Material Balance
(Nazaroff & Alvarez-Cohen, Section 1.C.2 + lots more here)
In the environment, we are not dealing only with local processes, such as
chemical reactions; we are mostly dealing with interconnected systems,
implying that the system under consideration is most often connected to
other systems through imports and exports.
We must therefore reckon with material balances in the presence of open
systems.
Inflow
Water
example:
Exchange
with the
atmosphere
Outflow
Inside:
Chemical reactions
Biological processes
Physical settling
Step 1: Choose a control volume.
This is an art rather than a science.
Rule 1: Boundaries must be defined clearly; you need to
know whether something is inside or outside.
Rule 2: The budget must yield practical information.
Step 2: Select the substance for which the budget is to be made.
Rule: Be specific (For example: Budget for S or SO2?)
Step 3: Consider all imports and exports.
Add sources, subtract sinks.
1
Examples of practical control volumes
plume
section
urban airshed
entire lake
Distinction
(Nazaroff & Alvarez-Cohen, Section 5.A)
Completely Mixed Flow Reactors (CMFRs)
are control volumes for which spatially uniform properties may be assumed.
Examples: A room in a building, a small pond, or an urban airshed.
Plug-Flow Reactors (PFRs)
are systems along which properties vary.
They need to be split into a series of sequential control volumes.
Examples: A river, an estuary, or a smokestack plume progressing downwind.
The analysis on the next slide applies only to CMFRs
or to a short segment of a PFR.
2
Mass balance for a CMFR
Qout , Cout
Accumulation
=  imports –  exports
+  sources –  sinks
amount at (t  dt )  amount at (t )
Accumulation 
duration dt
out
V C (t  dt )  V C (t ) d (VC )
dC


 V
dt
dt
dt
mass volume of carrying fluid
  Cin Qin
 imports  inlets
 volume 
time
inlets
mass volume of carrying fluid
  Cout Qout
 exports  outlets
 volume 
time
outlets
Qin Qin
Qin , Cin
 sources  S (to be specified according to application)
 sinks  (Decay constant)  (Amount present)
 KV C
dC
 V
  Cin Qin   Cout Qout  S  K V C
dt inlets
outlets
where V = volume of control volume (in L)
C = concentration of substance (in g/L)
Q = volumetric flux of fluid (in L/s)
S = sum of emissions (in g/s)
K = decay constant (in 1/s)
Particular cases
1. Steady state: Concentration remaining unchanged over time
dC
0 
dt
C
in
Qin  C
inlets
Q
out
 S  KV C  0  C 
outlets
C
Q
in
Qin  S
inlets
out
 KV
outlets
2. Conservative substance: No source (S = 0) and no decay (K = 0)
V
dC


  Cin Qin    Qout  C
dt inlets
 outlets

3. Isolated system: No import and no export (all Q’s = 0)
V
S   Kt
S 
dC
 S  KVC  C (t ) 
  Cinitial 
 e (if S is constant over time)
KV 
KV 
dt
3
Example of Material Balance
A lake contains V = 2 x 105 m3 of water and is fed by a river discharging
Qupstream = 9 x 104 m3/year. Evaporation across the surface takes away
Qevaporation = 1 x 104 m3/year, so that only
Qdownstream = 8 x 104 m3/year exits the lake in the downstream stretch of the river.
The upstream river is polluted, with concentration C = 6.0 mg/L.
Inside the lake, this pollutant decays with rate K = 0.12/year.
Qupstream
Qevaporation
Take volume V of the entire lake as
the control volume.
V K
Assume steady state
(= situation unchanging over time).
C
Qdownstream
Budget reduces to:
dC
 QupCup  QevapCevap  Qdown Cdown  KVC
dt
0  Qup Cup  (Qdown  KV ) C
Solution is:
C
Budget is:
V
Qup Cup
Qdown  KV

