Gas Absorption & Stripping in Chemical Engineering (Part 2/4)

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1. Introduction to Tray Columns
2. Equipment used in Tray Columns
 Tray Model (Sieve, Bubble-cap, valve)
 Tray Considerations
 All Other
3. Design & Model
 Counter-Current Dilute Model
 Equilibrium Line
 Operation Line
 Other Cases
4. Number of Equilibrium Stages
 Theoretical Number of Stages
 Graphical Method
 Algebraic Method (Kremser Equation)
 Stage Efficiency
 Flooding, Diameter & Pressure Drop

 The function of a tray is to facilitate contact between the vapor phase and liquid
phase so that mass transfer between the 2 phases can take place.
 A physical tray is designed to act as an equilibrium stage
 Single-Stage Operation vs. Multiple-Stage Operation

 Typical Absorption Units may have from 1-20 stages
 Each tray can be considered as being made up of 3 sections:
 weir, bubbling area, and downcomer
 Typically, 3 common types of trays:
 Sieve
 Valve
 Bubble-cap

 We will discuss the operating
principles of 3 common types of
trays:
 Sieve
 Valve
 bubble-cap

 The vapour passes through a large number of
"holes" known as perforations (sieves) and emerges
through the liquid in a vertical direction.
 Advantages:
 Simplest design
 No mechanical moving parts.

 There is no liquid seal and it is the passage of vapour
that effectively prevents the loss of liquid through the
sieves.
 Disadvantages
 Weeping
 can occur at low vapour flow and/or high liquid rates
 the liquid height on the tray exceeds the tray pressure drop.
 Fouling

 Typical layout for sieve tray follows:
 square hole pitch
 equilateral triangular hole pitch (centre-to-centre hole spacing)
 Typical sieve sizes
 1/16-inch to 1-inch.
 Small holes have better turndown characteristics because
they reduce tray weeping and therefore increase tray
capacity.
 In terms of cost
 larger holes are cheaper as the holes can be punched
 there are fewer holes to be punched.
 Smaller holes are more expensive as drilling may be
required.
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 A valve tray is a flat perforated plate, with
each perforation fitted with a movable disk
(the "valve").

 The valves will move up or down in response to changing vapour flow rates.
 At normal flow rate
 the valve is roughly in the middle position
 At low vapour rates
 the disk settles over the perforation and covers it to avoid liquid weeping.
 The valves should be heavy enough to prevent excessive opening at low vapour flowrates.
 As the vapour rate is increased, the disk rises vertically
 The upward movement of the disk is restricted either by retaining legs or a cage.

 The perforations and disks may be circular or rectangular.
 The disks (valves) are the movable component of the tray.

 Advantages
 Valve trays offer larger operating range and greater capacity compared to sieve trays.
 No weeping
 Disadvantages
 Fouling: the valves may get stuck due to the build-up of sludge or corrosion products
 "Sticking " valves reduce the tray open area  can lead to premature flooding.
 valve trays are more expensive than sieve trays.

 Bubble-cap consist a slotted cap on a central riser.

 The gas flows up through the riser, reverse flow under the cap,
passes downward through the annulus between riser and cap,
and finally passes into the liquid through a series of openings
or "slots" in the lower side of the cap.
 This device has a built-in liquid seal (i.e. the riser) which
prevents liquid drainage at low gas rates.
 This design does not rely on the velocity of the upcoming
vapour to hold the liquid on the tray.

 Disadvantages
 high cost
 Complexity
 Advantages
 used where very low vapour rates have to be handle
 Adequate residence time is necessary for separation and/or chemical reaction
 A positive liquid seal is essential at all flow rates.

 Spacing
 Sizing
 Layout
 Materials
 Manways
 Operation
 Maintenance
 Costs
 Summary

 Tray spacing is usually set to allow easy access for maintenance.
 Typical spacing: 18 - 24-inch
 Max Spacing  to 36-inch
 24-inch spacing  easier to maintain, less fouling and corrosive service.
 Increasing tray spacing necessarily will:
 add to the column height requirements.
 Larger tray spacing may be required where the trays require elaborate support structures
(e.g. beams)
 It may interfere with vapour disengagement.

 The tray manufacturers have their own sizing
handbooks
 Size and materials
 Models/Constructs
 Based on Diameter

 The cross-flow tray is the most commonly used.
 simple in design
 economic in construction.
 Passes:
 single-pass  Common
 2 or more pass trays  Uncommon (high loads)

 Whenever possible, the number of passes should not
exceed 2.
 This is because increasing the tray passes shortens the
path length.
 Shorter path lengths  reduces the tray efficiency.
 Multi-pass trays are also more expensive.
 Multiple-pass trays are sensitive to liquid and/or vapour
mal-distribution if the flow paths are not symmetrical.

 Typically stainless steel
 Types: 304, 316 or 410.
 The main factors affecting the choice of materials of
construction of tray parts are:
 Compatibility with the chemicals processed
 Compatibility with the column materials of construction
 Anticipated rate of corrosion
 Procedure and expected frequency of cleaning
 Cost (e.g. maintenance, replacement, effects on plant operations,
etc)

 Tray manways allow maintenance workers and inspectors
to travel from one tray to another.
 Min. Size: 16 x 20-inch.
 For multi-pass trays
 one tray manway should be provided for each tray pass, as the
central downcomers restrict access from one side to the other.

 This refers to the range of vapour and liquid rates over which the tray will operate
satisfactorily.
 The ratio of the highest to the lowest flowrates is often referred to as the "turndown" ratio.

 Bubble-cap trays have a positive liquid seal and can therefore operate efficiently at
very low vapour rates.
 Sieve trays rely on the flow of vapours through the holes to hold the liquid on the
tray, thus cannot operate at very low vapour rates.
 Valve trays are intended to give greater flexibility than sieve trays at a lower cost
than bubble-caps Thus, bubble-cap trays have the widest operating range, followed
by valve tray, and sieve tray.

 Pressure Drop
 This factor will be important in vacuum operations.
 The tray pressure drop will depend on the detailed design
of the tray, but in general, sieve trays give the lowest
pressure drop, followed by valves, with bubble-caps giving
the highest.

 Maintenance
 For dirty services, bubble-caps are not
suitable as they are most susceptible to
plugging.
 Sieve trays are the easiest to clean.

