Fractional Distillation using Enthalpy-C PDF

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Fractional Distillation using Enthalpy-Concentration Method  molal flow rates need not to be constant  using enthalpy as well as material balances When the operating lines are plotted with the equilibrium curve on the x-y diagram, the McCabeThiele step-by-step construction can be used to compute the number of ideal plates. However, unless Ln and Lm are constant, the operating lines are curved and can be plotted only if the change with concentration in these internal streams is known. Enthalpy balances are required in the general case to determine the position of a curved operating line. The enthalpy-concentration diagram requires the following data at a constant pressure: (1) heat capacity of the liquid as a function of temperature, composition, and pressure. (2) heat of solution as a function of temperature and composition. (3) latent heats of vaporization as a function of composition and pressure or temperature. (4) boiling point as a function of pressure, composition and temperature. The diagram at a given constant pressure is based on arbitrary reference states of liquid and temperature, which is usually taken as the boiling point of the lowerboiling component A. 85

The saturated liquid enthalpy is H x  x AC PLATb  T0   (1 x A )C PLB Tb  T0   H sol (1) = boiling temperature of mixture at xA Tb T0 = reference temperature CPLA = liquid heat capacity of A = liquid heat capacity of B CPLB Hsol = heat of mixing, usually ignored The saturated vapor enthalpy is

H y  y A H VA  C PVAT d  T0   (1  y A) H VB  C PVBT d T0 

Td T0 CPVA CPVB HVA

(2) = dew temperature of mixture at yA = reference temperature = vapor heat capacity of A = vapor heat capacity of B = latent heat of A at the reference temperature T0 & calculated from that at the normal boiling temperature TbA.

H VA  C PLATbA  T0   HVAb  C PVATbA  T0  (3) HVB  C PLBTbB  T0  HVBb  C PVBTbB  T0  (4) That is, the liquid is first changed from temperature T0 to Tb and then evaporated at Tb , the vapor temperature is finally changed back from Tb to T0.

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Example D6. Enthalpy-concentration plot for benzene-toluene Prepare an enthalpyconcentration plot for benzenetoluene at 1 atm pressure. Equilibrium and physical property data are given.

For the first point, we select pure toluene (xA=0). For liquid toluene at the boiling point of 110.6oC, H x  x A CPLA Tb  T0   (1 x A )CPLB Tb  T0   H sol  0  (1  0)(167.5)(110.6  80.1)  5109 kJ / kmol For the saturated vapor enthalpy of pure toluene, we first need to calculate ΔHVB at T0=80.1oC: HVB  CPLB TbB  T0   HVBb  C PVB TbB  T0   167.5(110.6 80.1) 33330 138.2(110.6  80.1)

Vapor pressure (kPa) T (K) 353.3 358.2 363.2 368.2 373.2 378.2 383.8

Benzene 101.32 116.9 135.5 155.7 179.2 204.2 240.0

Solution. We choose a reference temperature of T0=80.1oC so that the liquid enthalpy of pure benzene (xA=1) at the boiling point is zero.

Toluene 46.0 54.0 63.3 74.3 86.0 101.32

Component

Boiling point (oC)

Benzene (A) Toluene (B)

80.1 110.6

Mole fraction of benzene at 101.325 kPa xA yA 1.000 1.000 0.780 0.900 0.581 0.777 0.411 0.632 0.258 0.456 0.130 0.261 0 0

Heat capacity (kJ/kmol K)

Latent heat of vaporization (kJ/kmol)

Liquid Vapor 138.2 96.3 30820 167.5 138.2 33330 87

 34224 kJ/kmol Hence, at xA = 0,

H y  y A H VA C PVA T d  T0   (1 y A) H VB  C PVBTd T0 

= 0 + (1-0)[34224+138.2(110.6-80.1)] =38439 kJ/kmol For pure benzene, xA = yA = 1, T = T0=80.1oC, Hx = 0,

H y  y A H VA C PVA T d  T0   (1 y A) H VB  C PVBT d T0 

=1[30820+96.3(80.1-80.1)] + 0 = 30820 kJ/kmol Selecting xA = 0.5, the boiling point Tb = 92oC and the dew point for yA = 0.5 is Td = 98.8oC from the T-x-y plot. H x  xA C PLA Tb  T0   (1 x A )C PLB Tb  T0  = 0.5(138.2)(92-80.1)+(1-0.5)(167.5)(92-80.1) = 1820 kJ/kmol H y  y A H VA C PVA T d  T0   (1 y A) H VB  C PVBT d T0 

= 0.5[30820+96.3(98.8-80.1)] +(1-0.5)[34224+138.2(98.8-80.1)] = 34176 kJ/kmol 88

By the similar procedure, at xA = 0.3 and y A = 0.3, Hx = 2920 kJ/kmol and H y = 36268 kJ/kmol at xA = 0.8 and y A = 0.8, Hx = 562 kJ/kmol and H y = 32380 kJ/kmol Enthalpy-concentration data for benzene-toluene mixture at 1 atm.

