[distillation] - separation tower design [daniel.r.lewin]

054410 PLANT DESIGNLECTURE NINE

Daniel R. Lewin, Technion1

Separation Tower DesignPLANT DESIGN - Daniel R. Lewin1 9 -

054410 Plant DesignLECTURE 9:

SEPARATION TOWER DESIGN

Daniel R. LewinDepartment of Chemical Engineering

Technion, Haifa, IsraelRefs: Seider, Seader and Lewin (2004), Chapters 14 and 16

Seader and Henley “Separation Process Principles” (1998), Chaps. 6 and 7Kister, “Distillation Design” (1992), Chaps. 6 and 7


Lecture Objectives

Familiar with the constraints affecting the performance of trayed distillation column.Able to estimate the efficiency of a trayed distillation columnAble to compute the optimal diameter of a trayed distillation column.Able to define all of the dimensions of a distillation column, including the minimum wall thickness.

After this lecture, you should be:

b) http://lorien.ncl.ac.uk/ming/distil/distil0.htm

For a review of distillation, see:a) Multimedia section on HYSYS-Separations




Distillation Column Design OverviewSteps involved:

Selection of operating pressure, to allow the usage of cooling water for condenser, if possible.

R,I S,IT,IN N N= +Short-cut method used to estimate RR, number of ideal stages, , and location of feed tray.

Rigorous solution of material and energy balances to meet the number of specifications = DOFs.

R,A R,I 0 S,A S,I 0N N E and N N E= =Estimate tray efficiency, E0, and number of actual trays:Estimate tower height, diameter, and wall thickness.

It is assumed that you are familiar with steps , and . This lecture focuses on steps and .

Selection of operating pressure, to allow the usage of cooling water for condenser, if possible.

R,I S,IT,IN N N= +Short-cut method used to estimate RR, number of ideal stages, , and location of feed tray.

Rigorous solution of material and energy balances to meet the number of specifications = DOFs.


Focus of this lecture…

The focus of this lecture is on the additional details required to permit the mechanical design of multicomponent separation towers.




A Look Inside a Distillation Column

Outlet weir

Downcomer

Active tray area

Liquid

Vapor

a a a a


Bubble-caps, Valves or Sieves…

Bubble-cap tray Valve tray Sieve tray

245Typical turndown ratioHighestHighestLowestVapor capacityLowestHighestHighestEfficiencyLowestIntermediateHighestPressure drop

1.01.22.0Relative costSievesValvesBubble-caps




Bubble-cap tray




Valve tray




Sieve tray



Adverse vapor/liquid flow conditions can cause:

Foaming Entrainment FloodingWeeping/dumpingDowncomer flooding

Tray Performance Constraints




Tray Performance ConstraintsFoaming

Foaming refers to the expansion of liquid due to passage of vapor or gas, caused by high vapor flow rates.Although it provides high interfacial liquid-vapor contact, excessive foaming often leads to liquid buildup on trays. In some cases, foaming may be so bad that the foam mixes with liquid on the tray above. Whatever the cause, separation efficiency is always reduced.


Tray Performance ConstraintsEntrainment

Caused by excessively high vapor flow rates.Entrainment refers to the liquid carried by vapor to the tray above.It is detrimental because tray efficiency is reduced: lower volatile material is carried to a plate holding liquid of higher volatility.Excessive entrainment can lead to flooding.




Tray Performance ConstraintsFlooding

Flooding is brought about by excessive vapor flow, causing liquid to be entrained in the vapor up the column.The increased pressure from excessive vapor also backs up the liquid in the downcomer, causing an increase in liquid holdup on the plate above.Depending on the degree of flooding, the maximum capacity of the column may be severely reduced.Flooding is detected by sharp increases in column differential pressure and significant decrease in separation efficiency.


Tray Performance ConstraintsWeeping/Dumping

Caused by excessively low vapor flow.The pressure exerted by the vapor is insufficient to hold up the liquid on the tray. Therefore, liquid starts to leak through perforations.Excessive weeping will lead to dumping - the liquid on all trays will crash (dump) through to the base of the column (via a domino effect) and the column will have to be re-started.Weeping is indicated by a sharp pressure drop in the column and reduced separation efficiency.




