Sunday, 16 February 2014

Production of adipic Acid

Introduction

Adipic Acid

Adipic acid is one of the most important aliphatic acids with the following chemical formula ; COOH(CH2)4COOH. little of adipic acid occurs naturally but it is produced on large scale all around the world

Physical chemical properties

under the normal conditions  (room tenperature and atmospheric pressure) adipic acid is an odorless,sour tasting solid present as awhite powder.the melting point of thr acid is 152c. it is soluble in polar solvents the solubility of adipic acid in water grows with increasing temperature

Application of adipic acid

adipic acid is an important chemical intermediate the main use of aa is a manufacture of nylon6,6 . nylon 6,6 is mainly used for production of fibers (fishing lines tires carpets home furnishing and in tough fabrics for backpacks parachutes luggage and business cases) and of resins. nylon resins sre used in electrcal connectors auto parts and items such as self lubricating bearings gears and cams

although thr main use of adipic acid is thr above written there are more ways of using adipic acid as can be seen from the figer.

Production of adipic acid

in 2006 the global AA capacity was around 2.8 million metric tons per year. the overall growth for AA is about 3% per year but the demand is growing faster in the year 2010 it was expected to be about 5-6% the most rapidly growing sector is the nylon one where during the past decade it was between 8-10% per year 

Current industrial processes

adipic acid is produced by a two step oxidation process. the first one is oxidation of cyclohexane with air the second one is an oxidation of 

a) a mixture of cyclohexane/cyclohexanol (called KA oil  ) or

b) onoy of cyclohexanol with nitic acid 

First step: Oxidation of cyclohexane with air

cyclohexane is obtained by the hydrogenation of benzene of from naphtha fraction in small amounts oxidation of cyclohexane was first performed in the yer 1940 by Dupont. the process is performed at temperatures between 150-180c under the pressure of 10-20 atm in the presence of Co or Mn organic salts . it is a two steps process oxidation and deperoxidation step. the rate limiting step of the process is the synthesis of hydroperoxide its concentration is optimized by carrying out the oxidation in passivated reactors and in the absence of transition metal complexes in order to avokd thr decomposition of hydroperoxide. the deperoxidation step in done in the second reactor , wher the catalyst amount and reaction coneition are optimixed allowing the OL/ONE  ratio to be controlled . for the fact that both of the products are more reactive than the starting material , conversion of cyclohexane is kept low only about 5-7% in order to avoid consecutive reactions . the selectivity to KA oil is about 75-80% the by products are carboxylic acids amd cyclohexayl hydroperoxide . the unconverted cyclohexane is recycled scientific design company has developed a variation of this step , it consist in the addition of substantial quantities of anhydrous mera boric acid as alurry in cyclohexane to the first of a staged serier of oxidation reactors . no other catalydt is necessary . boric acid reacts with cyclohexanol to give a borate ester that stabilizes the products . conversion is as high as formed is easily hydrolysed by hot water to boric acid and cyclohexanol . after distillation a 99.5% OL/One mixture is obtained

CHAPTER2:

PROCESS DESCRIPTION:

Reactor 1:

in commercial use two approaches predominate the air oxidation of cyclohexane process; cobalt catalyzed oxidation and borate promoted oxidation. a third method the high peroxide process has found limited commercial use cobalt catalyzed air oxidation of cyclohexane is the most widely used method for producing acipic acid cyclohexane is oxidized with air at 150c to 160 and 810 to 1013kpa (about 8-10 atm) in the presence of the cobalt catalyst in a sparged reactor or multistaged column contactor.

several oxidation stages are usually necessary to avoid over oxidation the KA mixture cyclohexane on oxidation gives cyclohexane remains unreacted. unreacted ( excess) oxygen is drawn out from the product.

Distillation column:

oxidizer effluent is distilled to recover unconverted cyclohexane then recycled to the reactor feed. the resultant KA mixture may then be distilled for improved quality before being sent to the nitric acid oxidation stage . this process yields 75 to 80% mole percent KA with a kketone to alcohol ratio of 1:2 &  water is completely removed KA mixtre consisting of ketonic alcoholic group and water is the bottom product.

Heat Exchanger

The lightly viscous mixture is then cooled from 150-85c using warer as a coolant.

Reactor 2:

the step in commercial production of adipic acid is nitic acid oxidation cyclohexanol and cyclohexanone mixture the reaction proceeds as follows:

cyclohexanol+nitric acid → adipic acid + NOX + H2O = HEAT

As the reaction is highly exothermic heat of reaction is usually dissipated by maintaining a high ratio (40:1) of nitric acid to KA mixture. nitric acid (50 to 60% ) and a copper vanadium catalyst are reacted with the KA mixture on a reactor vessel at 60 to 80c and .1 to .4 MPa. convresion yields of 92 to 96 % are attainable when usig high purity KA feed stock upon reeeaction nitric acid is reduced to nitrogen oxides NO2, NO ,N2O AND N2, the oxidation results into the formation of 94% adipic acid along with NOX, Salicylic acid m glutaric acid as by products m water and excess , HNO3

Vaccum Distillation Column 

for enhancing the concentration of adipic acid up to 95% vaccum distillation of the mixture is done to remove water produced is the reactor2.

CRYSTALLIZER:

Now mixture is sent for crystallization to achieve up to 98% pure crystals of adipic acid crystals of adipic acid are obtained with small amount of mother liquor 

CENTRIFUGE:

the mother liquor is separated from crystals here by washing with water in centrifuge for obtaining highly pure crystals of adipic acid crystals are sent to dryer.

DRYER:

Dryer is used to remove the water used for washing the mother liquor away from crystals in centrifuge. if water removal is not done then the crystals may remain impure and cause brittleness which are not suitable for handling and transportation. up to 99% pure crystals adipic acid are obtained , this purity is needed becayse these crystals are going to be used in the manifacturing of very important plastic Nylon

CHAPTER 3:

MATERIAL BALANCE:

MATERIAL BALANCE ON REACTOR 1:

REACTION:-
                         2C6H12  +  3/2 O2         80%                 C6H12O+C6H10O+H2O
BASIS:-
                1500 Kg/hr OF C6H12
                    TOTAL moles of C6H12= 1500/84 =17.86 K. mole
C6H12O(Cyclohexanol):-
                                  2k.moles of C6H12 reacts with= 1k. mole of C6H12O
                                1k.mole of C6H12 reacts with  = 1/2k. mole of C6H12O


17.86k.mole of C6H12 reacts with=1/2 * 17.86
                                                      =8.93 k.mole
As 80% conversion is there so;
                                       8.93 *  8 * 100 =714.285kg

C6H10O(Cyclohexanone):-
     2k.moles of C6H12 reacts with = 1k.mole of C6H10O
    1k.mole of C6H12 reacts with  =1/2k.mole of C6H10O
17.86k.moles of C6H12 reacts with=1/2 * 17.86

                                                           =8.93 k.mole

As conversion is 80% so,
                                         8.93 * .8 * 78 =700.112 kg

