Transcript Slide 1

Mass and Energy Balances –
Stripping Section and Partial Reboiler

The previous mass and energy balances apply only to the enriching
section.

At some point down the column, we will have a feed to one of the
equilibrium stages – the feed stage. At this feed stage, the enriching
section of the column ends.

At the feed stage we have the introduction of additional liquid and/or
vapor depending upon the nature of the feed stream.

Liquid from the feed stream will flow down the column and vapor from
the feed stream will rise up the column.

Consequently, the ratio of vapor to liquid in the enriching section
above the feed stage is generally different than that in the stripping
section below the feed stage because of the feed between these two
sections.
Lecture 12
1
Enriching or
Rectifying Section
Feed Stage
Stripping Section
Lecture 12
2
Mass and Energy Balances –
Stripping Section and Partial Reboiler
 While we have designated the vapor and liquid streams
in the enriching section as L and V, we will designate the
vapor and liquid streams in the stripping section using
an “underline” or V and L (in place of the “overbar” in
the text) to delineate them from those in the enriching
section.
 L/V < 1 in the enriching section.
 Conversely, L/V > 1 in the stripping section.
 Let’s look at the mass and energy balances for the
stripping section of the column with a partial reboiler.
Lecture 12
3
L Nn
L Nn
Stage N- n
∙
V Nn1
L N 3
V N2
V N 1
L N2
L N 1
∙
∙
∙
V Nn1
L N 3
Stage N-3
V N2
Stage N-2
V N 1
Stage N-1
VN
L N2
L N 1
∙
∙
Stage N-3
Stage N-2
Stage N-1
VN
Stage N
Stage N
LN
V N1
Stage N- n
LN
V N1
Stage N+1
Partial Reboiler
Stage N+1
Partial Reboiler
QR
QR
B
B
Lecture 12
4
Mass and Energy Balances –
Stripping Section and Partial Reboiler
Total
Mass Balance
Component
Mass Balance
Energy
Balance
Stage
(Partial Reboiler)
V N1  L N  B
V N1 y N1  L N x N  Bx B
V N1H N1  L N h N  Bx B  QR
N+1
V N  L N1  B
V N y N  L N1 x N1  Bx B
V N H N  L N1h N1  Bx B  QR
N
V N1  L N2  B
V N1 y N1  L N2 x N2  Bx B
V N1H N1  L N2 h N2  Bx B  QR N-1
V N2  L N3  B
V N2 y N2  L N3 x N3  Bx B
V N2 H N2  L N3 h N3  Bx B  Q R N-2
∙
∙
∙
V Nn1  L Nn  B
∙
∙
∙
∙
∙
∙
∙
∙
∙
∙
∙
∙
V Nn1 y Nn1  L Nn x Nn  Bx B
∙
∙
∙
∙
∙
∙
∙
V Nn1H Nn1  L Nn h Nn  Bx B  QR N- n
n = 0, 1, 2,…
Lecture 12
5
Constant Molar Overflow (CMO)
Assumption – Stripping Section

Just as we did for the enriching section, we will assume that for
every mole of liquid that vaporizes at an equilibrium stage, an
equivalent amount of vapor condenses, then the LN-n’s are
constant and the VN-m+1’s are constant in the column – the CMO
assumption.

We can then rewrite the component mass balance as:
V Nn1 y Nn1  L Nn x Nn  Bx B
V y Nn1  L x Nn  B x B
(CMO)
L
B
x Nn  x B
V
V
(CMO)
or rearranging
y N  n 1 
Lecture 12
6
Indices
 Let’s do an indices substitution. If we let
k = N-n-1; then k = N+1, N, N-1, N-2, …
then the previous equation can be rewritten as:
yk 
L
B
x k 1  x B
V
V
 Note that this allows us to arrive at the indices used by
Wankat, e.g., Eq. (5-14), which we can derive from this
equation.
Lecture 12
7
Stripping Section Operating Line

Just as we did for the enriching section, we can also drop
the indices from the CMO equation for the stripping section
noting that the vapor and liquid compositions, yk and xk-1,
represent the vapor and liquid compositions at equilibrium
at stage k.
y

