Introducing Chemical Engineering

8/13/2019 Introducing Chemical Engineering

http://slidepdf.com/reader/full/introducing-chemical-engineering 1/80

Chapter 1

INTRODUCTION



1

INTRODUCTION

1. Introducing Chemical Engineering

Chemical Engineering covers a wide scope that is difficult to put in one definition.

However, a good definition is that chemical engineering is concerned with the design,

construction, and operation of chemical process plants.

In general, chemical plants perform the following :

Basically, there are 3 distinctive functions for a chemical plant.

- Crude Oil - Distillation of Steam reforming

dehydration crude oil of N.G → H2, CO

desalting

- Dehydration and - Separation of

The operation in the

chemical plant may involve

physical steps (separation)

and/or chemical steps (reactions)

Products (o u t p u t)

RawMaterials

Ener

( i n u t )

Purification of

raw materials

Separation into

products

Chemical

conversion

1 2 (3)



2

sweetening of C1, C2, C3 and C4 - NH 3 synthesis

natural gas

- Fresh water from - Hydrotreating and

saline sources hydrocraking in

petroleum refining

Notice the difference between a physical separation process and a chemical

conversion step.

(Input) (Output) (Reactant) (Product)

2. Classification of Processes

Processes are classified according to different classifications as follows:

2.1. Physical Versus Chemical Processes

Physical processes include mixing, heat exchange (heating or cooling), condensation,

evaporation, separation and purification processes such as distillation, extraction, and

absorption. In these processes, there is no molecular change and thus no new species

formed or consumed. Chemical processes, on the other hand, involve chemical

reactions in which atoms are rearranged or redistributed to form new molecules or

products and some reactants are partially or totally consumed.

As shown in the diagram below, process industries usually consist of chemical

processes as the heart of the plant preceded and followed by physical processes :

M.S.F. NH 3

Synthesis(Reactor)

NH 3 N2 + H 2

FreshWater

Brine

Sea

Water



3

RawMaterial

2.2. Continuous Versus Batch Processes

Batch process - The feed is charged into the system at the beginning and

removed at one at the end .

Continuous processes - Inputs and outputs flow continuously throughout the

duration of the process.

Semi-batch processes - Also referred to as semi-continuous process, is any

process that is neither batch nor continuous.

Physical processtypically mixingor heat exchange

Chemical process

Physical processtypically heat exchange

or separationProducts



6

3.3. Functions of Chemical Engineers

(with the help of others)

Develop, design andengineer the complete

process (process design &the equipment)

Choose the properraw materials

Operate the plant efficiently,

safely and economically

Assure that the productsare in specifications



Chapter 2

DIMENSIONS AND UNITS



1

DIMENSIONS AND UNITS

Dimension is a property that can be measured, e.g. length (L), time (t), Mass (M), or

temperature (T). These are the basic dimensions relevant to chemical engineering.From these, other dimensions can be calculated by multiplying or dividing the basic

dimensions, such as velocity (L/t), volume (L 3), or density (M/L 3).

Units are specific values of dimensions that have been defined by convention, custom,

or law, such as grams for mass, seconds for time, and centimeters for length.

Treat the units as ALGEBRAIC symbols, i.e. you can add, subtract, or equate like

units but not unlike units.

3 kg + 5 kg = 8 kg 3x + 5x = 8x

15 cm - 12 cm = 3 cm 15x - 12x = 3x But

3 kg + 4 seconds = ? meaningless like 3x + 4y = ?

On the other hand, numerical values and their corresponding units may always be

combined by multiplication or division:

hkm

50h2km100

=

km360h4xh

km90 =

)quantityensionless(dim6xkm2km12

3 ft x 4 ft = 12 ft 2



2

1. System of Units

There are three systems of units. These are the CGS, SI, and the American

Engineering Systems of units. Each system of units has the following components

relevant to chemical engineering:

• Base units - units of length, mass, time, and temperature• Multiple units - multiples or fractions, e.g. cm, m, km.• Derived units - obtained by multiplying or dividing base units such as ft 3 and

3mkg

or defined equivalents, e.g. N = kg. 2s

m and 1 lb f = 32.2 2m

s

ft.lb

Table (1) shows a comparison of the various base and some selected derived units

used in the above mentioned systems of units. Table (2) gives more comprehensive

details.

Table (1)

Base Units

Quantity CGS SI American Eng.

Length centimeter (cm) meter (m) foot (ft)

Mass gram (g) kilogram

(kg)

pound mass (lb m)

Time seconds (s) seconds (s) seconds (s)

Temperature Kelvin (K) Kelvin (K) Rankine ( R )

Derived Units

Quantity CGS SI American Eng.

Volume liter ( l or 1) liter ( l or 1) cubic feet (ft 3)

Force dyne newton (N) pound force (lb f )

Pressure pascal (Pa) lb f /in2 (psi)

Energy, work erg gram-calorie joule (J)

Power watt (W) watt (W) horsepower (hp)



3

2. Force and Weight

From Newton’s law, F = ma (where F is force, m mass, and a is acceleration).

Units of force 2s

mkg in SI units, 2s

cm.g in CGS units and 2m

s

ft.lb in

American Eng.

For simplicity, use derived force unit.

1 newton (N) ≡ 1 2

s

mkg

1 dyne ≡ 1 2s

cmg

In American Engineering, the derived force unit lb f (pound-force) is defined as the

product of a unit mass (l lb m) and the acceleration of gravity at sea level and 40 o

latitude which is 32.174 2s

ft .

1 lb f ≡ 32.174 lb m 2s

ft.

Since both units of mass and force in American Engineering system are called pounds,there is some confusion about this. So, always distinguish them in your mind, lb m ≡/

lb f .

Force conversion factor, g c = N

s/m.kg1

2

or =dyne

s/mcg1

2

or =f

.2m

lbs/ft.lb

174.32



4

Thus, the equation for force in defined units is:

cg

maF =

The weight of an object is the force exerted on the object by gravitational attraction.

Weight,cg

mgW = . g is the acceleration of gravity and is equal to :

kg N

8066.9g/gs

m8066.9g c2

⇒=

gdyn

66.980g/gs

cm66.980g c2

⇒=

m

f c2 lb

lb1g/g

s

ft174.32g ⇒=

Note that g is acceleration due to gravity which varies according to position on earth

and height while g c is conversion factor and is constant. This is a source of confusion

in American Engineering units.

3. The Mole Unit

The number of moles = ,weightmolecular

mass so, depending on the units of mass used,

the unit of moles is determined. When grams are used for mass, the resulting mole

unit is gram-moles, or simply moles, mass in kilograms will give mole unit in kg-mole, mass in pound-mass will give moles in lb-moles and so on. In the CGS and SI

systems of units, the mole unit is gram-mole (mole or g-mole) while in the American

Engineering, the pound-mole (lb-mole) is used.



5

4. Conversion of Units

A measured quantity can be expressed in terms of any units having the appropriate

dimension. For example, velocity =.etc)hmin,s(

)mile,km,ft,cm(

time

length

Thus, the units of velocity may be ,h

km,

sft

etc.

Numerical value of velocity depends on units chosen.

BEWARE of errors due to improper use of units. Using units with your number help

guide you to the correct answer.

The Conversion Factor is a ratio used to convert a certain quantity with certain unit

to its equivalent in another unit.

e.g.( )

2

2

2

2

cm1mm100

cm1mm10

and,min1

s60,

cm100m1

=

For conversion to the new units, multiply the old unit by the conversion factor which

should be the equivalent ofunitoldunitnew

.

e.g. convert 500 g to kg

500 g x kg5.0g1000

kg1=

Alternatively, this may be done by using a dimensional equation as follows :

kg5.0g1000

kg1g500=

This should avoid mistakes, compare with



6

kgg

000,500kg1

g1000g500 2

=

of course this is meaningless, and certainly does not result in the desired units.

When converting a quantity having a compound unit

.etc,cm

g,

hkm

.g.e 3 , set up a

dimensional equation.

Table 3 gives the conversion factors you will need .

Example :

1. Convert a flow rate of 5 ft 3/s to its equivalent in m 3/day:

daym

233,12day1

h24h1min60

min1s60

ft1m028317.0

sft5 3

3

33

=

2. Convert a density of 1000 kg/m 3 to its equivalent in lb m/ft3 :

3m

3

3m

3 ft

lb4.62

ft3145.35m1

kg1lb20462.2

mkg1000 =

So, the dimensional equation has vertical lines set up to separate each ratio

(conversion factor), and these lines retain the same meaning as multiplication sign placed between each ratio. At any point in the dimensional equation, you can

determine the consolidated net units and see what conversions are still required.



7

Dimensional Homogeneity:

As a rule, every valid equation must be dimensionally homogeneous; that is, all

additive terms on both sides of the equation must have the same dimensions.

e.g. )s(tsmg)s/m(V)s/m(V 2o

+=

dimensionally homogeneous

while V = V o + g is not homogeneous

This is also consistent in its unit.

If an equation is dimensionally homogeneous but its term are inconsistent in units,

these terms and hence equation can be made consistent in units by multiplying by the

conversion factor.

Thus, dimensional analysis can also be used to help identify the dimensions of terms

in an equation.

The opposite of the above rule is not necessarily true. An equation may be

dimensionally consistent but invalid.

e.g. If M mass of an object, then M = 2 M is dimensionally homogeneous but

incorrect.

Example :

Consider the equation

D(ft) = 3t(s) + 4

1. If the equation is valid, what are the dimensions of the constant 3 and 4?



8

2. If the equation is consistent in its units, what are the units of 3 and 4?

Solution

1. For the equation to be valid, it must be dimensionally homogeneous, so that each

term must have the dimension of length. The constant 3 must therefore have the

dimension length/time , and 4 must have the dimension length.

