1. CHAPTER 2 Processes and process variables
Sem 1, 2015/2016
ERT 214 Material and Energy Balance / Imbangan Bahan dan Tenaga

2. 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

3. Process Variables Density Flow Rate Chemical Composition Pressure
Temperature
Moles & Molecular Weight
Mass & Mole Fractions
Part per million (ppm) & part per billion (ppb)
Concentrations
Average Molecular Weight

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

5. 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. Example
The density of carbon tetrachloride, CCl4 is 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
(conversion factor)

7. 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

8. 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 = substance/ref
substance at 20oC
SG=
0.6
20˚ 4˚
ref at 4oC

9. Example: SG = substance/ref A liquid has a SG of 0.50. Find
Density in g/cm3
Density in lbm/ft3
ρ =
0.5
1 g =
0.5 g/cm3
cm3
ρ =
0.5
62.43 lbm =
lbm/ft3
ft3
(4˚C) = 1 g/cm3
=1000 kg/m3
=62.43 lbm/ft3
SG = substance/ref

10. d) Volume occupied by 18 g of this liquid
(4˚C) = 1 g/cm3
=1000 kg/m3
=62.43 lbm/ft3
SG = substance/ref
Given : SG of 0.50
c) Mass of 3 cm3 of this liquid
3 cm3
0.5 g =
1.5 g
cm3
substance = SG x 1g/cm3
= 0.5 x 1g/cm3
d) Volume occupied by 18 g of this liquid
18 g
cm3 =
36 cm3
0.5 g

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. 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 orifice meter

13. Example:
The mass flow rate of n-hexane (p = glcm3) in a pipe is 6.59 g/s. What is the volumetric flow rate of the hexane?
Density, ρ = g/cm3
Mass flow rate, ṁ = 6.59 g/s
V’ =
6.59 g
cm3 =
10cm3/s s
0.659 g

14. 2. The volumetric flow rate of CCl4 (p = glcm3) in a pipe is cm3/min. What is the mass flow rate of
the CCI4 ?
Density, ρ = g/cm3
volumetric flow rate, v’ = cm3/min
V’ =
100cm3
1.595 g =
159.5 g/min
min
cm3

15. Chemical Composition Moles and Molecular Weight
Mass and Mole Fractions
Average Molecular Weight
Concentration
Parts per Million (ppm) & Part per Billion (ppb)

16. 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
Eg;
Compound : OKSIGEN
Atomic oksigen : O (atomic weight = 16)
Molecular oksigen : O2 (molecular weight = 32)
Moles= Mass / Molecular Weight
Unit for moles are g-mole, kmol,lb-mole etc
( g-mole is same as mol )

17. A gram-mole (g-mol or mol in SI unit) of a species is the amount of that species whose mass in gram is numerically equal to its 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.

18. Example 34 kg of ammonia (NH3 : MW=17) is equivalent to
molecular weight as a conversion factor
34 kg of ammonia (NH3 : MW=17) is equivalent to
4 lb-moles of ammonia is equivalent to
34 kg NH3
1 kmol NH3 =
2.0 kmol NH3
17 kg NH3
4 lb-mole NH3
17 lbm NH3 =
68 lbmNH3
1 lb-mole NH3
It is often helpful in mass-mole conversions to include the chemical formula in the dimensional equation, as illustrated above.)

19. Example : Conversion Between Mass & Moles
How many of each of the following are contained in g of CO2 (M = 44.01)? (1) mol CO2; (2) lb-moles CO2;
molecular weight as a conversion factor
100 g CO2
1 mol CO2 =
2.273 mol CO2
44.01 g CO2
2.273 mol CO2
1 lb-mol CO2 =
5.011 x 10-3 lb-mol CO2
453.6 mol
The same factors used to convert masses from one unit to another may be used to convert the equivalent molar units: there is 454 g/lbm, for example, and therefore there is 454 mol/lb-mole, regardless of the substance involved.
(convert lib-mole of a substance with molecular weight M to gram-moles.)

20. Each molecule of CO2 contains one atom of C, one molecule of O2, or two atoms of O. Therefore, each 6.02 x 1023 molecules of CO2 (1 mol) contains 1 mol C, 1 mol O2, or 2 mol O. Thus,
(3) mol C;
(4) mol O;
(5) mol O2;
2.273 mol CO2
1 mol C =
2.273 mol C
1 mol CO2
2.273 mol CO2
2 mol O =
4.546 mol O
1 mol CO2
2.273 mol CO2
1 mol O2 =
2.273 mol O2
1 mol CO2

21. (6) g 0; 6.02 x 1023 molecules (7) g O2; (8) molecules of CO2
molecular weight as a conversion factor
(6) g 0;
(7) g O2;
(8) molecules of CO2
4.456 mol O
16.0 g O =
72.7 g O
1 mol O
2.273 mol O2
32.0 g O2 =
72.7 g O2
1 mol O2
2.273 mol CO2
6.02 x 1023 molecules =
1.37 x 1024 molecules
1 mol

22. Example :
The molecular weight of a species can be used to relate the mass flow rate of a continuous stream of this species to the corresponding molar flow rate.
For example, if carbon dioxide (C02: M = 44.0) flows through a pipeline at a rate of 100 kg/h:
1. What is the molar flow rate for 100kg/h CO2 (M=44) fed to the reactor?
2. What is the corresponding mass flow rate of 850lb-moles/min CO2?
100kg CO2
1 kmol CO2 =
2.27 kmol CO2 h
44 kg CO2
850 lb-moles CO2
44 lbm CO2 =
lbm CO2
min
1 lb-moles CO2

23. 3. How many gram of O2 consist in 100g of CO2?
4. Find the number of molecules of CO2 in 100g of CO2?
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

24. 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

25. 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.

26. 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

27. Example: Converting Mass Fractions to Moles Fractions
A mixture of gases has the following mass composition:
O2 16%
CO 4%
CO2 17%
N2 63%
What is the molar composition?

28. Step 2: using the known mass fractions to calculate the mass of each component in the basis quantity, and converting these masses to moles.
Step 1 : basis
Mass= mass fraction X basis
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
Moles = mass/MW
Step 3: Mole fraction= mole each component/total numb of moles
Eg: for N2, YN2=2.250/3.279=0.686

29. 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:
Moles= mass/MW

30. Concentration Mass concentration (cA):
mass concentration of a component of a mixture or solution is the mass of this component per unit volume of the mixture (g/cm3 , Ibm/ft3 , kg/in.3 , ... ).
Molar concentration (CA):
molar concentration of a component is the number of moles of the component per unit volume of the mixture (kmol/m3 ,Ib-moles/ft3, ... ).
Molarity :
The molarity of a solution is the value of the molar concentration of the solute expressed in gram-moles solute/liter solution (e.g., a 2-molar solution of A contains 2 mol A/liter solution).

31. 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

32. 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)

33. 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)

34. 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.

35. Temperature
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;
(K= kelvin ; R=Rankine)
T (K) = T (˚ C)
T (˚R) = T (˚ F)
T (˚ R) = 1.8T (K)
T (˚ F) = 1.8T (˚ C) + 32

36. 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

37. Example Consider the interval from 20˚F to 80˚F
Calculate the equivalent temperature in ˚C and the interval between them
T (˚ F) = 1.8T (˚ C) + 32

38. Calculate directly the interval in ˚C between the temperature
Consider the interval from 20˚F to 80˚F
Calculate directly the interval in ˚C between the temperature

39. THANK YOU