San Diego State University
Department of Civil, Construction and Environmental
Engineering
ENVE 355 – Environmental Engineering
Lectures 3 and 4 – Mass Transfer
(Chapter 1.1 – 1.3)
Introductory Activity
• You have a glass of wine and a glass of water.
You take a spoonful from the glass of water
and add it to the glass of wine. You then take a
spoonful from the glass of wine and add it to
the glass of water. Which glass is more
contaminated with the other?
Use Units to Your Advantage
• A parameter expressed in units can be an
equation in itself
– Speed (Velocity); mph or ft/s or m/s
• Your car travels 60 miles in one hour, what is its speed (in
miles per hour)?
• You go to Europe and from Paris to Lyon is 450 km. If
you drive the same speed as you do here, how long will
it take you?
– Flow (Q) = Volume/Time or Velocity*Area
• Million gallons per day (mgd)
• Cubic meters per second (m3/s)
– This will get very complex in this class!
Units of Measurement
• International System of Units (SI) - Metric
International System Units (SI)
Quantity
Length
Mass
Temperature
Area
Volume
Energy
Power
Velocity
Flow rate
Density
SI Units
Meter
Kilogram
Celsius
Square meter
Cubic meter
Kilojoule
Watt
Meter/sec
Meter3/sec
Kilogram/meter3
SI Symbol
m
kg
o
C
m2
m3
kJ
W
m/s
m3/s
kg/m3
Units of Measurement
• U.S. Customary System (USCS) - Standard
U.S. Customary System (USCS)
Quantity
Length
Mass
Temperature
Area
Volume
Energy
Power
Velocity
Flow rate
Density
USCS Units
Foot
Pound
Fahrenheit
Square feet
Cubic feet
British thermal unit
British thermal unit/hr
Miles/hour
Cubic feet/sec
Pounds/cubic foot
SI Symbol
ft
lb
o
F
ft2
ft3
BTU
BTU/hr
mi/hr
ft3/s
lb/ft3
Units of Measurement
• Conversion Factors
International System Units (SI) x Conversion Factor =
Quantity
Length
Mass
Temperature
Area
Volume
Energy
Power
Velocity
Flow rate
Density
SI Units
Meter
Kilogram
Celsius
Square meter
Cubic meter
Kilojoule
Watt
Meter/sec
Meter3/sec
Kilogram/meter3
SI Symbol
m
kg
o
C
m2
m3
kJ
W
m/s
m3/s
kg/m3
1 gal = 3.785 L
1 ft3 = 7.48 gal
1 gal = 8.345 lb of water
Conversion Factor
3.2808
2.2046
1.8(oC) + 32
10.7639
35.3147
0.9478
3.4121
2.2369
35.3147
0.06243
U.S. Customary System (USCS)
USCS Units
Foot
Pound
Fahrenheit
Square feet
Cubic feet
British thermal unit
British thermal unit/hr
Miles/hour
Cubic feet/sec
Pounds/cubic foot
1 drop = 0.05 mL
1 mg/L = 8.345 lb/mil gal
SI Symbol
ft
lb
o
F
ft2
ft3
BTU
BTU/hr
mi/hr
ft3/s
lb/ft3
Units of Measurement
• Common Prefixes
– Used with metric
units only
– μ is 10-6 small amounts of
contaminants can harm you
– Mtons SO2 released in
Atmosphere (is 106)
Common Prefixes
Quantity
10-15
10-12
10-9
10-6
10-3
10-2
10-1
10
102
103
106
109
1012
1015
1018
1021
1024
Prefix
Symbol
femto
pico
nano
micro
milli
centi
deci
deka
hecto
kilo
mega
giga
tera
peta
exa
zetta
yotta
f
p
n
m
m
c
d
da
h
k
M
G
T
P
E
Z
Y
Units of Measurement
• Concentrations in Liquids
Conc. = Mass/Volume
– Mass of substance per unit of volume of
mixture.
• Mass = milligrams (mg) or grams (g)
• Volume = liters (L) or cubic meters (m3)
NOTE: 1 L = 0.001 m3
1000 L = 1 m3
mg/L or mg/mL or
3
g/m
Units of Measurement
• Concentrations in Liquids
Conc = mass (solid)/mass (liquid+solid)
– Mass of substance per mass of mixture.
• Mass = milligrams (mg) or grams (g)
• Parts per million (ppm) = mg/L (prove it!)
Parts per billion (ppb) = μg/L
Examples:
1. Concentration of pollutants in water
2. Concentration of chemicals in water
Units of Measurement
• Concentrations in Liquids
– Converting ppm or ppb to mass per volume
• 1 L of water has a mass of 1000 g
• Usually the volume of contaminant is
extremely small and can be ignored.