(Cdown  C )
(9  10 4 m 3 /yr)(6.0 mg/L)
 5.19 mg/L
(8 10 4 m 3 /yr)  (0.12/yr)(2  105 m 3 )
Variation on the preceding example
Un-aided (natural) remediation
Suppose that the source of pollution in the upstream river has now been eliminated.
The entering concentration in the lake has thus fallen to zero. Slowly, the concentration
of the pollutant decays in the lake because of its chemical decay (K term) and flushing
(Q term).
The equation becomes:
dC
 QupCup  QevapCevap  Qdown Cdown  KVC
dt
dC
Q

  down  K  C
dt
V


V
The solution to this equation is:
 Q

C (t )  Cinitial exp   down  K 
V




t   (5.19 mg/L) exp(0.52 t )

4
From this solution, we can find that:
It takes 1.33 years for the concentration
to drop by 50%.
If the acceptable concentration is 0.10 mg/L,
it takes 7.6 years.
Question: If 7.6 years is too long, what can be done?
Check time scales of the problem:
● Residency time = V/Q = (2 x 105 m3)/(8 x 104 m3/yr) = 2.50 years
● Decay time = 1/K = 1 / (0.12/yr) = 8.33 years
Conclusion: Decay is slow and flushing comparatively fast.
Flushing is primarily responsible for the natural cleaning of the lake.
Adding a chemical to speed up decay would only bring incremental
change, while increasing the flushing rate would have greater impact.
5
Mass balance for a PFR
(Nazaroff & Alvarez-Cohn, Section 5.A.3)
Most often when the Plug-Flow Reactor model is used,
the actual system under consideration (river, scrubber,
electrostatic precipitator) exhibits no perceptible
variation over time and contains no internal sources.
The only thing that is happening is a transformation
along the movement of the substance being traced.
Qin Qin
out
The formulation below holds in the case of a gradual removal of the substance, modeled
as a first-order reaction.
The material budget is most easily established if one takes the control volume as a little
piece dV of the moving material (thus containing C dV amount of the substance).
Following this little piece of material, there is no addition or removal from it; it is a mini
closed system, with the balance for the embedded substance reducing to:
dV
dC
  K (dV ) C
dt
The solution to this equation is:
C (t )  C (0) e  Kt
Connecting the end concentration Cout = C(t=) to the entrance concentration Cin = C(t=0) ,
we write:
Cout  Cin e  K
in which  = V/Q is the total transit time through the system
also called the residence time in the system.
Comparison between CMFR and PFR
(Mihelcic & Zimmerman, pages 122-125)
At equal system volume V and throughflow Q, the plug-flow reactor (PFR) is more efficient
than the completely mixed flow reactor (CMFR). Indeed:
CMFR
Cout
Q
V
Q Cin  Q Cout  KV C 
Cout
1

Cin 1  K
PFR
Q
V
Cout  Cin e  K

either way with  
V
Q
Cout
 e  K
Cin
So, why don’t we always use
PFRs instead of CMFRs?
6
Residence time
(Nazaroff & Alvarez-Cohen, page 212)
The concept of “residence time” (alternatively called “detention time”
or “retention time”) exists to answer the question:
How long does the “stuff” stay “there”?
Definition: Residence time = Containing volume divided by flow rate:

V
Q
Notation:  , “theta”, the Greek letter for “th”.
The dimension is time (T) because V/Q has dimensions of L3/(L3/T) = T.
While the standard unit is the second, typical residence times in the environment
are measured in days, weeks, months or even years, depending on the size of the
containing volume (ex. small pond to large lake).
Example of residence time
(Mihelcic & Zimmerman, Section 4.1.6, page 128)
Age of water at Niagara Falls?
Lake
Volume
(x 109 m3)
Outflow
(x 109 m3/yr)
Residence
time (years)
Superior
12,000
67
179
Huron
3,500
161
21.7
468
182
2.57
Erie
Total:
203.4
We are now in 2015.
Subtracting 203 years brings
us back to 1812.
What was happening in the
world in 1812?
7
12 February 1812:
Napoléon Bonaparte authorizes the usage of
Mesures usuelles, which became the basis of
the metric system.
Emperor Napoléon Bonaparte in his study.
(Painting by Jacques-Louis David)
8