8. ABSORPTION 8.1
Introduction
Absorption is one of the important gas – liquid contact operations in which a gaseous mixture is contacted with a solvent to dissolve one or more components of the gas preferentially and provide a solution of them in the solvent. Some of the applications of this operation are as follows: (i) Ammonia Ammonia is removed removed from coke – oven oven gas with water water (ii) Benzene and toluene vapors are removed using hydrocarbon oil from the coke –
oven gas. (iii (iii))
Hydr Hydrog ogen en sulf sulfid idee is remov removed ed from from natu natura rall lly y occu occurr rrin ing g hydr hydroc ocar arbo bon n gase gasess
with alkaline solutions. (iv)Ammonia and other water soluble harmful gases from air are removed using water. 8.2 8.2
Gas Gas sol solub ubil ilit ity y in in liq liqui uids ds at equi equili libr briu ium m
The equilibrium characteristics of gas solubility in liquids are generally represented as partial pressure of solute in gas (p*) vs mole fraction of solute in liquid (x). A typical gas solubility curve drawn at a particular temperature and pressure for different gases is shown in Fig 8.1. If the gas solubility is low, then the equilibrium pressure for that particular system is very high. The solubility of gas is significantly affected by the temperature. Generally absorption processes are exothermic and if the temperature is increased at equilibrium, equilibrium, the solubility of gases, but not always, will be decreased due to evolution of heat.
s a g n i e t u l o s f o e r u s s e r p l a i t r a P * p
A 40ºC A 30ºC
A 15ºC
B
20ºC B 10ºC
C 10ºC
x
Mole fraction of solute in liquid
Fig. 8.1: Solubility of gas in liquid 8.3
Ideal and Non – ideal liquid solutions
In an ideal solution, all the components present in the solution approach simila similarit rity y with with regar regard d to their their chemi chemica call nature nature.. When When the the gas gas mixtu mixture re is in equilibrium with an ideal solution, then it follows Raoult’s law. p* = Px
(8.1)
where p* is the partial pressure pressure of solute, P is the vapour pressure of solute at the same temperature and x is the mole fraction of solute in liquid. For non – ideal solutions, Henry’s law can be applied and is given as, y* =
p * P t
= m.x
(8.2)
where ‘m’ is Henry’s Henry’s constant, Pt is the total pressure and y* is the mole fraction of solute in gas. 8.4
Choice of solvent for absorption
The following properties are to be considered while choosing a particular solvent in any absorption system. (i) Gas solubility solubility: Solubility of the solute to be absorbed in solvent should be
relatively high, as it will decrease the quantum of solvent requirement.
186
(ii) Chemical nature: Generally solvent should be chemically similar in structure to
that of the solute to be absorbed as it will provide good solubility. (iii) Recoverability: Solvent should be easily recovered and as it will help in reusing it. (iv) Volatility: Solvent should have a low vapor pressure (i.e.) less volatile. (v) Corrosiveness: Solvent should not be corrosive to the material of construction
equipment. (vi) Cost and Availability: The solvent should not be costly (inexpensive) (inexpensive) and readily
available. (vii)Viscosity: Should have low viscous as it will reduce pumping and transportation costs. (viii)Toxi Toxic, c, Flam Flamma mabi bili lity ty and and Stab Stabil ilit ityy: Solvent should be non – toxic, inflammable, chemically stable and non – reactive. 8.5
Design of Isothermal Absorption towers
The design of isothermal absorption towers is based on material balance in them. The flow of streams could be either co current or counter current. The operation is either carried out as a single stage operation or as a multistage operation.
8.5.1
Single stage – one component transferred – counter current and Isothermal Operation
Consider a single stage isothermal absorber shown in Fig.8.2, where (1) and (2) refer the bottom and top sections of the equipment respectively. Gaseous mixture entering the absorber at the bottom is contacted counter currently with liquid solvent, entering from the top. L2, x2, LS,
G2, y2, GS,
X2
Y2 (2)
(1) L1, x1, LS,
187 G1, y1, GS,
Fig.8.2: Flow in a counter current absorber absorber
Let, G1 and L2 be the molar flow rates of entering binary gaseous mixture and liquid respectively in moles/ (area) (time). Let, G2 and L1 be the molar flow rates of leaving gaseous mixture and liquid respectively in moles/ (area) (time). Let, GS and LS be the molar flow rates of inert gas and pure liquid respectively in moles/ (area) (time). Let x, y be the mole fractions of solute in liquid and gas phases respectively. Let X, Y be the mole ratios of solute to inert component in liquid and gas phases respectively. In the gas phase, only one component is transferred and the other component remains as inert. Similarly in the liquid phase, solvent is the inert component. It is more convenient to represent the concentrations concentrations of solute in liquid and gas phases in terms of mole ratios, (X and Y) of solute to inert component. So, X=
x
(1 − x)
Likewise, x =
and Y =
X
(1 + X )
y
(1 − y )
and y =
GS = G1 (1 – y1) or G =
Y
(1 + Y ) GS
(1 - y)
(8.3) (8.4)
(8.5)
Writ Writing ing the mater material ial balan balance ce on solut solutee basis basis for the above above count counter er curre current nt operation, we get GSY1 +LSX2 = GSY2 +LSX1
(8.6)
Gs (Y1 – Y2) = LS (X1 – X2)
(8.7)
∴
(i.e.)
L s G s
=
( Y 1 − Y 2 ) ( X − X ) 1 2
(8.8)
188
Eq. (8.8) represents operating line for a single stage counter current absorber. The operating operating line is linear which passes through the coordinates (X1, Y1) and (X2 ,Y2) with a slope of (LS / Gs). Since the solute transfer is taking place from gas to liquid phase, the operating line always lies above the equilibrium curve, which is shown in the following Fig.8.3 Operating line Y1 LS/GS
Y
Y2 Y = f(X) X2
X1 X
Fig.8.3: Equilibrium curve and operating line in mole ratio basis.
Suppose, if the flow rates of gas and liquid streams are not considered on inert basis (i.e.) on mole fraction basis, then the operating line would be a non – linear one passing through the coordinates coordinates (x1, y1) and (x2,y2) as shown in Fig.8.4. It is also highly impossible to know the intermediate concentrations which will enable one to draw this operating curve passing through the terminal points (x1,y1) and (x2,y2). Hence, it is more preferable to obtain the linear operating line with the known terminal concentrations of the system as shown in Fig.8.4.
Operating line y1
y
y2 y*= f(x) x2
x1 x
189
Fig.8.4: Equilibrium curve and operating line in mole fraction basis. 8.5.2
Determination of minimum (LS/GS) ratio
In absorption, minimum (Ls/Gs) ratio indicates a slope for operating line at which the maximum amount of solute concentration is obtained in the final liquid. It will be achieved only at the presence of infinite number of stages for a desired level level of absorp absorpti tion on of solut solute. e. When When the opera operati ting ng line line is tange tangenti ntial al to the equilibrium curve, then there is no net driving force and the required time of contact for the concentration concentration change desired is infinite and an infinitely infinitely tall tower will result. This is highly uneconomical. So, the tower is operated at an (Ls/Gs) ratio of 1.2 to 2.0 times the minimum (L S/GS) ratio. 8.5.3
Steps involved in determining the (L/G) min
1. Plot X and and Y data data to to draw draw equili equilibriu brium m curve. curve. 2.
Locate the point A (X2,Y2)
3. From point point A draw tangent tangent to the the equili equilibriu brium m curve. curve. 4.
Determine the slope of this line which will be (Ls/Gs) min.
5.
Extend the line from Y1 to intersect this operating line which corresponds to the point [(X1) max,Y1]
6.
Determine (Ls/Gs) Actual and find the slope.
7.
Using the operating line equation, obtain (X1) Actual as shown in Fig.8.5
Y1
B
B
(Ls/Gs) min
(Ls/Gs) act
A: (X2, Y2) Y
Y2
B: (X1, Y1)
A
B : (X1 max, Y1) 0
X2
X1
(X1) max
X
Fig. 8.5: Minimum L/G ratio
In some cases, the equilibrium curve will be more or less a straight line or concave upward. In such cases the minimum (L/G) ratio can be determined as shown in the following Figures 8.6 (a) and Fig.8.6 (b)
190
Y1 Y
Y1
(LS/GS) min
(LS/GS) min
Y
Y2
Y2
X2
X
(X1) max
X2
Fig.8.6 (a)
X
(X1) max
Fig.8.6 (b)
Fig.8.6: Equilibrium curve and operating line for special cases
Therefore, (Ls/Gs) min Since (Ls/Gs)
min
=
(Y 1 − Y 2 ) ( X ) − X 2 1 max
is known, (X1)
max
(8.9)
can be determined as all the other quantities
in Eq. (8.9) are known. 8.5.4
Multista Multistage ge counterc countercurre urrent nt isotherma isothermall Absorptio Absorption n
Let us consider a multistage multistage tray tower containing Np number of stages as shown in Fig.8.7 Fig.8.7 where the suffix represents represents the tray number. number. The operation operation is isothermal and assuming that the average composition of gas leaving from a tray is in equilibrium with the average composition of liquid leaving from the same tray. The characteristics of the entering and leaving streams are also represented. The flow of streams is countercurrent. The liquid flows downwards and the gas upwards and only one component is transferred. The number of theoretical or ideal stages required for the desired operation in the tower is determined as follows: The liquid flows downwards and the gas The material balance on inert basis gives, GSY Np+1 + LSX0 = GSY1 +LSX Np
(8.10)
GS (Y Np+1 – Y1) = LS (X Np – X0)
(8.11)
(i.e)
L s G s
=
( Y NP +1 − Y 1 ) ( X − X ) NP 0
(8.12)
Eq. (8.12) represents a linear operating line for a multistage countercurrent absorber which passes through the coordinates (X0,Y1) and (X Np,Y Np+1) with a slope (LS/GS). Betwee Between n the equilibriu equilibrium m curve curve and operati operating ng line, line, a stepwis stepwisee construction is made to obtain the number of theoretical trays. The stepwise
191
construction is started from (X0, Y1) since it represents operating condition in plate number 1 (as per our convention). This is illustrated in Fig.8.8. L0, LS, X0, x0
G1, GS,Y1, y1
Y1 X1
Y2
X2
Y Np
Np – 1 Np
X Np
LS, L Np, X Np, x Np
GS, G Np +1, Y Np+1, y Np+1
Fig.8.7: Various streams in a counter current multistage tray tower Y N p+1
N
Operating line
P
P
n
Y
Equilibrium curve
1
Y1
X0
X
X Np
Fig.8.8: Stepwise construction for estimating the number of plates/stages
8.5.5
Analytical method to determine the number of trays.