• 1 ppm = 1 mg/L
– 1 mg = 10-3 g and 1 L = 103 g
– 1mg/L = 10-3 g/103 g = 1/106
Units of Measurement
• Concentrations in Liquids
– Converting ppm or ppb to mass per volume
• 1 L of water has a mass of 1000 g
• Usually the volume of contaminant is
extremely small and can be ignored.
• 1 ppb = 1 mg/L
– 1 mg = 10-6 g and 1 L = 103 g
– 1mg/L = 10-6 g/103 g = 1/109
Units of Measurement
• Concentrations in Liquids
– Parts per million ~ 1 drop in a 15 gallon
aquarium (approximately).
Units of Measurement
• Concentrations in Liquids (Example 1)
Units of Measurement
• Concentrations in Liquids
– Parts per billion = 1 drop in a 13,000 gallon
pool (approximately).
Units of Measurement
• Concentrations in Gases
Conc. = Volume/Volume
– Usually measured as volume of substance
per unit of volume of mixture.
Units of Measurement
• Concentrations in Gases
Conc = Mass/Volume
– Can be measured as mass of substance per
unit of volume of mixture.
• e.g. mg/m3
– Relationship between ppmv and mg/m3
depends on the pressure, temperature and
molecular weight of the pollutant.
Example:
Concentration of gaseous
pollutant in the atmosphere.
Units of Measurement
• Ideal Gas Law
PV = nRT
P = absolute pressure (atm)
V = volume (m3)
n = mass (mol)
R = ideal gas constant =
0.082056 L*atm*K-1*mol-1
T = absolute temperature (K)
Units of Measurement
• Ideal Gas Law
n = mass (mol)
mol = contains 6.02*1023 molecules
(Avogadro’s number)
The mass of 1 mol = its molecular mass in
grams
Exp. The molar mass of water (H2O)≈18 g/mol
Units of Measurement
• Ideal Gas Law
PV = nRT
T = absolute temperature (K)
K = oC + 273.15
Units of Measurement
• Ideal Gas Law
PV = nRT
P = absolute pressure (atm)
1 atm = 101.325 kPa (kiloPascals)
1 atm = 14.7 pounds per square inch (PSI)
100 kPa = 1 bar
100 Pa = 1 mbar (millibar)
Units of Measurement
• Ideal Gas Law
PV = nRT
P = absolute pressure (atm)
Must convert all pressure units to atm
because:
R = ideal gas constant =
0.082056 L*atm*K-1*mol-1
Units of Measurement
• Volume of an ideal gas (Example 2, Book 1.2)
– Find the volume that 1 mol of an ideal gas
would occupy at standard pressure of 1 atm
and a temperature of 0oC (this is STP).
Units of Measurement
• Volume of an ideal gas (Example 2)
Units of Measurement
• Volume of an ideal gas (Example 3, Book 1.2)
– Find the volume that 1 mol of an ideal gas
would occupy at standard temperature and
pressure conditions of 1 atm of pressure and
25oC temperature (this is U.S. regulatory
standard conditions).
Units of Measurement
• Volume of an ideal gas (Example 3)
V = 24.465 L
Units of Measurement
• Ideal Gas Law – Now Add Chemistry
molecular mass
1 mol of O = 16g
1 mol O2 = 32 g
(2 x 16.00)
Units of Measurement
• Ideal Gas Law – Now Add Chemistry
molecular mass
1 mol of O = 16g
1 mol H = 1 g
1 mol H2O = 18 g
(2*1 + 16)
Units of Measurement
• Converting ppm to mg/m3 (Example 3,
Book 1.3)
– The federal Air Quality Standard for carbon
monoxide is 9.0 ppm. (a) Express this
standard as a percent by volume;
(b) Express as in mg/m3 at 1 atm and 25oC.
– Carbon monoxide molecular formula = CO
– Molecular Masses
• O = 16
• C = 12
Units of Measurement
• Converting ppm to mg/m3 (Example 4)
(a)
(b)
Units of Measurement
• Special Units – This class will occasionally utilize
more specialized/alternate versions of these
units. E.g.
– Standard Cubic Foot (scf)– the amount of gas that
occupies 1 ft3 at standard temperature and pressure
– Pound mole (lb-mol) – 453.6 moles. Allows so that
you can use MWs directly with English units
– Million British Thermal Units (MMbtu) – refers to an
amount of energy, commonly used in engineering
design of thermal systems
Units of Measurement
• Special Units – Parts per million
– Parts per million (ppm) is a commonly used unit
that means two completely different things
depending on what phase the system is
– Liquid/Solid: Parts per million is stated on a weight
basis (ppmw), e.g. kg/million kg or mg/kg.
