MULTIPLE STAGE EQUILIBRIUM PROCESSES 1.
VAPOUR LIQUID EQUILIBRIA AND DISTILLATION
Let us consider a simple problem: A process produces a product B from reactant A with only limited conversion, so that the reactor outlet contains 60 mol% A and 40 mol% B. It is desirable to separate the reactant (A) from the product (B) in order to recycle the reactants and to obtain a high purity product. We will choose a basis of 100 kmol of reactor product:
A rich stream
100 kmol mixture 0.6 mol A/mol mixture 0.4 mol B/mol mixture B rich stream
Suppose that A has a lower boiling point than B. It is proposed that 60% (on a mole basis) of the mixture is boiled off to produce a vapour rich in A and a liquid rich in B. 60 kmol yA kmol A/kmol yB = (1-yA) kmol B/kmol 100 kmol 0.6 kmol A/kmol 0.4 kmol B/kmol
Boiling solution at TºC
40 kmol xA kmol A/kmol xB = (1-xA) kmol B/kmol
Heat
We thus have two unknowns: xA and yA. We have already ensured that the total mole balance is satisfied [input = output; 100 kmol = (60 + 40) kmol] so we can develop only one equation by performing a material balance for one of the components (A or B). Balance for A:
60 = 60 yA + 40 xA
Balance for B:
40 = 60 yB + 40 xB 40 = 60 (1 – yA) + 40 (1 – xA) 40 = 60 + 40 – 60 yA – 40 xA 60 = 60 yA + 40 xA
∴ ∴ ∴
1
Thus the problem cannot be solved without further information. In order to determine the composition of the streams we need to know how the composition of the vapour produced from a boiling mixture varies with the composition of the mixture. At its boiling point the vapour pressure of a liquid is equal to the ambient pressure. The vapour pressure of the mixture is the sum of the vapour pressure of each component. Dalton’s law gives the concentration of a component in terms of its vapour pressure (yA = PA/P). Thus, we need to know how the vapour pressure of a component varies with the composition of the mixture and temperature. For an ‘ideal solution’ (where equal forces act between molecules of {A and A}, {A and B} and {B and B}) we can use Raoult’s law. Raoult’s law states that the vapour pressure exerted by a component above an ideal solution is proportional to its mole fraction. The constant of proportionality is the vapour pressure of the pure component. 0
If the vapour pressure of pure A at temperature T is PA then the vapour pressure of A solution with a mole fraction xA mols A per mol of solution is Raoult’s law 0
PA = xA PA
Dalton’s law 0 PA xAPA yA = P = P
and
where P is the ambient pressure. Similarily for B: Raoult’s law 0
PB = xB PB
Dalton’s law 0 PB xBPB yB = = P P
and
0
PA
0
PB PA PB
0
xA
1
Raoult’s law: vapour pressures of ideal solutions (diagram shows behaviour at constant temperature)
2
Thus we can solve the problem above if the solution is ideal and we know the ambient pressure and the saturated vapour pressure of A or B. e.g. If A is hexane, B is octane, P = 760 mm Hg and the solution is boiling at 94°C at which 0 temperature PB = 288 mm Hg, determine xA and yA for the problem on page 1 above. 0 xBPB [Hint: use yB = P and the material balance equation] 0
0
xBPB (1 − xA)PB yB = (1 – yA) = = P P 0
∴
0
PB xAPB yA = 1 − P + P 0
Substituting into the material balance equation at the top of the page with P = 760 mm Hg and PB = 246 mm Hg 288 xA × 288 60 = 60 1 − 760 + 760 + 40 xA 60 . 288/760 = xA (60 . 288/760 + 40) ∴
xA = 0.362 kmol A/kmol yA = 1 – 40 xA/60 = 0.758 kmol A/kmol 0
0
If the data available is PA and PB (but not P) then we can still solve the problem: 0
0
e.g. Solve the problem above if PA = 1578 mm Hg and PB = 288 mm Hg and P is unknown. 0 0 xAPA xBPB [Hint: divide yA = to eliminate P] P by yB = P 0
xAPA Dividing yA = P by
0
xBPB yB = P to eliminate P:
0
yA xA PA yB = xB PB0
substituting for yB = (1 – yA) and xB = (1 – xA): 0
yA xA P A = 0 1 − yA (1 − xA) PB yA ASIDE: The term y B xA xB
Substituting for
0 P is sometimes called the relative volatility α. For ideal systems α = A0 PB