In some special cases such as dilute gaseous mixtures or solutions, the equilibrium curve is a straight line then the number of trays can be determined analyti analyticall cally y by using using Kremse Kremser-Br r-Brown own–Sou –Souders ders equation equation given given below below without without going in for a graphical method.
192
N p =
log[ (
y Np+1 − mx 0 y1 − mx 0
)(1 − 1/A) + 1/A
]
(8.13)
log A where, A is the absorption factor equals
L
mG
, and m is the slope of the equilibrium
curve. Absorption factor is defined as the ratio of the slope of the operating line to that of the equilibrium curve. If ‘A’ varies due to small changes in L/G from bottom to top of the tower, then the geometric mean value of A will be considered. Hence, Geometric mean value of A = A1 =
A2 =
L1 mG 1
L Np mG Np
≈ ≈
A1 A2
(8.14)
L0 mG 1
L Np mG NP +1
where A1 is absorption factor at the top of the tower and A2 is absorption factor at the bottom of the tower. For larger variations in A, graphical computations must be followed. 8.5.6 8.5.6
Signif Significa icance nce of Absorp Absorptio tion n fact factor or
If A<1, the operating line and equilibrium curve converge at the lower end of the tower indicating that the solubility of solute is limited even for a large number of trays provided. If A>1, any degree of separation is possible with adequate number of trays. However, However, as A increases increases beyond 1.0 for a fixed quantity of gas and a given degree of absorption, the absorbed solute is dissolved in a larger quantity of liquid and hence becomes less valuable. In addition to that, the number of trays also decrease leading to a lower cost of equipment. This leads to a variation in total cost of operation which will pass through a minimum. Hence, for an economical operation, the value of A has been estimated for various systems and found to be in the range of 1.2 to 2.0. 8.6
Design of Multistage Non – Isothermal absorber
Generally the absorption operations are exothermic. Hence, the solubility of gas decreases as temperature temperature of the liquid increases increases which in turn decreases the capacity of the absorber. absorber. When concentrated gaseous mixtures are to be absorbed
193
in solvent then the temperature effects have to be taken into account. If the heat liberated liberated is more, then cooling coils should be provided for an efficient efficient operation. Since the temperature is varying from tray to tray, it influences the concentration changes and as well as the flow rate of streams. Hence, energy balance should also be incorporated along with material balance to determine d etermine the number numbe r of trays. It is very very diffi difficul cultt to comput computee manua manually lly the the tray tray to tray tray calcu calcula latio tions. ns. A simple simple algorithm is developed for one ideal tray involving trial and error calculations and then programming to other trays for the determination of the number of trays. Consider a stage wise tray tower operating non – isothermally as shown in Fig.8.9. Total mass balance around the tower gives, G Np +1 + L0 = L Np +G1
(8.15)
Component balance gives, [G Np +1] y Np+1 + L0x0 = L Npx Np +G1y1
(8.16)
Energy balance gives, [G Np+1] HG,Np+1 + L0HL0 = L NpHLNp +G1HG1
(8.17)
L0 x0 HL0
G1, y1, HG1, tG1
tL0
G1 L1 ….
2
L2
Gn
Ln
G Np-1
L Np-1
1
G Np, y Np
n Np-1
Np
L Np
L Np
G Np+1
x Np
y Np+1
HL, Np
HG, Np+1
tL, Np
tG, Np+1
Envelope - I
Fig.8.9: Streams in a counter current multistage tray tower and envelope – I
194
where, H is the molal enthalpy of streams. Enthalpies can be determined using the available literature data with reference to some base temperature say, t0. (Pure state) HG = C pG, inert (tG – t0) (1 – y) + y [C pG solute (tG – t0) + λ 0]
(8.18)
HL = C pL, inert (tL – t0) (1 – x) + x [C pL solute (tL – t0)]
(8.19)
where Cp is the specific heat of the component and λ 0 is the latent heat of vaporization at reference temperature. Now let us consider the envelope en velope I, Mass and energy balance in envelope I give, Ln + G Np+1 = Gn+1 +L Np
(8.20)
Ln xn + [G Np+1] y Np+1 = Gn+1y n+1 +L Npx Np Ln HLn + [G Np+1] HG Np+1 = Gn+1HG n+1 +L NpHLNp
(8.21) (8.22)
Let n = N p − 1, L Np−1 + G Np+1 = G Np +L Np
(8.23)
L Np−1 x Np−1 + [G Np+1] y Np+1 = G Npy Np+L Npx Np
(8.24)
L Np−1 HL,Np−1 + [G Np+1] HG,Np+1 = G NpHG,Np+L NpHL,Np
(8.25)
To solve the above system of equations and determine the number of trays, the following procedure is used. 1.
Assume the top tray temperature, temperature, tG1. The other values like G Np+1, y 1, y Np+1, L 0, x 0, tL0 and tG Np+1 are known.
2.
Calculate GS from the relationship, GS = G Np+1 (1 - y Np+1)
3.
Calculate G1 from the relationship, G1 = GS/(1 – y1)
4.
Using Eq. (8.15), Calculate L Np.
5.
Find x Np from Eq. (8.16).
6.
Calculate HGNp+1, HL0 and HG1 using Eqs. (8.18) and (8.19)
7.
Find HLNp from Eq. (8. 17).
8.
Determine tLNp making use of Eq. (8.19).
9.
With this knowledge of the temperature of the last tray N p, the compositions can be determined by y* = (V.P/T.P)x or o r y* =m.x where V.P. is vapor pressure, press ure, T.P is total pressure and m is equilibrium constant. Hence y Np = (m) x Np
10. Now for the last tray, x Np, y Np, tLNp are known.
195
11. Find G Np =
G s
(1 − y Np )
12. Now calculate L Np – 1 using Eq. (8.23) 13. Find x Np–1 from Eq. (8. 24). 14. Calculate HLNp–1 using Eq. (8.25) 15. Find tLNp–1 from Eq. (8.19). 16. Now determine deter mine the composition, y Np–1 and G Np–1 as mentioned in step (9) and step
(11) respectively. 17. Simila Similarly rly calculate calculate for the next tray by taking n = Np–2 and startin starting g from step (12), by making use of material and enthalpy balances. 18. Final Finally ly,, the comput computat ation ion is stopp stopped ed on reach reaching ing the value value of ‘y1’ and also
satisfying the assumed tG1. If these two are not satisfied together, once again the iteration has to be started fresh by assuming a new temperature ‘tG1’. However, the values of ‘y1’ and ‘tG1’, are both satisfied, the number of trays are known from the computation values.
8.7
Design of Co – current absorber
In a co - current absorber both gas and solvent streams are entering parallel into the absorber as shown in Fig.8.10. L1 LS x1 X1
G1 GS y1 Y1
196
(1)
(2) L2
G2
LS
GS
x2
y2
X2
Y2
Fig.8.10: Co – current absorber
By writing material balance, LSX1 +GSY1 = LSX2 +GSY2 ∴
(ie)
-
(8.26)
Ls (X1 – X2) = GS (Y2 – Y1)
(8.27)
L s G s
(8.28)
=
( Y 1 − Y 2 ) ( X − X ) 2 1
Eq. (8.28) is the operating line equation for co – current absorption operation with the slope - (LS/GS) and this will be presented in the X – Y diagram of Fig 8.11. If the leaving streams are in equilibrium with each other, then the compositions are represented by (X2’ Y2’).
Operating line
Y1
Y
Equilibrium curve
Y2 Y2’
Slope = – (LS/GS)
X1 X X2’ 2 X
197
Fig.8.11: Equilibrium curve and operating line in a co current absorber. 8.8 8.8
Desi Design gn of of cont contin inuo uous us con conta tact ct equi equipm pmen entt for for abso absorp rpti tion on
Packe Packed d colum columns ns and spray spray towe towers rs fall fall in the the categ category ory of conti continuo nuous us contac contactt or diffe differen renti tial al conta contact ct tower towers. s. They They are diffe differen rentt from from stage stage wise wise contactors in the sense that the fluids are in continuous contact throughout the tower. So the liquid and gas compositions compositions change continuously with respect to the height of the tower. Consider a packed tower of unit cross sectional area as shown in the Fig.8.12. The characteristics of inlet and outlet streams are also indicated. Let ‘Z’ be the total height of the tower and ‘dZ’ be the differential height which is same as differential volume. ‘S’ is the total effective interfacial surface per unit tower cross section. s ection. Hence, Interfacial area (a).[A.Z] S= = Area of tower A ∴
(8.29)
dS = a.dZ
(8.30)
where where dS is the diffe differen renti tial al inter interfa facia ciall surfa surface ce in the the differ different entia iall volum volumee of packing.
G2 L2 LS x2 X2 (2)
GS y2 Y2
Z dz
(1) L1
G1
LS
GS
x1
y1
X1
Y1
198
Fig.8.12: Continuous counter current absorber.
