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11.6 Distillation and Absorption Efficiencies for tray and packed towers11.6.1 Tray EfficiencyIn all the previous discussions of theoretical trays or stages in distillation, we assumed that the vapor leaving a tray was in equilibrium with the liquid leaving.

However, if the time of contact and the degree of mixing on the tray are insufficient, the streams will not be in equilibrium. As a result the efficiency of the stage or tray will not be 100%. This means that we must use more actual trays for a given separation than the theoretical number of trays determined by calculation.

Three types of tray or plate efficiency are used:

Overall tray efficiency E0,

Murphree tray efficiency EM

Point or local tray efficiency EMP (sometimes called Murphree point efficiency).

11.6.2 Types of Tray Efficiencies

Overall tray efficiency: The overall tray or plate efficiency E0 concerns the entire tower and is simple to use, but it is the least fundamental. It is defined as the ratio of the number of theoretical or ideal trays needed in an entire tower to the number of actual trays used:

Murphree tray efficiency. The Murphree tray efficiency EM is defined as follows:

Point Efficiency.

Wherey`n=concentration of vapor at specific point in plate ny`n+1= concentration of vapor entering the plate n at the same pointy* = concentration of vapor that would be in equilibrium with xn`4. Relationship between tray Efficiencies

5. Estimation of Efficiencies of Tray and Packed Towers 5.1. Efficiency of Tray Towers. For estimating the overall tray efficiency of bubble-tray towers for distillation, the OConnell (01) correlation can be used (K2) with about a 10% error.The following equation for these data from Lockett (Li) can be used for sieve and valve trays as well, but predictions will be slightly conservative:

Where E0 is fraction efficiency, a is relative volatility of the two key components at the average tower temperature, and L is the molar average viscosity in cp of the liquid feed at the aver-age tower temperature of the top and the bottom. Most typical efficiencies are between 40 and 80%.To estimate the overall tray efficiency for absorption towers, the OConnell correlation(01) can be used. This correlation is represented by the equation (S3)

5.2. Efficiency of random-packed towers.For estimates for random packing, Equation. can be used to determine the HETP for second- and third-generation packings only (S2): HETP = 0.0180dP (SI)

HETP = 1.5dP (English)where HETP is in m and d is packing diameter in mm. In English units, HETP is in ft and d is in in. Also, for small-diameter towers, where the tower diameter D is less than 0.60 m (2 ft), HETP = D, but not less than 0.3 m (1 ft).

For vacuum service (S3), HETP = 0.0180dP + 0.15 (SI)

HETP = 1.5dP + 0.50 (English)5.3. Efficiency of Structured Packing in towers HETP = 100/a+0.1 (S.I) HETP = 100/a+0.33 (English)Please See Example 11.5.1 P72911.7.1 IntroductionIn the case before, the main assumptions in the method we made are that the latent heats are equal, sensible heat differences are negligible, and constant molal overflow occurs in each section of the distillation tower.But now in fractional distillation using enthalpy-concentration data, the molal overflow rates are not necessarily constant.The analysis will be made using enthalpy as well as material balances.11.7 Fractional distillation using enthalpy-concentration methodAn efficient separation can be achieved if the a > 1.05Efficiency of separation process highly depends on the reflux ratio and stages

RefluxManipulated variables711.7.2 Enthalpy-concentration DataAn enthalpy-concentration diagram for a binary vapor-liquid mixture of A and B takes into account latent heats, heats of solution or mixing, and sensible heats of the components of the mixture.The saturated liquid line in enthalpy h kJ/kg or kJ/kg mol is calculated by h = xA cpA (T-T0) + (1-xA)cpB (T-T0) + Hsolwhere,xA = weight or mole fraction A.T, T0 = boiling point of the mixture and ref. temp.in K . cpA , cpB = liquid heat capacity of comp. A and B. Hsol = heat of solution at T0 in kJ/kg mol.8The saturated vapor enthalpy line of H kJ/kg or kJ/kg mol of a vapor composition yA is calculated by,H = yA [A + cpyA (T-T0)] + (1-yA)[B + cpyB (T-T0)]where, A = cpA (TbA -T0) + Ab - cpyA (TbA -T0) B = cpB (TbB -T0) + Bb - cpyB (TbB -T0) Ab ,Bb = the latent heat of comp. A and B at normal the boiling point TbA. ,TbB respectively. cpyA , cpyB = vapor heat capacity of comp. A and B . T0 is equal to the boiling point of the lower boiling comp. A. This means A = Ab. Hence, only Bb must be corrected to B. 9Example 11.6-1: Enthalpy-Conc. Plot for Benzene-Toluene. Prepare an enthalpy-concentration plot for benzene-toluene at 1 atm pressure. Equilibrium data are given in Table 11.01-1 and Figs. 11.1-1 and 11.1-2. Physical property data are given in Table 11.6-1.cp (kJ/kg mol.K) Component Tb (C) Liquid Vapor Bb (kJ/kg mol)Benzene (A) 80.1 138.2 96.3 30820Toluene (B) 110.6 167.5 138.2 33330Table 11.6-1. Physical Property Data for Benzene and Toluene.10SolutionA reference temperature of T0 = 80.1 C will be used for convenience so that the liquid enthalpy of pure benzene (xA = 1.0) at the boiling point will be zero. (1) For the first point we will select pure toluene (xA = 0). For pure toluene at the boiling point of 110.6 C, with zero heat of solution and data from Table 11.6-1, h = xA cpA (T-T0) + (1-xA)cpB (T-T0) + Hsol h = xA cpA (T-80.1) + (1-xA)cpB (T-80.1) + 0 h = 0 + (1-0) 167.5 (110.6 80.1) = 5109 kJ/kg mol.

