Absorption stripping

Absorption & Stripping• Introduction • Graphical Methods • Packed Towers

SHR Chapter 6

Introduction

Trayed tower

Goals: • minimize mass transfer resistance to

achieve equilibrium on each tray • minimize bubble carry-over to tray below • minimize liquid entrainment to tray above • minimize “weeping” of liquid through

holes in tray

SHR §6.1

Types of Traysperforated valve cap bubble cap

Regimes in a Trayed Tower

“Spray” gas phase is continuous

(low liquid depths, high gas flow rates)

“Froth” gas passes through liquid as jets or a series of bubbles

“Emulsion” !

“Bubble” low gas flow rates - swarms of bubbles

“Cellular Foam” (think blowing bubbles

in chocolate milk)

SHR §6.1.1

Packed Towers

Stru

ctur

ed P

acki

ng

Uns

truc

ture

d Pa

ckin

gSHR §6.1.2

Other Configurations

Spray tower • very low pressure drop • use for absorption only when

solute is highly soluble in the liquid (e.g. SO2 in flue gas)

Bubble column • absorption • high pressure drop • use when solute is poorly

soluble in liquid • use when slow chemical

reactions occur that require long residence time

Centrifugal contactor • short residence time • compact

SHR §6.1.3

Heuristics & Design Considerations

Trayed towers • “reliable” design

• low liquid velocities

• liquid phase (the continuous phase) is typically mass-transfer limiting

Unstructured Packed towers • corrosive environments

• small towers (<2 ft diameter, <20 ft tall)

• when foaming problems tend to occur

• gas phase (continuous phase) is typically mass-transfer limiting

Structured packed towers • (expensive!)

• retrofit when trying to get more efficiency

• low pressure drop

• gas phase (continuous phase) is typically mass-transfer limiting

Analysis Approach

Trayed towers: • analyze each tray as an equilibrium problem ‣what assumption here???

• write coupled equations for mass & energy balances between trays

Other towers (packed, etc.): • Height Equivalent of a Theoretical Plate (HETP) ‣ effective height acts as one tray

‣ vendors of packing report this value

Graphical Methods

SHR §6.3

Some Terminology

Stripper Solute enters in liquid. Stripping agent enters

bottom of column.

L′ Molar flow rate of solute-free liquid V′ Molar flow rate of solute-free gas x mole fraction of solute in liquid y mole fraction of solute in gas X mole ratio of solute to solute-free liquid Y mole ratio of solute to solute-free gas

Assume that only solute is transferred from one phase to another (no vaporization of liquid or condensation of gas carriers).

What does XL′ and YV′ represent?

X =x

1� x

Y =y

1� y

Absorber Solute enters in gas.

Liquid absorbent enters from top of column.

Note different tray ordering convention...

Stage "i"

Xi

Xi-1 Yi

Yi+1

For the absorber:

Ki =yi

xi=

Yi/(1+Yi)

Xi/(1+Xi)

Streams leaving the tray are assumed to be in equilibrium

Mole Balances & Operating LinesSolute balance around arbitrary # of trays in the “top” section of the absorber:

X0L0 + Yn+1V

0 = XnL0 + Y1V

0

solute flow rate in solute flow rate out

Yn+1 = (Xn �X0)L0

V 0 + Y1

L0

V 0 = “slope” of operating line

(absorber)

(stripper)

What happens as ?L0/V 0 ! 1

Absorber Solute enters in gas.

Liquid absorbent enters from top of column.

We know these.

What happens as gets small?L0/V 0

Yn = (Xn+1 �X1)L0

V 0 + Y0

SHR §6.3.2

For an absorber, we typically know YN+1, X0 and Vʹ. Therefore, we get to choose Lʹ to achieve desired Y1.

design variable

Absorber: Minimum Flow RateYn+1 = (Xn �X0)

L0

V 0 + Y1

Absorber Solute enters in gas.

Liquid absorbent enters from top of column.

L0 =V 0 (YN+1 � Y1)

XN �X0

Over the whole tower (n=N):

L′min corresponds to equilibrium with XN and YN+1.

KN =yN+1

xN=

YN+1/(1+YN+1)

XN/(1+XN )

For dilute solutes (Y ≈ y and X ≈ x):

If X0 ≈ 0 then:

L

0min = V

0✓

yN+1 � y1

yN+1/KN � x0

◆

Corresponding equation for a

stripper:

L0min =

V 0 (YN+1 � Y1)YN+1/[YN+1(KN�1)+KN ] �X0

As V′ ↑, L′min↑.