– When referring to an aqueous system, assuming
water has a density of 1 kg/L, equivalent to 1
mg/kg
Units of Measurement
• Special Units – Parts per million
– Gaseous: ppm stated based on a volumetric basis,
e.g. L/ million L or mL/m3
– For ideal gasses, since volume is directly
proportional to number of moles of gas, this is
equivalent to mol/million moles
Units of Measurement
Aqueous Solution (V=1L, m=1 kg)
=1 mg Chemical A
=1 ppm Chemical A
Gaseous System (V=1000 L, m=f(T,P))
=1 mL Chemical A
=1 ppm Chemical A
Nomenclature and Variables
• Be prepared for some inconsistency with the same properties
being referred to by different variables in different equations
– Different Engineering Disciplines use different standards
• Civil Engineering: Q=Flow Rate, Chemical Engineering: 𝑉=Volumetric Flow Rate,
𝑚=Mass Flow Rate, Air Pollution Modeling: Q=Mass Emission Rate, Heat Capacity
– Changes in equations, references over time
• Residence time=detention time=retention time (Most of the time)
• BODu, uBOD, CBOD, COD (slight differences, but sometime interchangeable)
– Implied variables or quantities
• “There isn’t really an equation” - Stoichiometry
• Time and unit conversions (e.g. if I tell you a process runs constantly and give you an
hourly flow rate (L/hr), calculating a daily flow rate (L/day) should be doable from the
original equation only)
– Poorly written equations
• USEPA (no units in constants)
Materials Balance
• “Everything has to go somewhere”
– Law of conservation of mass: matter can
neither be created nor destroyed (but may
converted into other materials or energy)
Materials Balance
• Define the boundaries
– e.g. Can be a Lake, River or Airspace above
a city.
We can calculate the amount of material entering the atmosphere if we know the
amounts that went in, the transformations, and the waste streams to land and water
(Mass or Material Balance technique).
Materials Balance
• Write the materials balance equation
A substance entering a control volume
has 4 possible fates: leave, accumulate,
convert, generate. (See Equation 1.12)
Accumulation Rate = Input Rate-Output Rate + Reaction Rate
Reaction rate = [Decay rate] or [Generation rate]
Materials Balance
• Write the materials balance equation
Materials Balance
• This equation can also be expressed
mathematically as a differential equation
𝑑𝐶
=
𝑄𝑖 𝐶𝑖 −
𝑄𝑜 𝐶𝑜 + 𝑉𝑅𝑔
𝑑𝑡
Accumulation = Input Rate-Output Rate + Generation Rate
𝑉
V= Volume of System (L)
dC/dt= Change in concentration of chemical per change in time (mg/L-day)
Qi= Each flow rate entering the system (L/day)
Ci= Corresponding concentration of each input (mg/L)
Qo= Each flow rate entering the system (L/day)
Co= Corresponding concentration of each output (mg/L)
Rg= Generation rate for reaction (mg/L-day)
Can also replace the product Q*C = M or replace V*(dC/dt) with dM/dt when
dealing directly with mass
Materials Balance
• Simplify the materials balance equation
Conservative contaminant steady state
assumption:
Decay rate = 0
Accumulation rate = 0
Materials Balance
• Simplify the materials balance equation
Steady state assumption:
Examples: Total Suspend Solids in a lake
Heavy Metals in soils
CO2 in the atmosphere
Not Valid for: Radioactive Radon Gas
Decomposing Organic Waste
Materials Balance
• Steady-state Conservative System
Steady state assumption:
Materials Balance
• Steady-state Conservative System
Cs*Qs + Cw*Qw = Cm*Qm
Materials Balance
• Steady-State Conservative System (Example 5,
Book 1.4)
– A stream flowing at 10.0 m3/s has a tributary
feeding into it with a flow of 5.0 m3/s.
– The river’s concentration of Cl- upstream of
the junction is 20.0 mg/L and the stream’s
concentration of Cl- is 40.0 mg/L.
– Assume Cl- is a conservative substance and
that the system is completely mixed.
– What is the downstream Cl- concentration?
Materials Balance
• Steady-State Conservative System (Example 5)
– What is the downstream Cl- concentration?
Materials Balance
• Steady-State Conservative System (Example 5)
– What is the downstream Cl- concentration?
Step 1: Qs + Qw = Qm (Quick Answer)
10.0 m3/s + 5.0 m3/s = 15.0 m3/s
Since Concentration is based on
mass per volume you can’t add
Step 2: CsQs + CwQw = CmQm
Materials Balance
• Steady-State Conservative System (Example 5)
– What is the downstream Cl- concentration?
Materials Balance
• Steady-State Conservative System (Example 5)
– What is the downstream Cl- concentration?
