Mass Balance, Reactors and Kinetics
T1 – Title
T2 – Some definitions
Environmental Engineering and Sustainability
In our every deliberation we must consider the impact of our decisions on the next seven
generations. Iroquois Confederacy
We seek to meet the needs of the present without compromising the ability of future
generations to meet their own needs. World Commission on Environment and
Development, 1987
How do we do this? Mathematical models provide a means of quantifying the impacts of
our decisions.
What is a Model?
A mathematical model is an idealized formulation that represents the response of a
physical system to external stimuli. Chapra 1997, p. 10
T2 (continued) – Cannon example
Why Model?
Consider modeling as an alternative to build and measure, providing a more rational
basis for making water quality control decisions, such a basis to include a defensible,
credible, predictive framework, within the larger framework of cost-benefit analysis.
T3 – the bridge to nowhere
T4 – the bridge falling down
T5 – when and to what extent
Reactor Analogs
Models in environmental engineering are based on reactor theory, with three reactor
analogs commonly employed: the completely mixed flow reactor (CMFR), the batch
reactor and the plug flow reactor (PFR).
1
T6 – Reactor analogs in engineered systems
T7 – Reactor analogs in natural systems
Ask students to explore mass balances in their lives, e.g.


a soaking rain T8 – The Completely-Mixed Flow Reactor
buying a Lexus T9 – The Completely-Mixed Flow Reactor
Model Development
The Law of Conservation of Mass states that mass can neither be created nor destroyed,
but it may change form. This means that the change in mass in a control volume must
equal the mass entering minus the mass leaving plus or minus the mass reacting
(changing form).
The Law of Conservation of Mass forms the basis for one of the most important tools in
environmental engineering - the mass balance.
The mass balance provides a means for constructing a budget for a material (mass) which
is no more complicated than the soaking rain or Lexus example introduced previously.
Let's apply the principles evident in those examples to the case of chemical mass:
M t  t  M t  Min , t t  t  Mout , t t  t  M reacted , t t  t
The analysis is performed over the time period t. Moving the initial mass to the other
side and dividing by t yields:
M t  t  M t M in , t  t  t M out , t  t  t M reacted , t  t  t



t
t
t
t
The left hand side of the equation is the rate of change in chemical mass, i.e.
M t  t  M t M

t
t
and as t  0,
M dm

t
dt

The terms on the right hand side are each a mass flux ( m , units of mass per time), i.e. the
rate at which mass enters, exits or reacts within the system. This can then be written:


dm 
 min  mout  m reaction
dt
2
This is the governing equation for mass balances throughout environmental engineering.
It remains to identify an approach for quantifying the terms in this equation.
The Control Volume
We begin our exploration of the mass balance using the CMFR reactor analog.
T10 – The Completely-Mixed Flow Reactor
The mass balance is written for a control volume, i.e. a specific region of space which has
boundaries across which the mass flux in and mass flux out can be determined.
T11 – The Control Volume
The control volume for the CMFR is the entire reactor, taken to include the inlet and
outlet.
Terms in the Mass Balance
There are four terms in the mass balance:


dm 
 min  mout  m reaction
dt
the rate of mass accumulation, mass flux in, mass flux out and the rate of mass reaction.
Rate of mass accumulation
As mentioned earlier, it is concentration, not mass, which is of great importance in
environmental systems. For this reason, we write the mass balance in terms of
concentration, remembering that:
m  VC
and substituting:
bg
dm d VC

dt
dt
For our purposes, and in most environmental applications, it can be assumed that the
volume of the reactor is constant. Thus:
dm
dC
 V
dt
dt
Note that the assumption of constant volume requires that inflow equal outflow.
3
Mass flux in
The mass flux in is most commonly quantified as the product of the volumetric flow rate
(Q) and the chemical concentration in the inflow (Cin):

min  Qin  Cin
and given constant volume, Qin = Qout,

min  Q  Cin
Note the way that the units work out in this equation:
g m3 g


d d m3
Mass flux out
The mass flux out is calculated in a similar way:

mout  Qout  Cout
and again given constant volume, Qin = Qout,
m out  Qout  Cout
again, with units:
g m3 g


