DESIGN OF A PACKED DISTILLATION

COLUMN

NAME : Sandaruwan R.A.N.

INDEX NO : 090466M

DATE OF SUB: 01 / 08 / 2013

Task identification

According to the determination of packed column height through the number of ideal stages and the

HETP value, first the conditions and the objective of the column have to be clarrified using the data

provided by the customer. So the required data has been provided as follows.

System: carbon disulphide – carbon tetrachloride

Feed rate: 45kmol/hr Feed condition: saturated liquid

Feed composition: 40mol% carbon disulphide

Distillate composition: 90mol% carbon disulphide

Bottom product composition: 4mol% carbon disulphide

Selection of packing material

Principal requirements of a packing are,

Provide a large surface area for high interfacial area between the gas and liquid

Have an open structure, so low resistance to gas flow

Promote uniform liquid distribution on the packing surface

Promote uniform vapour gas flow across the column cross section

To satisfy these requirements many diverse types and shapes have been developed.

In this distillation column design, berl saddles types rings are used as packing material. Berl

saddles were developed to give improved liquid distribution compared to rasching rings. These are the

original type of saddle packing. They have a smaller free gas space but their aerodynamic shape is

better giving a lower pressure drop and higher capacity. They are usually made up by ceramic or metal

or carbon or plastic. The choice of material will depend on the nature of the fluids and the operating

temperature. Ceramic packing will be the first choice for corrosive liquids and low cost. Metal packing

are usually selected for non corrosive service. They have higher capacity and efficiency. Plastic are

normally polypropylene. It is inexpensive and most popular when temperature does not exceed 250˚F.

In the process industries, random packings re more commonly used.

So for this design plastic berl saddles are selected and random packing is used.

Packing size

In general the largest size of packing should be used upto 50mm. small sizes are more

expensive than the larger sizes. The size of packing used influences the height and diameter of a

column, the pressure drop and cost of packing. When packing size is increased, the cost per unit

volume of packing and the pressure drop per unit height of packing are reduced, but in other side

which will reduce the mass transfer efficiency. Reduced mass transfer efficiency results in a taller

column. Also use of too large a size in a smaller column can cause poor liquid distribution.

In this design packing size is selected as 25mm to maintain the column diameter below 0.3m.

Pressure drop across packing

Recommended design values, mm water per m packing

According to this, 60mm water per m of packing is selected as pressure drop.

Calculating equilibrium data

Equilibrium data calculation for Carbon disulphide – Carbon tetrachloride vapour – Liquid system

According to antion equation

𝑙𝑛 𝑃˚ = 𝐴 − 𝐵/(𝑇 + 𝐶)

Where, A, B, C - Constants in the antion equation

Vp - Vapour pressure (mmHg)

T - Temperature (K)

Antion constants for Chloroform and Benzene

Carbon disulphide (A) Carbon tetrachloride (B)

A 15.9844 15.8742

B 2690.85 2808.19

C -31.62 -45.99

Boling point of Carbon disulphide = 46˚C

Boling point of Carbon tetrachloride =77˚C

From Raoul’s law,

P = Pox Where, P - Partial pressure

Po – Vapour pressure

X – Molar fraction in the liquid phase

Applying Raoult’s law for A and B;

PA = P0

AxA

PB = PoBxB

From Daltons Law;

pA + pB = PT

Where, PT – Total pressure

pA + pB = PT = pO

AxA + poBxB

PT = 750 mm Hg (1bar)

xA + xB = 1

pA + pB = PT = p0

AxA + poB (1 - xA)

PT - p0

B

xA = ----------

p0

A - poB

From Dalton’s law;

PA = yAPT

yA = (pO

A/PT) xA

Sample calculation

At 46 ˚C

𝑙𝑛 𝑃˚𝐴 = 𝐴 − 𝐵/(𝑇 + 𝐶)

= 15.9844-2690.85/(319-31.62)

𝑙𝑛 𝑃˚𝐴 = 6.621

𝑃˚𝐴 = 750.695Hgmm

𝑙𝑛 𝑃˚𝐵 = 𝐴 − 𝐵/(𝑇 + 𝐶)

=15.8742-28.08.19/(319-45.99)

