch

Cookies baking √

Cloth washing √

Water storage √

Process of growing √

The flowing river √

Water filling into bottle √

Toilet flushing √

Balloon filling √

Water boiling in open flask √

By nature, batch & semi-batch processes are unsteady-state operations (why?)

Continuous processes may be either:

Steady-state, e.g. hydroelectricity

Transient, e.g. process start-up, shut down

Type of process operation

Steady state

- All the variables (i.e. temperatures, pressure, volume, flow rate, etc) DO NOT change with time

- Minor fluctuation can be acceptable

Unsteady state or transient

- Process variable DO change with time

because it would not be

possible to maintain all

process variables

constant with time

Steady-state vs. transient System

boundary

1000 kg H2O

Transient

100 kg/min 50 kg/min

1000 kg H2O

Steady state

100 kg/min 100 kg/min

Back to daily operations

Processes Steady-state Transient

Cookies baking (b) √

Clothes washing (b) √

Water storage (b) √ √

Process of growing (c) √

The flowing river (c) √ √

Water filling into bottle (sb) √

Toilet flushing (sb) √

Balloon filling (sb) √

Water boiling in open flask (sb) √

Test Yourself

Define type and operation of process given below

A balloon is filled with air at steady rate of 2 g/min

A bottle of milk is taken from the refrigerator and left on the kitchen

Water is boiled in open flask

Answer

Semibatch and unsteady state

Batch and unsteady state

Semibatch and unsteady state

Balances Law of Conservation of Mass states:

Mass can neither be created nor destroyed.

“So, we should be able to account for all the mass entering and leaving a

given system.” This is known as a material balance or mass

balance.

Material Balance -- A quantitative description of all materials that

enters, leaves, and accumulates in a system.

In its simplest form for balance equation:

Input

(enter

through

system

boundaries

+ Generation

(produced

within the

system)

-

Output

(leaves

through

system

boundaries

Consumption

(consumed

within system)

Accumulation

(buildup within

system) - =

input generation output consumption accumulation

Differential balances • balances that indicate what is happening in a system at an instant

time. • balance equation is a rate (rate of input, rate of generation, etc.)

and has units of the balanced quantity unit divided by a time unit (people/yr, g SO2/s).

• usually applied to a continuous process.

Integral balances • Balances that describe what happens between two instants of time. • balance equation is an amount of the balanced quantity and has the

corresponding unit (people, g SO2). • usually applied to a batch process, with the two instants of time

being the moment after the input takes place and the moment before the product is withdrawn.

Differential & Integral Balances

Simplified Rule for Material Balance

If the balanced quantity is TOTAL MASS, set generation = 0 and consumption = 0. Mass can neither be created nor destroyed.

If the balanced substances is a NONREACTIVE SPECIES (neither a reactant nor a product), set generation = 0 and consumption = 0.

If a system is at STEADY STATE, set accumulation = 0, regardless of what is being balanced.

Balances on Continuous Steady State Process

Steady state: accumulation = 0

IF balance on NONREACTIVE species or total mass; gen. = 0,

cons. = 0, balance equation become

input generation output consumption 0

input output

Integral Balances on Batch Process Ammonia is produced from nitrogen and hydrogen in a batch reactor. At time t =

0 there are n0 mol of NH3 in the reactor, and at a later time tf the reaction terminates and the contents of the reactor, which include nfammonia, are withdrawn. Between t0 and tf no ammonia enters or leaves through the reactor boundaries.

From GMBE: (no component enters or leaves the reactor during operation, input=0; output=0)

Generation – Consumption = Accumulation

For batch reactor:

Accumulation = Initial input – Final output

Final GMBE for batch process

Initial input + Generation = Final output – Consumption

Solving Material Balance

In material balances problems, equations involving

input and output stream variables are derived.

Solving the equations is simple, but deriving them

from a description of process is considered tricky.

Hence, procedure to derive a set of equations that

can be solved for unknown variables will be

outlined.

Strategy in Solving Material Balances

1 • Choose as basis of calculation

2

• Draw a flowchart and fill in all known variables values, including the basis of calculation. Then label unknown stream variables on the chart.

3 • Express what the problem statement asks you to determine in terms of the labeled variables.

4

• 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.

5 • Do the degree-of-freedom analysis.

6

• Solve the equations (if DOF=0).

• Calculate the quantities requested in the problem statement if they have not already been calculated.

7

• If a stream quantity or flow rate ng was given in the problem statement and another value nc was either chosen as a basis or calculated for this stream, scale the balanced process by the ratio ng/nc to obtain the final result.

Basis of Calculation

Balanced process can always be scaled.

Mean that material balance calculation can be performed on the basis of

any convenient set of stream amount or flow rate and the results can

afterward be scaled to any desired extent.

A basis of calculation is an amount (mass or moles) OR flow rate (mass or

molar) of one stream or stream component in a process. All unknown

variables are determined to be consistent with the basis.

if the amount or flow rate of a stream is given – use it as a basis for

calculation

If NO stream amount or flow rate are known, choose an arbitrary convenient

value ( i.e. 100 kg, 100 kmol/h) on the stream with KNOWN composition. If

mass fraction is known, choose total mass or mass flow rate as basis. If mole

fraction is known, choose a total moles or molar flow rate as basis

Strategy in Solving Material Balances

1 • Choose as basis of calculation

2

• Draw a flowchart and fill in all known variables values, including the basis of calculation. Then label unknown stream variables on the chart.