H2O(Water):-

2k.moles of C6H12 reacts with = 1 k.mole of H2O1k.mole of C6H12 reacts with =1/2 k.mole of H2O17.86k.moles of C6H12 reacts with = 1/2 * 17.86                                                           =8.93k.moleFor 80%conversion;                                     8.93 * .8 * 18=128.592kg2k.moles of C6H12 reacts with = 1.5 k.mole of O21k.mole of C6H12 reacts with =1/2 k.mole of O217.86k.moles of C6H12 reacts with = 1.5 * 17.86
 =13.395k.mole13.395k.mole of O2                   1k. 13.395k.mole mole of air                                                      0.21k.mole of O2                        63.78k.mole of airAs 25% excess air is supplied63.78 * 1.25=79.725k.mole of airSo O2 supplied;                        79.725 * .21=16.74k.mole                      16.74 * 32=535.75kg

O2 Unreacted

 O2actually unreacted(before excess)=13.395 * 32 * 0.21

                                                                 =85.73kg
                 O2 Unreacted from excess=535.75 * 0.2
                                                               =107.15kg
    Total unreacted=107.15+85.73=192.6kg


Material Balance on the Absorber


Reaction 1
             2NO + O2                     2NO2


Reaction 2
                 3NO2 + H2O               2HNO +  NO

From Reaction 1
                   NO=604.29/30=20.143k.moles
             2k.moles of NO react=1mole of  O2
             1k.moles of NO react=1/2mole of  O2
            20.143k.moles of NO react=1/2*20.143mole of  O2
                                                         =10.0715*30=302.145
         For 90% conversion=302.145*0.90=271.93

Material Balance on the vacuum Distillation


         ....................................

    Material Balance on the Reactor 2

Reaction
3C6H12O  +  2C6H10O  +  18HNO3                              5C6H10O4   +  12H2O  +  9NO + 9NO2

Moles of reactants
C6H12O   =   714.285/100   =   7.1428 kmol
C6H10O  =   700.112/98  =  7.144 kmol
HNO3  =  2700.1/63  =  42.85 kmol
C6H10O  ( M.W. = 146 )
3 kmol of C6H12O  reacts with = 5 kmol of C6H10O4
1 kmol of C6H12O  reacts with =  5/3
7.143 kmol of C6H12O  reacts with =  5/3 ( 7.143)
                                                        = 11.9 kmol

As conversion is 94% so 11.9 * 146 * 0.94 = 1633.56 kg
H2O   ( M.W. = 18 )
3 kmol of C6H12O   forms = 12 kmol of H2O

 1 kmol of C6H12O   forms  =  12/3
7.143 7 kmol of C6H12O   forms =  12/3 (7.143)
 Hence
12/3 * 7.143 * 18 * 0.94    = 483.43 kg
NO ( M.W.= 30 )
3 kmol of C6H12O forms = 9 kmol of NO
1 kmol of C6H12O forms   =  9/3
4.143 kmol of C6H12O forms = 9/3 (7.143)
Hence
9/3 * 7.143 * 30 * 0.94 = 604.29 kg

NO(M.W. = 46)
3 kmol of C6H12O forms = 9 kmol of NO2
1 kmol of C6H12O forms   =  9/3
4.143 kmol of C6H12O forms = 9/3 (7.143)
Hence
9/3 * 7.143 * 46 * 0.94 = 926.58 kg
HNO(M.W. = 63)
3 kmol of C6H12O react = 18 kmol of HNO3
1 kmol of C6H12O react = 18/3
7.143 kmol of C6H12O react = 18/3 (7.143)
Hence
18/3 * 7.143 *63 = 2700.05 kg
HNO3 unreacted
2700.05 * 0.06 = 162 kg

Material Balance on the Distillation column

As   C6H12    is completely removed from the top so;
H2O  =  300.1/0.9   = 333.44 kg
Where water is = 333.44-300.1=33.34kg
Water at the bottom of column = 128.592-33.34=95.25kg

Material Balance on the Flash Drum
...........
Material Balance on the Crystallizer
........................

Material Balance on the Centrifuge

Solubility of HNO3 in Water =completely
Solubility of G.A in Water=430g/lit
Solubility of G.A in Water=80g/lit
In M.L the Composition is ;
                                                              HNO3=4.1kg
                                                                G.A=40.3 kg
                                                                  S.A=22.3kg
So on the basis of solubility
0.43 kg of G.A emits with = 1 lit of water
1 kg of G.A emits with  =1/0.43
40.3 kg of G.A emits with =1/0.43 *40.3
                                               = 93.72 lit
0.08 kg of S.A emits with =1 lit of water
1 kg of S.A emits with = 1/0.08
22.3 kg of S.A emits with =1/0.08 *22.3
                                               =278.75 lit
3%extra water is taken
So total water=483.9lit
                       =0.4839m3
                        =0.4839m3*1000kg/m3
                         =483.9kg
As we require 99% A.A so

Material Balance on the Dryer

...............................................

CHAPTER 3: 

ENERGY BALANCE:

Energy balance  on the Reactor

Heat Inlet
Mass flow rate of cyclohexanol = 714.285  kg
Mass flow rate of cyclohexanone=700.112 kg
Mass flow rate of water=95.25 kg
Mass flow rate nitric acid=2700.05 kg
Specific heat capacity of cyclohexanol =2.4679kj/kg.k
Specific heat capacity of cyclohexanone =2.039 kj/kg.k
 Specific heat capacity of water =4.2 kj/kg.k
 Specific heat capacity of nitric acid =0.00174 kj/kg.k
      
                Q   =    m*cp*dt   =4209.697*8.70924*60 = 215728.8 kj

Heat of reaction  
                                            = 384000 kj
               Q t  =  619728.8 kj
Heat outlet


Mass flow rate of nitric acid=162 kg


Mass flow rate of water=578.67 kg
Mass flow rate of adipic acid=1633.156 kg
Mass flow rate of nitric acid=604.29 kg
Mass flow rate of nitrogen dioxide=926.58 kg
Mass flow rate of salisilic acid=120.8 kg
Mass flow rate of giutaric acid=183.44 kg
Specific heat capacity of nitric acid =.00174 kj/kg.k
Specific heat capacity of water=4.2 kj/kg.k
Specific heat capacity of adipic acid =2.44 kj/kg.k
Specific heat capacity of nitric acid =1.0057 kj/kg.k
Specific heat capacity of nitrogen dioxide =3.522 kj/kg.k
Specific heat capacity of salisilic acid =0.66 kj/kg.k
Specific heat capacity of giutaric acid =0.695 kj/kg.k
Q =  m * cp * dt  =  4208.936 * 12.52444 * 60 = 629637.9 kj

Energy balance on the Dryer
Heatt inlet
Mass flow rate of adipic acid= 1600.34kg
Mass flow rate of water=60.2kg
Specific heat capacity of adipic acid=0.92 kj/kg.k
Specific heat capacity of giutaric acid=4.184 kj/kg.k
  Q = m * cp * dt = 1615 * 5.104 * 75 = 115033.6 kj
Heat  at outlet
Mass flow rate of water=45.5kg
Latent heat of water=2258
Q=102739kj
Mass flow rate of adipic acid=1600.3kg
Mass flow of water=14.7kg
Q=m *cp *dt=1615 *5.104*55=14264.25kj

