Eskedar T. (AAiT-CED)

Tools for quantitative understanding of the behavior of
environmental systems.

For accounting of the flow of energy and materials into
and out of the environmental systems.
Material Balance
Energy Balance
Pollutant
Energy
modeling
production, transport, and fate
Conservation of Matter



The law of conservation of matter states that (without
nuclear reaction) matter can neither be created nor
destroyed.
We ought to be able to account for the “matter” at any
point in time.
The mathematical representation of this accounting
system is called a materials balance or mass balance.
Conservation of Energy



The law of conservation of energy states that energy
cannot be created or destroyed.
Meaning that we should be able to account for the
“energy” at any point in time.
The mathematical representation of this accounting
system we use to trace energy is called an energy
balance.

The simplest form of a materials balance or mass balance
Accumulation = input – output
input
Accumulation
Environmental System
(Natural or Device)
output
Control Volume
Consumer
goods
Accumulation
Solid
Waster
Food to
people
Mass rate of accumulation =
Mass rate of input – Mass rate of output
Selam is filling her bathtub but she forgot to put the plug in.
if the volume of water for a bath is 0.350 m3 and the tap is
flowing at 1.32 L/min and the drain is running at 0.32 L/min,
how long will it take to fill the tub to bath level? Assuming
Selam shuts off the water when the tub is full and does not
flood the house, how much water will be wasted? Assume
the density of water is 1,000 kg/m3
Qin = 1.32 L/min
Vaccumulation
We must convert volumes to masses.
Mass = (volume)(density)
Volume = (flow rate)(time) = (Q)(t)
Qout = 0.32 L/min








From mass balance we have
Accumulation = mass in –mass out
(Vacc)() = (Qin)()(t) - (Qin)()(t)
Vacc = (Qin)(t) – (Qin)(t)
Vacc = 1.32t – 0.32t
350L = (1.00 L/min)(t)
t= 350 min or 5.833 hr
The amount of wasted water is
 Waste water = (0.32)(350) = 112 L
Mass flow rate =
Mass balance 
Efficiency of a system
Mass
 (concentration)(flow rate)
time
dM
 cinQin  cout Qout
dt
dM dt Cin Qin  Cout Qout

Cin Qin
Cin Qin
m assin  m assout
OR   
m assin
The air pollution control equipment on a municipal waste incinerator
includes a fabric filter particle collector (known as a baghouse). The
baghouse contains 424 cloth bags arranged in parallel, that is 1/424 of the
flow goes through each bag. The gas flow rate into and out of the
baghouse is 47 m3/s, and the concentration of particles entering the
baghouse is 15 g/m3. In normal operation the baghouse particulate
discharge meets the regulatory limit of 24 mg/m3.
Calculate the fraction of particulate matter removed and the efficiency of
particulate removal when all 424 bags are in place and the emissions
comply with the regulatory requirements. Estimate the mass emission
rate when one of the bags is missing and recalculate the efficiency of the
baghouse. Assume the efficiency for each individual bag is the same as
the overall efficiency for the baghouse.
Cout = 24 mg/m3
Qout = 47 m3/s
Baghouse
Cin = 15 g/m3
Qin = 47 m3/s
Accumulation =
particle
removal
Hopper
dM
 cinQin  cout Qout
dt
𝑑𝑀
=
𝑑𝑡
(15,000mg/m3)(47m3/s)-(24mg/m3)(47m3/s)=703,872 mg/s
The fraction of particulates removed is
703,872𝑚𝑔/𝑠
703,872𝑚𝑔/𝑠
=
3
(15,000𝑚𝑔/𝑚 )(47𝑚𝑔/𝑠) 705,000𝑚𝑔/𝑠
The efficiency of the baghouse is
15,000 𝑚𝑔/𝑚3−24 𝑚𝑔/𝑚3
η=
15,000 𝑚𝑔/𝑚3
=99.84%
(100%)
= 0.9984
Cemission= ?
Qemission= 47 m3/s
Bypass
Cin = 15 g/m3
QBypass =(1/424)( 47 m3/s)
Baghouse
Cout= ?
Qout= (423/424)47 m3/s
Cin = 15 g/m3
423
Qin =(
)( 47 m3/s)
424
𝑑𝑀
𝑑𝑡= ?
Control Volume
A control volume around the baghouse alone reduces the number of
unknowns to two:
Because we know the efficiency and the influent mass flow rate, we can
solve the mass balance equation for the mass flow rate out of the filter.
Cin Qin  Cout Qout

