DESIGNING

 

REACTOR DESIGN

REACTOR DEFINITION:

 

Reactors or more precisely speaking chemical reactors are vessels designed to contain chemical reactions.

 

TYPES OF GAS LIQUID REACTORS:

 

Following are the four main types of gas liquid reactors.

 

  • Gas liquid continuous stirred tank reactor
  • Bubble column
  • Packed column
  • Plate columnREACTOR SELECTION:

     

    The selection of the gas liquid reactor to be used is as follows.

     

    Reasons for selecting gas liquid continuous stirred tank reactor:

     

    • Excellent gas and liquid mixing.

     

    • High mass transfer occurs.

     

    • High heat transfer occurs.

     

    • Good temperature control.

     

    • Gas spends more time in liquid phase due to stirring this facilitates reaction in liquid phase.

     

    • Stirring greatly decreases the probability of coalescence of bubbles.

     

    • Very less pressure drop.

     

    • High liquid hold up.

     

     

    Reasons for not selecting Bubble Column:

     

     

    • Non negligible pressure drop because the column is usually high.

     

    • Less mass transfer as compared to stirred tank reactor.

     

    • Less heat transfer as compared to stirred tank reactor.

     

    • There is more probability of coalescence of bubbles in thus type of reactor.

     

    • Less efficient gas liquid mixing as compared to stirred tank reactor

     

    • Problem of foaming can occur.

     

     

    Reasons for not selecting Plate Column:

     

     

    • Relatively high capital investment.

     

    • Less liquid holdup.

     

    • Flooding can occur.

     

    • Less efficient mass transfer as compared to stirred tank reactors

     

     

    Reasons for not selecting Packed Column:

     

     

    • Poor heat transfer.

     

    • High pressure drop.

     

    • Less efficient heat transfer as compared to mechanically stirred tank reactor.

     

    • Cost is more than as compared to mechanically stirred tank reactor

    IMPELLER SELECTION:

     

    There are two main types of impellers.

     

    1. Axial flow impellers
    2. Radial flow impellers

     

    Axial flow impellers:

     

    The impellers that generate currents parallel to the shaft of the impeller are called axial flow impellers.

     

    Radial flow impellers:

     

    The impellers that generate currents which flow tangentially or radially from the impeller blade are termed as radial flow impellers.

     

    There are three main categories of impellers for low to moderate viscosity systems.

     

    1. Propellers
    2. High efficiency impellers
    3. Turbines

     

     

    As the system of this process is of low to moderate viscosity, so we will have to choose from the categories described above.

     

    1. Propellers:

     

    A propeller is an axial flow high speed impeller for low viscosity liquids. Small propellers turn at about 1150 to 1750 rpm. And larger ones turn at bout 400 to 800 rpm. The direction of rotation is usually chosen to force the liquid downward, and the flow currents leaving the impeller continue until deflected by the floor of the vessel.As described above these are recommended for low viscosity liquids, but are not recommended for gas dispersion in liquid. Also note that they are only axial flow.

    So they are ruled out from the selection.

     

    1. High efficiency impellers:

     

    They are designed to provide more uniform axial flow and better mixing. These impellers are widely used for low to moderate viscosity liquids, but are not recommended for very viscous liquids or for gas dispersion. So they are also eliminated from the selection.

     

    1. Turbines:

     

    These push the liquid radially and tangentially. The currents they generate travel towards the vessel wall and than flow either upward or downward. Main types of turbines are shown below.

    Out of these only disk turbine and concave blade CD-6 impeller are used for gas dispersion. So the selection is to be made among them. Following is the selection between these two types of turbine impellers.

     

    • Disk turbine is useful for gas-liquid dispersion. But efficiency is not high.

     

    • Concave blade CD-6 disk turbine impellers are highly efficient. It is different from simple disk turbine in the manner that its blade are in a concave shape which cup the gas and disperse it more efficiently than simple disk turbine.

     

    • Concave blade CD-6 impeller show enhanced mass transfer for the same power/volume & superficial gas velocity than simple disk turbine.

     

    • There are two processes in gas dispersion which continuously oppose each other, that is dispersion and coalescence. CD-6 impeller decrease the probability of coalescence to a large extent by decreasing the probability of bubbles existing in the same portion of volume.

     

    • Also Concave blade CD-6 disk turbine impellers are quite economical

     

    • Concave blade CD-6 disk turbine impellers are being widely used in industries which require gas dispersion in liquid.

     

    Based upon all only these points Concave Blade CD-6 Impellers are   selected.

    SELECTION OF HEATING MEDIUM:

     

    As the reactions occurring in the reactor are overall endothermic therefore there is a need of  a heating medium. Dowtherm Q (transfer fluid contains a mixture of diphenylethane and alkylated aromatics) has been selected as the heating medium. Because of the following reasons.

     

    1. Compared to hot oils, it exhibits better thermal stability, particularly at the upper end of hot oils’ use range, and significantly better low-temperature pumpability.

     

    1. Does not corrodes the reactor jacket.

     

    1. Highly efficient.

     

    1. Suitable working temperature range.

     

    1. Considered as economical in the thermal fluid family.GAS LIQUID CONTINUOUS STIRRED TANK REACTOR

      DESIGN STEPS:

       

      • Calculate gas holdup.

       

      • Subtract the gas holdup from 1 to get liquid holdup.

       

      • Calculate the volume occupied by the liquid using overall residence time.

       

      • As liquid holdup is volume occupied by the liquid divided by total volume. From this calculate the total volume of reactor.

       

      • Give allowance of headspace to this total reactor volume

       

      • Assume a value of superficial gas velocity.

       

      • Divide the volumetric flowrate of entering gas with superficial gas velocity. To get area of reactor.

       

      • From this area calculate the diameter of the reactor.

       

      • Divide the volume of rector by area of reactor to get the height of reactor.

       

      • Now the ratio of height of rector to diameter of reactor should be equal to 1. This provides the right value of superficial gas velocity.

       

      • Now calculate the internal dimensions using standard shape factors.

       

      • Calculation of power absorbed by the impeller requires first the determination of power absorbed if only liquid phase is present

       

      • For this purpose determine Reynolds number against this Reynolds number see the value of power number.

       

      • From this power number determine the value of power absorbed by the impeller.

       

      • Now determination of aeration number is done. Against this aeration number see the ratio of power absorbed by the liquid to the power absorbed by the gas.

       

      • This ratio is than used to calculate the power absorbed in the gas liquid mixing.

       

      • This is than followed by the calculation of gas and liquid holdup.

       

      • Than bubble diameter and sparger type is determined.

       

       

      REACTOR JACKET DESIGN STEPS:

       

      • Gather all the standard suitable jacket specifications.

       

      • Calculate the inside heat transfer coefficient.

       

      • Calculate the outside heat transfer coefficient.

       

      • Calculate the overall heat transfer coefficient and compare it with standard values.

