Chap. 2 Processes and Process Variables

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Chapter 2Processes and

Process

Variables



Variables

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Content

DensityFlow RateChemical CompositionPressure

Temperature



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Process

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

ProcessUnitInput/Feed Output/Product

ProcessUnit Output/Product



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Density & Specific GravityDensity ( )

mass/volumeUnit: g/cm 3; kg/m 3; lb m /ft 3

Specific Volume

volume/ unit massinverse of densityUnit: cm 3 /g; m 3 /kg; ft 3 /lbm



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Density & Specific Gravity

Densities of pure solids and liquids areessentially independent of pressure andvary relatively slightly with temperature.

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



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Specific Gravity

Specific Gravity (SG)ratio of the density ( ) of the substance tothe 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 areference density; whereas the value isshowed below:

ref@ H2O(l) (4˚C) = 1.000

g/cm3

3



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Specific Gravity

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

mean that the specific gravity of asubstance at 20˚C with reference to water

at 4˚C is 0.6

SG= 0.6 20˚ 4˚



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Exercise

The density of CCl 4 is 1.595 g/cm 3; whatisa) Mass of 20 cm 3 of CCl4b) Volume of 6.20 lb m of CCl420 cm 3 1.595 g

= 31.9 gcm 3

6.20 lb m 454 g cm 3 = 1760

cm 3 1 lb m 1.595 g



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Exercise

A liquid has a SG of 0.50. Finda) Density in g/cm 3

b)Density in lbm/ft

3

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



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Solution

a) ρ = 0.5 1 g

= 0.5 g/cm 3 cm 3

b)

ρ =

0.5 62.43 lb m = 31.215 lb m/ft3

ft3

c) 3 cm 3 0.5 g= 1.5 g

cm 3

d)18 g cm 3 = 36 cm 3

0.5 g



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Flow RateFlow rate- the rate at which a material istransported 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 toconvert a known volumetric flow rate of aprocess stream to the mass flow rate of that stream or vice versa.



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Flow Rate

The mass flow rates of process streamsmust be known for many processcalculations, but it is frequently moreconvenient to measure volumetric flowrates than mass flow rate. Therefore, thedensity is used to convert volume flowrate to mass flow rate.

Flow meter is a device mounted in aprocess line that provides a continuousreading of the flow rate in the line.



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Chemical Composition

Topic will be covered:Moles and Molecular WeightMass and Mole Fractions

Average Molecular Weight

ConcentrationParts per Million (ppm) & Part per

Billion (ppb)



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Moles & Molecular Weight

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

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

Moles= Mass / Molecular Weight

Unit for moles are g-mole, kmol ,lb-mole



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Moles & Molecular Weight

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

The molecular weight may thus be usedas a conversion factor that relates the

mass and the number of moles of aquantity of the substance.

One gram-mole of any species contains6.02 x 10 23 (Avogadro’s number)



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Exercises

a) What is molar flow rate for 100kg/h CO 2 (M=44) fed to the reactor?

b) What is corresponding mass flow rate of 850lb-moles/min CO 2?

c) How many gram of O 2 consist in 100g of CO2?



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Solution

a) 100kg CO 2 1 kmol CO 2 = 2.27 kmol CO 2

h 44 kg CO 2 h

b)850 lb-moles CO 2 44 lb m CO 2 = 37 400 lb m CO 2

min 1 lb-moles CO 2 min

C)

100 gCO 2 1mol CO 2 1 mol O 2 32 g O 2 = 72.73 g O 2

44 g CO 2

1 molCO 2 1 mol O 2

d)100 gCO 2 1mol CO 2 6.02 x 10 23 Molecules = 1.37 x 10 24

Molecules44 g CO 2 1 mol CO 2



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Mass and Mole FractionsProcess streams consist of mixtures of liquids orgases, or solutions of one or more solutes in aliquid solvents.

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

Mass fraction: x A= mass of A / total massUnit: kg A/kg total; g A/g total; lb m A/lbm

total



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Mass and Mole Fractions

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

Note that the numerical value of mass or amole fraction does not depend on themass units in the numerator anddenominator as long as these units are thesame.