2. For consistency of units, the constants must be 3 ft/s and 4 ft.

Table 2.

SI units

Physical Quantity Name of UnitSymbol

for Unit * Definition of Unit

LengthMassTimeTemperatureAmount of substance

EnergyForcePowerDensityVelocityAccelerationPressureHeat Capacity

TimeTemperatureVolumeMass

Basic SI Units

metre, meterkilogramme, kilogramsecondkelvinmole

Derived SI Units

joulenewtonwattkilogram per cubic metermeter per secondmeter per second squarednewton per square meter, pascal

joule per (kilogram . kelvin)

Alternative Units

minute, hour, day, yeardegree Celsiuslitre, liter (dm 3)tonne, ton (Mg), gram

mkgsKmol

J NW

min, h, d, yoCLt, g

kg . m 2 . s -2 kg . m . s -2 → J . m -1 kg . m 2 . s 3 → J. s -1

kg . m -3 m . s -1 m . s -2

N . m -2, PaJ . kg -1 . K -1



9

American Engineering System Units

Physical Quantity Name of Unit Symbol

LengthMassForceTimeTemperature

EnergyPowerDensityVelocityAccelerationPressureHeat capacity

Basic Units

feet pound (mass) pound (force)second, hourdegree Rankine

Derived Units

British thermal unit, foot pound (force)horsepower

pound (mass) per cubic footfeet per secondfeet per second squared

pound (force) per square inchBtu per pound (mass) per degree F

ftlbm lb f s, hroR

Btu, (ft) (lb f )hplbm/ft3 ft/sft/s 2

lbf /in. 2 Btu/(lb m) (oF)



Chapter 3

PROCESS VARIABLES



1

PROCESS VARIABLES

1. Liquid and Solid Densities

Density is the ratio of mass per unit volume, for example, kg / m 3, g / cm 3, lb m/ft 3. It

has both numerical values and units. To determine the density of a substance, you

must find both its mass and its volume. If the substance is solid, the common method

to determine its volume is to displace a measured quantity of inert liquid. For

example, a known weight of a material can be placed into a container of liquid of

known weight and volume, and the final weight and volume of the combination be

measured. The density (or specific gravity) of a liquid is commonly measured with a

hydrometer which consists of a body of known weight and volume which is dropped

into a liquid and the depth to which it penetrates into the liquid is noted.

Specific gravity is a dimensionless ratio of two densities - that of substance (A) of

interest to that of a reference substance.

Specific gravity =

lb ft

lb ft

g cm

g cm

kg m

kg m

m

A

m

ref

A

ref

A

ref

3

3

3

3

3

3

=

=

(1)

The reference substance of liquids and solids is normally water.

Gas densities are quite difficult to measure; one device used is the Edwards Balance

which compares the weight of a bulb filled with air to the same bulb filled with the

unknown gas. The specific gravity of a gas is frequently referred to air but may be

referred to other gases.

In petroleum industry the specific gravity of petroleum products is usually reported in

terms of hydrometer called °API, defined as,



2

o

o

o

API

SpGr

= −141 5

60

60

1315.

. (2)

or

SpGr API

60

60

141 5

1315

o

o o

=

+

.

. (3)

The density of liquids and solids does not change very much with pressure. For

common substances data is reported in literature. When a liquid or solid is heated, it

normally expands (i.e. its density decreases). In most process applications, however, it

can usually be assumed with little error that solid and liquid densities are independent

of temperature, provided that no phase change occur. Similarly changes in pressuredon not cause significant changes in the liquid or solid densities; these substances are

therefore termed incompressible .

2. Flow Rate

The flow rate of a process stream may be expressed as a mass flow rate (mass/time) oras a volumetric flow rate (volume/time). Suppose a fluid (gas or liquid) flows in the

cylindrical pipe shown below, where the shaded area represents a section

perpendicular to the direction of flow.

Figure1. Flow of a fluid in a pipe

If the mass flow rate of the fluid is ‘ m’ (kg/sec), then every second m kilograms of

fluid pass through the cross section. If the volumetric flow rate of the fluid at the

given cross-section is ‘ v’ (m 3 /sec), then every second, v cubic meters of fluid pass

m (kg fluid /sec)v (m 3 fluid /sec)



3

through the cross-section. However, the mass flow rate m and volumetric flow rate v

are not independent quantities but are related through density.

3. Chemical Composition

3.1. Moles and Molecular Weight

The atomic weight of an element is the mass of an atom on a scale that assigns 12C (the

isotope of carbon whose nucleus contains six protons and six neutrons) a mass of

exactly 12. The molecular weight of a compound is the sum of atomic weights of the

atoms that constitute a molecule of the compound.

A gram-mole( gmole or mole in SI units system) of a species is the amount of that

species whose mass in grams is numerically equal to its molecular weight. Other types

of moles (e.g. kg-mole or kmol, lb-mole, ton-mole etc.) are similarly defined. Carbon

monoxide (CO), for example, has a molecular weight of 28. One mole of CO therefore

contains 28 grams; 1 lb-mole contains 28 lbs.; 1 ton-mol contains 28 tons).

The molecular weight of a substance, M, can be expressed as M kg/kmol, M g/gmol.,

M lbs/lb-mol etc. The molecular weight may thus be used as a conversion factor that

relates mass and number of moles of the quantity of the substance. Therefore,

34 kg ammonia (NH 3) are equivalent to:

34117

233

33kg NH

kmol NH kg NH

kmol NH

= (4)

3.2. Mass and Mole Fractions

Process streams occasionally contain one substance but more often consist of mixture

of liquids or gases or solutions of one or more solutes in liquid solvent. Following



4

terms may be used to define the composition of a mixture of substances including a

species A.

Mass Fraction

x massof A

Total masskg of Atotal kg

g of Atotal g

or lb of Atotal lb A

m

m

=

, (5)

Mole Fraction

y molesof A

Total moleskmol of Atotal kmol

mol of Atotal mol

or lb mol of Atotal lb moles A = −

−

, (6)

For example, if a solution contains 15% A by mass (x A = 0.15) and 20% B (x B =

0.20), calculation of the mass of A in 175 kg of solution:

175015

26kg solution kg A

kg of sol kg A

..

= (7)

similarly, the mass flow rate of A in a stream of solution flowing at the rate of 53 lb m/

hr is given as,

Flow rate of A = 53015

8 0lbhr

lb Alb

lb Ahr

m m

m

m..

= (8)



5

3.3. Conversion from a Composition by Mass to Molar Composition

Example : A mixture of gases has the following composition by mass:

O2 = 16%, CO = 4%, CO 2 = 17% N 2 = 63%Define the molar composition.

Solution:

Basis: 100 g of the mixture

n g total g O g total

mol O g O

mol O2100

016 132

052 2

2

=

=.

.

Similarly,

( )n mol CO =

=100 004128

0143. .

( )n mol CO2100 017

144

0386=

=. .

( )n mol N 2 100 0 631

282 25=

=. .

n n n n n mol T O CO CO N = + + + =2 2 2

3279.

Therefore,

yn

nmol O

total mol OO

T 2

2 053279

015 2= = =..

.

yn

n

mol CO

total mol COCO

T

= = =01433279

0044..

.

yn

n

mol CO

total mol CO

CO

T 2

2 0386

3279

012 2= = =.

.

.

yn

nmol N

total mol N N

T 2

2 2 253279

0 69 2= = =..

.

Σ yi = + + + =015 0 044 012 0 69 10. . . . .



6

3.4. Average Molecular Weight

The average molecular weight or mean molecular weight of a mixture, M (kg/kmol)

is the ratio of the mass of sample of the mixture (m total) to the moles (n total) of all thespecies in the mixture, i.e.

M m

n y M y M y M total

total = = + + +1 1 2 2 3 4

LLL (9)

or

M y M

where N total number of components

i ii

N =

=

=∑

1 (10)

Also,

1 1

1

2

2

3

3 1 M x M

x M

x M

x M

i

ii

N = + + + +

=∑LLL (11)

3.5. Concentration

The mass concentration of a component of a mixture or solution is the mass of this

component per unit volume of the mixture (g/cm 3, kg/m 3, lb m/ft 3). The concentration

can also be expressed as molar concentration which is the number of moles of the

component per unit volume of the mixture (gmol/cm 3, kmol/m 3, lbmol/ft 3). The

molarity of the solution is the value of the molar concentration of the solute expressed

in gram moles of solute per liter of solution. The density of a mixture of liquids can be

estimated by assuming that component volumes are additive: e.g. if 2 ml of liquid A

and 3 ml of liquid B are mixed, the resulting volume would be assumed to be 5 ml.

This leads to a simple averaging formula for the mixture density ( ρ);

1 1

1

2

2 ρ ρ ρ = +

x x (12)



7

where ρ 1 and ρ 2 are the densities of component A and B respectively while x 1 and x 2

are their mass fractions respectively.

4. Pressure

Pressure is the ratio of a force to the area on which it acts. Therefore, the pressure

units are force units divided by area units (N / m 2, dynes / cm 2, or lb f / ft 2). The SI unit

of pressure is N / m 2 and is known as Pascal.