Energy balance on  the heat exchanger
Heat at inlet
Mass flow rate of cyclohexanon=700.112kg
Mass flow rate of water=128.592kg
Mass flow rate of cyclohexane=300.0816kg
Mass flow rate of cyclohexanol=714.2kg
Q= m * cp *dt=1842.985 * 6.41 * 125=371863.85kj
Heat at outlet

mass flow rate of cyclohexanol=714.285kg
mass flow rate of cyclohexanon=700.112kg
mass flow rate of water=128.592kg
mass flow rate of cyclohexane=300.0816kg
Q= m *cp * dt=1842.985* 5.42 * 60=158072.65kj

CHAPTER 5: 

DESIGNING.
Heat Exchanger Design
Background of heat exchanger
Heat exchangers are found in most chemical or mechanical systems. It’s a device that transfers heat between one medium to another. Some  of  the  more  common  applications of  heat exchangers  are  found  in heating, air conditioning  systems,  radiators  on  internal  combustion engines, boilecondensers, and  as  preheaters  or coolers  in  fluid systems.These equipments are widely used in many fields such as, power plants, petrochemical plants, petroleum refineries, and natural gas processing.Because of heat exchangers come in so many shapes, sizes, makes, and models, they are categorized according to common characteristics.One common characteristic that can be used to categorize them is the direction of flow the two fluids have relative to each other.

The three categories are parallel flow, counter flow and cross flow as the following figure 


Figure : Three flow arrangements categories. 
·         Parallel flow exists when both the tube side fluid and the shell side fluid flow in the same direction
·         Counter flow exists when the two fluids flow in opposite directions. Each of the fluids enters the heat exchanger at opposite ends.
             Cross flow exists when one fluid flows perpendicular to the second fluid.

Types of Heat exchangers
There are many types of heat exchangers in the industrial fields and the common types ofheat exchangers are shell & tube heat exchanger and plate heat exchanger.

Plate heat exchanger 
                                  is uses metal plates to transfer heat between two fluids. This has a major advantage over a conventional heat exchanger in that the fluids are exposed to a much larger surface area because the fluids spread out over the plates.This facilitates the transfer of heat, and greatly increases the speed of the temperature change.
Shell & tube heat exchange
 is the most widespread and commonly used in the process industries. This type of heat exchanger consists of a set of tubes in a container called a shell.The fluid flowing inside the tubes is called the tube side fluid and the fluid flowing on the outside of the tubes is the shell side fluid.In CO2 capture plant the type of the heat exchanger between absorber and stripper is shell and tube heat exchanger to rise the temperature of rich amine and to reduce lean amine temperature.Before entering lean amine to the absorber, the lean amine needs to be cooled to the appropriate absorption temperature.This heat exchanger will save the energy of the streams lean-rich amines instead of using heating or cooling units.The design of a heat exchanger involves many important considerations such as, Tube and shell side flow calculations, Heat transfer area, Area and number of tubes, Overall heat transfer coefficient (U),Shell diameter, types of material construction …etc.
Selection Criteria:
Function
            To cool the organic liquids from 2450 C to 60 o C.
Design: 
          Three exchangers connected parallel having same specifications.
Flow arrangement 
         Counter current system
Fluid side selection:
         Organic fluids on shell side, organic fluids on tube side

DESIGN CALCULATIONS OF HEAT EXCHANGER

Inlet temperature of the process stream                      = T1 = 170oC
Outlet temperature of the process stream                   = T2 = 85oC
Inlet temperature of the water                                                t1 = 25 o C
Outlet temperature of the water                                 t2 = 55 o C
Mass flow rate of the process stream                          m = 2035.8 Kg/hr
Mass flow rate of the Water                                       m = .81904 Kg/hr
Cp of process stream                                                  Cp = 2.14 Kj/Kg  o C
Cp of  water                                                                Cp = 4.18 Kj/Kg o C
Heat duty
               Q= m Cp ∆T
                    = 2035.8*2.14*85/3600
             = 103.34kW
 Mass flow rate of cooling water

 =103.30/ 4.18*(55-25) = 81904 Kg/hr
LMTD

                                                           = 52.810 oC
Assumed Calculations
                                       Value of UD Assumed
                                                     UD     =   400 W/m 2C
                                                       R      = (T1-T2)/ (t2-t1  = 2.833
                                                       S       = (t2-t1)/ (T1- t1)  = 0.35
R & S intersect at fig 12.19 RC VOL 6 so from the value of Ft is 0.83
Del Tm = 0.83*52.810  = 43.83oC
Assumed Ud = 400 W/m2.oC         (rc vol6 table 12.1)
Heat Transfer Area

A= Q/UD∆T
                                                                 =103.3 /43.83*400   = 5.8902m2       
                                                                  =103.3 /43.83*400   = 5.8902m2       
TUBE DIMENSIONS
O.D. = 20 mm
I.D. = 16 mm

Length of tube = Lt = 4.83 m
 No. of tubes  = Ao/At  = 5.8902/.303=53
Pt = 1.25 ´ 0.020= 0.025m
Tube Bundle Diameter
                                          Db = O.D* (Nt/K)1/n
                                     For square pitch pattern
                                                n=2.291
                                                K=0.156              (table 12.4 rc vol.6)
                                                                                       = 0.020*(20/.156)1/2.291
                                                                 = 0.2163m m
No. of tubes in centre row
Nr = Db/Pt
                                                                    = 0.3175/0.03125=10.1
Split ring floating head is selected. (From Figure 12.10 , volume 6)
                        BDC = 53mm=0.053m
Shell Diameter, Ds = Db+BDC= 0.21638+0.053= 0.2686 m
side heat transfer coefficient Shell
 Baffle spacing, lb = DS/5= 268.66/5= 53.72m m=.05372m
       As =  = 0.0028879 m2
Gs = M/As= 2035.8/0.0028879*3600= 195.945 kg/(s.m2)
Re, Reynold's number=Gs.do/µ= 469.8
jh = 2.4x10-1         (Fig. 12.31, Vol. 6)
Pr = Cpµ/k= 0.004675
hoc = (jH/do).k.Re.Pr1/3= (2.4x10-1/0.020)x0.2758x4.7026x102 (0.004675)1/3
       = 260.243W/m .o C
Fn, Tube row correction factor
(BC) baffle cut is 25% of shell dia     (rc vol.6)
Baffle cut height = 0.25xDs= 0.06716 m
Height between baffle tips = 0.1350 m
Ncv = 0.1350/0.025= 5.4
From Fig 12.32, Vol. 6
Fn =0.98
Fw, Window factor
Hb = (Db/2)-Ds(0.5-0.25)= 0.04115 m
Bundle cut, = Hb/Db=0.19 or 19%
Ra' = 0.16 (Fig 12.41, Vol. 6)
Tubes in one window, Nw = NtxRa'= 20x0.19 = 3.8
Rw = (2x3.8)/20 = 0.40
Fw  = 1.06 (Fig. 12.33, Vol. 6)
Fb, Bypass correction factor
Ab = lb(Ds-Db)= 53.732(268.66-216.88)* 10-6= 0.002782m2
α = 1.35 (for turbulent flow from Vol. 6)
Fb = exp[-α.(Ab/As)(1-2(Ns/Ncv)1/3]
     = 0.71006 (from Fig. 12.34, Vol. 6)
Fl, Leakage correction factor
     Fl = 1-βl(Atb+2Asb/Al)