       

      • Determine the pressure drop in the reactor jacket.DESIGN CALCULATIONS:

         

        • Step 1 :

                               Reactor Volume calculations

         

         

        • GAS HOLDUP(ЄG)

         

         

                                            ЄG = 0.25(QVGNR2/σ)0.45  

                                                    

                                   ( Ref : Hassan and Robinson (1997) )

         

         

        QVG = Volumetric flowrate of gas entering = 0.031m3/s

        NR = Rotational speed = 150 rpm = 2.5 rps

        σ = Surface tension = 0.07 N/m

         

        ЄG = 0.40

         

         

        • LIQUID HOLDUP (ЄL):

         

         

        ЄL = 1 – ЄG

         

        So putting the values e get

         

        ЄL = 0.60

         

         

         

        • VOLUME:

         

         

        ЄL = VL/VT

         

        Where

        VL = volume occupied by liquid = Flowrate of liquid entering multiplied (QL) multiplied by overall residence time(tr).

        As residence time(tr) is as follows

         

        tr = 1 hr = 60 minutes = 3600 seconds

         

        So

         

        VL = QL * tr = 0.00087 * 3600 = 3 m3

         

         

        VT = Total reactor volume =?

         

         

        Putting the values in the above equation we get

         

        VT = 5.2 m3

         

        According to the book “Chemical Process Equipment Design And

        Economics by H.Silla (2003)

         

        • If Volume < 1.9 m3 than 15 % allowance is must
        • If volume > 1.9 m3 than 10 % allowance is must

         

        Because VT > 1.9 m3. So with allowance of 10 %

         

         

        Total Reactor Volume = VR = 6 m3

         

         

        • Step 2:

                        Diameter and Height Calculation

         

        • SUPERFICIAL GAS VELOCITY(VSG):

         

        Assume a superficial gas velocity

         

        VSG = 0.01 m/s

         

        • AREA (A):

         

                                              A = QVG / VSG

         

        Putting the values we get.

        A = 3 m2

         

        • DIAMETER(DR ):

         

         

                                              A = (π/4) DR2

         

         

        From this equation we calculate diameter.

         

         

        DR = 2.00 m

         

        • HEIGHT(HR):

         

                                              HR = VR / A

         

        Putting the values we get.

         

        HR = 2.00 m

         

         

        Optimum HR / DR ratio of a gas liquid CSTR is 1. Reference : book

        Chemical Reactors by Tremobuze and Euzen. And also in my case

         

                                              HR / DR = 1

         

         

        • Step 3 :

         

                           Internal Dimension calculations

         

          

        According   to the book “CHEMICAL REACTORS” by

        Trembouze and Euzen we have.

        • Distance from the bottom of rector to the bottom of impeller = H1 = DR/3 = 0.67m
        • Height of the impeller blade = H3 = DR/5 = 0.13 m
        • Width of the impeller blade = L3 = DR/4 = 0.17m
        • Diameter of the impeller = DA = DR/3 = 0.67 m
        • Width of the baffles = DR/12 = 0.17 m

         

         

        • Step 4 :

         

        Power absorbed by the impeller

         

        • First let us calculate the power absorbed by the liquid stream only = PL

         

        • Reynolds number = Re = (ρL*NR*DA2)/µL

         

        Nomenclature

        • ρL = Average liquid density = 1103 Kg/m3
        • µL = Average liquid viscosity = 0.00065 Kg/m.s
        • NR = Rotational speed of impeller = 150 rpm = 2.5 rps

         

        Substituting the values we get

         

        Re = 1882356

         

        Now against this Reynolds number, Power number from figure 4.20 book  “Chemical Reactors” by Trembouze and Euzen is.

         

        NP = 6.5

         

        Now as power number is

         

        NP = PL / (ρL*NR3*DA5)

         

        Substituting the values and calculating PL

         

        PL = 4.0 hp/m3

         

        Aeration number = NQG = QVG / ( NR*DA3)

         

        Putting the values we get

         

        NQG = 0.042

         

        Now from figure 4.21 page # 276 book “chemical reactors“

         

        We have against this NQG ratio PLG/PL equal to

         

        PLG/PL = 0.58

         

                  PLG = 2.5 hp / m3

         

        • Step 7:

                                     Liquid holdup

            

                                                     LH = 3.14 m3

         

        • Step 8:

                                   Gas holdup

         

                                                    GH = 2.07 m3

         

         

        • Step 9:

         

                                  Sparger calculations and selection

         

        For this first we will have to calculate bubble diameter

         

         

        According to Jiang et al (1995) bubble diameter is given as below.

         

         dB2 = 8.8 * ((VSGL/σ)-0.04) * ((ρL* σ3)/gµL4) -0.12) * ((ρLG) 0.22) * (σ/g ρL)

         

        Where

         

        dB = bubble diameter

         

        Substituting the values we get

         

         dB = 3 mm

         

        According to data given by Trembouze et al if bubble diameter

        comes this small than sintered metal sparger is used.

         

        Selected Sparger: Sintered metal sparger

         

         

        • Step 10:

         

         

                           Reactor jacket calculations

         

           

         

         

        Jacket Specifications

         

        • Jacket type = Spiral baffle jacket
        • Height of jacket = 1.36 m
        • Spacing between reactor and jacket = 75 mm = 0.075m
        • Pitch = 200 mm= 0.2 m
        • Tenetring = 200 oC = 473 K
        • Tleaving = 190 oC = 463 K

         

        • Calculation of overall heat transfer coefficient:

         

        • Heat transfer coefficient at the outside wall of the reactor:

          

             Using the following equation to calculate the heat transfer coefficient at the outside wall of reactor (from Chemical Engineering by Coulson and Richardson volume 6).

         

             Nu = C * Re 0.8 * Pr 0.33       

         

        Where

        Re = Reynolds number

        Pr = Prandtl number

         

        C = constant = 0.023 (Reference: Chemical Engineering by Coulson and Richardson volume 6)

         

        Re = (ρ * v * de) / µ      eq (E)

         

        Where

         

        ρ = density of Dowtherm Q = 833.1 Kg/m3

        v = Velocity with which dowtherm is moving in jacket = 8.7 m/s

        de= hydraulic mean diameter = 0.109 m

        µ = viscosity of dowtherm = 0.000323 Pa.s

         

        Substituting the values in equation E

         

        We get

         

        Re = 2440250

         

        And

         

        Pr = Cp * µ / Kf      eq (F)

         

        Where

        Cp = Specific heat = 2.193 KJ/Kg .K

        Kf = Thermal conductivity = 0.0974 W/m.  K

         

        Substituting the values in equation F we get

         

        Pr = 7.3

         

        Substituting all these values in eq (D)

         

        Nu = 5702

         

        As

        Nu = ho de / Kf    eq ( G )

         

        Equating the value of Nusselt number in the equation above we get

         

        ho = 5091 W / m2 .C

         

        • Heat transfer coefficient at the inside wall of reactor

         

        Nu = 1.10 * Re .62  * Pr .33

         

        Re = Reynolds number = ρ * NR * DA 2/ µ = 1882356

        Pr = Prandtl number = Cp µ / K = 11

         

        Nu = 18646.3

         

        From definition of Nusselt number we have

         

        hi = 1448.30 W/m2.oC

         

        • OVERALL HEAT TRANSFER COEFFICIENT:

         

                                         1/U = (1/hi) + (1/ho) + (Xw/K)

        Where

        Xw = wall thickness=.01 m

        K = Thermal conductivity = 22 W/m. oC

         

        U = 760 W/m2 oC

         

        Also from Chemical Engineering by Coulson and Richardson Volume 6

        Figure 12.1 page # 639 U is between 750-1000 W/m2.oC

         

         

        • Calculation of pressure drop:

         

               From equation # 12.18 chemical engineering by Coulson and

               Richardson Volume 6 it is given as.

