5 Reactor Vessels
Motivation
Fully-mixed reactors
Single & multiple tanks with pulse, step and
continuous inputs
Dispersed flow reactors with pulse and
continuous inputs
Examples
Introduction
m&
Qin
Qout
A tank, reservoir, pond or reactor with controlled in/out flow
If Qin = Qout = Q then tres = τ = t* = V/Q
Used interchangeably
Generally more interest in what comes out than what’s inside reactor
Applications
Natural ponds and reservoirs
Engineered systems (settling basins,
constructed wetlands, combustion
facilities, chemical reactors, thermocyclers…)
Laboratory set-ups: simple configurations
with known mixing so you can
predict/control the fate processes
Wastewater Treatment Plants
Secondary settling
Primary settling
Waste stabilization
Constructed Wetlands
Cell 6
Cell 5
Cell 7
Cell 12
Potash Evaporation Ponds
Southern shoreline Dead Sea
Classification
Flow: continuous flow or batch
Spatial structure: well-mixed or 1-, 2-,
3-D
Loading: continuous, intermittent (step)
or instantaneous (pulse)
Single reactor or reactors-in-series
Initially look at single continuously stirred tank reactor
(CSTR)
Well-mixed tank, arbitrary input
c
m& = Qin cin
t
c
dV
= Qin (t ) − Qout (t )
dt
d (cV )
= cin (t )Qin (t ) − c(t )Qout (t ) + riV
dt
t
Conservation of volume
Conservation of mass
Well-mixed tank
c
m& = Qin cin
c
t
t−τ
dc cin (t ) − c
=
− kc
dt
t*
c(t ) = co e
−( t / t *+ kt )
t
If Qin = Qout = const, V = const; t* =
V/Q; and ri = -kc
t / t*
+ ∫ cin (τ )e −[(t −τ ) / t*+ k (t −τ ) ]d (τ / t*)
0
IC
Convolution integral (exponential filter that
credits inputs w/ decreasing weight going
backwards in time)
Roadmap to solutions (3x2)
m& = Qin cin
m& = Qin cin
m& = Qin cin
Inputs
t
Pulse
t
Step
Continuous
Reactors
Single
t
Multiple
Pulse input to single tank
m& = Qin cin
Pulse
May have practical significance—e.g. instantaneous spill
More commonly used as a diagnostic; Produces stronger
gradients than other types of inputs
Pulse input, single tank, cont’d
m& = Qin cin
Mo
M o −( t / t *+ kt )
e
c(t ) =
V
co = Mo/V (not initial conc.)
Pulse
c(t )
= e −(t / t*+ kt )
co
Know t* and co;
Plot c/co vs t/t*
Determine kt* = > k
WE 5-2
Pulse input, single tank, cont’d
m& = Q in c in
Pulse
General case:
c(t )
= e −( t / t*+ kt )
co
Special cases:
c(t )
= e − kt
co
c(t )
= e −t / t* = e −tQ / V
co
No flow;
batch reactor
No loss; just
flushing
In practice never exactly wellmixed
Time from Injection of Tracer (h)
Count Rate with Floating Detector in
Outlet Bay (counts/s)
1200
4
8
12
16
20
24
28
Data versus theory tells
how close to well mixed
Ways to mix
800
„
Perfectly
mixed system
400
0
„
8 March
1973
7 March
16
20
„
Mid
Night
4
8
12
16
20
Time of Day (hrs)
White, (1974)
Figure by MIT OCW.
Directed discharge and
intake
Baffles
Stirrers
May need to avoid overmixing (benthic flux
chambers)
Benthic Flux Chambers
Slight diversion: Residence time
properties of reactor vessels
Not necessarily well-mixed
But known residence time
Co = Mo/V
t* = V/Q
c/co
cV/Mo=coutV/Mo
m& = Qin cin
1
Mo
t
t/t*
1
tQ/V
Residence time distribution
Unit impulse response
Recall from Chapter 4
m&
m*
f* = -dm*/dt = coutQ
1
1
0
t
Rate of injection
0
t
Mass remaining in system
∞
∫
t* = V / Q
Mo 1
co =
=
V
V
f * = cout Q =
0
Mass leaving rate
∞
f * dt = 1 ⇒
0
cout ⎛ t ⎞
∫0 co d ⎜⎝ t * ⎟⎠ = 1
∞
∫ f * tdt = t *
cout Q cout
=
co V co t *
t
0
∞
∞
or
Oth
Moment
1
f *tdt = 1
∫
t *0
cout ⎛ t ⎞ ⎛ t ⎞
⇒∫
⎜ ⎟d ⎜ ⎟ = 1
co ⎝ t * ⎠ ⎝ t * ⎠
0
1st
Moment
RTD, cont’d
0th and 1st moments of normalized distribution
(cout/co vs t/t*) are both unity
For well mixed tank, k=0 (CSTR)
cout
c
=
= e −t / t *
co
co
c/co
1
∞
−t / t *
e
∫ d (t / t*) = 1
o
∞
−t / t *
e
∫ (t / t*)d (t / t*) = 1
o
1
t/t*
Area under curve and center of mass
are both one
Pulse input to tanks-in-series
m& = Qin cin
V
Mo
∞
n=∞
t
Plug Flow
Reactor
2.0
dci ci −1 − ci
=
− kci
dt
t*
n = 40
20
15
C
10
C0
Fully-Mixed Tank
5
1.0
c
nn ⎡ t ⎤
=
co (n − 1)! ⎢⎣ nt * ⎥⎦
.k = 0
a=0
n −1
e −(t / t*+ kt )
Co = Mo/nV; t* = V/Q = res. time of
each tank; total res time = nt*
Figure by MIT OCW.