Ct= 0.0008    (rc vol 6)
      = 0.001490 m2

Cut = 25%, θ = 2.1 (Fig. 12.41, Vol. 6)
Cs=0.0032m (table 12.5)
        = 0.002695 m2
AL/As = 0.5167
ΒL = 0.34 (Fig 12.35, Vol.6)
FL = 1-βl(Atb+2Asb/Al)= 1.08
hs = hoc x Fn x Fb x Fw x Fl
hs = 260.243 x 0.98 x 1.06x .71006 x 1.08
hs =207.314 W/m2 .o C
AL/As = 0.5167
Shell side Pressure Drop
Cross flow zone
∆Pc = ∆Pi F'b F'L
Re = 654.5, for square pitch
jf = 8 x 10-2 (Fig. 12.36, Vol. 6)
us = .2153 m/s
 ∆Pi, ideal tube bank pressure drop
∆Pi = 73.051 N/m2
F'b, bypass correction factor for pressure drop
(α=4)  for turbulent flow
F'b= exp[-α.(Ab/As)(1-2(Ns/Ncv)1/3]
F'b =0.3373
F'L, leakage factor for pressure drop
β'l = 0.57( Fig. 12.38, Vol.6)
F'l = 1-β'l(Atb+2Asb/Al)
F'l = 0.7069
∆Pc = ∆Pi x F'b x F'l
∆Pc = 73.051 x 0.3373 x0.7069 = 17.41 N/m2
Window zone
∆Pw = F'l(2+0.6Nwu)ρuz2/2
For baffle cut 0.25, Ra = 0.16 (Fig. 12.41, Vol. 6)
Aw = (πDs2/4 x Ra) - (Nwπdo2/4)= 0.008123m2
uw = 0.076m/s
uz = 
 = 0.1279m/s
Nwu = Hb/Pt’= 41.154/25= 1.64
∆Pw = F'l(2+0.6Nwu)ρuz2/2= 15.734 N/m
End zone
∆Pe = ∆Pi[(Nwu+Ncv/Ncv)]F'b
       = 34.693 N/m2
Total pressure drop
Number of baffles, Nb = 4.8/0.005373 - 1
Nb = 48.25= 87
∆Ps = 2∆Pe + ∆Pc(Nb - 1) + Nb x ∆Pw
            =156.00532  N/m2 or 0.2271Psi
Tube side Pressure drop
Re =  8108.52 (for tube side)
jf =  8.9 x 10-2 ( Fig. 12.24, Vol. 6)
Np, Number of passes = 2
∆Pt= Np[8jf(L/di)+2.5]ρut2/2
∆Pt = 2[8 x 8.9*10-2(4.88/0.016)+2.5]995*040952/2
∆Pt = 4041.74 N/m2 or 0.58609 Psi
 Tube side Co-efficient
Mean temperature = (55+25)/2 = 40oC
Tube cross-sectional area = π/4 x d02= 201 mm2
As 2 passes are used so number of tubes per pass = 20/2 = 10
Total flow area = 10*201*10-6= 0.00201 m2
Mass velocity = .81904/0.00201= 407.48 kg/s.m2
ρ = 995 kg/m3
Water linear velocity, ut = 407.48/995 = 0.4095 m/s
The equation for heat transfer co-efficient below has been adapted from data given by Eagle and Ferguson
hi = 4200(1.35+0.02t)ut0.8/di0.2
hi = 4200(1.35+0.02x40)0.40950.8/160.2
hi = 2491.79 W/m2.oC

Overall Design Co-efficient


kw = 45 (stainless steel)
1/hod = 0.0005 m2.o  C/W
1/hid = 0.0002 m2.C/W               (Table 12.3, 12.4, Vol.6)

Uo = 438 W/m2. o C
Tube side Pressure drop
Re =  pdu/µ (for tube side)
     =16827.42
jf =  4.3 x 10-3 ( Fig. 12.24, Vol. 6)
L/di= 1.83*10^3/21.8= 83.9
Np, Number of passes = 2
∆Pt= Np[8jf(L/di)+2.5]ρut2/2
∆Pt = 2[8 x 4.3*10-3*83.9+2.5]995*0.622/2
∆Pt = 2060 N/m2 or 0.298 psi
Heat Transfer Cofficient
                                                          hio = hi ´ ID/OD
           hio =2468 W/m2 0C
Clean Overall Coefficient
                                                                       Uc  =1232.7 W/m2 0C

Design Overall Coefficient Calculated
 Dirt factor = Rd = 0.0014

Centrifuge
·         It includes the following steps.
·         Brief Introduction
·         Classification of centrifuges
·         Selection of centrifuge
·         Description of selected centrifuge
·        Design consideration

Brief Introduction
A centrifuge is most often used for the separation of particles from solutions according to their size, shape, density, viscosity of the medium and rotor speed. These machines utilize the natural separation realities present in a high-speed circular G-force environment. Like a high-powered clothes dryer, these exceedingly fast machines spin in order to separate materials from one another. The denser materials separate from the less dense during the centrifugation process. The term “centrifuge” encompasses a wide variety of process equipment used for many different applications in the chemical process industries. Although these units may look different and play key roles in much unrelated processes.

Classification of Centrifuges
Centrifuges are classified according to the mechanism used for solids separation.
·         Sedimentation centrifuges
·         Filtration centrifuges      
In sedimentation centrifuges the separation is dependent on a difference in density between the solid and liquid phases (solid heavier).In filtration centrifuge the separate the phases by filtration. The walls of the centrifuge basket are porous, and the liquid filters through the deposited cake of solids and is removed.

These two categories can be further split into subcategories,

Tubular bowl

High-speed, vertical axis, tubular bowl centrifuges are used for the separation ofimmiscible liquids, such as water and oil, and for the separation of fine solids. The bowl is driven at speeds of around 15,000 rpm (250 Hz) and the centrifugal force generated exceeds 130,000N.