𝑙𝑛 𝑃˚𝐵 = 5.588

𝑃˚𝐵 =267.2Hgmm

PT - p0

B

xA = ---------- yA = (pOA/PT) xA

p0

A - poB

=0.999 = 0.998

Equilibrium data

Temperature

(˚C) P˚A P˚B XA YA

46 750.71 267.25 1.00 1.00

47 775.48 277.47 0.95 0.98

48 800.89 288.00 0.90 0.96

49 826.95 298.86 0.85 0.94

50 853.68 310.04 0.81 0.92

51 881.07 321.55 0.77 0.90

52 909.15 333.41 0.72 0.88

53 937.92 345.61 0.68 0.85

54 967.40 358.17 0.64 0.83

55 997.59 371.09 0.60 0.80

56 1028.52 384.37 0.57 0.78

57 1060.19 398.04 0.53 0.75

58 1092.61 412.09 0.50 0.72

59 1125.80 426.54 0.46 0.69

60 1159.76 441.38 0.43 0.66

61 1194.52 456.63 0.40 0.63

62 1230.08 472.30 0.37 0.60

63 1266.45 488.39 0.34 0.57

64 1303.64 504.92 0.31 0.53

65 1341.68 521.88 0.28 0.50

66 1380.57 539.29 0.25 0.46

67 1420.32 557.16 0.22 0.42

68 1460.95 575.50 0.20 0.38

69 1502.47 594.30 0.17 0.34

70 1544.89 613.59 0.15 0.30

71 1588.22 633.37 0.12 0.26

72 1632.49 653.65 0.10 0.21

73 1677.69 674.44 0.08 0.17

74 1723.84 695.74 0.05 0.12

75 1770.96 717.57 0.03 0.07

76 1819.06 739.93 0.01 0.02

77 1868.16 762.83 -0.01 -0.03

Equilibrium curve

Reflux ratio

Determination of minimum reflux ratio (Rm)

q – line equation

In this case feed is a saturated liquid so feed is at its boiling point

If Feed at boiling point q = 1 and q – line is passing through the point of (Xf , Xf)

𝑋𝐹= 0.4

fqq x. 1q

1 x.

1q

qy

feed theofheat latent Molar

feed of mole 1 vapourise torequiredHeat q

Top operating line equation

Dn1n .x

R1

1.x

R1

Ry

But, at minimum reflux ratio we can rewrite top operating line equation as follows,

Dn1n .xRm1

1.x

Rm1

Rmy

Top operating line is passing through the point of (XD , XD)

XD = 0.9

In minimum reflux ratio intersection of the top operating line and Q line should be at the curve.

According to the graph,

Intercept of the top operating line = 0.414

Gradient of the top operating line = 0.54

Therefore Rm1

Rm

= 0.54

Minimum reflux ratio, Rm = 1.174

Operating reflux ratio

When we operate high reflux ratio our plates are less than high reflux ratio. Then construction

cost is less. But we have to spend lot of money on operating cost and steams. So in industry reflux

ratio is between 1.2-1.5 of minimum reflux ratio. Here distill concentration is high so I think operating

reflux ratio should be 1.3Rm

R = 1.3Rm

Then operating reflux ratio = 1.3 * 1.174

R = 1.5262

HETP Value

For the design of packed distillation columns it is simpler to treat the separation as a stage

process and use the concept of the height of an equivalent equilibrium stage to convert the number of

ideal stage required to a height of packing.

The height of an equivalent equilibrium stage, called height of a theoretical plate (HETP), is the height

of packing that will give the same separation as an equilibrium stage. The HETP for a given type and

size of packing is essentially constant and independent of the system physical properties.

HETP value for a packing size of 25 mm can be assumed as a 0.46 m.

Number of ideal stages required

Then top operating line equation,

𝑦𝑛+1= 0.6041 𝑥𝑛 + 0.3563

q = 0.4

No of ideal stage required = 12-1

= 11

Feed tray location is above the 5th

tray from the top of the tower

Column height

Column height = number of stages * HETP

= 11* 0.46 m

= 5.06 m

Column height = 5.1m

Column diameter

Material balance

F = D + W

Material balance for carbon disulphide

F.𝑥𝑓 = D𝑥𝑑 + W𝑥𝑤

F.𝑥𝑓 = D𝑥𝑑 + (F-D)𝑥𝑤

D = 𝐹(𝑥𝑓−𝑥𝑤 )

𝑥𝑑−𝑥𝑤

D = 45(0.4−0.04)

