CHAPTER 5CHEMICAL DESIGN 5.1 PIPE REACTOR 5.1.1 Introduction The reactor is the heart of most processes; developments in process technology often centre on improvements in the design and operation of the reactor. Subsequent stages in the chemical process are usually concerned with the separation of various chemicals from the desired product, followed by the final purification stages. The treatment of reactor design in this section will be restricted to a discussion of the selection of the appropriate reactor type for a particular process, and an outline of the steps to be followed in the design of a reactor. The design of an industrial chemical reactor must satisfy the following requirements: 1. The chemical factors: the kinetics of the reaction. The design must provide sufficient residence time for the desired reaction to proceed to the required degree of conversion. 2. The mass transfer factors: with heterogeneous reactions the reaction rate may be controlled by the rates of diffusion of the reacting species; rather than the chemical kinetics. 3. The heat transfer factors: the removal, or addition, of the heat of reaction. 4. The safety factors: the confinement of hazardous reactants and products, and the control of the reaction and the process conditions. The production of ammonium nitrate solution is carried out by a neutralization process between nitric acid and ammonia in a pipe reactor. The pipe reactor or turbular reactor is a length of pipe having a small diameter and containing no inside stuffing. 2 The concept of the plug flow reactor or pipe reactor denotes an ideal tubular reactor, in which all fluid elements travels in one direction with exactly the same speed. A PFR may be used for both liquid phase and gas phase reactions, for both laboratory-scale investigation of kinetics and a large-scale production. The reactor itself may consist of an empty tube or vessel, or it may contain packing or a fixed bed of particles. The output of a chemical reactor depends, among other things, on the residence time (the average length of time the material is in the reactor), the temperature, and the fluid dynamics. These conditions may make designing a single reactor to handle both the initial and final flow rates undesirable. However, sometimes this can be done by installing a vertical reactor and initially maintaining it only partially full. A pipe reactor is similar to a CSTR in being flow reactor, but is different in its mixing characteristics. It is different from batch reactor in being a flow reactor, but it similar in the progressive change of properties, with position replacing time. The flow through the vessel both input and output streams, is continuous but not necessarily at constant rate. There is no axial mixing of fluid inside the vessel. There is complete radial mixing of fluid inside the vessel; thus, the properties of the fluid, including its velocity, are uniform in this plane. This pipe reactor is designed to produce ammonium nitrate with a capacity of 100000 MT per year. The process is considered exothermic reaction. There are a few advantages by using this pipe reactor such as: a) Low investment cost with no preneutralizer and low recycle ratio. b) A lower operating cost with no low electric power consumption and a high efficiency of ammonia consumption. c) A high adaptability in the use of feed stocks- nitric acid from various origin and concentrations can be used. d) A higher operating flexibility and stability e) A low environment impact because of the low emission of gases. 3 5.1.2 CHEMICAL DESIGN FOR PIPE REACTOR 5.1.2.1 Chemical Reaction Below is showed the main reaction between ammonia and nitric acid occurring in the reactor to produce ammonium nitrate. N H 3 ( g ) + HN O3 ( aq ) → N H 4 N O3 (aq) ∆ H=−145 kJ ( gmole ) To ensure the validity of the reactions and equations used in designing the reactor. These assumptions that been made are: 1. The reaction is in steady state condition. 2. The reaction rate is valid for the above reaction 3. Irreversible reactions. 4. Density of the component remain constant throughout the reactor, ε = 0. 