Tray Performance ConstraintsDowncomer Flooding

Caused by excessively high liquid flow and/or a mismatch between the liquid flow rate and the downcomer area. This can be avoided by ensuring that the downcomer back-up (level) is below 50% of the tray spacing. This can be checked by performing tray sizing using a process simulator.If necessary, design multipass trays (see later).


Tray Efficiency EstimationThe actual number of trays required for a particular separation duty is determined by the efficiency of the plate.Any factors that cause a decrease in tray efficiency will also change the performance of the column.Tray efficiencies are affected by fouling, wear and tear and corrosion, and the rates at which these occur depends on the properties of the liquids being processed. Thus appropriate materials should be specified for tray construction.




Empirical Efficiency Estimation

O’Connell correlation: ( ) 0.2450.492 10%O LE −= µ α ±viscosity

relative volatility at average column conditions of key component

Lµ = ⎫⎪α = ⎬⎪⎭


Example 1: Tray Efficiency CalculationEstimate the tray efficiency for the simulated column shown in the table below.

1.945LHα =




Example 1: Tray Efficiency CalculationSolution.The average column temperature is (70 + 309)/2 = 190 oF. The closest match to this temperature is at stage 8, at which the viscosity is 0.133 cP (note that the viscosity does not change all that much over the entire column).

( )( )

0.245

0.245

Hence, 0.492

0.492 0.133 1.9450.69

O LE −

−

= µ α

= ×

=Given that the estimate is subject to ±10% error, a reasonable estimate would be 0.62. Thus, the total number of trays will be:

18/0.62 = 29 trays29 7 18 11 trays in the rectifying section29 11 18 18 trays in the stripping section

× =⎧⎨ × =⎩


Tray Section CapacityDefining column diameter.

Most of the factors that affect column operation are due to vapor flow conditions: either excessive or too low. Vapor flow velocity is dependent on column diameter. Weeping determines the minimum vapor flow required while flooding determines the maximum vapor flow allowed, hence column capacity.If the column diameter is not sized properly, the column will not perform well. Not only will operational problems occur, the desired separation duties may not be achieved.




Estimating Flooding VelocityThe flooding velocity is computed based on a force balance on a suspended liquid droplet. This is the critical velocity at which liquid droplets become suspended, a result of a perfect balance between gravitational, buoyant and drag forces (Sounders and Brown, 1934):

( )3

6

dpL

gravity

gπρ

fu

drag buoyancy

gravity1 2

L Gf

Gu C ⎛ ⎞ρ − ρ

= ⎜ ⎟ρ⎝ ⎠

4where

3p

D

d gC

C⎛ ⎞

= ⎜ ⎟⎝ ⎠

( )3

6

dpG

buoyancy

gπ−ρ ( )22

40

2fdp

GD

drag

uC π

ρ− =

Solving for flooding velocity:


Estimating Flooding Velocity

1, for non-foaming systems (e.g., most distillation applications)0.5-0.75, for foaming systems (e.g., absorption with heavy oils.

FF⎧⎪⎪= ⎨⎪⎪⎩

( )ah

a ah h

1, for A A 0.15 A A 0.5, for 0.06 A A 0.1 HAF

≥⎧= ⎨ + ≤ <⎩

In practice, C is treated as an empirical parameter determined using experimental data.

SB F HASTC C F F F=

( )0.2and 20 , liquid surface tension [dyne/cm]STF = σ σ =

where CSB is an empirical function of the ratio:( )LG G LF L G= ρ ρ




Estimating Flooding Velocity

( )1 2

0.220 L GSB F HAf

Gu C F F ⎛ ⎞ρ − ρ

= σ ⎜ ⎟ρ⎝ ⎠


Tower inside cross-sectional area, AT, is computed at a fraction f (typically 0.75-0.85) of the vapor flooding velocity, uf :

Tray Section Capacity

( ) ( ) GTf dG fu A A= − ρ (14.10)

( ) ( )

