1. PTT108/4 MATERIALAND ENERGY BALANCE
SEM 11 (2012/2013)
Chapter 2
Week 2
Processes and Process Variables

2. Part per million (ppm) & part per billion (ppb)
Density
Temperature
Processes and
Process Variables
Flow rate
Pressure
Chemical composition
Moles & molecular weight
Part per million (ppm) & part per billion (ppb)
Mass & mole fractions
Concentration
Average molecular weight

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

4. Density & Specific VOLUME
Mass per unit volume of the substance
Unit g/cm3; kg/m3; lbm/ft3
Specific Volume
Volume occupied by a unit mass of the substances
Inverse of density
Unit cm3/g; m3/kg; ft3/lbm
Densities of pure solids and liquids are essentially independent of pressure and vary relatively slightly with temperature.

5. 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
Solution: a)
20 cm3
1.595 g =
31.9 g
cm3
6.20 lbm
454 g
cm3 =
1760 cm3
1 lbm
1.595 g b)

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 = /ref
The reference most commonly used for solids and liquids is water at 4˚C:
(4˚C) = g/cm3
= 1000 kg/m3
= lbm/ft3

7. Specific Gravity Specific Gravity (SG) SG is dimensionless.
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. Specific Gravity TEST YOURSELF
To get the density of a substance, multiply the SG value to the value of reference density.
TEST YOURSELF
If the specific gravity of a liquid is 2.00, find its density in the following units:
g/cm3
kg/m3
lbm/ft3
Solution:
2.00 g/cm3
2.00 x 103 kg/m3
lbm/ft3

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 Density in g/cm3 ρ = 0.5 1 g = 0.5 g/cm3 cm3
Density in lbm/ft3
ρ =
0.5
62.43 lbm =
lbm/ft3
ft3
Mass of 3 cm3 of this liquid
3 cm3
0.5 g =
1.5 g
cm3
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.
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.

12. FLOWmeter Flow meter Device mounted in a process line that provides a
continuous reading of the flow rate in the line.
Two commonly used flow meter:
a) Rotameter
b) Orifice meter

13. MOLES & MOLECULAR WEIGHT
Atomic weight of an element Mass of an atom relative to carbon
isotope 12C having a mass of exactly 12
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 ( g-mole is same as mol {SI unit} )

14. 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.

15. EXERCISE
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?

16. SOLUTION
What is the molar flow rate for 100kg/h CO2 (M=44) fed to the reactor? a)
100kg CO2
1 kmol CO2 =
2.27 kmol CO2 h
44 kg CO2
What is the corresponding mass flow rate of 850lb-moles/min CO2? b)
850 lb-moles CO2
44 lbm CO2 =
lbm CO2
min
1 lb-moles CO2
How many gram of O2 consist in 100g of CO2? C)
100 g CO2
1mol CO2
1 mol O2
32 g O2 =
72.73 g O2
44 g CO2
1 mol CO2
Find the number of molecules of CO2 in 100g of CO2? d)
100 g CO2
1mol CO2
6.02 x 1023 Molecules =
1.37 x 1024 Molecules
44 g CO2
1 mol CO2

17. Mass & 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
UNIT must be the SAME!
Mole fraction: yA= moles of A/ total moles
Unit: kmol A/kmol total; lb-moles A/lb-mole total
The mass percent of A is 100xA and the mole percent of A is 100yA

18. HOW to Convert 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

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

20. solution Basis: 100g of mixture Component Mass Fraction Mass MW Moles
Mole Fraction i Xi mi Mi ni yi O2 CO
CO2 N2
Total

21. 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

22. 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:

23. EXERCISE Calculate the average molecular weight of air
from its approximate molar composition of 79 % N2, 21 % O2 and
from its approximate mass composition of 76.7 % N2, 23.3 % O2
Solution:
29 kg/kmol
29 g/mol

24. concentration Mass concentration (cA): Molar concentration (CA):
Molarity :

25. concentration 0.02 molar solution of NaOH means:
A solution containing 0.02 mol NaOH/L
5 L of this solution contains:
If a stream of this solution flows at a rate of 2 L/min, the molar flow rate of NaOH is: 5L
0.02 mol NaOH =
0.1 mol NaOH L
2 L
0.02 mol NaOH =
0.04 mol NaOH
min L

26. Part per million (ppm) &
part 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 .

27. pressure
A pressure is the ratio of a force to the area on which the force acts (P= F/A).
Pressure units: dynes/cm2, lbf/in2 or psi, (N/m2 is called a Pascal (Pa) SI unit)
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)

28. 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)

29. 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.
Many pressure-measuring devices give the gauge pressure of a fluid, or the pressure relative to atmospheric pressure.

30. exercise Pressure below the surface of a fluid
What is the pressure 30.0 m below the surface of a lake? Atmospheric pressure (the pressure at the surface) is 10.4 m H20, and the density of water is kg/m3. Assume that g is m/s2.
Solution:
Ph = Po + ρgh
Ph =
10.4 m H20
1.013x105 N/m2 +
kg/m3
9.807 m
30.0 m
1 N
10.33 m H20 s2
1 kg.m/s2 =
3.96 X 105 N/m2 (Pa) =
396 kPa

31. 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;
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)