In most investigations we are concerned with absolute pressure. Most pressure

measuring devices read the difference between absolute pressure and atmospheric

pressure existing at the gauge and this is referred to as gauge pressure . This is shown

graphically in the figure below:

Figure 2. Illustration of terms used in pressure measurement

Pressure aboveAtmospheric Pressure Ordinary Pressure

gauge reading

Ordinary vacuum gauge readsdifference between atmosphericand absolute pressure

AtmosphericPressure

Absolute pressure thatis less than atmospheric

pressure

ZeroPressure



8

From the principles of hydrostatics, it is possible that pressure may be expressed as

head of a particular fluid - i.e. as a height of a hypothetical column of this fluid that

will exert the given pressure at its base if pressure at the top were zero. You can thus

speak of a pressure of 14.7 lb f /in 2 as equivalent of 33.9 ft of water (33.9 ft of H 2O) or

76 cm of mercury (76 cm Hg)

∆ ∆ P g h= ρ (13)

where,

ρ = density of fluid

g = acceleration due to gravity

∆h = head of the fluid.

If ∆ P = 2.0 ×10 5 Pa, the corresponding head of mercury can be calculated using the

above equation:

∆h N

m

mkg

kg N

mmm

mm= ×

= ×2 0 1013600 9 807

1015 105

2

3 33.

..

5. Temperature

The temperature is a measure of the degree of hotness or coldness of the body. There

are two commonly used scales for measuring temperature, namely the Fahrenheit and

Celsius scales. The Celsius scale was formerly called the Centigrade scale.

Until 1954 each of these scales was based on two fixed and easily duplicated points,the ice point and the steam point. The temperature of the ice point is defined as the

temperature of a mixture of ice and water which is in equilibrium with saturated air at

the pressure of 1 atm. The temperature of the steam point is the temperature of water

and steam which are in equilibrium at the pressure of 1 atm. On the Fahrenheit scale,



9

these two points are assigned the number 32 and 212 respectively. On the Celsius

scale the respective points are numbered 0 and 100. Letters ° F and ° C denote the

Fahrenheit and Celsius scales respectively.

In 1954 the Celsius scale was redefined in terms of a single fixed point and the

magnitude of the degree. The triple point of water (the state where the solid, liquid

and vapor phases of water exist together in equilibrium) assigned the value 0.01 ° C.

On the scale the steam point is experimentally found to be 100.0 ° C. Thus there is

essential agreement between the old and new temperature scales.

The absolute scale of temperature related to Celsius scale is referred to as the Kelvin

scale designated as K.

K C = +o

27315. (14)

The absolute scale related to the Fahrenheit scale is referred to as the Rankine scale

designated as R. The relation between the two scales is

R F = +o

459 67. (15)

The following relationships may be used to convert a temperature expressed in one

defined scale unit to its equivalent in another:

( )T K T C = +( ) .o

27315 (16)

T R T F ( ) ( ) .= +o

459 67 (17)

( )T R T K ( ) .= 18 (18)

( ) ( )T F T C o o

= +18 32. (19)



Chapter 4

MATERIAL BALANCES AND APPLICATIONS



1

MATERIAL BALANCES AND APPLICATIONS

4.1. Introduction

Material balances are important first step when designing a new process or analyzing anexisting one. They are almost always prerequisite to all other calculations in the solution of

process engineering problems.

Material balances are nothing more than the application of the law of conservation of mass,which states that mass can neither be created nor destroyed. Thus, you cannot, for example,specify an input to a reactor of one ton of naphtha and an output of two tons of gasoline orgases or anything else. One ton of total material input will only give one ton of total output,i.e. total mass of input = total mass of output.

A material balance is an accounting for material. Thus, material balances are often comparedto the balancing of current accounts. They are used in industry to calculate mass flow rates ofdifferent streams entering or leaving chemical or physical processes.

4.2. The General Balance Equation

Suppose propane is a component of both the input and output streams of a continuous processunit shown below, these flow rates of the input and output are measured and found to bedifferent.

q in (kg propane/h) q out (kg propane/h)

If there are no leaks and the measurements are correct, then the other possibilities that canaccount for this difference are that propane is either being generated, consumed, oraccumulated within the unit.

A balance (or inventory) on a material in a system (a single process unit, a collection of units,or an entire process) may be written in the following general way:

Process unit



2

Input + generation output consumption = accumulation(entersthroughsystem

boundaries)

(producedwithinsystem

boundaries)

(leavesthroughsystem

boundaries)

(consumedwithinsystem)

(buildupwithinsystem)

This general balance equation may be written for any material that enters or leaves any process system; it can be applied to the total mass of this material or to any molecular oratomic species involved in the process.

The general balance equation may be simplified according to the process at hand. Forexample, by definition, the accumulation term for steady-state continuous process is zero.Thus the above equation becomes:

Input + generation = output + consumption

For physical process, since there is no chemical reaction, the generation and consumptionterms will become zero, and the balance equation for steady-state physical process will besimply reduced to:

Input = Output

4.3. Balances on Single and Multiple Physical Systems

4.3.1. Procedure for Material Balance Calculations

In material balance problems, you will usually be given a description of a process, the valuesof several process variables, and a list of quantities to be determined. In order to be trainedon using a systematic procedure to solve material balance problems, you are advised tofollow the steps summarized below:

1. Draw and label the process flow chart (block diagram). When labeling, write the values ofknown streams and assign symbols to unknown stream variables. Use the minimumnumber possible of symbols.

2. Select a basis of calculation. This is usually the given stream amounts or flow rates, if nogiven then assume an amount of a stream with known composition.



3

3. Write material balance equations. Note in here the maximum number of independentequations you can write for each system is equal the number of species in the input andoutput streams of this system. Also note to first write balances that involve the fewestunknown variables.

4. Solve the equations derived in step 3 for the unknown quantities to be determined.

Notes

i. Minimize the symbols assigned to unknown quantities by utilizing all the given processspecifications and using laws of physics.

ii. After doing calculations on certain basis, you may scale up or scale down (convert to new basis) while keeping the process balanced. This is done by multiplying all streams (except

mass or mole fractions) by the scale factor which is equal to the ratio of the new streamamount or flow rate to the old one. You can only scale between mass amount or flow ratesregardless of units used but not from mass to molar quantity or flow rate.

The examples below will illustrate the procedure of balances on physical processes:

EXAMPLE: Balance on a mixing unit

An aqueous solution of sodium hydroxide contains 20% NaOH by mass. It is desired to produce an 8% NaOH solution by diluting a stream of the 20% solution with a stream of purewater.

1. Calculate the ratios (g H 2O/g feed solution) and (g product solution/g feed solution).

2. Determine the feed rates of 20% solution and diluting water needed to produce2310 lb m/min of the 8% solution.

Solution

We could take a basis of 2310 lb m product/min, but for illustrative purposes and to haveneater numbers to work with let us choose a different basis and scale the final results.

Basis: 100 g Feed Solution



4

Draw and label the flowchart, remembering that the amount of the product stream is nowunknown.

(Since the known stream amount is given in grams, it is convenient to label all unknownamounts with this unit.)

There are two unknowns - Q1 and Q2 - and since there are two substances - NaOH and H 2O -in the input and output streams, two balances may be written to solve for them. The totalmass balance and the water balance involve both unknowns, but the NaOH balance involvesonly one.

NaOH Balance

( ) ( )

( )( ) g2500.080g1000.20

out NaOHgin NaOHg

22 =⇒=

⇓

=

QQ

It is a good practice to write calculated variable values on the flowchart as soon as they areknown for ease of use in later calculations; at this point, 250 would therefore be written in

place of Q2 on the chart.

Total Mass Balance

OHg150

g250

g100

21

2

21

=

=

=+

⇓

⇓

Q

Q

QQ

100 g

0.20 g NaOH/g0.80 g H 2O/g

Q2 (g)

0.080 g NaOH/g0.920 g H 2O/g

Q1 (g H 2O)



5

The desired ratios can now be calculated:

solutionfeedg

OHg1.5

solutionfeedg100

O)H(g 2150

211 =

⇒

QQ

solutionfeedg productg

2.5solutionfeedg100

O)H(g 25022

2 =

⇒

QQ

The scale factor is obtained as the true flow rate of the product stream divided by the ratecalculated on the assumed basis.

g/minlb

9.24 productg250n product/milb2310 mm

=

Feed Solution Flow Rate

minsolutionfeedlb

924g

/minlb9.24g100 mm =

Dilution Water Flow Rate

minOHlb

1386g

/minlb9.24g150 2mm =

Check: (924 + 1386) lb m/min = 2310 lb m/min



6

EXAMPLE: Scale up of a separation process flowchart

A 60 - 40 mixture (by moles) of A and B is separated into two fractions. A flowchart of the process is shown here.

It is desired to achieve the same separation with a continuous feed of 1250 lb-moles/h. Scalethe flowchart accordingly.

Solution

The scale factor is

molh/lbmoles

5.12mol100

h/lbmoles1250=

The masses of all streams in the batch process are converted to flow rates as follows:

Feed: specified)(ashmoles-lb

1250mol

moles/h-lb12.5mol100=

Top product stream: (50.0)(12.5) = 625 lb-moles/h

Bottom product stream: (12.5)(12.5) = 156 lb-moles A/h

(37.5)(12.5) = 469 lb-moles B/h

The units of the mole fractions in the top product stream may be changed from mol/mol to lb-mole/lb-mole, but their values remain the same. The flowchart for the scaled-up process isshown here.

100.0 mol0.60 mol A/mol0.40 mol B/mol

50.0 mol

0.95 mol A/mol0.05 mol B/mol

12.5 mol A37.5 mol B

625 lb-moles/hr0.95 lb-mole A/b-mole0.05 lb-mole B/b-mole1250 lb-moles/hr

0.60 lb-mol A/lb-mole0.40 lb-mol B/lb-mole

156 lb-moles A/hr469 lb-moles B/hr



7

EXAMPLE: Material balances on a distillation column

A mixture containing 45% benzene (B) and 55% toluene (T) by mass is fed to a distillationcolumn. An overhead stream of 95 wt% B is produced, and 8% of the benzene fed to thecolumn leaves in the bottom stream. The feed rate is 2000 kg/h. Determine the overhead flowrate and the mass flow rates of benzene and toluene in the bottom stream

Solution

Basis : Given Feed Rate

The labeled flowchart is as follows.