1/2

TTd

4D1 A A Gf

Gfu

⎡ ⎤= ⎢ ⎥

π − ρ⎢ ⎥⎣ ⎦ (14.11)

( )2T T

T

Substituting A D 4 into Eq.(14.10) and solving for D :

= π

( )d

T

0.1 , 0.10.1A 0.1 , 0.1 1.0

A 90.2 , 1.0

LG

LGLG

LG

FF F

F

≤⎧ ⎫⎪ ⎪−⎪ ⎪= + ≤ ≤⎨ ⎬⎪ ⎪

≥⎪ ⎪⎩ ⎭

( )LG G LF L G= ρ ρ




Selection of Multipass Trays


Example 2: Tray Diameter CalculationCompute the diameter of a valved-distillation column with the following data - Liquid phase: = 7.1 dyne/cm, L = 215,000 lb/hr, ρL =32.4 lb/ft3, Vapor phase: G = 244,000 lb/hr, ρG =1.095 lb/ft3.Solution.FLG = (215,000/244,000)(1.095/32.4)0.5 = 0.162

FF = 1 (no foaming), FHA = 1 (valves), so:From slide 9-23 , for 24” tray spacing, CSB = 0.09 m/s

( ) ( ) ( )0.2 32.24 1.0950.09 7.1 20 1 1 0.39 m/s 4,610 ft/hr1.095fU −

= = =

Assuming operation at 80% flooding: ( )TdA A 0.1 0.1 9 0.107.LGF= + − =

( )( ) ( )

1/2

T4 244,000

D 9.3 ft0.8 4,610 1 0.107 1.095⎡ ⎤

= =⎢ ⎥π −⎣ ⎦Note: (a) In general, diameters in rectifier and stripper may differ.

(b) If DT < 2 ft, use a packed column.

Slide 9-23




Example 2: Tray Diameter CalculationFor this large diameter column, we should consider installing a multipass tray. Recall from data: L = 215,000 lb/hr and ρL =32.4 lb/ft3

= 4.33 lb/gal( )Volumetric flow rate 215,000/60 / 4.33 828 gpm= =

828

9.3


Compute the tray diameter for the simulated column shown in the table on slide 9-18 . Assume valve trays and light hydrocarbon service.

Example 3: Tray Diameter Calculation

Solution.The first thing we need to do is to identify the critical tray in both the rectifier and stripping sections, defined as the trays in which the loads for each section are maximized.

FF = 1 (no foaming), FHA = 1 (valves) Slide 9-18

Rectifier Section - based on Stage 3Stripper Section - based on stage 19




Example 3: Tray Diameter Calculation

FLG = (85,360/121,184)(2.478/27.944)0.5 = 0.2098

( ) ( ) ( )0.2 27.979 2.4780.21 3.3 20 1 1 0.47 m/s 5,551 ft/hr2.478fU −

= = =

Assuming operation at 80% flooding: ( )= + − =TdA A 0.1 0.1 9 0.112.LGF( )

( ) ( )⎡ ⎤

= =⎢ ⎥π −⎣ ⎦

1/2

T,R4 121,184

D 3.97 ft0.8 5,551 1 0.112 2.478

Rectifier Section (based on Stage 3).

Stripper Section (based on stage 19).FLG = (185,434/129,112)(3.614/27.191)0.5 = 0.5236

( ) ( ) ( )0.2 27.191 3.6140.16 2.84 20 1 1 0.277 m/s 3,272 ft/hr3.614fU −

= = =

=TdA A 0.147. Hence, ( )( ) ( )

⎡ ⎤= =⎢ ⎥π −⎣ ⎦

1/2

T,S4 129,112D 5.40 ft

0.8 3,272 1 0.117 3.614

Since the difference more than 20%, note that the rectifier diameter is 4 ft and the stripper diameter is 5.5 ft (to nearest ½’).