 Cost: Bubble-caps are appreciably more
expensive than sieve or valve trays, and the
relative cost will depend on the material of
construction used.
 Due to its simple design, sieve trays are
normally the cheapest.

 In summary
 sieve trays are the cheapest and are satisfactory for most applications.
 Valve trays should be considered if the specified turndown ratio cannot be met with sieve
trays.
 Bubble-caps should only be used where very low vapour rates have to be handled and a
positive liquid seal is essential at all flow rates.

 The bubbling area is the place where vapour-liquid
contact takes place.
 Vapour flows through the openings on the tray from
below, and into the liquid flowing across the tray,
forming a foaming
 Foaming:
 turbulent mixture.
 high interfacial area for efficient vapour-liquid mass
transfer is produced.

 Above the liquid (before the next tray) is the vapour disengagement space, where
the vapour separates from the liquid after contact and continue its up-flow to the
next tray above.
 Ideally, the vapour should carry no liquid droplets (entrainment) to the tray above.
 The liquid overflows the weir into the tray below, its flow path being guided by the
downcomer.

 The Weir will…
 Maintain desired liquid level on the tray.
 Typical weir height is between 2 - 4 inches.
 Low weirs are frequently used in low pressure or vacuum
columns.
 Notched (rectangular or V-shaped) weirs are commonly
used for low liquid loads.
 The higher the liquid level
 the higher the tray pressure drop
 more liquid hold-up on the tray

 Downcomers are used to guide liquid flow:
 from an upper tray to a lower tray.
 The most common is the segmental-type
 Straight
 sloped (angled).

 The straight, segmental, vertical downcomer is widely
used as it provides good utilization of column area for
downflow and has cost and simplicity advantage.
 Sloped downcomer
 vapour-liquid disengagement is difficult (e.g. due to
foaming).
 slightly larger active area for vapour-liquid contact
 more expensive.

 A downcomer must be sufficiently large to allow liquid
to flow smoothly without choking.
 Sufficient time must also be provided in the
downcomer to allow proper vapour
disengagement from the down-flowing liquid, so that
the liquid is relatively free of vapour by the time it
enters the tray below.
 Inadequate downcomer area will lead to downcomer
choking, whereby liquid backs up the downcomer into
the tray above and eventually flood the column.

 Counter-current gas absorption is discussed:
 as it was widely used in the industry.
 The main differences between the two
configurations will be highlighted.
 Note that for counter-current operation, the
gas enters the column or tower from below as
leaves at the top, while liquid enters from
the top and flows in opposite direction and
exits from the bottom.

 We will be concerned primarily with counter-
current gas absorption.
 The gas flows upwards
 The liquid flows downwards.
 Inside the column where there is vapour-liquid
contact, mass transfer by absorption occurs
 there is a transfer of solute(s) from the gas phase to
the liquid phase.

 Notations
 In terms of mole fraction and total flowrates
 y : mole fraction of solute A in the gas phase
 x : mole fraction of solute A in the liquid phase
 G : total molar flowrate of the gas stream (gas flux),
kg-moles/m2.s
 L : total molar flowrate of the liquid stream, kg-
moles/m2.s
 Gy and Lx are the molar flowrates of A in the gas
and liquid respectively (kg-moles A/m2.s) at any
point inside the column.

 Inside the column:
 mass transfer takes place as the solute (component A) is
absorbed by the liquid.
 The quantities of L and x (for the liquid side) and G
and y (for the gas side) varies continuously:

 Moving up the column:
 component A is continuously being transferred from the
gas phase to the liquid phase.
 Limit is equilibrium conditions
 decrease in the total gas flowrate
 a decrease in the concentration of A in the gas phase.
 Moving down the column
 increase in the total liquid flowrate
 increase in the concentration of A in the liquid phase.
 Thus,

 For dilute systems:
 the solute content is small relative to the non-soluble
inerts and non-volatile liquid.
 Thus, we can assume:
 G1 = G2 = G = constant
 L1 = L2 = L = constant

 The relationship between:
 L, x, G and y
 Is given by the operating line equation.
 The operating line equation is obtained by material
balance around the colum:

 Math:
 At steady-state: IN = OUT
 Thus, G(y) + L1(x1) = L(x) + G1(y1)
 Using the dilute system assumptions, we simply
the equation and obtain:
 Gy = L(x) + G(y1) – L(x1)
 Re-arranging:

 Since L and G are assumed to be approximately
constant, the operating line is a straight line of
the form:
 y = mx + c, with the gradient of L / G, the liquid-to-
gas ratio.
 The operating line connects the 2 end points -
point 1 (x1 , y1) that represents conditions at
the bottom of the column, and point 2 (x2 , y2)
that represents conditions at the top of the
column.

 The operating line equation for counter-current gas absorption (for dilute system) is:
 We've assumed that L and G are approximately constant for dilute systems.
 The operating line is a straight line of the form y = mx + c, with a gradient of L / G, the
liquid-to-gas ratio.
 It connects the 2 end points - point 1 (x1 , y1) that represents conditions at the bottom
of the column, and point 2 (x2 , y2) that represents conditions at the top of the column.

 Draw the Operation line given the following conditions:
 The Stack gas has about 45% of CO which must be absorbed
 The specifications according to Environmental Policies are max. of 7% content
 There are several Solvents, the most convinient and readily availbale:
 SOLVAMAX
 CO content 3%
 There are no limits on the max CO content, but a 22% CO content has been recommended by senior
operaitons

 Step 1
 Identify Bottom Column points
 x1 = 0.22
 y1 = 0.45

 Step 2
 Identify Top Column points
 x2 = 0.03
 y2 = 0.07

 Step 3
 Draw Operation Line
 Slope = L/G
 Slope = (y1-y2)/(x1-x2)
 L/G = (0.45-0.07)/(0.22-0.07)
 L/G = 2.53

 For dilute solution, the equilibrium solubility line is
also straight, as represented by Henry's Law,
 y = mx
 where m is the Henry's Law constant which is
also the gradient of the line

 Independent of the Absorber Unit Operation
 Dependent on the system (gas-liquid)
y = mx
y = mx
DILUTE!