Saturated Liquid Saturated Vapor Mole Enthalpy, Hx, Mole Enthalpy, Hy, fraction, xA (kJ/kmol) fraction, yA (kJ/kmol) 0 5109 0 38439 0.30 2920 0.30 36268 0.50 1820 0.50 34716 0.80 562 0.80 32380 1.00 0 1.00 30820

Enthalpy-concentration plot for benzene-toluene mixture at 1 atm.

The tie line represents the enthalpies and composition of the liquid and vapor phases in equilibrium. 89

Distillation in Enriching Section of Tower Material balances (5) Vn 1  Ln  D (6) Vn 1 yn 1  Ln x n  Dx D By rearranging Eq. (6) we obtain the operating line L Dx D y n 1  n xn  (7) Vn1 Vn1 Ln and V n+1 may vary throughout the tower so Eq. (7) will not be a straight line on an xy plot.

Making an enthalpy balance, (8) Vn 1H y ,n 1  L nH xn  DH xD  q c where qc is the condenser duty, kJ/h. An enthalpy balance for the condenser only gives (9) qc  V1 Hy1  LHxD  DHxD 90

Substitute Eq. (9) into Eq. (8) we have Vn1H y, n1  LnH xn  V1H y1  LH xD  V n 1  D H xn  V1H y1  LH xD

Distillation in Stripping Section of Tower (10)

Eqs. (7) & (10) are the final working equations for the enriching section. In order to plot the operating line Eq. (7), the terms Vn+1 and Ln must be determined from Eq. (10). If the reflux ratio is set, V1 and L are know. The values of Hy1 and HxD can be determined by Eqs. (1) & (2) or from an enthalpy-concentration plot. If a value of xn is selected, it is a trial-and-error solution to obtain Hy,n+1 since yn+1 is unknown. The steps to follow are given below. 1. Select a value of xn. Assume Vn+1 = V1 = L+D, then Ln = Vn+1 - D. 2. Use Eq. (7) to calculate the approximate value of yn+1. 3. Using this yn+1 to obtain Hy,n+1 and xn to obtain Hxn . Substitute these values into Eq. (10) and solve for the new Vn+1. Obtain new Ln from Eq. (5) 4. Substitute the above values into Eq. (7) to get the new yn+1. 5. If the newly calculated yn+1  the assumed value, repeat steps 3-4. Usually one iteration is enough. Assume another xn and repeat steps 2-5. 6. Plot the curved operating line for the enriching section.

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Material balances Lm  Vm1  B Lm xm  Vm1ym 1  Bx B so that the operating line is L Bx y m1 m xm  B Vm 1 Vm1

(11) (12) (13)

Making an enthalpy balance, V m1H y, m1  Vm1  B H xm  BH xB  qr (14) where qr is the reboiler duty, kJ/h. An enthalpy balance for the whole distiller gives (15) qr  DH xD  BHxB  qc  FH F The final working equations are Eqs. (13) and (15). Using a method similar to that of the enriching section to solve the equations.

Component Benzene Toluene

T b (oC) 80.1 110.6

CP (kJ/(kmol K) Liquid Vapor 138.2 96.3 167.5 138.2

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ΔHvb (kJ/kmol) 30820 33330

Example D7. Distillation Using EnthalpyConcentration Method A liquid mixture of benzene-toluene is to be distilled in a fractionating tower at 101.3 kPa pressure. The feed of 100 kmol/h is liquid, containing 45 mol % benzene and 55 mol % toluene, and enters at 54.4oC. A distillate containing 95 mol % benzene and 5 mol % toluene and a bottoms containing 10 mol % benzene and 90 mol % toluene are to be obtained. The reflux ratio is 1.755. The average heat capacity of the feed is 159 kJ/(kmol K) and the average latent heat is 32099 kJ/kmol. Use enthalpy balances to calculate the flow rates of the liquid and vapor at various points in the tower and plot the curved operating lines. Determine the number of theoretical stages needed.