         

                                       ΔP = 8*Jf*(L/de)*ρ*v2/2

        Where

        Jf = friction factor

         

        Putting the values we get

         

        ΔP = 17412 Pa

        = 0.2 atm

        = 2.5 psi

        MECHANICAL DESIGN OF REACTOR

        • Step 1:

        Design Pressure

         

             Pdesign = Poperating + 0.075Poperating + Phydrostatic

         

        Pdesign = 29.2 atmospheres = 2958.7 K Pa

         

         

        • Step 2:

        Design Temperature

         

              Tdesign = 509 K

         

         

        • Step 3:

                                 Material Selection

         

        Following factors must be considered in selecting a suitable material of construction for the reactor.

         

        1. Mechanical properties
        2. Corrosion resistance
        3. Availability
        4. Cost

         

        Selection:

        The selected material is Zirconium (Zircadyne 702). The reasons for selection are as follows.

         

        1. The criteria of selection that overrides all others from chemical engineering point of view is corrosion. And whenever hot acids especially acetic acid because of its high corrosivity is present than Zirconium is used.
        2. It has great mechanical properties
        3. It is now easily available
        4. Its cost is not very high. It is very similar in price with high Nickel steel.

         

             Selected Material = Zirconium (Zircadyne 702)       

         

         

        • Step 4:

                                 Design Stress (f)

         

         

                f = 369300000 N/m2

        = 369.3 N/mm2

         

        • Step 5:

                                 Wall thickness

         

              e = PiDi/ (2*f-Pi)

         

              Where

        e = minimum wall thickness

        Pi= internal pressure

        Di=internal diameter

         

        Substituting the values in the above equation we get

        e = 0.008 m

        = 8 mm

         

        Giving 2mm corrosion allowance we get

         

                                            efinal = 10 mm

        = 0.001 m

         

        • Step 6:

                                 Head thickness    

                                  

                                   Ellipsoidal head is chosen

         

              ehead=PiDi/ (2*J*f-0.2*Pi)

         

              J = joint factor = 1

        Substituting the values in the above equation we get

        ehead = 8mm

        = 0.008 m

         

        Observe the similarity between the values of wall thickness and head thickness. This indicates that the choice of ellipsoidal head is correct.

         

         

        • Step 7:

        Weight of vessel:

         

        It is given as

         

        Wv = CvπρmDmg(Hv+0.8Dm) t

         

        Where

         

        Wv = Weight of vessel

        Cv   = A factor to account for manways, internal support etc. = 1.08

        Hv   = Length of cylindrical section = 2 m

        g   = Acceleration due to gravity = 9.8m/s

        ρm = Density of material of construction

        Dm = Mean diameter of the vessel = Di + e = 2.01 m

         

        After giving 2 mm corrosion allowance it becomes

         

        ehead = 10mm

         

        Substituting the values in the above equation we get

         

        Wv = 15081.4 N

         

        • Step 8:

                           Direct Stress

         

        It is due to the weight of the vessel and its contents

         

        σw = W/π* (Di + t)*t

         

        Substituting the values we get

         

        σw = 1428335 N / m2

        = 1.4 N/mm2

        • Step 9:

                          Principal stresses

         

         

        σ 1  =  (1/2)*(σh+σz+sqrt((σh-σz)^2+4ĩ^2))= longitudinal

        σ 2  =  (1/2)*(σh+σz-sqrt((σh-σz)^2+4ĩ^2))=circumferential 

        σ 3  =   0.5(P)=radial   

         

         Where

        ĩ = Torsional shear stress is very small and is usually neglected

        σz=Total longitudinal stress = σL + σw = 147 N / mm2

         

        Substituting the values in the above equations we get

         

                                            σ1=292 N/mm2

        σ2=147 N/mm2

        σ3= 1.48N/mm2

         

        Step 10:

                          Vessel support

         

        The method used to support a vessel depends on the size, shape and weight of the vessel; the design temperature and pressure; the vessel location and arrangement; and the internal and external fittings and attachments.

         

        There are basically three types of support for vessels.

         

        1. Saddle support
        2. Skirt support
        3. Bracket or lug support

         

        1. Saddle Support:

                                        These are used for horizontal vessels. As the reactor is not a horizontal vessel in this case, so this choice is ruled out.

         

        1. Skirt Support:

                                     These are used for tall vertical columns. As the H/D ration of reactor is 1 therefore it cannot be supported for this type of support.

         

        1. Bracket Support:

                                         These are used for all types of vessels. Therefore as the above two types of supports are not suitable for the reactor, so this type of support must e chosen.

         

        Thus the selected support is BRACKET SUPPORT.

        GAS LIQUID CONTINOUOUS STIRRED TANK REACTOR MECHANICAL DIAGRAM SHOWING DIMENSIONS:

        FLASH TANK DESIGN

        Introduction

         

        If we say that Chemical Engineering is nothing but the combination of art and science to design and control the separation equipment, it won’t be a lie. In a chemical industry, more than the 70% of total capital investment is incurred on separation and purification equipment. These stats might highlight the importance of separation equipment in chemical industry.

         

        Defining the problem:-

         

        In Cativa Process, one of the product streams is coming out from the reactor. This stream contains the Acetic Acid; which is our sole product, and the Iridium Catalyst Complex. We have to maintain some liquid level in the reactor as well so that we might use this liquid as the solvent for the incoming feed stream. The catalyst has to be recycled back to reactor for further utilization. So we need equipment that might separate out the product (not essentially all of it) and recycle back some fraction of Acetic Acid along with the catalyst. A little amount of water should also be maintained in the reactor as this is the requirement of the technology (Cativa Process) we are using. So up till now, we have successfully defined our problem. Let’s look for a solution to it.

         

         

        Looking for the solution:-

         

        Now there are a number of equipments that are available to us for this purpose. We need to have a look at the physical conditions of the stream. All the components are in liquid state at 110 oC and 27 atm pressure.

         

        We need to recycle some of the Acetic Acid and the catalyst back to reactor. Both of these are required to be there for further conversion. The feed mixture is in homogenous phase. This makes our choice quite simple. We can eliminate the possibility of a phase separator. One thing that must be kept in mind is that the solution has to be economical and quite effective. If we have a look at various industries; we find that most industries generate a second phase from this feed and recycle successfully some of the desired components in liquid state. This is quite an energy efficient process. Now let’s have a look at the possible choices that we have at our hand.