 Recall the Points:
 P; M
 Concentrations:
 xAL; xAi; xA*
 yAG; yAi; yA*
 P is any point of the system
 Operating line
 M is the “tie line”

 Points of Operation:
 B = Bottom (x1,y1)
 T = Top (x2,y2)
 P = Generic (x,y)
 Graphing them…

 Points of Operation:
 B = Bottom (x1,y1)
 T = Top (x2,y2)
 P = Generic (x,y)
 Tie Lines
B
T
P
Operation Line

 When these 2 lines are plotted on mole fraction
coordinates

 Any point P (x, y) on the operating line
represents gas-liquid contact
 analysis can be carried out using the 2-film theory
 The larger the distance between the operating
line and equilibrium line:
 the larger the concentration difference for mass
transfer
 the easier the separation.
Note:
Operating line for gas absorption lies above the equilibrium line.
Operating line for gas desorption lies below the equilibrium line.

 The minimum liquid rate will be required.
 This is known as the minimum liquid-to-gas ratio.
 The analysis is applicable to both tray and packed
column.
Note: Operating line for gas absorption lies above the equilibrium line.

 The inlet gas has a solute mole fraction of y1.
 The solute mole fraction is reduced to y2 at the outlet.
 By material balance for the solute in the gas:
 the amount to be removed is G ( y1- y2 )
 The least amount of liquid Lmin that can remove this
amount of solute is the minimum liquid rate
 often expressed in terms of a liquid-to-gas ratio, Lmin/G.
 Understanding the effect of reducing liquid rate
requires an analysis of the operating line equation.

 The condition at the top of the column (point D) is
known:
 x2 the mole fraction of entering liquid
 the mole fraction of gas leaving y2
 Point D is fixed

 The mole fraction of gas entering y1  known.
 The mole fraction of liquid leaving x1
 Depends on the liquid rate used (L)
 For the same amount of solute to be removed:
 using a larger quantity of liquid will result in smaller value
of x1
 The same is truth in vice versa
 When the liquid rate is modified:
 the condition at the bottom of the column varies along the
horizontal line through y1.

 Recall that the operating line has a gradient of L/G.
 By reducing the liquid rate:
 The slope decreases (L/G)
 The exit concentration x1; increases
 The operating line rotates around point D as L is
decreased
 e.g. from line DE to DF (shifts )
 The operating line has moves to the equilibrium line.
 When this happens, the driving force for mass transfer
is smaller
 i.e. the absorption process becomes more difficult.

 At point M:
 the operating line intersects the equilibrium line
 Driving force is zero
 The liquid rate can’t be decreased
 Hence, the liquid rate at this point of equilibrium is known
as the minimum liquid rate, Lmin
 At minimum liquid:
 the outlet liquid concentration is a maximum, x1(max) .

 The minimum liquid rate results in infinite column height
 infinite number of trays or packed height required for
separation
 at zero driving force
 The minimum liquid rate, Lmin can be calculated from the
gradient of the operating line:
 Given y1:
 x1(max) can also be calculated using Henry's Law.

 Math review:
 For straight line  two points
 If “M” and “D” are given
 x2  Depends on solvent, typically 0
 y2  typically given as specification, i.e. 300 ppm
max
 y1  typically fixed as our gas feed; must be lowered
 x1  calculated via Henry’s (for min. reflux)
 Slope can be calculated
 Lmin/G
 G  given Gas
Liq
Liq
Gas
y1 x1
x2 y2

 Typically, the actual liquid rate to be used:
 is specified as multiples of the minimum liquid rate
 e.g. 1.5Lmin or 3.3 Lmin
 If the liquid rate for absorption is initially unknown:
 The minimum liquid rate must be calculated FIRST
Step (III)

 Given the conditions:
 G, y2,x2 and y1
 Get the values of:
 x1*
 Lmin/G
 actual x1

 Step 1
 Identify Top of Column
 (x2,y2)

 Step 2
 Identify (y1)
 Identify x1*; the equilbirium condition of y1
 This is the ZERO driving force  Point “M”
 Best case scenario

 Step 3
 Draw Operating line from:
 Top of column  M
 Slope is always L/G
 This is Lmin/G

 Step 4
 Recall that we know:
 G; gas flow rate
 Conditions at top of column (x2,y2)
 (x1*) Equilibrium conditions for y1  Best Case
 Get Lmin from slope
 Lmin/G = Slope
 Lmin/G = (y1-y2)/(x1*-x2)
 Lmin = (y1-y2)/(x1*- x2) x G

 Step 5
 From Lmin, typically, L operaiton is given:
 1.3x the Lmin; 2x the Lmin
 Let it be 1.5xLmin
 L = 1.5x(Lmin)

 Step 6
 Get x1, the real operation point
 Given:
 L/G = (y1-y2)/(x1-x2)
 Analytically:
 x1 = (y1-y2)x(G/L)+x2

 A gas mixture has 3% v/v pentane vapor (all other inert gas)
 Solvent is an special oil. Free from pentane
 97% of Pentane is required to be removed
 G = 50 kmol/h; (48.5; 1.5)
 Equilibrium line
 Y = 0.25x/(1+0.75x)
 (a) Calculate the Min. Required Solvent Flow (Lmin)
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 1) Get equilibrium line Y = 0.25x/(1+0.75x)

 Step 2)
 Get (Top Column) Data (X0,Y1)
 X0 = 0 (solvent is free)
 Y1 = (3%left)x(1.5kmol/h)/(50kmol/h)
 Y1 = 0.0009
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 2)
 Get (Top Column) Data (X0,Y1)
 X0 = 0 (solvent is free)
 Y1 = (3%left)x(1.5kmol/h)/(50kmol/h)
 Y1 = 0.0009
 (X0,Y1) = (0, 0.0009)

 Step 3) Get the Bottom Conditions
 Bottom condition:
 XN = X* (based on y(N+1))
 Locate the point 
 Y = 0.25x/(1+0.75x)
 (0.03)=0.25x/(1+0.75x)
 (0.03) (1+0.75x)=0.25x
 0.03+0.0225x=0.25x
 0.03 = (0.25-0.0225)x
 x= 0.1209
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 4) Get Lmin
 Draw Line (RED)
 Slope: (Y1-YN+1)/(X*-X0)
 Slope: (0.030-0.009)/(0.1209-0)
 Slope = 0.17369
 Slope = Lmin/G
 Lmin/G = 0.17369
 Lmin/50kmol/h = 0.17369
 Lmin = 0.17369 x(50kmol/h)
 Lmin = 8.68 kmol/h

 Absorption
 Co-Current
 Counter-Current*** Main Study
 Stripping
 Co-Current
 Counter-Current

 The operating line has negative slope.
 There is no minimum liquid-to-gas ratio
 Equilibrium point will be achieved anyways
 To produce an exit liquid and gas streams at
equilibrium (xe, ye) on the equilibrium curve:
 an infinitely tall column must be used.
 It is less efficient than counter-current operation.