yA = 1 – 40 xA/60
from the material balance 3
0
60 − 40 xA xA P A = 0 40 xA (1 − xA) PB 0
0
(1.5 – xA)(1 – xA)PB = xA2 PA 0
0
0
0
xA2(PA – PB) + 2.5 PB xA – 1.5 PB = 0 0
−2.5 PB ±
xA =
0
0
0
0
(2.5 PB)2 + 6 PB(PA − PB) 0
0
2(PA − PB) −2.5 . 288 +
xA =
(2.5 . 288)2 + 6 . 288 (1578 − 288) = 0.363 as before (nearly!). 2(1578 − 288)
In general if we select the operating pressure, we do not know the boiling point of the solution as we do not know its composition. We thus need to consider how the boiling point of the solution varies with composition, and thus determine the composition it generates as above. For ideal solutions, the total vapour pressure above the solution P is thus 0
0
P = PA + PB = xA PA + xB PB 0 0 = xA PA + (1 − xA)PB Rearranging: 0
xA =
P − PB 0
0
PA − PB
Thus if vapour pressure data (as a function of temperature) is available for the two components, then this equation can be used to determine the boiling point of any liquid mixture at given pressure P or the composition of any mixture boiling at a given temperature.
4
e.g.
Hexane and Octane P = 760 mm Hg Mixture boiling point = 94°C 0 Vapour pressure of pure hexane at 94°C PA = 1578 mm Hg 0 Vapour pressure of pure octane at 94°C PB = 288 mm Hg
xA =
760 − 288 = 0.366 1578 − 288 0
xAPA 0.327 × 1819 yA = = = 0.760 P 760
Data is often calculated for a range mixtures between pure A and pure B which are plotted on a T – x – y diagram, or if temperatures are not important an x-y diagram. Note that these diagrams are normally drawn for the more volatile component, so on the T-x-y diagram the vapour (y) line is above the liquid (x) line and on the x-y diagram the equilibrium line is above the diagonal.
5
T - x - y plot for hexane and octane at 1 atmosphere total pressure 130 125 120 115
Temperature (degrees C)
110 105 100
y (mol hexane per mol vapour)
95 90 85 x (mol hexane per mol liquid)
80 75 70 65 0
0.1
0.2
0.3
0.4
0.5
0.6
x or y , mol fraction of hexane in liquid/vapour
6
0.7
0.8
0.9
1
Vapour-liquid equilibrium for hexane/octane at P=1 atm. 1
0.9
0.8
y (mol fraction hexane in vapour)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
x (mol fraction hexane in liquid)
7
0.8
0.9
1
For a non-ideal system we will need to use the graphical data with the material balance, requiring a graphical solution. The values of yA and xA must (i)
satisfy the material balance equation 60 = 60 yA + 40 xA
AND (ii)
must lie on the equilibrium line on the x-y diagram.
Thus if we plot the material balance line: 60 = 60 yA + 40 xA solution will be the point which crosses the equilibrium x-y line.
on the x-y diagram, the
Although this simple process can only achieve a crude separation, it is often used in industry. The process is sometimes called ‘equilibrium distillation’ or ‘Rayleigh distillation’. Heating is rarely used however, as the same effect can be achieved by dropping the pressure so that the boiling point of the solution falls to below its actual temperature.
A rich stream
P>1 atm 0.6 mol A/mol mixture 0.4 mol B/mol mixture
P = 1 atm
B rich stream
8
MULTI-STAGE DISTILLATION The separation achieved in the example above was rather poor. In practice we want a high purity product and very little of the product in the recycle. First consider the more volatile component A. We could condense the vapour produced and boil it once again to obtain a higher concentration of A, and this could be repeated any number of times to increase the concentration of A.
100 kmol mixture 0.6 mol A/mol mixture 0.4 mol B/mol mixture
Boiling solution
Heat
Boiling solution
cooling water
Boiling solution
cooling water
Heat
Heat
This will certainly produce a stream concentrated in A, but what are the practical disadvantages of this approach? Expensive in both cooling water and heat. Many side streams need handling with varying composition. Very little concentrated A generated, most A lost in liquid streams. High capital cost (many process stages).
