ENVE-2110
Lecture#6
BATCH REACTOR
 Radon 222 is a naturally occurring radioactive gas formed
by the decay of radium-226, a trace element in soil and rock.
The radioactive decay of radon can be described by the
elementary reaction
radon  222  polonium
 218  alphaparticle
The rate constant for this reaction is k=2.1x10-6 s-1 independent
of temperature. At time t=0 a batch reactor is filled with air
containing radon at concentration C0. How does the radon
concentration in the reactor change over time?
 For t>0 only one reaction influences the concentration
of radon (CRn). The change in radon concentration is
related to the reaction rate by expression
r 
 dC Rn
b/c consumed

dt
 Since radon decay is an elementary reaction, we expect the rate to
be first-order (always true for radioactive compounds)
r   kC Rn
Differential equation can be solved by direct integration and
rearrangement (see notes from Thursday and today), we obtain
C Rn ( t )  C 0 exp(  kt )
The radon concentration decays exponentially towards zero.

CSTR
 Consider a 10 x 106 lake fed by a polluted stream
having a flow rate of 5.0 m3/s and pollutant
concentration of 10.0 mg/L. There is also a sewage
outfall that discharges 0.5 m3/s of wastewater having a
concentration of 100 mg pollutant/L. The stream and
sewage wastes have a decay rate coefficient of
0.20/day. Assuming the pollutant is completely mixed in
the lake and assuming no evaporation or other water
losses or gains, find the steady state pollutant
concentration in the lake.
DRAW A SCHEMATIC
Sewage outfall
Qw = 0.5 m3/s
Cw = 100 mg/L
Incoming stream
Outgoing stream
Qout =?
Cout = ?
Qin = 5.0 m3/s
Cin = 10.0 mg/L
Inputrate
 outputrate
 kCV
V
dC
dt
 Q in C in  Q w C w  Q out C out  kCV
AT STEADY STATE
V
dC
dt

dM
dt
0
Completely mixed

Q total  Q out  Q in  Q w
Input rate –output rate –kCV = accumulation rate = 0
 Thus,
Q out C out  kC out V  Q in C in  Q w C w
C out (Q out  kV )  Q in C in  Q w C w
C out 
C out 

(Q in C in  Q w C w )
Q out  kV
(Q in C in  Q w C w )
(Q in  Q w )  kV
 Insert values
C out 
(5 * 10 )  (0.5 * 100 )
(5  0.5)  0.2 * 10 * 10
6

100
5.5  (( 0.2 * 10 * 10 ) * (
6
1
24
)*(
1
3600

))
100
28 .65
 3.5 mg / L
PFR
 Every year herons, seagulls, eagles, and other birds mass
along a 4.75 km stretch of stream connecting to a lake and
to the ocean to catch the fingerling salmon as they migrate
downstream to the sea. The birds are efficient fishermen
and will consume 10,000 fingerlings per km of stream each
hour regardless of the number of salmon in the stream. In
other words, there are enough salmon; the birds are only
limited by how fast they can catch and eat the fish. The
stream’s average cross-sectional area is 20 m2 and salmon
move downstream with the stream’s flow rate of 700 m3/min.
If there are 7 fingerlings per m3 in the water entering the
stream, what is the concentration of salmon that reach the
ocean when the birds are feeding?
DRAW A SCHEMATIC
LAKE
Qin = 700 m3/min
Cin=7 fish/m3
OCEAN
L
L = 4.75 km
A = 20 m2
Qout=700 m3/min
Cout=?
 Since the birds eat the salmon at a steady rate that is
not dependent on the concentration of salmon in the
stream, the rate of salmon consumption is zero-order.
k 
10 ,000 fish * km
1
2
* hr
20 m * 1,000 m / km
1
 0.50 fish * m
For a steady-state PFR
C out  C in  kt

3
* hr
1
 Insert values
t 
V
Q
2

4.75 km * 20 m * 1000 m / km
3
700 m / min* 60 min/ hr
 2.26 hrs
 Solve equation by inserting values:
C out  7 fish / m  (0.5 fish / m hr * 2.26 hr )  5.9 fish / m
3
2
3
OTHER EXAMPLES
 EXAMPLE 1:
A pipe from municipal wastewater treatment plant
discharges 1.0 m3/s poorly treated effluent containing
5.0 mg/L of phosphorous compounds reported as mg
P/L into the river with an upstream flow rate of 25 m3/s
and background phosphorous concentration of 0.010
mg P/L. What is the resulting concentration of
phosphorous in mg P/L in the river just downstream of
the plant flow?
DRAW A SCHEMATIC
Qu, Cu
Qd, Cd
River River
Treatment plant
With pipe discharge.
 Determine the phosphorous concentration downstream
of the discharge pipe, Cd. To find Cd, use the standard
mass balance equation with steady-state conditions
and without chemical formation or decay:
dM
dt
dM

dt
  in   out  r V
 (Q u C u  Q e C e )  Q d C d  0  0

INPUT VALUES
Cd 
Q uC u  Q e C e
Qd

(0.01 * 25 )  (5 * 1)
(25  1)
 0.20 mg / L
 EXAMPLE 2:
The CSTR shown below is used to treat industrial waste,
using a reaction that destroy the pollutant according to
first-order kinetics, with k=0.216/day. The reaction
volume is 500 m3, the volumetric flow rate of the single
inlet and exit is 50 m3/day, and the inlet pollutant
concentration is 100 mg/L.
[a] What is the outlet concentration after treatment?
[b] Calculate the hydraulic retention time.
Qin, Cin
Qout, Cout
SOLUTION
 An obvious C.V. is the tank itself. The problem is asking
for a single, constant outlet concentration, and all
problem conditions are constant. Therefore, this is a
steady-state problem. The mass balance equation with
first-order decay term is
dC
dt
  kC
and
dM
dt
  in   out  rV
Therefore,

 C
Q in C in  Q out
 VkC  0
out
SOLVE
 Solve for Cout and assume Qin=Qout=Q
C out 
QC
3
in
Q  kV

(100 mg / L )( 50 m / d )
50 m / d  (0.216 / d * 500 m )
3
3
[b] solve for hydraulic retention time:
 
V
Q

500 m
3
3
50 m / d
 10 d
 32 mg / L