         

        Possible Choices Available:-

         

        We have our feed in liquid state in which catalyst is homogenously dissolved. We want some of the Acetic Acid, little amount of water and the catalyst recycled back to reactor. We’ll make use of equipment that can generate the vapor phase without expenditure of much of external energy and then successfully recycle the desired components back to reactor. One choice looks obvious. It’s the Flash Drum. There are other possible alternatives available to us, likewise Knockout Drum, Horizontal Flash Drum or the spherical one. All have their own characteristics and are used in specific situations. We’ll make use of Vertical Flash Drum.

         

        Construction of a Flash Drum:-

         

        When feed is flashed in a Flash Drum, vapor and liquid mixture is generated. As this mixture enters the drum, the surface area is increased, due to which pressure drop is generated. Right at eh entrance of the feed, there’s a splash plate in the drum. This splash plate directs the vapor and liquid flow downwards. This way the effect of gravity is enhanced. The liquid settles down at the bottom while the vapors with little momentum, change their path and rise up the vessel. At the top of the vessel, there’s a mist eliminator. Actually when vapors rise up the vessel, small liquid droplets also accompany them. The phenomenon of splashing is avoided by the use of splash plate. So our splash plate is serving two major purposes. First it helps us to avoid the splashing of liquid. Secondly, it directs the vapor liquid mixture downwards which in turn enhances the effect of gravity. Due to this effect, liquid is separated out of vapor. One thing should be kept in mind is that most of the impaction process takes place at the splash plate. So it has to be mechanically sound so that it can handle all the impact. Now there are two kinds of mist eliminators.

        • Vane type Mist Eliminator
        • Mesh Eliminators

        Vane type mist eliminator consists of metallic plates arranged closely to each other. Vapors with small liquid droplets rise. The plates are arranged in such a manner that they provide a zigzag path to the incoming vapor and liquid droplets. Droplets due to inertia and large momentum strike the plates and are captured at the surface while the vapors change their path accordingly and escape the eliminator. The phenomenon is referred to as Impaction and the size increase of droplets is called as Coalescence. Hence vapors are collected at the top of the vessel. A vane type mist eliminator is shown in the following figure.

        Now in mesh mist eliminators, a metallic or plastic wire mesh with a diameter ranging 0.006 to 0.011 in is used. The phenomenon is the same; impaction on the wire and then captured. Mist escapes the wire while droplets are captured at the surface where they coalesce and fall down as large drops. A mesh mist eliminator is shown in the following figure:

        There’s a radial vane vortex breaker shown at the bottom of the vessel. The purpose of this vortex breaker is to avoid the phenomenon of Vortex Formation. There are a couple of causes that induce the vortex formation in the drum. The first one is the earth’s rotational speed. Due to the earth’s rotational speed, anticlockwise vortex is observed in Northern Hemisphere while a clockwise motion is observed in Southern Hemisphere. Second reason is the introduction of feed in the vessel tangentially. Whenever feed is entered tangentially, vortexes are formed. Third reason is the vapors. Whenever there’s a two phase mixture and they differ in their velocity; then the fluid with lesser velocity and high density would start the rotational motion (Vortex Formation). In our case, we are handling a vapor-liquid mixture. Vapors are at a higher speed in the vessel while the liquid are a bit slower due to the impaction with the splash plate. So the vapors would induce the vortex to the liquid. The formation of vortexes brings some disadvantages to the system. Our system with vortex formed, experiences:

        • Loss of valuable vapors
        • Downstream equipment damage
        • Loss of flow
        • Erroneous liquid level readings resulting in poor control
        • Vibrations caused by unsteady two phase flow.

        The formation of vortexes is shown in the following figure:

        To avoid the vortex formation, we should avoid the usage of a tangential feed line. Secondly, we can use a vortex breaker to get rid of vortexes. Following types of vortex breaker are usually used in the industry:

        • Flat plate vortex breaker
        • Crosses
        • Radial vane or gratings

        We are using a Radial Vane Vortex Breaker. A vortex breaker is stationary and it doesn’t move. If it starts the motion with the vortex then it wouldn’t break the vortex rather it would just weaken it. To break the vortex and get rid of it, we’ll have to fix the vortex breaker and make it stationary.

         

        Why use Vertical Flash Drum?

         

        Let’s carry out the process of elimination to justify our choice. We can simply rub aside the choice of Knockout Drum as it is used wherever there’s gas in the feed stream. In our stream there are no gases. We have only liquid phase. So we will not go for the Knockout Drum. Now we are left with Horizontal, Spherical and Vertical Flash Drums. Horizontal Drums are used when we have to handle a large liquid flow rate. But in our case we’ll see that the liquid flow rates wouldn’t be that huge. Instead we’ll have to deal with a high amount of vapor flow rate. Also Walas carried out a survey and in his book “Chemical Process Equipment Selection and Design” writes that out of every ten chemical industries; seven are making use of Vertical Flash Drums. The choice is made due to the economy and the ease with which we can handle the flow rates. A design engineer is required to start designing a Vertical Flash Drum by default and then after the design is complete we have a look at the L/D (length to Diameter Ratio) to decide which configuration to use. So we’ll follow the same procedure. We’ll design a Vertical Flash Tank and then would analyze the L/D ratio obtained to determine which configuration to use. Just remember one rule of thumb; for large liquid flow rates, we’ll use Horizontal Flash Drum and for small liquid flow rate, you’ll go for a vertical configuration. You can start designing any one of these and then the final decision would rest upon the L/D ratio of the drum. So don’t bother. Just start your computer software and begin designing any configuration. Let’s start the design of Vertical Flash Drum. Before the process of designing, we’ll see what exactly flashing is.

         

        Throttling:-

         

        When a fluid (liquid or a liq/vapor mixture) at high temperature and high pressure experiences sudden reduction in pressure, then some of the liquid is vaporized and the phenomenon is referred to as Throttling. During the process the temperature of the feed stream doesn’t change that much and in such a case the process is called as Adiabatic Flashing. Actually for an ideal gas or a fluid behaving likewise an ideal gas, there’s no temperature drop. But in real fluids, little temperature drops have been observed. These temperature drops are due to the Joule-Thomson Effect and the frictional loss. Since there’s no appreciable change in the kinetic and potential energy; and also there’s no shaft work or heat transferred, therefore the eq:

        Δ (H + u2/2 + gz) = Q + Ws reduces to ΔH = 0.

        We know that the enthalpy depends upon the temperature of fluids. Since there’s no change in enthalpy so theoretically there will be no change in the temperature of the fluid stream. Usually for real fluids, a very little temperature drop is observed. In our case, the feed is at 110 oC and the pressure is 27 atm. We’ll suddenly reduce the pressure of the liquid stream and this would ultimately generate a vapor phase without the expenditure of any external energy. There will be ignorable temperature drop. All the beauty of equipment lies in this phenomenon. We are generating a second phase without expanding any external energy. But we know that energy is always conserved. We have generated the vapors on the expense of the pressure of the incoming feed So although the process of throttling makes us lose some of the energy contents of the feed stream, yet we get more benefits. Now the problem comes out to be the selection of the valve.