There are three unknowns on the chart - D , w B, and wT - and therefore three equations areneeded. We are entitled to write only two material balances since two species are involved inthe process; the third equation must therefore come from additional given information (theamount of benzene in the bottom stream.) The latter relation is.

chart)theonof placein72(WriteB/hkg72

kg/h]00)[(0.45)(200.080B/h)(kg

feed)in(B0.080stream bottominB

BB

B

ww

w

=

⇓

=⇓

=

Toluene and total mass balances each involve two unknowns, D and wT, but a benzene balance involves only one, D.

Benzene Balance

( )( )

chart)theonit(Writekg/h870D

B/hkg72

D0.95h

Bkg20000.45

B

B

=

⇓

=⇓

+=

w

w

2000 kg/h

0.45 kg B/kg0.55 kg T/kg

D kg/h

0.95 kg B/kg0.05 kg T/kg

Contains 8% of the Bin the feed

wB kg B/kgwT kg T/kg



8

Total Mass Balance (A toluene balance could be used equally well)

T/hkg1060

kg/h72

kg/h870D

Dh

kg2000

T

B

TB

=

⇓

=

=⇓

++=

w

w

ww

EXAMPLE: Two Unit Distillation Process

A labeled flowchart of a continuous steady-state two-unit distillation process is shown below.Each stream contains two components, A and B, in different proportions. Three streamswhose flow rates and/or compositions are not known are labeled 1, 2 and 3.

Calculate the unknown flow rates and compositions of streams 1, 2, and 3.

Solution

The systems about which balances might be taken are shown on the following representationof the flowchart.

40 kg/h0.900 kg A/kg0.100 kg B/kg

30 kg/h0.600 kg A/kg0.400 kg B/kg

30 kg/h0.300 kg A/kg0.700 kg B/kg

100 kg/h

0.500 kg A/kg0.500 kg B/kg

1 2 3

40 kg/h0.900 kg A/kg0.100 kg B/kg

30 kg/h0.600 kg A/kg0.400 kg B/kg

30 kg/h0.300 kg A/kg0.700 kg B/kg

100 kg/h

0.500 kg A/kg0.500 kg B/kg

Q1 (kg/h) Q2 (kg/h)

x1 (kg A/kg) x2 (kg A/kg)1- x1 (kg B/kg) 1- x2 (kg B/kg)

Q3 (kg/h)

x3 (kg A/kg)1- x3 (kg B/kg)



9

The outer boundary encompasses the entire process. Two of the interior boundaries surroundthe individual process units, and the fourth boundary encloses a stream junction point.

Basis: Given Flow Rates

There are two unknowns in the streams that enter and leave the total process, Q3 and x3, andsince there are two independent components in these streams (A and B) we may write two

balances.

Overall Mass Balance

( ) ( )

kg/h60

hkg

3040hkg

30100

3

3

=

⇓

++=+

Q

Q

Overall Balance On A

( )( ) ( )( ) ( )( ) ( )( ) ( )( )

A/kgkg0.0833

60300.600400.900300.3001000.500

3

3

=

⇓

++=+

x

x

To determine the flow rate and composition of a connecting stream, we must write balanceson a subsystem whose boundary intersects this stream. Of the three such boundaries shown inthe flowchart, the middle one (about the stream junction) would not be a good one to use atthis point since its input and output streams contain four unknown quantities ( Q1, x1, Q2, x2),while the boundaries about the process units each intersect streams that contain twounknowns.

Let us choose the boundary about unit 1 for the next set of balances. There are twounknowns, Q1 and x1, in the streams that intersect this boundary, and up to two balances may

be written.

Total Mass Balance on Unit 1

h/kg60

hkg

40h/kg100

1

1

=

⇓

+=

Q

Q



11

EXAMPLE: 2SO 2+O 2→ 2SO 3. What is the stoichiometric coefficient of SO 2?

Solution The number that precede SO 2 is 2. Therefore, stoichiometric coefficient of SO 2 is2.

In a stoichiometric equation, the number of atoms in both sides of the equation must be balanced. In this example, the number of atoms of S and O are 2 and 6, respectively, in bothsides of equation.

Ratio of stoichiometric coefficients of two species is known as stoichiometric ratio .

EXAMPLE: 2SO 2+O 2→ 2SO 3. What is the stoichiometric ratio of SO 2 to SO 3?

Solution stoichiometric ratio of SO 2 to SO 3 =2 mole of SO reacted

2 mole of SO produced = 12

3

If proportion of chemical species fed to a reactor is same as the stoichiometric ratio, thenchemical species combine in stoichiometric proportion , otherwise one or more species will

be in excess of the other. The chemical compound which is present less than itsstoichiometric amount, will disappear first. This reactant will be the limiting reactant and allthe others will be excess reactants. Fractional and percentage excess are given by thefollowing formulas.

fractional excess =n n

nS

S

−

percentage excess = 100n nnS

S ×−

wheren = number of moles fed

ns = number of moles corresponding to stoichiometric amount

EXAMPLE: 100 moles of SO 2 and 100 moles O 2 are fed to a reactor and they reactaccording to 2SO 2+O 2→ 2SO 3. Find the limiting reactant, excess reactant,fractional excess and percentage excess?

Solution

ratio of SO 2 to O 2 fed = 100/100 = 1

stoichiometric ratio of SO 2/O2 = 2/1=2

Therefore, SO 2 is fed less than the stoichiometric proportion (or stoichiometric ratio). SO 2 isthe limiting reactant. The other reactant (O 2) will be the excess reactant .



12

n = number of moles of excess reactant (O 2) fed = 100

ns = stoichiometric amount of O 2 to react with 100 moles of the limiting reactant SO 2 = 50

Therefore, fractional excess = 0 .150

50100n

nn

S

S =−

=−

percentage excess = 100%100n

nnS

S =×−

4.4.2. Fractional Conversions, Extent of Reaction, Chemical Equilibrium

In many cases, chemical reactions do not go to completion and only a fraction will beconverted. Therefore, fractional and percentage conversions are used. They are defined asfollows,

fractional conversion (f) =mole reacted

mole fed to the reactor

percentage conversion = f ×100

fraction unreacted = (1-f)

EXAMPLE: 200 moles of SO 2 and 100 moles O 2 are fed to a reactor. Only 100 moles ofSO 2 react according to 2SO 2+O 2→2SO 3 Find fractional conversion, percentageconversion and fraction unreacted?

Solution

fractional conversion of SO 2 (f) =mole reacted

mole fed to the reactor =100/200=0.5

percentage conversion = f ×100=0.5 ×100=50%

fraction unreacted = (1-f)=1-0.5=0.5

When a reaction is not complete, remaining amount in the reactor will be given by

n ni io i= + β ζ



14

Solution

Basis : 100 mol Feed

The feed to the reactor contains

( )

( )

( )( )( )( ) excessinisO

1.51.5/1HC/O

1.6416.4/10.0HC/O

excessinis NH11/1H/C NH

1.2012.0/10.0H/C NH

mol16.4air mol

Omol0.210air mol78.0O

mol12.0 NH

mol10.0HC

2stoich632

in632

3stoich633

in633

2in2

in3

in63

⇒==

==

⇒==

==

==

=

=

⇓

Since propylene is fed in less than the stoichiometric proportion relative to the two otherreactants, propylene is the limiting reactant.

To determine the percentages by which ammonia and oxygen are in excess, we must firstdetermine the stoichiometric amounts of these reactants corresponding to the amount of

propylene in the feed (10 mol).

( ) 363

363stoich3 NHmol0.10

HCmol1 NHmol1HCmol0.10

NH ==

( ) 263

263stoich2 Omol0.15

HCmol1Omol5.1HCmol0.10

O ==

( ) ( )

( )[ ] 3

stoich3

stoich303 NH

NHexcess20%100%/101012

100%)(NH

)(NH NHexcess%

3

=×−=

×−

=

OHmoln

NHCmoln

Nmoln

Omoln

NHmolnHCmoln

2OH

33 NHC

2 N

2o

3 NH

63HC

2

33

2

2

3

63

100 mol

0.100 mol C 3H6/mol0.120 mol NH 3/mol0.780 mol air/mol

(0.21 mol O 2/mol air0.79 mol N 2/mol air)



16

00.3n

1n

n

00.2n

00.1n

total

H

CO

OH

CO

2

2

2

m)equilibriuat presentCOof molsof (number

=

ξ=

ξ=

ξ−=

ξ−=

e

e

e

e

from which

/3.00ξy

/3.00ξy

)/3.00ξ(2.00y

)/3.00ξ(1.00y

eH

eCO

eOH

eCO

2

2

2

=

=

−=

−=

Substitution of these expressions in the equilibrium relation (with K = 1.00) yields

00.1)00.2)(00.1(

2

=ξ−ξ−

ξ

ee

e

This may be rewritten as a standard quadratic equation and solve to yield ξ e = 0.667. Thisquantity may in turn be substituted back into the expressions for y i(ξ e ) to yield

0.222y0.222,y0.444,y0.111,y222 HCOOHCO ====

The limiting reactant in this case is CO (verify). At equilibrium,

0.3330.6671.00n CO =−=

The fractional conversion of CO at equilibrium is therefore

f CO = (1.00 - 0.333) mol reacted/(1.00 mol fed) = 0.667

4.4.3. Multiple Reactions, Yield and Selectivity

In a chemical process, our objective is to produce a certain product (desired product), butthere may be several unwanted reactions which will produce undesirable by products.Therefore, we must maximize the production of a desired product in the process. Twoquantities, yield and selectivity , are used for this purpose and they are defined as follows,



17

yield =

completelyreactsreactantlimitingtheandreactionssideno werethereif formed productdesiredof moles

formed productdesiredof moles

selectivity =moles of desired product formed

moles of undesired product formed

When we have multiple reactions, the remaining amount or flow rate will be given by

jij j

ioi nn ξβ+= ∑

where i = compound i j = reaction jξ j = extent of reaction for the j th reactionβ ij = + ν i, stoichiometric coefficient of a product i in the j th reaction

= -ν i, stoichiometric coefficient of a reactant i in the j th reaction= 0, inert

EXAMPLE: Consider the following pair of reactions.