 Important points to note:
 Mass transfer from the liquid-phase to the gas-
phase.
 Sometimes gas stripping units go by the name
of regenerators.
 Analysis is similar to counter-current gas
absorption.
 Important difference
 operating line lies below to equilibrium line

 Recall:

 For the given process…
 Stage 0 = pure Solvent
 Stage 1 = last gas interaction
 Stage x = inbetween interactions
 Stage N = last liquid interaction
 Stage N+1 = pure Gas

 The number of theoretical trays can be
determined graphically using a method similar to the
McCabe-Thiele Method used in continuous distillation.
 We are following the consistent nomenclature of
using:
 subscript "1" to refer to conditions at the bottom of the
packed column
 subscript "2" to refer to conditions at the top of the
packed column.

 Note the slight difference in the way the triangles are
drawn:
 Start from point 1 and work the way down towards point 2
 Draw triangles between the operating line and equilibrium line.
 In the above example, 5 triangles are drawn.
 Therefore 5 theoretical trays are required for the separation.
 For the case of gas absorption, the last triangle represents a
theoretical tray, not a reboiler.
 Analysis for the changes in gas phase and liquid phase
compositions is similar to the distillation process

 For example, consider tray (b)
 where the liquid concentration changed:
 from xa at the inlet
 to xb at the outlet
 the gas composition changed
 from yc at the inlet
 to yb at the outlet.
 And
 ( yc - yb ) showed the decreases in gas concentration as it
passed through tray (b),
 ( xb - xa ) showed the increase in liquid concentration as it
passed through tray (b).
 The larger the triangle, the more effective the
separation.

 It is desired to absorb 95% of acetone by water from a mixture of
acetone and nitrogen containing 1.5% of the component in a
countercurrent tray tower.
 Total gas input is 30 kmol/hr and water enters the tower at a
rate of 90 kmol/hr. The tower operates at 27ºC and 1 atm.
 The equilibrium relation is Y=2.53X
 (a) Determine the number of ideal stages necessary for the
separation using graphical method.
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

Approach:
1. Dilute vs. Non-Dilute
 Verify if L and G can be considered constant
 If 5% > change, assume non-dilute
 If 5% < change, assume dilute
2. Define/Write given data (Flows/concentrations)
3. Get Equilibrium Data  Henry’s Law (Straight Line)
4. Get Operation Line  (From top/bottom conditions)
5. Calculate Total Number of Equilibrium Stages

 Step 2:
Basis : 1 hour
G N+1( ) = 30kmol
Y N+1( ) = 0.015
Lo = 90kmol
Inlet :
Moles acetone in = (30kmol / h)´ (0.015) = 0.45 kmol Ac
Moles nitrogen in = 30 - 0.45( ) kmol = 29.55 mol Inert
Outlet :
Moles acetone leaving 95% absorbed( ) = 0.45kmol ´ 1- 0.95( ) = 0.0225 kmol
Gs = 29.55 kmol » 30kmol
Ls = 90 kmol
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 2:
y1 = 0.0225kmolAc / 30kmolGas = 7.5x10-4
y(N+1) = 0.15
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 3: Equilibrium line
 Construction of equilibrium line (Y=2.53X):
 X: 0 0.00100 0.00200 0.00300 0.00400 0.00500
 Y: 0 0.00253 0.00506 0.00759 0.01012 0.01265
x y
0 0
0.001 0.00253
0.002 0.00506
0.003 0.00759
0.004 0.01012
0.005 0.01265
0.006 0.01518
0.007 0.01771
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
0.02
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008
y
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 4: Get Operation line
 Get XN
Gs(y(N+1) )- y1= Ls(XN - X0 )
30kmol(0.015)- 7.5x10-4
= 90kmol(XN - 0)
XN =
30kmol(0.015)- 7.5x10-4
90kmol
XN = 4.99x10-3
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Solution by graphical method.
 Construction of operating line PQ:
 Identify Points
P(X0,Y1);Q(XN ,Y(N+1) )
P(0,7.5x10-4
);Q(4.00x10-3
,0.015)
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
0.02
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008
y
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Identify Points
 Draw Line
P(X0,Y1);Q(XN ,Y(N+1) )
P(0,7.5x10-4
);Q(4.00x10-3
,0.015)
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
0.02
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008
y
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Identify Points
 Draw Line
 Get Stages  McCabe Method
P(X0,Y1);Q(XN ,Y(N+1) )
P(0,7.5x10-4
);Q(4.00x10-3
,0.015)
0
0.002
0.004
0.006
0.008
0.01
0.012
0.014
0.016
0.018
0.02
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008
y
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 A = L/(mG) or 1/m x (L/G)
 If A <1
 Infinite
 If A >1
 Possible

 For dilute system:
 the operating line is straight
 equilibrium line are straight
 There is an algebraic way to relate:
 Henry’s Law
 Operaiton Line
 Number of Stages

 In this case, the number of theoretical stages required for a given separation can be
calculated using the Kremser-Brown-Souders Equation as shown:
 Absorption
 where A = L /mG is the absorption factor and is assumed constant.
 L/G = slope of operation line
 m = Henry's Law constant = slope of the equilibrium line.
y* = m(x)

 Stripping
 where the stripping factor S = 1/A, is assumed constant.
 A = L/G = slope of operation line
 m = Henry's Law constant = slope of the equilibrium line.