Disc bowl


The conical discs in a disc bowl centrifuge split the liquid flow into a number of very thin layers, which greatly increases the separating efficiency. Disc bowl centrifugesused for separating liquids and fine solids, and for solids classification
Scroll discharge
  In this type of machine the solids deposited on the wall of the bowl are removed by a scroll (a helical screw conveyer) which revolves at a slightly different speed from the bowl. Scroll discharge centrifuges can be designed so that solids can be washed and relatively dry solids be discharged.
Solid bowl batch centrifuge
                               The simplest type; similar to the tubular bowl machine type but with a smaller bowl length to diameter ratio (less than 0.75). The tubular bowl type is rarely used for solids concentrations above 1 per cent by volume. For concentrations between 1 to 15 per cent, any of the other three types can be used. Above 15 per cent, either the scroll discharge type or the batch type may be used, depending on whether continuous or intermittent operation is required.
Filtration centrifuge
It is convenient to classify centrifugal filters into two broad classes, depending on how the solids are removed; fixed bed or moving bed.In the fixed-bed type, the cake of solids remains on the walls of the bowl until removed manually, or automatically by means of a knife mechanism. It is essentially cyclic in operation.In the moving-bed type, the mass of solids is moved along the bowl by the action of a scroll (similar to the solid-bowl sedimentation type); or by a ram (pusher type); or by a vibration mechanism; or by the bowl angle. Washing and drying zones can be incorporated into the moving bed type.
Selection of Centrifuge
There are following factors which must be considered in the selection of centrifuge for aprocess
a)    Physical Properties of Materials
The characteristics of the solids and liquids handled in a process will influence centrifuge selection.
Specific Gravities of the Solids and Liquids
If the solids are lighter than the liquid, a decanting centrifuge is not         an option. If the specific gravities are very close, but the solids are  slightly heavier, a decanting centrifuge may be considered, but only if   either the particle size or the centrifugal force improves the settling of the solids
Particle Size
 Coarse solids with particle size greater than 100 um are generally best suited for filtering type centrifuges. Finer solids that measures less than 100 um are best handled in sedimentation centrifuges.
Centrate Clarity
Decanting centrifuge provides the best clarity of all centrifuge  types. Filtering centrifuges typically are not used where centrate clarity is the principle process requirement because they use either a filter medium or a screen
a)    Process requirements
 Continuous centrifuges should be considered when the following criteria are important
Pressure and Temperature
Continuous centrifuges are used when higher operating pressures are required. Continuous decanters have been used in operations where pressures were as high as 90 psig and temperatures as high as 175 C.Batch centrifuges are restricted to lower pressure applications. There does not appear to be a restriction or preference based on temperature
Flow Rate
The higher the solid flow rate, the greater the tendency to use continuous centrifugation. As a rule of thumb, batch centrifuges are applicable for loadings of up to 1 ton/hr.
Solid Concentration
Continuous filtering centrifuges, with the possible exception of the screen bowl design, prefer thick feed slurry (typically 50% by weight). There are two reasons for this: continuous filtering centrifuges may be limited by the amount of liquid to be centrifuged. Thus the overall recovery of solids is generally higher with continuous filtering centrifuges when the feed has a high solids concentration. So Scroll Conveyor & Bowl Centrifuge is selected

Parameters Required For Designing

 1: Volumetric Flow Rate                                          

 2:Length OF centrifuge                                            

3:Volume OF bowl                                                   

 4:Residence Time                                                      

 5:Relative Centrifugal Force                                     

 6:Settling Velocity Of Prticles


Volumetric Flow Rate:
V= m/ρAVG                                                                          

m=2150.97 kg/hr                                                                                

ρavg = Xm ρ A.A + Xm ρS.A + Xm ρG.A + Xm ρN.A                              
V=1280.069 kg/m


Length Of Centrifuge:

  L/D = 1.8
D = 1300 mm     ,,,,  D=1.3 m
L= 1.8 * 1.3= 2.34 m
L/D: Refrence; From Patent ucdc 66.067 . 55/57002.237
D: Mcab - Smith                , p#1067, CH#29
3:Volume Of Bowl:
  V = ( π*D2 L/4)
  = (3.14)*(1.3)2*(2.34)/4
V = 3.14 m3
Residence Time
t = (vol of liquid in the bowl)/(volumetric flow rate
VL = πL( r 2 2– r1)
V L= 3.14 * 2.34 ( (0.65) 2– (0.39)2 )
VL   = 1.98 m  3
t = (1.98) / (2.15
t = 54 min
VL   : Refrance; Mcab – Smith , P# 1069
Relative Centrifugal Force
RPM of centrifuge = n = 1000 rev/min = 16.7 rev/sec 
RCF = FC / g
RCF = ( rω 2) / g
ω = angular velocity of bowl
 = 2 * π * n
 = 2 * 3.14 * 16.7
= 104.9 rev/ssec
RCF = (104.9) 2* (0.65) / 10
RCF = 715.26
Refrance; Decanter Centrifuge Hand Book By Alan Records , P # 203

Settling Velocity Of Particles
V S= DP2 ( ρS – ρL ) ωr 2/ 18μ
D P= 60 * 10-6 m
ω = 104.9 rev / sec
μAVG = Xm waterwater + XmN.AN.A + Xm G.A* μ G.A + Xm S.A* μS.A
μAVG  = 7.27 * 10-4 Pa Sec
VS = (60*10-6)2 * (1360 -998) * (104.9)2 * (0.65) / (18) * (7.72*10-4)
VS  = 0.67 m/sec
 Refrance; Perry, sec 19….95 , 6th edition , MCCAB SMITH, P#1069 

Dryer
Calculation of heat duty   
Qt=    Q1  +Q2
Qt = total heat requirements
     Q1 = Heat needed to raise product to discharge temperature               
     Q2 = Heat needed to remove moisture
    Q1 =mscps∆T +mwlcpw∆T
  ms=mass flow of solid entering
cps=specific heat of solid(20 OC)
Mwl=mass flow of water leaving
cpw=specific heat of water(60 OC)
Q1 =mscps∆T +mwlcpw∆T
       =(1600.3*2.4*40)+(14.7*1.453*40)
       =154483.2kj/kg
Q2 =mw [cpw∆T+ λ+ Cpv∆T’]               (3)
 λ =   Latent heat of vaporization   
Cpv =   vapour  specific heat                                 
∆T’ =   Ta2 --Ts2        =20  
  Q2 = 1148.4 kj/kg
 Qt= 155631.6  kj/kg 

Outlet temprature of air
NTU = ln (Tg1-Tw)/(Tg2-Tw)
  Where
 NTU = 1.5—2.5
 Tg1=inlet temp of air
 Tw=wet bulb temp
 Tg2=? (outlet temp of air)
Tg2 =  158 oF  or 70 oC  

Mass of air needed
Ma =   Qt /Cp dt
  dt=air inlet temp – air outlet temp
  Cp =specific heat of air(150 oC)
  Ma =   155631.6/.9923* 80 
         =    1960.49 kg

Area of dryer
A =    1960.49 kg
Area=   Ma  / G
   G= Mass velocity of air (1000 lb/ft2 *hr)
                       =  0.801 m2
    ( Ref; terephtalic-250acid-design-2520 of- 2520 eq ) 
 Diameter of dryer
D = (4A / π )1/2
    = (4*0.801 / 3.146 )1/2    =1.0094 m   OR     3.3117 ft

Volume of dryer
Qt = Ua V(dt)LMTD    
   V = Qt   /Ua (dt)LMTD …………….(A)
  Where
  Ua  =  15*G .16 /D
        = 15*1000 .16 /3.3117
        =  9.710023 BTU/ft 3 hr 
(dt)LMTD = dt2 – dt1/ln(dt2/dt1) 
                  = 10 - 130/ln(10/130) 
                   =  46.784 oC   or  116.21 oF
      Put all the values in eq (A)
          V  =  3.701m3  or   130.73

Length of dryer
L =    V/A ……………(B)
                     V = volume of dryer
                     A = area of dryer
                        =3.701/.801
                         = 4.619m     or    15.154ft
Slope of drum
Slope of drum is kept from 0 to 5°. More the slope of the drying drum more   will be forward driving force but product abrasion will also increase. We have kept  3° slope.
Rotational speed of drum
Rotational speed of drum may be between 2 to 5rev/min 

Residence time
Tr   = k(Ln)/.5 / NDtanΦ            (C)
 Where,                                                                                                                                              
Tr = residence time ,( minute )  
L   = length of dryer ,15.154 ft        
  k   = constant      = 0.0433                                                                                                                                  
  N  = rpm    = 2                                             
D  =  dryer dia ,ft     = 3.3117 ft
Φ =  slope of dryer = 3 degrees                                                       
 After putting values in eq   ( C)
  Tr = 3.21 min    