         

        Selection of Valve:-

         

        No ordinary valve would be used for this purpose. We need such a valve that would handle a feed stream with such a high temperature and pressure and allow it to expand suddenly. The valve would allow only one sided flow of the stream. There are a number of options open to us. Globe Valve, Gate Valve, Butterfly Valve, Ball Valve etc are all at our disposal. But none of these is manufactured for the purpose of throttling. As we look for the best choice, we come to know that there’s a valve that is manufactured keeping in mind the sole idea of throttling. This is Lever sealed Plug Cock. The valve operates up to a temperature of 260 oC. It has plastic lining that makes it corrosion resistant. It has a tapered plug that is moved by a lever. The plug contains perforations just like a ball valve. As the feed stream passes through it, pressure drops from 27 atm to 1.4 atm. The temperature change is negligible. So after getting flashed, vapors are generated. The temperature of the stream remains more or less the same.

         

        Determination of Flash Temperature:-

         

        To determine the Flash temperature, we’ll have to determine the Dew

        Point and the Bubble Point Temperatures. To calculate the Bubble Point Temperature, we assume that all of our feed is saturated liquid. We assume a temperature and at that temperature, the K value for the component is determined. Then by multiplying this K value with the liquid weight fraction, we get the vapor fractions. The sum of these fractions should be unity in order to have the correct Bubble Point Temperature. So we see that it’s a hit-and-trial method. Similarly, we assume a temperature and at that temperature we determine the K value for the component. Then we assume that all of our feed is saturated vapor. So dividing these fractions with the K values, we get liquid fractions; whose sum should be unity. If the sum of liquid fractions is unity then our assumed Dew Point is correct. Taking the arithmetic average of this Bubble Point and Dew Point Temperatures, we get the Flash Temperature. K values for Iodomethane and Acetic Acid has been determined directly from the Himmelblau Software. The equation that this software uses is V.P = A- {B/(T + C)} Here A,B and C are empirical constants while T is the assumed temperature. By dividing this Vapor Pressure (V.P) by the total pressure, we get the K value. K values for Water and Methyl Acetate have been determined by using the empirical relation given in Perry’s Chemical Engineering Handbook. The relation is:

        V.P = exp [C1+C2/T + (C3*lnT) + (C4*TC5)] * 9.869233E-06 atm

        Here C1, C2, C3, C4 and C5 all are empirical constants and there value is given in Chemical Engineer’s Handbook by Perry. The Vapor Pressure thus obtained is divided by the total pressure to get the K value at the assumed temperature. The process of calculating Bubble Point and the Dew Point Temperature is given below:

         

        Compound Xi K at 101 oC K*Xi
        Acetic Acid 0.632 0.416 0.263
        Methyl Acetate 0.215 2.673 0.574
        Iodomethane 0.018 3.682 0.065
        Water 0.136 0.736 0.100
          1.000   1.002

         

        This employs that our Bubble Point Temperature is 101 oC.

         

        Compound Yi K at 119 oC Yi/K
        Acetic Acid 0.632 0.739 0.855
        Methyl Acetate 0.215 4.163 0.052
        Iodomethane 0.018 5.469 0.003
        Water 0.136 1.348 0.101
          1.000   1.011

         

        This determines our Dew Point Temperature which comes out to be 119 oC. Now:

        Flash Temperature = (101 + 119)/2

        = 110 oC

        Determining the Vapor and Liquid Flows:-

         

        Determination of Vapors going out and the liquid draining the drum is a result of some lethal calculations. These calculations are explained over here. First we make a material balance for a single component. It yields:

        Fxfi = Vyi + Lxi ………….Eq. I

        From Henry’s Law, we have:  xi = yi /K

        Putting this value in Eq.I, we get:

        Fxfi = Vyi + L (yi/K)……….Eq. II

        • yi = Fxfi / (V + L/K)

        Since F = V + L which employs that V = F – L, therefore;

        yi = Fxfi / (F- L + L/K)

        yi = xfi / {1 – L/F (1 – 1/K)}

        Also from Eq. II, we can write that

        yiV = Fxfi / (1 + L/VK)………..Eq. III

        Which employs that:    yi = (Fxfi / V) / (1 + L/KV)……..Eq. IV

        Now after determining yi’s, we can calculate xi’s by using the K values from the expression:         yi = Ki xi where Ki is determined by using the relation Ki = V.P/P

        V.P stands for Vapor Pressure at the specified Flash Temperature.

        Eq. III can be written in the form as:

         

         i=c                       i=c

        Σi=1 (yi V) = Σi=1 {Fxfi / (1 + L/KV)}……..Eq. V

         

        Procedure to be followed for Flash Calculations:-

         

        So simplifying all the procedure, we come to know that if we are to calculate V, L, yi’s and xi’s then we’ll have to follow these steps:

        1. Assume V.
        2. Calculate L = F – V
        3. Calculate L / V
        4. Look up for K values at Flash Temperature and Total Vessel Pressure
        5. Substitute values in;

                   i=c

        V = Σi=1 {Fxfi / (1 + L/KV)}

        If equality is obtained between the assumed V and the calculated V, then the assumed value is satisfactory.

        1.       Calculate yi’s from Eq. IV
        2.       Calculate xi’s from yi = Ki xi

        Now using this procedure the values of V, L, yi’s and xi’s have been calculated for the Flash Drum. The calculations are given below.

        Getting started with Design of Flash Drum:-

         

        Since our calculations are based upon an hour of operation, so we have the following amount of vapor and liquid flow rates;

        FL = 1480.196 kg/hr                pL = 961.55 kg/m3

                        Fv = 4664.308 kg/hr                pv = 2.654 kg/m3

        Vapor liquid separation factor, which is equal to (FL/Fv) / (pv/pL) ½; comes out to be 0.017. Using the graph, we notice that the Vapor Velocity Factor is equal to 0.35 m/sec.

        Maximum design vapor velocity is obtained by multiplying the vapor velocity factor with {(pL – pv)/ pL} 0.5. The value of velocity comes out to be

        Uv = Kv* {(pL – pv)/pL} 0.5

        Uv = 6.653 m/sec

        If we divide the vapor mass flow rate by the density, we get the volumetric flow rate. So

        VL = 0.488 m3/sec

        Dividing volumetric flow rate by the vapor velocity, we get the minimum cross sectional area of the drum. Hence

        Amin = VL/Uv

        Amin = 0.073 m2

        From this minimum cross sectional area, we can calculate the minimum diameter for the vessel. The minimum diameter is:

        Dmin = 0.306 m

        Actual internal diameter is obtained by adding 6in to this minimum diameter. Therefore

        D = 0.458 m

        For a vertical Flash Drum the surge time is in the range of 4 to 7 min and that for a horizontal vessel, it ranges between 7 to 12 min. Flash Drum used in Cativa Process has a surge time of 5 min. So multiplying this time with the liquid volumetric flow rate, we get the liquid volume held in the flash drum.