Materials Balance
• Steady-State Conservative System (Example 6,
Book Problem 1.7)
Materials Balance
• Steady-State Conservative System (Example 6,
Book Problem 1.7)
Materials Balance
• Steady-State Conservative System (Example 6,
Book Problem 1.7)
Materials Balance
• Steady-State Conservative System (Example 6,
Book Problem 1.7)
Batch System with
Non-conservative Pollutants
• Batch Systems
• No input nor output
• Non-conservative Pollutants Examples
• Radio active decay
• Bacterial growth in a closed water tank
• CO2 generation in a poorly ventilated
room
Batch System with
Non-conservative Pollutants
• Batch Systems
• No input nor output
-
Batch System with
Non-conservative Pollutants
• Generation and Decay Rates
• Zero Order
• 1st Order
• 2nd Order
Rate order is determined by its dependency
on the pollutant concentration.
Batch System with
Non-conservative Pollutants
• Zero Order Reactions
• Not dependent on concentration
– Example: evaporation of water from a bucket
r(C) = + k
r(C) = rate of reaction
+ k = generation
mass*volume-1*time-1
- k = decay
(e.g. mg*L-1*s-1)
Batch System with
Non-conservative Pollutants
• Zero Order Reactions Mass Balance
Batch System with
Non-conservative Pollutants
• First Order Reactions
• Is dependent on concentration
– Example: radio active decay (half-life
reactions)
r(C) = + k*C
r(C) = rate of reaction
+ k = generation
time-1
- k = decay
(e.g. s-1)
Batch System with
Non-conservative Pollutants
• First Order Reactions Mass Balance
Decay
Batch System with
Non-conservative Pollutants
• Second Order Reactions
• Is dependent on concentration2 (rare)
– Example: smog formation (H2O2 production)
.
.
» OH + OH = H O
2 2
r(C) = + k*C2
r(C) = rate of reaction
+ k = generation
-1*time-1
volume*mass
- k = decay
(e.g. L*mg-1*s-1)
Batch System with
Non-conservative Pollutants
• Second Order Reactions Mass Balance
Decay
CSTRs with
Non-conservative Pollutants
• Continuously Stirred Tank Reactors (CSTR)
• Completely mixed
• Has inflow and outflow
• Concentration (C) is uniform throughout
Examples:
1. Shallow lakes with inlet and outlet
2. Air in a well ventilated room
CSTRs with
Non-conservative Pollutants
• Reaction rates
• Total amount of substance = C*V
• Reaction rates must include V
Zero order = + V*k
First order = + V*k*C
Second order = + V*k*C2
CSTRs with
Non-conservative Pollutants
• CSTR Example (Example 7, Book 1.5)
A lake is fed by a polluted stream and a sewage
outfall. The stream and sewage wastes have a
decay rate coefficient (k) of 0.20/day (1st order
units).
Assuming complete mixing and no other water
losses or gains, what is the steady-state pollutant
concentration in the lake?
CSTRs with
Non-conservative Pollutants
• CSTR Example 7 (A polluted lake)
CSTRs with
Non-conservative Pollutants
• CSTR Example 7 (A polluted lake)
CSTRs with
Non-conservative Pollutants
• CSTR Example 7 (A polluted lake)
CSTRs with
Non-conservative Pollutants
• CSTR Example 7 (A polluted lake)
CSTRs with
Non-conservative Pollutants
• CSTR Example 7 (A polluted lake)
CSTRs with
Non-conservative Pollutants
• CSTR Example 8 (BOD in lakes)(Example 7, Book
Problem 1.12)
CSTRs with
Non-conservative Pollutants
• CSTR Example 8 (BOD in lakes)(Example 7, Book
Problem 1.12)
CSTRs with
Non-conservative Pollutants
Example 9: A wastewater contains contaminant "A" with an
initial concentration of 1200 mg/L. It is to be treated in a batch
reactor. The reaction of A to products is assumed to be first
order. The decay rate constant, k, is 2.5/day. Determine the
time required to convert 75 percent of A to products.
CSTRs with
Non-conservative Pollutants
Example 9: A wastewater contains contaminant "A" with an initial
concentration of 1200 mg/L. It is to be treated in a batch reactor. The reaction
of A to products is assumed to be first order. The decay rate constant, k, is
2.5/day. Determine the time required to convert 75 percent of A to products.
PFRs with
Non-conservative Pollutants
• Plug Flow Reactors (PFR)
• No mixing
• Reactions may still occur
• Usually thought of as a long pipe.
PFRs with
Non-conservative Pollutants
• Plug Flow Reactors (PFR)
• Similar to a conveyor belt with bottles
filled with liquid undergoing reactions
A series of
mini-batch
reactors
PFRs with
Non-conservative Pollutants
• Reaction Rates
• The same as batch reactors.
• t = hydraulic residence time = the amount
of time it takes to move through the PFR
• t = l/v = V/Q
–
–
–
–
l = length of the PFR
v = velocity moving through the PFR
V = volume of the PFR
Q = flow rate of the PFR