5. The temperature is assumed as constant (Isothermal reaction) 5.1.2.2 Effect of Temperature: Arrhenius Equation and Activation Energy The theory of the temperature effect on the reaction rate originated from the temperature effect on the equilibrium constant. It is known that: d ln K −H = Equation5.1 R 1 d T ( ) where K is an equilibrium constant, R is the gas constant H is the heat of reaction. Arrhenius, 1889 from Levenspiel (1999) proposed a similar expression for the rate constant k 4 d ln k A E A = Equation5.2 dT RT2 Where EA is a characteristic (molar) energy which is called as the energy of activation. Since (-rA) (hence kA) increases with increasing T almost every case, EA is a positive quantity. Integration of equation (5.2) on the assumption that EA is independent of T leads to or ln k A =ln A− k A = A exp EA Equation5.3 RT ( −ERT ) Equation5.4 A Where A is a constant referred to as the pre-exponential factor. Together, EA and A are called the Arrhenius parameters. Equations (5.2) to (5.4) are all forms of the Arrhenius equation. This equation is usefulness to represent experimental results for the dependence of kA on T. In this ammonium nitrate production, the equilibrium constant for the reaction can be calculated as shown below. Inlet temperature : 86.04 oC Outlet temperature : 180 oC Reaction temperature : 180 oC (453.15 K) From HYSYS, the value of EA is 3.2744 x 10-2 and value for A is 1.327 are inserted in the equation of 5.4. 5 k A = A exp ( −ERT ) A k A =1.327 exp ( −2 −3.2744 x 10 1.98588 x 453.15 ) k A =1.33 l mole . sec 5.1.2.3 Step Taken For Chemical Design of Reactor Below are the steps taken to determine chemical design specification for designing this pipe reactor: 1) Determination of specific reaction rate 2) Calculation of residence time by using Simpson’s Rule equation. 3) Determination of working volume of reaction and size of reactor 6 Start Determination of specific rate of reaction, interpretation of the rate Determination of residence time by using Simpson’s Rule Determination of volume of reactor End Figure 5.1: Algorithm for chemical design of reactor 5.1.2.4 Analysis of the Reactor Inlet and Effluent Analysis of the reactor inlet and effluent is important to determine the volume of the reactor. It also will give more of the properties needed for the reactor design, the mechanical strength needed for the reactor, also type of reactor suitable to be used. From the Design Project 1 (DP1), the material balance for the reactor as below: Table 5.1: Analysis data for reactor inlet and outlet Phase Temperature Pressure Stream In Stream Out Mixture 86.04 oC 600 kPa (Operating condition) Mixture 180 oC 540 kPa 7 Molar Flow Mass Flow Component 436 kgmole/h 1.418 x 104 kg/h 290.6kgmole/h 1.418 x 104 kg/h Molar Flowrate Molecular Mass Flowrate Mass (kgmole/h) Weight (kg/h) Fraction 222.11 145.45 17 63 3782.5 9165.4 0.2667 0.6263 145.45 80 11643 0.821 Properties Flowrate (kg/hr) Molar Flowrate Ammonia 3782.5 222.11 Nitric Acid 9165 145.45 Water 1233 68.448 (kmol/hr) Mass Fraction Density (kg/m3) 0.2667 616.07 0.6463 1523.64 0.0869 997.99 Reactant Ammonia Nitric Acid Product Ammonium Nitrate Total flowrate (kg/hr) = 3782.5 + 9165 + 1233 = 14180.5 kg/ hr Density mixture (kg/m3) = ∑ α i ρi Where α is mass fraction and ρ is density of each material Density mixture (kg/m3) = 1235.76 kg/m3 14180.5 Flowrate in m3/hr = ( hkgr ) kg 1235.76 3 m ( ) =11.48 m3/hr Initial Concentration for each component is: Nitric Acid : 12.67 kmol/m3 8 Ammonia, : 19.35 kmol/m3 Water : 5.96 kmol/m3 Ammonium nitrate : 0.000 kmol/m3 5.1.2.5 Determination Rate of Reaction The reactor is made of a main reaction, which is reaction to produce ammonium nitrate. The reactions are: N H 3 ( g ) + HN O3 ( aq ) → N H 4 N O3 (aq) From basic equation: A + B→ C The kinetic reaction of this reaction is: r c =k C nA C mB Equation 5.5 Where n and m are the order of kinetic reaction Equation 5.6 EA = Activation energy R = Ideal gas constant T = Temperature of the reaction Substituting 5.5 into 5.6 From the reaction between ammonia and nitric acid and produce ammonium nitrate: N H 3 ( g ) + HN O3 ( aq ) → N H 4 N O3 9 The reaction rate given by Gerhard Kramm et al., (1994) r=1.33 