Chapter 3 Process and Process Variables
Process is any operation or a series of operations which cause physical or chemical
changes in a substance or a mixture of substances.
input
(feed)
Process
Unit
output
(product)
Operating variables represent the amounts (mass flow rates), composition, and states
of substances that go in and come out of process units.
(ex) mass, volume, density, flow rate, mole fraction, composition, pressure,
temperature, etc.
In this chapter, definitions and units of the operating variables are
explained.
Process
Operating Variables
Equations
Components
Equipments
Energy requirement
(Units)
(q+W)
Flow rates
Streams
Energy balance equations
Mass balance equations
Compositions
Temperature
Pressure
Energy balance equations
3.1 MASS AND VOLUME
① Density (
) of a substance is the mass per unit volume of the substance:
② Specific volume (
) of a substance is the volume per unit mass of the
substance: Specific volume is the inverse of density.
(Example) The density of
The mass of
.
is
③ Specific gravity of a substance is the ratio of the density of the
substance (
) to the density of a reference substance (
specific condition.
The reference substance most commonly used for
solids and liquids is water at
.
) at a
The notion of
to water at
The SG of a substance at
is
with reference
(See Table B.1)
(Example 3.1-1) Mass, Volume and Density
(a) Calculate the density of mercury in
(b) Calculate the volume in
.
occupied by
of mercury.
[Solution] From Table B.1, for mercury
(a)
(b)
3.2 FLOW RATE
3.2a Mass and Volumetric Flow Rate
Flow rate of a material is the rate at which the material is transported
through a process line.
Mass flow rate is in mass per time: m (kg/s)
Volume flow rate in volume per time: V (m3/s)
pipe
fluid
Cross section perpendicular to the direction of flow
  m/V = m / V
Test Yourself on p46
1
2
(a) How do the mass flow rates of the gas at the inlet and outlet
compare?
(At steady state only)
From the conservation of mass : m1 = m2
(b) If the density of the gas is constant, how do the volumetric flow rates
at these two points compare?
m = uS, V = uS
if
, then uS =  uS, V = V
(c) What if the density decreases from inlet to outlet?
uS = uS --> uS = uS  V = V , V < V
3.2b
Flow Rate Measurement
Rotameter
Orifice Meter
3.3 CHEMICAL COMPOSITION
3.3a Moles and Molecular Weight
The atomic weight of an element is the mass of an atom on a scale that assigns
a mass of exact 12.
The molecular weight of a compound is the sum of the atomic weights of the atoms that
constitute the compounds.
(ex)
Example 3.3-1 Conversion between Mass and Moles
How many of each of the following are contained in
Carbon dioxide
The number of molecules in
mass flow rate
(Avogadro number)
molar flow rate
of CO2 ?
molar flow rate
mass flow rate
3.3b Mass and Mole Fractions and Average Molecular Weight
The following terms may be used to define the composition of a mixture
of a substance.
(1) Mass fraction: mostly used for liquid and solid mixtures
, percent (%)
(2) Mole fraction: mostly used for gas mixtures
, percent (%)
(3) Average molecular weight or mean molecular weight :
Air : 79 % N2 and 21% O2
Ave MW of Air?
= 0.79 x 28
+ 0.21 x 32 = 28.84 = 29
① With mole fractions and the molecular weight known:
Basis: 1 mole mixture
: mole fraction of
component in the mixture
: molecular weight of
component
: number of components in the mixture
[Proof] Basis of calculation
Total mass of 1 mol:
② With mass fractions and molecular weight known
: mass fraction of component
in the mixture
: molecular weight of component
: number of component in the mixture
[Proof] Basis of calculation:
(Example 3.3-3) Conversion from a Composition by Mass to a
Molar Composition
Basis :
, 1000 kg, ...
Set all the calculations in tabular form
(Example 3.3-4) Calculation of an Average Molecular Weight
Calculate the average molecular weight of air
(1) from its molar composition: 79% N2 and 21% O2
(2) from its mass composition: 76.7% N2 and 23.3% O2.