L
B
x  xB
V
V
Just as we derived the enriching section operating line (OL)
from the mass balances and assuming CMO, this equation is
the OL for the stripping section.
Lecture 12
8
Stripping Section Operating Line
L
B
y  x  xB
V
V
Stripping Section OL
 The stripping section operating line (OL) for a distillation
column (assuming CMO) is a linear equation with:
slope
y-intercept
L/V and
–(B/V)xB
 Note that the L/V ratio for the stripping section of a
distillation column will always be greater than one, L/V > 1,
since there will be a greater amount of liquid than vapor in
the stripping section below the feed stream.
Lecture 12
9
Alternative Stripping Section OL –
Liquid to Vapor Ratio
Stripping Section OL:
y
L
B
x  xB
V
V
From a mass balance around the reboiler,
B LV
B L
 1
V V
and substituting into the previous stripping section OL yields:
y
L 
L
x    1x B
V
V 
Lecture 12
Eq. (5-22)
10
Stripping Section OL and
y = x Intersection
If we substitute y = x into any of these OL’s, including Eq. (5-22), we
find that
y  x  xB
This is the intersection of the Stripping Section OL and the y = x line,
which is xB, the composition of the bottom stream.
Lecture 12
11
Distillation Column –
Stripping Section Operating Line
1.0
0.9
Equilibrium Curve
0.8
y mole fraction
0.7
Stripping Section OL
0.6
0.5
y=x
0.4
0.3
Slope = L/V
0.2
y-int = -(B/V)x B
0.1
= - (L/V-1)x B
0.0
0.0
0.1
xB
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
x mole fraction
Lecture 12
12
Feed Stage

At some point down the column, we introduce the feed at the
feed stage.

The phase and temperature of the feed affects the vapor and
liquid flow rates in the column.

If the feed is a liquid, then L > L.

If the feed is a vapor, then V > V.

The feed may also be flashed into the column yielding both vapor
and liquid – remember flash distillation!

Remember, however, L/V < 1 and L/V >1.

Let’s look at the feed stream and how we handle it…
Lecture 12
13
L f 1
Stage f-1, j
Lf
Stage f, j+1, k-1
L f 1
Stage f+1, k
Vf
F
V f 1
V f 2
Lecture 12
14
Mass and Energy Balances –
Feed Stage
Total Mass Balance
F  L f 1  V f 1  L f  Vf
Component Mass Balance
Fz F  L f 1x f 1  V f 1 y f 1  L f x f  Vf y f
Energy Balance
Fh F  L f 1h f 1  V f 1H f 1  L f h f  Vf H f
Lecture 12
15
Constant Molar Overflow (CMO)
Assumption – Feed Stage
 Just as we did for the enriching and stripping sections,
we will assume CMO for the feed stage and drop the
indices. We also add the liquid and vapor designations
for our enthalpies in the energy balance.
Total Mass Balance
FL V  L V
Eq. (5-15)
Component Mass Balance
FzF  Lx  Vy  Lx  Vy
Energy Balance
Fh F  LhL  VHV  Lh L  VH V
Lecture 12
Eq. (5-16)
16
Handling Feed Stream Conditions
 Since the nature (both phase and temperature) of the
feed affects the column’s liquid and vapor flows, we need
to derive a method for handling these various types of
possible feeds.
 It would be useful to derive such a method that allows
us to readily incorporate a parameter that accounts for
the condition of the feed stream.
 We will start with the total mass and energy balances
around the feed stage…
Lecture 12
17
Some Manipulations…
The energy balance, Eq. (5-16) can be rearranged to:
FhF  (L  L)hL  (V  V)HV  0
If we solve the mass balance, Eq. (5-15), for V – V
VV  LLF
and substitute into the previous equation, we have, after some rearranging,
(L  L)HV  (L  L)hL   F(HV  h F )
or
(L  L)(HV  h L )  F(HV  h F )
and one final rearrangement yields the relationship:
L  L HV  h F

F
HV  h L
Lecture 12
Eq. (5-17)
18
“Quality” q
We define the left-hand side of Eq. (5-17) as the “quality”, q
q
L  L HV  h F

F
HV  h L
Eq. (5-17)
It can also be shown from the previous material balances that
q  1
V  V HV  h F

F
HV  h L
The quality, q, is
q
liquid flow rate below the feed  liquid flow rate above the feed
feed rate
q
vaporenthalpyon thefeed plate feed enthalpy
vaporenthalpyon thefeed plate liquid enthalpyon thefeed plate
The quality, q, is the fraction of feed that is liquid. This is analogous to the q that we saw defined
for flash distillation – remember that we assume that the feed is adiabatically flashed to the column
pressure!
Lecture 12
19
OL Intersection
1.0
0.9
Equilibrium Curve
0.8
y mole fraction
0.7
0.6
0.5
Enriching Section OL
0.4
0.3
Stripping Section OL
0.2
0.1
0.0
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
x mole fraction
Lecture 12
20
Another Mass Balance –
OL Intersection
At the feed stage, the enriching section OL and the stripping section OL
must intersect. These OL’s can be written as:
Vy  Lx  Dx D
Enriching Section OL
Eq. (5-26)
Vy  Lx  Bx B
Stripping Section OL
Eq. (5-27)
We can represent this point of intersection by subtracting the stripping
section OL from the enriching section OL (essentially a simultaneous
solution). Doing so and grouping terms yields:
y(V V)  (L  L)x  DxD  Bx B
Lecture 12
Eq. (5-29)
21
Some Further Manipulations –
General Feed Line
The component mass balance around the column yields
Dx D  Bx B  Fz F
Substituting this mass balance into the difference of the OL’s yields, upon
rearrangement,
y
(L  L)
F
x
zF
(V  V)
(V  V)
Feed Line
Eq. (5-30)
This equation is linear and in the form of an operating line.
It is one of the various forms, as we shall see, of the feed line and is the
most general form.
Lecture 12
22
Some Further Manipulations –
Another Feed Line
The total mass balance around the feed stage yields
FL V  L V
Combining this mass balance with the previous feed line yields, upon rearrangement,
LL
1
F
y
x
zF
L