Cin Qin
Solving for CoutQout
423
3
3
CoutQout =(1- η)CinQin= (1-0.9984)(15,000mg/m )(47m /s)( )
424
=1,125mg/s
Effluent
Bypass
From baghouse
𝑑𝑀
=CinQin from bypass +CinQin from baghouse-CemissionQemission
𝑑𝑡
Because there is no accumulation in the junction
𝑑𝑀
=0 and the mass balance equation
𝑑𝑡
CoutQout=CinQin from bypass +CinQin from baghouse
3
3
= (15,000mg/m )(47m /s)(1/424)+1,125=2788 mg/s
The concentration in the effluent is
𝐶𝑜𝑢𝑡𝑄𝑜𝑢𝑡
2,788 mg/s
=
= 59 mg/m3
3
𝑄𝑜𝑢𝑡
47𝑚 /𝑠
The overall efficiency of the baghouse with missing bag is
15,000 𝑚𝑔/𝑚3−59 𝑚𝑔/𝑚3
η=
(100%)= 99.61%
15,000 𝑚𝑔/𝑚3

A storm sewer is carrying snow melt containing 1.200
g/L of sodium chloride into a small stream. The
stream has a naturally occurring sodium chloride
concentration of 20 mg/L. If the storm sewer flow rate
is 2.00 L/min and the stream flow rate is 2.0 m3/s,
what is the concentration of salt in the stream after
the discharge point? Assume that the sewer flow and
the storm flow are completely mixed, that the salt is a
conservative substance (it does not react) and that
the system is at steady state.


kinetic reactions : reactions that are time
dependent.
Reaction kinetics: the study of the effects
of temperature, pressure, and
concentration on the rate of a chemical
reaction.

The rate of reaction, ri, the rate of formation or
disappearance of a substance.
𝑑𝑀

𝑑𝑡


𝑑(𝑖𝑛)
𝑑𝑡
𝑑(𝑜𝑢𝑡)
+r, r=-KCn
𝑑𝑡
=
−
Homogenous reactions. single phase reactions
Heterogeneous reactions : multiphase reactions (between
phases surface)
 ri = kf1(T,P);f2([A],[B], …)
Rate constant
Concentration of reactant
Assuming that the pressure and temperature are constant
aA + bB
cC
Rate of reaction  rA = - k[A]α[b]β = k[C]γ
Rate of reaction  rA = - k[A]α[b]β = k[C]γ
order of reaction = α + β,
the order with respect to reactant A is α, to B is β, and to product C is γ.
rA = -k
zero-order reaction
rA = -k[A]
first-order reaction
rA = -k[A2]
second-order reaction
rA = -k[A][B]
second-order reaction
batch reactors and flow reactors.
fill-and-draw
Unsteady state
material flows into, through, and
out of the reactor
IDEAL REACTORS
REAL REACTOR
Conserved system: where no chemical or biological
reaction takes place and no radioactive decay occurs for
the substance in the mass balance.
 Steady-state:
Input rate = Output rate  Accumulation =0

Decay rate = 0
Accumulation rate = 0
Stream
Qs
Cs
Qw
Wastes
Cw
Qm
Mixture
Cm
Q = flow rate
C = concentration
CsQs + QwCw = QmCm
For non-conservative substances
Accumulation rate = input rate – output rare ±
transformation rate

With first-order reactions
Total mass of substance = concentration x volume

 when V is a constant, the mass rate of decay of the substance is
first-order reactions can be described by r = -kC=dC/dt,

A well-mixed sewage lagoon is receiving 430 m3/d of
sewage out of a sewer pipe. The lagoon has a surface
area of 10 ha and a depth of 1.0m. The pollutant
concentration in the raw sewage discharging into the
lagoon is 180 mg/L. The organic matter in the sewage
degrades biologically in the lagoon according to firstorder kinetics. The reaction rate constant is 0.70 d-1.
Assuming no other water losses or gains and that the
lagoon is completely mixed, find the steady-state
concentration of the pollutant in the lagoon effluent.
Decay
Cin = 180 mg/L
Qin = 430 m3/d
Sewage
Lagoon
Ceff = ?
Qeff = 430 m3/d
Control volume
Accumulation=input rate – output rate – decay rate
Assuming steady-state condition, accumulation = 0
input rate = output rate + decay rate
CinQin = CeffQeff + (K)(Clagoon)(V)
Ceff=1.10mg/ L
dM/dt = ?
d (in) d (out )

0
dt
dt
dM
  kCV
dt
dM
dC

V
dt
dt
dC
  kC
dt
C  Co e
 kt
Mass balance for each plug element
dM
d (in) d (out )
d (C )


V
dt
dt
dt
dt
No mass exchange occurs across the plug boundaries, d(in) and
d(out) = 0
dM
dC

V
dt
dt
dC
  kC
dt
Cout  Cin e
 kt
Cin  Cout e
 k
Residence time
L=length
The residence time for each plug:
( L) ( A) V


