5-1CHAPTER 5 CHEMICAL DESIGN 5.1 DESIGN OF ESTERIFICATION REACTOR (R-101) 5.1.1 Selection of Reactor Esterification process is a process that involves heating and boiling of the reactants in liquid phase. The reactants will vaporize and sent to the separation units. The industrial chemical processes convert raw materials to products, and the reactor is often considered as the heart of the plant. The selection and design of the reaction units are essential for the economic success of a chemical industry, imposing final yields and conversions. The following table shows the comparison between the chemical reactor available for the reaction. Table 5.1: Comparison on the type of reactor with its criteria evaluation Criteria Batch reactor Semi-batch reactor Suitable for phase Gas phase, liquid phase, liquid-solid Gas phase, liquid phase, liquid-solid Usage Advantages Small scale production Intermediate or one shot production Pharmaceutical Fermentation High conversion per unit volume Continuous stirred tank reactor Gas-liquid, solidliquid, liquid phase Hydrolysis Chlorination Reactive distillation (Esterification) Control of raw materials (limit When agitation is required Series configurations for different concentration stream Continuous operation Plug flow reactor Primarily Gas Phase Large Scale Fast Reactions Homogeneous Reactions Heterogeneous Reactions Continuous Production High Temperature High Conversion per 5-2 for one pass Flexibility of operation-same reactor can produce one product one time and a different product the next Easy to clean unwanted side product formation) Good temperature control Easily adapts to two phase runs Good control Simplicity of construction Low operating (labor) cost Easy to clean Disadvantages High operating cost Product quality more variable than with continuous operation Not applicable for continuous process not famously applied in industry Lowest conversion per unit volume By-passing and channelling possible with poor agitation Unit Volume Low operating (labor) cost) Continuous Operation Good heat transfer Undesired thermal gradients may exist Poor temperature control Shutdown and cleaning may be expensive Since the process involved is reversible. It is not easy to control the parameters inside the vessel with this tubular reactor. Therefore, from these comparisons, CSTR is the most suitable for the esterification reaction. However, due to the process specifications, some modifications have to be made on the reactor to meet the desired requirements. A stirrer is not required to mix the reactants. 2-ethylhexyl acrylate is a compound that can easily form through boiling; thus money shouldn’t have to be spent on the stirrer or agitator. The reactor that is chosen for this plant comprises of the following features: (i) A process vessel to provide a reservoir of boiling liquid for the esterification column and to provide for surge protection. 5-3 (ii) An internal coil as the cooling element in the vessel to assist the exothermic reaction. (iii) Use of stirrer as to ensure the uniform mixing and temperature throughout the reactor. The type of reactor that been choose is Continuous Stirred Tank Reactor (CSTR) including jacket there. The rationales of CSTR as our chemical reactor because: (i) The composition is constant. To maintain its composition of the products, the reactor needs to be operated in continuous process and CSTR is the best reactor. (ii) CSTR is the reactor that easy to maintain the temperature which is 110℃ or in other words is easy to maintain the temperature control because it‘s perfectly mixed. (iii) To maintain high conversion. The amount of the product is quite huge (100000MT/yr). The benefits used CSTR are the right reactor for huge amounts of productions and can maintain high conversion which is 59%. The power will depend on the degree of agitation required and will range from about 0.2kW/m3 for moderate mixing to 2kW/m3 for intense mixing. (iv) Cheap to construct. CSTR is cheap to construct. It‘s very economic where the cost of construction can be minimized. 5.1.2 Sizing of the Reactor 5.1.2.1 Assumptions in Calculations (i) The reactor behaves as an ideal CSTR. (ii) Perfect mixing. (iii) The composition of the components is constant and the same as the product stream, since it is operated as a continuous process. (iv) There is no temperature gradient inside of the reactor (isothermal). 5-4 5.1.2.2 Determination of Reaction Rate The stoichiometric equation of esterification process: k1 C3H4O2 (l) + C8H18O (l) ↔ C11H20O2 (g,l) + H2O (g,l) AA k2 EHOL EHA W By assuming an elementary reaction, the order of the above reaction is figured out to be as second order (Pawel Nowak, 1999). The process is reversible, hence the reaction rate R will be: R = k1 AA EHOL − k 2 EHA W (5.1.1) In order to determine R, the concentration of all four components at the inlet and outlet streams of the reactor need to be calculated first. Table 5.2: Properties of the Inlet and Outlet Stream Inlet Component Molar flow, kmol/hr outlet Mole fraction Molar flow, kmol/hr Mole fraction Acrylic acid 129.3798 0.5386 32.234 0.2949 2-ethylhexanol 79.6725 0.3317 12.7076 0.0958 2-EHA 15.5231 0.0646 73.2903 0.5523 Water 14.8413 0.0618 13.3743 0.1008 Source: Hysys 5-5 Table 5.3: Constant Value for Density Calculation at Various Temperatures Range of Name Formula C1 C2 C3 C4 Temperature, k ρ at 383.15K, kmol/m3 Acrylic acid C3H4O2 1.2414 0.25822 615 0.30701 286.15-615 4.0356 2-ethylhexanol C8H18O 0.5573 0.2714 583 0.29538 180-583 1.7196 2-EHA C11H20O2 NA NA NA NA NA 4.8139 water H2O 4.97 0.278 647 0.187 333.15403.15 16.0953 Source: Perry’s handbook, 2001 liquid density, ρ , V ′ component , kmol = m3 C2 C1 T(K) 1+ C C4 3 m3 mol fraction component × molar flow component = s ρ component Vmixture = compound = Vcompound number of mol of component Vmix (5.1.2) 5.1.3 5.1.4 (5.1.5) The density of each component as the function of temperature can be determined by following: 1.2414 kmol = 4.0356 3 383.15 m 0.2582 1 + 0.30701 615 0.5573 kmol ρ EHOL = = 1.7196 3 383.15 m 0.2714 1 + 583 0.29538 kmol ρ 2EHA = 4.8139 3 m 4.97 kmol ρ W = = 16.0953 3 383.15 m 0.278 1 + 647 0.187 ρ AA = 5-6 At the inlet stream of reactor The volumetric flow of each component can be determined by following: kmol 3 hr = 31.9531 m V AA = kmol hr 4.0356 3 m kmol 0.3317 × 239.4167 hr m3 V 2EHOL = = 46.1820 kmol hr 1.7196 3 m kmol 3 0.0646 × 239.4167 hr = 3.2128 m V 2EHA = kmol hr 4.8139 3 m kmol 3 0.061 × 239.4167 hr = 0.9074 m V AA = kmol hr 16.0953 3 m 0.5386 × 239.4167 Therefore, the volumetric flow for mixture of compound is Vmix = 82.2553 m3 hr From this answer, the concentration of each compound can be determined by following: 129.3798 kmol = 1.5729 82.2553 m3 79.6725 kmol 2 − EHOL = = 0.9686 82.2553 m3 15.5231 kmol 2 − EHA = = 0.1887 82.2553 m3 14.8413 kmol AA = = 0.1804 82.2553 m3 AA = Where else, at the outlet of reactor: The volumetric flow of each component can be determined by following: kmol 0.2429 × 239.4167 hr m3 V AA = = 14.4103 kmol hr 4.0356 3 m 5-7 kmol 0.0958 × 239.4167 hr m3 V 2EHOL = = 13.3381 kmol hr 1.7196 3 m kmol 0.5523 × 239.4167 hr m3 V 2EHA = = 27.4683 kmol hr 4.8139 3 m kmol 0.1008 × 239.4167 hr m3 V AA = = 1.4994 kmol hr 16.0953 3 m Therefore, the volumetric flow for mixture of compound is Vmix = 56.7161 m3 hr From this answer, the concentration of each compound can be determined by following: 32.234 kmol = 0.5683 56.7161 m3 12.7076 kmol 2 − EHOL = = 0.2241 56.7161 m3 73.2903 kmol 2 − EHA = = 1.2922 56.7161 m3 13.3743 kmol AA = = 0.2350 56.7161 m3 AA = According to Levenspiel (2002), rate of reaction R based on unit volume of reacting fluid is given by, R= 1 dN moles of product formed Moles of reactant consumed = = V dt Volume of fluid Time Volume of fluid time where 5.6 N = concentration V Hence, R = d component dt (5.7) For this esterification process, by referring to the stoichiometric equation: R=− d AA d EHOL d EHA dW =− = = dt dt dt dt 5-8 From the calculation of concentrations, it is found that d AA kmol = 0.5683 − 1.5729 = 1.0046 3 dt m h d EHOL kmol − = 0.2241 − 0.9074 = 0.6833 3 dt m h d EHA kmol = 1.2922 − 0.1887 = 1.1035 3 dt m h dW kmol = 0.2358 − 0.1804 = 0.0554 3 dt m h − The value of the reaction rate is different to each component; hence, the highest rate of reaction is chosen in order to form the desired product. It can be concluded that in order to obtain the desired concentration of the products, the rate of reaction that should be used is 1.1035 kmol / m3 hr. By inserting the value of R into eq. 8.1, the reaction constant k2 can be obtained. The reaction constant k1 is obtained from Pawel Nowak, 1999. Even though the catalyst referred by him is sulphuric acid, yet, it is can be equate to this esterification process which employed PTSA as the catalyst. The difference and advantages of having PTSA as the catalyst compared to sulfuric acid for this process had been discussed in Design Project 1 in Chapter 1. Semibatch reactor can also be considered as a continuous reactor; hence the prediction of the order of the esterification reaction can also be said to be second order. 