 Chloroform is to be absorbed by a scrubber.
 Water is selected due to its properties as a good solvent, assume no content of
solute
 Initially. 200 ppm of gas is present; Ideal conditions are 10 ppm or less
 The total Gas of the room is approx. 1000 kmol
 The L:G ratio is 187.3:1
 M = 141 (ppm/ppm)
 (a) Find total ideal stages
https://www.youtube.com/watch?v=dByYrj7-
tYQ&feature=youtu.be&list=PL61BFC1C064B40049

 Step 1) Get All data required
 Material balances
 Flow Rates
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…
G = 1000kmol / h
Ratio -187.3:1
L0 = 187.3x(L0 ) = 187.3x1000kmol / h = 187300kmol / h
x0 = 0
y1 = 10ppm
y(N+1) = 200ppm
xN : N / A
y*

 Step 2:
 Get the equilibrium line & y*
 Henry’s constant, 141….
 Note that this is in ppm/ppm not mol/mol...
 For y* (x2 valuated in equation equilbirium)
y = mx
y = 141x
y*
=141x(0) = 0

 Step 3:
 Calculate “A” factor
 A = L/(mG)
A =
L
mG
A =
187300kmol / h
141x(1000kmol / h)
A = 1.32836

 Step 5:
 Apply Kremser Equation
N=
ln 1-
mG
L
æ
èç
ö
ø÷
yN+1 - y*
y1 - y*
æ
èç
ö
ø÷ +
mG
L
é
ë
ê
ù
û
ú
ln
L
mG
æ
èç
ö
ø÷
N=
ln 1-
1
A
æ
èç
ö
ø÷
yN+1 - y*
y1 - y*
æ
èç
ö
ø÷ +
1
A
é
ë
ê
ù
û
ú
ln A( )

 Step 5:
 Apply Kremser Equation
N=
ln 1-
1
1.326
æ
èç
ö
ø÷
200 - 0
10 - 0
æ
èç
ö
ø÷ +
1
1.326
é
ë
ê
ù
û
ú
ln 1.326( )
N =
ln((0.24585)(20)+1.328))
ln(1.326)
N =
ln(6.244)
ln(1.326)
N=6.46

 Step 2) Get XN
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…
inlet = outlet
(L0 )x0 + (GN+1)yN+1 = G1y1 + LN xN
xN =
(L0 )x0 + (GN+1)yN+1 - G1y1
LN
xN =
(187300kmol / h)(0)+ (1000kmol / h)(200ppm)- (1000kmol / h)(10ppm)
187300kmol / h
XN =
(1000kmol / h)(200ppm)- (1000kmol / h)(10ppm)
187300kmol / h
XN = 1.014

 Step 3) Get Equilibrium Equation/Graph

 Step 4) Locate Top/Bottom conditions
 X0 = 0; y1 = 10
 xN = 1.1; YN+1 = 200
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 5) Draw Operation Line
 X0 = 0; y1 = 10
 xN = 1.1; YN+1 = 200
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 6) Apply Graph. Method
1
2
3
4
5
6…
6

http://www.learncheme.com/simulations/separations/operation-of-an-absorption-column

http://www.learncheme.com/simulations/cdf-
files/StrippingColumnOperation.cdf?attredirects=0&d=1

 A gas stream comprising of air and vapor of an organic compound is to be scrubbed in an
absorption tower for separation of organic compound by absorption in oil.
 The operation is countercurrent.
 Given:
 Mol. wt. of oil: 250 kg/kmol;
 Inlet concentration of vapor of organic compound in gas stream: 5% (by volume)
 Targeted (or desired) removal of organic vapor: 95%
 Flow rate of gas stream: 1000 m3 /h at 1.2 bar and 30ºC
 Mol wt of organic vapor: 80 kg/kmol
 Vapor pressure of organic vapor at 30ºC: 0.125 bar.
 You can assume that the system obeys Raoult’s law.
 If the inlet oil to the absorption column does not contain any trace of organic vapor
initially, answer the following:

 (A) Calculate the minimum flow rate (mass) of oil to the column for desired removal
of organic vapor.
 (B) Calculate the number of theoretical stages using Kremser’s equation if the
absorption factor A = L/mG = 1.4.Check out Full COURSE:

 Step 1A. Set-up the equilibrium
 If raoult’s … Let 1 = organic vapor, Let 2 = non-vapor
x1Pº1= y1xPT
x(0.125bar) = y(1.2bar)
y =
0.125bar
1.2bar
x
y = 0.1042x

 Step 2a. Calculate Conditions of top/bottom column
 Bottom:
 Gas Flow  Ideal Gas Law
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…
PV= nRT
n =
PV
RT
=
(1.2x105
Pa)(1000m3
h )
8.314
Pa im3
mol - K
æ
èç
ö
ø÷ x 30 + 273K( )
= 47,635.26 mol/h
n=47.6 kmol/h

 Get all other required data…
 From material balance:
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…
GN+1 = 47.6kmol / h
x0 = 0
yN+1 =0.05
Inlet -Vapor :
(GN+1)(yN+1) = (47.6kmol / h)(0.05) =2.38kmol/h
95%inLiq
5%Gas = (0.05)x2.38kmol/h=0.119kmol/h
y1 =
0.119kmol/h
47.6kmol / h
= 0.0025
y1 = 0.0025

 For Lmin…  In equilibrium
y = 0.1x
x =
y
0.1
=
0.0025
0.1
= 0.025
x*
= 0.025
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Step 3A. Calculate slope for Lmin…
 For the operation line:
m =
Lmin
G
m =
Dy
Dx
=
yN+1 - y1
x*
- x0
=
0.05 - 0.0025
0.025 - 0
= 1.9
Lmin
G
= 1.9

 Step 4A. Calculate Lmin
 For the operation line:
 Convert to mass
Lmin
G
= 1.9
Lmin = 1.9xG=1.9x(47.6kmol / h)
Lmin = 90.44kmol/h
Lmin(mass) = (90.44kmol/h)x(250kg/kmol)
Lmin(mass)=22600 kg/h

absorption factor A = L/mG = 1.4.
 Step 1B. Get Kremser Equation.
N=
ln 1-
mG
L
æ
èç
ö
ø÷
yN+1 - y*
y1 - y*
æ
èç
ö
ø÷ +
mG
L
é
ë
ê
ù
û
ú
ln
L
mG
æ
èç
ö
ø÷
N=
ln 1-
1
A
æ
èç
ö
ø÷
yN+1 - y*
y1 - y*
æ
èç
ö
ø÷ +
1
A
é
ë
ê
ù
û
ú
ln A( )
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

absorption factor A = L/mG = 1.4.
 Step 1B. Get Kremser Equation. L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…
N =
ln 1-
1
A
æ
èç
ö
ø÷
yN+1 - y*
y1 - y*
æ
èç
ö
ø÷ +
1
A
ln(A)
N =
ln 1-
1
1.4
æ
èç
ö
ø÷
0.05 - 0
0.0025 - 0
æ
èç
ö
ø÷ +
1
1.4
é
ë
ê
ù
û
ú
ln(1.4)
N =
ln 0.2857x
0.05 - 0
0.0025 - 0
æ
èç
ö
ø÷ + 0.71428
é
ë
ê
ù
û
ú
0.3364
N =
1.860
0.3364
N = 5.535
N » 6

 Concept of theoretical stages
 The stages that we have been studying so far
 Real Stages
 In engineering, not all physical stages act as equilibrium stages
 The actual number of stages increases
 The efficiency of mass transfer is expressed as the ratio of the actual change in
mole fraction to thechange that could occur if equilibrium were attained
 Eª= dy/dy*
 Real chang / Theoretical change
 Real < Theoretical

 We define the overall tray efficiency EO as:
 EO is applied throughout the whole column, i.e. every tray is assumed to have the
same efficiency.
 The advantage is that it is simple to use, but it must be bear in mind that in actual
practice, not all the trays have the same efficiency.