DESIGN OF DISTILLATION COLUMN
In industry it is common practice to separate a liquid mixture by distillating the components, which have lower boiling points when they are in pure condition from those having higher boiling points. This process is accomplished by partial vaporization and subsequent condensation
CHOICE BETWEEN PLATE AND PACKED column
Vapour liquid mass transfer operation may be carried either in plate column or packed column. These two types of operations are quite different. A selection scheme considering the factors under four headings.
i)                    Factors that depend on the system i.e. scale, foaming, fouling factors, corrosive systems, heat evolution, pressure drop, liquid holdup.
ii)                  Factors that depend on the fluid flow moment.
iii)                Factors that depends upon the physical characteristics of the column and its internals i.e. maintenance, weight, side stream, size and cost.
iv)                Factors that depend upon mode of operation i.e. batch distillation, continuous distillation, turndown, and intermittent distillation.
The relative merits of plate over packed column are as follows:
i)                    Plate column are designed to handle wide range of liquid flow rates without flooding.
ii)                  If a system contains solid contents, it will be handled in plate column, because solid will accumulate in the voids, coating the packing materials and making it ineffective.
iii)                Dispersion difficulties are handled in plate column when flow rate of liquid are low as compared to gases.
iv)                For large column heights, weight of the packed column is more than plate column.
v)                  If periodic cleaning is required, man holes will be provided for cleaning. In packed columns packing must be removed before cleaning.
vi)                For non-foaming systems the plate column is preferred.
vii)              Design information for plate column are more readily available and more reliable than that for packed column.
viii)            Inter stage cooling can be provide to remove heat of reaction or solution in plate column.
ix)                When temperature change is involved, packing may be damaged.
x)                  Plates are mostly used for large diameter more than 0.6m
For this particular process, “DME, Methanol, Water ”, I have selected plate column because:
i)                    System is non-foaming.
ii)                  Temperature is high.
Diameter is greater than 0.6 meter

CHOICE OF PLATE TYPE
 There are four main tray types, the bubble cap, sieve tray, ballast or valve trays and the counter flow trays. I have selected sieve tray because:
i)                    They are lighter in weight and less expensive. It is easier and cheaper to install.
ii)                  Pressure drop is low as compared to bubble cap trays.
iii)                Peak efficiency is generally high.
iv)                Maintenance cost is reduced due to the ease of cleaning.
 DESIGNING STEPS OF DISTILLATION  COLUMN
·         Calculation of Minimum Reflux Ratio Rm.
·         Calculation of Optimum 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.
·         Determination of Flow Parameters
·         Calculation of the Height of the column

For feed

component
Mass Flow rate(kg/hr)
Molecular wt
Kg mole
Mole fraction
C6H12
300.14
84.16
3.565
0.1425
C6H12O
714.285
100.152
7.1315
0.2855
C6H10O
700.112
98.15
7.133
0.2549
H2O
128.592
18
7.`44
0.2680

 For Top product
component
Mass Flow rate(kg/hr)
Molecular wt
Kg mole
Mole fraction
C6H12
265
84.16
3.154
0.589
C6H12O
15
100.152
0.1497
0.02729
C6H10O
20
98.15
0.203
0.0379
H2O
33.2       
18
1.844
0.

For Bottom product

component
Mass Flow rate(kg/hr)
Molecular wt
Kg mole
Mole fraction
C6H12
35
84.16
0.4166
0.0212
C6H12O
699
100.152
6.978
0.3558
C6H10O
680
98.15
6.928
0.3552
H2O
95.25    
18
5.2916
0.3698


Volatility Calculation
Top product:-
                                                                    αA =PAXB/PBXA
                                                                 αA =0.97877*0.02797/0.0394*0.589
                                                                 αA =1.775
Bottom  product:-
                                                                    αB =PAXB/PBXA
                                                                 αB =2.142*0.3558/35.938*0.0212
                                                                   αB =1.04
Average Relative Volatility:-
                                                          αavg  =  
                                                           αavg   =
                                                             αavg  = 1.35
Key Component
Heavy Key component         C6H120
Light key component             C6H12
Minimum Reflux Ratio
1-q= αi xi
αi-θ
αi= The relative volatility of component i with respect to some reference component, usually the heavy key
Rm=αxid .
α-θ
Rm= Minimum ReflUX Ratio
Θ=0.75      
1=(1.775)(0.1427)/()
Rm+1=1.8725-1
Rm=0.87
Rm= Minimum Reflux Ratio
α= Relative Volatility
xid= Concentration of component / in the tops at minimum reflux
q= Thermal condition of feed
Θ2-1.75 θ+0.7004=0
Θ=0.75
We take θ=0.75
Rm+1 = 1.8725-1
Rm=0.87

Minimum Number of Plates

Minimum  number  of  plates  are  calculated  by  using  Fenske’s equation
Nmin=ln[(xd/xb)LK(xd/xb)HK]

ln (α)






    where
Nmin= Minimum Number of Plates xd=Mole Fraction of distillate product xb= Mole Fraction of bottom product LK=Light Key
HK=Heavy Key
α=Relative Volatility

Nmin=ln[(0.589/0.02797(0.3558/0.012)]

                                                                                                                                       ln (1.35)

Nmin=20

Optimum Reflux and Ideal Number of Plates
Optimum reflux ratio and number of plates are calculated by using Gilliland

Empirical relation.


R
R-Rm

R+1
(n+1)-(nm+1)

n+2
n
1
0.065
0.55
46
1.5
0.252
0.40
34
2
0.376
0.33
30
2.5
0.465
0.28
28
3
0.532
0.22
24
3.5
0.584
0.20
23
4
0.626
0.17
22
4.5
0.66
0.16
21
5
0.668
0.15
20



 Reflux Ratio, R= 3
 Ideal Number of Plates
 N= 24





Efficiency of Plates






Efficiency of Plate
                          
Eo  =  51-32.5log10(0.256*1.1066

       

Eo  =  0.68

Actual Number of Plates
 Eo= Number of Ideal Stag
      Number of Real Stages


Number of Real Stages= 24/0.68
= 34
Recommended Number of Plates= 33
Feed Tray Location

Feed tray location is found by using Kirkbride relation.




                              CHAPTER 6:
       
                         COST ESTIMATION

        

                      


EQUIPMENT
COST($)
REACTOR
76000
DISTILLATION COLUMN
87000
HEAT EXCHANGER
43000
CRYSTALLIZER
41700
CENTRIFUGE
220300
DRYER
23860
Total
491860

            

                                                    CHAPTER NO 6:
                                                 
                                 
                          HAZOP STUDY




Introduction

Traditionally, safety in the design of chemical plants relied upon the application of codes of practice, design codes and checklists based on the wide experience and knowledge of professional experts and specialists in the industry. However, such approaches can only cope with problems that have arisen before, with the increasing complexity of modern plant, these traditional approaches are likely to miss other issues which need to be considered at the design stage of a project.
Hazard and operability studies (HAZOPs) were developed by ICI during the 1960s as a technique to overcome this problem and to systematically identify potential hazards and operability problems in new designs for chemical and petrochemical plant, in both batch and continuous processes. HAZOPs can also be used for modifications and for the review of existing processes.