1.1035 = 0.85 1.5729 0.9686 − k 2 1.2922 0.2358 k 2 = 0.6306 m6 kmol2 hr 5.1.2.3 Determination of Reactor Volume C3 H4 O2 + C8 H18 O ⟷ C11 H20 O2 + H2 O eq 1.8 a 2C2 H4 O2 + C8 H18 O → C6 H8 O4 + C8 H16 + H2 O eq 1.8(b) C3 H6 O2 + C8 H18 O → C11 H22 O2 + H2 O eq 1.8 c 5-9 Since there is no presence of 2-ethylhexyl Acetate in the first reactor, therefore the third reactions can be ignored. t=0 Co X=0 Vo Fo t=tf C X=Xf V F B+A↔C+D −rA = k1 CA CB − k 2 CC CD 2E+A → G+H+D −rA = k 3 CA CE I+A→J+D −rA = k 4 CA CI At 383.15K, the specific reaction rate constant is k1 = 0.85 m6 kmol2 hr k 2 = 0.6306 m6 kmol2 hr 3 m6 k 3 = 9.2083E − 5 kmol2 hr 3 m6 k 4 = 6.5983E − 4 kmol2 hr The reaction constant is determined from several journals that relates towards each reaction in which the try error calculation is made and tested with hysys simulation. (Pawel Nowak,1999 ;Sami Ali et.al., 2000; V. Ragaini, 2007;Basf Technical Data Sheet, 1997) The component mole balance equation, thus: input = output + disappearance by reaction + accumulation FA0 − FA + (−rA )V = 0 5-10 −rA = FA − FA0 V FA = FA0 (1 − XA ) −rA = FA0 XA V Once the component mole balance for each species has been derived, the subsequent step is to determine the net rate of formation of each species that relative to component A which is 2-ethylhexanol. riJ riA riB riC riD riE riG riH riI = = = = = = = = −ai −bi ci di ei gi hi ii ji Since the reaction occur in the liquid phase which assigned as incompressible and newtonian liquid, the specific volume is constant and independent of pressure and temperature, thus: rj = dCj dτ Fj Cj = vo (i) Rate of disappearance of 2-ethylhexanol: rA = −rA1 − rA2 − rA3 rA = dCA FAO X = −k1 CA CB − k 3 CA CE − k 4 CA CI + k 2 CC CD = dt V dFA = −k1 CA CB − k 3 CA CE − k 4 CA CI + k 2 CC CD dV dFA 1 = 2 −k1 FA FB −k 3 FA FE −k 4 FA FI + k 2 FC FD dV vo Initial condition for this particular ordinary differential equation is: kmol kmol , C = C = 14.6999 , B Bo m3 m3 kmol kmol kmol CEo = 17.7778 3 , CC = 4.8226 , CD = 55.3987 , 3 m m m3 kmol CI = 13.5211 m3 At t = 0, CA = CAo = 6.4252 5-11 kmol kmol , FB = FBo = 129.3798 , hr hr kmol kmol kmol FC = 15.5231 , FD = 14.8413 , FE = 0.016 , hr hr hr kmol FI = 0.096 hr At V = 0, FA = FAo = 79.6725 (ii) Rate of disappearance of acrylic acid: rB = −k1 CA CB + k 2 CC CD rB = dCB FBO X = −k1 CA CB + k 2 CC CD = dt V dFB = −k1 CA CB + k 2 CC CD dV dFB 1 = 2 −k1 FA FB + k 2 FC FD dV vo Initial condition for this particular ordinary differential equation is: kmol kmol , CB = CBo = 14.6999 3 , 3 m m kmol kmol CC = 4.8226 , CD = 55.3987 3 m m3 kmol kmol At V = 0, FA = FAo = 79.6725 , FB = FBo = 129.3798 , hr hr kmol kmol FC = 15.5231 , FD = 14.8413 hr hr At t = 0, CA = CAo = 6.4252 (iii) Rate of disappearance of 2-ethylhexyl acrylate: rC = k1 CA CB − k 2 CC CD rC = dCC FCO X = k 1 CA CB − k 2 C C CD = dt V dFC = k 1 CA CB − k 2 CC CD dV dFC 1 = 2 k 1 F A F B − k 2 CC CD dV vo 5-12 Initial condition for this particular ordinary differential equation is: kmol kmol kmol , CB = CBo = 14.6999 3 , CC = 4.8226 , 3 m m m3 kmol CD = 55.3987 m3 kmol kmol At V = 0, FA = FAo = 79.6725 , FB = FBo = 129.3798 , hr hr kmol kmol FC = 15.5231 , FD = 14.8413 hr hr At t = 0, CA = CAo = 6.4252 (iv) Rate of disappearance of water: rD = rD1 + rD2 + rD3 rD = dCD FDO X = k 1 CA CB + k 3 CA CE + k 4 C A CI − k 2 CC CD = dt V dFD = k 1 CA CB + k 3 CA CE + k 4 CA CI − k 2 CC C D dV dFD 1 = 2 k1 FA FB +k 3 FA FE + k 4 FA FI − k 2 FC FD dV vo Initial condition for this particular ordinary differential equation is: kmol kmol kmol , CB = CBo = 14.6999 3 , CC = 4.8226 , 3 m m m3 kmol kmol kmol CD = 55.3987 , C = 17.7778 , C = 13.5211 Eo I m3 m3 m3 kmol kmol At V = 0, FA = FAo = 79.6725 , FB = FBo = 129.3798 , hr hr kmol kmol kmol FC = 15.5231 , FD = 14.8413 , FE = 0.016 , hr hr hr kmol FI = 0.096 hr At t = 0, CA = CAo = 6.4252 (v) Rate of disappearance of acetic acid: rE = −2k 3 CA CB rE = dCE FEO X = −2k 3 CA CB = dt V dFE = −2k 3 CA CB dV dFE 1 = 2 −2k 3 FA FB dV vo 5-13 Initial condition for this particular ordinary differential equation is: kmol kmol , CB = CBo = 14.6999 , 3 m m3 kmol kmol = 79.6725 , FB = FBo = 129.3798 , hr hr At t = 0, CA = CAo = 6.4252 At V = 0, FA = FAo (vi) Rate of disappearance of diacrylic acid: rG = k 2 CA CB rG = dCG FGO X = k 2 CA CB = dt V dFG = k 2 CA CB dV dFG 1 = 2 k 2 FA FB dV vo Initial condition for this particular ordinary differential equation is: kmol kmol , CB = CBo = 14.6999 , 3 m m3 kmol kmol = 79.6725 , FB = FBo = 129.3798 , hr hr At t = 0, CA = CAo = 6.4252 At V = 0, FA = FAo (vii) Rate of disappearance of octene: rH = k 3 CA CB rH = dCH FHO XA = k 3 CA CB = dt V dFH = k 3 CA CB dV dFH 1 = 2 k 3 FA FB dV vo Initial condition for this particular ordinary differential equation is: kmol kmol , CB = CBo = 14.6999 , 3 m m3 kmol kmol = 79.6725 , FB = FBo = 129.3798 , hr hr At t = 0, CA = CAo = 6.4252 At V = 0, FA = FAo 5-14 (viii) Rate of disappearance of propionic acid: rI = −k 4 CA CI rI = dCI FIO XA = −k 4 CA CI = dt V dFI = − k 4 CA CI dV dFI 1 = 2 −k 4 FA FI dV vo Initial condition for this particular ordinary differential equation is: kmol kmol , CI = CIO = 13.5211 3 3 m m kmol kmol = 79.6725 , FI = 0.096 hr hr At t = 0, CA = CAo = 6.4252 At V = 0, FA = FAo (ix) Rate of disappearance of octyl propionate: rJ = k 4 CA CI rJ = dCJ FJO XA = k 4 CA CI = dt V dFJ = k 4 CA CI dV dFJ 1 = 2 k 4 FA FI dV vo Initial condition for this particular ordinary differential equation is: kmol kmol , CI = CIO = 13.5211 3 3 m m kmol kmol = 79.6725 , FI = 0.096 hr hr At t = 0, CA = CAo = 6.4252 At V = 0, FA = FAo Euler method is applied in order to solve ODE for every rate of reaction of particular species. The general formula of euler method is: yi+1 = yi + f xi , yi h 5-15 Thus, the ODE of molar flor rate is a function of volume of the reactor for every species is: dFA 1 = 2 −k1 FA FB −k 3 FA FE −k 4 FA FI + k 2 FC FD dV vo dFB 1 = 2 −k1 FA FB + k 2 FC FD dV vo dFC 1 = 2 k 1 FA FB − k 2 FC FD dV vo dFD 1 = 2 k1 FA FB +k 3 FA FE + k 4 FA FI − k 2 FC FD dV vo dFE 1 = 2 −2k 3 FA FB dV vo dFG 1 = 2 k 3 FA FB dV vo dFH 1 = 2 k 3 FA FB dV vo dFI 1 = 2 −k 4 FA FI dV vo dFJ 1 = 2 k 4 FA FI dV vo Sample of solution for the first iteration with step size, h = 1 m3 is: Volumetric flow rate of the reaction mixture inside the reactor is taken as constant, Thus: 24.7983 m3 hr 79.6725 kmol 129.3798 kmol 0.016kmol 0.096kmol = , FB0 = , FE0 = , FI0 = , hr hr hr hr kmol kmol FC = 15.5231 , FD = 14.8413 hr hr m 3 k1 = 0.85 kmol. hr m 3 k 2 = 0.6306 kmol. hr m 3 k 3 = 9.2083E − 5 kmol. hr m 3 k 4 = 6.5983E − 4 kmol. hr vo = FA0 5-16 yi+1 = yi + f Vi , Fi h 2-ethylhexanol: f Vi , Fi = dFA dV FA 1 = FA 0 + f 0, FA0, FB0 , FE0 , FI0 h FA 1 = 79.6725 kmol hr 1 + − 0.85 × 79.6725 × 129.3798 24.79832 − 9.2083E − 5 × 79.6725 × 0.016 − 6.5983E − 4 × 79.6725 × 0.096 + 0.6306 × 15.5231 × 14.8413 FA 1 = 65.66 Therefore, at V= 1m3, FA= 65.66 kmol kmol hr . hr Acrylic acid: f Vi , Fi = dFA dV FB 1 = FB 0 + f 0, FA0, FB0 , FE0 , FI0 h FB 1 = 129.3798 kmol hr 1 + − 0.85 × 79.6725 × 129.3798 24.79832 − 9.2083E − 5 × 79.6725 × 0.016 − 6.5983E − 4 × 79.6725 × 0.096 + 0.6306 × 15.5231 × 14.8413 FB 1 = 115.37 Therefore, at V = 1m3, FA = 65.66 kmol hr kmol hr . The sample calculation for other component is shown in appendix. From the calculation of ordinary differential equation by using Euler method, it is found that the largest volume require for reaction is 17.01m 3. Since we are dealing with liquid reactants in the vessel, it is a common practice to assume that the liquid level in the vessel is about 50%. This is to allow for enough space of vaporization and also for the safety purpose. 5-17 Vactual = Table 5.4: 17.01 m3 = 34.02 m3 0.5 Ratio of Length to Diameter of Vessel as a Function of Pressure P (atm) 0-16 17-33 >33 L/D 3 4 5 Hence, for a vessel operates at 0.29 atm, L =3 D (5.1.9) Vcylinde r = πrvessel 2 Lvessel = Vvessel (5.1.10) By inserting equation (5.1.8) into equation (5.1.9), to obtain, 3 Vvessel = 6πrvessel rvessel Vvessel = 6π 1 3 Dimensions of the reactor are calculated as following: rvessel 34.02 = 6π 1 3 = 1.2175 m Dvessel = 2rvessel = 2.435 m Lvessel = 3Dvessel = 7.305 m 5.1.2.4 Determination of Space Time Space time require to achieve the conversion of 59% of 2-ethylhexyl acrylate is τ= V = vo 17.01 m3 = 0.6859 hr = 41.16 min m3 24.7983 hr It is means that 41.16 minutes is required to process one reactor volume of the feed, which is 17.01 m3 of the reactants. 5-18 5.1.3 Summary of Chemical Design Table 5.5: Summary of Chemical Design of Esterification Reactor 1 (R-101) SPECIFICATION DATA CHEMICAL DESIGN Identification Type of reactor Continuous Stirred Tank Reactor Design orientation Vertical Production 100,000 MT 2-ethylhexyl acrylate / year Physical Condition Volume of reactor, m3 34 Diameter of reactor, m 2.435 Height of reactor, m 7.305 Type of insulator Rockwool Reaction Type of reaction Esteriferication process Raw materials 2-ethylhexanol, acrylic acid & mixture of other product from absorber Order of reaction Second order Residence time, min 41.16 min Type of catalyst PTSA 5-19 5.2 DESIGN OF ESTERIFICATION REACTOR (R-102) 5.2.1 Introduction Chemical reactors, particularly for continuous processes, are often custom designed to involve multiple phases, different geometries and various regime of momentum, heat and mass transfer. There are so many configurations, involving different combinations of these attributes that attempts to develop generalized reactor models have met with limited success. Most of the process simulators provide four kinds of reactor models including a stoichiometric model that permits the specification of reactant conversions and extents of reaction for one or more specified reactions; a model for multiple phases in equilibrium, where the approach to equilibrium for individual reactions can be specified; a kinetic model for a continuous-stirred-tank reactor (CSTR) that assumes perfect mixing of homogeneous phases and a kinetic model for a plug-flow tubular reactor (PFTR or PFR), for homogeneous phases and assuming no backmixing (dispersion). These ideal models are used in the early stages of process synthesis, when the details of the reactor designs are less important, but reactor effluents and heat duties are needed. 5.2.2 Selection of Reactor In order to produce 2-EHA, variety of reactor types can be used. This includes a batch reactor, a tubular reactor and a continuous stirred tank reactor (CSTR). However, it is important for us to choose the most suitable reactor to be used for the process plant by giving consideration on practicality, production and cost of installation and operation. After the decision on the type of the reactor has been made, a chemical and mechanical design of the reactor can be developed. As the first esterification reactor (R-101), the type of second reactor R-102 used is also the continuous stirred tank reactor. The actual purpose for involving two esterification reactors in in this 2-EHA plant is due to the factor of residence time. In order for an esterification process to achieve higher conversion, it is very important to ensure that the residence time inside the reactor where reactions taking place, is suitable since it is a reversible reaction. Through this cascade arrangement, we also want to increase the conversion of the reactions involved. For this 2-EHA plant, the conversion of second esterification reactor 5-20 is about 80%. It will also have no stirrer or agitator and also no motor power since the reactions taking place in a homogeneous liquid phase and the reaction is well-mixed. R-102 is comprised with features such as to provide for surge protection as well as to have an internal coil as the cooling element in the vessel. 