9
Solution - multistage distillation: cooling water
Boiling solution
Boiling solution
Boiling solution
Feed
total condenser
Boiling solution
reflux
Boiling solution
Boiling solution
Distillate or top product
Distillation column Feed
Boiling solution
Heat
steam
Bottom product
Distillation trays or plates
10
reboiler
By selecting the number of trays in a distillation column we can design the column to achieve a desired separation. Thus a general distillation problem is where the feed flowrate and composition is known and a column must be designed to achieve desired top and bottom product compositions. E.g. 100 kmol/h of a mixture of 60 mol% hexane and 40 mol% octane must be separated to achieve a top product containing 95 mol% hexane and a bottom product containing 95% octane. The first step in solving this problem is to carry out an overall material balance for the column to determine the top and bottom product flow rates.
D F 100 kmol/h
B
Total balance
F=D+B 100 = D + B
Balance on hexane
F xF = D xD + B xB 60 = 0.95 D + 0.05 B
Substituting
B = 100 – D
from the total balance
60 = 0.95 D + 5 – 0.05 D D = 55/0.9 = 61.11 kmol/h B = 100 – 61.11 = 38.89 kmol/h
These overall balances should always be carried out first when tackling distillation problems.
11
Now we must consider what is happening on each tray, in order to determine the vapour and liquid flows between the trays. Liquid and Vapour Flow Rates Consider how much vapour will be produced from a tray in the section of the column above the feed (called the enrichment section). This amount of vapour produced is related to the amount of heat supplied to the tray. Heat is supplied in the form latent heat from condensing vapour flowing from the tray below. Ln+1
Vn
Vn-1
Ln
The nomenclature is defined in the figure above: The flowrate of liquid and vapour leaving tray n are called Vn and Ln respectively. Assuming: (i)
There are no heat losses from the column
(ii)
The difference in temperature between adjacent trays is negligible
(iii)
The latent heat of the two components are equal
If this latent heat is λ, we can write a heat balance as:
λ Vn-1 = λ Vn ∴
Vn-1 = Vn
In words, for every mole of vapour condensed, one mole of vapour boils off. A total material balance for the tray gives: Vn-1 + Ln+1 = Vn + Ln ∴
Ln+1 = Ln
For every mole of liquid flowing to tray n, one mole of liquid flows from tray n to tray n-1.
12
Now consider the condenser and reflux. The vapour from the condenser is completely condensed so that the resulting liquid has the same composition. This liquid is split to produce the top product and to provide liquid for the top tray. Thus these three streams, the vapour leaving the top tray, the top product and the reflux all have the same composition, xD. V n , yN = xD
tray N Vn yN-1
Ln xN+1 = xD
D xD
Ln xN
Notice that we have used Vn and Ln to denote the vapour and liquid flowrates since we have shown above that these flowrates are constant throughout the enrichment section. We have also extended the nomenclature to include the composition: the mole fraction of the more volatile component (MVC) in the liquid leaving tray n is xn, while the mole fraction of the MVC in the vapour generated on tray n is yn. Initially we will assume that these two streams (the vapour and liquid leaving the tray) are in equilibrium, i.e. the vapour leaving a tray is assumed to have the composition that would be produced by boiling the liquid which is leaving the same tray. This means that (xn,yn) lies on the equilibrium (x,y) line on the x-y diagram. As the condenser at the top of the column is a total condenser the vapour flowrate Vn is equal to the sum of the reflux and the distillate flowrate: Vn = D + Ln Thus in the enrichment section the vapour flow rate is greater than the liquid flowrate.
13
Now consider the feed tray:
Ln
Vn
F
Vm Lm
Here we denote the vapour and liquid flow rates in the section of the column below the feed (called the stripping section) as Vm and Lm. If the feed is a liquid at its boiling point, the heat balance gives Vm = Vn and the material balance gives: Lm = Ln + F For each tray below the feed tray, by the same argument used above in the enrichment section, the vapour and liquid flow rates are constant. Thus the liquid flow rate in the stripping section is greater than that in the enrichment section, while the vapour flow rates in the two sections are equal. If the feed is a vapour at its boiling point, then the heat balance gives: F + Vm = Vn and the material balance gives: Lm = Ln In this case the liquid flowrates in the two sections are equal, while the vapour flow rate in the enrichment section is greater than in the stripping section.