(u ) ( A) Q
Cout
( L)
V
ln
 k  k
 k
Cin
(u )
Q
A wastewater treatment plant must disinfect its effluent before
discharging the wastewater to a near-by stream. The
wastewater contains 4.5 x 105 fecal coliform colony-forming
units (CFU) per liter. The maximum permissible fecal coliform
concentration that may be discharged is 2,000 fecal coliform
CFU/L. It is proposed that a pipe carrying the wastewater be
used for disinfection process. Determine the length of the pipe
required if the linear velocity of the wastewater in the pipe is
0.75 m/s. Assume that the pipe behaves as a steady-state
plug-flow system and that the reaction rate constant for
destruction of the fecal coliforms is 0.23 min-1.
Cin=4.5x105 CFU/L
U=0.75 m/s
Cout=2,000 CFU/L
U=0.75 m/s
Using the steady-state solution on the mass-balance equation, we obtain
ln
Cout
L
 K
Cin
u
2000CFU / L
L
1
ln 
 23 min
5
4.5 x10 CFU / L
(0.75m / s )(60 s / min)
Solving for the length of pipe, we have
𝐿
-3
-1
ln(4.44x10 )=-0.23min
45𝑚/𝑚𝑖𝑛
L=1,060m
A contaminated soil is to be excavated and treated in a
completely mixed aerated lagoon. To determine the time
it will take to treat the contaminated soil, a laboratory
completely mixed batch reactor is used to gather the
following data. Assuming a first-order reaction, estimate
the rate constant, k, and determine the time to achieve
99 % reduction in the original concentration.
Time (d)
1
16
Waste Concentration (mg/L)
280
132

Using the 1st and 16th day, the time interval t= 16-1 = 15 d
Ct
 kt
e
Co
132m g / L
 k (15 d )
e
280m g / L
Solving for k, we have k = 0.0501 d-1
To achieve 99 % reduction the concentration at
Time t must be 1 – 0.99 of the original concentration
Ct
 0.05 ( t )
 0.01  e
Co
t = 92 days
Concentration, Cin
0
Time
0
Time
0
Time
Step increase
Step decrease
pulse/spike increase
Concentration, Cin
0
Time
Ct
 kt
e
Co
1/k
Time
1/k
Time
Decay
Formation
C1
Concentration, Cin
Concentration, Cin
C1
C0
-0.37C0 + 0.63C1
C0
C0
0
t=
Time
Effluent concentration
Concentration, Cin
Concentration, Cin
0
Time
Influent concentration
C0
0.37C0
For balanced flow (Qin = Qout) and no reaction, the mass
balance becomes
dM
 cinQin  cout Qout
dt
Where M = CV. The solution is


 t
Ct  C0 exp 
 

Where  = V/Q


 t
  C1 1  exp 

 






Flushing of nonreactive contaminant from a CMFR by a
contaminant-free fluid
Which means Cin = 0 and the mass balance becomes
dM
 cout Qout
dt
Where M = CV. The initial concentration is C0=M/V
For time t  0 we obtain

 t
Ct  C0 exp 
 




Before entering an underground utility vault to do
repairs, a work crew analyzed the gas in the vault and
found that it contained 29mg/m3 of H2S. Because the
allowable exposure level is 14 mg/m3 the work crew
began ventilating the vault with a blower. If the volume
of the vault is 160 m3 and the flow rate of contaminantfree air is 10 m3/min, how long will it take to lower the
H2S level to that will allow the work crew to enter?
Assume the manhole behaves as CMFR and that H2S
is nonreactive in the time period considered.

For balanced flow (Qin = Qout) and first-order reaction the
mass balance becomes
dM
 CinQin  Cout Qout  kC outV
dt
Where M = CV. By dividing with Q and V we have
dC
1
 Cin  Cout   kC out
dt

For stead-state conditions dC/dt=0
Cout
Co

1  k
Cout
OR
Decay
Co

1  k
Formation
Co
0
Time
Concentration, Ceff
Concentration, Cin
Co
0
Time
A step decrease in influent concentration (Cin=O)
for non-steady-state conditions with first-order decay
dM
 0  Cout Qout  kC outV
dt
Where M = CV. By dividing with Q and V we have
dC  1

   k Cout
dt  

Cout
  Co
 Co exp 
  1  k
 
t 
 
Concentration, Cin
Concentration, Cin
C0
0
Time
Influent concentration
C0
0
Time
Effluent concentration
A chemical degrades in a flow-balanced, steady-state
CMFR according to first-order reaction kinetics. The
upstream concentration of the chemical is 10 mg/L and the
downstream concentration is 2 mg/L. Water is being treated
at a rate of 29 m3/min. The volume of the tank is 590 m3.
What is the rate of decay? What is the rate constant?


For a first-order reaction, the rate of decay, r =-kC, thus we
have to solve for kC from
dM
 CinQin  Cout Qout  kC outV
dt
For steady-state, dM/dt = 0 and for balanced flow, Qin =
Qout
CinQin  Cout Qout
(10m g / L  2m g / L)(29m3 / min)
r  kC 

V
580m3

r=kC=0.4

For a first-order reaction in a CMFR
Cout

Co

1  k
The mean hydraulic detention time is
3
V
580m
 
 20 min
3
Q 29m / min

Solving for the rate constant we get
k
(Co / Cout )  1

(10 mg / L / 2mg / L)  1

 0.20 min 1
20 min
Report format
20 to 30 pages
Cover page: Title and Group member names
Table of content
References
Last Date of Submission: June , 2011