5.2.3 Reactor Sizing In reactor sizing, the first important is to list the all assumptions that we to take into account before proceed to another step. The assumptions as followed: (i) As the reactions taking place in a homogeneous liquid phase, the reactor contents are assumed to be perfectly mixed. (ii) The fluid elements do not all have the same residence time in the reactor. There is residence time distribution. (iii) The composition and temperature are uniform throughout the reactor volume and equal to the composition and temperature of the reactor effluent, since it is operated in a continuous mode of operation. (iv) The intermediates and by-products formed together in the esterification process present in small amount but they are taken into consideration in determining the rate of reaction of this process. 5.2.4 Determination of Reaction Rate, R Figure 5.1: Components at Outlet Stream 8 5-21 Reactions occurred in R-102 as below: Main Reaction: C3 H4 O2 + C8 H18 O (AA) ↔ C11 H20 O2 + H2 O (EHOL) (2-EHA) (W) By-product reactions: 2C2 H4 O2 + C8 H18 O (DAA) ↔ (EHOL) C6 H8 O4 + C8 H16 + H2 O (DAA) (Oc) (W) As we discuss in Design Project 1, we have one main reaction and one by-product reaction. All the reactions are reversible reaction which the order of the main reaction is figured out to be second order as well as for the by-product reaction. The catalyst used in the reaction is PTSA or para-toluene sulfonic acid. In order to determine R, the concentration of all components at the inlet and outlet stream of reactor R-102 need to be calculated. All the required properties are taken from HYSYS simulation data. All the values we take as 4 decimal places for more exactly results. Table 5.6: Properties of the Inlet and Outlet Stream of R-102 Composition Composition Inlet Outlet S6 S8 AA 0.1558 AC Component Molar Molar Flow(kgmol/hr) Flow(kgmol/hr) Inlet S6 Outlet S8 111.6 98.61 0.1071 17.3873 10.5611 0.0000 0.0000 0.0000 0.0000 P 0.0003 0.0002 0.0355 0.0197 DAA 0.0068 0.0078 0.7589 0.7692 EHOL 0.0733 0.0436 8.1803 4.3000 2-EHA 0.6888 0.7901 76.8701 77.9118 EHAC 0.0000 0.0000 0.0000 0.0000 EHOP 0.0000 0.0000 0.0000 0.0000 OC 0.0020 0.0017 0.2232 0.1676 W 0.0730 0.0494 8.1468 4.8713 5-22 Average molar density is taken from HYSYS which is 4.915 kgmol/m3. Volume, V (m3/s) can be calculated using equation: Vcomponent mol fraction total molar flowrate component (5.2.1) Concentration (kmol/m3) of the components can be calculated using equation: Component 5.2.4.1 Flowrate component Vmixtures (5.2.2) Inlet of Reactor R-102 (S6) Table 5.7: Volume and Concentration of Component in S6 Component Volume (m3) Concentration (kgmol/m3) AA 3.5376 0.7658 AC 0.0000 0.0000 P 0.0068 0.0015 DAA 0.1544 0.0334 EHOL 0.6643 0.3603 2-EHA 15.6400 3.3855 EHAC 0.0000 0.0000 EHOP 0.0000 0.0000 OC 0.0454 0.0098 W 1.6575 0.3588 5-23 5.2.4.2 Outlet of Reactor R-102 (S8) Table 5.8: Volume and Concentration of Component in S8 Component Volume (m3) Concentration (kgmol/m3) AA 2.1488 0.5418 AC 0.0000 0.0000 P 0.0040 0.0010 DAA 0.1565 0.0395 EHOL 0.8747 0.2206 2-EHA 15.8518 4.0 EHAC 0.0000 0.0000 EHOP 0.0000 0.0000 OC 0.3411 0.0086 W 1.9911 0.2500 According to Octave Levenspiel (2002), rate of reaction, R based on unit volume of reacting fluid is given by: 1 dN Moles of product formed Moles of reac tan t consumed R ( Volume of fluid)(Time ) V dT ( Volume of fluid)(Time ) (5.2.3) N Where Concentration V Hence R d [component ] dt For this esterification process, according to stoichiometric equations: −d[AA] −d[EHOL] d[2 − EHA] d[W] = = = dt dt dt dt −d[AC] −d[EHOL] d[EHAC] d[W] R2 = = = = dt dt dt dt −d[DAA] −d[EHOL] d[DAA] d[OC] d[W] R3 = = = = = dt dt dt dt dt −d[P] −d[EHOL] d[EHOP] d[W] R4 = = = = dt dt dt dt R1 = (5.2.4) 5-24 From the value of concentration for each component in inlet and outlet stream of R102, we found that: Table 5.9: Reaction Rate for Each Component Component Rate (kgmol/m3h) AA 0.224 AC 0.0000 P 0.0187 DAA 0.0061 EHOL 0.1397 2-EHA 0.6145 EHAC 0.0000 EHOP 0.0000 OC 0.0012 W 0.1088 We take average values of each reaction rate as:0.224 + 0.1397 + 0.6145 + 0.1088 kgmol = 0.2718 4 m3 h 0.0061 + 0.1397 + 0.0061 + 0.0012 + 0.1088 kgmol R3 = = 0.0524 4 m3 h R1 = As we can see, the reaction rate for the main reaction is the highest value which is 0.2718 kgmol m3h And the by-product reaction rate is about 0.0524 5.2.5 kgmol m 3h Determination of Reactor Volume and Dimensions In CSTR reactor, we can use the formula below to get the volume of reactor X A V ; XA0 ≠ 0 FA 0 RA (5.2.5) 5-25 Table 5.10: Volume of Component in Reactor R-102 Composition Composition Inlet Outlet S4 S5 AA 0.1558 AC Component Molar Molar Flow(kgmol/hr) Flow(kgmol/hr) V FA 0 X A R Inlet S4 Outlet S5 111.6 98.61 0.1071 17.3873 10.5611 3.1154 0.0000 0.0000 0.0000 0.0000 0.0000 P 0.0003 0.0002 0.0355 0.0197 1.2877 x 10-6 DAA 0.0068 0.0078 0.7589 0.7692 2.7921 x 10-3 EHOL 0.0733 0.0436 8.1803 4.3000 0.8939 2-EHA 0.6888 0.7901 76.8701 77.9118 28.6495 EHAC 0.0000 0.0000 0.0000 0.0000 0.0000 EHOP 0.0000 0.0000 0.0000 0.0000 0.0000 OC 0.0020 0.0017 0.2232 0.1676 2.4636 x 10-4 W 0.0730 0.0494 8.1468 4.8713 0.7074 R=0.2718 Only refer to 2-EHA; V2−EHA = 76.8701(0.7901 − 0.6888) = 28.6495m3 0.2718 Since the process involved is in homogeneous liquid phase reaction, it is a common practice to assume that ??????? = 1.5?2−??? . This is for safety purposes on the esterification reactor and allows enough space for vaporization inside the reactor. So, the actual volume as below: Vactual = 1.5 28.6495 = ?? ?? From Coulson & Richardson's Chemical Engineering Design Book Volume 6, we are commonly use cylindrical vessel and the optimum L/D of the vessel is about 3, in range of pressure 0-16 atm. 5-26 Table 5.11: Ratio of Length to Diameter of Vessel as a Function of Pressure P (atm) L/D 0-16 3 17-33 4 >33 5 Operating pressure of R-102 is: 29kPa = 0.2862 atm. This pressure is below 1 atm pressure of atmosphere. So, this is proved our L/D for this esterification reactor is 3. Vcylinder rvessel 2 L vessel Vvessel (5.2.6) Vvessel = 2rvessel (5.2.7) Lvessel Lvessel = Dvessel 2rvessel 3 So, we get Vvessel = 6πrvessel and radius of the vessel become as equation below; 1 V rvessel = ( vessel 6π)3 (5.2.8) Vvessel = Vactual = 42.9743m3 1 rvessel = (42.9743 6π)3 = 1.3161m From the value of radius, the diameter and length of the vessel can directly obtain. Diameter of vessel: Dvessel = 2rvessel = 2 1.3161 = 2.6322m Length of vessel: Lvessel = 3Dvessel = 3 2.6322 = 7.8966m So, height or length of esterification reactor R-102 is 7.8966m 5.2.6 Determination of Space Time Space time is defined as the time required to process one reactor volume of feed. It is determined using the following equation: ( time ) C A0 C A R where R = 0.2718 (5.2.9) 5-27 Table 5.12: Space Time Required for Each Component Component Time/hr AA 0.8241 P 0.0018 DAA 0.0244 EHOL 0.5140 2-EHA 2.2609 EHAC 0.0000 EHOP 0.0000 OC 0.0044 W 0.4000 From the above table, the space time for 2-EHA is the highest value which is 2.2600 hours. So, taking the largest value of space time; Hence, Ʈ = 2.2609 x 60 =135.654 min From the value of space time obtained, it shows that 136 minutes is required to process one reactor volume of feed stream (S6). 5.2.7 Summary of Chemical Design Table 5.13: Type of reactor Summary of Esterification Reactor 2 Design Continuous Stirred Tank Reactor Shape of reactor Straight cylindrical reactor Volume of reactor 43 m3 Radius of reactor 1.3161m Diameter of reactor 2.6322m Height of reactor 7.8966m Space time 135.654 min 5-28 5.3 DESIGN OF DISTILLATION COLUMN (T-102) 5.3.1 Introduction Distillation is a method used to separate the components of a liquid solution, which depends upon the distribution of these various components between a vapor and a liquid phase. All components are present in both phases. Feed mixture to the distillation column contains more than two components, thus it is called as multicomponent distillation. Before commencing the design of multicomponent distillation, two key components are selected. Key components are the two components of the feed mixture between which one likes to make sharp separation. Two key components are further divided as light key component and heavy key component. Component with small boiling point and more volatile is called light key while component which has large boiling point and less volatile component is called heavy key. In this production of Etyhlhexyl Acrylate, there are also contains Diacrylic Acid, Octylpropionate and Ethylhexyl Acetate as the by-products. Therefore this distillation column is required to remove Diacrylic Acid and Octylpropionate as the heavy product from the raw material which is Acrylic Acid so that the raw can be recycle back. In this design purposes, the light key and the heavy key were determined. The light key in this column is Acrylic Acid whilst the heavy key in the column is Diacrylic Acid which is the byproduct. 5-29 Table 5.14: Boiling Point of Each Component Component Boiling Point Acrylic Acid, AA 141 Acetic Acid, Ac 117 Propionic Acid, P 141 Diacrylic acid, DAA 205 2-Ethylhexanol, EHOL 184 2-Ethylhexyl Acrylate, EHA 216 Ethylhexyl Acetate, EHAc 199 Octylpropionate, EHOP 232 Octenes, Oc 122 Water, W 100 The characteristics required in the chosen types of distillation column are the separation objective satisfied in the column, the cost of construction and the design of the selected distillation column The design of a distillation column can be divided into the following steps: i) Specify the degree of separation required: set product specifications. ii) Select the operating conditions: batch or continuous: operating pressure. iii) Select contacting device: plates or packing. iv) Determine the stage and reflux requirement: the number of equilibrium stages. v) Column size: number of real stages and diameter. vi) Design the column internals: plates, distributors, packing supports. vii) Mechanical design: vessel and internal fitting. 