 The overall efficiency has been found to be a complex function of the following:
 Geometry and design of the contacting trays
 Flow rates and flow paths of vapour and liquid streams
 Compositions and properties of vapour and liquid streams
 Values of EO can be predicted by any of the following 4 methods:
 Comparison with performance data from industrial columns for the same or similar systems
 Scale-up of data obtained with laboratory or pilot plant columns
 Use of empirical efficiency models derived from data on industrial columns
 Use of semi-theoretical models based on mass and heat transfer rates

 The Murphree Tray Efficiency EM , is based on a semi-theoretical models that assumes:
 the vapour between trays is well-mixed (uniform composition)
 the liquid in the downcomers is well-mixed (uniform composition)
 the liquid on the tray is well mixed and is of the same composition as the liquid in the
downcomer leaving the tray.
 It is defined for each tray according to the separation achieved on each tray.
 This can be based on either the liquid phase or the vapor phase.
 For a given component, it is equal to the change in actual concentration in the phase,
divided by the change predicted by equilibrium condition.

 When Murphree Efficiency is constant in all trays:
 Operation line and Equilibrium lines are straight
 Eº = efficiency
 EMG = murphree gas-phase tray efficiency
 EMGE = Murphree gas-phase tray efficiency; corrected for entertainment

 Values for the overall efficiency EO for absorbers and strippers are typically low,
often less than 50%.
 Examples of performance data for industrial bubble cap columns are here:

 Consider a tray scrubber with:
 Murphree Eff EMGE = 0.75
 avg. A = 1.5
 (a) Estimate Overall Tray efficiency
 (b) If scrubber requires 6.34 ideal stages, calculate total real trays required

 A) Tray eff.
 Step 1. Get Equation
 Step 2. Substitute Equation
E0 =
ln 1+ 0.75(1/1.5 -1)[ ]
ln(1/1.5)
E0 =
ln(0.75)
ln(0.666)
E0 = 0.7095
E0 =
ln 1+ EMGE (1/ A -1)[ ]
ln(1/ A)

 B) Ideal Trays
 Step 1. Get E0 and subtitute
E0 =
equilibrium trays
real trays
real trays =
6.34
0.7095
real trays = 8.935 » 9

 A sieve-tray tower is being designed for a gas absorption process.
 The entering gas contains 1.8% (molar) of A, the component to be absorbed.
 The gas should leave the tower containing no more than 0.1 % (molar) of A.
 The liquid to be used as absorbent initially contains 0.01 % (molar) of A.
 The system obeys Henry’s law with m = y/x = 1.41.
 Bottom of the tower  L/V= 2.115
 Top of tower  L/V = 2.326.
 The Murphree efficiency is constant at EMGE = 0.65.
Example 5.1 Number of Real Sieve Trays in an Absorber
Benitez – Mass Transfer Operations

 (a) Estimate the number of trays required.
 (b) Estimate the height of the tower, given Tower Diameter approx. Dtower = 1.5 m

 (a) Step 1. Get A in top/bottom.
 Calculate the absorption factor for the molar liquid-to-gas ratio given at the extremes of
the column and the slope of the equilibrium curve:
 A1 = 2.115/1.41 = 1.50;
 A2 = 2.32/1.41 = 1.65.
 (a) Step 2. Get Geomtric Average of top/bottom conditions
Ageo.avg = (A1iA2)0.5
= (1.5x1.65)0.5
Ageo.avg =1.5732

 (a) Step 3. Get required data
 From the concentration data given:
 TOP:
 y1 = 0.001
 xo = 0.0001.
 BOTTOM
 yN+1 = 0.018
 xN = no need * (mass balance required, no need for Kremser)
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 (a) Step 4. Substitute in Kremser Equation
N =
ln
yN+1 - mx0
y1 - mx0
1-
1
A
æ
èç
ö
ø÷ +
1
A
é
ë
ê
ù
û
ú
ln A
N =
ln
0.018 -1.41x(0.0001)
0.001-1.41x(0.0001)
1-
1
1.5732
æ
èç
ö
ø÷ +
1
1.5732
é
ë
ê
ù
û
ú
ln(1.5732)
N =
ln
0.017859
0.000859
(0.36435)+ 0.6356
é
ëê
ù
ûú
ln(1.5732)
N =
ln(8.210)
ln(1.5732)
N = 4.6464
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Now, for “real number of trays”
 Step 5. Get Equilibrium equation
 Step 6. Given Murphree, substitute for E0
E0 =
ln 1+ EMGE (1/ A -1)[ ]
ln(1/ A)
E0 =
ln 1+0.65
1
1.5732
-1
æ
èç
ö
ø÷
é
ë
ê
ù
û
ú
ln
1
1.5732
E0 =
ln(0.7640)
ln(0.63564)
E0 =0.5924

 Step 7. Calculate Real Stages
 Therefore, the number of real trays required
 Use 8 trays, since it is not possible to specify a fractional number of trays.
E0 =
equilibrium stages
real stages
0.5924=
4.6464
real stages
real stages =
4.6464
0.5925
= 7.842
real stages = 7.842 » 8

 (b)For a tower diameter of 1.5 m; use:
 Step 1. Use table to calculate approx. tray spacing
 Select t=0.6 m approx.