Solution: F = 100 kg mol/h, xF = 0.45, xD = 0.95, xB = 0.10, R = 1.5Rm = 1.5(1.17) = 1.755, D = 41.2 kg mol/h, B = 58.8 kmol/h The feed enters at 54.5oC so q = 1.195. The flows at the top of the tower are calculated as follows. R = L/D = 1.755; L = 1.755D = 1.755(41.2) = 72.3; V1 = L+D = 72.3+41.2 = 113.5 The latent heats of vaporization of benzene and toluene at the reference temperature of 80.1oC are HVA = HVAb = 30820 kJ/kmol 93

HVB  C PLBTbB  T0  HVBb  C PVBTbB  T0  = 167.5(110.6-80.1) + 33330 - 138.2(110.6-80.1) = 34224 kJ/kmol The saturation temperature (dew point) at the top of the tower for y1 = xD = 0.95 is 82.3oC. Hy1 = 0.95[30820+96.3(82.3-80.1)] +(1-0.95)[34224+138.2(82.3-80.1)] = 31206 kJ/kmol The boiling point of the distillate D is obtained as 81.1oC. HxD = 0.95(138.2)(81.1-80.1) + (1-0.95)(167.5)(81.1-80.1) = 139 kJ/kmol

Step 1: select xn = 0.55 and guess yn+1 from Eq. (7) yn1 

Ln Dx D 72 .3 41. 2 xn   ( 0. 95)  0. 695 ( 0. 55)  Vn1 113. 5 Vn1 113. 5

Step 2: using the Figure in Example D6 for xn = 0.55 obtain Hxn = 1590 and for yn+1 = 0.695, Hy,n+1 = 33240. Substituting into Eq. (10) Vn+1 (33240) = (Vn+1 - 41.2)1590 +113.5(31206) - 72.3(139) Vn+1 =109.5 Using Eq. (5), 109.5=Ln +41.2 or Ln =68.3 Step 3: substituting into Eq. (7), 41. 2 68 .3 ( 0. 55)  yn1  ( 0. 95)  0. 700 109. 5 109. 5 94

This calculated value is close enough to the approximate value of 0.695 so that no further trials are needed. Select other values of xn and calculate the responding yn+1, plot the curved operating line in the enriching section. The condenser duty is obtained from Eq. (9) qc  V1 Hy 1  LHxD  DHxD

Selecting ym+1 = yB = 0.207 and using Eq. (13), an approximate value of xm = xN is obtained. L Bx ym1 m xm  B Vm1 Vm 1 58. 8 191. 8 xN  ( 0. 10) 0 .207  133. 0 133. 0 Solving xN =0.174. From the Figure in example D6 for xN =0.174, HxN =3800, and for yB = 0.207, Hyb =37000.

 113 .5 (31206 ) 72 .3 (139 ) 41.2 (139 )  3526100 kJ / h

For xB = 0.10, HxB = 4350 from the Figure in page 89. The feed is at 54.5oC, using Eq. (1), we have HF = 0.45(138.2)(54.5-80.1) + (1-0.45)(167.5)(54.5-80.1) = -3929 kJ/kg mol Using Eq. (15), qr = 41.2(139) + 58.8(4350) + 3526100 - 100(-3929) = 4180500 kJ/h Making a material balance below the bottom tray and around the reboiler we have LN  B  Vb (16) Enthalpy balance Vb H yb  Vb  BH xN  BHxB  qr (17) From the equilibrium diagram we find that for xB = 0.10, yB = 0.207, which is the vapor composition leaving the reboiler. For the equimolal overflow in the stripping section, L m  L n  qF  72 .3  1 .195 (100 )  191.8 (18) Vm 1  Vn 1  (1  q )F  113 .5  (1  1.195 )100 133. 0 95

Substituting into Eq. (17), Vb (37000)  Vb  58.83800 58.8(4350)  4180500 Solving Vb = 125.0. Using Eq. (16) we get LN = 183.8. Substituting into Eq. (13) and solving for xN. 58. 8 183 .8 0 .207  xN  ( 0. 10) 125. 0 125. 0 x N = 0.173 This value is quite close to the approximate value of 0.174. Assuming other values of ym+1 and solving for xm using the same procedure, the curved operating line for the stripping section can be constructed. The number of theoretical stages is determined to be 10.4, compared with 9.9 steps using the constant flow method.

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Ponchon and Savarit method Rectifying section We rearrange Eq. (8) as V n1 H y, n1  L nH xn  DH xD  q c  DQ ' (16) DH xD  qc where Q'  D The left hand side is the net rate flow of heat upwards through the enriching section, while the right hand side is constant. This means that the net rate of flow of heat is constant and independent of the tray number in the rectifying section.

Solid line: enthalpy-concentration method Dashed line: constant molal overflow The solid line is slightly above the dashed line.