5.3.2 Chemical Design Calculation The separation of liquid mixtures by distillation is depends on the differences in volatility between the components. This is known as continuous distillation. Vapour flows up the column and the liquid counter-currently down the column. The vapour and liquid are brought into contact on plates. Part of the condensate from the condenser is returned to the top of the column to provide liquid flow above the feed point (reflux), 5-30 and part of liquid from the base of the column is vaporized in the reboiler and returned to provide the vapor flow. The purpose of distillation column is to separate component mixture from the heavy product in the bottom stream. 5.3.2.1 Estimation of the Number of Stages To estimate number of stages, condenser and reboiler temperature, and procedures are required for calculating dew and bubble points. By definition, a saturated liquid is at its bubble point (any rise in temperature will cause a bubble vapor to form), and a saturated vapor is at its dew point (any drop in temperature will cause a drop of liquid to form). Dew point and bubble point can be calculated from knowledge of the vaporliquid equilibrium for the system. In term of equilibrium constants, the bubble point is defined by the equation: Bubble point = Dew point = yi = xi = k i xi = 1.0 yi / k i = 1.0 In order to calculate the bubble point and dew point temperature, Antoine equation had been used to find vapor pressure (P o) ?? ?? = ? + ? (? + ?) Table 5.15: COMPONENT (?. ?. ?) Constant for Antoine’s Equation A B C D Acrylic Acid, AA 39.8372 -6587.1 -3.2208 5.23 x 10-7 Acetic Acid, Ac 61.3409 -6768.88 -6.72663 4.8427 x 10-6 Propionic Acid, P 86.4985 -8721.62 -10.0255 5.7246 x 10-6 Diacrylic acid, DAA 75.3232 -9672.9 -8.18819 4.620 x 10-18 2-Ethylhexanol, EHOL 13.3461 -2773.45 -1.524 1.98 x 10-10 2-Ethylhexyl Acrylate, EHA 68.6752 -8572.08 -7.9506 5.5064 x 10-3 Ethylhexyl Acetate, EHAc 134.23 -11591 -17.441 1.0301 x 10-5 Octylpropionate, EHOP 83.585 -9809.7 -10.0089 6.3913 x 10-3 Octenes, Oc 79.896 -7273.7 -9.6982 7.0132 x 10-6 Water, W 65.9278 -7227.53 -7.17695 4.0313 x 10-6 5-31 Ki = (i) Po P (5.3.2) Dew Temperature (Top Column) ( xi = yi /k i = 1.0 ) By using trial and error method, the dew point temperature calculated is 351.67 K with operating pressure 4 kPa. COMPONENT yi = xd Pi (kPa) ki xi = yi/ki Acrylic Acid, AA 0.4250 3.041334 0.826759 0.514055 Acetic Acid, Ac 0.0001 3.730926 1.956473 5.11 x 10-5 Propionic Acid, P 0.0008 4.110384 1.027596 0.000779 Diacrylic acid, DAA 0.0029 3.853586 0.963397 0.00301 2-Ethylhexanol, EHOL 0.0456 1.691066 0.970852 0.046969 2-Ethylhexyl Acrylate, EHA 0.2453 3.778172 1.534781 0.159827 Octenes, Oc 0.0080 4.07118 1.748034 0.004577 Water, W 0.2723 3.805498 0.951374 0.286218 Σ xi Total (ii) Bubble Point (Bottom Column) ( yi = 1.0015 k i xi = 1.0 ) By using trial and error method, the dew point temperature calculated is 405.23 K with operating pressure 11 kPa. COMPONENT xi Pi (kPa) ki yi = ki xi Acrylic Acid, AA 0.1076 3.154945 0.78874 0.0308 Propionic Acid, P 0.0002 4.165577 1.04139 7.57 x 10-5 Diacrylic acid, DAA 0.0076 3.931056 0.98276 0.00271 2-Ethylhexanol, EHOL 0.0654 1.868122 1.82076 0.67940 2-Ethylhexyl Acrylate, EHA 0.8132 3.852236 0.96305 0.28479 Octenes, Oc 0.0007 4.11914 1.02978 0.00026 Water, W 0.0054 3.86641 0.96660 0.00189 Total Σ yi 1.000 5-32 5.3.2.2 Minimum Number of Stages Fenske (1932) equation can be used to calculate minimum number of stages required at total reflux. Bottom and top stream temperature is calculated by the use of Antoine equation and Raoult’s Law in order to get value of average relative volatility. Those equations are fairly applied for multicomponent systems. Minimum number of stages (Fenske) is given as: x log x LK HK Nm = D xHK xLK W log(αL,av ) (5.3.3) Where, xLK,D = mol fraction of light key in distillated xHK,D = mol fraction of heavy key in distillated xLK,W = mol fraction of light key in bottom stream xHK,D = mol fraction of heavy key in bottom stream αL,av = average relative volatility of light key Average Relative Volatility, αL,av 5.3.2.3 It is in need to identify the light and heavy key to determine average relative volatility. Acrylic acid is needed to recover much at distillate; therefore it is a light key. Average relative volatility, αL,av can be determined by knowing top and bottom column temperature. αL,av = αLD aLW (5.3.4) The complete composition of the inlet and outlet streams for the distillation column is shown in Table below. 5-33 Table 5.16: Composition of Outlet and Inlet Streams Feed (F) Distillate (D) Bottoms (W) Component xF x FF yD = xD yDD xW xWW Acrylic Acid 0.181263 20.3212 0.42500 11.0627 0.1075578 9.2585 Acetic Acid 0.000025 0.0028 0.00008 0.0022 0.0000070 0.0006 Propionic Acid 0.000311 0.0349 0.00076 0.0199 0.0001743 0.015 Diacrylic Acid 0.006532 0.7323 0.00289 0.0753 0.0076325 0.657 Ethylhexanol 0.060802 6.8165 0.04562 1.1876 0.0653920 5.6289 Ethylhexyl Acrylate 0.681335 76.3839 0.24533 6.3859 0.8131804 69.998 Ethylhexyl Acetate 0.000002 0.0002 0.00000 0 0.0000023 0.0002 Octylpropionate 0.000009 0.001 0.00000 0 0.0000116 0.001 Octene 0.002372 0.2659 0.00797 0.2074 0.0006796 0.0585 Water 0.067349 7.5504 0.27234 7.0889 0.0053625 0.4616 TOTAL 1 112.11 1 26.03 1 86.08 The relative volatility αi for each individual component in a multicomponent mixture can be defined in a manner similar to that binary mixture. By referring to HYSYS, the K values for both temperatures are taken and αi values are calculated using formula given below, αi = Ki (5.3.5) K Base There are involve ten components in the mixture, however we are considering only heavy and light key for the calculation of relative volatility. Diacrylic Acid is taken as the base and the heavy key while Acrylic Acid as the light key component. The distillate and bottom composition also K i and αi values for both temperatures are as follows: (i) At top temperature, 100.1 ºC ??? = ??? ? ? (???. ?℃) ?? Acrylic Acid (L) 0.42500 3.3851 1.2847 Diacrylic Acid (H) 0.00289 2.6349 1 Component 5-34 (ii) At bottom temperature, 132.4 ºC ??? ? ? (???. ?℃) ?? Acrylic Acid (L) 0.1076 6.197 2.1157 Diacrylic Acid (H) 0.0076 2.9291 1 Component Substitute all known values into the equation in order to find the minimum number of stages, αL,av = αL,av = 1.2847 2.1157 log Nm = αLD αLW = 1.649 0.4250 0.0076 × 0.00289 0.1076 log (1.649) ?? = ?. ?? ?? ≈ ? ??????????? ?????? 5.3.2.4 Minimum Reflux Rate Underwood equations, 1−q = αi xiF αi − θ (5.3.6) The K i values obtained from HYSYS and the calculated αi values are shown in table below. 5-35 Table 5.17: Composition, K values and Relative Volatility of Components Component ??? ??? ?? ?? ??? Acrylic Acid 0.181263 0.42500 4.7911 6.1266 9.2585 Acetic Acid 0.000025 0.00008 14.3653 18.3696 0.0006 Propionic Acid 0.000311 0.00076 6.615647 8.4598 0.015 Diacrylic Acid 0.006532 0.00289 0.782014 1.0000 0.657 Ethylhexanol 0.060802 0.04562 1.661265 2.1243 5.6289 Ethylhexyl Acrylate 0.681335 0.24533 0.593689 0.7592 69.998 Ethylhexyl Acetate 0.000002 0.00000 1.088463 1.3919 0.0002 Octylpropionate 0.000009 0.00000 0.303445 0.3880 0.001 Octene 0.002372 0.00797 12.49824 15.9821 0.0585 Water 0.067349 0.27234 28.75233 36.7670 0.4616 By substituting q = 1.0 for feed at the boiling point 6.1266 × 0.181263 1.0000 × 0.006532 2.1243 × 0.060802 + + 6.1266 − θ 1.0000 − θ 2.1243 − θ 0.7592 × 0.681335 15.9821 × 0.002372 36.7670 × 0.067349 + + + 0.7592 − θ 15.9821 − θ 36.7670 − θ 1.110526 0.006532 0.129162 0.51727 0.03791 2.476221 0= + + + + + 6.1266 − θ 1.0000 − θ 2.1243 − θ 0.7592 − θ 15.9821 − θ 36.7670 − θ 1−1= 0= Assume θ=1.50 for the first trial θ (Assumed) 1.110526 6.1266 − θ 0.006532 1.0000 − θ 0.129162 2.1243 − θ 0.51727 0.7592 − θ 0.03791 2.476221 15.9821 − θ 36.7670 − θ 1.50 0.240031 -0.01306 0.20689 -0.69826 0.002618 0.101024 -0.1608 1.55 0.242653 -0.01188 0.224903 -0.65411 0.002627 0.104391 -0.0914 1.60 0.245333 -0.01089 0.246351 -0.61521 0.002636 0.107758 -0.0240 1.62 0.246422 -0.01054 0.256121 -0.60092 0.00264 0.109105 0.002835 ∑ (Sum) The final value of θ =1.62 is substitute into Rm + 1 = αi xiD αi − θ (5.3.7) 5-36 Rm + 1 = 6.1266 × 0.425 1.0000 × 0.00289 2.1243 × 0.04562 0.7592 × 0.24533 + + + 6.1266 − 1.62 1.0000 − 1.62 2.1243 − 1.62 0.7592 − 1.62 15.9821 × 0.00797 36.7670 × 0.27234 + + 15.9821 − 1.62 36.7670 − 1.62 ? ? = ?. ???? 5.3.2.5 Operating Reflux Rate and Number of Theoretical Stages By using equation below, R = 1.5R m (5.3.8) R = 1.5 0.2755 = 0.413 R 0.413 = = 0.292 R + 1 0.413 + 1 Rm 0.2755 = = 0.22 R m + 1 0.2755 + 1 Refer to figure below, Figure 5.2: Erbar-Maddox Correlation between Reflux Ratio and Number of Stages (Rm based on Underwood method) (Source: Christie John Geankoplis. 2003. Transport Processes and Separation Process Principles. Fourth Edition) 5-37 From the figure, Nm = 0.33 N N= 5 = 15.63 0.32 N ≈ 16 theoretical stages – 1 reboiler N = 15 theoretical stages 5.3.2.6 Feed Point Location By using emperical equation given by Kirkbride (1994), log Nr Ns = 0.206 log B D xf,HK xf,LK xb,LK xd,HK 2 (5.3.9) Where, Nr = number of stages above the feed, including any partial condenser Ns = number of stages below the feed, including the reboiler B = molar flow bottom product D = molar flow top product Xf, HK = concentration of the heavy key in the feed Xf, LK = concentration of the light key in the feed Xd, HK = concentration of the heavy key in the top product Xb, LK = concentration of the light key in the bottom product Nr log Ns = 0.206 log 73.729 24.324 0.0078 0.10729 Nr = 1.029 Ns Number of stages excluding the reboiler is 15. 0.0457 0.0054 2 5-38 Nr + Ns = 15 Ns = 10 − Nr Ns = 15 − 1.029 Ns Ns = 7.39 ≈ 7 The feed-point location is at the 7th stage from the bottom. 5.3.2.7 (i) Approximate Column Sizing Physical Properties The properties consider in this design are liquid flow rate, LW, vapor flow rate, VW , liquid density, ρL and vapor density, ρv. This physical properties evaluated at the system temperature by using HYSYS generated data or by manual calculations the from mass and energy balance data. The useful properties data are from HYSYS, mass and energy balance data is given as below: Liquid flow rate, LW = 14410 kg/hr (4.003 kg/s) Vapor flow rate, VW = 2292 kg/hr (0.6367 kg/s) Liquid density, ρL = 801.7 kg/m3 Vapor density, ρv = 0.1167 kg/m3 Data evaluated are at system temperatures and pressures. (ii) Plate Spacing The overall height of column will depend on the plate spacing. Plate spacing ranging from 0.5 to 1.0 m is normally used. Its spacing depends on column diameter and operating condition. In this multicomponent distillation column, the plate spacing was assumed to be 0.5 m as it is in the range of 0.5 m to 1.0 m recommended by Coulson and Richardson’s, Chemical Engineering, Volume 6. 