SHR §6.3.3

Why are these in equilibrium?

L0min = V 0KN · (fraction absorbed)

V 0min =

L0

KN· (fraction stripped)

This is the “best” we can achieve given the inlet constraints.

Examples

Given: feed stream composition & flow rates, recovery (Y1)

Find: XN.

Given: feed stream composition & flow rates,

solvent loading (XN) Find: Y1.

Given: feed stream composition, gas flow rate, solvent loading (XN), and recovery (Y1),

Find: solvent flow rate.

Yn+1 = (Xn �X0)L0

V 0 + Y1

X

Y

XNX0

YN+1

Y1 Y = KX

L'

V'

XY

XNX0

YN+1

Y1Y =

KX

L'

V'X

Y

XNX0

YN+1

Y1

Y = KX

L'

V'

Number of “Equilibrium” StagesSHR §6.3.4

1. Locate the point for the solvent feed (X0) and desired Y1 on the graph.

2. Determine operating line from Vʹ and Lʹ. 3. March off to determine the stages

(assuming each stage is in equilibrium)

Stage "i"

Xi

Xi-1 Yi

Yi+1

For the absorber:

• Streams leaving the tray are assumed to be in equilibrium

• Streams passing one another are on the operating line.

Yn+1 = (Xn �X0)L0

V 0 + Y1

equilibrium relates Yi and Xi.

Operating line:

What happens to the number of stages as Lʹ/Vʹ

approaches ∞ or its minimum?

Algebraic ApproachYn+1 = (Xn �X0)

L0

V 0 + Y1

Operating line:

Given: X0, Y1, Lʹ/Vʹ, 1. K1 = Y1/X1 → solve for X1.

2. Find Y2 from the operating line.

3. K2 = Y2/X2 → solve for X2.

4. Find Y3 from the operating line. ⋮

Must have a model for Ki.

If Ki is not a function of composition: 1. Calculate Ki at given T and P. 2. Follow steps outlined above.

If Ki is a function of composition: 1. Guess Xi. (note that Yi is known from

operating line). 2. Calculate Ki.

3. Update guess for Xi and return to step 2 if not converged.

SHR §6.4 presents an alternative to this formulation.

Number of Stages for StrippersYn = (Xn+1 �X1)

L0

V 0 + Y0

Stage EfficiencyComplex function of: • tray design/geometry

• fluid dynamics on trays

Typically less than 50% efficient (10%-50%) • trays are not at equilibrium!

• more viscous liquids typically lead to lower efficiencies (inhibit mass transfer)

logEo

= 1.597� 0.199 log

✓KM

L

µL

⇢L

◆

� 0.0896

log

✓KM

L

µL

⇢L

◆�2

Empirical correlation for stage efficiency

Data over a wide range of column diameters, pressures, temperatures and liquid viscosities.

Eo

⌘ Nt

Na

# theoretical (equilibrium) stages

# actual stages

Other (less empirical) methods exist - see SHR §6.5.4.

Packed ColumnsSHR §6.7

SHR §6.7

Analysis Options

Option 1: graphical techniques • HETP is known ‣HETP = (height) / (number of theoretical equilibrium stages)

‣Use methods previously discussed to get number of trays/stages ‣ solve for height given number of stages

• HETP is typically found empirically & supplied by packing vendors. !

Option 2: rate-based techniques • Use mass transfer coefficients (and a few hefty assumptions)

• See SHR §6.7 for more details.

lT = HETP ·Nt

Operating Lines

x

in

L

in

+ yV` = xL` + y

out

V

out

y = x

✓L

V

◆+ y

out

� x

in

✓L

V

◆

Solute mole balance:

For dilute solutions, V and L are approximately constant:

Packed absorber operating line

y = x

✓L

V

◆+ y

in

� x

out

✓L

V

◆

Solute mole balance:

For dilute solutions, V and L are approximately constant:

xL` + y

in

V

in

= x

out

L

out

+ yV`

Packed stripper operating line

Here, x and y are bulk compositions.