(2) For the saturated vapor enthalpy line, we first must calculate B at the reference temperature T0 = 80.1 C .B = cpB (TbB -T0) + Bb - cpyB (TbB -T0) = 167.5(110.680.1) + 33330138.2(110.6-80.1) = 34 224 kJ/kg mol.

11(3)To calculate H, at yA = 0. H = yA [A + cpyA (T-T0)] + (1-yA)[B + cpyB (T-T0)] = 0 + (1.0-0) [34224 + 138.2 (110.6-80.1)] = 38 439 kJ/kg mol.

(4) For pure benzene, xA = 1.0 and yA = 1.0, since T = T0 = 80.1, using , h = xA cpA (T-80.1) + (1-xA)cpB (T-80.1) + 0h = 0

12(4)For the saturated vapor enthalpy, at T = 80.1.H = 1.0[30820 + 96.3(80.1-80.1)] + 0 = 30 820

(5) Selecting xA = 0.50, the boiling point Tb = 92 C and the temperature of saturated vapor for yA = 0.50 is 98.8 C from Fig. 11.1-1. Using equation for the saturated liquid enthalpy at the boiling point.h = 0.5(138.2)(92-80.1)+(1-0.5)167.5(92-80.1)=1820Also, for yA = 0.50, the saturated vapor enthalpy at 98.8 C is H = 0.5[30820 + 96.3(98.8-80.1)] + (1-0.5)[34224 +138.2(98.8-80.1)] = 34 716

(6) Selecting xA = 0.30 and yA = 0.30, h = 2920 and H = 36268. These values are tabulated in Table 11.6-2 and plotted in Fig. 11.6-1.13DISTILLATIONSaturated Liquid Saturated VaporxA Enthalpy, hyA Enthalpy, H 051090384390.3029200.30 362680.5018200.50 347160.805620.80 323801.0001.00 30820

Table 11.6-2: Enthalpy-Concentration Data for Benzene-Toluene Mixtures at 101.325 kPa total pressure.14DISTILLATIONEnthalpy of mixture, H or h(kJ/kg mol mixture)00.20.40.60.81.0Mole fraction benzene, xA or yA. 010000200003000040000H vs yA (saturated vapor) h vs xA (saturated liquid) tie line Fig. 11.6-1: Enthalpy-concentration plot for example 11.6-1 15The enriching-section operating line, mass balance

Arrangements;

However, the Vn+1 and Ln may vary, equation will not become a straight line

Distillation in Enriching Section of Tower.16Making an enthalpy balance (eliminating the assumption of constant molal overflow in McCabe-Thiele method)

Enthalpy balance for condenser

Eliminating the terms qc

17In order to plot the operating line, the terms Vn+1 and Ln must be determined first.

If the reflux ratio is set, V1 and L are known.

The values H1 and hD can be determined from an enthalpy-concentration plot.

If the value of xn is selected, it is a trial-and-error solution to obtain Hn+1 since yn+1 is not known.

18Distillation in Stripping Section of Tower.

The stripping-section operating line

Making an enthalpy balance,

Using the similar steps to that for enriching section to solve for stripping section. Assume equimolal overflow.

19A liquid mixture of benzene-toluene is being distilled using the same conditions as in example 11.4-1 except that a reflux ratio of 1.5 times the minimum reflux ratio is to be used. The value of Rm = 1.17 from example 11.4-2 will be used. Use enthalpy balances to calculate the flow rates of the liquid and vapor at various points in the tower and plotted the curved operating lines.

Determine the number of theoretical stages needed.Example 11.6-2: Distillation Using Enthalpy-Concentration Method.