A → 2B (desired)

A → C (undesired)

100 moles of A are fed to a batch reactor and the final product contains 10 mol of A, 160 molof B and 10 mol of C. Calculate (1) percentage yield of B, (2) the selectivity of B relative toC and (3) the extents of the reactions.

Solution

Percentage Yield

moles of desired product (B) formed = 160

moles of desired product formed if there were no side reactions and the limiting reactant

reacts completely =reactedAof mole1

producedBof moles2 Aof moles100 × = 200 moles

yield (%) = 160/200*100 = 80%

Selectivity

moles of desired product (B) formed = 160



18

moles of undesired product (C) formed = 10

selectivity =moles of desired product formed

moles of undesired product formed = 160/10 = 16.

Extent of Reactions

j ij j

ioi nn ξβ+= ∑

for two reaction system as above, 2211ioi nn ξβ+ξβ+= ii

applying this equation for B, 120160 ξ+= , this gives ξ1 = 80

for C, 2010 ξ+= this gives ξ2 = 10.

check: for A,

2110010 ξ−ξ−= , 10 100 80 10 10= − − =

EXAMPLE: Yield and Selectivity in a Dehydrogenation Reactor

The reactions

4262

24262

CH2HHC

HHCHC

→++→

take place in a continuous reactor at steady state. The feed contains 85.0 mole% ethane(C2H6) and the balance inerts (I). The fractional conversion of ethane is 0.501, and thefractional yield of ethylene is 0.471. Calculate the molar composition of the product gas andthe selectivity of ethylene to methane production.

Solution

Basis: 100 mol Feed

The outlet component amounts in terms of extents of reaction are as follows:

0.850 C 2H6 0.150 I

100 mol

I)(moln)CH(moln

)H(moln)HC(moln)HC(moln

5

44

23

422

621



19

n1 (mol C 2H6) = 85.0 - ξ1 - ξ2 n2 (mol C 2H4) = ξ1

n3 (mol H 2) = ξ1 - ξ2 n4 (mol CH 4) = 2 ξ2

n5 (mol I) = 15.0

Ethane Conversion

2162

62

62

621

85.0HCmol42.4

fedHCmol85.0fedHCmol

unreactedHCmol0.501)(1.000n

ξ−ξ−==

−=

Ethylene Yield

( ) 142422

62

4262

HCmol40.0HCmol85.00.471n

mol85.0HCmol1HCmol1fedHCmol85.0

ethylene possibleMaximum

ξ===⇓

==

Substituting 40.0 for ξ1 in (1) yield ξ2 = 2.6 mol. Then

I10.7%,CH3.7%,H26.7%,HC28.6%,HC30.3%

mol140.015.0)5.237.440.0(42.4n

Imol15.0n

CHmol5.22n

Hmol37.4n

424262

tot

5

424

2213

⇓

=++++=

=

=ξ=

=ξ−ξ=

Selectivity = (40.0 mol C 2H4)/(5.2 mol CH 4)4

42

CHmolHCmol

7.7=

4.4.4. Atomic and Molecular Balances

In a chemical process, molecules are either generated (produced) or consumed. Therefore,one should take into account the amounts (moles) generated, consumed, fed and remaining inmolecular balances .



20

EXAMPLE: In a steady state process, 100 moles ethane (C 2H6) react to produce ethylene(C 2H4) and hydrogen (H 2) according to C 2H6→ C2H4 + H 2. Product gas shows40 moles of hydrogen remaining. Perform molecular balances for all species.

Solution:

40 mol of H 2 q1 mol of C 2H6

100 mol C 2H6 q2 mol of C 2H4

Then, the molecular balance of H 2 is as follows,

input + generated (produced) = output + consumed (reacted)

input = 0.0

output = 40

consumed = 0

0 + generated (produced) = 40+ 0 ⇒ H 2 (generated) = 40 moles.

Molecular balance of C 2H6 is as follows,


input = 100generated = 0

output = q 1

100 + 0 = q 1 + consumed ⇒ C2H6 (consumed) = 100-q 1

H2 (generated) = 40 moles=C 2H6 (consumed) = 100-q 1 ⇒ q1 = 60 moles

Molecular balance of C 2H4 is as follows,


input = 0

output = q 2

consumed = 0

0 + generated = q 2 + 0 ⇒ C2H4 (generated) = q 2

C2H4 (generated) = q 2 = H 2 (generated) = 40 moles ⇒ q2 = 40 moles



21

According to conservation principle, atoms can neither be created (produced) nor destroyed(consumed). Therefore, in atomic balances there is no generation or consumption terms.Simply, input = output.

EXAMPLE: For the same problem as above, do atomic balances?

Solution

Atomic carbon balance: 100 × 2 = q 1× 2 + q 2× 2

Atomic hydrogen balance 100 × 6 = 40 × 2 + q 1× 6 + q 2× 4

Solving these two atomic balance equations, you will get the same answer as above.

EXAMPLE: Combustion of Methane

Methane is burned with oxygen to yield carbon dioxide and water. The feed contains 20mole% CH 4, 60% O 2, and 20% CO 2, and a 90% conversion of the limiting reactant isachieved. Calculate the molar composition of the product stream using (1) balances onmolecular species, (2) atomic balances, (3) the extent of reaction.

Solution

Basis : 100 mol Feed

OH2COO2CH 2224 +→+

Since a 2:1 ratio of O 2 to CH 4 would be stoichiometric and the actual ratio is 3:1, CH 4 is thelimiting reactant and O 2 is in excess.

Before the balances are written, the given process information should be used to determinethe unknown variables or relations between them. In this case, the methane conversion of90% tells us that 10% of the methane fed to the reactor emerges in the product, or

0.200 CH 4 0.600 O 2 0.200 CO 2

100 mol

O)H(moln)CO(moln

)O(moln)CH(moln

2OH

2CO

2O

4CH

2

2

2

4



22

44CH CHmol2.0fed)CHmol(20.00.100n4

=×=

Now all that remains are the balances. We will proceed by each of the indicated methods.

1. MolecularBalances

2CO

CO4

242

44

4

COmol38n

nCHmol1 producedCOmol1reactCHmol18

fedmolCOmol0.200mol100

output)generation(input

reactCHmol18fedmol

reactmol0.900fedCHmol20.0reactedCH

2

2

=

⇓

=+

=+

==

BalanceCO 2

OHmol36n

nreactCHmol1

producedOHmol2reactCHmol18

output)n(generatio

2OH

OH4

24

2

2

=

⇓

=

= BalanceO H 2

22O

4

24O

2

Omol24Omol36)(60n

reactCHmol1reactOmol2reactCHmol18

nmol

Omol0.600mol100

n)consumptiooutput(input

2

2

=−=

⇓

+=

+= BalanceO 2

In summary, the output quantities are 2 mol CH 4, 24 mol O 2, 38 mol CO 2, and 36 molH2O, for a total of 100 mol. (Since 3 moles of products are produced for every 3 moles ofreactants consumed, it should come as no surprise that total moles in = total moles out.)The mole fractions of the product gas components are thus:

0.02 mol CH 4/mol,0.24 mol O 2/mol,0.38 mol CO 2/mol,0.36 mol H 2O/mol



23

2. Atomic Balances

Referring to the flowchart, we see that a balance on atomic carbon involves only oneunknown (

2COn ), and a balance on atomic hydrogen also involves one unknown ( O2Hn ),

but a balance on atomic oxygen involves three unknowns. We should therefore write theC and H balances first, and then the O balance to determine the remaining unknownvariable. All atomic balances have the form input = output. (We will just determine thecomponent amounts; calculation of the mole fractions then follows as in part 1.)

C Balance

2CO

2

2CO

4

4

2

2

4

4

COmol38n

COmol1Cmol1COmoln

CHmol1Cmol1CHmol2.0

COmol1Cmol1COmol20.0

CHmol1Cmol1CHmol20.0

2

2

=⇓

+=

+

H Balance

OHmol36n

OHmol1Hmol2OHmoln

CHmol1Hmol4CHmol2

CHmol1Hmol4CHmol20

2OH

2

2OH

4

4

4

4

2

2

=⇓

+

=

O Balance

2O

222O

2

2

2

2

Omol24n

O)(1)Hmol(36(2))COmol(38(2))Omol(nOmol1Omol2COmol20

Omol1Omol2Omol60

2

2

=⇓

++=

+

confirming the results obtained using molecular balances.

3. Extent of Reaction

As discussed earlier, for all reactive species

ξβ+= iinout nn

For the species in this problem, we may write



25

4.5.2. Recycle and Purge

A material such as an inert gas or impurities which enter with the feed will remain in therecycle stream. This material will accumulate and the process will never reach steady state.To prevent this buildup, a portion of the recycle stream must be withdrawn as a purge

stream .