Finite-Rate Mass Transfer (Back to Two-Film Theory)

Often we don’t know the surface area for mass transfer from all of the packing.

r = mass transfer rate per unit volume,

a = surface area per unit volume of packing

mol/(m3·s)

r = Ja = k

y

a(y � yI)

= k

x

a(xI � x)

y = yI �k

x

a

k

y

a

(x� xI)

relative resistance of mass transfer between

the two phases

Gas Interface Liquid

liquid film composition

gas film composition yI or pI

xI or cI

bulk liquid composition

bulk gas compositiony or p

x or c

x*

y*

r = K

y

a (y � y

⇤) = K

x

a (x⇤ � x)Overall mass transfer coefficient approach:

1

Ky

a=

1

ky

a+

K

kx

a1

Kx

a=

1

kx

a+

1

Kky

a

J = ky(y � yI)

y = yI �k

x

a

k

y

a

(x� xI)AB line:

Making Connections...

y = x

✓L

V

◆+ y

out

� x

in

✓L

V

◆

Applicable for small x, y.

Gas Interface Liquid

liquid film composition

gas film composition yI or pI

xI or cI

bulk liquid composition

bulk gas compositiony or p

x or c

x*

y*

Operating line:

Mole fraction solute in liquid, x

Mol

e fra

ctio

n so

lute

in g

as, y

Equilibrium

curve, y=Kx

Operating line

A

B

CF

E

D(y⇤, x)

(yI, xI)

(y, x⇤)

(y, x)y

=x

✓ L

V

◆ +y

o

u

t

� x

i

n

✓ L

V

◆

N

A

= K

x

(x⇤A

� x

Ab)

= K

y

(yAb � y

⇤A

)

1

Kx

=1

KA

ky

+1

kx

1

Ky

=1

ky

+K

A

kx

x

⇤A =

yAb

KALiquid mole fraction

Gas mole fraction y

⇤A = xAbKA

From Mass-transfer notes on two-film theory:

What do AE and AF represent?

comes from equating “r” in

bulk and interface.

HOG - NOG (Getting the height)Material balance over dl:

Change in gas phase: V (y + dy) - VyTransfer to liquid phase: Kya (y - y*) S dl �V dy = Kya(y � y⇤)S dl

KyaS

V

Z lT

0dl =

KyaSlTV

=

Z yin

yout

dy

y � y⇤

lT =V

KyaS| {z }HOG

Z yin

yout

dy

y � y⇤| {z }

NOG

HOG: Height of a (gas) transfer unit (HTU) NOG: Number of (gas) transfer units (NTU)

For y* = Kx (constant K), and linear operating line (dilute solute),

A = L/(KV )

Given Ky (overall gas-phase MTC), flow rates (L/V), and K, we can get lT.

NOG =A

A� 1ln

⇢(A� 1) (y

in

�Kx

in

)

A (yout

�Kx

in

)+ 1

/A

�

A - “Absorption factor” see SHR §5.4.1 for derivation of A.

What leaves the gas, goes to the liquid:

HOG, NOG, Nt & HETPHOG: Height of a (gas) transfer unit (HTU)

NOG: Number of (gas) transfer units (NTU)

Nt: Number of theoretical stages

HETP: Height-equivalent of a theoretical plate

A = L/(KV )

Nt = NOGA

1�Aln (1/A)

Mole fraction solute in liquid, x

Mol

e fra

ctio

n so

lute

in g

as, y

Equilibrium

curve, y=Kx

Operating line

A

B

CF

E

D(y⇤, x)

(yI, xI)

(y, x⇤)

(y, x)y

=x

✓ L

V

◆ +y

o

u

t

� x

i

n

✓ L

V

◆

Note: if operating & equilibrium curves are parallel then L/V = K.

Then what is the relationship between HETP and HOG?

HETP = HOGA

1�Aln (1/A)

Example

L, xin V, yout

V, yin L, xout

3,500 lbmol/hr water

2,500 lbmol/hr 2% ethylene oxide in

inert gas

T=30 ºC P=20 atm

!• Two 12’ sections of 1.5” metal Pall ring packing • 4’ diameter column • K = 0.85 for ethylene-oxide • kya = 200 lbmol/(h-ft3) • kxa = 165 lbmol/(h-ft3)

Find HETP for this packing.

HOG =V

KyaS

HETP = HOGA

1�Aln (1/A) A = L/(KV )

1

Ky

a=

1

ky

a+

K

kx

a