exp ( ) −3.2744 x 10−2 C 0.81 C 0.4 1.98588 x 406.17 HN O N H 0.4 r=1.33 C 0.81 HN O C N H 3 3 3 3 5.1.2.6 Interpretation of the Reaction Rates 0.4 r=1.33 C 0.81 HN O C N H 3 3 Equation 5.7 C HNO3 =C HNO3 O −C HNO3 O X HNO3 Equation 5.8 C NH3=C NH3 O−C HNO3 O X HNO3 Equation 5.9 Substituting equation 5.8 and 5.9 into 5.7, therefore: C HNO3 O −C HNO 3O X HNO 3 ¿ ¿ C NH3 O −C HNO3 O X HNO3 ¿ ¿ r =1.33 ¿ The reaction rate based on the kinetic experimental data under integral regime will be inserted to the performance equation under Levenspiel, 1999 pg. 111:- 10 1 −r A (¿) d X A XA τ =C A 0∫ ¿ 0 C HNO3 O −C HNO3O X HNO3 ¿ ¿ C NH3 O −C HNO3 O X HNO3 ¿ ¿ 1.33¿ 1 ¿ ¿ 0.99 τ=C HN O 3 ∫¿ 0 12.67−12.67 X HNO3 ¿ ¿ 19.35−12.67 X HNO3 ¿ ¿ 1.33 ¿ 12.67 ¿ ¿ 0.99 τ= ∫ ¿ 0 5.1.2.7 Calculation of Residence Time, of reactor The values of reaction rates and graph of reaction rate versus conversion below is calculated by using Microsoft excel. Table 5.2: Data for Calculated Residence Time, s XHNO3 0.00 0.05 0.10 CHNO3/-rNH4NO3 0.37239 0.39340 0.41671 XHNO3 0.55 0.60 0.65 CHNO3/-rNH4NO3 0.76513 0.85009 0.95504 11 0.44272 0.47194 0.50496 0.54258 0.58582 0.63600 0.69494 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.70 0.75 0.80 0.85 0.90 0.95 0.99 1.08799 1.26196 1.49973 1.84528 2.39683 3.43237 23.57769 CHNO3/rNH4NO3 versus Conversion, XHNO3 25.00000 20.00000 15.00000 CHNO3/rNH4NO3 10.00000 5.00000 0.00000 0 0.2 0.4 0.6 0.8 1 1.2 XHNO3 Figure 5.2: Graph of CHNO3/rNH4NO3 versus Conversion, XHNO To evaluate the area under graph, Simpson’s Rule is used. 0.99 τ =C HNO3 O ∫ ( rNH 41 NO 3 ) d X HNO3 h= 0 b−a n F=f o h = 3 [F+ L+ 4 R+2 E] 12 L=f n n−1 R= ∑ f i=f 1 +f 3+ …+f n−1 i=odd n−2 E= ∑ i =even f i=f 2 +f 4 +…+ f n−2 Let take n= 5 h= 0.99−0 5 ¿ 0.198 N X Fn 0 0 0.372395 1 0.198 0.4707 2 0.396 0.631692 3 0.594 0.941161 F=f o=0.372395 L=f n=23.57769 R=f 1 +f 3 = 0.470700 + 0.941161 = 1.411861 E=f 2 + f 4=0.631692+1.779774=2.411466 0.99 τ =C HNO3 O ∫ 0 (r 1 NH 4 NO 3 )d X HNO 3 4 0.792 1.779774 5 0.990 23.57769 13 0.198 [ 0.372395+23.57769+ 4 ( 1.411861 ) +2 ( 2.411466 ) ] 3 = ¿ 2.3 s 0.99 τ =C HNO3 O ∫ 0 (r 1 NH 4 NO 3 )d X HNO 3 =C HNO3 x 2.3 O τ =12.67 x 2.3 s ¿ 29.14 s 5.1.2.8 Calculation of the Volume, V of the reactor To find the volume of the pipe reactor, τ C HN O V= τ × F HN O C HN O = 3 V F HN O 3 3 3 V= 29.14 145.45 kmol hr × × 12.67 hr 3600 s V =0.093 m 3 Height and radius of the vessel Consider allowance for the reactor, therefore the volume for the pipe reactor is 0.0795 m3 .Volume of the cylinder, can be calculated from equation: 2 V =π r L Radius of the reactor to be taken as 0.2 m, therefore, the height of the reactor is, 14 0.093 L= 3 2 2 m =π × 0.065 m × L 0.093 m 3 2 2 π × 0.065 m L=7.007 m L=7 m Villard et. al state that the diameter of most the pipe flow reactor generally used are between 200 mm and 10 mm and a length ranging between 1.5 m and 10 m. the ratio of length/diameter is usually not below than 50. So the diameter and length from the calculation are valid. Figure 5.3: Illustrate diagram for reactor sizing 5.1.2.9 Flow In and Out of the Reactor Ammonia will be introduced into the pipe reactor along the horizontal lines. For nitric acid, the reactant will flow into the pipe reactor at the right angles or from above of the reactor. The product stream will flow directly into the separator for separation process. The temperature and pressure at the feed are at 86 oC and 600kPa respectively while for the outlet stream, the pressure and temperature are 180oC and 540kPa. 15 Nitric Acid Product to separator Ammonia Figure 5.4: Diagram for Pipe Reactor 5.2 GAS LIQUID SEPARATOR (V-101) 5.2.1 Chemical Design of Gas Liquid Separator The main purpose of this gas liquid separator vessel is to increase the concentration of ammonium nitrate into 94%. The separator also act as knockout drum were the pressure of the inlet stream is reduced from 6 bar to 5.4 bar. Vertical Separator Figure 5.5: Stream Involve in Vertical Separator 16 Bellow are the summary equation used for vertical separator design: Summary equation for vertical separator design: Vv'=vvA Equation 5.10 kv = 0. 