20SolutionThe given data are as follows; F = 100 kg mol/h, xF = 0.45 , xD = 0.95,xW = 0.10, R = 1.5(1.17) = 1.755, D = 41.2 kg mol/h, W = 58.8 kg mol/h.

the feed enters at 54.4 C and q = 1.195. The flows at the top of the tower are calculated as follows.L/D = 1.755;L = 1.755(41.2) = 72.3;V1 = L+D = 72.3 + 41.2 = 113.5.The saturation temp. at the top of the tower for y1 = xD = 0.95 is 82.3 C from Fig. 11.1-1. Using equation,H1 = 0.95[30820+96.3(82.3+80.1)]+(1-0.95)[34224+138.2(82.3-80.1)] = 31 206. 21The boiling point of the distilled D is obtained from Fig.11.1-1 and is 81.1 C. The enthalpy hD then,hD = 0.95(138.2)(81.1-80.1)+(1-0.95)(167.5)(81.1-80.1)= 139

Following the procedure outlined for the enriching section: step 1, a value of xn = 0.55 is selected. Assuming a straight operating line, an approximate value of yn+1 is obtained.,

22 step 2, using Fig. 11.6-1, for xn = 0.55, get hn = 1590 and for yn+1 = 0.695, get Hn+1 = 33240. Substituting into equation and solving,

Then, calculate Ln109.5 = Ln + 41.2,Ln = 68.3

step 3, substituting into enriching operating line

23This calculated value of yn+1 = 0.7 is sufficiently close to the approximate value of 0.695 so that no further trials are needed.

Selecting another value for xn = 0.70 and solving for

Then,Ln = 110.8-41.2 = 69.6Substituting into,

24In Fig. 11.6-3, the points for the curved operating line in the enriching section are plotted. This line is approximately straight and is very slightly above that for constant molal overflow.

The condenser duty is calculated. qC = 113.5(31206)-72.3(139)-41.2(139) = 3 526 100 kJ/h

To obtain the reboiler duty qR , values for hW and hF are needed. Using Fig. 11.6-1 for xW = 0.10, hW = 4350. The feed is at 54.5 C. Using equation to calculate hF, hF = 0.45(138.2)(54.5-80.1)+(1-0.45)(167.5)(54.5-80.1) = -3929Then,qR = 41.2(139)-58.8(4350)+3526100-100(-3929)= 4180500 25Making a material balance below the bottom tray and around the reboiler,LN = W + VW

Rewriting equation for this bottom section,VWHW = (VW +W)hN + qR WhWFrom the equilibrium diagram, for xW = 0.10, yW = 0.207, which is the vapor composition leaving the reboiler. For equimolal overflow in stripping section,Lm = Ln + Qf =72.3+1.195(100)=191.8Vm+1 = Vn+1 (1-1.195)100 = 133.0Selecting ym+1 = yW = 0.207, and value of xm = xN is obtained.

26Solving, xN = 0.174. From Fig. 11.6-1 for xN = 0.174, hN = 3800, and for yW = 0.207, HW = 37000. Substituting into VWHW = (VW +W)hN + qR WhW VW (37000)=(VW + 58.8)(3800)+4180500-58.8(4350)Solving , VW = 125.0. Then calculated LN = 183.8. Substituting into equation to solve for xN,

This value of 0.407 is sufficiently close to the approximate value of 0.412 so that no further trials are needed.

The two points calculated for the stripping section are plotted in Fig. 11.6-3. This stripping line are also approximately straight and is very slightly above the operating line for constant molal overflow.

Using the operating line for the enthalpy balance method, the number of theoretical steps is 10.4. For the equimolal method 9.9 steps are obtained.

27000.20.40.60.81.00.20.40.60.81.0Mole fraction in liquid, x Mole fraction in vapor, yxFFig. 11.6-3: Plot of curved operating line for example 11.6-2q-line28ExampleA feed containing 40 mole percent n-hexane and 60 percent n-octane is fed to a distillation column. A reflux ratio of 1.2 is maintained. The overhead product is 95 percent hexane and the bottoms 10 percent hexane. Find the number of theoretical stages and the optimum feed stage. Assume that a total condenser is used. The column is to operate at 1 atm.

The equilibrium data are as follow:

x00.10.30.50.55o.71.0y00.360.700.850.900.951.029EXAMPLEMOLE FRACTION OF HEXANEENTHALPHYSAT. LIQUIDSAT. VAPOR07000157000.16300154000.35000147000.54100139000.73400129000.93100116001.030001000030Obtain enthalpy-composition diagram Fix the feed point F, and product points D and B using stream compositions and enthalpies Use the overhead product enthalpy and the reflux ratio to find the adjusted enthalpy of the overhead. Plot it as point D', on a vertical line with point D. Construct the overall enthalpy line from point D' through the feed point. It intersects a vertical line drawn through point B at point B'. Plot point V1. For a total condenser, the composition entering the condenser is the same as the overhead product, so this point will be vertically above point D on the saturated vapor curve. Follow the tie line from point V1 to the saturated liquid curve. This intersection will be point L1. Construct an operating line connecting points D' and L1. The intersection of the operating line with the saturated vapor curve will be point V2. Repeat the two preceding steps until one of the V or L points is to the left of the overall enthalpy line. Once it is crossed, construct operating lines using points Li and B'. When xi is less than xB, construction is finished

Summary of Procedure31DISTILLATIONPlot the enthalpy concentration data.

Plot the point for feed and product

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