d d m3
But note that since the chemical concentration is the same anywhere in a CMFR,
including at the outlet, we can refer to Cout as simply C:

mout  Q  C
Rate of mass reaction
This term quantifies gains or losses of a chemical as a result of biological or chemical
reactions. Chemical reaction rates are typically expressed in terms of concentration, not
mass, as it is concentration which often drives the reaction:
4

mrxn  V 
dC
dt rxn only
note the dimensional character of this term:
g
g
 m3  3
d
m d
Some materials (e.g. chloride) are termed conservative, i.e. they do not participate in
biological or chemical reactions. In this case, the expression for the rate of mass
accumulation (V·dC/dt) does not include a reaction term:

mrxn  V 
dC
0
dt rxn only
Our consideration of materials that do participate in such reactions evolves from the
Kinetic Rate Law:

mrxn  V 
dC
 V  k Cn
dt rxn only
where k is the reaction rate coefficient and n is the reaction order.
We will consider two reaction orders, although there are others (e.g. second order, mixed
order):

Zero Order: here the rate of reaction is constant and independent of the
concentration of reactant. For a zero order decay:

mrxn  V 
dC
 V  k  Cn   V  k  C0  V  k
dt rxn only
where k has units of M∙V-1∙T-1. Integrating the dC/dt portion of the equation
permits prediction of the time course of reaction:
dC
  k  C , integrating , Ct  C0  k  t
dt
and a plot of C versus t yields a straight line with a slope, k.
T12 – Example of zero order kinetics, oxygen depletion in Dollar Bay.
5

First Order: here the rate of reaction is not constant and depends on the
concentration of reactant. For a first order decay:

mrxn  V 
dC
  V  k  C n   V  k  C1   V  k  C
dt rxn only
where k has units of T-1. Integrating the dC/dt portion of the equation permits
prediction of the time course of reaction:
dC
  k  C , integrating , Ct  C0  e  k t
dt
and a plot of lnC versus t yields a straight line with a slope, k.
T13 – Example of first order kinetics, radioisotope decay; half-life calculation
The magnitude of the reaction rate coefficient, k, can be influenced by a variety of
environmental factors, most notably temperature. The effect of temperature on the
reaction rate coefficient is calculated using a 'theta function':
k T  k 20  T20
The value of k increases with increasing temperature with  ranging from 1.00 (no
temperature effect) to ~1.2 (very strong temperature effect).
T14 – Temperature and kinetics
T15 – Example of first order kinetics, radioisotope decay; half-life calculation
Putting it all Together
The original mass balance expression:


dm 
 min  mout  m reaction
dt
can thus be written as:
V
F
I
G
HJ
K
dC
dC
 Q  Cin  Q  C  V 
dt
dt
rxn only
6
Assuming a first order decay (but remembering that there is also growth and reactions of
different orders), can be further written as:
V
dC
 Q  Cin  Q  C  V  k  C
dt
Steady state versus Non-steady state
Before trying our first application of the mass balance, we need to address a key concept:
steady state. It can be seen from the above equation that changes in flow, inlet
concentration, or reactivity result in changes in chemical concentration within the reactor.
If these conditions remain constant for a sufficiently long period, they come into balance
and the concentration and mass within the control volume remain constant. This
condition is termed steady state and since concentration is not changing:
dm
dC
 V
0
dt
dt
T16 –CMFR: discuss how the ‘QC’ and VkC’ terms allow the reactor to ‘catch up’.
If the flow, inlet concentration, and/or reactivity are changing, then the system is
considered non-steady state (dynamic or time-variable) and:
dm
dC
 V
0
dt
dt
This has important implications in solving the mass balance equation.
CMFR: Steady State - Conservative
The steady-state conservative case for a CMFR with a single inlet and outlet is quite
straightforward:
V
dC
 Q  Cin  Q  C  V  k  C
dt
0  Q  Cin  Q  C
and
C  Cin
and has relatively little application. Of more interest is the case where two inlet streams
meet with a single outlet.
7
CMFR: Steady State – Conservative – Multiple Sources
Consider the above example, but with two (or more) inputs:
V
dC
 Q1  Cin ,1  Q2  Cin ,2  Q  C  V  k  C
dt
Simplifying for the steady-state (because the time it takes to mix is relatively short),
conservative case and solving for C:
C
Q1  Cin ,1  Q2  Cin ,2
Q
This particular mass balance, i.e. one for a conservative substance at steady state, is often
referred to as a mixing basin calculation and is commonly employed wherever two
chemical streams join together. The approach can be applied to reactive materials as
well, because the residence time of the mixing basin is so short that there is no change
due to reaction.
T17 – Mixing basin example – Sonora River
Let’s consider the case of a river which receives a chloride input from a wastewater
treatment plant. The river has a flow (Qup) of 9 m3∙s-1 and a chloride concentration (Cup)
of 2 mg∙L-1 and the treatment plant has a flow (Qin) of 1 m3∙s-1 and a chloride
concentration (Cin) of 100 mg∙L-1. Determine the chloride concentration in the mixing
basin.
Cmb 
Qup  Cup  Qin  Cin
Qup  Qin