0.9−0.04 kmol/hr

D = 18.837 kmol/hr

Then,

W = (45 – 18.837)kmol/hr

= 26.163 kmol/hr

Material balance for rectifying section

L + D = G

R = 𝐿

𝐷

1.5262 = 𝐿

18.837

L = 28.75 kmol/hr

So,

28.75 + 18.375 = G

G = 47.125kmol/hr

Distillate (D) = 18.837 kmol/hr

Down ward liquid flow rate (L) = 28.75 kmol/hr

Upward vapour flow rate (G) = 47.125 kmol/hr

Material balance for stripping section

W + 𝑉 ′ = 𝐿′

According to BOL,

𝑦𝑚+1 = 𝐿′

𝐿′−𝑤 𝑥𝑚 −

𝑤

𝐿′−𝑤𝑥𝑤

Gradient = 1.55 and W = 26.163kmol/hr

𝐿′

𝐿′−𝑤 = 1.55

𝐿′ = 73.732kmol/hr

𝑉 ′ = 73.732 – 26.163

𝑉 ′ = 47.569kmol/hr

Bottom product (W) = 26.163kmol/hr

Upward vapour flow rate in stripping section (𝑉 ′ ) = 47.569kmol/hr

Downward liquid flow rate in stripping section (𝐿′) = 73.732kmol/hr

Vapour liquid equilibrium mixture for carbon disulphide and carbon tetrachloride

For rectifying section

Composition of the 4th

tray (above plate to the feed)

𝑦5= 0.62

𝑥4 = 0.43

So, temperature of the 4th tray = 333K

Composition of the 1st tray (top plate to the feed)

𝑦1 = 0.9

𝑥0 = 0.9

Temperature of the 1st tray = 324K

Average temperature of the striping section = 333+324

2 K

= 328.5K

Average carbon disulphide composition of vapour (y) = 0.760

Average carbon disulphide composition of liquid (x) = 0.665

Density

Molecular weight of C𝑆2= 0.0761kg/mol

Molecular weight of C𝐶𝑙4= 0.1538kg/mol

For rectifying section,

Density of C𝑆2 at 328.5K = 1207.566kg/𝑚3

Density of C𝐶𝑙4 at 328.5K = 1526.596 kg/𝑚3

For liquid

Weight fraction of C𝑆2= 0.665×0.0761

0.665×0.0761+0.335×0.1538

= 0.4955

Basis 1000kg,

Volume of C𝑆2=0.4955×1000

1207.566

= 0.41𝑚3

Volume of C𝐶𝑙4= 0.5045×1000

1526.596

= 0.33𝑚3

Density of liquid mixture = 1000

0.41+0.33

= 1351.35kg/𝑚3

For vapour mixer

According to, PV = nRT

ρ = 𝑃𝑀

𝑅𝑇

Average molecular weight of the vapour stream = 0.76×0.0761+0.24×0.1538

= 0.09475kg/mol

ρ = 1×105×94.75

8.314×328.5

Density of vapour mixture = 3.47kg/𝑚3

For stripping section

Composition of 6th

plate (below plate to the feed tray)

𝑦6= 0.58

𝑥5 = 0.38

Temperature of the 5th tray = 334 K

Composition of 11th

plate (lowest plate of tower)

𝑦12 = 0.07

𝑥11 = 0.06

Temperature of the 11th plate = 347 K

Average temperature for stripping section = 334+347

2

= 340.5 K

Average carbon disulphide composition of vapour (y) = 0.325

Average carbon disulphide composition of liquid (x) = 0.22

Density of C𝑆2 at 340.5K = 1187.885 kg/𝑚3

Density of C𝐶𝑙4 at 340.5K = 1503.208kg/𝑚3

Similar to above calculation,

For liquid mixer

Weight fraction of the C𝑆2 = 0.1225

Density of liquid mixture = 1455.8kg/𝑚3

For vapour mixer

Average molecular weight of the mixer = 0.12855kg/mol

Density of vapour mixer = 4.541kg/𝑚3

Column diameter for rectifying section

Gas flow rate = 47.125 kmol/hr = 47.125×0.09475×103

3600 kg/s

= 1.24 kg/s

Average molecular weight of liquid mixer = 0.665×0.0761+ 0.335×0.1538

= 0.1021kg/mol

Liquid flow rate = 28.75kmol/hr = 28.75×0.1021×103

3600kg/s

= 0.8154kg/s

Gas density at 328.5K = 3.47kg/𝑚3

Liquid density at 328.5K = 1351.35kg/𝑚3

𝐹𝐿𝑉= 𝐿𝑤

𝑉𝑤

ρ𝑣

ρ𝐿

= 0.8154

1.24

3.47

1351.35

= 0.033

Pressure drop correlation graph

Pressure drop = 60mm water per m of packing

According to the graph,

𝐾4 = 2.2

At flooding, 𝐾4= 5.6

Percentage flooding = 2.2

5.6 = 0 .627

62.7% satisfactory

Viscosity calculation

Log [viscosity] = [VISA] × [1

𝑇−

1

𝑉𝐼𝑆𝐵]