What are HAZOPs

No (not, none)                                     none of the design intent is achieved
More (more of, higher)                        quantitative increase in a parameter
Less (less of, lower)                            quantitative decrease in a parameter
As well as (more than)                        An additional activity occurs
Part of                                               Only some of the design intention is achieved
Reverse                                             logical opposite of design intention occurs
Other than (other)                           complete substitution, another activity takes
                                                          Place
Other useful guide words include

Where else                                           applicable for flows, transfers, sources and
 destinations
Before/after                                        the step (or some part of it) is effected out of
                                                            Sequence
Early/late                                           the timing is different from the intention
Faster/slower                                      the step is done/not done with the right timing

The deviations from the intended design are generated by coupling the guide word with a variable parameter of characteristic of the plant or process, such as reactants, reaction sequence, temperature, pressure, flow, phase, etc. in other words:

rigorous questioning, prompted by a series of standard ‘guidewords’ applied to intended design.

Standard guidewords and their generic meanings


No (not, none)                                     none of the design intent is achieved
More (more of, higher)                        quantitative increase in a parameter
Less (less of, lower)                            quantitative decrease in a parameter
As well as (more than)                        An additional activity occurs
Part of                                               Only some of the design intention is achieved
Reverse                                             logical opposite of design intention occurs
Other than (other)                           complete substitution, another activity takes
                                                          Place
Other useful guide words include

Where else                                           applicable for flows, transfers, sources and
 destinations
Before/after                                        the step (or some part of it) is effected out of
                                                            Sequence
Early/late                                           the timing is different from the intention
Faster/slower                                      the step is done/not done with the right timing

The deviations from the intended design are generated by coupling the guide word with a variable parameter of characteristic of the plant or process, such as reactants, reaction sequence, temperature, pressure, flow, phase, etc. in other words:
    

GUIDEWORD + PARAMTER = DEVIATION

 For example, when considering a reaction vessel in which an exothermic reaction is to be undertaken and one of the reactants is to be added stepwise, the guideword “more” would be coupled with the parameter “reactant” and deviation generated would be isothermal “runway”.
Systematic examinations are made of each part of a plant or process using
these guidewords. Having examined one part of the design and recorded any potential hazards and operability problems associated with it, the study progresses to focus on the next part of the design or the next step in the operation. The examination is repeated until the whole plant has been studied for all major modes of operation. Recommendations for changes in design, operating procedure, materials or for referral outside for further consideration can then be made to overcome the problems which have been identified.
The approach described above will generate hypothetical deviations from the design intention.

The success or failure of study depends on four aspects :
The accuracy of the design drawings and other data used as the basis for the study
The technical skills and expertise of the team
The ability of the team to use the approach as an aid to their imagination in visualizing possible deviations , causes and consequences and
The ability of the team to maintain a sense of the of proportion , particularly

When assessing the seriousness of the hazards which are identified.
Careful thought must, therefore be given to preparative work, team composition , keeping of records and so on.

Preparation for carrying out hazop

The amount of preparation for a HAZOP depends upon the size and complexity of the plant. Typically , the data required consist of various drawings in the form of line diagrams ,flow sheets .plant layout , isometrics and fabrication drawings, operating instructions , instrument sequence control charts ,logic diagrams and computer programmes. Occasionally there are plant manuals and equipment manufacturer manuals. The data must be accurate and sufficiently comprehensive. In particular, for existing plant , line diagrams must be checked to ensure they are up to date and that modifications have not been made since the plant was constructed.

Composition of the team to carry out a hazop
HAZOPs are normally carried out by a multi-disciplinary team , including chemical engineers and chemists , with members being chosen for their individual knowledge and experience in design , operations, maintenance or health and safety. A typical team would have between 4 and 7 members, each with a detailed knowledge of the way in which the plant is intended to operate. The technique allows experts in the process to bring their knowledge and expertise to bear systematically so that problems are less likely to be missed. HAZOP is not a technique for bringing fresh minds to work on a problem.
HAZOP studies generate recommendations for design changes. The team should have the authority to agree changes there and then , as progress is slow if every change has to be referred elsewhere for a decision.
It is also essential that the team leader is an expert in the HAZOP technique. The team leader ‘s role is to ensure the team follows the procedure. He or she needs to be skilled in leading a team of people who may not normally be responsible to him or her and the sort of person who pays meticulous attention to detail.the team should have a secretary to prepare notes after each meeting and to circulate them before the next.
It is recommended that the team leader should be an independent person,
i.e., this should not be somebody who is closely associated with the plant under study. The team leader must have sufficient technical knowledge to guide the study properly but should not expected to make his technical contribution.it is beneficial if team members have had some training in the HAZOP technique.

Record keeping

It is usual to record each step of a HAZOP for all the physically meaningful deviations or, if a subset is used, to include those requiring an action plus those which consider significant but required no action because the existing protection was demed adequate. A particularly useful type of record is the “Hazard file”.
This would normaly include:
A copy of the dat(flow sheets, original and final process and instrument diagrams,running instructions,bar sheets,models etc. used by the team during the examinations sessions and marked by the study leader to show that they have been examined.
A copy of all the working papers , questions , recommendations , re-designs etc , produced by the team and others as a result of the study. And
Confirmation that all the agreed actions have been carried out .

The file should be retained on the plant to provide a source of information for future use and in the event that changes are subsequently contemplated by the operating personnel.

 

Assumptions

In a HAZOP study ‘operability’ is as important as hazard and in most cases more operating problems are identified than hazards. The HAZOP technique can therefore enable companies to use resources more effectively and become more efficient as well as safer. It must be remembered , however, theat the use of the HAZOP technique comes too late for fundamental change in design. All that can usually be done is to add on equipment or procedures to control the hazards that have been identified.
The techniques assumes a good level of general management competence , in particular that the plant will be operated and maintained in the manner assumed by the design team and in accordance with good management and engineering practices .If these assumptions are not valid , then HAZOPs are of little value. It is no use identifying hazards if nothing is to be done to manage and control the consequent risk; it is of little use installing trips and alarms if they are going to be badly maintained. The time spent on a HAZOP would be better spent improving management awareness and commitment to provision of maintenance schedules, systems for controlling plant modifications, and so on.

Resource implications


HAZOPs are time consuming and on a continuous plant can take , perhaps , up to half a day per main item of plant            (e.g. still, furnace, reactor, pipeline or transfer line, etc. ) depending on whether the plant is similar to an existing one or completely new. For batch plants both the preparation of data and the study itself will take longer , especially where the same equipment is used to manufacture a range of different products.Meetings are usually restricted to 3 hours , 2 or 3 days per week , to give the team time to attend to their other duties and because people ‘s concentration wanes after 3 hours at a stretch. Many HAZOP studies can be completed in 5 to 10 meetings, although for a small modification only one or two meetings may be necessary. However, for a large project it may take several months even with 2 or 3 teams working in parallel on different sections of the plant.
HAZOPs have major resource implications which should not be under estimated. if HAZOPs are to be introduced to any organization for the first time , it may be appropriate to apply the technique to one or two problems to find out whether it is useful and can be applied successfully.If so technique can grow naturally and be applied to larger projects. For major projects, there may be six(or eight) different hazard studies. A HAZOP study is often the third of these studies. Inherent safety is usually covered in the second study.