 Step 2. Calculate Total height of trays
 Total tray height will be approx. 4.8 m
Z = N( )x t( )
Z = 8trays( )x 0.6m /tray( )
Z = 4.8m
L0
x0
G1
y1
GN+1
yN+1
LN
XN
N=1
N=2
N
N=…

 Typical operation:
 (a) Normal Operation
 (b) Flooding
 (c) Priming condition
 (d) Coning condition
 (e) Weeping
 (f) Dumping
(a) (b) (c)
(d) (e) (f)

 All mechanical aspects which occur in a column are referred
as Effects in Tower
 These mechanical problems caused by the physical
properties and the mechanism by which column is operated
by the control valves and inlet and outlet streams flow rates
 The structure and internal design also considered in this
concept
 There must be a correct liquid flow as well as a gas flow

 It occurs in a column due to high-pressure drop.
 At the same gas flow rate, the pressure drop in a tower being
irrigated with liquid is greater than the dry tower.
 Operating velocity in a tower is usually equal to the flooding
velocity.
 This effect can be well understood as simple as liquid filling up
from the bottom of the column to the top and exhausted out from
the top inlet of the column.
 The point at which this effect occurs the velocities of which a
column is operated is called as flooding velocities.
 Downcomer and space between the trays are completely filled up
by the liquid than the tower is said to be flooded, due to high-
pressure drop due to increased flow rates of the streams.

 Effects due to flooding:
 Tray efficiency falls
 The liquid may force out of the exit pipe at the tower top
 Overall tray efficiency is defined as the ratio of a number of
real trays required to the number of ideal trays required.
 Channelling is most severe in towers packed with stacked
packing. Wetted wall tower experiment are used to determine
the volumetric coefficient of two interacting phases.

 Priming in a gas-liquid column might be desirable from point
efficiency considerations.
 It is an exaggerated condition of liquid entrainment.
 The packed column provides substantially smaller liquid hold-
up as compared to plate column.
 Outlet weirs (provided on the plate in a plate column)
maintain a desired liquid level on the plate.
 Inadequately large weir height may cause all of the foregoing
 a common weir height for absorbers and strippers is 3 to 4 inch.
 The binary liquid-liquid system has two degrees of freedom.
 Due to high gas velocity, liquid from the bottom trays is
carried away along with the vapor to the top trays.

 Coning
 occurs due to low liquids flow velocities when compared to
gas which results in pushing of the liquid away from the tray
openings.

 Weeping is due to at low gas velocity which is not equal to
liquid flow velocity, and the liquid is not enough resisted to
hold on the tray pass from the downcomers.
 The complete liquid will flow through the openings in the
tray itself.
 Weeping occurs when gas velocities (in a plate column) are
too low.
 Most of the liquid is rained down from tray openings and
some through the downcomer.

 In the event of severe weeping, no liquid reaches the
downspouts.
 Complete liquid drops down by the tray opening only.
 The gas hold up is defined as the fraction of the liquid-
gas mixture occupied by the gas.
 Weber number is defined as ratio of shear forces to
inertial forces and ratio of inertial forces to surface
forces. Absorption factors is defined as mE/R.

 The tower diameter and, consequently, its cross-sectional area must be sufficiently
large to handle the gas and liquid rates within the satisfactory region (Summary Fig)
 The superficial velocity of the gas VGF
 volumetric rate of gas flow Q, per net cross-sectionnal area for gas flow in the space
between trays
*C = empirical constnat

 The value of the empirical constant C depends on
 tray design
 tray spacing
 flow rates
 liquid surface tension
 foaming tendency.
 It is estimated according to the empirical relationship
 Seader and Henley, 1998; Fair, 1961

 FST = surface tension factor = (σ/20)^0.2
 σ = liquid surface tension, dyn/cm
 FF = foaming factor = 1.0 for non-foaming systems
 for many absorbers may be 0.75 or even less (Kister, 1992)
 FHA = 1.0 for Ah/Aa, > 0.10, and 5 (Ah/Aa) + 0.5 for Ah/Aa < 0.1
 Ah/Aa = ratio of vapor hole area to tray active area
Lf the value of X is in the range 0.01 to 0.10  X = 0.10 in equation (4-31).

 Typically, the column diameter D is based on a specified fractional approach to
flooding
 Qg = volumetric flow rate of gas m3/s
 f = flooding factor (0.5-0.8)
 VGF = superficial velocity of gas m/s
 Ad = area taken by the perforations on a sieve tray; m2
 At= total cross-section area, m2.
Typical values for Ad/At

 Depends on diameter

 When molasses are fermented to produce a liquor containing ethanol, a C02-rich vapor
containing a small amount of ethanol is evolved.
 The alcohol will be recovered by countercurrent absorption with water in a tray tower.
 Gas inlet: rate of 180 kmol/h, at 303 K and 110 kPa.
 Mol. Composition 98% CO, and 2% ethanol.
 The required recovery of the alcohol is 97%.
 Solvent  Pure liquid water
 303 K, Rate = 151.5 kmol/h
 50% above the minimum rate required for the specified recovery
ρl = 986 kg/m3
ρg = 1.923 kg/m3
Benitez, 2nd, Mass Transfer Operations, Example 4.6
based on Example 4.4

 Design a sieve-tray column for the ethanol absorber
 Recommendations:
 For alcohol absorbers, Kister (1992) recommends a foaming factor FF = 0.9.
 The liquid surface tension is estimated as σ = 70 dyn/cm.
 Take do = 5 mm on an equilateral-triangular pitch 15 mm between hole centers,
 Punched in stainless steel sheet metal 2 mm thick.
 Design for an 80% approach to the flood velocity.

 Solution:
 Get VGF 
 Get C 
 Get FST, FF, FHA, CF

 For the surface tension factor:
 Given
 σ = 70 dyn/cm

 FF = Foaming factor…
 Given as 0.90
 FF = 0.90

 For FHA:
 Since X < 0.1; Ah/Aa = 0.1
 FHA =1 X = L’/G’(ρg/ρl)^0.5
X = (0.804/2.202)x((1.923/986)^0.5)
X = 0.016
Eth. Absorbed = 180 kmol/h * (2%)x(97%abs.) x 46 g/mol = 160.6 kg/h
L’ = [(151.5kmol/h)*(18g/mol)+160.6kg/h )/(3600s/h) = 0.804kg/s
G’ = (180 kmol/h x 44 g/mol)/(3600s/h) = 2.202 kg/s

 For CF:
 Assume t = 0.5 m, since it is common ranger for 1m diameter
 Get X, since X < 0.1, use X = 0.1
 After X, get alpha and beta
 alpha =(0.0744*0.5+0.01173) = 0.0489
 beta =(0.015) = 0.0302
 Get CF
 CF = (0.0489)xlog[1/(0.1)]+ 0.0302
 CF = 0.0791 m/s