As shown in Fig.8.12, the quantity of solute A passing through the differential section is G.y moles/ (area) (time). The rate of mass transfer is d (G.y) moleA/ (differential volume) (time). Since NB = 0, NA/ (NA + NB) = 1.0. The molar flux of A is obtained by applying the original basic flux equation, Rate of absorption of solute ' A' Interfacial area
NA =
=
d (Gy ) adZ
(1 − y i ) (1 − y )
= FG ln
(8.31)
d (Gy) can be written as d (Gy) =d
G s y (1 − y )
(8.32)
Since one component is transferred, G and y vary throughout the tower.
G s y (1 − y )
(i.e.) d
=
G s dy
=
(1 − y ) 2
Gdy
(8.33)
(1 − y )
Substituting Eq (8.33) in Eq (8.31), rearranging and integrating we get,
y1
Z
∫
Gdy
∫ F a(1 − y) ln[(1 − y ) /(1 − y)]
(8.34)
It is more convenient to write, y – y i = [(1 – yi) – (1 –y)]
(8.35)
Z =
dZ =
0
y 2
G
i
The numerator and denominator of Eq. (8.34) can be multiplied by the right and left hand sides of Eq. (8.35) respectively to obtain Z =
y1
G (1 − y ) iM dy
∫ F a(1 − y)(1 − y )
y2
G
(8.36)
i
where (1 − y ) iM is logarithmic mean of (1 –yi) and (1 –y) Z =
G F G a
y1
(1 − y ) iM dy
∫ (1 − y)( y − y ) ≈ H
N tG
tG .
(8.37)
i
y2
where H tG. is height of a gas transfer unit and NtG is number of gas transfer units. Thus, H tG
=
G F G a
=
G k y a (1 − y ) iM
=
G k G aP t (1 − y ) iM
(8.38)
In terms of other individual mass transfer coefficients,
199
N tG is simpli simplifi fied ed furthe furtherr by substi substitut tutin ing g the arithm arithmeti eticc avera average ge instea instead d of
logarithmic average of (1 –y) iM Hence, (1 − y ) iM
=
(1 − y i ) − (1 − y )
(1 − yi ) (1 − y )
≈
(1 − y i ) + (1 − y ) 2
ln y1
N tG =
(1 - y) iM dy
y1
dy
∫ [(1 − y)( y − y ) ∫ ( y − y ) i
y2
=
i
y2
+
(8.39)
1 (1 − y 2 ) ln 2 (1 − y1 )
(8.40)
Similarly, when the above mentioned relations have been applied for liquid compositions we obtain Z =
x1
L
∫
(1 − x ) iM dx
F L a x2 [(1 − x )( x i
− x)]
≈ H tL . N tL
(8.41)
where H tL. is the height of liquid transfer unit, N tL is the number of liquid transfer units and (1 –x) iM is logarithmic mean of (1 – x) and (1 –xi) On simplification we get, H tL. =
L FL a
=
L k x a(1 − x) iM
(8.42)
and x1
N tL
= ∫
dx
( xi − x ) x2
+
1 (1 − x1 ) ln 2 (1 − x 2 )
(8.43)
The above Eqs. (8.38), (8.40), (8.42) and (8.43) can be used to determine the height of the tower. With the known quantities, HtG or HtL can be easily determined. But NtG
and NtL can be determined only through graphical method. For this plot
1 ( y − y i )
against y and the area under the curve will give N tG . The values values of y and yi can be evaluated evalu ated by b y drawing a line between equilibrium equilib rium curve curv e and operating line with the slope (−k xa/k ya) where y and yi are points of intersection of this line on operating line and equilibrium curve respectively. 8.8.1
Overall transfer units
200
In some cases where the equilibrium curve is straight and the ratio of mass transfer coefficients coefficients is constant, it is more convenient to make use of overall mass transfer coefficients. The height of the tower can be expressed in such cases as Z = NtoG.HtoG N toG
(8.44)
(1 − y ) * M dy
y1
=
G F OG a
N toL
= ∫
H toL
=
≈
+
y 2
x1
8.8. 8.8.2 2
dy
∫ (1 − y)( y − y*) ∫ ( y − y*)
=
y 2
H toG
y1
=
G K y a(1 − y ) * M
dx
( x * − x ) x2 L
=
F oL a
+
=
1 (1 − y 2 ) ln 2 (1 − y1 ) G
K G aP t (1 − y ) * M
1 (1 − x1 ) ln 2 (1 − x 2 )
(8.45)
(8.46)
(8.47)
L
(8.48)
K x a(1 − x) * M
Dilu Dilute te solu soluti tion onss
For Dilute solutions or gaseous mixtures, the above equations become much much simp simple ler. r. The The seco second nd term term in Eq. Eq. (8.4 (8.45) 5) and and in Eq. Eq. (8.4 (8.47) 7) beco become mess negligible. Hence, y1
N toG
=
dy
∫ ( y − y*)
x1
or N toL
y 2
= ∫ x2
dx
( x * − x)
(8.49)
If the equilibrium curve in terms of mole fractions is also linear over the entire range of x, then y* = m.x + C
(8.50)
If the solutions are dilute, there won’t be variations in L/G ratio throughout, and the operating line can be considered as a straight line so that the driving force (y – y*) is also linear. In such cases, Eq. (8.43) is simplified to N toG
=
( y1 − y 2 ) ( y − y*) M
(8.51)
where (y – y*)M is logarithmic average of the concentration differences at the terminals of the tower. Therefore, ( y − y*) M
=
( y1 − y1 *) − ( y 2 − y 2 *) ( y − y1 *) ln 1 ( y 2 − y 2 *)
(8.52)
201
and HtoG = 8.8.3
G
G K y a
or
(8.53)
K G a P t
Dilute solutions using Henry’s law
In dilute solutions, if Henry’s law is applied, then y* = m.x
(8.54)
The operating line can be written in a linear form as (y – y2) = (L/G) (x – x2)
(8.55)
Eliminating x between Eqs. (8.54) and (8.55) and substitution of y* in Eq. (8.49) gives, y1 − mx 2 1 1 1 - + y 2 − mx 2 A A
ln
NtoG =
1 1 - A
(8.56)
where A is the absorption factor = L/mG The overall height of transfer units can also be expressed in terms of individual phases, HtoG = HtG + (mG/L) HtL or Ht0L = HtL + (L/mG) HtG
8.9
(8.57)
Stripping or Desorption
When mass transfer occurs from liquid to gas, i.e., the solute is removed from the liquid solution by contacting with a gas, then the operation is called Desorption or Stripping. 8.9. 8.9.1 1
Oper Op erat atin ing g line line for for str strip ippe perr
The schematic representation of operating lines for both countercurrent and co-current operations of a stripper are shown in Fig.8.13 and Fig.8.14.
Y1 Y
Slope (LS/GS)
Y NP+1 X NP 202
X Fig.8.13: Equilibrium curve and operating line in a counter current stripper.
Y
Y2 Slope – (LS/GS) Y1
X2 X1 X Fig.8.14: Equilibrium curve and operating line in a co current stripper. Analyt Analytica icall rela relatio tion n to to det deter ermin minee numb number er of plate platess .
8.9.2 8.9.2
N p =
x 0 − y Np+1 /m 1 1 log 1 - + − x y /m Np Np+1 S S
(8.58)
log S where S is the stripping factor, S =
mG L
For dilute solutions, if Henry’s law is applied, NtoL =
ln [
− y1 /m ](1 − A) + A x 1 − y1 /m 1− A
x2
(8.59)
Worked Examples:
1)
An air- NH3 mixture containing 5% NH3 by volume is absorbed in water using a packed tower at 20ºC and 1 atm pressure to recover 98% NH3. Gas flow rate is
1200
kg . Calculate a) minimum mass flow rate of liquid. b) NTU using 1.25 hr m 2
times the minimum liquid flow rate c) Height of packed column using K Ga = 128 kg atm. The equilibrium relation is y = 1.154x where, x, y are expressed in hr m 2 mole fraction units.