Results of hazop


On completion of a HAZOP study the likely outcomes are:

Some improvements in operating/maintenance procedures, control programs and instructions which may already have been implemented together with minor (low cost) hardware alterations. These will have been put in hand as parts of the study are completed.
Some proposed changes may await the result of a more detailed quantitative assessment.
Major recommendations will have yet to be implemented, possibly awaiting capital sanction. And
The team members will already have both a better understanding of the plan/process and a better appreciation of potential hazards and risks than if the study had not been carried out.

Benefits of hazop

The circumstances when HAZOPs are likely to produce benefits are:

·         During the design or installation of any new plant or process, or major modification to an existing one.
·         When there are novel hazards such as environmental hazards and quality or cost issues associated with the operation.
·         Following a major incident involving fire , explosion , toxic release etc. and
·         To justify why a particular code of practice , guidance note or industry code is not to be followed.

Cautions


Even the most rigrous HAZOP cannot be relied upon to foresee every hazard and some accidents may well occur in the future .
When an accident occurs on a plant which has undergone a HAZOP study , several questions of particular significance should be asked :

·         Had the set of conditions (deviations) which led to the incident been considered by the HAZOP study team ? if not , could the team reasonably have been expected to have done so ? and
·         If such deviations and their causes had been considered , had the team made a reasonable judgement of the likely frequency of the events and had concluded that they were unlikely to occur and thus posed ‘ acceptable risks ?

In such circumstances it is clearly important to document all the outcomes of study in order to answer these questions .

Conclusion

HAZOPs are an essential tool for hazard identification and have been used successfully to improve the safety and operability of both new and existing chemical plant. The technique is not confined to the chemical and pharmaceutical industries and has also been used successfully in a number of other industries , including the off-shore oil and food industries.


               CHAPTER NO 7:

INSTRUMENTATION AND PROCESS                                   CONTROL

INSTRUMENTATION AND PROCESS CONTROL

No plant can operate until it is adeaqetly intrumented . the monitoring of temperature , pressure mfloqm level is necessary in almost every process in order to operate plant in proper manner.

OBJECTIVES OF CONTROL SYSTEM
Objectives are
Ø  To ensure the stability of the process
Ø  To reduce the effect of external disturbances
Ø  To optimize the overall process
Ø  To keep the process variables within limit
Ø  To give desire purity
Ø  To Maintain the production composition
To monitor and controle the operation of process there are seven hardware elements as
Ø  Process
Ø  Measuring device
Ø  Tranceducer
Ø  Transmission lines
Ø  Controller
Ø  Final control element
Ø  Recorder
PROCESS
It represent the material equipment together with physical and chemical oprations

MEASURING DEVICE
The instruments by which process is controlled  and und required enviroment is created  for process.
TRANSDUCER
It is interface between process and control system and its job is to convert the sensor signal into control system.

 TRANSMISSION LINES
These lines carry signal from transducer to control system. In modern systems transmission lines are not used as they reauire large investment in capital cost. Now in modern system signal are transferred by electrical signals which don’t require any transmission lines.

CONTROLLER
Controller is the brain of whole plant. Its objective is to compare output signal and set point and give correction to the system as required.

FINAL CONTROLE ELEMENT
Final control element work on the instruction of controller.Final control element  mostly is control valve.

RECORDER
Recorder , records the all process variable for monitoring purpose.

GENERAL CONTROL SYSTEM
Following are the  general control system
Ø  Open loop system
Ø  Close loop system
Ø  Feed backward system
Ø  Feed forward system
Ø  Combined control system
Ø  Cascade control system
Ø  Ratio control system
Ø  Proportional control
Ø  Proportional integral control
Ø  Proportional derivative control
Ø  Pid control

MEASRING DEVICES

In any chemical process we usally measure
Ø  Temperature
Ø  Pressure
Ø  Flow rate
Ø  Level

PIPING NETWORK

The American National Standards Institute (ANSI) and the American Petroleum Institute (API) have established detailed standards for the most widely used components for piping systemMany of these standards contain pressure-temperature ratings that will be assistance to engineers in their design functions.
Some of the specific requirements for piping system have been included in the Occupational Safety & Health Administrations (OSHA) requirements.

SELECTION OF PIPING MATERIALS

·         General aspects that need to be evaluated when selecting piping materials are:
·         Possible to exposure to fire with respect to the loss in strength.
·         Susceptibility of the pipe to brittle failure or thermal shock failure when exposed to fire.
·         Ability of thermal insulation to protect the pipe from fire.
·         Susceptibility of pipe & joints to corrosion or adverse electrolytic effect.
·         Suitability of packing, seals, gaskets and lubricants used on joints & connections.
·         Refrigeration effect during sudden loss of pressure with volatile fluids.
·         Compatibility with the fluid handled.

Metal Piping Materials
Possible Material Precautions and Alternatives
Metal
                                         
Iron(Cast, Malleable & High Silicon)
Lack of ductility & sensitivity to thermal & mechanical shock
Carbon steel & Several low- alloy steels
Embrittlement when handling alkaline or strong caustic fluids, conversion of carbides to graphite during long time exposure to temperature above 4270C
High-alloy stainless steels
Stress corrosion cracking of austenitic stainless steels exposed to chlorides & other halides; intergranular corrosion of austenitic stainless steel after long time exposure to temperature between 4270C and 8710C
Nickel & Nickel-base alloys
Grain boundary attack of both materials not containing chromium when exposed to sulfur at temperatures above 3150C
Aluminum & Aluminum alloys
Corrosion from concrete, mortar, lime, plaster, & other alkaline materials; intergranular attack of alloys 5154, 5087, 5083 and 5456 when exposed to temperature above 650C
Non-Metal Piping Materials

Thermoplastics
Minimum & Maximum operating temperature limits between -340C and 2100C for thermoplastic piping material and -1980C 2600C for thermoplastics used in linings, special precautions when used for transporting compressed air or other gases
Reinforced Thermosetting resins
Minimum & Maximum operating temperature limits between 290C and 930C




COSTS FOR PIPING AND PIPING SYSTEMS AUXILIARIES

Piping is a major item in the cost of all types of process plants. These costs in a fluid-process plant can run as high as 80% of the purchased equipment cost or 25% of the fixed-capital investment. There are essentially two basic methods for preparing piping cost estimations:
·         The percentage of installed equipment method.
·         The material and labor takeoff method.


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3 comments:

  1. Valuable for information.. Is there any further reading you would recommend on this?

    Ally
    High Speed Tubular Bowl Centrifuge

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  2. A catalytic oxidizer, or catalytic converter, is an automobile component
    found in the exhaust system. It oxidizes (burns) carbon monoxide and hydrocarbons, using platinum and palladium/rhodium as catalysts, while reducing (separating the oxygen atom rather than adding it) nitrogen oxides to create nitrogen. This greatly reduces toxic tailpipe emissions and reduces smog.

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  3. Optimum information indeed...
    Few of the data is missing...is there complete data somewhere regarding Adipic acid production???that u can share the missing points as well

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