For carbon disulphide

VISA = 274.08

VISB = 200.22

T = 328.5k

Log [viscosity] = [274.08] × [1

328.5−

1

200.22]

µ1= 0.292 mNs/𝑚2

For carbon tetrachloride

VISA = 540.15

VISB = 290.84

T = 328.5k

Log [viscosity] = [540.15] × [1

328.5−

1

290.84]

µ2= 0.6125 mNs/𝑚2

Average viscosity of mixer

µ𝑚𝑖𝑥 = ( 𝑥𝑖 × µ𝑖

13)3

µ𝑚𝑖𝑥 = (0.665 × 0.2921

3 + 0.335 × 0.61251

3)3

= 0.3821 mNs/𝑚2

𝑉𝑤 = [𝐾4ρ𝑣(ρ𝐿−ρ𝑣)

13.1F𝑃(µ𝐿/ρ𝐿)0.1]

For this type packing material (plastic), 𝐹𝑝= 170

𝑉𝑤 = [2.2×3.47(1351.35−3.47)

13.1×170(0.3821/1351.35)0.1]

= 10.46kg/𝑚2𝑠

Column area required = 1.24

10.46 = 0.1185𝑚2

Diameter = 4

𝜋× 0.1185

Diameter of the rectifying section= 0.388m

Column diameter for stripping section

Gas flow rate = 47.569kmol/hr = 47.569×0.12855×103

3600 kg/s

= 1.7 kg/s

Average molecular weight of liquid mixer = 0.22×0.0761+ 0.78×0.1538

= 0.1367kg/mol

Liquid flow rate = 73.732kmol/hr = 73.732×0.1367×103

3600kg/s

= 2.8kg/s

Gas density at 340.5K = 4.541kg/𝑚3

Liquid density at 340.5K = 1455.8kg/𝑚3

𝐹𝐿𝑉= 𝐿𝑤

𝑉𝑤

ρ𝑣

ρ𝐿

= 2.8

1.7

4.541

1455.8

= 0.092

From the graph, 𝐾4 = 1.6

At flooding, 𝐾4= 3.8

Percentage flooding = 1.6

3.8 = 0 .6488

64.88% satisfactory

Viscosity calculation

Similar to above calculation, T = 340.5K

For carbon disulphide

µ1= 0.273mNs/𝑚2

For carbon tetrachloride

µ2= 0.536mNs/𝑚2

Average viscosity of mixer,

µ𝑚𝑖𝑥 = (0.22 × 0.2731

3 + 0.78 × 0.5361

3)3

= 0.468 mNs/𝑚2

𝑉𝑤 = [3.8×4.541(1455.8−4.541)

13.1×170(0.468/1455.8)0.1]

= 24.635kg/𝑚2𝑠

Column area required = 1.7

24.635 = 0.069𝑚2

Diameter = 4

𝜋× 0.069

Diameter of the stripping section= 0.296m

Feed tray location

Feed is given to the 7th tray from the bottom,

So, it can be assumed as feed is fed into column at 7.5th

stage.