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Procedure to Convert from MassFractions to Moles Fractions

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

2. using the known mass fractions tocalculate the mass of each component in

the basis quantity, and converting thesemasses to moles.

3. taking the ratio of the moles of each



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Exercise

A mixture of gases has the following masscomposition:

O2 16%CO 4%

CO2 17%N2 63%

What is the molar composition?



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Solution

Basis: 100g of mixtureCompone

nt

Mass

Fraction

Mas

s MW Moles

Mole

Fractioni xi m i Mi n i yi

O 2 0.16 16 32 0.500 0.152CO 0.04 4 28 0.143 0.044CO 2 0.17 17 44 0.386 0.118N2 0.63 63 28 2.250 0.686

Total 1.00 100 3.279 1.000



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Average Molecular Weight

The average molecular weight (or meanmolecular weight) of a mixture, (kg/kmol,lbm/lb-mole, etc.), is the ratio of the mass of a

sample of the mixture (m t) to the number of moles of all species (n t) in the sample.

If yiis the mole fraction of the

ith the

component of the mixture:

If x i is the mass fraction of the ith component of

M

componentallii2211 My.....MyMyM

componentall i

i

2

2

1

1

Mx

.....Mx

Mx

M



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Exercise

If 100 lb m /min of A (M A =2) and 300 lb m /minof B (MB=3) flow through pipes, find

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



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Solution

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

b) 0.333 mole A/mole; 0.667 mole B/mole

c) 100 lb m A/mind) 100 lb-mole B/mine) 400 lb m /min

f) 150 lb-moles/min



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Concentration

Mass concentration (c A ):

Molar concentration (C A ):

Molarity :

mixtureof VolumeAof mass

c A

mixtureof VolumeAof moles

C A

Liter inmixtureof VolumeAof moles

Molarity



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

To express the concentrations of tracespecies in gases or liquids

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

ppm i= y i x 10 6

ppb i = y i x 109

15 ppm SO 2 in air meaning that:every million moles of air contains 15 moles of



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Pressure A pressure is the ratio of a force to the area onwhich the force acts (P= F/A).

Pressure units: N/m 2, dynes/cm 2, lb f /in 2, psi, Pa.

The fluid pressure may be defined as the ratioF/A, where F is the minimum force that would

have to be exerted on a frictionless plug in thehole to keep the fluid from emerging.F (N)P (N/m 2 ) A (m 2 )

P (N/m 2 )

F (N)

A (m 2 )



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Pressure

Hydrostatic pressure of the fluid- the pressure Pof 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 givenpressure at the top were zero.The equivalence between a pressure P(force/area) and the corresponding head P h (height of a fluid) is given by:



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Atmospheric, Absolute & Gauge Pressure

Relationship between absolute pressure and gaugepressure is:

P absolute = P gauge + P atmosphere

The atmosphere pressure can be thought of as thepressure at the base of a column of fluid (air) located atthe 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 standardpressure of 1 atmosphere.

The fluid pressure referred to so far are all absolute

pressures, in that a pressure of zero corresponds to aperfect vacuum.



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Temperature of a substance in a particular state of aggregation (solid, liquid, or gas) is a measure of theaverage kinetic energy possessed by the substancemolecules.

Some temperature measuring devices based onsubstance properties include electrical resistance of aconductor ( 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 atemperature expressed in one defined scale unit to itsequivalent in another;

T (K) = T (˚ C) + 273.15 ˚ ˚



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Conversion Factor for Interval Temperature

Consider the temperature interval from

0˚C to 5˚C: 9 Fahrenheit and 9 Rankine degree in thisintervals5 Celsius and 5 Kelvin degree in this interval

K

C

R

F

K

R

C

F

11

;11

;18.1

;18.1



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Example

Consider the interval from 20˚F to 80˚F a) Calculate the equivalent temperature in

˚C and the interval between them b) Calculate directly the interval in ˚C

between the temperature



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Solutiona)

b)

C T T T

C C F T

C C F T

F T C T

3.33)7.6(6.26

6.268.13280

)80(

7.68.13220

)20(

8.132)()(

12

2

1

C F

C F

F

C F T C T

3.338.1

1)2080(

8.1

1)()(

Chap. 2 Processes and Process Variables

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