2
n=1
0
.0
.5
1.0
1.5
Dimensionless Time, t/nt*
2.0
Pulse input to tanks-in-series
∞
n=∞
Area under curve and center
of mass are both one
2.0
∞
c(t / nt*) ⎛ t ⎞
∫0 co d ⎜⎝ nt * ⎟⎠ = 1
∞
c(t / nt*) ⎛ t ⎞ ⎛ t ⎞
∫0 co ⎜⎝ nt * ⎟⎠d ⎜⎝ nt * ⎟⎠ = 1
Plug Flow
Reactor
.k = 0
a=0
n = 40
20
15
C
10
C0
Fully-Mixed Tank
5
1.0
2
n=1
Limits of plug flow and fully
well-mixed
0
.0
.5
1.0
1.5
Dimensionless Time, t/nt*
Figure by MIT OCW.
2.0
Advantages of Plug Flow?
Advantages of Plug Flow?
1
e-kt (this function multiplies k=0 sol’n)
0.5
1
1.5
t/nt*
Everything “cooks” the same time. The mean residence
time of a water parcel is always nt* (by definition) but
under plug flow all parcels reside for nt*
This advantage obviously applies for continuous and step
injections as well
Step input, single and multiple
m& = Q in c in
t
Step
c(t ) = co e
− ( t / t *+ kt )
c(t )
1
=
cin
(t * κ ) n
[
cin
+
1 − e −(t / t*+ kt )
1 + kt *
]
⎡
(κt ) n −1 ⎞⎤
(κt ) 2
−κt ⎛
⎟⎟⎥
+ ... +
⎢1 − e ⎜⎜1 + κt +
(n − 1)! ⎠⎦
2
⎝
⎣
κ = k + 1/ t *
Multiple reactors, k = 0
Response of n tanks to step input, n = 1, 2, 4, 8, and 16
1
0.9
Normalized concentration, c/cin
0.8
0.7
0.6
0.5
n=1
0.4
16
0.3
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
1.2
Normalized time, t/ntstar
Figure by MIT OCW.
1.4
1.6
1.8
2
WE 5-3 Thermal storage tank
T2
T1
Day
Collector
Demand
Night
T (t ) − T1
(κt ) 2
(κt ) n −1 ⎞⎤
1 ⎡
−κt ⎛
⎟⎟⎥
1 − e ⎜⎜1 + κt +
=
+ ... +
n ⎢
2
(n − 1)! ⎠⎦
T2 − T1
(t * κ ) ⎣
⎝
100% t T2 − T (t )
R(t ) =
dt
∫
0
∆T0
nt *
t = start of the “day”)
(when tank has cold
water at T = T1 and you
want to keep T = T1
Percent of theoretical cooling (or
heating) potential, a measure of
hydraulic efficiency of storage system
Thermal storage tank, cont’d
% of total theoretical cooling
potential recovered
100
m=5
m = ∞ (plug - flow)
50
4 Horiz. Barriers
m = l (well - mixed)
0
0
4 Vert. Barriers
No Barriers
1.0
Non - dimensional Time t / (V/Q)
Figure by MIT OCW.