9  2  1 100
9 1

118
mg
 11.8
10
L
The power of this approach is evident in this more difficult example dealing with
industrial wastes.
Steady State - Conservative Example: Chloride in 9 Mile Creek
T18 – Chloride in 9 Mile Creek – background
T19 – Chloride in 9 Mile Creek - calculation
Here, the terms available to us in the previous mixing basin calculation are not all known.
Instead, we know the upstream flow (3.5 m3∙s-1) and chloride concentration (40 mg∙L-1)
and the mixing basin flow (4.2 m3∙s-1) and chloride concentration (1500 mg∙L-1) and can
back-calculate the input from the waste beds.
8
Beginning with the basin mixing basin equation:
Cmb 
Qup  Cup  Qin  Cin
Qup  Qin
and given that Qmb = Qup + Qin:
QmbCmb  Qup  Cup  Qin  Cin
4.2  1500  3.5  40  6160
The input flow is 4.2 – 3.5, yielding a wastebed chloride concentration of:
Qin  Cin 6160
mg

 8,800
Qin
0.7
L
To this point we have considered a conservative substance and multiple feeds in a CMF
reactor. While it might seem logical to continue with the CMF reactive case, we will
delay that for a bit and turn our attention to the other major reactor analog, the plug flow
reactor.
The Plug Flow Reactor
The plug flow reactor is typically applied to columns (in engineered systems) and rivers
(in natural systems). The quantitative basis for modeling the PFR begins with a simpler
system, the batch reactor.
Batch Reactor
T19 – Batch reactor
A reactor which has no inlet our outlet is termed a batch reactor. Mathematically:
V
F
IJ
G
HK
dC
dC
 Q  Cin  Q  C  V 
dt
dt
V
rxn only
F
I
G
HJ
K
dC F
dC I
 GJ
dt H
dt K
dC
dC
 V
dt
dt
rxn only
rxn only
9
and for a first order decay:
dC
  k C
dt
and integrating:
C t  C0  e  k  t
Consider the shape of the C = ƒ(t) curve.
T20 – Concentration = f (t) in a batch reactor
Plug Flow Reactor
One might think of the a plug flow reactor as a train of batch reactors. In a PFR,
complete mixing occurs in a radial direction, but there is no mixing in the axial direction.
Thus each 'plug' of fluid is considered to be a separate entity (mini-batch reactor) as it
flows down the pipe.
T21 – Single batch reactor in a pipe
T22-33 – Train of batch reactors in a pipe
Consider the shape of the C = f(x) curve for the train of reactors. The mass balance for a
PFR with first order decay is as follows:
V
F
I
G
HJ
K
dC
dC
 Q  Cin  Q  C  V 
dt
dt
V
rxn only
dC
  V k C
dt
C t  C0  e  k  t
The influent and effluent concentrations are related by the hydraulic retention time of the
pipe, i.e. substituting  for t in the equation above:
Cout  Cin  e  k 
10
and  is related to the length of the pipe (L) and the fluid velocity ():

L

where fluid velocity is a function of flow (Q) and the cross-sectional area of the pipe (A):

Q
A

and thus:
and thus:

LA
Q
LA
Q
Note that the apparent time dependence here is really a position dependence which is
governed by the velocity and downstream distance in the pipe:
t
L

where  is the fluid velocity and L is the downstream distance. Thus:
C L  C0  e
 k
L

Go To Water Quality
BOD and River DO
CMFR: Steady State - 1st Order Decay
T35 – CMF Reactor
The steady state case with first order decay for a CMFR is very common in both natural
(lakes) and engineered (treatment units) systems:
V
dC
 Q  Cin  Q  C  V  k  C
dt
At steady state, the mass balance is:
0  Q  Cin  Q  C  V  k  c
11
and solving for C:
C  Cin 
Q
Q V k
Steady State - 1st Order Decay Example: Portage Lake Treatment Plant
The Portage Lake Treatment Plant receives an inflow of 8000 m3·d-1 with a pollutant
concentration of 250 mg·L-1. The first unit operation removes the pollutant according to
first order kinetics with a rate constant of 0.5 d-1. The volume of the tank is 2500 m3.
Calculate the effluent pollutant concentration.
C  C in 
Q
8000
mg
 250 
 216
Q  Vk
8000  2500  0.5
L
Another way of looking at this application is to consider the tank size required to effect a
particular reduction in pollutant concentration.
Calculate the tank size (m3) required to achieve a 75% reduction in the effluent
concentration (i.e. an effluent concentration of 63 mg·L-1):
For a 75% reduction, C  0.25 Cin
0.25 C in  C in 
0.25 
Q
Q  Vk
Q
Q  Vk
0.25 Q  0.25 V  k  Q
V
0.75 Q
0.75 8000

 48000 m3
k
0.5 0.25
Effect of Kinetics and Volume on Css
Note in the steady state solution to the CMFR equation,
C  Cin 
Q
Q V k
that the effluent concentration can be reduced through increases in V and k. Why is this?
How can we influence these parameters?
12
Effect of Changes in Kinetics
An increase in the reaction rate coefficient leads to a reduction in the effluent
concentration and improved treatment because the reaction proceeds more rapidly.
Kinetics Example: Portage Lake Treatment Plant
Some reactions, such as the conversion of ammonia to nitrate, are highly sensitive to
temperature. Calculate the % removal for the case above (V = 48,000 m3) for
temperatures of 4°C and 20°C if  = 1.08.
At 20°C, k = 0.5 d-1, removal efficiency is 75% and the effluent has C = 63 mg·L-1, as
calculated previously.
At 4°C:
k4  k20 T  20  0.5 1.08(420)  0.5  0.29  0.15
C  Cin 
Q
8000
mg
 250 
 132
Q V k
8000  48000  0.15
L
Removal efficiency has dropped from 75% to 47% due to temperature impacts.
Effect of Changes in Volume
An increase in the reactor volume leads to a reduction in the effluent concentration and
improved treatment because the waste remains in the reactor for a longer time.
Hydraulic Retention Time
The volume effect is quantified through a concept termed the hydraulic retention time (,
units of time; also called detention time and residence time), i.e. the average period which
a volume of water spends within the reactor.
 
V
Q
Since this is the time available for reactions to proceed,  influences the degree of
treatment.
13
Hydraulic Retention Time Example: Portage Lake Treatment Plant
In an earlier example, we increased treatment efficiency from ~15% removal to ~75%
removal by changing the tank volume from 2500 to 48000 m3. What was the
corresponding change in hydraulic retention time?
1 
V1
2500

 0.33 d
Q
8000
2 
V2
48000

 6d
Q
8000
Increases in hydraulic retention time will result in reduced effluent concentrations and
improved treatment.
Thus we can see that Css, i.e. the effluent concentration in a CMF reactor at steady state,
can be influenced by retention time (e.g. reactor volume) and kinetics (e.g. temperature).
Time Variable Solution
When we choose the steady state solution to be mass balance equation, it is assumed that,

the system will reach steady state in a time period meaningful to the
application at hand and