L
L

L




1 

 1

 F 
 F 
or, from the definition of quality, q:
y
q
1
x
zF
q 1
1 q
Feed Line
Lecture 12
Eq. (5-35)
23
Feed Line

The previous equation is the feed line for the column in
terms of quality q.
y
q
1
x
zF
q 1
1 q
Feed Line
Eq. (5-35)

This should look familiar – it is the same as the operating
line that we obtained from the mass balances for flash
distillation!

We can use the conditions of the feed to determine q from
its enthalpy relationship:
q
L  L HV  h F

F
HV  h L
Lecture 12
Eq. (5-17)
24
Feed Line Equations

By inspection from the results of our flash distillation operating
lines, the feed line can also be expressed in terms of fraction of
feed vaporized, f = V/F. This, as well as the other feed line
equations, are summarized below:
y
(L  L)
F
x
zF
(V  V)
(V  V)
Eq. (5-30)
y
q
1
x
zF
q 1
1 q
Eq. (5-35)
1 f
1
x  zF
f
f
Eq. (5-34)
y
Lecture 12
25
Feed Line and OL Intersection
 Remember that we derived these feed line
equations from the intersection of the
enriching section and stripping section OL’s.
 It can be shown that the feed line also
intersects the OL’s at their intersection – all
three lines intersect at the same point.
 We will need to use this intersection point
in our solutions…
Lecture 12
26
OL and Feed Line Intersection
Simultaneous solution of the enriching section and stripping section OL’s
and feed line yields their intersections, xI and yI:
L 
 L
1   x D    1 x B
 V
V 
xI 
L L

V V
yI 
 L
 q  11   x D  z F
 V
xI 
q  1 L   q
V
x Dq
L 0 /D
yI 
q
1
L 0 /D
L
 L
x I  1  x D
V
 V
zF 
Lecture 12
Eq. (5-38)
27
Possible Feed Stream Conditions
 We assume that the incoming feed is
adiabatically flashed to the column pressure,
Pcol.
 We can have 5 possible feed stream conditions
for a given feed composition zF:





Subcooled liquid feed if TF < Tbp
Saturated liquid feed if TF = Tbp
Two-phase feed if Tbp <TF < Tdp
Saturated Vapor if TF = Tdp
Superheated Vapor if TF > Tdp
Lecture 12
28
Saturated Liquid Feed – Given TF = Tbp
L  L (L  F)  L
q

1
F
F
or since hF = hL
HV  h F 0  h F
q

1
HV  h L 0  h L
Note that q = 1.
Lecture 12
29
Saturated Vapor Feed – Given TF = Tdp
LL LL
q

0
F
F
or since HV = hF,
HV  h F HV  HV
q

0
HV  h L HV  h L
Note that q = 0.
Lecture 12
30
Two-Phase Feed – Given f
f is the fraction of feed vaporized.
L  L (L  L F )  L L F
q


F
F
F
V  (V  VF )
VF
VV
q  1
 1
 1
F
F
F
q  1
VF
 1 f
F
Note that 0 < q < 1.
Lecture 12
31
Two-Phase Feed – Given TF
Since we assume CMO, all HV’s and
hL’s are constant.
H V  h F H V(sat'd vapor)  h F(feed temp)
q

H V  h L H V(sat'd vapor)  h L(sat'd liquid)
all determined at zF.
Note that HV > hF > hL, 0 < q < 1.
Lecture 12
32
Subcooled Liquid Feed – Given c
c is the amount of V condensed.
c
q
L  L (L  F  c)  L
c

 1
F
F
F
Note that q > 1.
Lecture 12
33
Subcooled Liquid Feed – Given TF < Tbp
c
q  1
C PL (Tbp  TF )
λ
Note that TF < Tbp, q > 1.
Lecture 12
34
Superheated Vapor Feed – Given v
v is the amount of L vaporized.
v
L  L (L  v)  L
v
q


F
F
F
Note that q < 0.
Lecture 12
35
Superheated Vapor Feed – Given TF > Tdp
q
C PV (Tdp  TF )
λ
v
Note that TF > Tdp, q < 0.
Lecture 12
36
Possible Feed Lines
Lecture 12
37
End of Lecture 12
Lecture 12
38