(Solution)
(1) Basis: 100 mol air
Component
Molecular Weight
Moles
Mass(g)
N2
28
79
79*28
O2
32
21
21*32
100
2880
Total
Average molecular weight = 28.8 g/g mol
(2) Basis: 1000 g air
Component
Molecular Weight
Mass (g)
Moles
N2
28
767
767/28
O2
32
233
233/32
1,000
35
Total
Average molecular weight = 1000 g/35 mole = 28.6
3.3c Concentrations
(1) Mass concentration: Mass/volume
Mass of a component per unit volume of the mixture
(2) Molar concentration: Moles/volume
The number of moles of a component per unit volume of the mixture
(kmol/cm3, lb-mol/ft3, …)
(3) Molarity (M): Moles solute/L solution
(4) Molality (m): Moles solute/kg solvent . Why use molality ?
Normality (N): for acids, bases, ionic compounds, Molarity of proton (H+ ) or
electrons in a solution.
HCl: 0.5 M HCl= 0.5 N HCl
H2SO4: 0.5 M H2SO4 = 0.5*2 N H2SO4
(Example)
0.02 M NaOH
If a stream of this solution flows at a rate of
,
(Example) Assume 10 wt % NaCl in Water ( 1 kg/L)
Molarity of NaCl? = moles NaCl/L solution = 1.71 mol NaCl/L solution
Molality of NaCl? = moles of NaCl/1kg solvent (H2O) = 1.71/0.9 =
<Solution>
(1) Basis: 1000 g Mixture of NaCl and H2O
Components
Molecular wt
Mass
Moles
NaCl
58.45
100 g
100/58.45 = 1.71
H2 O
18
900 g
900/18 = 50
M NaCl = 1.71/1 L = 1.71 M
m NaCl = 1.71/ 0.9 kg H2O
=
1.9 m NaCl
3.3d Parts per Million (ppm) and Parts per Billion (ppb) for Trace
Species
1 ppm = 1 particle (molecule) among 1 millions of total particles (molecules)
1 ppb = 1 particle (molecule) among 1 billions of total particles (molecules)
(1) Liquid mixtures: mass ratio
ppm = x (mass fraction) x 106, ppb = x (mass fraction) x 109
(Ex) 125 ppb phenol in water
Mass fraction of phenol ?, mg phenol/kg liquid ?, mg phenol/L liquid ?
Basis: 1000 g mixture
Component
MW
Mass
mol
---------------------------------------------------------------------------------------------Phenol (TB.1)
94.11
0.125 mg
0.0013 mmol
H2 O
18
1000 g
55.6 mol
1 ppm = 1 mg/kg solution
Exact concentration of phenol in ppb from definition = 1.3x10-6x109/(1.3x10-6+55.6) =
23.3 ppb
<Dilute solution in liquids>: 1 ppm = 1 mg/L
10 ppm phenol = 10 mg phenol/1 L water
(2) Gas mixtures: mole ratio
ppm = y (mole fraction) x 106, ppb = y (mole fraction) x 109
(Ex) 15 ppm SO2 in air
Mole fraction of SO2 = 15 x 10-6
mg SO2/L air?
3.4 PRESSURE
3.4a Fluid Pressure and Hydrostatic Head
Pressure is the ratio of a force to the area on which the force acts:
The fluid pressure (Figure 3.4.1) for a fluid contained in a closed
vessel or flowing through a pipe. F may be defined as the ratio of 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.
Hydrostatic pressure is defined as the force exerted on the base of
the base area A. (Figure 3.4.2)
F equals the force on the top surface plus the weight of the fluid in
the column.
Head of a particular fluid;
The height of a hypothetical column of the particular fluid that would
exert the given pressure at its base if the pressure at the top were
zero.
(Ex) 1 atm (
)
(Example 3.4-1) Calculation of a Pressure as a Head of Fluid
Express a pressure of 2.00 x 105 Pa in terms of mm Hg.
200,000 Pa * 1 atm/100,000 Pa * 760 mm Hg/1 atm = 1520 mm Hg
(Example 3.4-2) Pressure below the Surface of a Fluid
[Solution]
(1) Pressure in Pa
(2) Pressure in Head
3.4b Atmospheric Pressure, Absolute Pressure, and Gauge Pressure
Atmospheric Pressure (Patm ) is the pressure of the atmosphere at the base of
a column of air located at the point of measurement.
Density of air
acceleration of gravity
The standard pressure of 1 atmosphere (760 mmHg) is the atmospheric
pressure at sea level.
Absolute pressure (Pabs)
A zero pressure corresponds to a perfect vacuum .
Gauge pressure (Pgauge ) is measured by many pressure-measuring devices
A zero gauge pressure corresponds to an atmospheric pressure.