EXAMPLE: Dehydrogenation of Propane

Propane is dehydrogenated to from propylene in a catalytic reactor

26383 HHCHC +→

The process is to be designed for a 95% overall conversion of propane. The reaction productsare separated into two streams: the first, which contains H 2, C 3H6, and 0.555% of the propanethat leaves the reactor, is taken off as product; the second stream, which contains the balanceof the unreacted propane and 5% of the propylene in the product stream, is recycled to thereactor. Calculate the composition of the product, the ratio (moles recycled)/(moles freshfeed), and the single-pass conversion.

Solution

Basis : 100 mol Fresh Feed

Note: In labeling the feed stream to the reactor, we have implicitly used balances on propane and propylene about the stream junction.

In terms of the labeled variables, the quantities to be calculated are

Mole Fractions of Product Stream Components: ...,321

1

QQQQ

++etc.

Recycle Ratio:100

21 r r QQ +

Reactor Separation

Unit

Fresh feed100 mol C 3H8 100 + Q

r 1 mol C

3H

8Qr 2 mol C 3H6

Q r 1 mol C 3H8

Q r 2 mol C 3H6

P 1 mol C 3H8

P 2 mol C 3H6

P 3 mol H 2

Q1 mol C 3H8

Q2 mol C 3H6

Q2 mol H 2



26

Single-Pass Conversion: % )Q(

P )Q(

r

r 100100

100

1

11 ×+

−+

We must therefore calculate the values of Q1, Q2, Q3, Q r1 , Q r2 and P 1.

Let us first examine the overall process.

831

83

183

HCmol5

0.95HCmol100

HCmol10095%conversionOverall

=

⇓

=−

⇒=

Q

Q

Overall Propane Balance: input = output + consumption

reactedHCmol95C

)reactedHCmol(CHCmol5HCmol100

83

838383

=⇓

+=

Overall Propylene Balance: output = generation

632

83

63832

HCmol95

reactedHCmol1formedHCmol1reactedHCmol95

=

⇓

=

Q

Q

Similarly, overall H 2 balance yields

Q3 = 95 mol H 2

The analysis of the product is, therefore,

totalmol195

H%7.48Hmol95

HC%7.48HCmol95

HC%6.2HCmol5

22

6363

8383

⇒

We may now proceed to the analysis of the interior streams including the recycle stream.First, we summarize the information given in the problem statement.



27

632r

6H3Cmol952

22r

831

8H3Cmol51

11

HCmol75.405.0

HCmol900P00555.0

=⇒=

=⇒=

=

=

QQQ

P Q

Q

Q

Next, write balances around the separation unit (which is nonreactive, so that input = outputfor all species.)

C 3 H 8 Balance About Separation Unit

831r

831

831

1r 11

HCmol895

HCmol5

HCmol900

=

=

=⇓

+=

Q

Q

P

QQ P

We now have all the variable values we need. The desired quantities are

feedfreshmolrecyclemol

00.9

mol75.4

mol895

feedfreshmol100

recyclemol)(ratiocycleRe

2r

1r

2r 1r

=

=⇓

+=

Q

Q

QQ

9.55%

HCmol900HCmol895

100%)(100

)(100

conversion passSingle

831

831r

1r

11r

==

⇓

×+

−+=−

P Q

Q

P Q

EXAMPLE: Recycle and Purge in the Synthesis of Ammonia

The fresh feed to an ammonia production process contains 24.75 mole % nitrogen, 74.25mole% hydrogen, and the balance inerts (I). The feed is combined with a recycle streamcontaining the same species, and the combined stream is fed to a reactor in which a 25%single-pass conversion of nitrogen is achieved. The products pass through a condenser inwhich essentially all of the ammonia is removed, and the remaining gases are recycled.However, to prevent buildup of the inerts in the system, a purge stream must be taken off.The recycle stream contains 12.5 mole% inerts. Calculate the overall conversion of nitrogen,the ratio (moles purge gas/mole of gas leaving the condenser), and the ratio (moles freshfeed/mole fed to the reactor).



28

Solution

Basis : 100 mol Fresh Feed .

N 2 + 3H 2 → 2NH 3

Strategy: Overall N, H, I balances ⇒ n6, n7, x ⇒ overall conversion of N 2

n6, 25% single pass conversion ⇒ n1, n4

n7, x, n 4, n2 balance around purge-recycle split point ⇒ n8

H2, I balances around recycle-fresh mixing point ⇒ n2, n3

Overall I balance: (0.0100)(100) = 0.125 n 7 n7 = 8.00 mol purge gas

/mol Nmol0.219

NHmol46.0n

3n(2))(0.8758.0025)(2)(100)(0.74

n(2))(8.0075)(2)(100)(0.24

2

36

6

6

==

⇒

+−=

+=

x x:balance H Overall

x:balance N Overall

92.9%100%fedmol24.75

effluentinmol00)(0.219)(8.fedmol24.75 =−

= xbalance N Overall 2

24

21

32

321

Nmol69.00)(0.75)(92.n

reactor tofeed Nmol92.0n

NHmol46.0 Nmol1

NHmol2

fedmol1

reactmol0.25fed Nmoln:

==

=⇓

=Conversion PassSingle25%

n7(mol purged)n8(mol recycled)

Reactor Condenser

0.125 I x N2

0.875 - x H2

n3 (mol. I)n4 (mol N 2)n5 (mol H 2)

0.125 I x N2

0.875 - x H2

n6 (mol NH 3)n4 (mol N 2)n5 (mol H 2)n6 (mol NH 3)

n3 (mol I)

n1 (mol N 2)n2 (mol H 2)n3 (mol I)

0.2475 N 2

0.7425 H 2 0.0100 I

100 mol



29

N 2 balance around purge-recycle split point: 69.0 = (0.219)n 8 + (0.219)(8.00)

⇓ n8 = 307 mol recycled

H 2 balance around mixing point: (100)(0.7425) + (307) (0.875 - 0.219) = n 2

⇓ n2 = 276 mol H 2

I balance around mixing point: (100)(0.01) + (307)(0.125) = n 3 n3 = 39.4 mol I

025.03078

8condenser leavingmoles

purgemoles =+

=

reactor tofedmolfeedfreshmoles

245.04.3927692100

=++

EXAMPLE: Recycle and Purge in the Synthesis of Methanol

Methanol may be produced by the reaction of carbon dioxide and hydrogen.

OHOHCHH3CO 2322 +→+

The fresh feed to the process contains hydrogen and carbon dioxide in stoichiometric proportion, and 0.5 mole% inerts (I). The reactor effluent passes to a condenser, whichremoves essentially all of the methanol and water formed, none of the reactants or inerts. Thelatter substances are recycled to the reactor. To avoid build-up of the inerts in the system, a

purge stream is withdrawn from the recycle. The feed to the reactor contains 2% inerts, andthe single-pass conversion is 60%. Calculate the molar flow rates of the fresh feed, the totalfeed to the reactor, and the purge stream for methanol production rate of 1000 mol/h.

Solution

Basis: 100 mol Combined Feed to the Reactor

As a general rule, the combined feed to the reactor is a convenient stream to use as a basis ofcalculation for recycle problems, provided that its composition is known. Since in this

process the reactants are fed in stoichiometric proportion and they are never separated fromeach other, they must be present in stoichiometric proportion throughout the process; that is,(CO 2/H2) = 1/3. The feed thus contains 2 mol I (2% of 100 mol), and 98 mol CO 2 + H 2, ofwhich 24.5 mol are CO 2 (1/4 of 98) and 73.5 mol are H 2.



30

Take a moment to examine the chart labeling. In particular, notice that we have built in thefacts that CO 2 and H 2 are always present in a 1:3 ratio and that the compositions of thegaseous effluent from the condenser, the purge stream, and the recycle stream are allidentical. The more of this sort of information you can incorporate in the chart labeling, theeasier the subsequent calculations become. Let us outline the solution, referring to theflowchart.

1. Calculate n1, n2, and n3, from the feed to the reactor, the single-pass conversion and reactor balances.

2. Calculate n4, from a total mole balance about the condenser, and then x4 from a CO 2 balance about the condenser.

3. Calculate n0, and n6, from balances on total moles and I about the recycle-fresh-feedmixing point. (Two equations, two unknowns.) Then calculate n5, from a mole balanceabout the recycle-purge split point.

4. Scale up the calculated flows of fresh feed, combined reactor feed, and purge streams bythe factor (1000/ n2). The results will be the flow rates in mol/h corresponding to amethanol production of 1000/h.