Reflux Ratio The ratio of the reflux stream flow rate (Ln) to the top product flow rate (D) can be varied during the operation of the column. This ratio is a very important parameter and is called the reflux ratio. Reflux ratio
R=
LN+1 Ln D =D 14
Thus Ln = R D and Vn = (1 + R)D
Material Balances for Each Tray Let us consider the condenser and the top two trays of the column: V n , yN = xD
tray N Ln xN+1 = xD
Vn , y N-1
D xD
tray (N – 1) Ln xN
LN xN
Vn yN-2
LN xN-1
e.g. 100 kmol/h of a mixture of 60 mol% hexane and 40 mol% octane must be separated to achieve a top product containing 95 mol% hexane and a bottom product containing 95% octane. We now need to specify the reflux ratio R. We will choose R=2 (the choice of the value of R will be discussed later). Carry out a material balance for tray N, and using the x-y diagram, determine xN and yN-1. [HINT: remember (xN, yN) is on the equilibrium line.] Material balance equation
Vn yN-1 + Ln xD = Vn xD + Ln xN Ln Vn − Ln yN-1 = V xN + V xD n
n
substituting for Vn and Ln 15
yN-1 =
xD R xN + 1+R 1+R
2 0.95 yN-1 = xN – = 0.667 xN + 0.317 3 3 Using the x-y diagram, (xN,yN) = (xN,xD) so when y = 0.95 we can read off xN as 0.751. Thus
yN-1 = 0.667 × 0.75 + 0.317 = 0.818
Mark the point (xN,yN-1) on the x-y diagram. Now carry out the material balance for tray N–1 and using the x-y diagram determine xN-1 and yN-2. Material balance equation
Vn yN-2 + Ln xN = Vn yN-1 + Ln xN-1 Vn xN-1 − Ln xN Ln yN-2 = V xN-1 + V n
n
From the material balance on tray N we can substitute for Vn xN-1 − Ln xN Vn xN-1 − Ln xN Ln yN-2 = V xN-1 + V n
yN-2 =
n
Ln Vn − Ln Vn xN-1 + Vn xD
[ASIDE we could obtain this equation directly by considering an overall material balance for the top two trays.] Substituting for Ln and Vn as before: xD R yN-2 = 1 + R xN-1 + 1 + R yN-2 = 0.667 xN-1 + 0.317 As before, using the x-y diagram, (xN-1,yN-1) = (xN-1,0.818) so when y = 0.818 we can read off xN as 0.442. yN-2 = 0.667 × 0.457 + 0.317 = 0.612 Now plot the point (yN-2, xN-1) on the x-y diagram.
16
Notice that the equation between yN-2 and xN-1 is the same equation as that between yN-1 and xN. This equation can be derived by considering a material balance for any tray n in the enrichment section and the top section of the column:
D xD
N N-1 . .
. . n+1 n
Vn yn-1
Ln xn
1
y
Vn yn-1 = Ln xn + D xD Ln D yn-1 = V xn + V xD n
yn-1 =
xD 1 + R
n
xD R xn + 1+R 1+R
0
xD
x
1
This is the Enrichment Section Operating Line (ESOL). The composition on each tray can be constructed by stepping between the ESOL and the equilibrium line. Below the feed tray, the the ESOL no longer applies and we use the Stripping Section Operating Line (SSOL).
Vm ym
Lm xm+1
m m-1 . .
Material balance
Lm xm+1 = Vm ym + B xB Lm B ym = V xm+1 – V xB m
. . 2 1
For a liquid feed B xB
Vm = Vn ym =
17
m
and
SSOL
L m= L n + F
RD+F B x – m+1 (1 + R)D (1 + R)D xB
Consider the reboiler and bottom tray (tray 1). Unlike the condenser, the solution is only partially boiled, so that the vapour flowing to the first tray has a different compostion to the liquid leaving the first tray and going to the reboiler. In fact the reboiler acts like an equilibrium tray in the column. Vm y1
tray 1 Lm , x2
V m , yR
reboiler
Lm x1
LN xN
heat
B xB
The vapour leaving the reboiler is assumed to in equilibrium with the bottom product, so (xB,yR) lies on the equilibrium line, xB = 0.05 gives yR = 0.195 from the x-y diagram. The material balance for the reboiler gives Lm x1 = Vm yR + B xB (1 + R)D B x1 = R D + F yR + R D + F xB x1 =
3 × 61.11 38.89 0.195 + 0.05 2 × 61.11 + 100 2 × 61.11 + 100
x1 = 0.170 The vapour leaving tray 1 is in equilibrium with x1 = 0.170, so from the equilibrium line on the x-y diagram, y1 = 0.504. Now we are stepping between the SSOL and the equilibrium line.