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Using Eq. (5) to eliminate Vn+1 in Eq. (6), we have Ln  D  yn 1  Ln xn  DxD L n  y n1  x n   D x D  y n 1  (17) Hence, Ln x D  y n1    internal reflux ratio y n1  x n  D

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(18)

This can be explained by the lever rule. A vapor of Vn+1 of composition y n+1 is separated into a distillate D of xD, and a liquid Ln of xn. (Vn+1, yn+1) is at the fulcrum of a beam, balanced by the distillate (D, xD) and liquid (Ln, xn). In order for the beam to be “balance”, the moments about the fulcrum must be equal. Hence, we have Eq. (17). A similar analysis for Eq. (16) gives  Ln  DH y, n 1  LnH xn  DQ ' Ln H y, n1  H xn  D Q' H y, n1 or Q'  H y, n1 Ln   internal reflux ratio (19) H D H















y ,n 1

xn



Again this obeys the lever rule.

Combining Eqs. (18) & (19), we have Q 'H y, n 1  L n  x D  y n1    D  yn 1  xn  H y,n 1  H xn 

Since the heat and mass balances apply to any tray, then at the top plate, n = 0, we have line length Q 'H y ,1  x  y1  Q 'H y ,1  L   R 0  D D  y1  xD  H y ,1  H xD  line length H y ,1  H xD

(20)

On the H-x-y diagram, this is a straight line through the points (Hy,n+1, y n+1) at Vn+1, (Hx,n, xn) at L n, and (Q’, xD) at Δ D as linked above and shown in the figure below. ΔD is termed the difference point. 99

The line length H y,1  H xD is the latent heat of vaporization at a distillate composition xD. If the reflux ratio is specified, then the length of Q’ – Hy,1 can be calculated, and hence Q’ determined.

100

Equilibrium data are used to determine the liquid and vapor compositions leaving the first theoretical plate. The graphical solution of Ponchon and Savarit is continued using the equilibrium data and the H-x-y diagram until the reboiler composition is reached. In the following case, a total of 4 theoretical plates is shown, 3 in the column and 1 for the reboiler having an enthalpy HR and composition xR.

Stripping section The same analytical techniques used in the rectifying section can be applied to the stripping section and the entire column.

We rearrange Eq. (14 ) as Lm H xm  Vm1 H y, m1  BH xB  qr  BQ' ' where Q” = HxB – qr /B, by eliminating B (B=Lm – Vm+1 ),  y  x  H y,m1  Q " Lm  m 1 B  (21) x  x  H  Q " V



m 1

B

m



x, m

On the H-x-y diagram, this is a straight line passing throught (Hy,m+1, y m+1) at V m+1, (H x,m, x m) at L m, and (Q”, xB) at ΔB as shown in the figure above. ΔB is also termed the difference point. B is W in the figure. Eq. (21) applies to every plate in the stripping section. On the H-x-y diagram, this is a line from VZ+1 (vapor leaving the reboiler and entering the bottom tray Z of the 101

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tower) to ΔB, intersects the saturated liquid enthalpy curve at LZ, the liquid leaving the bottom plate. Vapor VZ leaving the bottom plate is in equilibrium with the liquid LZ and is located on the tie line Z. The number of theoretical plates in the stripping section can be determined from the H-x-y diagram by alternatively constructing lines to ΔB and tie lines. Each tie line is a theoretical plate.

The complete column

The overall mass balance is F=D+W The more volatile component mass balance is FxF = DxD + WxW The heat balance is FHF = QC + DHyD + WHxW – QR = DQ’+WQ” Eliminating F yields D  x F  xW  H F  Q"   W x D  x F  Q 'H F  This is a straight line on the H-x-y diagram passing through (Q’, xD) at ΔD, (HF, xF) at F, and (Q”, xW) at ΔW and is shown in the above figure for a case of liquid feed below its boiling point (cold liquid). The procedure to determine the number of theoretical plates is summarized below. 1. Locate the feed enthalpy and composition 2. Locate xD and xW 3. Locate ΔD by computation of Q’ or for a specific reflux ratio as R = L0/D 4. The line ΔD, F is extended to x = xW and thus locates ΔW whose coordinates can be used to calculate QR. 5. The number of theoretical plates is then determined by application of the straight line relationship on the H-x-y diagram and the equilibrium data of the y-x diagram.

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104

Example D8. Distillation Using Ponchon Savarit Method

Re-calculate the number of theoretical stages required for example D7 using Ponchon Savarit method. xF = 0.45, HF = -3929 kJ/kg mol xD = 0.95, x B = 0.10 DH xD  qc 41.2(139)  3526100   85724 kJ/kmol Q'  41.2 D

Q” = HxB – qr/B =435 -4180500/58.8 = -66747 kJ/kmol or in the H-x-y diagram, draw the line from ΔD (xD=0.95, Q’=85724) through (xF=0.45, HF=-3929) to intersect xB=0.10 gives Q” = -66747 kJ/kmol from ΔB. The number of theoretical plates is then determined by application of the straight line relationship on the H-x-y diagram and the equilibrium data of the y-x diagram. 10.4 plates, or 9.4 plates plus 1 reboiler are needed. The feed is at plate 6.

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