60% Single-Pass Conversion

( )

OHmol7.14

OHCHmol7.14

generationoutput:OH,OHCHonBalances

reactCOmol7.145.24600.0

Hmol4.293COmol80.9)5.24(400.0

23

32

23

2

21

21

=

=

=⇓

=

=

==

n

n

nn

Reactor Condensern0 mol

n5 mol

x4 mol CO 2 /mol

3 x4 mol H 2 /mol

(1-4 x4) mol I/mol

n5 mol

x4 mol CO 2 /mol

3 x4 mol H 2 /mol

(1-4 x4) mol I/mol

n4 mol CO 2 /mol x4 mol CO 2 /mol3 x4 mol H 2 /mol

(1-4 x4) mol I/mol

100 mol

0.249 mol CO 2 /mol

0.746 mol H 2 /mol

0.005 mol I/mol

24.5 mol CO 2

75.5 mol H 2

2 mol I

n1 mol CO 2

3n1 mol H 2

n2 mol CH 3OHn3 mol H 2O2 mol I

n2 mol CH 3OH

n3 mol H 2O



31

Mole Balance About Condenser

mol2.41

for substitute

42

4

1

432321

=

⇓

++=++

n

n

nnnnnn

CO 2 Balance About Condenser

mol/COmol2379.0)2.41/80.9( 24

441

==

⇓

=

x

xnn

Mole Balance About Mixing Point

n0 + n6 =100

I Balance About Mixing Point

20484.000500.0

2379.0

2)41(00500.0

60

4

460

=+

=⇓

=−+

nn

x

xnn

Solving the above two equations simultaneously yields

no = 65.4 mol fresh feedn6 = 34.6 mol recycle

Mole balance About Purge Takeoff

purgemol6.66.342.415

654

=−=

⇓

+=

n

nnn

The required scale factor is

( )[ ] 17.142

2 h03.8mol/h/mol1000 −

=

⇒ 6 nn



32

The desired flow rates are therefore

65.4 × 68.06 = 4450 mol fresh feed/h100.0 × 68.03 = 6803 mol/h fed to reactor

6.6 × 68.03 = 449 mol/h purge

4.6. Combustion Reactions and Their Material Balances

4.6.1. Complete and Partial Combustion

Combustion is a rapid reaction of fuel with oxygen. Combustion products are CO 2, NO, CO,H2O, and SO 2. In a combustion reaction if CO is formed, then the reaction is incomplete andreferred as incomplete combustion or partial combustion. During a complete combustion ofa fuel, carbon will be oxidized to CO 2, hydrogen will be oxidized to H 2O, and sulfur will beoxidized to SO 2.

C + O 2 → CO 2 H2+½O 2 → H2O

C + ½O 2 → CO, S+O 2 → SO 2

Complete and incomplete combustion of C 3H8 are given by the following chemical reactions,complete C 3H8 +5O 2 → 3CO 2 + 4H 2O

incomplete C 3H8 + 7/2O 2 → 3CO + 4H 2O

4.6.2. Wet and Dry Basis

Product gas that leaves the combustion chamber is called stack or flue gas . Composition of aflue gas is given on a wet (including water ) or dry basis (excluding water).

EXAMPLE: Suppose a stack gas contains equimolar amounts of CO 2, N 2 and H 2O. Find thecomposition on wet and dry basis?

Solution

Suppose we have n moles of each, then



33

composition of CO 2 on wet basis = 100 basison wetmolestotal

COof moles 2 × = (n/3n) ×100 = 33.33%

composition of CO 2 on dry basis = 100

basisdryonmolestotal

COof moles 2 × = (n/2n) ×100 = 50%.

EXAMPLE: A stack gas contains 60 mole% N 2, 15% CO 2, 10% O 2, and the balance H 2O.Calculate the molar composition of the gas on a dry basis.

Solution

Basis: 100 mol Wet Gas

gasdrymolOmol118.0

8510

gasdrymolCOmol

176.08515

gasdrymol Nmol

706.08560

gasdrymol85Omol10

COmol15 Nmol60

2

2

2

2

2

2

=

=

=

⇓

EXAMPLE: Dry Basis ⇒ Wet Basis

An Orsat analysis (a technique for stack gas analysis) yields the following dry basiscomposition:

%10O

%11CO%14CO

%65 N

2

2

2

A humidity measurement shows that the mole fraction of H 2O in the stack gas is 0.07.Calculate the stack gas composition on a wet basis.



34

Solution

Basis: 100 lb-moles Dry Gas

gasdrymolelbOHmolelb

0753.0gaswetmolelb/gasdrymolelb93.0

gaswetmolelb/OHmolelb07.0

gaswetmolelbgasdrymolelb

93.0gaswetmolelbOHmolelb

07.0

22

2

−−=

−−−−

⇓

−−

⇔−

−

gaswetmolelb5.107Omolelb00.10

)100.0)(100(

COmolelb0.11)110.0)(100(

COmolelb0.14)140.0)(100(

Nmolelb0.65gasdrymolelb Nmolelb650.0gasdrymolelb100

OHmoleslb53.7gaswetmolelb

OHmolelb0753.0gasdrymolelb100

2

2

22

22

−−=

−=−=

−=−

−−

−=−

−−

The mole fractions of each stack gas component may now easily be calculated:

etc...,.gaswetmoleslbOHmoleslb

0.070gaswetmoleslb107.5OHmoleslb7.53

y 22H 2 −

−=−−=O

4.6.3. Theoretical and Excess Air

Theoretical oxygen is the amount needed for complete combustion of reactants to form CO 2 and H 2O. Air that contains the theoretical amount of oxygen is called theoretical air .Theoretical air does not depend on how much of a reactant is converted. The difference

between the amount of air initial fed and the theoretical air is known as excess air. Therefore, percentage excess air is defined as,

% excess air = {[Air (fed) - Air (theoretical)]/Air (theoretical) }×100.

EXAMPLE: Theoretical and Excess Air

One hundred mol per hour of butane (C 4H10) and 5000 mol per hour of air are fed into acombustion reactor. Calculate the percent excess air.



35

Solution

First, calculate the theoretical air from the feed rate of fuel and the stoichiometric equationfor complete combustion of butane.

hair mol

3094Omol

air mol76.4h

Omol650)air (

hOmol

650

HCmol

requiredOmol5.6

hHCmol100

)O(

OH5CO4O2

13

HC

2

2th

2

104

2104ltheoretica2

222104

==

=

=

+→+

Hence

%61.6100%309430945000

100%(air)(air)(air)

air excess%th

thfed=×

−

=×

−

=

If instead you had been given 61.6% excess air, you could have calculated

(air) fed = 1.616 (air) th = 1.616 (3094) = 5000 mol/h.

4.6.4. Combustion Material Balances

When doing material balances for combustion reactions, one should consider balances of thefollowings:

a. inert N 2 at both inlet and outlet

b. unreacted fuel

c. excess oxygen

d. combustion products CO 2, H 2), CO etc.

Suppose 15% excess of oxygen is fed,

oxygen fed = 1.15 × theoretical oxygen

nitrogen fed = (79/21) × oxygen fed

air fed = (100/21) × oxygen fed



36

EXAMPLE: Combustion of Ethane

Ethane (C 2H6) is burned with 50% excess air. The percentage conversion of the ethane is90%; of the ethane burned, 25% reacts to form CO and the balance to form CO 2. Calculatethe composition of the fuel gas and the ratio of water to dry fuel gas.

Solution

Basis: 100 mol Ethane Fed

OH3CO2O25

HC

OH3CO2O27

HC

2262

22262

+→+

+→+

Note:

1. Since no product stream mole fractions are known, subsequent calculations are easier ifindividual component amounts rather than a total amount and mole fractions are labeled.

2. Known information about the composition of air has been used to label the incoming N 2

stream

=

2179

76.3 .

3. If the ethane reacted completely, q1 would be omitted.

4. Since excess air is supplied, O 2 must appear in the product stream.

Theoretical O 2

mol350HCmol1Omol50.3HCmol100

62

262

=

O2 Fed ( )( ) fedOmol5253505.1 2

==Q

N 2 Fed (3.76)(525) = 1974 mol N 2 fed

Q mol O 2 3.76 Q mol N 2

100 mol C 2H6

OHmol

COmol

COmol

Nmol

Omol

HCmol

26

25

4

23

22

621

q

q

q

q

q

q

50% excess air



38

Atomic O Balance

22

2

2

2

2

2

22

2

2

Omol232

OHmol1Omol1OHmol270

COmol1Omol2COmol135

COmol1Omol1OCmol45

Omol1Omol2)Omol(

Omol1Omol2Omol525

OutputInput

=

⇓

+

+

+=

=

q

q

The analysis of the stack gas is now complete. Summarizing:

totalmol2666

OHmol270

gasdrymol2396

COmol135

COmol45

Nmol1974

Omol232

HCmol10

2

2

4

23

22

621

=+

=

=

=

=

=

6

5

q

q

q

q

q

q

Hence the flue gas composition on a dry basis is

molCOmol

0.0563gasdrymol2396

COmol135y

molCOmol

0.019gasdrymol2396

COmol45y

mol

Nmol0.824

gasdrymol2396

Nmol1974y

molOmol

0.0970gasdrymol2396

Omol232y

mol

HCmol0.00417

gasdrymol2396

HCmol10y

225

4

223

222

62621

==

==

==

==

==



39

and the mole ratio of water to dry flue gas is

gasfluedrymolOHmol

113.0gasfluedrymol2396

OHmol270 22 =

EXAMPLE: Combustion of a Natural Gas

A natural gas of an unknown composition is burned with air. An analysis of the product gasyields the following results

0.130 mole H 2O/mole wet gas

Orsat analysis of flue gas: 1.5% CO6.0% CO 2 8.2% O 2

84.3% N 2

Calculate the ratio of hydrogen to carbon in the gas, and speculate on what the gas might be.

Solution

Basis: 100 mol Dry Flue Gas

Since the composition of the fuel is unknown, label it as though carbon and hydrogen wereentering separately.

From the given mole fraction of water,

OHmol9.14gasdrymol870.0OHmol130.0gasdrymol100

22 ==wQ

qH mol HqC mol C

qO2 mol O 2

3.76 qO2 mol N 2

mol/ Nmol843.0

mol/Omol082.0

mol/COmol060.0

mol/COmol015.0

gasdrymol100

2

2

2

Qw mol H 2O



40

Atomic Carbon Balance

COmol1Cmol1

gasfluemolCOmol015.0gasfluemol100

c =q

( )( )( ) Cmol5.7106.0100

COinC 2

=+4 4 84 4 76

Atomic H Balance

CmolHmol

3.97Cmol7.5Hmol29.8

whichfrom

Hmol29.814.92OHmolHmol2O)H(mol

C

H

2

2wH

==

=×==

qq

Qq

The natural gas may therefore be written with the formula (CH 3.97 )n. Since there is only onehydrocarbon for which the ratio of H to C is close to 3.97 - that is, CH 4 - we may conclude inthis case that the natural gas is essentially pure methane, with perhaps trace amounts of otherhydrocarbons. [If we had obtained, say, qH/qC ≈ 2, we could have gone no further than towrite the fuel as (CH 2)n: from the information given, there would have been no way todistinguish between C 2H4, C 3H6, a mixture of CH 4 and C 2H2, etc].



