Download Chapter 3. Analysis of Environmental System 3.1 Analysis of a

Chapter 3. Analysis of Environmental System
3.1 Analysis of a Simple Environmental System
A simple environmental system can be analyzed using simple differential equations. Let
us look over a basic mechanism arising in environmental system analysis.
3.1.1 Analysis of a Lake
We may assume that a lake is a Completely Mixed Flow Reactor (CMFR). Then the
phenomena may be described in a differential equation as follows.
Fig. 3.1 Schematic diagram of a lake
If we write a mass balance for the lake regarding a pollutant which is being discharged
into the lake, it may be expressed as the following.
V
dC
(3.1.1)
 QC0  W1  W2  W3  ......  QC  VkC
dt
where V : volume of a lake (m 3 )
C0 : concentration of a pollutant
( Example : P, N ( g / m 3 ))
Q :
flow rate(m 3 / day)
C : concentration of a pollu tan t
in lake( g / m 3 )
W1 , W2 , ... : disch arg ed amount of pollu tan t
( g / day)
k : reaction coefficien t (day1 )
Rearranging Eq.(3.1.1),
V
dC
 (Q  Vk )C  QC 0  W1  W2  ......
dt
(3.1.2)
Dividing both sides of Eq.(3.1.2),
dC  Q
Q 1

   k C   (W1  W2  ......)
dt  V
V V

(3.1.3)
Eq.(3.1.3) may be solved without any difficulty if we follow the process as we have
done so far. Especially, when discharging rate of a pollutant into the lake is constant, it
is much easier because the condition is steady state, dC/dt = 0.
3.1.2 Analysis of Chemical Reaction Using Differential Equations
Chemical reactions arising in an environmental system may be solved using differential
equations. Let us take a look various chemical reactions.
ka
A  B
(3.1.4)
kb
As shown in reaction (3.1.4), a chemical reaction may be expressed using a differential
reaction. This chemical reaction is a reversible reaction where A is converted to B, and
at the same time B is converted to A.
dC A
  A  k a C A  kb C B
dt
(3.1.5)
dC B
  B  k a C A  kbCB
dt
(3.1.6)
If we solve the equations, (3.1.5), and (3.1.6), simultaneously, we may understand the
way which the chemical is converting from A to B. In equilibrium state, reaction
coefficients, Ka, and Kb are the same, and
where
dC A dC D

 0.0 .
dt
dt
C A : Concentration of A (mg/l)
C B : Concentration of B (mg/l)
t : time (min)
 A : reaction reate of A (mg/l  min)
 B : reaction rate of B (mg/l  min)
k a : reactio rate coefficient of A(min -1 )
k b : reaction rate coefficient of B(min -1 )
3.2
Chemical Reactions
A general type of chemical reaction taking place in environmental system may
be expressed like Eq.(3.2.1).
k1
aA  bB 
cC  dD
k
(3.2.1)
2
Reaction equation (3.2.1) means that a moles of material A and b moles of material B
react chemically to produce c moles of material C, and d moles of material D. At the
same time, c moles of material C and d moles of material D react to produce a mole of
material A, and b moles of material B.
To understand the way of changing of the materials with time, their chemical reaction
may be expressed in differential equations as the followings, which are mass balances of
each species.
1 d [ A]

  k1[ A]a [ B ]b  k 2 [C ]c [ D ]d
a dt
(3.2.2)
1 d [ B]

  k1[ A]a [ B ]b  k 2 [C ]c [ D ]d
b dt
(3.2.3)
1 d [C ]

 k1[ A]a [ B ]b  k 2 [C ]c [ D ]d
c dt
(3.2.4)
1 d [ D]

 k1[ A]a [ B ]b  k 2 [C ]c [ D ]d
d dt
(3.2.5)
From Eq. (3.2.2) ~ Eq(3.2.5), we may deduce the following relationships in Eq.(3.2.6).
1 d [ A]

a dt

1 d [ B]

b dt

1 d [C ]

c dt

1 d [ D]

d dt
(3.2.6)
In reaction formula (3.2.1), according to a, b, c, and d, coefficients of Eq. (3.2.2.) ~ Eq.
[3.2.5] and order of reactions are determined. The followings are the some examples.
1. 0th Order Reaction
k
A

product
(3.2.7)
A mass Balance equation for the reaction (3.2.7) is as follows.
d [ A]
 k[ A]0
dt

d [ A]
 k
dt
(3.2.8)
When the type of changing of material A is like Eq. (3.2.8), the reaction is called 0th
order reaction.
2. 1st order reaction
k
A

product
(3.2.9)
Mass Balance for the reaction (3.2.9) may be written as follows.
d [ A]
 k[ A]1
dt
(3.2.10)
When material A is changing by Eq.(3.2.10), the reaction is called 1st
order reaction.
3. 2nd order reaction
k
2A 