        Liquid Volume = VL * 300

        Liquid Volume = 0.128 m3

        Since the vessel is cylindrical, therefore its volume is equal to 3.145*(radius) 2*height. Using this relation, we can determine the liquid height in the vessel. The liquid height comes out to be:

        Liquid Height = 0.779 m

        Now both H. Silla and Coulson have suggested the following formula for determining the vapor height in the vessel. This formula is:

        Vapor Height = 1.5* D + 0.4

        Vapor Height = 1.087 m

        Now by adding the liquid and vapor heights, we can determine the total internal height of the vessel. Thus

        Total Height = 1.866 m

        The L/D ratio for the Flash Drum comes out to be 4.072 which is a satisfactory value. This ratio actually determines the type of vessel. It tells us that whether we should go for a horizontal vessel or a vertical one. If the value of L/D ratio is between 3 and 5, then a vertical flash drum is used. If its value exceeds 5, then a horizontal vessel should be employed.

         

        Material of construction:-

         

        Though material of construction is the part of mechanical design of the equipment but we can predict about it. Since we are dealing with acidic, corrosive fluid; therefore we’ll have to look for a material that is corrosion resistant. We come across two important choices that are corrosion resistant as well as economical. The flash drum can either be manufactured from Stainless Steel or we may make use of Aluminium. We can use either of the materials. Both have good mechanical strength, quite resistant to corrosion and are also cheap. Most of the heat transfer equipment in industry is made up from Aluminium Alloys. We are not that concerned with the heat transfer over here, so stainless steel is recommended as the priority material of construction.

        DESIGN OF DISTILLATION COLUMN

        Introduction:

        The separation of liquid mixtures into their various components is one of the major operations in the process industries, and distillation, the most widely used method of achieving this end, is the key operations in any oil refinery. In processing the demand for purer products, coupled with the need for greater efficiency, has promoted continues research into techniques of distillation. This process of getting pure products is accomplished by partial vaporization and subsequent condensation.

         

        Distillation:

        “Process in which a liquid or vapour mixture of two or more substances is separated into its component fractions of desired purity, by the application and removal of heat”

        TYPES OF DISTILLATION COLUMNS;

        There are basically two types of distillation columns used in industries.

         

        • Batch columns
        • Continuous columns

        There selection criteria depends upon total number of stages and reflux ratio. As it is shown that when a large number of plates are used, then continuous distillation has the lowest reflux requirements and hence operating costs. If a smaller number of plates are used and high purity product is not required, then batch distillation is probably more attractive.

         

         

        Batch Columns:

        In batch distillation the more volatile component is evaporated from the still which therefore becomes progressively richer in the less volatile constituent. Distillation is continued, either until the residue of the still contains a material with an acceptably low content of the volatile material, or until the distillate is no longer sufficiently pure in respect of volatile content. In batch operation, the feed to the column is introduced batch-wise. That is, the column is charged with a ‘batch’ and then the distillation process is carried out. When the desired task is achieved, a next batch of feed is introduced. Most distillation processes operate in a continuous fashion, but there is a growing interest in batch distillation, particularly in the food, pharmaceutical, and biotechnology industries. The advantage of this separation process is that the distillation unit can be used repeatedly, after cleaning, to separate a variety of products. The unit generally is quite simple, but because concentration are continuously changing, the process becomes more difficult to control.

         

         

        Continuous Distillation:

        In contrast to batch columns, a continuous feed is given to the column. No interruptions occur unless there is a problem with the column or surrounding process units. They are capable of handling high throughputs and are the more common used. I will put light only on this type of distillation column.

         

        CHOICE BETWEEN PLATE AND PACKED COLUMN

         

        The choice between use of tray column or a packed column for a given mass transfer operation should, theoretically, be based on a detail cost analysis for the two types of contactors. However, the decision can be made on the basis of a qualitative analysis of relative advantages and disadvantages, eliminating the need for a detailed cost comparison.

        Which are:

        • Because of liquid dispersion difficulties in packed columns, the design of tray column is considerably more reliable.
        • Tray columns can be designed to handle wide ranges liquid rates without flooding.
        • If the operation involves liquids that contain dispersed solids, use of a tray column is preferred because the plates are more accessible for cleaning.
        • For non-foaming systems the plate column is preferred.
        • If periodic cleaning is required, man holes will be provided for cleaning. In packed columns packing must be removed before cleaning.
        • For large column heights, weight of the packed column is more than plate column.
        • Design information for plate column is more readily available and more reliable than that for packed column.
        • Inter stage cooling can be provided to remove heat of reaction or solution in plate column.
        • When temperature change is involved, packing may be damaged.
        • Random-packed columns generally are not designed with diameters larger than 1.5 m, and diameters of commercial tray column are seldom less than 0.67m.

         

        As my system is non foaming and diameter calculated is larger than 1.5m so I am going to use tray column.

        Also as average temperature calculated for my distillation column is higher that is approximately equal to 98oc. So I prefer Tray column.

         

        PLATE CONTACTORS:

        Cross flow plate are the most commonly used plate contactor in distillation. In which liquid flows downward and vapours flow upward. The liquid move from plate to plate via down comer. A certain level of liquid is maintained on the plates by weir. Other types of plate are used which have no down comer (non-cross flow) the liquid showering down the column through large opening in the plates (called shower plates). Used when low pressure drop is required.

        Three basic types of cross flow trays used are

        • Sieve Plate (Perforated Plate)
        • Bubble Cap Plates
        • Valve plates (floating cap plates)

        I prefer sieve plate because:

        (1) Their fundamentals are well established,        entailing low risk.

        (2) The trays are low in cost relative to many other types of trays.

        (3) They can easily handle wide variations in flow rates.

        (4) They are lighter in weight. It is easier and cheaper to install.

        (5) Pressure drop is low as compared to bubble cap trays.

        (6) Peak efficiency is generally high.

        (7) Maintenance cost is reduced due to the ease of cleaning.

        Factors Affecting Selection of Trays:

            

        • Relative Cost of plate will depend upon material of construction used.

         

         

        For mild steel, the ratio of cost between plates is

         

         

        Sieve plate     :    valve plate    :    bubble-cap plate

        3.0           :         1.5           :             1.0

         

        • There is little difference in Capacity Rating of the three types (the column diameter required for a given flow rate).

        Sieve tray     >      valve tray    >      bubble-cap tray

              

        • Operating Range means the range of liquid and vapour flow rates which must be above the weeping conditions and below the flooding conditions. Operating range flexibility comparison is.

        Bubble cape tray >   Valve tray   >   Sieve tray

         

        Sieve plate depends on the vapours flow through the holes to hold the liquid on the plate, and cannot operate at very low vapour flow rates. But with good design, sieve plate gives satisfactory operating range.

         

        • The Plate pressure drop will depends on the detailed design of plate but, in general, sieve plate gives the lowest pressure drop, followed by valves, with bubble-caps giving the highest.