Estimating Column Pressure DropTypically, tray pressure drop for flow of vapor in a tower is between 0.05-0.15 psi/tray.For a sieve tray, the head loss is due to the friction for vapor flow through the tray perforations, the holduo of the liquid, and the loss due to surface tension:

t dh h h hσ= + + total pressure drop [in] dry tray pressure drop [in] equivalent head on tray [in] pressure drop due to s.t. [in]

t

d

hhhhσ

=

=

=

=




Estimating Column Pressure DropDry sieve tray pressure drop is computed using a modified orifice equation:

0

0

hole velocity [ft/s]depends on percent hole area and the ratio of tray thickness

to hole diameter. Range: 0.65-0.85. Typical value: 0.73.

uC

=

−

2

20.186 o Gd

o L

uhC

⎛ ⎞⎛ ⎞ ρ= ⎜ ⎟⎜ ⎟ ρ⎝ ⎠⎝ ⎠


Estimating Column Pressure Drop

( )

=φ =

= −

⎛ ⎞ρ= = ⎜ ⎟ρ − ρ⎝ ⎠=

0.91

1 2

weir height [in]effective relative froth density (ht. of clear liquid/froth height)exp 4.257

capacity parameter [ft/s]

superficial vapor velocity [ft/s] based on

w

e

S

GS a

L G

a

h

K

K u

u

( )

= −= === + −

a

a

W T

L

active bubbling area, AAL weir length [in] (for 0.1, taken as 73% of D )q liquid flow rate across tray [gal/min]C 0.362 0.317 exp 3.5

T d

Td

W

A AA A

h

Equivalent height of clear liquid holdup on tray: 2 3

Le w

w e

qh h CL

⎡ ⎤⎛ ⎞= φ +⎢ ⎥⎜ ⎟φ⎢ ⎥⎝ ⎠⎣ ⎦




Estimating Column Pressure DropAs the gas emerges from the tray perforations, the bubbles must overcome surface tension. The pressure drop due to the surface tension is given by the difference between the pressure inside the bubble and that due to the liquid:

( )max

6L B

hg Dσ

σ=

ρ

Generally, the maximum bubble diameter, DB(max), may be taken as the tray hole diameter.


Example 4: Estimating Tray ∆P

Solution.

Estimate the tray vapor pressure drop for a 1m diameter absorber equipped with sieve trays. Given: hw = 2”, DH = 3/16”Liquid phase: σ = 70 dyne/cm, L = 2,883 kg/hr, ρL = 986 kg/m3

Vapor phase: G = 7,920 kg/hr, ρG = 1.92 kg/m3.

At the bottom of the tower, vapor velocity based on the total cross-sectional area of the tower is:

( ) ( )27,920 3,600 1.46 m/s1.92 1 4

=π

For a 10% hole area, based on total cross-section of the tower:

01.46 14.6 m/s 47.9 ft/s0.1

u = = =

2

247.9 1.92Hence, 0.186 1.56"0.73 986dh ⎛ ⎞ ⎛ ⎞= =⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠




Example 4: Estimating Tray ∆PSolution (cont’d).Taking weir length as 73% of DT gives LW = 0.73 m = 28.7”

= =×L

2,883 60Liquid flow rate in gpm, q 12.9 gpm986 0.003785

Ad/AT = 0.1, Aa/AT = 0.9, so ua = 1.46/0.9 = 1.62 m/s = 5.32 ft/s.

( )( )⎡ ⎤= φ + φ =⎣ ⎦2 3Hence, 0.67"e w L w eh h C q L

( )( )⇒ = ρ ρ − ρ =1 2 0.235 ft/sS a G L GK u

( )⇒ φ = − =0.91 exp 4.257 0.32e SK

( )C 0.362 0.317 exp 3.5 0.362Wh= + − =


Example 4: Estimating Tray ∆PSolution (cont’d).Assuming DB(max) = DH = 3/16” = 0.00476 m

σ = 70 dynes/cm = 0.07 N/m = 0.07 kg/s2 and g = 9.8 m/s2

( )max

6Hence, 0.000913 m 0.36"L B

hg Dσ

σ= = =

ρ

Thus, total head loss/tray, σ= + +

= + +

=

t dh h h h1.56 0.67 0.362.59"