18
The intersection of the SSOL and ESOL occurs at the ‘feed line’. For a liquid feed this is at x = xF, while for a vapour feed this is at y = xF. PROOF: ESOL
Vn yn-1 = Ln xn + D xD
SSOL
Lm xm+1 = Vm ym + B xB
Intersection at (xi,yi), so (xi,yi) lies on both the ESOL and SSOL xm+1 = xn = xi
Thus
and
ESOL
Vn yi = Ln xi + D xD
SSOL
Lm xi = Vm yi + B xB
ym = yn-1 = yi
Adding these two equations: (Vn – Vm)yi = (Ln – Lm)xi + D xD + B xB Substitute
F xF = D xD + B xB
and for a liquid feed Vm = Vn
and
Lm = Ln + F
0 yi = – F x + F xF for a vapour feed
Vm + F = Vn
xi = xF
and
Lm = Ln
F yi = 0 xi + F xF
yi = xF
1
1
y
y
xD 1 + R
0
xD 1 + R xB
x
xF
xD
1
0
xB
x
xF
Liquid feed Vapour feed Construction of the operating lines
19
xD
1
Vapour-liquid equilibrium for hexane/octane at P=1 atm. 1 N
0.9 N–1 0.8
y (mol fraction hexane in vapour)
0.7 N–2 0.6 1 0.5
0.4
0.3
R 0.2
0.1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
x (mol fraction hexane in liquid)
20
0.8
0.9
1
We are now in a position to determine the number of plates required in a distillation column to achieve a given separation. This involves a graphical construction which must be the best known graphical procedure in Chemical Engineering and is called the McCabe-Thiele Method. A detailed procedure is shown on the separate sheet. The construction for the problem above is shown below, giving 4 equilibrium stages plus the reboiler.
Vapour-liquid equilibrium for hexane/octane at P=1 atm. 4
1
3 0.9
0.8 2
y (mol fraction hexane in vapour)
0.7
0.6
0.5
0.4
1
xD = 0.317 1 + R 0.3
0.2 R 0.1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
x (mol fraction hexane in liquid)
21
0.8
0.9
1
Choice of the Reflux ratio. Decreasing the reflux ratio means decreasing the liquid flow rate to the top tray from the condenser (Ln) and hence the liquid flow rates in throughout the column are also decreased. Since Vn = Ln + D, this also implies decreasing the vapour flow throughout the column. Thus if the reflux ratio is decreased, the amount of vapour generated in the reboiler and condensed in the condenser are decreased. By implication the operating costs associated with steam and cooling water are reduced when the reflux ratio is reduced. But what happens to the number of trays required to achieve a given separation when the reflux ratio is reduced? Consider the example above, but with the reflux ratio R reduced to a value of 0.5. The construction is shown on the next page, with the SSOL intercept now at 0.633. The number of trays required is increased from 4 to 5 trays. This means that the capital cost of the column will increase. The choice of the reflux ratio is thus a balance between operating and capital costs. Minimum Reflux Ratio If the reflux ratio is reduced too much, it will no longer be possible to achieve the separation required. The minimum reflux ratio is the value of R at which the number of trays approaches infinity. This occurs when the intersection of the two operating lines occurs on the equilibrium line. 1
1
xD y 1 + Rmin
y
xD 1 + Rmin
0
xB
x
xF
xD
1
0
xB
Liquid feed
x
xF
xD
1
Vapour feed
For the example above Rmin can be obtained from the x-y diagram: xD 1 + Rmin = 0.81
Rmin = 0.173 The optimum reflux ratio is often expressed as a function of Rmin. Typically the optimum reflux ratio is around 1.3 Rmin.
22
Vapour-liquid equilibrium for hexane/octane at P=1 atm. 1
5 4
0.9 3 0.8
xD = 0.633 1 + R
y (mol fraction hexane in vapour)
0.7 2 0.6
0.5
0.4 1
0.3
0.2 R 0.1
0 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
x (mol fraction hexane in liquid)
23
0.8
0.9
1