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Chapter 4Total Reflux and Minimum Reflux Ratio

a. Total Reflux. In design problems, the desired separation is specified and a column isdesigned to achieve this separation. In addition to the column pressure, feed

conditions, and reflux temperature, four additional variables must be specified.Specified Variables Designer Calculates

Case A1.xD2.xB3.External reflux ratioL0/D4.Use optimum feed plate

D,B: distillate and bottoms flow ratesQR, QC: heating and cooling loadsN: number of stages,Nfeed: optimum feed plateDC: column diameter

xD,xB= mole fraction of more volatile component A in distillate and bottoms, respectively

The number of theoretical stages depends on the reflux ratio R= L0/D. As Rincreases, theproducts from the column will reduce. There will be fewer equilibrium stages needed since

the operating line will be further away from the equilibrium curve. The upper limit of thereflux ratio is total reflux, orR= . The rectifying operating line is given as

yn+1=1

R

R +xn +

1

1R +xD

When R= , the slop of this line becomes 1 and the operating lines of both sections of the

column coincide with the 45o line. In practice the total reflux condition can be achieved by

reducing the flow rates of all the feed and the products to zero. The number of trays required

for the specified separation is the minimum which can be obtained by stepping off the trays

from the distillate to the bottoms.

Figure 4.4-8Minimum numbers of trays at total reflux.

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y , VN-1 N-1

x , LN N

1

2

y x1 2

N

N-1

yN xN+1

x1y0

Totalreboiler

Totalcondenser

Figure 4.4-9Distillation column operation at total reflux.

The minimum number of equilibrium trays can also be approximated by Fenske equation,

Nm=( )

1log

1

log

D B

D B

ave

x x

x x

(4.4-22)

In this equation ave= (1B)1/2

where 1 is the relative volatility of the overhead vapor and

B is the relative volatility of the bottoms liquid. We can derive Fenske equation using the

notation shown in Figure 4.4-9 where stages are numbered from the bottom up. The vapor

leaving stageNis condensed and returned to stage Nas reflux. The liquid leaving stage 1 is

vaporized and returned to stage 1 as vapor flow. For steady state operation with no heat loss,heat input to the reboiler is equal to the heat output from the condenser. From material

balance, vapor and liquid streams passing between any pair of stages have equal flow rates

and compositions, for example, VN-1= LNand yN-1=xN. In general, molar vapor and liquid

flow rates will change from stage to stage unless the assumption of constant molar overflow

is valid. At stage 1, the equilibrium relation is written as

y1= K1x1 (4.4-23)

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From the material balance

y1=x2 (4.4-24)

Combine Eqs. (4.4-23) and (4.4-24)

x2= K1x1 (4.4-25)

Similarly for stage 2

y2= K2x2 (4.4-26)

Combine Eqs. (4.4-25) and (4.4-26)

y2= K2K1x1 (4.4-27)

The procedure can be repeated to stageNwhere

yN= KNKN-1K2K1x1 (4.4-28)

Similarly for the less volatile component i

1 yN= Ki,NKi,N-1Ki,2Ki,1 (1 x1) (4.4-29)

Dividing Eq. (4.4-8) by Eq. (4.4-9), we have

1

N

N

y

y= NN-121

1

11

x

x (4.4-30)

In this equation k =,

k

i k

K

K = relative volatility between the two components on stage k.

Rearranging Eq. (4.4-30) we obtain

1

Ny

x=

1

1

1

Ny

x

min

1

N

k

k

=

or 11

Nx

x

+ = 1

1

1

1

Nx

x

+

min

1

N

k

k

=

(4.4-31)

SincexN+1=xD,x1=xB, and assuming constant relative volatility, Eq. (4.4-31) becomes

minN =

1

D

D

xx

1 B

B

xx

(4.4-32)

Solving for the minimum number of equilibrium trays gives

Nmin=( )

1log

1

log

D B

D B

x x

x x

(4.4-33)

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Eq. (4.4-33) is the Fenske equation (4.4-22) where = ave= (1B)1/2

Nm=( )

1log

1

log

D B

D B

ave

x x

x x

(4.4-22)

b. Minimum reflux ratio. As the reflux ratio is reduced, the distance between theoperating line and the equilibrium curve becomes smaller. The minimum reflux ratio

Rmis the limiting reflux where the operating line either touches the equilibrium curve

or intersects the equilibrium curve at the q-line. The minimum reflux ratio will

require an infinite number of trays to attain the specified separation of xD and xB.

Figure 4.4-10 shows an equilibrium plate nwith streams Ln-1 and Vn+1 entering and

streams Lnand Vn leave the plate. If the two steams Ln-1and Vn+1are at equilibrium

there will be no net mass transfer between the liquid and vapor streams. The

equilibrium curve will touch or intersect the operating line at this point.

n

n-1

n+1

Ln-1

Vn

Ln

Vn+1

Figure 4.4-10Equilibrium plate nwith vapor and liquid streams.

Given q,xDandxF, the feed line is fixed and the upper operating line depends on the reflux

ratio R. At total reflux, the operating line coincides with the 45oline. AsRis decreased, the

slope of the enriching operating line R/(R + 1) is decreased. The operating line will rotate

clockwise around the point (x = xD,y = xD) until it is tangent to the equilibrium curve or it

intersects the q-line at the equilibrium point whichever comes first. The location where the

operating line touches or intersects the equilibrium curve is called the pinch point. The

enriching operating line at minimum reflux is then defined.

yn+1=

1

m

m

R

R +

xn +1

1mR +

xD

The minimum reflux Rm can be obtained from either the intercept of the slope of the

enriching operating line. The operating flux ratio is between the minimum Rm and total

reflux. Usual value is between 1.2Rmto 1.5Rm. Figure 4.4-11 shows the pinch points for case

1 where the operating line intersects the equilibrium curve and case 2 where the operating

line touches the equilibrium curve.

