1. Chapter 2 Processes and Process Variables

2. Content
Density
Flow Rate
Chemical Composition
Pressure
Temperature

3. Process
Process
Unit
Input/Feed
Output/Product
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
Formation of process flow sheet/layout
Specification of individual process unit
Associated operating variables
Operate: running day-to-day process

4. Density & Specific Gravity
mass/volume
Unit: g/cm3; kg/m3; lbm/ft3
Specific Volume
volume/ unit mass
inverse of density
Unit: cm3/g; m3/kg; ft3/lbm

5. 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

6. 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:
(4˚C) = g/cm3
= 1000 kg/m3
= lbm/ft3

7. 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=
0.6
20˚ 4˚

8. Exercise The density of CCl4 is 1.595 g/cm3; what is
Mass of 20 cm3 of CCl4
Volume of 6.20 lbm of CCl4
20 cm3
1.595 g =
31.9 g
cm3
6.20 lbm
454 g
cm3 =
1760 cm3
1 lbm
1.595 g

9. Exercise A liquid has a SG of 0.50. Find Density in g/cm3
Density in lbm/ft3
Mass of 3 cm3 of this liquid
Volume occupied by 18 g of this liquid

10. Solution ρ = 0.5 1 g = 0.5 g/cm3 cm3 a) ρ = 0.5 62.43 lbm =
lbm/ft3
ft3 b)
3 cm3
0.5 g =
1.5 g
cm3 c)
18 g
cm3 =
36 cm3
0.5 g d)

11. 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.

12. 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

13. 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)

14. 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 )

15. 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.

16. Exercises
What is the molar flow rate for 100kg/h CO2 (M=44) fed to the reactor?
What is the corresponding mass flow rate of 850lb-moles/min CO2?
How many gram of O2 consist in 100g of CO2?
Find the number of molecules of CO2 in 100g of CO2?

17. Solution a) 100kg CO2 1 kmol CO2 = 2.27 kmol CO2 h 44 kg CO2 b)
850 lb-moles CO2
44 lbm CO2 =
lbm CO2
min
1 lb-moles CO2 C)
100 g CO2
1mol CO2
1 mol O2
32 g O2 =
72.73 g O2
44 g CO2
1 mol CO2 d)
100 g CO2
1mol CO2
6.02 x 1023 Molecules =
1.37 x 1024 Molecules
44 g CO2
1 mol CO2

18. 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

19. 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.

20. Procedure to Convert from Mass Fractions to Moles Fractions
assuming as a basis of calculation a mass of the mixture (e.g. 100 kg or 100 lbm)
using the known mass fractions to calculate the mass of each component in the basis quantity, and converting these masses to moles.
taking the ratio of the moles of each component to the total number of moles

21. Exercise A mixture of gases has the following mass composition: O2 16%
What is the molar composition?

22. 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
2.250
0.686
Total
1.00
100
3.279
1.000

23. 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:

24. Exercise
If 100 lbm/min of A (MA=2) and 300 lbm/min of B (MB=3) flow through a pipe, find
Mass fractions of A and B
Mole fractions of A and B
Mass flow rate of A
Molar flow rate of B
Total mass flow rate
Total molar flow rate
of the mixture.

25. Answers 0.25 lbm A/lbm; 0.75 lbm B/lbm
0.333 mole A/mole; mole B/mole
100 lbm A/min
100 lb-mole B/min
400 lbm/min
150 lb-moles/min

26. Concentration Mass concentration (cA): Molar concentration (CA):
Molarity :

27. 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

28. 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)

29. 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)

30. 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, 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.

31. 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)
T (˚R) = T (˚ F)
T (˚ R) = 1.8T (K)
T (˚ F) = 1.8T (˚ C) + 32

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

33. Example Consider the interval from 20˚F to 80˚F
Calculate the equivalent temperature in ˚C and the interval between them
Calculate directly the interval in ˚C between the temperature

34. Solution a) b)

35. ANY QUESTION?