Chapter 2

Processes and Process Variables

Content

Density Flow Rate Chemical Composition Pressure Temperature

Process

Process- any operation that cause a physical or chemical change in a substance. Can consist of several process unit.

Chemical/bioprocess engineering is responsible to design and operate the process. Design

o Formation of process flow sheet/layouto Specification of individual process unito Associated operating variables

Operate: running day-to-day process

Process Unit Input/Feed Output/Product

Process Unit Output/Product

Density & Specific Gravity Density ()

mass/volume Unit: g/cm3; kg/m3; lbm/ft3

Specific Volume volume/ unit mass inverse of density Unit: cm3/g; m3/kg; ft3/lbm

Density & Specific Gravity

Densities of pure solids and liquids are essentially independent of pressure and vary relatively slightly with temperature.

The density of a substance can be used as a conversion factor to relate the mass and the volume of the substances

Specific Gravity

Specific Gravity (SG) ratio of the density () of the substance to the

density of a reference (ref) substance at a specific condition:

SG = substance/ref or substance = SG * ref

density of water at 4˚C is used as a reference density; whereas the value is showed below:

ref@H2O(l) (4˚C) = 1.000 g/cm3

= 1000 kg/m3

= 62.43 lbm/ft3

Specific Gravity

SG is dimensionless. To get the density of a substance, multiply the SG value to the value of reference density.

means that the specific gravity of a substance at 20˚C with reference to water at 4˚C is 0.6

SG= 0.620˚

4˚

Exercise

The density of CCl4 is 1.595 g/cm3; what is

a) Mass of 20 cm3 of CCl4b) Volume of 6.20 lbm of CCl4

20 cm3 1.595 g= 31.9 gcm3

6.20 lbm 454 g cm3

= 1760 cm3

1 lbm 1.595 g

Exercise

A liquid has a SG of 0.50. Finda) Density in g/cm3

b) Density in lbm/ft3

c) Mass of 3 cm3 of this liquidd) Volume occupied by 18 g of this liquid

Solution

a) ρ =0.5 1 g

= 0.5 g/cm3

cm3

b) ρ =0.5 62.43 lbm

= 31.215 lbm/ft3

ft3

c)3 cm3 0.5 g

= 1.5 gcm3

d)

18 g cm3

= 36 cm3

0.5 g

Flow Rate Flow rate- the rate at which a material is

transported through a process line.

Can be expressed as : mass flow rate (mass/time) volumetric flow rate (volume/time)

The density of a fluid can be used to convert a known volumetric flow rate of a process stream to the mass flow rate of that stream or vice versa.

Flow Rate

The mass flow rates of process streams must be known for many process calculations, but it is frequently more convenient to measure volumetric flow rates than mass flow rate. Therefore, the density is used to convert volume flow rate to mass flow rate.

Flow meter is a device mounted in a process line that provides a continuous reading of the flow rate in the line.

Two commonly used flow meter are rotameter and orifice meter

Chemical Composition

Topics that will be covered: Moles and Molecular Weight Mass and Mole Fractions Average Molecular Weight Concentration Parts per Million (ppm) & Part per

Billion (ppb)

Moles & Molecular Weight

Atomic weight of element- mass of an atom based on carbon isotope 12C

Molecular weight of compound- sum of the atomic weights of atoms that constitute a molecule of the compound

Moles= Mass / Molecular Weight

Unit for moles are g-mole, kmol,lb-mole etc ( g-mole is same as mol )

Moles & Molecular Weight

If the molecular weight of a substance is M, then there are M kg/kmol, M g/mol, and M lbm/lb-mole of this substance.

The molecular weight may thus be used as a conversion factor that relates the mass and the number of moles of a quantity of the substance.

One gram-mole of any species contains 6.02 x 1023 (Avogadro’s number) molecules of that species.

Exercises

a) What is the molar flow rate for 100kg/h CO2 (M=44) fed to the reactor?

b) What is the corresponding mass flow rate of 850lb-moles/min CO2?

c) How many gram of O2 consist in 100g of CO2?

d) Find the number of molecules of CO2 in 100g of CO2?

Solution

a)100kg CO2 1 kmol CO2 = 2.27 kmol CO2

h 44 kg CO2 h

b)850 lb-moles CO2 44 lbm CO2 = 37 400 lbm CO2

min 1 lb-moles CO2 min

C)100 g CO2 1mol CO2 1 mol O2 32 g O2 = 72.73 g O2

44 g CO2 1 mol CO2 1 mol O2

d)100 g CO2 1mol CO2 6.02 x 1023 Molecules = 1.37 x 1024

Molecules44 g CO2 1 mol CO2

Mass and Mole Fractions Process streams consist of mixtures of liquids or gases, or

solutions of one or more solutes in a liquid solvents.

The following terms may be used to define the composition of a mixture of substances, including a species A.

Mass fraction: xA= mass of A / total mass

Unit: kg A/kg total; g A/g total; lbm A/lbm total

Mole fraction: yA= moles of A/ total moles

Unit: kmol A/kmol total; lb-moles A/lb-mole total

Mass and Mole Fractions

The percent by mass of A is 100xA, and the mole percent of A is 100yA.

Note that the numerical value of mass or a mole fraction does not depend on the mass units in the numerator and denominator as long as these units are the same.