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Figure 4.4-11The pinch points for minimum reflux.

Example 4.4-4 ----------------------------------------------------------------------------------

A distillation column receives a feed that is 40 mole % n-pentane and 60 mole % n-hexane.

Feed is saturated liquid with a flow rate of 2,500 lbmol/hr. The column is at 1 atm. A

distillate of 90 mole % n-pentane is desired. A total condenser is used. Reflux is a saturated

liquid. Bottoms from the reboiler is 98 mole % n-hexane. Determine the minimum number of

equilibrium trays and the minimum reflux ratio.

Data: Vapor pressure, Psat, data: ln Psat=AB/(T+ C), where Psatis in kPa and Tis in K.

Compound A B C

n-pentane (1) 13.9778 2554.6 36.2529

n-hexane (2) 14.0568 2825.42 42.7089

Heat of evaporation for n-pentane, C5= 11,369 Btu/lbmol, CpL,C5= 39.7 Btu/lbmoloF

Heat of evaporation for n-hexane, C6= 13,572 Btu/lbmol, CpL,C6= 51.7 Btu/lbmoloF

Solution ------------------------------------------------------------------------------------------

(a)Minimum number of equilibrium trays

The equilibrium data for n-pentane and n-hexane at 1 atm are listed in Table 4.4-4. The data

were generated with the Matlab codes listed in Table 4.4-5 assuming ideal solution.

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Table 4.4-4Equilibrium data for n-pentane and n-hexane system at 1 atm.x= mole fraction of n-pentane in the liquid

y = mole fraction of n-pentane in the vapor

x= 0.00000 , y= 0.00000, T(K) = 342.06

x= 0.05000 , y= 0.12705, T(K) = 339.40

x= 0.10000 , y= 0.23699, T(K) = 336.91

x= 0.15000 , y= 0.33263, T(K) = 334.58x= 0.20000 , y= 0.41626, T(K) = 332.39

x= 0.25000 , y= 0.48975, T(K) = 330.32

x= 0.30000 , y= 0.55462, T(K) = 328.38

x= 0.35000 , y= 0.61214, T(K) = 326.53

x= 0.40000 , y= 0.66335, T(K) = 324.79

x= 0.45000 , y= 0.70911, T(K) = 323.14

x= 0.50000 , y= 0.75016, T(K) = 321.56

x= 0.55000 , y= 0.78711, T(K) = 320.07

x= 0.60000 , y= 0.82048, T(K) = 318.64

x= 0.65000 , y= 0.85070, T(K) = 317.28

x= 0.70000 , y= 0.87816, T(K) = 315.97x= 0.75000 , y= 0.90317, T(K) = 314.72

x= 0.80000 , y= 0.92601, T(K) = 313.53

x= 0.85000 , y= 0.94692, T(K) = 312.38

x= 0.90000 , y= 0.96610, T(K) = 311.28

x= 0.95000 , y= 0.98374, T(K) = 310.22

x= 1.00000 , y= 1.00000, T(K) = 309.20

The minimum number of trays required is obtained by stepping off between the 45o

line and

the equilibrium curve fromxB= 0.1 toxD= 0.9. The answer is 4.1 equilibrium trays.

Figure E-1Minimum numbers of trays at total reflux.

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The minimum number of trays can also be estimated by Fenske equation,

Nm=( )

1log

1

log

D B

D B

ave

x x

x x

=( )

0.9 1 0.1log

1 0.9 0.1

logave

=( )

( )

log 81

logave

In this equation ave= (DB)1/2where D is the relative volatility of the overhead vapor and

Bis the relative volatility of the bottoms liquid.

Figure E-2Distillation column operation at total reflux.

From the notation in Figure E-2, at the top yN = xD = 0.9 0.9032, xN = 0.75000. At the

bottomsx1=xB= 0.1,y1= 0.237 (equilibrium data from Table 4.4-4).

D=( ) ( )

/

1 / 1

N N

N N

y x

y x =

( ) ( )

0.9032 / 0.7500

1 0.9032 / 1 0.7500 = 3.1102

B=( ) ( )

1 1

1 1

/

1 / 1

y x

y x =

( ) ( )

0.237/ 0.100

1 0.237 / 1 0.100 = 2.7955

y , VN-1 N-1 x , LN N

1

2

y x1 2

N

N-1

yN xN+1

x1y0

Totalreboiler

Totalcondenser

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ave= (DB)1/2

= (3.11022.7955)1/2

= 2.9487

Nm=( )

( )

log 81

logave

=( )

( )

log 81

log 2.9487= 4.1

(b)Minimum reflux ratio.

Figure E-3The pinch point for minimum reflux.

Since feed is saturated liquid, the feed-line is vertical and intersects the equilibrium curve atthe point x = 0.4, y = 0.6633. The enriching operating line for minimum reflux passes

through this point (x= 0.4,y = 0.6633) and the point (x=xD= 0.9,y = 0.9). The slope of the

rectifying operating line is given by

1

m

m

R

R +=

0.9000 0.6633

0.9000 0.4000

= 0.4734

The minimum reflux ratio is then

Rm= 0.47341.0000 0.4734

= 0.90