B
product
(3.2.11)
Mass Balance equation of the chemical reaction for Eq.(3.2.11) can be expressed as the
followings.
1 d [ A]

 k[ A]2
2 dt
(3.2.13)
d [ B]
 k[ A]2
dt
(3.2.13)

When chemical reaction for (3.2.11) is expressed like Eq. (3.2.12), and Eq. (3.2.13), the
reactions are 2nd order reaction.
As shown so far, order of chemical reaction may be determined by type of chemical
reaction. In water chemistry, chemical reaction rates generally follows to
Eq.(3.2.2)~Eq.(3.2.5). However, the reaction rate may not follow the rule which we
have investigated in combustion environment related with air pollution.
3.3
Application
of
Differential
Equations
in
Solving
Environmental System
Problem 1>
A process reaction in a completely mixed lake follows first order kinetics with a rate
constant K=0.1day -1. The influent concentration in stream is Cio and the volume of
the Lake in which the process occurs in 4,000m3. Determine the detention time
required in the lake to promote a 80% conversion of reaction.
Sol)
V

Q
V
(det ention time)
dC
 W  QC  kVC
dt
 QCio  QC  kVC
At steady state,
V
dC
0
dt
QCio  QC  kVC  0
Q
Q
Cio  C  kC  0
V
V
Q
(Cio  C )  kC
V
1
(Cio  C )  kC


Cio  C
kC
80% Conversion C  0.2Cio
 

Problem 2>
Cio  C
C  0.2Cio
 io
kC
0.1  0.2  Cio
0. 8
 40 (day)
0.02
A first order reversion reaction
k1
A  B
k2
1) It is known that
when CA=1mole/liter ,
 rA 
CB=0.5mole/liter and
d  A
 0.2(mole /(liter  sec)).
dt
2)  rA  0.9(mole /(liter  sec))
when CA=3.5mole/liter ,
Find k1 and k2.
Sol)
CB=1.5mole/liter.
If we write mass balance for the reaction,
d  A
1
1
 rA   k1  A  k 2 B   0.2,
dt
when C A  1, C B  0.5
 k111  k 2 0.51  0.2
 k1  0.5k 2  0.2
d  A
1
1
 rA   k1  A  k 2 B   0.9,
dt
when C A  3.5, C B  1.5
 k1 3.51  k 2 1.51  0.9
 3.5k1  1.5k 2  0.9
 k1  0.5k 2  0.2  (1)
 3.5k1  1.5k 2  0.9  (2)
(1)  3  (2)
 0.3  0.5k1
 k1  0.6 (sec1 )
Plugging k1 to (1)
0.4  0.5k 2
Problem 3>
 k 2  0.8 (sec1 )
The reaction
K
A

2B
Is observed as a 3/2reaction. When the reaction starts, CA= 2 moles/l
CB=0.0 moles/l. Plot CB and CB vs. time. Where
Sol)
d  A
3/ 2
  k  A
dt
rA 
Separaing var iables
d  A
  kdt
A3 / 2
d  A
  A   k  dt
3/ 2
 2 A
1 / 2
A  2
  kt  C
at t  0
 C
2 A
1 / 2
2
 2
2
 kt  2
kt  2
2
2
 2 
  A  

 kt  2 
A
1 / 2

k=0.05sec-1(mole/l)-1/2.
and
In
aA  bB reaction,
relationship between reaction rates may be written as,
1
1
rA   rB
a
b
Therefore in the given reaction A  2B, reaction rate can be written,
1 d  A
1 d B 

1 dt
2 dt
 d A   2  d B 
1
2 A  [ B ]  C
When t  0,
A  2,
and B   0
 C4
B   4  2A
 2 
 B   4  2

 kt  2 
2