Procedure to Convert from Mass Fractions to Moles Fractions

1. assuming as a basis of calculation a mass of the mixture (e.g. 100 kg or 100 lbm)

2. using the known mass fractions to calculate the mass of each component in the basis quantity, and converting these masses to moles.

3. taking the ratio of the moles of each component to the total number of moles

Exercise

A mixture of gases has the following mass composition:

O2 16%

CO 4%CO2 17%

N2 63%

What is the molar composition?

Solution

Basis: 100g of mixture

Component Mass Fraction Mass MW Moles Mole Fraction

i xi mi Mi ni yi

O2 0.16 16 32 0.500 0.152

CO 0.04 4 28 0.143 0.044

CO2 0.17 17 44 0.386 0.118

N2 0.63 63 28 2.250 0.686

Total 1.00 100   3.279 1.000

Average Molecular Weight

The average molecular weight (or mean molecular weight) of a mixture, (kg/kmol, lbm/lb-mole, etc.), is the ratio of the mass of a sample of the mixture (mt) to the number of moles of all species (nt) in the sample.

If yi is the mole fraction of the ith the component of the mixture:

If xi is the mass fraction of the ith component of the mixture:

M

component all

ii2211 My.....MyMyM

component all i

i

2

2

1

1

M

x.....

M

x

M

x

M

1

Exercise

If 100 lbm/min of A (MA=2) and 300 lbm/min of B (MB=3) flow through a pipe, find

a) Mass fractions of A and Bb) Mole fractions of A and Bc) Mass flow rate of Ad) Molar flow rate of Be) Total mass flow ratef) Total molar flow rate

of the mixture.

Answers

a) 0.25 lbm A/lbm; 0.75 lbm B/lbm

b) 0.333 mole A/mole; 0.667 mole B/molec) 100 lbm A/min

d) 100 lb-mole B/mine) 400 lbm/min

f) 150 lb-moles/min

Concentration Mass concentration (cA):

Molar concentration (CA):

Molarity :

mixture of Volume

A of masscA

mixture of Volume

A of molesCA

Literin mixture of Volume

A of molesMolarity

Parts per Million (ppm)& Parts per Billion (ppb)

To express the concentrations of trace species in gases or liquids

May refer to mass ratios (usual for liquids) or mole ratios (usual for gases)

ppmi= yi x 106

ppbi = yi x 109

15 ppm SO2 in air means:

every million moles of air contains 15 moles of SO2

mole fractions of SO2 in air is 15 x 10-6

Pressure A pressure is the ratio of a force to the area on which the

force acts (P= F/A).

Pressure units: N/m2, dynes/cm2, lbf/in2, psi, Pa.

The fluid pressure may be defined as the ratio F/A, where F is the minimum force that would have to be exerted on a frictionless plug in the hole to keep the fluid from emerging.

F (N)P (N/m2) A (m2)

P (N/m2)

F (N)

A (m2)

Pressure

Hydrostatic pressure of the fluid- the pressure P of the fluid at the base of the column

P = Po + ρgh

Head pressure- the height of a hypothetical column of the fluid that would exert the given pressure at its base if the pressure at the top were zero.

The equivalence between a pressure P (force/area) and the corresponding head Ph (height of a fluid) is given by:

P (force/area) = ρ fluid g Ph (head of fluid)

Atmospheric, Absolute & Gauge Pressure

Relationship between absolute pressure and gauge pressure is:

Pabsolute = Pgauge + Patmosphere

The atmosphere pressure can be thought of as the pressure at the base of a column of fluid (air) located at the point measurement (e.g. at sea level)

A typical value of the atmospheric pressure at sea level, 760.0 mm Hg, has been designated as a standard pressure of 1 atmosphere.

The fluid pressure referred to so far are all absolute pressures, in that a pressure of zero corresponds to a perfect vacuum.

The pressure relative to atmospheric pressure is called the gauge pressure.

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.

Some temperature measuring devices based on substance properties include electrical resistance of a conductor (resistance thermometer), voltage at the junction of two similar metals (thermocouple), spectra of emitted radiation (pyrometer), and volume of a fixed mass of fluid (thermometer).

The following relationship may be used to convert a temperature expressed in one defined scale unit to its equivalent in another;

T (K) = T (˚ C) + 273.15T (˚R) = T (˚ F) + 459.67T (˚ R) = 1.8T (K)T (˚ F) = 1.8T (˚ C) + 32

Conversion Factor for Interval Temperature

Consider the temperature interval from 0˚C to 5˚C: 5 Celsius and 5 Kelvin degree in this interval 9 Fahrenheit and 9 Rankine degree in this

intervals

K

C

R

F

K

R

C

F

1

1;

1

1;

1

8.1;

1

8.1

Example

Consider the interval from 20˚F to 80˚Fa) Calculate the equivalent temperature in ˚C and

the interval between themb) Calculate directly the interval in ˚C between the

temperature

Solution

a)

b)

CTTT

CCFT

CCFT

FTCT

3.33)7.6(6.26

6.268.1

3280)80(

7.68.1

3220)20(

8.1

32)()(

12

2

1

CF

CF

F

CFTCT

3.338.1

1)2080(

8.1

1)()(

ANY QUESTION?