Recalling that ρL = 986 kg/m3 = 0.0356 lb/in3

Thus, the tray vapor pressure drop = htρL = 0.092 psi




Complete Column Sizing

Skirt

Stripping Section

Rectifying Section 4 ft

2×Nr ft

2×Ns ft

≥ 10 ft

Sump

Disengagement

Dr ft

Ds ft

Maximum height of column = 175’, Maximum L/D ratio = 30


ASME Pressure Vessel CodeIn the absence of wind and earthquake conditions and excluding vacuum operation:

14,7501%Cr, 0.5%Mo Steel, SA-387B to 800 oF with H2

13,1001%Cr, 0.5%Mo Steel, SA-387B to 900 oF with H2

15,0001%Cr, 0.5%Mo Steel, SA-387B -20-750 oF with H2

13,750CS SA-285, grade C-20-650 oF no H2

S (psig)Recommended materialConditions

12 (16.60)2 1.2

Idp

d

P DTS E P

⋅=

⋅ ⋅ − ⋅ wall thickness [in] to withstand internal pressure internal design pressure [psig]

inside shell diameter [ft] maximum allowable stress at design temp. [psig] weld integrity ( 0.85 f

p

d

I

TPDSE E

===== = or wall thicknesses < 1.25". A value of 1 is used for thicknesses more than 1.25")




ASME Pressure Vessel CodeFor vertical vessels, the vessel walls need to withstand wind load, computed using:

( ) 2

2

0.22 18 o

wo

D LTSD

+=

where Do is outside shell diameter (inches), L is vessel height (tangent to tangent length, in inches), and the factor of 18 allows for the column cage ladders, which adds additional effective diameter to the column.When there is wind load, the girth seam must withstand the combined load of the wind and the internal pressure, the latter computed using: 12

2 0.4g I

gg

P DT

SE P=

+An estimate for the required wall thickness at the bottom of the tower is then: Tb = Tw + Tg


ASME Pressure Vessel CodeTo estimate the vessel thickness (assumed constant), use the average of the top and bottom thicknesses, plus the corrosion allowance, Tc, usually 0.125". Thus the values of wall thickness are computed as follows:

s p cT T T= +HORIZONTAL VESSELS.

VERTICAL VESSELS. ( )0 5s p cbT . T T T= + +




ASME Pressure Vessel CodeAt low pressures, wall thickness computed using the above equations may be too small to give sufficient rigidity to vessels. The minimum wall thickness below should be used.

3/86-85/164-6

1/210-127/168-10

1/4Up to 4Minimum value for tp [in]DI [ft]

Finally, the values computed need to be rounded up to the nearest standard plate thickness, as given by the table below:

1/831/162

1/4>3

1/321Rounding increment [in]Ts up to [in]


Example 5: Wall Thickness Calculation

( ) 2

2

0 22 10 12 18 175 123 10 12 1 2113 750 10 2 13 750 1 0 0 4 123b

.T . ", , . .× + × ×

= + =× × × + ×

Compute the wall thickness for a distillation column with height175 ft and inside diameter 10 ft. The operating pressure is 110 psia and 150 oF at the bottom of the tower and 100 psia and 120 oF at the top. Material of construction is CS.Solution. Design basis: Pd = 1.2×max P = 1.30×(110-14.7) = 123 psig

Td = max T + 50 oF = 200 oF.

120 123 0.635"2 13,750 0.85 1.2 123pT ×

= =× × − ×

Using Eq. (16.60) assuming CS shell:

The vessel thickness at the bottom of the tower is:

Thus: ( ) ( )0 5 0 5 1 21 0 635 0 125 1 049s p cbT . T T T . . . . . "= + + = + + =

Rounding up, this gives Ts = 1.0625” (1 1/16”)




Familiar with the constraints affecting the performance of trayed distillation column.Able to estimate the efficiency of a trayed distillation column.Able to compute the optimal diameter of a trayed distillation column.Able to define all of the dimensions of a distillation column, including the minimum wall thickness.

SummaryAfter reviewing the materials of this lecture, you should be:

[distillation] - separation tower design [daniel.r.lewin]

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