 Get C 
 C = (1.285)(0.90)(1)(0.0791) = 0.09147 m/s
 Get VGF
 VGF = (0.09147)x((986-1.923)/1.923))^0.5 = 2.07 m/s

 Now, get the diameter via the correlation
 Qg = 1.145 m3/s
 f = 80%  0.80
 VGF = 2.07 m/s
 (1-A/A) = (1-0.01) = 0.90
 Pi = 3.1416
 Get Diameter!
 D = [(4x)(1.145)/(0.8*2.07*0.9*31.14)]0.5
 D = 0.99 m = 1m*
*as predicted in the tray spacing!
Qg:
PV = nRt
Q = nRT/P
Q =(180kmol/h)x(8.314Pam3/molK)(303K)/(101kPa)
Q = (50mol/s)x(8.314)(303)/(11000) = 1.145 m3/s

 Typical tray pressure drop for flow of vapor in a tower is from 0.3 to 1.0 kPa/tray.
Pressure drop (expressed as head loss) for a sieve tray is due to friction for vapor
flow through the tray perforations, holdup of the liquid on the tray, and a loss due
to surface tension:
 Ht = total head loss/tray, cm liq.
 Hd = dry tray head loss, cm liq
 Hl = equivalent head of clear liquid on tray, cm liq
 Hσ = head loss due to surface tension, cm liq

 The dry sieve-tray pressure drop is given by a modified orifice equation (Ludwig,1979)
 Hd dry tray head loss, cm liq.
 Ah = area taken by the perforations on a sieve tray; m2.
 Aa = Active Area of Tray m2
 C0 = orifice coefficient  correlation
 V0 = hole velocity, in m/s ,
 l = tray thickness; m
 d0 = perforation diameter in a sieve plate; m.
(Wankat, 1988):

 The equivalent height of clear liquid holdup on a tray depends on weir height, hW,
liquid and vapor densities and flow rates, and downcomer weir length
 hW = weir height, cm (typical values are from 2.5 - 7.5 cm)
 Φe = effective relative froth density (height clear liquid/froth height)
 Φe = exp (-12.55Ks^(0.91))
 Ks = capacity parameter, m/s 
 va = superficial gas velocity based on tray active area, m / s
 qL = liquid flow rate across tray, m3/s
 Cl = 50.12 + 43.89 exp(-1.378xhW)

 As the gas emerges from the tray perforations:
 the bubbles must overcome surface tension.
 The pressure drop due to surface tension is given by the difference between the pressure
inside the bubble and that of the liquid according to the theoretical relation
 Where,
 g= gravity 9.8m/s2
 σ = surface tension
 d0 = perforation diameter in a sieve plate; m.

 Estimate the tray gas-pressure drop for the ethanol absorber of Exercises#11 & #12
 Assume:
 weir height hW = 50 mm. = 0.05 m = 5 cm
 Weir length = 0.720m
 Ah = 0.062 m2
 Ah/Aa = 0.101
 RECALL:
Benitez, 2nd, Mass Transfer Operations, Example 4.7
based on Example 4.4 and 4.6
QG = 1.145 m3
/ s

 Step 1. Get total head loss equation
 Ht = total head loss/tray, cm liq.
 Hd = dry tray head loss, cm liq
 Hl = equivalent head of clear liquid on tray, cm liq
ht = hd + hl + hs

 Step 2. Get hd
 From 
 Get hd
 Get C0 for hd
 Ensure d0/l > 1.0
 d0/l= 5 cm /2 cm = 2.5
ht = hd + hl + hs

 Step 3.
 For C0,
 Step 4. Get Vo
C0 = (0.85032)- (0.04231)x(2.5)+ 0.0017954(2.5)2
C0 = 0.7557
V0 =
QG
Ah
=
1.145m3/ s
0.062m2
( )
æ
è
ç
ö
ø
÷
V0 = 18.48 m / s

 Step 5. Get hd
hd = 0.0051x
18.48m
s
0.756
æ
èç
ö
ø÷
2
1.923kg
m3( )
995kg
m3
986kg
m3
æ
èç
ö
ø÷ (1- (0.101)2
)
hd = 5.853cm

 Step 6. Get hl
 Get ht 
 Get hl
 Get Ks
 Step 7. Calculate va
 From Ah/Aa = 0.101;  Aa =0.615 m2
 Va = (1.14 m3/s) / (0.615 m2) = 1.863 m/s

 Step 8.
 Get Ks
 Get phi 
 Get Cl
 Get ql
Ks = va
rG
rL - rG
æ
èç
ö
ø÷
1
2
= 1.863m
s( )
1.923kg
m3
986kg
m3 -1.923kg
m3
æ
èç
ö
ø÷
0.5
Ks = 0.082m / s
fe = e-12.55(0.0820.9
)
= 0.2667
Cl = 50.12 + 43.89e(-1.378x5)
Cl = 50.1646
qL =
804kg / s
986kg
m3
qL = 0.000815m3
/ s

 Step 9. Get hl
hl = fe hw + Cl
qL
Lwfe
æ
èç
ö
ø÷
2
3
é
ë
ê
ê
ù
û
ú
ú
hl = 0.2667 5 + 50.16
0.000812qL
0.72x0.2667
æ
èç
ö
ø÷
2
3
é
ë
ê
ê
ù
û
ú
ú
hl = 1.5cm

 Step 10. Get hσ
g = 9.8 m
s2
rL = 986 kg
m3
s = 70 dyne / cm = 0.07 N / m
hs =
6x(0.07N / m)
9.8 m / s2( ) x(986kg
m3 )x 0.005m( )éë ùû
hs = 0.0087m
hs = 0.87 cm

 Step11. Get total h
 Step 12. Convert from head units to Pressure units
 From head to pressure  x gravity  x density
ht = 5.85cm +1.5cm + 0.87 m( )
ht = 8.22cm of liq. /tray
dP = rL( )x(ht )x(g) = (986kg / m3
)x(8.22cm /tray)x
1m
100cm
æ
èç
ö
ø÷ x(9.8m / s2
)
dP = 794.28 Pa /tray
dP = 0.79 kPa /tray

 Step 13. Calculate total height
 If total trays  N = 8
Pdrop = N( )x(dP tray)
Pdrop = = 8tray( )x(0.79 kPa /tray)
Pdrop = = 6.32kPa

 Based on Ex. 11, 12, and 13

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