203
L2, x2, LS, X2
G2, y2, GS, Y2
(2)
(1) L1, x1, LS, X1
G1, y1, GS, Y1
Fig. 8.15 (a) Example 1 Solution:
Data: y1= 0.05, Pt = 1 atm, T = 20oC and X2 = 0 Gas flow rate = 1200
kg hr m 2
Average molecular weight of mixture= (0.05×17) + (0.95×28.84) = 28.25 1200 28.25
G1 =
= 42.478
k mole . hr m 2
Gs = G1 (1-y1) = 42.478 (1-0.05) = 40.354
k mole hr m 2
Y2 = 0.02 × 0.0526 = 0.001052 Y1=
y1 0.05 = 1 − y1 1 − 0.05
= 0.0526
k mole NH3 k mole dry air
Y2 = 0.001052 y2 =
Y2 1 + Y2
=
0.001052 1.001052
G2 =
GS 1 − y2
=
40.354 1 - 0.00105
= 0.00105 = 40.396
k mole hr m 2
y =1.154x Y 1+ Y Y=
=
1.154X 1+ X
1.154X 1 − 0.154X
204
X 0.01 0.02 Y 0.0116 0.0232 For minimum liquid flow rate
0.03 0.0348
0.04 0 .0 4 6 4
0.05 0.058
y = 1.154 x, then y 1 = 1.154 x1 0.05 = 1.154 x1, so x1 = 0.0433 X1 =
x1 k mole NH3 = 0.04526 1 − x1 k mole water
Fig. 8.15 (b) Example 1
(This can also be obtained from graph)
LS min = ( Y1 − Y2 ) = ( 0.0526 − 0.001052 ) = 1.139 ( 0.04526 − 0) GS ( X1 − X2 ) k mole (Ls) min = 40.354 × 1.139 = 45.969 hr m 2 Mass of minimum water = 45.969 × 18 = 827.44
kg hr m 2
LS actual = 1.25 LS min = 1.25 ×1.139 = 1.42375 GS GS Again,
LS actual = ( Y1 − Y2 ) ( X1 − X2 ) GS
205
1.42375 =
( 0.0526 − 0.001052) ( X1 − 0 )
X1 = 0.0361
Hence, x1 =
X1 (1 + X1)
= 0.0349
y1* = m x1 = 1.154 × 0.0349 = 0.0403 y2* = m x2 = 0 NTU =
( y1 − y2 ) ( y − y *) lm
( y − y *) lm =
NTU = (c)
[ ( y1 − y1 *) − ( y2 − y2 *) ] [ ( 0.05 − 0.0403) − ( 0.001052 − 0) ] = = 3.89 ×10 −3 ( 0.05 − 0.0403) ( y1 − y1 *) ln ln ( 0.001052 ) ( y2 − y2 *)
( 0.05 − 0.001) 3.89 ×10 −3
= 12.581 ≈ 13
Average gas flow rate =
HTU =
G K GaPt
=
41.437 128
( G1 + G2 ) 2
=
( 42.478 + 40.396) 2
= 41.437
kmole m 2 hr
= 0.3237m
Height of the tower, Z = NTU × HTU = 12.581 × 0. 3237 = 4.073 m 2)
Air containi containing ng methanol methanol vapor vapor (5-m (5-mole ole %) is scrubbed scrubbed with with water water in a packed packed tower at 26ºC and 760 mm Hg pressure to remove 95 % of the methanol. The entering water is free of methanol. The gas-phase flow rate is 1.22
liquid-phase liquid-phase rate is 0.631
k mole and the m 2sec
k mole . If the overall height of a transfer unit based on m 2sec
the liquid phase resistance is 4.12m, determine NTU and the overall liquid phase mass transfer coefficient. The equilibrium relation is p = 0.280 x, where p is partial pressure of methanol methan ol in atmospheric a tmospheric and x is mole fraction of methanol meth anol in liquid. Solution:
L2, x2, LS, X2
G2, y2, GS, Y2
(2)
(1) L1, x1, LS, X1
206 G1, y1, GS, Y1
Fig. 8.16 Example 2
y1 = 0.05, T = 26°C, Pressure = 760 mm Hg, Y1 =
y1 (1 − y1)
=
0.05 0.95
= 0.0526
Y2 = 0.05 × 0.0526 = 0.00263 Gas flow rate = 1.22
k mole , m 2sec
Liquid flow rate = 0.631
k mole m 2sec
HtoL = 4.12m Equilibrium relationship is: p = 0.280 x where p = partial pressure x = mole fraction of methanol in liquid. L2 = LS =
0.631 k mole = 0.0351 2 18 m sec
(Assuming entering water is pure) Average molecular weight = G1 =
1.22 28.998
= 0.0421
( 0.05 × 32) + ( 0.95 × 28.84) 1
= 28.998
k mole m 2sec
GS = G1 (1- y1) = 0.0421(1-0.05) = 0.04
k mole m 2sec
Equilibrium relation is: p = 0.280 x pt y =0.280 x
(pt = total pressure, x = mole fraction of liquid)
1×y = 0.280 x y = 0.280 x
LS = ( Y1 − Y2 ) GS ( X1 − X2 )
207
y2 =
Y2 1 + Y2
=
0.00263 (1 + 0.00263)
= 0.00262
X2 = 0 (assuming pure water enters the reactor)
LS = ( Y1 − Y2 ) GS ( X1 − X2 ) (i.e.) (i.e)
0.0351 ( 0.0526 − 0.00263) = ( X1 − 0 ) 0.04
Therefore, X1 = 0.0569 L1 = Ls (1+ X1) = 0.0351 (1.0569) = 0.0371
k mole m 2sec
L Avg = (L1×L2)0.5 = (0.0371 × 0.0351)0.5 = 0.0361
x1 =
X1 (1 + X1)
=
0.0569 (1 + 0.0569)
k mole m 2sec
= 0.0539
We have, y1* = 0.280 x1 x1 = 0.0539 x2=0.0 y1= 0.05 y2 = 0.00262 x1* =
y1 0.280
=
0.05 0.280
x2* =
y2 0.280
=
0.00262 0.280
( x * − x)lm =
= 0.1786 = 0.00936 [ (0.1786 - 0.0539) − (0.00936 - 0)]
[ ( x1 * − x1 ) − ( x 2 * − x 2 )] ( x1 * − x1 ) ( x 2 * − x 2 )
ln
=
(0.1786 - 0.0539) (0.00936 - 0)
ln
= 0.04455 NtoL =
( x1 - x2 ) ( 0.0539 − 0) = = 1.21 0.04455 ( x * -x ) lm
HtoL = 4.12m HtoL =
LAvg K La
Therefore, K La =
0.0361 k mole = 8.76 ×10 −3 2 (Δx) 4.12 m sec
208
3)
An air- NH3 mixture containing 20-mole % NH3 is being treated with water in a
packed tower to recover reco ver NH3. The incoming gas rate is 1000
kg . The hr m 2
temperature is 35ºC and the total pressure is 1 atm. Using 1.5 times the minimum water flow rate, 95% of NH 3 is absorbed. If all the operating conditions remain unchanged, how much taller should the tower be to absorb 99% of NH3? Henry’s law is valid and ye = 0.746 x. Variations in gas flow rate may be neglected. Solution: L2, x2, LS, X2
G2, y2, GS, Y2
(2)
(1) L1, x1, LS, X1
G1, y1, GS, Y1
Fig. 8.17 (a) Example 3
Data: y1 = 0.2 kg hr m 2 Temperature = 35°C, pressure = 1 atm Gas flow rate (incoming) = 1000
HTU = 1 m. (LS) actual = 1.5 × (LS) min Assuming incoming water to be pure, its flow rate L1 is LS
k mole hr m 2
Equilibrium relation = ye = 0.746 x Y1 =
y1 (1 − y1)
=
0.2 (1 - 0.2)
= 0.25
95% Ammonia is absorbed Y2 = (1-0.95) × 0.25 = 0.0125
209
y2 =
Y2 1 + Y2
=
0.0125 (1 + 0.0125)
= 0.0123
Average molecular weight = [(0.2×17) + (0.8×28.84)] = 26.472 1000 26.472
G1 =
= 37.776
k mole m 2 hr
GS = G1 (1- y1) = 37.776(1-0.2) = 30.221
k mole . hr m 2
For minimum liquid flow rate. y1* = 0.746 x1 x1 =
0.2 0.746
X1 =
x1 = 0.3663 1 − x1
= 0.2681
LS min = ( Y1 − Y2 ) = ( 0.25 − 0.0125) = 0.648 GS ( X1 − X2 ) ( 0.3663 − 0) (Assuming pure water enters, X2 = 0) We can also obtain this graphically for which X-Y data has to be computed. y1* = 0.746 x1 Y (1 + Y )
= 0.746
X (1 + X )
Fig. 8.17 (b) Example 3
Y=
0.746X (1 + 0.254X )
X
0
0.1
0.2
0.3
0.4
210
Y
0
0.072 0.14 0.20 8 2 8 From the graph X1,max = 0.3663 (which is same as by calculation also)
0.271
LS min = 0.648 GS LS actual = 1.5 LS min = 1.5 × 0.648 = 0.972 GS GS LS actual = ( Y1 − Y2 ) = ( 0.25 − 0.0125) ( X1 − X2 ) ( X1 − 0) GS X1 = 0.2443 x1 =
X1 (1 + X1)
= 0.1963
y1* = 0.746 x1 y1* = 0.746 ×0.1963 = 0.1464
( y − y *) lm =
NTU =
[ ( y1 − y1 *) − ( y2 − y2 *) ] [ ( 0.2 − 0.1464) − ( 0.0123 − 0) ] = = 0.0281 ( 0.2 − 0.1464 ) ( y1 − y1 *) ln ln ( 0.0123 - 0) − ( ) y 2 y 2 *
( y1 − y2 ) ( 0.2 − 0.0123) = = 6.68 ( y − y *) lm 0.0281
Z = HTU × NTU = 1 × 6.68= 6.68m Now if 99% of NH 3 is absorbed, Y2 = 0.25× 0.01 =0.0025 y2 =
Y2 1 + Y2
= 0.0025
LS min , GS
For
y1* = 0.746 x1 X1 0.2 = = 0.2681 (1 + X1) 0.746 x1 = 0.3663 X1 = 1 − x1 LS min = ( Y1 − Y2 ) = ( 0.25 − 0.0025) = 0.6755 GS ( X1 − X2 ) ( 0.3663 − 0) LS actual = 1.5 × 0.6755 = 1.013 GS LS actual = ( Y1 − Y2 ) = ( 0.25 − 0.0025) ( X1 − X2 ) ( X1 − 0) GS x1 =
