PTT108/4 MATERIALAND ENERGY
BALANCE
SEM 11 (2012/2013)
Chapter 2
Week 2
Processes and Process Variables
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
Input/Feed
Process
Process
Unit
Output/Product
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
 Specification of individual process unit
 Associated operating variables
running day-to-day process
Density ()
 Mass per unit volume of the substance
 Unit
g/cm3; kg/m3; lbm/ft3
m

V
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.
The density of CCl4 is 1.595 g/cm3; what is
a) Mass of 20 cm3 of CCl4
b) Volume of 6.20 lbm of CCl4
Solution:
a)
20 cm3
1.595 g
cm3
b)
6.20 lbm
= 31.9 g
454 g
cm3
1 lbm
1.595 g
=
1760 cm3
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:
ref@H2O(l) (4˚C)
=
=
=
1.000 g/cm3
1000 kg/m3
62.43 lbm/ft3
Specific Gravity (SG)
 SG is dimensionless.
SG=
0.6
20˚
4˚
 means that the specific gravity of a substance at
20˚C with reference to water at 4˚C is 0.6
 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:
a) g/cm3
b) kg/m3
c) lbm/ft3
Solution:
a) 2.00 g/cm3
b) 2.00 x 103 kg/m3
c) 124.86 lbm/ft3
A liquid has a SG of 0.50. Find
a)
b)
c)
d)
Density in g/cm3
Density in lbm/ft3
Mass of 3 cm3 of this liquid
Volume occupied by 18 g of this liquid
a)
Density in g/cm3
ρ=
0.5
1g
cm3
= 0.5 g/cm3
b) Density in lbm/ft3
0.5 62.43 lbm
ρ=
ft3
c)
=
31.215 lbm/ft3
Mass of 3 cm3 of this liquid
3 cm3
0.5 g
cm3
=
1.5 g
d) Volume occupied by 18 g of this liquid
18 g
cm3
0.5 g
=
36 cm3
Flow rate
The rate at which a material is transported through a process line
Can be expressed as :
 (mass/time)
 mass flow rate, m
 volumetric flow rate, V (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.
  m / V  m / V
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
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
Atomic weight of an element
Molecular weight of compound
Mass of an atom relative to carbon
isotope 12C having a mass of exactly 12
Sum of the atomic weights of atoms that
constitute a molecule of the compound
Moles= Mass / Molecular Weight
m
n
M
Unit for moles are g-mole, kmol,lb-mole ( g-mole is same as mol {SI unit} )
• 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.
m
n
M
• One gram-mole of any species contains 6.02 x 1023
(Avogadro’s number) molecules of that species.
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 850lbmoles/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?
a)
What is the molar flow rate for 100kg/h CO2 (M=44) fed to the reactor?
a)
b)
b)
a)
C)
a)
d)
100kg CO2
h
1 kmol CO2
=
2.27 kmol CO2
44 kg CO2
h
What is the corresponding mass flow rate of 850lb-moles/min CO2?
850 lb-moles CO2
min
44 lbm CO2
=
37 400 lbm CO2
1 lb-moles CO2
min
How many gram of O2 consist in 100g of CO2?
100 g CO2
1mol CO2
44 g CO2
1 mol O2
32 g O2
=
72.73 g O2
1 mol CO2 1 mol O2
Find the number of molecules of CO2 in 100g of CO2?
100 g CO2
1mol CO2
6.02 x 1023 Molecules
44 g CO2
1 mol CO2
= 1.37 x 1024
Molecules
• 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
HOW to Convert 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
A mixture of gases has the following mass
composition:
O2
CO
CO2
N2
16%
4%
17%
63%
What is the molar composition?
Basis: 100g of mixture
Component
Mass Fraction
Mass
MW
Moles
Mole Fraction
i
Xi
mi
Mi
ni
yi
O2
CO
CO2
N2
Total
mi
mtotal
mi
Mi
xi .mtotal
Basis: 100g of mixture
ni
ntotal
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
• 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:
M  y1M1  y 2 M 2  ..... 
y M
i
all component
i
• If xi is the mass fraction of the ith component of the mixture:
1
x
x
xi
 1  2  .....  
M M1 M 2
all component M i
Calculate the average molecular weight of air
(a) from its approximate molar composition of
79 % N2, 21 % O2 and
(b) from its approximate mass composition of
76.7 % N2, 23.3 % O2
Solution:
(a) 29 kg/kmol
(b) 29 g/mol
• Mass concentration (cA):
mass of A
cA 
Volume of mixture
• Molar concentration (CA):
CA 
moles of A
Volume of mixture
• Molarity :
moles of A
Molarity 
Volume of mixture in Liter
• 0.02 molar solution of NaOH means:
A solution containing 0.02 mol NaOH/L
• 5 L of this solution contains:
5L
0.02 mol NaOH
=
0.1 mol NaOH
L
• If a stream of this solution flows at a rate of 2 L/min, the molar flow rate of
NaOH is:
2L
0.02 mol NaOH
min
L
=
0.04 mol NaOH
min
 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
.
• 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)
A (m2)
P (N/m2)
A (m2)
F (N)
P (N/m2)
• 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)
•
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.
•
Many pressure-measuring devices give the gauge pressure of a fluid, or the
pressure relative to atmospheric pressure.
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 1000.0 kg/m3. Assume that g is 9.807 m/s2.
Solution:
Ph = Po + ρgh
Ph =
10.4 m H20
1.013x105 N/m2
10.33 m H20
= 3.96 X 105 N/m2 (Pa)
= 396 kPa
+
1000.0 kg/m3 9.807 m 30.0 m 1 N
s2
1 kg.m/s2
•
•
•
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 (˚R)
T (˚ R)
T (˚ F)
= T (˚ C) + 273.15
= T (˚ F) + 459.67
= 1.8T (K)
= 1.8T (˚ C) + 32
Conversion Factor for Interval Temperature
1.8 F 1.8 R 1 F 1C
;
;
;
1C
1K 1 R 1K
• 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
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
Solution
a)
T ( F )  32
1.8
 20  32 
T1 (20 F )  
C  6.7C
 1.8 
T (C ) 
 80  32 
T2 (80 F )  
C  26.6C
1
.
8


T  T2  T1  26.6  (6.7)  33.3C
b)
1C
1.8 F
1C
 (80  20) F 
 33.3C
1.8 F
T (C )  T ( F ) 