         

         

         

        Operation of Typical distillation Column:

        The operation of typical distillation column may by followed by figure. The column consists of a cylindrical structure divided into sections by a series of perforated trays which permit the upward flow of vapour. The liquid reflux flows across each tray, over a weir and down a down comer to the tray below. The vapour rising from the top tray passes to condenser and then through an accumulator or reflux drum and a reflux divider, where part is withdrawn as the overhead product D and the remainder is returned to the top tray as reflux R.

        In the bottom there is reboiler which is used to give heat to the system. Liquid from the bottom of distillation column is fed to the reboiler which vaporises the in coming liquid. These vapours in turn move towards the bottom plate interact with the liquid over that plate. Due to which partial condensation of vapours occur. Also partial vaporization of liquid occurs too. That is less volatile component condensed first and more volatile component vaporizes first. This phenomenon occurs on each plate. Causing enrichment on each plate.

         

        A schematic of a typical distillation unit with a single feed and two product streams is shown below.

        FACTORS AFFECTING DISTILLATION COLUMN OPERATION

         

                              Vapour Flow Conditions

         

        Adverse vapour flow conditions can cause:

        • Foaming
        • Entrainment
        • Weeping/dumping
        • Flooding

         

        è Foaming

         

        Foaming refers to the expansion of liquid due to passage of vapour or gas. Although it provides high interfacial liquid-vapour contact, excessive foaming often leads to liquid build-up on trays. In some cases, foaming may be so bad that the foam mixes with liquid on the tray above. Whether foaming will occur depends primarily on physical properties of the liquid mixtures, but is sometimes due to tray designs and condition. Whatever the cause, separation efficiency is always reduced.

         

        è  Entrainment

         

        Entrainment refers to the liquid carried by vapour up to the tray above and is again caused by high vapour flow rates. It is detrimental because tray efficiency is reduced: lower volatile material is carried to a plate holding liquid of higher volatility. It could also contaminate high purity distillate. Excessive entrainment can lead to flooding.

         

         

        è  Weeping/Dumping

         

        This phenomenon is caused by low vapour flow. The pressure exerted by the vapour is insufficient to hold up the liquid on the tray. Therefore, liquid starts to leak through perforations. Excessive weeping will lead to dumping. That is 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.

         

        è  Flooding

         

        Flooding is brought about by excessive vapour flow, causing liquid to be entrained in the vapour up the column. The increased pressure from excessive vapour also backs up the liquid in the down comer, causing an increase in liquid hold-up 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.

         

        Reflux Conditions:

        Minimum trays are required under total reflux conditions, i.e. there is no withdrawal of distillate. On the other hand, as reflux is decreased, more and more trays are required.

         

         

        Feed Conditions:

        The state of the feed mixture and feed composition affects the operating lines and hence the number of stages required for separation. It also affects the location of feed tray.

         

        State of Trays:

        Remember that the actual number of trays required for a particular separation duty is determined by the efficiency of the plate. Thus, 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.

         

        Column Diameter:

        Vapour flow velocity is dependent on column diameter. Weeping determines the minimum vapour flow required while flooding determines the maximum vapour flow allowed, hence column capacity. Thus, if the column diameter is not sized properly, the column will not perform well.

         

        DESIGNING STEPS OF DISTILLATION COLUMN

        • Calculation of Minimum number of stages.Nmin
        • Calculation of Minimum Reflux Ratio Rm.
        • Calculation of Actual Reflux Ratio.
        • Calculation of theoretical number of stages.
        • Calculation of actual number of stages.
        • Calculation of diameter of the column.
        • Calculation of weeping point.
        • Calculation of pressure drop.
        • Calculation of thickness of the shell.
        • Calculation of the height of the column.Process Design:

          Using Vapors Liquid Equilibrium Data

          Temperature of feed = 119 o C

          Temperature of top product =71 o C

          Temperature of bottom product = 118 o C

          From Material Balance:

           

           Component

           

          Feed

          Fraction

          xf

            Bottom

          Fraction

            xb

          Top

          Fraction

            xd

           

          (1) Methyl Iodide

           

          (2) Acetic Acid

           

          (3)Methyl Acetate

           

          (4) Water

           

          0.074

           

          0.65

           

          0.215

           

          0.065

           

          0

           

          0.99

           

          0

           

          0.01

           

          0.212

           

          0.0005

           

          0.62

           

          0.167

           

          Heavy Key Component = Acetic Acid

          Light Key Component = Water

          Calculation of Minimum no. of Plates:

          The minimum no. of stages Nmin is obtained from Fenske relation which is,

          Nmin  =        LN[(xLK/xHK)D(xHK /xLK)B]

          LN (αLK/HK) average

           

          To find average geometric relative volatility of light key to heavy key:

           

          Average geometric relative volatility = 1.53

          So,

           

          Nmin = 24 (reboiler is excluded)

           

           

          Calculation of Minimum Reflux Ratio Rm:

          Using Underwood equation,

           

          As feed is entering as saturated vapors so, q = 0

          By trial, q = 1.68,

          Using equation of minimum reflux ratio,

           

          Putting all values we get,

          Rm = 4.154

          Actual Reflux Ratio:

          The rule of thumb is:

          R   = (1.2 ——- 1.5) R min

          R   = 1.5 R min

          R   = 6.23

           

           

          Theoretical no. of Plates:

          Gilliland related the number of equilibrium stages and the minimum reflux ratio and the no. of equilibrium stages with a plot that was transformed by Eduljee into the relation;

           

           

           

          From which the theoretical no. of stages to be,

          N= 39

           

          Calculation of actual number of stages:

          Overall Tray Efficiency:

          Using O’Connell’s Correlation overall tray efficiency can calculated using average viscosity and relative volatility evaluated at average temperature.

          Eduljee has expressed the O’ Connell graphical method to a mathematical relation,

           

          α avg =average relative volatility of light key component  = 1.75

          μ avg = molar average liquid viscosity of feed evaluated at average            temperature of column

           

          Average temperature of column = (118+71)/2 = 95 oC

          Feed viscosity at average temperature = mavg = 0.3850

          mNs/m2

          So,

          Eo = 56.60%

          So,

          No. of actual trays = 39/0.566 = 68

           

          Location of feed Plate:

          The Kirk bride method is used to determine the ratio of trays above and below the feed point.

           

           

          From which,

          Number of Plates above the feed tray = ND = 47

          Number of Plates below the feed tray = NB = 21

           

          Determination of the Column Diameter:

           

                     Top Conditions         Bottom Conditions
                 Ln =155.13 Kgmol/hr

          Vn = 180.02 Kgmol/hr

          T = 71 0C

          ρV =  3.469 Kg/m3

          ρL = 1123.93 Kg/m3

                 Lm = 226.8  Kgmol/hr

          Vm = 180.02 Kgmol/h r

          T = 118 0C

          ρV = 1.87 Kg/m3

          ρL = 938.85 Kg/m3

           

          Because liquid flow rates are greater at top so based upon bottom liquid flow rates.