the inputs to the calculation will remain constant until that steady state is
reached.
Let’s examine those assumptions.
T36 – Change from Css,1 to Css,2
If those conditions are not met, a time-variable solution is required,
V
dC
 Q  Cin  Q  C  V  k  C
dt
integrating for a step change from Css1 to Css2 yields,
F
G
H
1 I
F
F
IJ
 k J t
G
G
H
H
K
C t  Css1  e
 Css2  1  e  K
1
   k t
I
J
K
Let's look at this equation in its parts and then as a whole.
T37 – Time variable response
14
Response Time
The decision regarding a choice between the steady-state and time variable solution is
guided by a knowledge of how long it takes to reach steady state, i.e. the response time.
Response time, defined as the period required for a system to reach equilibrium (steady
state) with a new set of inputs or kinetics, is influenced by both the hydraulic residence
time and the kinetics of the target substance. Because a system never perfectly achieves a
steady state (100% of the Css value), it is necessary to establish a criterion for the
response time. In practice, we say that a system is at steady state when it has covered
95% of the distance between Css,1 and Css,2 (note that 90 or 99 or 99.9% could be used as
well.
Re-arranging the time variable solution,
F
G
H
F1 I
F
G k J t
G k IJ t
C t  Css1  e  H K  Css2  1  eH K
1
I
J
K
to yield,
Ct  Css ,2
Css ,1  Css ,2
1 
  k  t

 e 
and recognizing that the system has covered 95% of the distance from C ss,1 to Css,2 when
the left-hand side of the equation equals 0.05, gives the response time expression,
t95% 
 ln 0.05
3

1
1
k
k


Note that for a conservative substance, we can say that steady state is achieved after three
retention times:
t95% 
3
 3 
1

and that for a reactive substance it will reach steady state more quickly. Now we are
prepared to select a steady state or time-variable approach.
15
Selection of SS or Time-Variable Approach
As introduced previously the selection of an approach depends on the time necessary to
achieve steady state (as reflected in t95%) and the constancy of the inputs over that
interval. Let’s look at these factors for engineered and natural systems.
Treatment Facilities
Evaluation of treatment performance is conducted on a monthly or weekly (and
occasionally, daily) basis. As can be seen in T37,
T37 – Response time
the hydraulic retention times for most unit operations in water and wastewater treatment
are short and the response times are on the order of hours to days. This meets the criteria
for a steady state solution.
Inputs to treatment plants vary seasonally, but are reasonably constant over the response
time of the unit operations. An exception to this is diurnal flows which are
accommodated with equalization basins. This also meets the criteria for a steady state
solution.
Natural Systems
Evaluation of water quality conditions in lakes is typically considered on an annual basis.
While this would suggest that a steady state solution is valuable, the hydraulic retention
time of lakes varies widely and thus so does the utility of a steady state approach:
Onondaga Lake (0.25 yr), Lake Ontario (8 yr), Lake Michigan (136 yr), Lake Superior
(179 yr). Here, that application is lake and chemical (due to kinetics) specific.
Seasonal variation in inputs and kinetics must also be considered. This is especially
important with respect to spring runoff and biological features and depends on the
specific case.
Comparison of CMFR and PFR
Efficiency
Consider the case of PF and CMF reactors having the same volume (V), inflow
(Q), influent concentration (Cin), and reaction kinetics (k). The PFR is more efficient in
removing the chemical and thus the PFR effluent concentration would be lower than that
for the CMFR. This occurs because the influent to the CMFR is immediately mixed with
the more dilute reactor contents, reducing C and minimizing the value of the reaction
term:
V  k C
16
[P] PFR vs. CMFR efficiency; Mihelcic, Figure 4-8, p. 159
Therefore a smaller (and thus more economical) PF reactor would be required to meet the
same effluent standards.
Influent spikes
Many pollution control systems encounter fluctuations or spikes in their influent
concentrations. These spikes would simply travel down a PFR, but the complete mixing
in a CMFR tends to dampen the impact of spikes on the effluent concentration.
[P] PFR vs. CMFR – sensitivity to spikes; Mihelcic, Figure 4-9, p. 160
Selection of a Reactor
The selection of a PFR or CMFR in engineered systems is guided by efficient
utilization of tank volume (favoring PFRs) and sensitivity to inlet spikes (favoring
CMFRs). In natural systems, the choice is based on which best describes the lake or
river.
17