Geyer and Golay (1983),
Adams (1986)
Continuous input single & multiple
m& = Q in c in
From step input solutions for large t
Continuous
c
1
=
cin 1 + kt *
cout
1
=
cin
(1 + kt*) n
Continuous input, single tank
Continuous input single & multiple
cout
1
=
cin
(1 + kt*) n
(knt* = 1; constant total volume)
n
kt*
cout/cin
1
1
0.5
2
0.5
0.44
5
0.2
0.40
10
0.1
0.386
100
0.01
0.37
infinity
0
e-1
Dispersed flow reactor
W
x
L
Engineer delayed response (--> plug flow) by making the reactor long
& narrow. Density stratification should also be minimal (see Sect 5.4.1)
∂c ⎞
∂c 1 ∂ ⎛
∂c
=
+u
⎜ E L A ⎟ − kc
∂x ⎠
∂x A ∂x ⎝
∂t
(Small) entrance and exit sections may be treated differently
Pulse input
Qin = Qout = const; A = const; k = 0 and pulse input at x =0
2
⎡
⎤
⎛
Pe
2⎞
Pe
⎛
⎞
⎜
⎟
+ µi ⎟
µ i ⎜ sin µ i + µ i cos µ i ⎟
⎢
⎥
⎜
∞
4
cL
Pe
t
2
⎛
⎞
⎠ exp ⎢ − ⎝
⎠
⎥
= 2∑ ⎝
⎜
⎟
⎢ 2
co
Pe
⎛ Pe 2
⎝ t * ⎠⎥
i =1
2⎞
⎜⎜
+ Pe + µ i ⎟⎟
⎢
⎥
⎢⎣
⎥⎦
⎠
⎝ 4
co = Mo/V; Pe = UL/EL
⎛ µ i Pe
µ i = cot ⎜⎜ −
⎝ P 4µ i
−1
⎞
⎟⎟
⎠
Dispersed flow, pulse input
2.8
∞
p=∞
A lot like tanks-inseries
Greater n or Pe
(lower EL) => plug
flow
Pe ~ 2n-1 ( WE 5-5)
Elongation
sometimes done
with baffles
Plug Flow
Reactor
2.0
35
CL
C0
25
Fully-Mixed Tank
1.0
15
10
6
4
2
P=0
0
0
.5
1.0
1.5
2.0
t/t*
Figure by MIT OCW.
Dispersed flow vs tanks-in-series
2.8
∞
∞
n=∞
p=∞
Plug Flow
Reactor
2.0
2.0
35
CL
C0
1.0
n = 40
20
15
C
25
Fully-Mixed Tank
Plug Flow
Reactor
.k = 0
a=0
10
C0
Fully-Mixed Tank
15
5
1.0
10
6
4
2
2
n=1
P=0
0
0
.5
1.0
1.5
t/t*
Figure by MIT OCW.
2.0
0
.0
.5
1.0
1.5
Dimensionless Time, t/nt*
Figure by MIT OCW.
2.0
Dispersed flow reactor
2.8
∞
p=∞
Area under curve and center
of mass again are both one
Plug Flow
Reactor
2.0
∞
c(t / nt*) ⎛ t ⎞
∫0 co d ⎜⎝ nt * ⎟⎠ = 1
35
CL
C0
25
Fully-Mixed Tank
1.0
∞
15
c(t / nt*) ⎛ t ⎞ ⎛ t ⎞
∫0 co ⎜⎝ nt * ⎟⎠d ⎜⎝ nt * ⎟⎠ = 1
10
6
4
2
Limits of plug flow and fully
well-mixed
P=0
0
0
.5
1.0
t/t*
1.5
2.0
Figure by MIT OCW.
(Idealized) effect of central baffle
Pe =
UL
Q A 1
QA
=
=
E L HB B E L HE L B 2
0.01U 2 B 2 UB 2
~
EL ≅
u* H
H
Pe =
UL ULH LH
~
~ 2
2
E L UB
B
If EL = const, B decreases by 2x, so
Pe increases by 4x; real increase
could be even more
L increases by 2x and B decreases
by 2x, so Pe increases by 8x.
CV
Mo
2
1
0
0
0.5
1
tQ
V
I
1
tQ
V
II
A
1
CV
Mo 0.5
0
0
0.5
B
Can you place the UIR
(A-D) with the schematic
cooling pond (I-IV)?
1
CV
Mo
0.5
0
0
0.5
1
tQ
V
III
1
tQ
V
IV
C
CV
Mo
3
2
1
0
0
0.5
D
Figure by MIT OCW.
Cerco, 1977
WE5-5 Dispersed flow vs
tanks-in-series
River: L = 10 km; B = 20 m; H = 1 m; Q = 10m3/s; S = 10-4
What are equivalent values of Pe and n?
u* ≅
gHS = 10 ⋅ 1 ⋅ 10 −4 = 0.032m / s
u = Q/BH =0.5 m/s
E L = 0.01U B / Hu* ≅ 0.01 ⋅ 0.5 ⋅ 20 /(1 ⋅ 0.032) ≅ 30m / s
2
2
2
2
Pe = UL / E L = 0.5 ⋅ 10 4 / 30 = 167
n ≅ P / 2 = 83.