5-39 (iii) Column Diameter Value of column diameter is majorly depending on the vapor flow rate. The vapor velocity must be below that which would cause excessive liquid entrainment or highpressure drop. The equation given below, which is based on the well-known Souders and Brown equation, Lowenstein (1961), can be used to estimate the maximum allowable superficial vapor velocity, uv ρ − ρv = ( −0.171 l2t + 0.27 lt − 0.047 ) L ρv 1 2 (5.3.10) Where, uv = maximum allowable vapor velocity, based on the gross (total) column cross-sectional area, m/s lt plate spacing (range 0.5 – 1.0) = Therefore, uv = (−0.171 0.52 + 0.27 (0.5) − 0.047 ) 801.7 − 0.1167 0.1167 1/2 uv = 3.75 m/s Hence, the column diameter can be calculated by using this equation: Dc = 4 Vw π ρv uv (5.3.11) Based on the equation above, V w is the maximum vapor rate in unit of kg/s. Therefore, column diameter with maximum vapor rate of 0.6367 kg/s is: Dc = 4 (0.6367) π 0.1167 (3.75) Dc = 1.36 m 5-40 For safety consideration, the approximate diameter was increased 50% more than the calculated value, as it deals with vapor, which is in high pressure. Dc = 1.5 1.36 = 2.04 m (iv) Plate Contactors Sieve plate is the simplest type of cross flow plate. The vapor passes up through perforating on the plates, and the liquid is retained on the plate by the vapor flow. The perforating are usually small holes, but larger holes and slots are used. The sieve plate is used because of its cheapest price. (v) Liquid Flow Arrangement Before deciding liquid flow arrangement, maximum volumetric liquid rate were determined by using equation below, VL = LW ρL (5.3.12) VL = 4.003 kg/s = 4.993 × 10−3 m3 /s 801.7 kg/m3 Figure 5.3: Selection of liquid-flow arrangement (Source: R. K. Sinnott, John Metcalfe Coulson, John Francis Richardson, Chemical Engineering Design. Volume 6.Third edition) 5-41 Based on the values of volumetric flow rate and column diameter, D c, it is shown that the liquid flow type that can be considered as a single pass cross flow. (vi) Plate Layout The value of downcomer area, active area, hole area, hole size, and weir height were determined based on above value calculated, trial plate layout column area determined by using the equation below, Total column cross − sectional area, Ac = Ac = π D2 4 (5.3.13) π (2.042 ) 4 A c = 3.27 m2 Downcomer area was found by assuming 15 % of column area and using equation below, Downcomer Area, Ad = 0.15 Ac = 0.15(3.27 m2) = 0.4905 m2 Net area and active area were determined by using equations below, Net Area, An Active area, Aa = Ac – Ad = 3.27 – 0.4905 = 2.78 m2 = Ac – 2Ad = 3.27 – 2(0.4905) = 2.29 m2 Hole Area, AH are determine with trial value of 10% active area by equation below, Hole Area, Ah = 0.10(Aa) = 0.10(2.29) = 0.229 m2 5-42 The height of the weir determines the volume of liquid on the plate and it is important factor in determining the plate efficiency. A high weir will increase the plate efficiency but at the expense of a higher plate pressure drop. Weir Length, l w was calculated by referring Figure 3.3 as shown below from Coulson and Richardson’s, Chemical Engineering, Volume 6, page 526 which was determined based on the value of the ratio of Ad/Ac to get the ratio of lw/ Dc. The weir height determined and other dimensions are as below: Weir Height, hw = 50 mm (standard) Hole diameter, dh = 5 mm (standard) Plate Thickness = 5 mm (standard) Figure 5.4: Relationship Between Downcomer Area and Weir Length (Source: R. K. Sinnott, John Metcalfe Coulson, John Francis Richardson, Chemical Engineering Design. Volume 6.Third edition) 5-43 Ad 0.4905 = × 100% = 15% Ac 3.27 From the figure above, when Ad/Ac is equal to 15%, the value of lw/Dc is 0.81. Therefore, the weir length, lw, is: lw = 0.81 2.04 m = 1.65 m The flooding condition fixes the upper limit of vapor velocity. A high vapor velocity is needed for high plate efficiencies, and the velocity will normally be between 70 to 90 percent of that which could cause flooding. For design, a value of 80 to 85 percent of the flooding velocity should be used. The flooding velocity can be estimated from the correlation given by Fair (1961): ρL − ρv ρv uf = K1 (5.3.14) Where, uf = flooding vapor velocity (m/s) based on net column cross-sectional area, An K1 = a constant obtained from Figure 8.4 The liquid-vapor flow factor FLV in Figure 5 is given by: FLV = Lw Vw ρv ρL (5.3.15) Where, Lw = liquid mass flowrate (kg/s) Vw = vapor mass flowrate (kg/s) FLV = (4.003) 0.1167 = 7.585 × 10−2 (0.6367) 801.7 5-44 At FLV = 7.585 × 10-2 and plate spacing = 0.5 m, the K 1 value is been determined as 8.9 x 10-2 or 0.089. Figure 5.5: Flooding Velocity, Sieve Plate (Source: R. K. Sinnott, John Metcalfe Coulson, John Francis Richardson, Chemical Engineering Design. Volume 6.Third edition) Substitute K1 value into the correlation given by Fair (1961): uf = 0.089 801.7 − 0.1167 = 7.376 m/s 0.1167 5-45 (vii) Entrainment Evaluation Entrainment can be estimated from the correlation given by Fair (1961), as shown in Figure 3.5 below, from Coulson and Richardson’s, Chemical Engineering, Volume 6, page 523, which gives the fractional entrainment ψ (kg/kg gross liquid flow) as a function of the liquid-vapor factor, FLV, with the percentage approach to flooding as a parameter. Actual % of flooding = uv × 100% uf Actual % of flooding = (5.3.16) 3.75 × 100% = 50.84% 7.376 Figure 5.6: Entrainment Correlation for Sieve Plates (Fair, 1961) (Source: R. K. Sinnott, John Metcalfe Coulson, John Francis Richardson, Chemical Engineering Design. Volume 6.Third edition) 5-46 From the figure above, value of fractional entrainment, Ψ is 0.013 and it is below the initial guess of 0.1. Thus this entrainment is acceptable. (viii) Number of Holes With dh (hole size) is equal with 5mm (preferred size), π d2h 4 Area of one hole, AH = AH = (5.3.17) π (5 × 10−3 )2 = 1.96 × 10−5 m2 4 Therefore, number of holes can be calculated using equation below: Number of holes = Number of holes = Hole Area, Ah Area of one hole, AH 0.229 m2 = 11684 units 1.96 × 10−5 m2 (ix) Column Size The column height will be calculated based on the equation given below. The equation determines the height of the column without taking the skirt or any support into consideration. Its determination is based on condition in the column. Column Height = (num. of stages – 1)(tray spacing) + (tray spacing × 2) + (number of stages – 1) (plate thickness) = (15 – 1) (0.5) + (0.5 × 2) + (15 – 1)(0.005) = 8.07 m Including 10% of safety factor, = 1.1 (8.07) m = 8.877 m ≈ 8.9 m 5-47 The overall height from the calculation is 8.07 m, but in a real construction it will be added slightly about 10% for safety consideration because of vapor and liquid area at top and bottom column respectively. The space for vapor and liquid are required if uncertain condition occur in the column, such as over flooding, over vapor pressure or upset in reaction situation. However another 2m for clearance height was allowed therefore the total column height is 11m. 5.3.3 Summary of Distillation Column (T-102) Design Table 5.18: Summary of Chemical Design Spec Minimum Reflux Rate Reflux Rate Value 0.2755 4.13 Theoretical Stages 15 stages Feed Point Location 7th stage Plate Spacing 0.50 m Column Diameter, Dc 2.04 m Column Height, H 11.00 m Liquid Flow Arrangement Number of Holes (Tray) Single Pass Crossflow 11684 units 5-48 5.4 DESIGN OF FINISHING COLUMN (T-103) 5.4.1 Introduction This final distillation column is one of the important equipment in the production of 2Ethylhexyl Acrylate. It consist 2 feed streams and 2 outlet streams. The feed stream is coming from the first distillation column and cracking column to re-distillate the component. Since the heavy component have been remove at the first distillation column, only Acrylic Acid, Diacrylic Acid, 2-Hexanol, 2-EHA, Cis-3-Octene and Water being fed to this column. However, in this design, both feed stream have been combining into one feed stream only in order to make it easier to calculate. This column is installed to separate between the main products from other component based on their boiling points. The higher boiling point is less volatile component, therefore it being discharge at the bottom outlet as the liquid. Since the 2EHA is the less volatile component, it will be removing at the bottom product as the nearly pure product. The top component is recycling back to the process to avoid any wastage of the raw material. This column operates at pressure and temperature 2 bar and 117.4 oC, at top and 9 bar 30.63oC, at the bottom part. With boiling point of 2-EHA is about 216oC while water at 1000C, make differences that will allow the separation occur. The separation done with purity about 90%, hence there is still slightly number of Acrylic Acid, 2-EHA, Cis-3-Octene and Water in the main product. 