1 ft/s (0.03045 m/s) — with no mist eliminator kv = 0.35 ft/s (0.107 m/s) — with a mist eliminator A = TI D2/4 LL A = VL' ts — where the minimum value of LL is 2 ft (0.610 m) 3 < ts < 5 min L = LL+1.5D+1.5ft or L = 8.5 ft (2.59 m) — whichever is larger If L/D < 3.0, then recalculate L so that L/D > 3.0 by letting L/D = 3.2. If L/D >5 use a horizontal separator. Figure 5.6: A vertical gas liquid separator - Splash Plate: prevent the corrosion of vessel wall opposite the feed point. Radial vane vortex breaker: ensure gases are not entrained with the exiting liquid. 17 Summary equation for horizontal separator design: Vv' = 0.5vvA Equation 5.11 kv = 0.10 ft/s (0.0305 m/s) — with no mist eliminator kv = 0.35 ft/s (0.107 m/s) — with a mist eliminator A = 7iD2/4 — minimumD = 5.5 ft(1.67m) 0.5LA = VL'ts 7.5 <ts< lOmin If L/D < 3.0, then recalculate L so that L/D > 3.0 by setting L/D = 3.2. If L/D > 5.0, then recalculate D so that L/D < 5.0 by setting L/D = 4.8 5.2.2 Diameter and Length of Gas Liquid Separator First, we assume to design a vertical gas liquid separator. This is because in this vessels requred less space for installation compare to the horizontal vessel and have high efficiency for separating vapor phase and liquid phase product by only utilize their specific gravity. Uvapor max = K[(rL -rv )/ rv ]0 5 where U = Velocity, ft/sec r = Density of liquid or vapor, lbs/ft3 K = System constant (WL/Wv)(rv/rL)0.5 where W = Liquid or vapor flow rate, lb/sec For a horizontal vessel KH = 1.25Kv. (WL/Wv)(rv/rL)0.5 = ( 3.3528kgs-1/0.585 kgs-1)/(2.7493/1225)0.5 = 0.273 18 Figure 5.7: Graph shown the value of Kv From graph, Kv = 0.3ft/s = 0.3ft/s.1m/3.28ft = 0.09 m/s Uvapor max = 0.09[(1225-2.7849)/2.7849]0.5 = 1.88543m/s Minimum Cross sectional area for vapor, Av = Vmix/Uv = 0.2106m3/s ÷ 1.88543m/s = 0.11147m2 Assume that vapor occupied 20% from the total area during full liquid volume Total cross sectional Area, Atotal = 0.11147m2 / 0.2 = 0.5573m2 Minimum internal diameter, Dmin = (4x 0.5573m2 /π)0.5 = 0.8426m Dmin in Standard Pipe size : 33.25in = 0.8447m Liquid volume requirement = 0.00274m3/ s x 5 min x 60 s = 0.82109m3 Liquid level holdup, LL = (VLts)/ Atotal = (0.00274m3 s-1.300s)/ 0.5573m2 = 1.475 m Vessel Height, L = LL+1.5D+1.5ft = LL+1.5D+0.457m = 1.475 m + 1.5(0.8445) + 0.457m = 3.197m Standard Height Design available: 10ft 6in. = 3.200m 19 L/D = 3.2/0.8445 = 3.7892 is in 3<L/D <5 Design is acceptable because the length and diameter ratio is acceptable. The length and diameter ratio is an indicator to sizing any vessel design in order to standardize with the standard vessel design. 5.3 HEAT EXCHANGER 5.3.1 Introduction Heat exchanger is a device to change the heat between two fluids that have different temperature. The two fluids did not mix each other because they were separated by a solid medium. Heat exchangers are widely used in engineering application like air conditioning, power production, waste heat recovery and chemical processing. 5.3.1.1 Flow arrangement Heat exchangers may be classified according to their flow arrangement. There are two main flow arrangements which are parallel-flow and counter-current flow. In counter flow heat exchangers the fluids enter the exchanger from opposite ends. This flow arrangement usually used for liquid-liquid, condensing and gas cooling applications. Units are usually mounted vertically when condensing vapour and mounted horizontally when handling high concentrations of solids. Figure 5.8: Counter-current flow arrangement Parallel flow also known as co-current flow. In parallel-flow heat exchangers, the two fluids enter the exchanger at the same end, and travel in parallel to one another to the 20 other side. Compared both flow arrangements, the counter current design is most efficient, in that it can transfer the most heat from the heat transfer medium. So that, we choose counter current flow as our heat exchanger flow arrangement for production of Ammonium Nitrate. Figure 5.9: Co-current or parallel flow arrangement 5.3.1.2 Types of Heat Exchanger There are many types of heat exchanger in industry. The types chosen based on the function of the heat exchanger itself. Heat exchangers are classified according