Height of the feed location from the bottom of column = 7.5*HETP

= 3.45 m

Condenser heat load

Latent heat calculation

Latent heat of C𝑆2 = 351kJ/kg

Latent heat of C𝐶𝑙4 = 194kJ/kg

𝐿𝑣,𝑏 𝑇𝑏 𝑇𝑐

C𝑆2 26711.1 319.3 552

C𝐶𝑙4 29837.2 349.8 556.35

For top products

Temperature of top plate = 322K

Carbon disulphide,

𝐿𝑣 = 26711.1× [552−322

552−319.3]0.38

= 26593kJ/kmol

Carbon tetrachloride,

𝐿𝑣 = 29837.2× [556.35−322

556.35−349.8]0.38

= 31303.8kJ/kmol

Latent heat of mixtures

For rectifying section,

𝐿𝑣,𝑚𝑖𝑥 = 26593×0.9 + 31303.8×0.1

= 27064.08 kJ/kmol

Therefore condenser heat load = G× 𝐿𝑣,𝑚𝑖𝑥 = 47.125×27064.08

= 1275.4MJ/hr

= 354.3kW

Boiler heat load calculation

Using equation,

𝑉𝑚𝐻𝑣 ,𝑚 + W𝐻𝑤 = 𝐿𝑚−1𝐻𝐿 ,𝑚−1 + 𝑄𝑅

𝐻𝑣 = 𝐻𝐿 + 𝜆𝑚𝑖𝑥

𝐻𝐿 = (𝐶1 + 𝐶2𝑇 + 𝐶3𝑇2 + 𝐶4𝑇

3)𝑑𝑇𝑇2

𝑇1

𝐶1 𝐶2 𝐶3 𝐶4

C𝑆2 85600 -122 0.5605 -0.001452

C𝐶𝑙4 -752700 8966.1 -30.394 0.034455

Average temperature of stripping section, T = 340.5K

Calculating the latent heat of mixture

For stripping section

Temperature of bottom plate = 348K

For Carbon disulphide,

𝐻𝐿 = (𝐶1 + 𝐶2𝑇 + 𝐶3𝑇2 + 𝐶4𝑇

3)𝑑𝑇𝑇2

𝑇1

= (85600 − 122T + 0.5605𝑇2 − 0.001452𝑇3)𝑑𝑇340.5

298

= 2397.325kJ/kmol

𝐿𝑣 = 26711.1× [552−340.5

552−319.3]0.38

= 25758.8kJ/kmol

Carbon tetrachloride,

𝐻𝐿 = (−752700 + 8966.1T − 30.394𝑇2 + 0.034455𝑇3)𝑑𝑇340 .5

298

= 5671.523kJ/kmol

𝐿𝑣 = 29837.2× [556.35−340.5

556.35−349.8]0.38

= 30340.7kJ/kmol

For stripping section,

𝐿𝑣,𝑚𝑖𝑥 = 25758.8×0.22 +30340.7×0.78

Latent heat of mixture = 29332.7 kJ/kmol

Enthalpy of liquid mixture(𝐻𝐿 ,𝑚𝑖𝑥 ) = 2397.325×0.22 + 5671.523×0.78

= 4951.2 kJ/kmol

Enthalpy of carbon disulphide, 𝐻𝑣= 2397.325 + 29332.7

= 31730kJ/kmol

Enthalpy of carbon disulphide, 𝐻𝑣= 5671.523 + 29332.7

= 35004.2kJ/kmol

Enthalpy of vapour mixture (𝐻𝑣,𝑚𝑖𝑥 ) = 31730×0.325 + 35004.2×0.675

= 33940.1 kJ/kmol

For bottom product (residue),

Temperature = 348K

For Carbon disulphide,

𝐻𝐿 = (85600 − 122T + 0.5605𝑇2 − 0.001452𝑇3)𝑑𝑇348

298

= 2778.2kJ/kmol

Carbon tetrachloride,

𝐻𝐿 = (−752700 + 8966.1T − 30.394𝑇2 + 0.034455𝑇3)𝑑𝑇348

298

= 6703.6kJ/kmol

Enthalpy of liquid mixture of residue(𝐻𝑤 ) = 2778.2 ×0.04 + 6703.6×0.96

= 6546 kJ/kmol

Substituting to equation,

𝑉𝑚𝐻𝑣,𝑚 + W𝐻𝑤 = 𝐿𝑚−1𝐻𝐿 ,𝑚−1 + 𝑄𝑅

47.569×33940.1 + 26.163×6546 = 73.732×4951.2 + 𝑄𝑅

𝑄𝑅 = 1420.7MJ/hr

= 394.6kW

Heat load of the reboiler = 394.6kW

Final data

Packing material plastic

Packing size 25mm

Pressure drop 62water mm/ m of packing

Reflux ratio 1.5262

Number of ideal stages 11

Column height 5.1m

Diameter of rectifying 0.388m

Diameter of stripping 0.296m

Feed tray location 3.45m from bottom

Condenser heat load 354.3kW

Reboiler heat load 394.6kW

Referencess

http://www.engineeringtoolbox.com/fluids-evaporation-latent-heat-d_147.html

http://www.britannica.com/EBchecked/topic/94993/carbon-disulfide-CS2

http://encyclopedia2.thefreedictionary.com/Carbon+Tetrachloride

CULSON & RICHARDSON’S, CHEMICAL ENGINEERING, Volume 6, 3rd

edition, R K

Sinnott

CULSON & RICHARDSON’S, CHEMICAL ENGINEERING, Volume 2, 5th edition

PERRY’S CHEMICAL ENGINEERS HAND BOOK, 8th edition, DON W. GREEN, ROBERT

H. PERRY

DISTILLATION DESIGN, HENRY Z. KISTER