Enthalpy-concentration Enthalpy-concentration Diagram
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McCabe Thiele method assumes constant molar flow rate because it considers equal latent heat of vaporization.
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Here we consider varying molar flow rate by solving simultaneous material and energy balances.
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In this case, the operating lines for the enriching and stripping section will be determined from simultaneous solution of mass and energy balance equations.
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To facilitate the solution of the heat balance equation, an enthalpy diagram can be constructed and used.
Enthalpy diagram diagram
The liquid enthalpy 1
h = x A Cp A (T − T o ) + (1 − x A )Cp B (T − T o ) + ΔH sol
The vapor enthalpy 2
H = y A [λ A + Cv A (T − To )] + (1 − y A )[λ B + CvB (T − T o )]
Correction for latent heat
λ A = Cp A (T bA − T o ) + λ bA − Cv A (T bA − T o )
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λ B = Cp B (T bB − T o ) + λ bB − Cv B (T bB − T o )
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Where T T bA T bB To λ bA
is the boiling point for the mixture The boiling point of pure A The boiling point of pure B reference temperature latent heat of pure A at T bA bA
Example: Create the enthalpy diagram for Benzene-Toluene mixture
Benzene Toluene
T b, C
Cp, kJ/Kgmole K
Cv, kJ/kgmole K
80.1 110.6
138.2 167.5
96.3 138.2
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λ b kJ/Kgmole 30820 33330
Let the reference temperature be 80.1 oC, thus the latent heat of A (Benzene) does not need to be corrected. For component B:
λ B = 167.5(110.6 − 80.1) + 33330 − 138.2(110.6 − 80.1) = 34224 kJ/kgmole Create the liquid enthalpy line: x A = 0,
T = 110.6
h = 0 + (1 − 0) (176.5)(110.6 − 80.1) = 5109 kJ/kg mole x A = 0.3,
equilibrium diagram
T = 98 oC
h = 0.3(138.2)(98 ─ 80.1) + (1 − 0.3) (176.5)(98 − 80.1) = 2920 kJ/kg mole x A = 0.5,
equilibrium diagram
T = 92 oC
h = 0.5(138.2)(92 − 80.1) + (1 − 0.5) (176.5)(92 − 80.1) = 1820 kJ/kg mole x A = 0.8,
equilibrium diagram
T = 84 oC
h = 0.8(138.2)(84 − 80.1) + (1 − 0.8) (176.5)(84 − 80.1) = 562 kJ/kg mole x A = 1.0,
equilibrium diagram
T = 80.1 oC
h = 1.0(138.2)(80.1 − 80.1) + 0 = 0 kJ/kg mole
Create the vapor enthalpy line y A = 0,
T = 110.6
H = 0 + (1 − 0) [33330 + 138.2(110.6 − 80.1)] = 38439 kJ/kg mole y A = 0.3,
T=
H = 0.3[30820 + 96.3( T – 80.1)] + (1 − 0.3) [34224 + 138.2(T − 80.1)] = 36268 kJ/kg mole
y A = 0.5,
T = 98.8
H = 0.5[30820 + 96.3( 98.8 – 80.1)] + (1 − 0.5) [34224 + 138.2(98.8 − 80.1)] = 34716 kJ/kg mole
y A = 0.8,
T=
H = 0.8[30820 + 96.3( T – 80.1)] + (1 − 0.8) [34224 + 138.2(T − 80.1)] = 32380 kJ/kg mole
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y A = 1.0,
T = 80.1
H = 1.0[30820 + 96.3( 80.1 – 80.1)] + 0 = 30820 kJ/kg mole 40,000
y 30,000 p a h t 20,000 n E 10,000 0 0
0.2
0.4
0.6
0.8
1.0
x Column Design
V1 H1 y1
Enriching Section 1
The Mass balance: V n +1 = Ln + D
5 2
The component balance y n +1 =
Ln V n +1
x n +
6
Dx D
V2 y2
L1 x1
V3 y3
L2 x2
Vn+1 Hn+1 yn+1
Ln hn xn
n
V n +1 F zf
The enthalpy balance: 7
V n +1 H n +1 = Ln hn + Dh D + Qc
To eliminate the condenser duty is by heat balance around the condenser: 8
Qc = V 1 H 1 − ( L + D) h D
Substituting for Qc in equation 7 gives: 9
V n +1 H n +1 = Ln hn + V 1 H 1 − Lh D
Inserting the mass balance equation 5 into equation 9: 10
V n +1 H n+1 = (V n +1 − D) hn + V 1 H 1 − Lh D
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Qc
L hD xD
D hD xD
Stripping section The Mass balance:
F zf
Lm = V m+1 + B
11 m
The component balance ym +1 =
Lm V m +1
xm −
Lm hm ym
Vm+1 Hm+1 ym+1
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Bx B
m+1
V m +1
The enthalpy balance: V m +1 H m +1 = Lm hm − Bh B + Qr
Vm ym
N
Lm+1 xm+1
V N y N
L N x N
13
Substituting the mass balance equation 11 into equation 13: V m+1 H m+1 = (V m+1 + B) hm − Bh B + Qr
L N x N
y N H N Qr
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Overall heat balance: 15
Qr = Dh D + Bh B + Qc − Fh f
Substituting for Qr into the heat balance equation 14: 16
V m +1 H m +1 = (V m +1 + B ) hm + Dh D + Qc − Fh f
The feed condition The q line remain the same