211
X1 = 0.2443 x1 =
X1 (1 + X1)
= 0.1963
y1* = 0.746 ×0.1963 = 0.1464
[ ( y1 − y1 *) − ( y2 − y2 *) ] [ ( 0.2 − 0.1464) − ( 0.0025 − 0) ] = = 0.01667 ( 0.2 − 0.1464 ) ( y1 − y1 *) ln ln ( 0.0025 - 0) ( y2 − y2 *) ( y1 − y2 ) ( 0.2 − 0.0025) NTU = = = 11.847 ( y − y *) lm 0.01667 ( y − y *) lm =
Z = NTU × HTU =11.847 × 1 = 11.847 m. In first case, when 95% of NH3 was absorbed, Z = 6.68 m Increase in length of tower = 11.847 – 6.68 = 5.168 m So, when 99% of NH 3 is to be absorbed, the tower should be 5.168 m taller than that needed for 95% NH3 absorption, or 77.36% taller. 4)
An effluent gas containing 12% C6H6 is to be scrubbed in a packed column, operating at 43ºC and 1 atm. pressure. The column is to be designed for treating 15 m3 of entering gas per hour per m2 of column column cross-secti cross-section, on, such such that that the exit gas will will contain 1% benzene. 1% benzene. The solvent so lvent for fo r scrubbing scr ubbing is mineral oil which whic h will enter the
top of the column at a rate of 28
kg and a benzene content of 1%. Determine Determine hr m 2
the height of the column assuming height of transfer unit to be 0.75m The equilibrium concentration at the operating conditions may be estimated as y* = 0.263x. where x and y are in mole fraction units. Solution:
L2, x2, LS, X2
G2, y2, GS, Y2
(2)
(1) L1, x1, LS, X1
G1, y1, GS, Y1
Fig. 8.18 Example 4
212
y1 = 0.12,
T = 43°C,
pressure = 1 atm
m3 Gas flow rate = 15 hr m 2 y2 = 0.01 Solvent is mineral oil L2 = 28
k mole , hr m 2
x2 = 0.01,
HTU = 0.75 m Equilibrium relation is y* = 0.263 x Assuming the gas mixture to be ideal; P1V1 P2V2 = T1 T2
=
(1×15) ( 273 + 43)
=
(1× V2 ) 273
V2 = 12.9589 m3 (at S.T.P) (or) V2 =
12.9589 22.414
= 0.5782k moles
k mole hr m 2
G1 = 0.5782
GS = G1 (1- y1) = 0.5782(1-0.12) = 0.5088
LS = L2 (1- x2) = 28(1- 0.01) = 27.72
k mole . hr m 2
k mole . hr m 2
LS = ( Y1 − Y2 ) GS ( X1 − X2 ) Y1 =
y1 (1 − y1)
=
0.12 (1 - 0.12)
= 0.1364
Y2 =
y2 (1 − y2 )
=
0.01 (1 - 0.01)
= 0.0101
X2 =
x2 (1 − x1)
=
0.01 (1 - 0.01)
= 0.0101
LS = 27.72 = ( 0.1364 − 0.0101) ( X1 − 0.0101) GS 0.5088 X1=0.01242 x1 =
X1 (1 + X1)
= 0.0123
213
y1* = mx1 y1* = 0.263 × 0.0123 = 0.00323 y2* = 0.263 × 0.01 = 0.00263
( y − y *) lm =
NTU =
[ ( y1 − y1 *) − ( y2 − y2 *) ] [ ( 0.12 − 0.00323) − ( 0.01 − 0.00263) ] = = 0.0395 ( 0.12 − 0.00323) ( y1 − y1 *) ln ln ( 0.01 - 0.00263) ( y2 − y2 *)
( y1 − y2 ) ( 0.12 − 0.01) = = 2.786 ≈ 3 ( y − y *) lm 0.0395
Height of tower, Z = NTU × HTU =2.786 × 0.75= 2.0895 m. 5)
An air- NH3 mixture containing 5% NH3 is being scrubbed with water in a packed tower to recover 95% NH3. G1= 3000 kg/hrm2, Ls= 2500
kg . Tower is hr m 2
maintained at 25ºC and 1 atm pressure. Find NTU and Height of the tower. Equilibrium relation is y* = 0.98x, where x and y are mole fraction units. K Ga =
65
k mole hr m 3atm
Solution:
y1 = 0.05, Y1 =
0.05 (1 - 0.05)
= 0.0526
Y2 = 0.05 × 0.0526 = 0.00263 y2 =
Y2 1 + Y2
=
0.00263 (1 + 0.00263)
= 0.00262
Entering gas flow rate = 3000
kg hr m 2
L2, x2, LS, X2
G2, y2, GS, Y2
(2)
(1) L1, x1, LS, X1
G,y,G,Y 1
1
S
1
214
Fig. 8.19 Example 5
LS = 2500
kg , hr m 2
K G a = 65
k mole hr m 3atm
T = 25°C,
Pressure = 1 atm,
Equilibrium relation = y* = 0.98x Average molecular weight = G1 =
3000 28.248
LS =
2500 18
= 106.20
= 138.89
( 0.05 ×17 ) + ( 0.95 × 28.84) 1
k mole m 2 hr
k mole m 2 hr
GS = G1 (1- y1) = 106.2 (1-0.05) = 100.89
G2 =
GS 1 − y2
=
= 28.248
100.89 (1 - 0.00262 )
= 101.16
k mole . hr m 2
k mole hr m 2
LS = ( Y1 − Y2 ) = 138.89 = ( 0.0526 − 0.00263) ( X1 − 0) GS ( X1 - X2 ) 100.89 Therefore, X1 =0.0363 x1 =
X1 (1 + X1)
=
0.0363 (1 + 0.0363)
= 0.035
y1* = 0.98 x1 y1* = 0.98 × 0.035 = 0.0343 x2 = 0 ; y2* = 0
( y − y *) lm =
NTU = Gavg =
[ ( y1 − y1 *) − ( y2 − y2 *) ] [ ( 0.05 − 0.0343) − ( 0.00262 − 0) ] = = 0.0073 ( 0.05 − 0.0343) ( y1 − y1 *) ln ln ( 0.00262 - 0) ( y2 − y2 *)
( y1 − y2 ) ( 0.05 − 0.00262) = = 6.486 ( y − y *) lm 0.0073
( G1 + G2 )
HTU =
2 Gavg K GaPt
=
=
(106.2 + 101.16)
103.68 ( 65 ×1)
2
= 103.68
kmole m 2 hr
= 1.595m
215
Z = NTU × HTU = 6.486 × 1.595 = 10.346 m. 6)
An air-C air-C6H6 mixture containing 5% benzene enters counter current absorption
tower where it is absorbed with hydrocarbon oil. Gs = 600
k mole . Solubility hr
follows Raoult’s law. Temperature at 26.7ºC and 1 atm is operating condition. Average molecular weight of oil is 200. Vapor pressure of benzene at 26.7ºC is103 mm Hg. Find
i)
(Ls) min to recover 90% of entering C6H6
ii) Number of theoretical stages if 1.5 times the minimum liquid rate used. iii) The concentration of solute in liquid. Solution: L2, x2, LS, X2
G2, y2, GS, Y2
(2)
(1) L1, x1, LS, X1
G1, y1, GS, Y1
Fig. 8.20 (a) Example 6
Data: y1 = 0.05,
GS = 600
k mole , hr
T = 26. 26.7°C, °C,
Average mo molecular we weight of of oi oil = 200,
Pressu essure re = 1 atm
pA = 103 mm Hg
According to Raoult’s law, p* = pA xA y* = Y1 =
y1 (1 − y1)
=
p Pt
0.05 (1 - 0.05)
=
( pAxA ) Pt
=
103 x = 0.1355x 760
= 0.0526
Y2 = (0.1×0.0526) = 0.00526 y2 =
Y2 1 + Y2
=
0.00526 (1 + 0.00526)
= 0.00523
216
X2 = 0. (Assuming (Assuming pure oil enters) enters) We have, y* = 0.1355x Y (1 + Y )
= 0.1355
Therefore, Y = X Y
0 0
X (1 + X )
0.1355X (1 + 0.8645X )
0.1 0.012
0.2 0.023
0.3 0.0322
0.4 0 .0 4 0 3
5 8 From the graph, we can get, X 1, max = 0.54
0.5 0.047
0.6 0.0535
3
LS min = ( Y1 − Y2 ) GS ( X1, max − X2 ) For minimum flow rate of oil,
Fig. 8.20 (b) Example 6
LS min = ( Y1 − Y2 ) = ( 0.0526 − 0.00526 ) = 0.0877 ( 0.54 − 0) GS ( X1 − X2 ) (LS) min = 0.0877 × 600 = 52.62
k mole . hr
(LS) actual = 1.5 (LS) min (LS) actual = 1.5 × 52.62 = 78.93
k mole hr
LS actual = ( Y1 − Y2 ) = 78.93 = ( 0.0526 − 0.00526 ) ( X1 − X2 ) 600 ( X1 − 0) GS 217
X1 = 0.36 (which is same same as from graph) x1 =
X1 (1 + X1)
=
0.36 (1 + 0.36)
= 0.265
The number of stages by stepwise construction is 6. 7)
It is is desir desired ed to to absor absorb b 95% 95% of aceto acetone ne from from feed feed mix mixtur turee of acet acetone one and air air containing 2 (mole) % of acetone using a liquid flow rate of 20 % more than the kg . The gas mixture enters at 25ºC and 1 atm hr
minimum. Gas flow rate is 450
pressure, which is the operating operatin g condition. cond ition. Equilibrium Equilibr ium relation is y* = 2.5x. Find i) Flow rate of water ii) Number of theoretical plates iii) The operation is carried out counter currently.