5-49 T-103 6 Recycle Re-DC 10 Figure 5.7: Distillation Column (T-103) Assuming the feed streams is combined into one feed stream only. The conditions of each stream are as follow: Table 5.19: Condition and Composition of Each Stream Stream Feed 9 10 Vapor 0.51225 0 0 Temperature (K) 376.5 303.8 390.4 Pressure (atm) 0.06415 1.97E-02 8.88E-02 Molar Flow (Kgmole/hr) 111.45 28.55 82.9 Mass Flow (Kg/hr) 16612 2585 1.40E+04 Mass Molar Mass Molar Fraction Fraction Fraction Fraction Fraction Fraction Acrylic Acid 0.089 0.182 0.427 0.4274 4.17E-02 0.0979 Acetic Acid 0.000 0.000 8.20E-05 0.0001 2.00E-06 0 Propionic Acid 0.000 0.000 7.68E-04 0.0008 6.80E-05 0.0002 Diacrylic Acid 0.001 0.001 2.64E-03 0.0026 0.00E+00 0 2-Hexanol 0.053 0.061 4.84E-03 0.0484 5.00E-02 0.0655 2-EHA 0.847 0.685 2.58E-01 0.2579 9.07E-01 0.8326 2-Ethylhexyl Acetate 0.000 0.000 1.00E-06 0 2.00E-6 0 Composition Mass Molar 5-50 Octylpropionate 0.000 0.000 1.00E-06 0 0.00E+00 0 Cis-3-Octene 0.002 0.002 7.73E-03 0.0077 3.61E-04 0.0005 Water 0.009 0.068 2.55E-01 0.255 3.47E-04 0.0033 5.4.2 Determine Number of Plate Determine the number of plate is one of important tool for the distillation column design. Since this process is multicomponent process, the Fenske’s equation should be applied. This equation takes consideration of the average volatility which respect to the key component, heavy and light key. 5.4.2.1 Determination of Bubble Point and Dew Point By definition, a saturated liquid is at its bubble point (any rise in temperature will cause a bubble vapor to form), and a saturated vapor is at its dew point (any drop in temperature will cause a drop of liquid to form). Dew point and bubble point can be calculated from knowledge of the vapor-liquid equilibrium for the system. In term of equilibrium constants, the bubble point is defined by the equation: Bubble point: Σ yi = Σ yi / Ki = 1.0 (5.4.1) Dew point : Σ yi = Σ yi / Ki = 1.0 (5.4.2) Using Antoine equation to find vapor pressure (P 0) ln P kPa = A − Ki = B (T + C) (5.4.3) Pi P (5.4.4) Table 5.20: The Antoine Constant of Each Component Component A B C Acrylic Acid 39.8372 -6.59E+03 0 Acetic Acid 61.3409 -6.77E+03 0 Propionic Acid 86.4985 -8.72E+03 0 Diacrylic Acid 75.3232 -9.67E+03 0 2-Hexanol 13.3461 -2.77E+03 -130 5-51 2-EHA 68.6752 -8.57E+03 0 2-Ethyl Acetate 134.23 -1.16E+04 0 Octylpropionate 83.585 -9.81E+03 0 Cis-3-Octene 79.896 -7.27E+03 0 Water 6.59E+01 -7.23E+03 0 (i) Calculating Bubble Point at Feed Stream Temperature = 376.5K Operating pressure =6.5 kPa Table 5.21: The Compositions in Feed Stream Component yi=xd pi (kPa) ki yi/ki Acrylic Acid 0.182 38.305 5.893 0.031 Diacrylic Acid 0.001 52.465 8.071 0.000 2-Hexanol 0.061 24.597 3.784 0.016 2-EHA 0.685 46.362 7.133 0.096 Cis-3-Octene 0.002 42.690 6.568 0.000 Water 0.068 43.130 6.635 0.010 ∑yi/ki y i / ki 0.54 value close to 1.0, therefore this temperature is accepted. So, bubble temperature is 376.5 K. (ii) Calculating Dew Point at Top Stream Temperature = 303.8 K Operating pressure = 2 kPa Table 5.22: Component Acrylic Acid The Compositions in Top Stream yi=xd pi (kPa) ki yi/ki 0.4274 43.157 2.158E+01 1.981E-02 Diacrylic Acid 0.0026 60.360 3.018E+01 8.615E-05 2-Hexanol 0.0484 29.304 1.465E+01 3.303E-03 2-EHA 0.2579 53.116 2.656E+01 9.711E-03 5-52 Cis-3-Octene 0.0077 49.047 2.452E+01 3.140E-04 Water 0.255 2.453E+01 1.039E-02 49.064 ∑yi/ki k x i i 0.43 Value is close to 1.0, therefore this temperature is accepted. So, bubble temperature is 303.8 K. (iii) Calculating Bubble Point at Bottom Stream Temperature = 390.4 K Operating pressure = 9 kPa Table 5.23: Component Acrylic Acid The Compositions in Bottom Stream yi=xd pi (kPa) ki 0.0979 3.757E+01 4.175E+00 2.345E-02 Diacrylic Acid 0 5.126E+01 5.695E+00 0.000E+00 2-Hexanol 0.0655 2.400E+01 2.666E+00 2.457E-02 2-EHA 0.8326 4.534E+01 5.038E+00 1.653E-01 Cis-3-Octene 0.0005 4.173E+01 4.636E+00 1.078E-04 Water 0.0033 4.223E+01 4.692E+00 7.033E-04 ∑yi/ki k x i i yi/ki 0.45 Value is close to 1.0, therefore this temperature is accepted. So, bubble temperature is 390 K. 5.4.2.2 Calculate the Relative Volatility, α The equilibrium vaporization constant k is defined for a compound by k = yi /xi (5.4.5) where, yi = mole fraction of component i in vapour phase xi = mole fraction of component i in liquid phase 5-53 The relative volatility, α which is needed in the calculation, is defined as αi,j = ki/kj (5.4.6) where i and j represent the components to be separated. Ideal system, Raoult’s law, Pi = Pixi (5.4.7) The relative volatility of two components can be expressed as the ratio of their k value, α = KLK / KHK where,KLK KHK (5.4.8) = light components = heavy components Table 5.24: Relative Volatility of Each Component Top Bottom Average α α α Acrylic Acid 1.30E+01 9.91E+00 11.352 Diacrylic Acid 1.21E+00 2.15E+00 1.613 2-Hexanol 2.16E+00 2.54E+00 2.339 2-EHA 1.00E+00 1.00E+00 1.000 Cis-3-Octene 3.75E+01 2.09E+01 27.972 Water 7.38E+01 4.99E+01 60.652 Component 5.4.2.3 Calculate The Minimum Reflux Ratio, Rm Table 5.25: Summary of Data to Calculate the Minimum Reflux Ratio α fi di Acrylic Acid 11.35 20.32 12.20 8.12 Diacrylic Acid 1.61 0.08 0.07 0.00 2-Hexanol 2.34 6.82 1.38 5.43 2-EHA (HK) 1.00 76.38 7.36 69.02 Cis-3-Octene 27.97 0.27 0.22 0.04 Water (LK) 60.65 7.55 7.28 0.27 111.41 28.52 82.88 Component bi 5-54 From Colburn (1941) and Underwood (1948), as stated in Coulson Richardson (page 675) αi xid = Rm + 1 αi − θ (5.4.9) Where, Rm = minimum reflux ratio Xi,d = concentration of component i in the tops at minimum reflux Θ = the root of the equation; αi xi,f = 1−q αi − θ (5.4.10) With xi,f = the concentration of component I in the feed, and q depends on the condition of the feed. As the feed is at its boiling point, q =1 αi xi,f = 1−q= 0 αi − θ Using try and error method, adjusting the θ value; Table 5.26: Summary of Calculation of the Θ, Using Try and Error Method Try xi,f α α.xi,f θ=2 θ =3.77 Acrylic Acid 0.182 11.352 2.070 0.22 0.273 Diacrylic Acid 0.001 1.613 0.001 0.00 -0.001 2-Hexanol 0.061 2.339 0.143 0.42 -0.100 2-EHA 0.685 1.000 0.685 -0.69 -0.247 Cis-3-Octene 0.002 27.972 0.067 0.00 0.003 Water 0.068 60.652 4.108 0.07 0.072 Component 0.000 Σ =7.075 0.03 (close enough) 5-55 Using the value of θ, from previous trial and error method, to find the R m value; Table 5.27: Summary of Calculation the Rm Value xi,f α α.xi,f α.xi,f /(αi – θ) acrylic acid 0.4274 11.352 4.852 0.640 DMMaleate 0.0026 1.613 0.004 -0.002 2-Hexanol 0.0484 2.339 0.113 -0.079 2-EHA 0.2579 1.000 0.258 -0.093 cis-3-octene 0.0077 27.972 0.215 0.009 water 0.255 60.652 15.466 0.272 Component 0.747 Hence, the minimum reflux ratio Rm + 1 = 0.747 Rm = -0.253 Rm −0.253 = = −0.4 Rm + 1 0.747 Neglecting the negative sign give Rm = 0.253 5.4.2.4 Determining the Minimum Stages Since R = (1.5) Rm = (1.5) (-0.4) = -0.38 R = −0.27 R+1 From Erbar – Maddox correlation: Nm N = 0.4 (5.4.11) 5-56 Minimum number of stages can be calculated using Fenske Equation: Nm = log 7.28 69.02 7.36 0.27 log 0.93 = 8.41 (5.4.12) Therefore, N = Nm /(0.4) N = 8.41 (0.4) N = 21 Number of stages including reboiler is 21 stages. 5.4.2.5 Determining The Feed Location An estimate can be made by using the Fenske equation to calculate the number of stages in the rectifying and stripping section separately, but this requires an estimate of the feed-point temperature. An alternative approach is to use the empirical equation given by Kirkbride (1944): Nr N Log s = 0.206 log B x f , HK D x f , LK xb , LK x d , HK 2 (5.4.13) Where, Nr = number of stages above the feed, including any partial condenser Ns = number of stages below the feed, including the reboiler x f , HK = concentration of the heavy key in the feed x f , LK = concentration of the light key in the feed xd , HK = concentration of the heavy key in the top product xb , LK = concentration of the light key if in the bottom product B = molar flow of bottom product D = molar flow of top product 5-57 Xb,LK = 0.27/82.88 = 3.3 x 10-3 Xd,HK = 7.36/28.52 = 0.26 ?? log = 0.206 log ?? 82.90 28.55 0.685 0.068 3.3 × 10−3 0.26 2 Nr N s = 0.33 Where actual number of stages is 20 (excluding reboiler), Nr Ns = 20 0.33Ns + Ns = 20 Ns = 15 Therefore, the feed is at stages 15 from the bottom column (excluding reboiler). 5.4.2.6 Pressure Drop Pressure at top column = 2 kPa Pressure at bottom column = 9 kPa Total pressure drop in the column is = (9 - 2) kPa = 7 kPa 5.4.2.7 Calculation for Density and Relative Molar Mass (RMM) Table 5.28: Molecular Weight, Density and Composition of Each Component in Each Stream Feed, mol Top product, Bottom Product, fraction Mol frac. Mol fraction 1059.33 0.182 0.4274 0.0979 144.13 1158.13 0.001 0.0026 0 2-Hexanol 130.23 836.76 0.061 0.0484 0.0655 2-EHA 184.28 888.71 0.685 0.2579 0.8326 112.21 724.98 0.002 0.0077 0.0005 18.02 997.99 0.068 0.255 0.0033 MW Density acrylic acid 72.06 DMMaleate cis-3octene Water 5-58 (i) Calculation for Relative Molar Mass (RMM) RMM = Σ (Component mole fraction x Molecular weight ) (5.4.14) RMM of Feed = 0.182(72.06) + 0.001 (144.13) + 0.061(130.23) + 0.685(184.28) + 0.002(112.21) + 0.068(18.02) =148.88 kg / kmol RMM of Top Product = 0.427(72.06) + 0.0026 (144.13) + 0.0484(130.23) + 0.2579(184.28) + 0.0077(112.21) + 0.255(18.0) = 90.43 kg / kmol RMM of Bottom Product = 0.0979(72.06) + 0.0655(130.23) + 0.8326(184.28) + 0.0005(112.21) + 0.0033(18.02) = 169.13 kg / kmol (ii) Calculation for Density Liquid Density, ρL = Σ ( Component mole fraction x Density of component ) Vapor density, 1. V = MW TSTP POP x x TOP PSTP STP V (5.4.15) (5.4.16) Bottom product Liquid density, L = 0.0979(1059.33) + 0.0655(1158.13) + 0.8326(888.71) + 0.0005(724.98) + 0.0033(997.99) = 902.31 kg / m3 Vapor density, V = (184.28/22.4) kg/m3 (273 K / 390.4K) (0.09 bar / 1.0 bar) = 0.5176 kg / m3 5-59 2. Top product Liquid density, L = 0.4274(1059.33) + 0.0026(1158.13) + 0.2579(888.71) + 0.0077(724.98) + 0.255(997.99) = 1297.34 kg / m3 Vapor density, V = (184.28 / 22.4) kg/m3 (273 K / 303.8K) (0.02 bar / 1.0 bar) = 0.1479 kg / m3 5.4.2.8 Determination of Column Dimensions (i) Determination of Column Diameter The principal factor that determines the column diameter is the vapor flow rate. The equation given below, which is based on the well-known Souders and Brown equation, Lowenstein (1961), can be used to estimate the maximum allowable superficial vapor velocity, and hence the column area and diameter: u v (0.171l 2 t ( L V 0.27lt 0.047) V 1/ 2 (5.4.17) Where, uv = maximum allowable vapor velocity, based on the gross (total) Column cross-sectional area, m/s lt = plate spacing, (range 0.5 – 1.0) uv = −0.171(0.62 902.31 − 0.1479 + 0.27 0.6 − 0.047) 0.1479 1 2 Uv = 4.2018 m/s The column diameter can be calculated using equation: Dc = 4Vw v u v Where, Vw = the maximum vapor rate, kg/s (5.4.18) 5-60 Therefore, column diameter is: Dc = 4 × 0.7181 π 0.1479 4.2018 Dc = 1.2129 m ≈ 1.22 m (ii) Determination of Column Area Column area can be calculated using equation: Area = πr 2 π(1.222 ) = = 1.167 m2 ≈ 1.2 m2 4 4 (5.4.19) (iii) Flooding Velocity Flooding velocity can be estimated from the correlation given by Fair (1961): U f Ki L V V (5.4.20) Where, Uf = flooding vapor velocity, m/s, based on the net column cross-sectional area Ki = a constant obtain from figure 5.5 (page 5-44) The liquid-vapor flow factor, FLV is given by: FLV LW VW V L Where, LW = liquid mass flow rate, kg/s VW = vapor mass flow rate, kg/s FLV = 0.7181 0.1479 = 2.3608 3.8944 902.31 (5.4.21) 5-61 From the figure 5.5, value of Ki = 0.095 (with plate spacing estimate as 0.6 m) At the bottom; Uf = 0.095 At the Top; Uf = 0.095 902.31 − 0.5176 = 3.9653 m/s 0.5176 1297.34 − 0.1479 = 8.897 m/s 0.1479 For design, a value of 80 to 85 per cent of the flooding velocity should be used. Therefore; Bottom Uf =3.9653 m/s (0.85) = 3.3705 m/s Top Uf =8.897 m/s (0.85) = 7.562 m/s (iv) Maximum Volumetric Flow-rate Maximum volumetric flow-rate = Vm (RMM) ρl (5.4.22) Above the feed point, vapor flow rate will be: Vn D( R 1) (5.4.23) = 28.55 kmol/h (1 + 0.38) = 39.399 kmol/h Liquid down flow will be: Vn = Ln D Ln = Vn D = (39.399 – 28.55) kmol/h = 10.849 kmol/h (5.4.24) 5-62 Below the feed point, liquid flow rate will be: Lm = Ln F (5.4.25) = 10.849 + (111.45) kmol/h = 122.30 kmol/h Vapor flow rate will be: Vm = Lm W (5.4.26) = (122.30 – 82.9) kmol/h = 39.399 mol/h Therefore, maximum volumetric flow rate: 39.399(169.13) = 7.385 m3 /s 902.31 39.399(90.43) Top = = 2.746 m3 /s 1297.34 Bottom = volumetricflowrate Net area required = floodingvelocity (5.4.27) 7.385 = 2.191 m2 3.3705 2.746 Top = = 0.363 m2 7.562 Bottom = Liquid flow pattern, Lm ( RMM ) L Maximum volumetric liquid rate = (5.4.28) 3 Bottom = 122.30 169.13 m = 22.92 902.31 s The choice of plate type (reverse, single pass or multiple pass) will depend on the liquid flow rate and column diameter. An initial selection can be made using Figure 5.3 (page 5-40). From the figure, it show that cross-flow (single pass) plate can be used with the column diameter = 1.22 m and liquid flow rate = 23.92 m3/s. 5-63 (v) Provisional Plate Design Column diameter, DC = 1.22 m Column area, AC = 1.2 m2 Down comer area, Ad = 0.12 x 1.2 m2 (at 12%) = 0.144 m2 = Ac - Ad = 1.2 – 0.144 = 1.056 m2 = Ac - 2Ad = 1.2 – (2 x 0.144) = 0.912 m2 = 0.0912 m2 (take 10% of Aa as first trial) Net area, An Active area, Aa Hole area, Ah (vi) Weir Length Ad 0.144 × 100% = × 100% = 12% Ac 1.2 From figure 5.4 (page 5-42); lW = 0.76 x 1.2 m = 0.912 m (at 12%) Take weir height, hw = 12 mm (for vacuum operation) Hole diameter,dh = 6 mm Plate thickness = 3 mm (5.4.29) Area of 1 hole, Alh πdh 2 π(0.0062 ) = = = 2.8274 × 10−5 m2 4 4 (5.4.30) Number of holes per plate, Nh = Ah 0.0912 = = 3225.54 holes ≈ 3226 holes Alh 2.8274 × 10−5 Check weeping (Enough vapor to prevent liquid flow through hole) (5.4.31) 5-64 LW ( RMM ) Minimum liquid rate = 3600s = (5.4.32) 0.7181(169.13) = 0.0337 kg/s 3600 Minimum liquid rate = 0.7 x 0.0337 kg/s (at 70% turn-down ratio) = 0.0236 kg/s (vii) Weir Liquid Crest The height of the liquid crest over the weir can be estimated using the Francis weir formula. For a segmental down comer this can be written as: Lw 750 l (l w ) = how 2/3 (5.4.33) Where, lW = weir length Lw = liquid flow rate, kg/s L = liquid density how = 750 0.7181 902.31 × 0.912 2/3 = 6.8488 mm liquid At minimum rate, clear liquid depth, how hw = (6.8488 + 12) mm liquid = 18.8488 mm liquid From the figure 11.32 [weep-point correlation (Eduljee, 1959); sinnot 2003], When how hw =18.8488 K 2 = 27.91 (viii) Weep Point The lower limit of the operating range occurs when liquid leakage through the plate holes becomes excessive is known as weep point. The vapor velocity at the weep point 5-65 is the minimum value for stable operation. Minimum vapor velocity through the holes based on the holes area can be calculated using Eduljee (1959) equation: U h (min) K 2 0.9(25.4 d h ) ( v )1 / 2 (5.4.34) Where, Uh = minimum vapor velocity through the holes based on the holes area dh = hole diameter, mm K2 = a constant, dependent on the depth of clear liquid on the plate, Uh = 27.91 − 0.9(25.4 − 6) = 14.525 m/s 0.51761/2 Actual minimum vapor velocity, = minimum vapor rate 0.7 1.3874 m = = 10.6486 Ah 0.0912 s (5.4.35) So, minimum operating rate will be below the weep point. (ix) Plate Pressure Drop Dry plate drop; The pressure drop through the dry plate can be estimated using expression derived for flow through orifices: 2 U v hd 51 h Co L Where, Uh = Uh = velocity through the holes volumetric flowrate 2.191 = = 24.024 m/s hole area, Ah 0.0912 Co = orifice coefficient (can be find from figure 11.36) (5.4.36) 5-66 To find the percent perforated area (using figure 11.35); With lp = 3.0 lp Ah = 0.9 Ap dh = 0.9 2.5 6 2 (5.4.37) 2 = 0.1563 × 100% = 15.63% Plate thickness, 3 = = 0.5 Hole diameter, dh 6 (5.4.38) From figure 11.36, the orifice coefficient, Co is 0.7778 Therefore, dry plate drop: hd = 51 24.024 0.7778 2 0.1479 = 5.547 mm liquid 1297.34 (x) Residual Head Residual head can be calculating using Hunt et al (1955) equation: hr = 12.5 × 103 12.5 × 103 = 13.85 mm liquid ρl 902.31 (5.4.39) (xi) Total Drop The total plate drop is: ht hd (hw how ) hr ℎ? = 5.547 + 6.8488 + 12 + 13.85 = 38.246 ?? ?????? (xii) Down Comer Liquid Back-up Down comer pressure loss: hap = hw − 10mm = 6.8488 − 5 = 1.8488 Where, hap is the height of the bottom edge of the apron above the plate. Area under apron, (5.4.40) 5-67 Aap = hap lw = 1.8488 × 0.912 = 1.6861 m2 Where Aap (5.4.41) is the clearance area under downcomer. As this is less than Ad = 1.306 m2, thus equation can be used to calculate the head loss in down comer: L hdc 166 wd L Am 2 (5.4.42) Where, hdc = head loss in down comer, mm Lwd = liquid flow rate in down, kg/s Am = either the down comer area downcomer hdc = 166 0.7181 902.31 × 1.6861 Aap Ad or the clearance area under the ; whichever is the smaller 2 Back-up in the down comer, ( = 0.00004 mm hb ), hbc hw how ht hdc (5.4.43) hdc = (6.8488 + 12 + 38.246 + 0.00004) hdc = 57.095 mm liquid or 0.057 m Check, 1 0.118 m < 2 (plate spacing + weir height) 0.061 < 1 0.6 + 0.012 ? = 0.612? 2 So, tray spacing is acceptable, (to avoid flooding) 5-68 5.4.2.9 Check Entrainment Entrainment can be estimated from the correlation given by Fair (1961). The percentage flooding is given by: % flooding = Un actual velocity (based on net area) (5.4.44) Uf un = uf 3.3705 1.3873m = = An 2.4295 s % flooding = 5.4.2.10 (5.4.45) 1.3873 × 100% = 41.16% 3.3705 Trial Layout Use cartridge type construction. Allow 50 mm unperforated 50 mm wide calming zone. 0.912 mm θθ Figure 5.8: Trial layout of plate 50mm Perforated area From figure 11.34 (Sinnott, 2000) 1.22mm 5-69 ?? 0.912 = = 0.748 ?? 1.22 Therefore, (5.4.46) c = 1000 Angle subtended at plate edge by unperforated strips, = (180 –100) = 800 Mean length, unperforated edge strips, = (Dc –0.05) x x θ/100 (5.4.47) = (1.22– 50 x10-3) m x x 80/100 = 3.054 m Area of unperforated edge strips, = (50 x 10-3)m x (3.054 m) = 0.1527 m2 Mean length of calming zone, approx. = 0.912 + (50 x 10-3) =0.962 m Area of calming zones, = 2 (0.962 x (50 x 10-3)) = 0.0962 m2 Total area available for perforation, A p : = Active area – (Area of unperforated edge + Area of calming zones) 2 = 0.912 m – (0.1527 + 0.0962) m (5.4.48) 2 = 0.6631 m2 Ah 0.0912 = = 0.1375 Ap 0.6631 From figure 2.9 (Sinnott, 2000): lp d h = 2.12 (satisfactory, range normally within 2.5 to 4.0) (5.4.49) 5-70 5.4.2.11 Column Size The column height is calculated using equation below, without any consideration of the any skirt or support need to attach to it. Column Height = (no of stage – 1) (tray spacing) + (tray spacing x 2) + (no of stage – 1) (thickness of plate) = (21 – 1)(0.612) + (0.612x 2) + (21-1)(0.003) = 13.524 m 5.4.2.12 Plate Specification Plate no. = 1 (from bottom column) Plate ID = 1.22 m Hole size diameter = 6 mm Hole pitch = 12.5 Δ Active holes = 3226 Turn down ratio = 70% at max.Liquid Plate Material = stainless Downcomer material = stainless Plate spacing = 0.6 m Plate thickness = 3 mm Plate pressure drop = 42.4721 mm liquid (5.4.50) 5-71 5.4.3 Summary of Chemical Design Table 5.29: Item Summary of Chemical Design Specification Symbol Value Unit Operating Pressure P 9 kPa Operating Temperature T 117.2 o Number of plate N 21 Trays Column Design Type of tray C Sieve tray Reflux ratio Rm 0.253 Feed tray position Ns1 15 Flow pattern Reverse flow Column Dimensions Internal Column Diameter Dc Tray spacing 1.22 m 0.612 m Column internal area Ac 1.2 m2 Column height hc 13.524 m Net area An 2.191 m2 Donwcomer area Ad 0.144 m2 active area Aa 0.912 m2 weir length lw 0.912 m Weir height hw 12 m Plate dimensions Material of plate Diameter Stainless steel Dp Plate thickness 1.22 m 3 mm Hole diameter Dh 6 mm Holes area Ah 0.0912 m2 Area of 1 hole A1h 2.8274 x 10-5 m2 Number of hole 3226 5-72 5.5 DESIGN OF HEAT EXCHANGER (E-104) 5.5.1 Introduction Heat exchanger is specifically to denote equipment in which heat is exchanged between two process streams. This section is discussed to select and design a suitable heat exchanger after the refining stage (T-103). The objective of this heat exchanger is to reduce high temperature from heavy removal column at 117.2 oC to 20oC, so it can achieve the product temperature in demanding. This section contains the operating criteria, the equipment selection, and the result of the thermal design. 5.5.2 Selection of Equipment There are many types of heat exchanger. Basically, the most popular types of heat exchanger used are: i) Shell and tube heat exchanger ii) Air cooled heat exchanger iii) Plate fin heat exchanger Shell and tube heat exchanger type is selected because it is most widely used and can be designed for virtually any application. Furthermore, it is relatively cheap as compared to the air-cooled and plate fin heat exchanger and it is sufficient for this application. In addition, this type of heat exchanger provides fairly large ratios of heat transfer area to volume and weight, easily cleaned and tubes can be replaced and well-established design procedures. 5.5.3 Selection of Shell-Tube Type of Heat Exchanger Referring to the Tubular Exchanger Manufacture Association (TEMA) classification, the heat exchangers are as follows: 5-73 Table 5.30: Selections on types of shell Type Reason Front and stationary head Type A: Channel and Removal cover without Types removal cover break the flanges Shell types Type E: one pass shell The most commonly used Type M: Floating head with Used extensively in backing devices petroleum service Rear ends, head types Internal Floating Head Exchanger is selected due to following advantages: i) Allowing differential movement between shell and tube and complete tube bundle withdrawal. ii) Separate the shell and tubes side fluid at the floating head end. iii) Access to the tubes and at the stationary end is obtained by removing the stationary head cover or complete head. iv) The inside of the tubes may be cleaned in situ and complete bundle may remove for cleaning the outside of the tubes or repairs. v) The splits-backing ring floating head type holds a smaller number of tubes than fixed tube sheet and U-tubes types having the same shell diameter. 