41

References

1. Felder, R. and Rousseau, R. W., Elementary Principles of Chemical Processes, 2 nd ed.,John Wily & Sons, 1986.

2. Himmelblau, D. M., Basic Principles and Calculations in Chemical Engineering, 4 th ed.,

Prentice Hall, New Jersey, 1982.

3. Perry, R. H. and Green, D., Perry’s Chemical Engineering Handbook, 6 th ed., McGrawHill, New York, 1984.

4. Al-Amer, M. J., Session 1 of a Short Course on “Introduction to Chemical Engineeringfor Non-Chemical Engineers”, KFUPM, Dhahran, 1998.



Chapter 5

PVT CORRELATIONS



1

PVT CORRELATIONS

1. Ideal Gas

To solve most problems that involve gases, an expression is needed that relates

specific volume to temperature and pressure so that if any two of three properties are

known, the third can be calculated. An equation that relates the quantities (mass or

moles) and volume of a gas to its pressure and temperature is call Equation of State .

The simplest and most widely used equation of state is the ideal gas law which is

adequate for many engineering calculations over a wide range of conditions. At high

pressures and low temperatures, however, all gases deviate from ideal behavior andmore complex equations of state are needed to describe PVT relations.

1.1. Ideal Gas Law

The ideal gas law can be derived from the kinetic theory of gases by assuming that gas

molecules have negligible volume, exert no force on one another and collide

elastically with walls of their container. This law can be expressed as:

P V n R T = (1)

where,

P = Absolute pressure of the gas

V = Volume of the gas

n = Number of moles of the gas

R = The gas constantT = Absolute temperature of the gas

The above equation can be written for one mole of the gas, as

P v R T = (2)



2

The gas constant, R, has the units of (Pressure)(volume)/(mole)(temperature). Since

the product of pressure and volume has units of energy, R may also be expressed as

(energy)/(mole)(temperature). The units of R are listed on the last page of Felder and

Rousseau in various system of units.

1.2. Standard Conditions

The reference temperature of 0 °C (273 K, 492 R) and a pressure of one atmosphere

(1 atm) are commonly referred to as standard temperature and pressure. The gas law

gives,

P V n R T s s s s= (3)

or, R P

T V n

s

s

s

s

=

(4)

however, ( ) ( )V n

litersmol

STP ft

lbmol STP s

s

= =22 4 3593

.

The value of gas constant R can therefore be calculated from equation (4).

1.3. Ideal Gas Mixtures

Let us consider a mixture of ideal gases and assume that the mixture itself can be

considered an ideal gas. Let us further assume that the mixture consists of components

A and B and that initially the components are separated and exist at temperature and

pressure of the mixture as shown schematically in the figure below.



3

Figure 3. The Amagat’s rule

We can then write the equation of state for the individual component on a molal basis

as follows:

P V n R T A A= (5)

PV n R T B B= (6)

At the same temperature and pressure, if the components are mixed together in one

container, then

P V n R T = (1)It therefore follows that

V V

nn

y A A A= = (7)

This equation simply means that volume fraction (V A/V) is equation to mole fraction.

Further,

n n n A B= + (8)

or PV RT

PV RT

PV RT

A

A

B

B

= + (9)

Gases A + BVolume V

T P

Gas AVolume V A

Gas BVolumeVB

T P P

T



5

Further,

n n n A B= + (8)

RT

V P

RT

V P

RT

PV B A+= (14)

i.e. A P P P += (15)

where each partial pressure is evaluated at the temperature and volume of the mixture.The above equation is called Dalton’s law of Partial Pressure .

2. Real Gas

As the temperature of the gas decreases and the pressure increases, the ideal gas law

discussed earlier provides increasingly poor description of the gas behavior. It

becomes essential to turn to other equations of state which may satisfactorily represent

the PVT behavior.

One approach to represent the PVT behavior of real gases is to use equations of state

that contain number of constants whose value differ for different gases. Another

approach is to correct the deviation from ideal behavior by including a term called the

compressibility factor in the ideal gas law. The resulting equation is called the

compressibility factor equation of state.



6

2.1. Compressibility Factor Equation of State

An equation of state that retains the simplicity of ideal gas law but describes the PVT

behavior of real gases over a much wider rage of conditions is of the following form:

PV Z n R T = (16)

P v Z R T = (17)

The coefficient Z is called the compressibility factor and the equation is called the

compressibility equation of state. A value of Z = 1 corresponds to ideal gas behavior.

The compressibility factor depends on the gas temperature and pressure and differs for

different gases at a given T and P.

2.2. Estimation of Compressibility Factors

2.2.1 Law of Corresponding States

It is possible to construct a single universal chart for the estimation of compressibility

factors for all gases if independent variables other than P and T are used. Figure 5

(Figure 5.4 in Felder and Rousseau)) shows such a plot called generalized

compressibility chart. The variables P and T and replaced by P r = (P/P c), and T r =

(T/T c) where P c and T c are critical pressure and temperature respectively of the gas.

The basis for the generalized compressibility chart is the experimentally observed fact

that the value of certain physical properties of gases, such as compressibility depend

on a greater extent on how near the gas is to its critical state. This suggests that a plot

of Z versus reduced temperature (T r ) and reduced pressure (P r ) should be

approximately be the same for all substances which indeed is the case. This assertion

is known as law of corresponding states .



7

2.3. Real Gas Mixtures

The techniques that is both simple and reasonably accurate is Kay’s Rule which usesthe compressibility factor equation of state:

PV Z n R T m= (18)

P v Z R T m= (19)

with Z m as the mean compressibility factor.

In order to determine Z m for the mixture of gases, A, B, C …… with the composition

of mole fraction y A, y B, y C ……the pseudocritical constants of the mixture as average

values of critical constants of the mixture components are calculated as follows:

Pseudocritical Temperature

T y T y T y T c A cA B cB C cC ' = + + + LLL (20)

Pseudocritical pressure

P y P y P y P c A cA B cB C cC ' = + + + LLL (21)

If the temperature and pressure of the mixture are known, pseudo reduced temperature

T T

T r

c

' '

= and pseudo reduced pressure P P

P r

c

' '

= are calculated and used to obtain



8

the value of Z m for the mixture using the generalized compressibility charts. This

value of Z m can be used to calculate V or v as desired from Eq. (18) or (19).

4. Selected Instruments For Process Variable

Measurement

4.1. Fluid Pressure Measurement

Several mechanical devices are used for the measurement of fluid pressure. The most

common of these devices is a Bourdon Gauge which is a hollow tube closed at one

end and bent into a C congifuration. The open end of the tube is exposed to the fluid

whose pressure is to be measured. As the pressure increases, the tube tends to

straighten, causing a pointer attached to the tube to rotate. The position of the pointer

on the calibrated dial gives the gauge pressure of the fluid. Figure 6 shows the

schematic diagram of the Bourdon gauge.

Figure 6. Bourdon Gauge

P (psi)

P



9

A manometer is a U-shaped tube partially filled with a liquid of known density. When

the ends of the tube are exposed to different pressures the fluid levels drops in the high

pressure arm and rises in the low pressure arm. The difference between the pressures

can be calculated from the measured difference between the liquid level in each arm.

Manometers are used in different ways as shown in Figure 7 (Figure 3.4-4 in Felder

and Rousseau). In each arm P 1 and greater than P 2.

Figure 7a shows an open ended manometer. One end is exposed to the fluid whose

pressure is to be measured and the other end to the atmosphere. Figure 7b shows a

differential manometer, used to measure the pressure difference between two points in

a process line. Figure 7c shows a sealed end manometer which has a vacuum enclosedat one end. If the open end of the sealed-end manometer is exposed to the atmosphere

(P1 = P atm) the device functions like a barometer.

4.2. Temperature Measurement

The temperature of a substance in a particular state of aggregation (solid, liquid or

gas) is a measure of the average kinetic energy possessed by the substance molecules.

Since the energy cannot be measured directly, the temperature must be determined

indirectly by measuring some physical property of the substance whose value depends

on the temperature in a known manner. Such properties and the temperature

measuring devices based on them include electrical resistance of the conductor

( Resistance Thermometer ) voltage at the junction of two dissimilar metals

(Thermocouple ), spectra of emitted radiation ( Pyrometer ), and volume of fixed massof fluid ( Thermometer ).

4.3. Flow Rate Measurement



A flow meter is a device mounted in a process line that provides a continuous reading

of flow rate in the line. Two commonly used flow meters - The Rotameter and the

Orifice meter are shown in Figure 8 (Figure 3.2-1 in Felder and Rousseau).

The rotameter is a tapered vertical tube containing float, the larger the flow rate, the

higher the float rises in the tube. The orifice meter is an obstruction in flow channel

with a narrow opening through which the fluid passes. The fluid pressure drops from

the upstream side of the orifice to the downstream side. The pressure drop varies with

the flow rate; the greater the flow rate the larger is the pressure drop.

Introducing Chemical Engineering

Documents

Transcript of Introducing Chemical Engineering