           

          Flow Parameter:

           

           

          FLV = Liquid Vapor Factor = 0.0562

           

          Capacity Parameter:

          Assumed tray spacing = 18 inch (0.5 m)

          From Fig (15-5) Plant Design and Economics for                 Chemical Engineering, sieve tray flooding capacity,

          Csb = 0.0760 m/Sec

          Surface tension of Mixture = σ = 18.35 dynes/Cm

           

           

          Vnf=1.6740 m/sec

          Assume 90% of flooding then

          Vn=0.9Vnf

          So, actual vapor velocity,

          Vn=1.507 m/sec

          Net column area used in separation is An =mv/Vn

          Volumetric flow rate of vapors = mv

          mv = (mass vapor flow rate /(3600) vapor density)

          mv = 2.1184m3/sec

          Now, net area = mv/Vn = 1.4061m2

          Assume that downcommers occupies 15%of cross sectional Area (Ac) of column thus:

          Ac = An + Ad

          Where, Ad = downcommer area

          Ac = An + 0.15(Ac)

          Ac = An / 0.85

          Ac=1.6542 m2

          So Diameter of Column Is

          Ac =π/4D2

          D = (4Ac/π)

          D = 1.4513 meter = 5ft (based upon bottom conditions)

           

          Liquid flow arrangement:

          In order to find liquid flow arrangement first find maximum liquid volumetric flow rate

          So liquid flow rate =

          (Liquid mass rate)/ (3600) (Liquid    density)

          Max Liquid Rate Is At the bottom of column so using “Lm” values

          So Maximum liquid flow rate = 0.0050m3/sec

          So from fig11.28 cross flow single pass plate is selected

          Provisional Plate Design:

          Column Diameter Dc= 1.4513 m

          Column Cross-sectional Area(Ac)= 1.6542 m2

          Down comer area Ad    = 0.15Ac = 0.2481 m2

          Net Area (An) = Ac – Ad =1.406 m2

          Active area Aa=Ac-2Ad = 1.1580 m2

          Hole area Ah take 10% Aa = 0.1 × 1.1580 = 0.0462 m2

           Weir length

          Ad / Ac = 0.248 / 1.654 = 0.15

          (From figure 11.31 volume 6) Coulson and Richardson’s

          Lw / dc                  =          0.80

          Lw    = 1.452*0.80

             = 0.733 m

          Weir length should be 60 to 85% of column diameter which is satisfactory

          Take weir height, hw     =      50 mm

          Hole diameter, dh          =      5 mm

          Plate thickness              =      5 mm

          Check Weeping:

           

          where Umin is the minimum design vapor velocity.

          The vapor velocity at weeping point is the minmum velocity for the stable operation.

          In order to have K2 value from fig11.30 we have to 1st find how(depth of the crest of liquid over the weir)

          Where how is calculated by following formula:

          how=750{[Lm/lw*ρ]2/3}

           

          Maximum liquid rate    Lm   = 4.73 kg/sec

          Minimum Liquid Rate At 70% turn down ratio = 3.3Kg/sec

          At Maximum rate ( how)= 19.95 mm Liquid

          At Minimum rate (how) = 15.72 mm Liquid

          hw + how = 50 + 15.7245 = 66 mm Liquid

          From fig 11.30, Coulson and Richardson Vol.6

          K2 = 30.50

          So,

          U (min) = 8.89 m/sec

          Now maximum volumetric flow rate (vapors)

          Base = 2.12 m3/sec

          Top   = 1.14 m3/sec

          At 70% turn down ratio

          Actual minimum vapor velocity =minimum vapor rate / Ah

          = 12.81 m/sec

                  So minimum vapor rate will be well above the weep point.

           

          Plate Pressure Drop (P.D):

          Consist of dry plate P.D (orifice loss), P.D due to static head of liquid and residual P.D (bubbles formation result in energy loss + froth formed in operating plates)

          Dry Plate Drop:

          Max. Vapor velocity through holes (Uh) = Max Volumetric Flow Rate / Hole Area = 18.30 m/sec

          Perforated area Ap (active area) = 1.158 m2

          Ah/Ap = 0.100

          From fig. 11.34(for plate thickness/hole diameter) = 1.00

          We get,                            Co = 0.840

          This equation is derived for orifice meter P.D

          hd=  48.1 mm Liquid

          Residual Head (hr):

          hr = (12.5*103 / ρL)

          = 13.312 mm Liquid

          So,

                                                        

          Total pressure drop = 48.1 + (50 + 19.95) + 13.32

          ht = 131.35 mm liquid

          Total column pressure drop Pa, (N/m2) = (9.81*10-3) htρLN

          =      82771.6 Pa

          Down comer Liquid Backup:

          Caused by P.D over the plate and resistance to flow in the downcomer it self.

          hb = (hw+ how) + ht + hdc

          where hdc is equal to to head loss in downcomer.

          The main resistance to flow in downcomer will be caused by constriction in the downcomer outlet, and head loss in the down comer can be estimated using the equation given as,

          where Lwd is the liquid flowrate in downcomer, kg/sec

          and Aap is the clearance area under the downcomer, m2

          Aap =hapLw

           

          Where hap the height of bottom edge of apron above the plate.

          hap = hw – (5 to 10 mm)

          hap = 40.00 mm

          so,

          Area under apron “Aap” = .0464 m2

          As this is less than area of downcomer Ad so using Aap values in above formula.

          So,

          hdc = 1.95 mm

          As a result,

          hb   = 203.24 mm

          = 0.203 m

          hb < ½ (Tray spacing + weir height)

          0.203 < 0.25

                                      So tray spacing is acceptable

           

          Check Residence Time:

          Sufficient residence time should be allowed in the downcomer for the entrained vapors to disengage from liquid stream to prevent aerated liquid being carried under the downcomer.

          tr =Ad hbc ρL/L(max)

          tr = 10.02 sec

          It should be > 3 sec. so, result is satisfactory

           

          Check Entrainment:

          (un) actual velocity (based on net area) = (maximum volumetric flow rate at base Vm / net area An)

          (un) actual velocity = 1.51 m/sec

          Velocity at flooding condition Uf = 1.67 m/sec

          So Percent flooding =un/uf = 0.90 = 90%

          Liquid flow factor FLV = 0.0562

          From fig. 11.29 Coulson vol.6 fractional entrainment ψ can be found out.

          Fractional entrainment (ψ) = 0.0750

          Well below the upper limit of (ψ) which is 0.1. Below this the effect of entrainment on efficiency is small.

           

          No of Holes:

          Area of 1 Hole = (π/4) Dhole2 = 0.0000196 m2

          Area of N Holes = 0.1158 m2

          So,

          Number OF Holes = 5900

          Height of Distillation Column:

          Height of column Hc= (Nact-1) Hs+ ∆H

           

          No. of plates = 68

          Tray spacing Hs = 0.50 m

          ∆H= (1-1.5% of total height) for liquid hold up and vapor disengagement

          ∆H=0.55 m

          Total thickness of trays = 0.005*68 = 0.34 m

          So,

          Height of column = (68-1)*.50+ 0.55+0.34

          = 35 meters

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