5.5.4 Selection of Shell or Tube side for the Fluids Fluid in tubes side: Chilled Water (cold fluid) Fluid in shell: 2-Ethylhexyl Acrylate It is advisable that the cold fluid (water) flows in the tube and the hot fluid (2-Ethylhexyl Acrylate) flows in the opposite directions in the annular space between the two tubes. With this arrangement, the outer surface of the equipment will be at the lowest possible temperature and therefore, heat loss to surrounding is minimal. The water should be in the tube side rather than shell side because water is very corrosive material compared to 2-Ethylhexyl Acrylate. 5-74 5.5.5 Selection of Tubes The following tube dimension has been selected as in table below. Table 5.31: Selection of Tube Selected tube dimensions Reasons It provides an adequate heat transfer Tube length selected: 4.50 m surface area and reduce the shell diameter. Hence, lower the cost exchanger. Outer diameter selected, do: 22.23 mm Inner diameter selected, di : 19.74mm BWG number It is common and standard tube used : 18 (Source: Christie John Geankoplis. 2003. Transport Processes and Separation Process Principles. Fourth Edition) Triangular tube arrangement was selected because it will give higher transfer rate. 5.5.6 Assumption for Heat Exchanger Calculation Counter flow arrangement is selected. For LMTD involved, the following assumptions must be followed: i) Overall heat transfer coefficient is constant ii) Constant heat capacity iii) Isothermal phase change iv) Adiabatic operation Following assumptions are made in the derivation of the temperature correction factor Ft. i) Equal heat transfer areas in each pass ii) A constant overall heat-transfer coefficient in each pass iii) The temperature of the shell-side fluid in any pass is constant across any cross section iv) There is no leakage of fluid between shell passes 5-75 5.5.7 Calculation of Heat Exchanger The design is mainly to calculate the heat transfer coefficient of shell and tube side and to calculate the pressure drop for both sides. Table 5.32: Item Shell and Tube Process Condition Unit Shell-Side (Hot) Tube-side (Cool) Hydrocarbon Mixture Water Temperature range o 117.2-20 10-20 Mean Temperature o C 68.6 15 kJ/kg.oC 2.3357 4.1950 Ns/m2 0.00043 0.00131 0.1050 0.5815 Kg/m 802.2 999.73 Kg/h 14020.00 75784.71 Specific heat Dynamic viscosity Thermal Conductivity Density Flow rate C o W/m. C 3 5.5.7.1 Overall Coefficient, U From R.K.Sinnott, Chemical Engineering Design. Volume 6, (Refer Appendix A11) For shell and tube heat exchanger, the overall coefficient will be in range 250-750 W/m.oC. So, the estimation overall coefficient is 600 W/m.oC. It is still within the range. 5.5.7.2 Exchanger Type and Dimensions Assume: one shell pass and 2 tube passes T1 t2 T2 t1 T1 = 117.2 oC t1 = 10 oC T2 = 20 oC t2 = 20 oC 5-76 The logarithmic mean temperature, ∆ ??? : ∆Tlm = T 1 −t 2 −(T 2 −t 1 ) (5.5.1) T −t ln T 1 −t 2 2 1 Where, T1 = Inlet shell side fluid temperature, oC T2 = Outlet shell side fluid temperature, oC t1 = Inlet tube side fluid temperature, oC t2 = Inlet tube side fluid temperature, oC ∆Tlm = 117.2 − 20 − 20 − 10 117.2 − 20 ln 20 − 10 = ??. ?? oC The usual practice in the design of shell and tube heat exchanger is to estimate the ‘true temperature differences’ from the logarithmic mean temperature by applying a correction factor to allow for true counter-current flow. True temperature difference, ∆Tm = ∆Tlm × Ft Where, (5.5.2) Ft = Temperature Correction Factor To find value of Ft , two dimensionless temperature ratios is calculated: R = T1 − T2 t 2 − t1 R = 117.2 − 20 (5.5.3) 20 − 10 = 9.72 S = t 2 − t1 S = 20 − 10 T1 − T2 (5.5.4) 117.2 − 20 = 0.0932 From R.K.Sinnott, Chemical Engineering Design. Volume 6, (Refer Appendix A12): Ft = 0.94 Hence, ∆Tm = 38.34 × 0.94 = ??. ?? ?? 5-77 5.5.7.3 Heat Load in Shell and Tube Side Shell side: Q = ws × Cps × t 2 − t 1 (5.5.5) Where, ws = Flow rate of shell-side, kg/s Cps = Specific heat of shell-side, kJ/kgoC Q= 14020 × 2.336 × 20 − 10 3600 = ???. ???? ?? Tube side: wt = Q (5.5.6) Cp t × t 2 −t 1 Where, Cpt = Specific heat of tube-side, kJ/kgoC wt = 884.1549 4.2 × 20−10 = ??. ?? ??/? 5.5.7.4 Heat Transfer Area, Ao Heat transfer area, Ao = Q (5.5.7) U × ∆T m Where, Q = heat load, kW U = Overall coefficient, W/moC ∆Tm = True temperature difference, oC Ao = 884.1549 × 10 3 600 ×36.04 = ??. ?? ?? 5-78 5.5.7.5 Number of Tubes Area of one tube = π × do × L (5.5.8) Where, do = outside diameter of tube-side, m L = Length of tube, m Area of one tube = π × 22.23 × 10−3 × 4.50 = ?. ???? ?? Number of tubes = Ao Area of one tube (5.5.9) = 40.88 0.3143 = ??? So, for 2 passes, tube per pass = 128 2 = ?? 5.5.7.6 Bundle and Shell Diameter From R.K.Sinnott, Chemical Engineering Design. Volume 6, (Refer Appendix A13): Tube arrangement: Triangular pitch (for 2 tube passes) k1 = 0.249 n1 = 2.207 Bundle diameter, Db = do × number of tubes k1 = 22.23 × 65 0.249 1 n1 (5.5.10) 1 2.207 = ???. ???? ?? From Coulson & Richardson’s, Volume 6, Figure 12.10 (Refer Appendix A14): For a split-ring floating head exchanger, the typical shell clearance = 56 mm Shell Diameter, Ds = Db + bundle diametrical clearance = 378.7560 + 56 = ???. ???? ?? (5.5.11) 5-79 5.5.7.7 Tube-side Heat Transfer Coefficient Tube cross-sectional area, As = π 4 × d2i (5.5.12) Where, di = inside diameter of tube side, m Tube cross-sectional area, At = π 4 × 0.01974`2 = ?. ?????? ?? Total flow area = tube per pass × tube cross − sectional area (5.5.13) = 65 × 0.000306 = ?. ????? ?? Volumetric flow rate, V = wt ρt = 21.051 (5.5.14) 999.73 = 0.0211 m3 s Water linear velocity, uT = V At (5.5.15) = 0.0211 0.01989 = ?. ??? ? ? According to Coulson and Richardson, 1999, velocity for the process fluid is 1.0 - 2 m/s. Hence, velocity for the process fluid in this plant is acceptable. Using equation by Eagle and Ferguson (1930) from Coulson & Richardson’s, Volume 6: Tube-side heat transfer coefficient, hi = 4200 (1.35 +0.02t)u 0.8 t Where, ℎ? = inside coefficient for water, W/m2 oC t = water temperature, oC ut = water velocity, m/s di = tube inside diameter, mm d 2i (5.5.16) 5-80 hi = 4200 (1.35 +0.02(10))(1.06 0.8 ) 19.742 = ????. ???? ? ?? . ?? 5.5.7.8 Shell side Heat Transfer Coefficient Taking baffle spacing as 0.4 from the shell diameter Baffle spacing, lB = 0.4 × 434.7560 (5.5.17) = ???. ??? ?? For triangular arrangement, tube pitch, pt = 1.25 do Pt pt = 1.25 do = 1.25 22.23 = 27.7875 mm Cross flow Area, As = pt − do × Ds × lB pt (5.5.18) Where ?? = tube pitch, m ?? = Shell diameter, m ?? = Baffle spacing, m As = 0.02779 − 0.02223 × 0.434 × 173.902 0.02779 = 0.01512 m2 For an equilateral triangular pitch arrangement: Equivalent diameter, de = de = 1.10 0.02223 1.10 do (p2t − 0.917d2o ) 0.027792 − 0.917(0.022232 ) = ?. ????? (5.5.19) 5-81 Mass velocity, Gs = = ws 3600 × 14020 3600 1 As × 1 0.01512 = 257.5517 kg s. m2 Reynolds number, Res = Gs × de μs (5.5.20) Where, ρs = density of shell-side, kg/m3 Gs = mass velocity of shell-side, kg s. m2 de = equivalent diameter, mm μs = dynamic viscosity of shell-side, Ns/m2 Res = 257.5517 × 0.01578 0.00043 = 9476.2282 Prandtl number, Prs = Cps × 103 × μs ks (5.5.21) Where, Cps = specific heat of shell-side, kJ/kgºC μs = dynamic viscosity of shell-side, Ns/m2 ks = thermal conductivity of shell-side, W/mºC Prs = 2.3357 × 103 × 0.00043 0.105 = ?. ???? According to R.K.Sinnott, Chemical Engineering Design. Volume 6, baffle cut of 20 to 25% will be the optimum, which will give good heat-transfer rates without excessive drop. Hence, baffle cut with 25% was taken. From Coulson & Richardson’s, Volume 6, (Refer Appendix A15): Res = 9476.2282 → Heat transfer factor, jn = 0.0082 Shell-side heat transfer coefficient, hs = k s de × jh × Res × Prs 2 5-82 = 0.105 0.01578 × 0.0084 × 9476.2282 × 9.54302 = ????. ???? ? ?? ?? 5.5.7.9 Overall Coefficient, Uo From the table 12.2 in Coulson & Richardson’s, Volume 6 (Refer Appendix A16): : 6000 W/m2 ºC Fouling factors Coefficient; hid = Chilled Water : 5000 W/m2 ºC hod = Organic Solvent From table 12.6 in Coulson & Richardson’s, Volume 6 (Refer Appendix A17): Thermal conductivity of stainless steel, kw = 16 W/m ºC 1 Uo = 1 ho + 1 h od + d o ln do di + 2k w do di × 1 h id + do di × 1 hi (5.5.22) 0.02223 0.02 ln 0.01974 1 1 1 0.02223 1 0.02223 1 = + + + × + × Uo 1096.4070 5000 2 16 0.01974 6000 0.01974 3997.9467 1 = 0.001664 Uo Uo = ???. ??? ? ?? ?? Percentage of error = U o ,calc − U assumption = × 100% U assumption 600.974 − 600 600 × 100% = ?. ???? % < 30% Hence, Heat transfers Area: Acalc = = Q U × ∆T m 884.155× 10 3 600.974 × 36.043 = ??. ???? ?? 5-83 5.5.8 Pressure Drop 5.5.8.1 Tube side ∆Pt = Np 8jf L μ di μw + 2.5 ρu 2t (5.5.23) 2 Where, Np = number of tube passes jf = friction factor ρt = density of tube-side, kg/m3 L = length of one tube, mm ut = velocity of tube-side, m/s From R.K.Sinnott, Chemical Engineering Design. Volume 6, (Refer Appendix A18): Re =16831.53783→ Friction Factor, jf = ?. ?? ∆Pt = 2 8 0.01 4.5 0.01905 + 2.5 1.16 2 997.7 2 ∆Pt = ?????. ???? ? ?? = ?. ??? ??? = ??. ? ??? In accordance with Coulson’s and Richardson, 1999, Liquid : < 1 mN s m2 ± 35 kN m2 5.5.8.2 Shell Side ∆Ps = 8jf Ds L ρu 2s μ de lB 2 μw Where, L = tube length lB = baffle spacing −0.14 (5.5.24) 5-84 From R.K.Sinnott, Chemical Engineering Design. Volume 6, (Refer Appendix E9): Friction Factor, jf = ?. ??? Velocity in shell, us = = Gs ρ 257.5517 802.2 = 0.3211 m/s Refer to Coulson and Richardson,1999, velocity in the shell side need to be within 0.3 – 1.0 m/s. From the calculation of shell’s velocity, it is within the range. ∆Ps = 8 0.045 0.434 4.5 0.0142 0.1735 = ?????. ???? ? ?? = ?. ???? ??? = ??. ? ??? 802.2 0.3211 2 2 5-85 5.5.9 Summary of Chemical Design Table 5.33: Summary of Heat Exchanger Design Parameters Process condition: Heat load, Heat transfer Coeffiecent, Uo,calc 884.1548 kW 600.974 W m2 o C Tube side: Cooling Water Inlet temperature,t1 10 oC Outlet temperature, t2 20 oC Flow rate, wwater 75875.033 kg/h Outside diameter, do 22.23 mm Inside diameter,di 19.74 mm Length of tube, L 4.5 m Pitch, pt Number of tubes, Nt Passes, nt Pressure drop, ∆?? 27.7875 mm 128 2 31.9 kPa Shell side: Hydrocarbon mixture Inlet temperature, T1 Outlet temperature, T2 Flow rate, ws Shell inside diameter, Ds Passes,ns Pressure drop, ∆?? 117.2 20 14020 kg/h 434.7560 mm 65 35.4 kPa