Negative gauge pressure: Absolute pressure is lower than atmospheric
pressure.
-76 cm Hg - absolute pressure = 76 -76 = 0 (complete vacuum), P = 0
- 10 cm Hg  P = 76 -10 = 66 cm Hg
3.4c Fluid Pressure Measurement
◎ Two types of common pressure gauges
Bourdon Gauge (Figure 3.4-3) utilizes straightening of a hollow tube
closed at one end and open at the other end which is exposed to the
fluid. Pressure range : from vacuum to 7000 atm.
Manometer is a U-shaped tube partially filled with a fluid of known
Density (Figure 3.4-4).
Pressure difference causes the liquid level difference.
Open-end manometer measures gauge pressure
Differential manometer measures pressure difference between
the two points.
Sealed-end manometer measures absolute pressure.
Equations for manometers
(1) General manometer equation
(3.4-5)
(2) Differential manometer equation
(3) If either 1 or 2 is a gas,
(Example 3.4-3) Pressure Measurement with Manometers
1. Differential Manometer in a process line containing water.
= (f -w ) g h = 40 dynes/cm2
2. Open-End Mercury Manometer
3.5 TEMPERATURE
(1) The temperature of a substance is a measure of the average kinetic energy
possessed by the substance molecules.
(2) Temperature is determined indirectly by measuring some physical property of
the substance whose value depends on temperature.
Resistance thermometer – electrical resistance of a conductor.
Thermocouple – voltage at the junction of two dissimilar metals.
Thermocouple electricity generation
https://www.youtube.com/watch?v=xaT2hqHgLdY
Pyrometer – spectra of emitted radiation.
Bulb thermometer – Volume of a fixed mass of fluid.
(3) Temperature scales
Celsius (centigrade) – freezing point of water (
point (
)
Absolute zero is
Fahrenheit – For H2O (l),
is 32F and
) is
, and the boiling
at 1 atm.
.
is 212F.
Kelvin – absolute zero is 0. The size of one degree is the same as a Celsius
degree.
Rankine – absolute zero is 0. The size of one degree is the same as a
Fahrenheit degree.
(Source: David M. Himmelblau, "Basic Principles and Calculations in Chemical
Engineering", 6th ed., Prentice Hall, 1996)
Relations between temperature scales
* Temperature differences in various units in the numerators
(3.5-1)
(3.5-2)
(3.5-3)
(3.5-4)
(ex)
T difference of 100℃ corresponds to T difference of 180℉.
T difference of 100℃ corresponds to T difference of 100K.
T difference of 100K corresponds to T difference of 180R.
Relations between temperature intervals (
) in denominators.
Conversion factors for temperature intervals:
<Property changes over unit temperature difference>
From Figure, Pc (J/oC) = 1.8 PF (J/oF) or
Pc (J/oC) = PF (J/oF) x 1.8 oF/oC = 1.8 PF (J/oF)
(Example 3.5-2) Temperature Conversion
Consider the interval from
to
.
1.
2.
(Example 3.5-3) Temperature Conversion and Dimensional Homogeneity
The heat capacity of ammonia is
Determine the expression for Cp in
in terms of
[Solution]
1. Substitute T (℃) for
using equation (3.5-4).
.
2. Convert the desired temperature interval using Equation (3.5-5).
Using Basis for Engineering Calculation versus
Using Equations for Mathematical Calculation
Find the composition of air by mass fraction given that mole fractions of O2 and
N2 are 0.21 and 0.79, respectively. Molecular weights of O2 and N2 are 32 and
28, respectively.
1. Mathematical method
(1) Assignment of unknown variables
x = mass fraction of O2 , y = mass fraction of N2
(2) Set up equations satisfying the given conditions
x + y =1,
(x/32)/(x/32 + y/28) = 0.21
(3) solve the equations:
x = 0.23, y = 1-x = 0.77
2. Chemical Engineering Methods Using a Basis of Calculation
(1) Set up a proper basis of analysis
Basis: 100 kg mol Air
(2) Analyze question and given conditions in a table.
component
kg mol
mw
kg
mass fraction
-----------------------------------------------------O2
21
32
672
672/2900
N2
79
28
2228
2228/2900
-----------------------------------------------------Total
100
2900
(3) Answer the question from the table.
mass fraction of O2 = 672/2900 = 0.23
mass fraction of N2 =
2228/2900 = 0.77
1