3 • Express what the problem statement asks you to determine in terms of the labeled variables.

4

• 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.

5 • Do the degree-of-freedom analysis.

6

• Solve the equations (if DOF=0).

• Calculate the quantities requested in the problem statement if they have not already been calculated.

7

• If a stream quantity or flow rate ng was given in the problem statement and another value nc was either chosen as a basis or calculated for this stream, scale the balanced process by the ratio ng/nc to obtain the final result.

Flowcharts

A flowchart is drawn using boxes or other symbols to represent the

process units and lines with arrows to represent inputs and outputs

It must be fully labeled with values of known and unknown process

variables at the locations of the streams

Fresh feed

100 mols C3H8

P1 mols C3H8

P2 mols C3H6

P3 mols H2

product

Q1 mols C3H8

Q2 mols C3H6

Q3 mols H2

Qr1 mols C3H8

Qr2 mols C3H6

100 + Qr1 mols C3H8

Qr2 mols C3H6

Reactor separator

Flowcharts Labeling

Write the values and units of all known stream variables at the

locations of the streams on the flowchart.

For example, a stream containing 21 mole% O2 and 79% N2 at

320˚C and 1.4 atm flowing at a rate of 400 mol/h might be

labeled as:

400 mol/h

0.21 mol O2/mol

0.79 mol N2/mol

T = 320˚C, P = 1.4 atm

2 Ways to Label Process Stream

60 kmol N2/min

40 kmol O2/min

0.6 kmol N2/kmol

0.4 kmol O2/kmol

100 kmol/min

3.0 lbm CH4

4.0 lbm C2H4

3.0 lbm C2H6

0.3 lbm CH4/lbm

0.4 lbm C2H4/lbm

0.3 lbm C2H6/lbm

10 lbm

total amount or flow rate of the

stream with the fractions of each

component

amount or flow rate for

each component

Flowcharts Labeling

Assign algebraic symbols with units to unknown stream variables [such

as m (kg solution/min), x (lbm N2/lbm), and n (kmol C3H8)]

mol/h

0.21 mol O2/mol

0.79 mol N2/mol

T = 320˚C, P = 1.4 atm

n

400 mol/h

y mol O2/mol

(1-y) mol N2/mol

T = 320˚C, P = 1.4 atm

Consistent on Symbol Notation !!!

gasin fraction moles y

liquidin moles)or (massfraction component x

rate flow volumeV

volumeV

rate flowmolar n

moles n

rate flow mass m

mass m

Flowcharts Labeling

Try to reduce the number of unknown by using any relationship

information given

If that the mass of stream 1 is half that of stream 2, label the masses

of these streams as m and 2m rather than m1 and m2.

If you know that mass fraction of nitrogen is 3 times than oxygen,

label mass fractions as yg O2/g and 3yg N2/g rather than y1 and y2.

When labeling component mass fraction or mole fraction, the

last fraction must be 1 minus the sum of the others

Balance are not written on volumetric qualities

If volumetric flow rate of a stream is given, you still need to label

the mass or molar flow rate of this stream

Class Exercise

Two methanol-water mixture are contained in

separate flask. The first mixture contains 40wt%

methanol and the second flask contains 70wt%

methanol. If 200g of the first mixture combined with

150g of the second, what are the mass and

composition of the product.

Solution

Mixing

200 g

150 g

m g 0.40 g MeOH/g

0.60 g H2O/g

0.70 g MeOH/g

0.30 g H2O/g

x g MeOH/g

(1-x) g H2O/g

How many unit operation?

How many input stream?

How many outlet stream?

How many component?

Any reaction involve?

Any relationship information given?

What being asking?

etc….

Two methanol-water mixture are

contained in separate flask. The first

mixture contains 40wt% methanol

and the second flask contains 70wt%

methanol. If 200g of the first

mixture combined with 150g of the

second, what are the mass and

composition of the product.

Solution

Balance on TOTAL MASS (Input=Output)

200 g + 150 g = m g

m g= 350 g

Balance on MeOH (Input=Output)

x= 0.529 gMeOH/g

(1-0.529) g H2O/g = 0.471 g H2O/g

200 g 0.40 gMeOH +

150 g 0.70 gMeOH =

350 g x g MeOH

g g g

GMBE for BATCH process and NONREACTIVE;

Initial INPUT = Final OUTPUT

Class Exercise

An experiment on the growth rate of certain organism 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 rate of 20 cm3/min

B: Air (21% O2 and 79% N2)

C: Pure O2 with a molar flow rate one-fifth of the molar flow rate of stream B

The output gas is analyzed and is found to contain 1.5 mole% water.

Draw and label the flowchart of the process.

Solution

Evaporation

20 cm3 H2O (l)/min

mol H2O/min 2n

mol O2/min 1200.0 n

mol air/min 1n

mol/min 3n

0.21 mol O2 /mol

0.79 mol N2 /mol

0.015 mol H2O/mol

y mol O2 /mol

(0.985-y) mol N2/mol

A

B

C

Try This. Distillation column produce ethyl alcohol

Desired product at the top of distillation column

Waste at the bottom of distillation column

Given data

The fresh feed to distillation column is at 1000 kg/hr with 10

wt% of ethyl alcohol and water at 90 wt%

The mass flow rate for top product is at the ratio of 1 of 10

compared with the fresh feed

The composition % of top column is at 60 wt% of ethyl alcohol

Draw and label the flowchart of the process.