Solution:
y Np+1 = 0.02
y Np + 1 (1 − y Np + 1)
Y Np+1 =
= 0.0204
Y1 = 0.0204 × 0.05 = 0.00102, y1 = 0.00102 (LS) actual = 20% more than (LS) min Gas flow rate (entering) = 450 T = 2 5 ° C,
hr
Pressure = 1 atm,
Average molecular weight = G Np+1 =
kg
450 29.42
= 15.296
y* = 2.5 x
( 0.02 × 58) + ( 0.98 × 28.84) 1
= 29.42
k mole hr
k mole GS = G Np+1(1 - y Np+1) = 15.296 (1 – 0.02) = 14.99 hr
LS min , GS
For
Y (1 + Y ) Y=
= 2.5
Equilibrium relation is, y* = 2.5x
X (1 + X )
2.5X (1 − 1.5X ) X Y
0.0 0.0
0.002 0.005
0.004 0.01
0.006 0.0151
0.008 0.0202
0.01 0.0254
218
X Np =
x Np (1 − x Np )
=
0.008 = 0.00806 (1 − 0.008)
LS min = ( Y Np + 1 − Y1) = ( 0.0204 − 0.001) ( X Np − X0 ) ( 0.0086 − 0) GS (Assuming pure water enters, so X0 = 0) = 2.4069
LS actual = LS min × (1 + 0.20) = 2.4069 ×1.2 = 2.888 GS GS LS actual = ( Y Np + 1 − Y1) = ( 0.0204 − 0.001) ( X Np − X0 ) ( X Np − 0) GS X Np = 0.00672
LS actual GS
= 2.888
k mole (LS) actual = 2.888 × 14.8205 = 42.802 hr 0.00672 = 0.667 ii) X Np = 0.00672, x Np = (1 + 0.00672 ) (LS) = L Np (1 - x Np) 42.802 (1 − 0.00667 )
L Np =
= 43.089
k mole hr
GS = G1 (1- y1) G1 =
14.8205 (1 - 0.01)
= 14.835
k mole hr
(Assuming pure water enters, so L0 = LS)
N p =
8)
A1 =
L0 42.802 = mG1 ( 2.5 ×14.835)
A2 =
L Np 43.089 = mG Np + 1 ( 2.5 ×15.213)
A=
A1 A2 =
= 1.154 = 1.133
(1.154 × 1.133) = 1.143
( y Np + 1 − mx0 ) 1 1 1 1 × 1 − + log ( 0.02 − ( 2.5 × 0) ) × 1 − + 1.143 1.143 = 9.1 ≈ 9 ( y1 − mx1) A A = ( 0.001 − 0)
log
logA
log1.143
A soluble gas is absorbed in water using a packed tower. The equilibrium relation is Ye = 0.06 Xe. Hx = 0.24 m, Hy = 0.36 m Find HtoG Solution:
Given
219
X2 = 0, X1 = 0.08, Y2 = 0.0005, Y1 = 0.009, where X and Y are mole ratios. Ye =0.06Xe, X X (1 + X ) Y = 0.06X Y y= (1 + Y ) y m= x x=
0 0
0.02 0.0196
0.04 0.038
0 .0 6 0 .0 5 7
0.08 0.074
0 0
0.0012 0.0012
0.0024 0.0024
0.0036 0.0036
0.0048 0.0048
---
0.0612
0.0632
0.0632
0.0649
Fig. 8.21 Example 8
Average ‘m’ = 0.063
mG Hx L
HtoG = Hy +
Hx = 0.24 m, Hy = 0.36 m, X1 = 0.08, Y1 = 0.10, X2 = 0, Y2 = 0.005
LS = ( Y1 − Y2 ) = ( 0.1 − 0.005) = 1.1875 ( 0.08 − 0) GS ( X1 - X2 ) mG Hx = 0.36 + 0.063×1 × 0.24 = 0.3727m L 1.1875
HtoG = Hy +
(Since absolute flow rates are not available, we have taken the flow rates on solute free basis)
220
9)
Acetone Acetone is to be recovere recovered d from from a 5% 5% acetone acetone air mixture mixture by scrubbin scrubbing g with with water in a packed tower using counter current flow. Both liquid and gas rates are
0.85, 0.5
k mole kN k mole -4 partial pressure difference and the . K a =1.5×10 G sec m 2 m 2 m 2sec
gas film resistance controls the process. What should be the height of tower to remove 98 % acetone? Equilibrium data in mole fractions are, as follows: x y
0.0099 0.0076
0.0196 0.0156
0.036 0.0306
0.04 0 .0 3 3 3
Solution:
y1 = 0.05; Y1 =
y1 (1 − y1)
L2 = 0.85
=
0.05 (1 - 0.05)
k mole , m 2sec
= 0.05263
K Ga = 1.5 × 10- 4
k mole kN sec m 2 m 2
kg sec m 2
Gas flow rate = 0.5
Y2 = 0.05263 × 0.02 = 0.001053 y2 =
0.001053 (1 + 0.001053)
= 0.00105
x y m= X=
Y=
‘m’average =
y x x
0.0099 0.0076 0.7677
0.0196 0.0156 0.7959
0.036 0.0306 0.85
0.04 0 .0 3 3 3 0.8325
0.01
0 .0 2
0.037
0.042
0.0077
0.0158
0.0316
0.0344
(1 − x ) y
(1 − y )
0.7677 + 0.7959 + 0.85 + 0.8325 4
= 0.8115
Hence the equilibrium relation will be y* = 0.8115x Average molecular weight = G1 =
( 0.05 × 58) + ( 0.95 × 28.84) 1
= 30.298
k mole 0.5 = 0.0165 sec m 2 30.298
GS = G1 (1- y1) = 0.0165(1-0.05) = 0.0157
k mole sec m 2
221
G2 = Gs(1+Y2) = 0.0157 × (1+0.001053) = 0.01572
Assuming pure water enters, so L2 = LS =
k mole sec m 2
k mole 0.85 = 0.0472 sec m 2 18
LS = ( Y1 − Y2 ) = 0.0472 = ( 0.05263 − 0.001053) ; ( X1 − 0) GS ( X1 - X2 ) 0.0157 x1 =
X1 (1 + X1)
=
0.01716 (1 + 0.01716)
= 0.01687
y1* = mx1 = 0. 0.8115 115 × 0. 0.01687 1687 = 0. 0.01369 1369,,
( y − y *) lm =
NTU =
y2* = 0 (since x2=0)
[ ( y1 − y1 *) − ( y2 − y2 *) ] [ ( 0.05 − 0.01369 ) − ( 0.00105 − 0) ] = = 0.00995 ( 0.05 − 0.01369 ) ( y1 − y1 *) ln ln ( 0.00105 - 0) ( y2 − y2 *)
( y1 − y2 ) ( 0.05 − 0.00105) = = 4.92 ( y − y *) lm 0.00995 k mole k mole ; G = 0.01572 2 sec m 2 sec m 2
G1 = 0.0165
G average = G = 0.01611 HTU =
X1 = 0.01716
G K Ga
=
k mole sec m 2
0.01611 [(1.5 ×10 −4 ) × (1.013×10 2 ) ]
= 1.06
Z = NTU × HTU = 4.92 × 1.06 = 5.216 m. (Alternative method)
We can also draw the equilibrium curve and operating line (on mole ratio basis) and evalu evaluate ate the the
dY between the limits Y1= 0.001 and Y2 = 0.052 0.0525 5 ( Y − Y *)
∫
graphically. The values of Y and Y* have been presented below.
Y*
Y
1
( Y − Y *) 0.000 0.001 0.0025 0.005 0.0075 0.01 0.01125
0.001 0.005 0 .0 1 0.0175 0 .0 2 7 5 0.03625 0.04125
1 00 0 2 50 133.3 80 50 38.1 3 3 .3 3
222
0.0125 0.014
0 .0 4 6 2 5 0.0525
29.6 25.97
The NOG thus calculated is 4.9, which is in close agreement with the value reported above.
Fig. 8.22(a) Example 9
223
Fig. 8.22 (b) Example 9
10)
A counter counter current current packed packed absorpti absorption on tower tower is to be desig designed ned to to handle handle a gas gas containing 5 % C6H6, 95 % air at 26.5ºC and 1 atm. At the top of the tower a non – volatile oil is to be introduced containing 0.2 % C6H6 by weight. The other data are as follows; LS = 2000
kg hr
Molecular weight of oil = 230 Vapor pressure of C6H6 at 26.5ºC = 106 mm Hg. m3 Volumetric flow rate of inlet gas = 1140 at 26.5ºC and 1 atm. hr k moles K ya = 34.8 (mole fraction). hr m 3 Mass velocity of entering gas = 1460
kg . hrm 2
Calculate height and diameter of packed tower for 90 % C6H6 recovery. Raoult’s law is valid.