Solution procedure 1. starting from x D draw the enriching operating line: a. select a value for x between x D and x f b. assume a constant molar flow rate; Ln= L, Vn+1 = V c. solve the component balance equation for yn+1 d. solve the heat balance for Vn+1 e. solve the mass balance for Ln f. resolve the component balance for yn+1 g. if new value of yn+1 equals the old value go step h else go back to step c h. repeat steps a-g for another value for x. 2. do the same for the stripping section starting from x B
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B hB xB
Example: Consider the separation of Benzene-Tolune mixture of example 11.4.1:
F = 100 kgmole/h R = 1.755; Tf = 54.5 oC
x f = 0.45 x D = 0.95; D = 41.2 kgmole/h; B = 58.8 kgmole/h
x B = 0.1
q = 1.195
The stripping section: L/D = 1.755 L = 1.755D = 72.3 kgmole/h V1 = L + D = 72.3 + 41.2 = 113.5 kgmolr/h Knowing x D and y1= x D using the boiling diagram gives T = 82.3 oC using the enthalpy diagram gives: H 1 = 31206 kJ/kg mole h D = 139 kJ/kg mole
Let x = 0.55 from component balance
y n +1 =
using the enthalpy chart:
72.3 113.5
0.55 +
41.2 113.5
0.95 = 0.695
at xn = 0.55 hn = 1590 at yn+1 = 0..695
H n+1
= 33240
from the heat balance equation: V n +1 33240 = (V n+1 − 41.2)1590 + 113.5(31206) − 72.3(139) V n+1 = 109.5
From the mass balance equation:
109.5 = Ln + 41. 2
Resolving the component balance equation: y n +1 =
68.3 109.5
Ln = 68.3
0.55 +
41.2 109.5
0.95 = 0.7
Since 0.7 is very close to 0.695 we stop the iteration and go for another value of x. before proceeding to stripping section: Qc = 113.5(31206) − 113.5(139) = 3526kJ / h
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At the feed temperature and composition: Either using the enthalpy chart for x f = 0.45 or enthalpy equation for x f = 0.45 & T f = 54.4 oC we get: h f = -3929 kJ/kgmole
using the enthalpy chart for x B = 0.1 we get h B = 4350 kJ/kgmolr using the heat balance around the re-boiler: Qr = 41.2(139) + 58.8(4350) + 3526 + 3929 = 4,180,500 kJ/h Knowing the feed condition we calculate the molar flow rates in the stripping section: Lm = Ln + qF = 72.3 + 1.195(100) = 191.8 kgmole/h V m+1 = V n+1 - (1−q)F = 113.5 − (1 − 1.195)(100) = 133.0 kgmle/h
Srtipping section:
Let ym+1 = 0.207 Using the component balance: Using the enthalpy chart:
at at
0.207 =
191.8 58.8 x m + 0.1 133 133
xm = 0.174 ym+1 = 0.207
xm = 0.174
hm= 3800 H m+1 = 37000
using the enthalpy balance equation: V m+1 37000 = (V m+1 − 58.8)380 + 58.8(139) + 3526 − 100( −3929)
V m+1 = 125
Using the mass balance:
Lm = Vm+1 + B
Using the component balance:
0.207 =
Lm = 183.8
183.8 58.8 x m + 0.1 125 125
Since the new value of x is close enough to old value
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xm = 0.173
Tray Efficiency
In McCabe Thiele method 1000% phase equiklibruum is assumed which is not totally correct: Overall efficiency Eo =
Theroectical (ideal) numer of stages actual number of stages
Murphee Efficiency E M =
yn − yn+1 yn* − yn+1
Point Efficiency E MP =
yn '− yn+1 ' yn* − yn+1 '
Correlations for the overall Efficiency
Drickamar and Bradford correlation E = 0.17 − 0.616 log(
∑ x
xf μL
μ / μL )
fi Li
feed composition viscosity at average tower temperature 28
μw
viscosity of water at 293 K
CHU correlation
log E = 1.67 + 0.3 log( L.V ) − 0.25 log(μ L α) + 0.3h L L, V liquid and vapor molar flow rates μL viscosity of the liquid feed α relative volatility of the key component hL effective submergence , the distance from the top of the slot to the weir lip plus the slot height.
Flooding velocity and Column Diameter
The maximum allowable vapor velocity in the column is:
vmax = K v (
σ 0.2 ρ L − ρv ) 20 ρv
σ is the surface tension of the liquid K v is obtained from charts ρL, ρv is the liquid and vapor densities
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half
Solution Procedure compute liquid density from appendix compute vapor density from ideal gas law: RT ρv = MP convert vapor molar rate to mass rate compute liquid molar rate from overall molar balance convert liquid molar rate into mass rate use V, L, ρL, ρV to compute K v use σ =70 to compute vmax compute column cross section area from V/vmax compute diameter from cross section area
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