A finite difference model using upwind differencing would want
to use a spatial grid size of order 104/83 or 120 m
2
Dispersed flow, continuous input
W
x
L
GE assuming steady state
dc
d 2c
u
= E L 2 − kc
dx
dx
Boundary conditions
Qcin
Qc
x =0
x = L−
−
dc ⎞
⎛
= ⎜ Qc − AE L
⎟
dx ⎠
⎝
= Qc
x = L+
x =0+
at x = 0 (Type III)
at x = L (Type II)
Dispersed flow, continuous input
Solution (0 < x < L)
c( x)
⎡αPe
⎛ Peξ ⎞⎧
(1 − ξ )⎤⎥ − (1 − α ) exp ⎡⎢αPe (ξ − 1)⎤⎥ ⎫⎬
= g o exp⎜
⎟⎨(1 + α ) exp ⎢
cin
⎦⎭
⎣ 2
⎦
⎣ 2
⎝ 2 ⎠⎩
ξ = x/L
go =
2
(1 + α )2 exp⎛⎜ αPe ⎞⎟ − (1 − α )2 exp⎛⎜ − αPe ⎞⎟
⎝ 2 ⎠
⎡ 4kL ⎛ 1 ⎞⎤
a = ⎢1 +
⎜ ⎟⎥
u ⎝ Pe ⎠⎦
⎣
⎝
2 ⎠
Solution (0 < x < L)
Plug flow:
greater loss at
small x Æ
lower
concentration
at large x
Note:
c(x=0) < cin
At outlet
Solution (x=L)
cL
= 2ag o exp(Pe / 2)
cin
For Pe --> 0 (well mixed)
cL
1
=
cin 1 + kt *
For Pe --> oo (plug flow)
cL
= e − kL / u = e − kt*
cin
WE 5-6 Continuous solar
disinfection
SODIS: simple household
treatment technology for
destroying pathogens
using UV and
temperature
Pioneered by EAWAG,
SANDEC and others;
studied by former MEng
students in Nepal
Continuous (or semicontinuous) operation
more convenient than
discrete bottles
Dispersed flow reactor response to
step input (eq 2.104), k = 0
Pe = 128
Pe= 2
Xanat’s SCSODIS
Infer 16 <
Pe < 128
Cout/Cin vs t/t* for various values of Pe (2, 4, 8, 16,
32, 64, 128) compared with measurements
(connected dots)
Flores (2003)
Continuous SODIS, cont’d
Measurements show 99% of pathogens killed for t* = 2 days
(cout/cin = 0.01). Use Eq. 5.40 to estimate first order removal rate
(k) for plausible values of Pe & compare with plug flow reactor
Example
Pe = oo, cL/cin =
cout/cin = 0.01
kt* = 4.6, k = 2.3 d-1
0.01
Continuous SODIS, cont’d
Pe
k (d-1)
Cout/cin
(Plug Flow)
Cout (Dispersed)
Cout (Plug flow)
16
2.92
0.0029
3.45
32
2.62
0.0053
1.89
64
2.47
0.0072
1.39
128
2.38
0.0086
1.16
oo
2.30
0.0100
1.00
Skimmer walls
Plan View
Withdrawal of lower layer water
tends to “suck down” upper
layer. Bernoulli’s equation used
to compute max flow, draw down
Intake Channel
W
Skimmer Wall
Design Condition For Full Effectiveness
Maximum flow per unit width
Skimmer Wall
ρ
q
ρ + ∆ρ
B
hr
Partially Effective Condition
q2
H
hc2
Critical Section
⎡ ⎛ 2 ⎞3 ⎤
q c = ⎢ g ′⎜ hr ⎟ ⎥
⎢⎣ ⎝ 3 ⎠ ⎥⎦
Fr =
ρ
hc1
Mixing q1
Selective withdrawal of water
from density stratified tank or
reservoir
ρ + ∆ρ
q
(g ′h )
3
r
1
2
⎛ 2⎞
=⎜ ⎟
⎝ 3⎠
3/ 2
hr
For larger Fr, partially effective
Figure by MIT OCW.
Partially-effective skimmer wall
0.6
Data
Horleman and
Elder (1965)
0.5
Delft Laboratory
(1973)
State Electricity
Commission of
Victoria (1973-74)
H/hr
2.6 to 7.4
λ = q1/q
2.5
λ = 1− Frc/Fr
3.0 to 6.0
1.09
1.65
1.7
1.6
1.2
1.7
1.75
1.65
0.4
3.0
H/hr = 0.0
2.0
1.8
1.6
λ = 0.99 λmax
0.3
1.4
λ
λmax = 1-
0.2
1
H/hr
1.2
0.1
0.0
0.0
H/hr = 1.1
0.4
0.8
1.2
1.6
2.0
Frc = (2/3)3/2
Fr = q (g' hr3)-1/2
Figure by MIT OCW.
Jirka (1979)