224
Solution:
Data m3 Gas flow rate = 1140 at 26. 26.5°C, 5°C, 1 atm atm hr
y1 = 0.05,
Pressure (pt) = 1 atm Liquid flow rate = 2000
kg hr
Molecular weight of oil = 230 Vapour pressure of C6H6 = 106 mm Hg k moles mole fraction hr m 2
K ya = 34.8
Mass velocity of inert gas = 1460
kg hrm 2
y2 = 0.05 × 0.1 = 0.005 According to Raoult’s law: p = pA xA Pt × y = pA× x y=
p Pt
=
( pAxA ) Pt
M average =
=
106 x = 0.1395x 706
( 0.05 × 78) + ( 0.95 × 28.84) 1
= 31.3
Mass velocity of incoming gas in moles = Y1 =
y1 (1 − y1)
=
0.05 (1 - 0.05)
k moles 1460 = 46.645 hr m 2 31.3
= 0.0526
Y2 = (0.1× 0.0526) = 0.00526 y2 =
Y2 1 + Y2
x2 = 0.002,
=
0.00526 (1 + 0.00526) X2 =
x2 (1 − x1)
Mass velocity of gas =
= 0.00523 =
0.002 (1 - 0.002)
= 0.00204
k moles 1460 = 46.645 hr m 2 31.3
m3 Volumetric flow rate of incoming gas = 1140 at 26.5°C and 1 atm hr
225
Assume that mixture follows ideal gas law, P1V1 P2V2 = T1 T2
=
(1×1140) (1× V2 ) = ( 299.5) 273
m3 V2 = 1039.132 . hr Molar flow rate =
1039.132 22.414
= 46.361
k moles hr
k moles . hr
G1 = 46.361
We know that y = 0.1395 x Y (1 + Y ) Y=
= 0.1395
X (1 + X )
0.1395X (1 + 0.8605X ) x x
X=
0.05 0.0526
0.1 0.111
0.15 0.176
0.2 0.25
0.25 0.333
0.0066
0.0128
0 .0 1 8 5
0.0238
0.0287
(1 − x )
Y
9 Area of cross section = πD 2 4 LS =
=
46.361 46.645
2000
D = 1.1249 m
8.696
=
230
volumetric flow rate mass velocity
k mole hr
GS = G1 (1- y1) = 46.361(1-0.05) = 44.043 GS G2 = 1 − y2
=
44.043 (1 - 0.00523)
= 44.275
k moles hr
k moles hr
LS = ( Y1 − Y2 ) = 8.696 = ( 0.0526 − 0.00526 ) ( X1 − 0.00204 ) GS ( X1 - X2 ) 44.043 Therefore, X1 = 0.242 x1 =
X1 ( 1 + X1 )
=
0.242 (1 + 0.242)
= 0.1948
y1* = mx 1 =0.1395 × 0.1948 = 0.0272 y2* = 0.1395 × 0.002 = 0.000279 226
[ ( y1 − y1 *) − ( y2 − y2 *) ] [ ( 0.05 − 0.0272) − ( 0.00523 − 0.00279) ] = = 0.01133 ( 0.05 − 0.00272) ( y1 − y1 *) ln ln ( 0.005 - 0.00279 ) ( y2 − y2 *)
( y − y *) lm =
NTU =
Gaverage
( y1 − y2 ) ( 0.05 − 0.005) = = 3.95 ≈ 4 ( y − y *) lm 0.01133
k moles G1 + G2 = = 45.318 hr 2
πD 2 Cross sectional area = 4 Diameter = 1.1249 m πD 2 Cross sectional area = 4 HTU =
G K ya
=
π × (1.1249 ) 4
=
45.318 [ 0.9938 × 34.8]
2
= 0.9938 m 2
= 1.31 m
Z = NTU × HTU = 4 × 1.31 = 5.14 m. 11)
It is desired to recover 98 % of NH3 from air – NH3 mixture containing 2% NH3 at 20ºC and 1 atm by scrubbing with water in a tower packed with 2.54 cm
stoneware Raschig rings. If the gas flow rate is 19.5
Kg at the inlet and min m 2
liquid flow rate is 1.8 times the minimum, estimate the height of the tower for a counter current operation. Absorption is isothermal. y* = 0.746x, where x and y
are mole fractions. K Ga = 1.04
k moles . min m 2atm
Solution:
y1 = 0.02, Y1 =
y1 (1 − y1)
=
0.02 (1 - 0.02)
= 0.02041
Y1 = 0.02041 × 0.02 = 0.00041 y2 =
Y2 1 + Y2
=
0.00041 = 0.00041 1.00041
Gas flow rate = 19.5
Kg min m 2
(LS) actual = 1.8 × (LS) min,
227
Equilibrium relation = y* = 0.746x, Y (1 + Y )
= 0.746
Therefore, Y =
X (1 + X ) 0.746X (1 + 0.254X )
X
0 0
0.010 0 .0 0 7 4 4
0.020 0.01484
0.025 0 .0 1 8 5
0.03 0.0222
0.746X (1 + 0.254X ) For minimum liquid flow rate, we can calculate using the equilibrium relationship Y=
or from the Graph x1 =
y1* = mx1, X1 =
0.0242 0.746
x1 0.0268 = 1 − x1 (1 − 0.0268)
= 0.0268
= 0.0275
Y1 =
y1 (1 − y1)
=
0.02 (1 - 0.02)
Y2 =
y2 (1 − y2 )
=
0.0004 (1 - 0.0004)
= 0.02041 = 0.0004002
From Graph also we get, X 1 = 0.0275
LS min = ( Y1 − Y2 ) = ( 0.02041 − 0.00041) = 0.7273 ( 0.0275 − 0) GS ( X1 − X2 ) LS actual GS
= 0.7273 × 1.8 = 1.309
LS actual = ( Y1 − Y2 ) = 1.309 = ( 0.02041− 0.00041) = X1 = 0.01528 ( X1 − X2 ) ( X1 − 0) GS x1 =
X1 (1 + X1)
=
0.01528 (1 + 0.01528)
= 0.01505
y1* = 0.746x1 = 0.746 × 0.01505 = 0.01123,
( y − y *) lm =
NTU =
y2* = 0
[ ( y1 − y1 *) − ( y2 − y2 *) ] [ ( 0.02 − 0.01123) − ( 0.00041 − 0) ] = = 0.00273 ( 0.02 − 0.01123) ( y1 − y1 *) ln ln ( 0.00041 - 0) ( y2 − y2 *)
( y1 − y2 ) ( 0.02 − 0.00041) = = 7.176 ( y − y *) lm 0.00273
Average Molecular weight of incoming gas =
( 0.02 ×17 ) + ( 0.98 × 28.84) 1
= 28.6
228
G1 = 19.5
Kg k moles 2 = 19.5/28.6 = 0.682 min m min m 2
Gs = G1(1-y1) =0.682 ×(1-0.02) = 0.6684
G2 =
GS 1 − y2
Gaverage
=
0.6684 (1 − 0.00041)
= 0.6687
k moles min m 2
k moles min m 2
k moles G1 + G2 = = 0.6754 min m 2 2 HTU =
G K Ga
=
0.6754 1.04
= 0.649 m
Z = NTU × HTU = 7.176× 0.649 = 4.657 m.
Fig. 8.23 Example 11
12)
CS2 – N2 mixture containing 7 % CS2 is to be absorbed by using absorption oil. m3 The gas mixture enters at 24ºC and I atm at a rate of 0.4 . The vapor content is sec to be brought down to 0.5 %. The oil enters free from CS 2. Raoults law is valid Determine (i) Minimum liquid / gas ratio. (ii)) For a liquid / gas ratio of 1.5 times the minimum determine the Kgs of oil entering the tower. The number of theoretical states required. Vapor pressure of CS2 = 346 mm Hg,
Molecular weight of oil = 180.
Solution:
Average molecular weight = (0.07×32) + (0.93×28) = 28.28
229
Gas flow rate = 0.4 m 3 P0V0 T0
=
P1V1 T1
V0 =
V1 0.4 m3 × T0 = × 273 = 0.3677 T1 0.297 sec
G1 =
0.3677 22.414
G1 = 59.06
=
k mole sec
k moles hr k moles hr
GS = 59.06(1 – 0.07) = 54.93 y = mx y=
346 760
Y (1 + Y )
= 0.455
(1 + Y )
∴
=
0.455X
1 + X − 0.455X 0.455X
Y=
X Y
X (1 + X )
(1 + X )
=
Y 1 Y
x = 0.455 x
=
(1 + 0.545X ) 0.455X
0.455X
(1 + 0.545X )
0 0
0.05 0.1 0.02 0.04 2 3 X1, max = 0.1775
0.15 0.063 1
0.2 0.082
LS min = ( Y1 − Y2 ) = ( 0.0753 − 0.005) = 0.396 ( 0.1775 − 0) GS ( X1, max − X2 ) LS actual GS ∴
LS = (0.594×54.93) = 32.63
0.594 = ∴
= (1.5×0.395) = 0.594
( 0.0753 − 0.005) ( X1, Act − X2 ) ( X1, Act - 0) ( 0.0753 − 0.005)
X1, Act =
( Y 1 − Y2 )
k moles kg = 32.63 × 180 = 5873.4 hr hr
=
=
0.594
0.1184
230
Number of theoretical stages: 5
Fig. 8.24 Example 12
1 3)
NH3 is absorbed from a gas by water in a scrubber under atmospheric pressure.
The initial NH3 content in the gas is 0.04
k mole . The recovery of k mole of inert gas
NH3 by absorption is 90 %. The water enters the tower free from NH3. Estimate (i) concentration of NH3 in the exiting liquid if the actual water used is 1.5 times minimum. (ii) Number of theoretical stages required. X
0.005
0 .0 1
0.0125
0.015
0.02
0.023
Y
0.0045
0.0102
0.0138
0.0183
0.0273
0.0 3 2 7
Where x and y are mole ratios. Solution:
LS = ( Y1 − Y2 ) GS ( X1, max - X2 ) LS min = ( Y1 − Y2 ) = ( 0.04 − 0.004 ) GS ( X1, max − X2 ) ( X1, max − 0) X1, max (from graph) = 0.027
LS min = 0.036 = 1.333 GS 0.027 LS actual GS
= 1.5 ×
LS min = 2 GS 231
2=
( 0.04 − 0.004) ( X1, act − 0)
X1, act =
( 0.04 − 0.004 ) 2
=
0.036 2
= 0.018
Fig. 8.25 Example 13
Concentration of Ammonia in exiting liquid: 0.018
k mole k mole of water
Number of theoretical stages required: 3 14)
Gas from petroleu petroleum m refin refinery ery has its concent concentrati ration on of of H2S reduced from 0.03 k mole H2S to 1 % of these these value by scrubbin scrubbing g with a solvent solvent in a counter counter k mole inert gas
232
current tower at 27ºC and 1 atm. Equilibrium relation is Y* = 2 X. where X and Y* are in mole ratios. ratios. Solvent Solvent enters free free of H2S and leaves at a concentration concentration of
0.013
k mole H2S k moles . If the the flow flow rate rate of inco incom ming ing gas gas is 55.6 55.6 , k mole of solvent hr m 2
Calculate the Height of absorber used if the entire resistance to mass transfer lies k moles . m 3 of tower volume × s × ΔY
in gas phase.. Take K ya = 0.04 Solution:
L2, x2, LS, X2
G2, y2, GS, Y2
(2)
(1) L1, x1, LS, X1
G1, y1, GS, Y1
Fig. 8.26 Example 14
X1 = 0.013;
X2 = 0;
Y1 = 0.03;
Y2 = 0.0003
Y 1* = 2
× 0.013 =0.026
y1 =
Y1 0.03 = = 0.029 1 + Y1 1.03
y2=
Y2 1 + Y2
=
Y2* = 0
0.0003 = 0.0003 1.0003
Inert gas flow rate = Gs = G 1(1-y1) = 55.6 × 0.971 = 54
G2 = Gs(1+Y2) = 54
G = (54.016
× 1.0003 = 54.016
× 55.6)0.5 = 54.6
k moles hr m 2
k moles hr m 2
k moles hr m 2
233
(Y1 − Y2 )
NTU =
=
(Y1 − Y2 ) ( ΔY ) lm
(Y1 − Y1*) − (Y2 − Y2 *) − (Y Y *) 1 1 ln − (Y2 Y2 *) (0.03 − 0.026) − (0.0003 − 0.0) ( ΔY ) lm = (0.03 − 0.026) =1.428 × 10-3 ln (0.0003 − 0.0) NTU =
HTU =
(0.03 − 0.0003) =20.79 (1.428 × 10 -3 )
G K y a
=
(54.6) = 0.379m (0.04 × 3600)
Height of tower = HTU
× NTU = 0.379 × 20.79 = 7.879 m
Exercise: 1)
An air - NH3 mixture containing containing 20% (mole) NH3 is being treated with water in a
packed tower to recover re cover NH3. Incoming gas rate = 1000
kg . Water used is 1.5 hrm 2
times minimum. The temperature is 35ºC and pressure at 1 atm. Equilibrium relation is y* = 0.746x. Where (x and y are mole fraction units). Find NTU for removing 95% NH3 in the feed. 2)
An air – SO2 mixture containing 5% SO2 is scrubbed with water to remove SO2 in
a packed tower. 20
k moles of gas mixture is to be processed, to reduce SO2 s
conc concen entr trat atio ion n at exit exit to 0.15 0.15%. %. If Ls actual ual is 2 Ls min, min, and equilib equilibrium rium relationship is y = 30x. HTU = 30 cms. Find height of packing to be used. 3)
It is desired to absorb 95% NH3 from a feed mixture containing 10% NH3 & rest
air. The gas enters the tower at a rate of 500
k moles . If water is used as solvent hr
at a rate of 1.5 times min, estimate (i) NTU, (ii) (L s) act. 4)
An air –SO2 mixture containing containing 5.5% SO2 is scrubbed with water to remove SO2.
500
kg of gas mixture is to be processed and the SO2 content in the exit should hr
be brought broug ht to 0.15 % Calculate Calcu late the height of packing required if the liquid used us ed is
234
2.5 times the minimum liquid rate. Dilute solution are involved in operation the equilibrium equilibrium lines given by y = 30x.where 30x.where x and y are mole fractions. The HTU is 30 cm. 5)
An air – NH3 mixture containing 5 % NH3 enters a packed tower at the rate of 500 k moles . It is desired to recover 95 % NH 3 using a liquid flow rate of 1.5 hr m 2 (minimum). Estimate the height of tower. HTU is 0.25 m. Fresh solvent enters the absorber. Equilibrium relation is y* = 1.08 x where x and y are mole fractions.
6)
A pa packed to tower is is to to be be de designed to to ab absorb SO SO2 from air by scrubbing the gas with water. The entering gas contains 20 % SO2 by volume and the leaving gas is to contain 0.5 % SO2 by volume. The entering water is SO2 free. The water flow is
twice the minimum. The airflow rate on SO 2 free basis is 975
kg . The hrm 2
temperature is 30ºC and 1 atm. y* = 21.8 x, where x and y are mole fractions. Find NTU 7)
NH3 is to be absorbed from air at 20ºC and 1 atm pressure in a packed tower using
water as absorbent. GS = 1500
0.003. K ya = 0.3
k moles . Determine the height of tower by N toG method. hr m 2 ( Δy )
X 0.0164 0.0252 Y 0.021 0.032 X and Y are mole ratios. 8)
k moles k moles , L = 2000 .y1 = 0.0825; y2 = S hr m 2 hr m 2
0 .0 3 5 9 0.042
0.0455 0.053
0.072 0.08
An air – NH3 mixture containing 20mole % NH3 is being treated with water in a
packed tower to recover NH3. The incoming gas rate is 700
kg . The water hrm 2
used is 1.5 times the minimum and enters the tower free of NH 3. Under these conditions 95 % NH3 is absorbed from the incoming feed. If all the operating conditions remain unchanged, how much taller the tower should be to absorb 99 % NH 3, under the given conditions y* = 0.75 x where x and y are mole fractions of NH3 in liquid and gas phase respectively.
235
9)
A packe packed d tower tower is to be designed designed to recover recover 98% Carbon Carbon dioxide dioxide from a gas gas mixture containing 10% Carbon dioxide and 90% air using water. The equilibrium
relationship is Y =14X, where y =
kg CO2 kg CO2 and X = . The kg dry air kg dry water
water to gas rate is kept 30% more more than the minimum value. value. Calculate Calculate the height of the tower if (HTU) OG is 1 meter. meter. (Ans:11.42 m) 10)
An air – NH3 mixture containing 6 % NH3 is being scrubbed with water to recover
90 % NH3. The mass velocities of gas and water are 3200
kg and 2700 hrm 2
kg . Operating conditions are 25ºC and 1 atm. Find NTU and height of tower. hrm 2
Given that, K Ga = 65
11)
k moles ,y* = 0.987 x. where x and y are mole fractions. hr m 3atm
m3 500 of a gas at 760 mm Hg and 35ºC containing 3 % by volume of toluene is hr absorbed using a wash oil as an absorbent to remove 95 % of toluene. The wash oil enters at 35ºC contains 0.5 % toluene and has an average molecular weight of oil 250. The oil rate used is 1.5 times the minimum. Wash oil is assumed to be ideal. Vapor pressure of toluene is 110 mm Hg. Find amount of wash oil used and number of theoretical stages.
12) 12)
Ammonia is rec reco over vered fr from a 10 % NH NH3 - air mixture by scrubbing with water in a packed tower at 20ºC and 1 atm. pressure such that 99 % of the NH 3 is removed. What is the required height of tower? Gas and water enter at the rate of 1.2 kg kg and 0.94 sec m 2 sec m 2
respectively. Take K Ga = 0.0008 kgmole/m3s. atm.
Equilibrium data as follows: x
0.021
0.031
0.042
0.053
0 .0 7 9
0.106
0.159
p (mmHg)
12
18.2
24.9
31.7
50
69.6
114
where ‘x’ is mole fraction of NH3 in liquid ‘p’ is partial pressure of NH3. 13)
An air air - acetone acetone mixture mixture,, contain containing ing 5% acetone acetone by by volume volume,, is to be scrubb scrubbed ed with water in a packed tower to recover 95% of the acetone. Airflow rate is 1400
236
kg m3 at 20ºC and 1 atmosphere. The water rate is 3000 .The equilibrium hr hr relation is Ye = 1.68X, where Ye and X are mole fractions of acetone in vapour and liquid respectively. The flooding velocity is 1.56 meter per second and the operating velocity is 25% of the flooding velocity. The interfacial area of the packing is 204 2 04
m2 m3
of packing and the overall mass transfer coefficient Ky is 0.40
k moles . Estimate the diameter and packed height of the tower hr m 2 mole fraction operating at 1 atmosphere. 14)
CO2 evolved during the production of ethanol by fermentation contains 1 mole ratio of alcohol. It is proposed to remove alcohol by absorption in water at 40ºC.
The water contains 0.0001-mole ratio of alcohol. 500
moles of gas is to be hr
processed. Equilibrium data: y = 1.05 x, where x and y are mole fractions. Calculate Calculate the water rate for 98 % absorption using 1.5 times minimum liquid rate and determine the number of plates. 15)
A gas str strea eam m conta contain ining ing a valu valuabl ablee hydroc hydrocar arbon bon (mo (mole lecu cular lar wei weight ght = 44) 44) and and air is to be scrubbed with a non – volatile oil (molecular weight = 300) in a tower placed with 2.54 cm Raschig rings. The entering gas analyses 10 mole % Hydrocarbon and 95 % of this Hydrocarbon is to be recovered. The gas stream
enters the bottom of the column at 2270
kg and the hydrocarbon free oil used is hr
1.5 times the minimum. Find N toG for this operation. operation. The equilibrium equilibrium data data is as follows: X Y
0 0
0.1 0.01
0.2 0 .0 2
0.3 0.06
0.4 0.118
0.458 0.2
where X and Y are mole mole ratios. (ii) If the flow rate of liquid is 4600
kg hr
estimate the number of transfer units needed and the solute concentration in mole fraction in leaving liquid? (Ans : (ii) 4, 0.322)
237
16)
A soluble gas is absorbed in water using a packed tower. The equilibrium relationship may be taken as y = 0.06x Terminal conditions Top Bottom x 0 0.08 y 0.001 0 .0 0 9 (x,y : Mole fraction of solute in liquid and vapor phase respectively) If the individual height of transfer units based on liquid and gas phase respectively are Hx = 0.24 m and Hy = 0.36 m,(i) What is the value of (HTU) OG and (ii) what is the height of packed section? (Ans: (i) 0.511 m and (ii) 1.833 m)
17)
An air- NH3 mixture containing 20-mole % NH3 is being treated with water in a
packed tower to recover NH3. The The incom incomin ing g gas rate is 1000 1000
kg . The hrm 2
temperature is 35ºC and the total pressure is 1 atm. The water flow rate is 3000 kg . 95% of incoming NH3 is to be absorbed. If all the operating conditions hrm 2 remain unchanged, how much taller should the tower be to absorb 99% of NH3? Henry’s law is valid and Henry’s constant is 0.746. Variations in gas flow rates may be neglected. (Ans : 58.15%)
238