to transfer processes into indirect and direct contact types. For an indirect-contact heat exchanger, the fluid streams remain separate and the heat transfers continuously through an impervious dividing wall or into and out of the wall or into transient manner. Direct-contact heat exchanger is when heat transfers continuously from the hot fluid to the cold liquid through a dividing wall not direct mixing of the fluids. Choosing the right heat exchanger requires knowledge of different type of heat exchanger as well as well as the environment in which the heat exchanger will operate. With sufficient knowledge of heat exchanger types and operating requirements, the best selection can be made in optimizing the process. In selecting a suitable heat exchanger for constructing the cooler, several exchanger types were considered. Below, in Table 3.1 are list of types and functions of each heat exchanger. Table 5.3: Types and Functions of Heat Exchanger in Industry 21 No Types Functions . 1. Double pipe heat The simplest type. Use for heating and exchanger cooling. Shell and tube heat Used for all application. 2. exchanger 3. Plate exchanger Use for heating and cooling. Plate-fin exchanger Use for heating and cooling. Spiral Use for heating and cooling. 4. 5. heat exchanger 6. Air cooled Cooler and condenser. Direct contact Cooling and quenching. Agitated vessels Use for heating and cooling. Fired heaters Use for heating and cooling. 7. 8. 9. 5.3.1.3 Selections of Heat Exchanger Typically in the manufacturing industry, several different types of heat exchangers are used for just the one process or system to derive the final product. In order to select an appropriate heat exchanger, one would firstly consider the design limitations for each heat exchanger type. Although cost is often the first criterion evaluated, there are several other important selection criteria which include: High/ Low pressure limits Thermal Performance Temperature ranges Product Mix (liquid/liquid, particulates or high-solids liquid) Pressure Drops across the exchanger Fluid flow capacity Clean-ability, maintenance and repair Materials required for construction Ability and ease of future expansion 22 5.3.1.4 Basic Principles of Design The principal design parameters that are evaluated at this stage involve all the operating conditions and basic exchanger configurations. Others parameters also includes are tubes and shell conditions (diameter, length, layout, baffles), coolant flow rate, pressure drop at shell and tube side, heat transfer rate and lastly heat transfer area. 5.3.1.4.1 Design Criteria for Process Heat Exchangers The criteria that a process heat exchanger must satisfy are easily enough stated if we confine ourselves to a certain process. The criteria include: 1) The heat exchanger must meet the process requirements. This means that it must effect the desired change in thermal condition of the process stream within the allowable pressure drops. 2) The heat exchanger must withstand the service conditions of the environment of the plant which includes the mechanical stresses of installation, start up, shutdown, normal operation, emergencies and maintenance. Besides, the heat exchanger must also resist corrosion by the environment, processes and streams. This is mainly a matter of choosing materials of construction, but mechanical design does have some effect. 3) The heat exchanger must be maintainable, which usually implies choosing a configuration that permits cleaning and replacement. In order to do this, the limitations is the positioning the exchanger and providing clear space around it. Replacement usually involves tubes and other components that may be especially vulnerable to corrosion, erosion, or vibration. 4) The cost of the heat exchanger should be consistent with requirements. Meaning of the cost here implement to the cost of installation. Operation cost and cost of lost production due to exchanger malfunction or unavailable should be considered earlier in the design. 5) The limitations of the heat exchanger. Limitations are on length, diameter, weight and tube specifications due to plant requirements and process flow. 5.3.1.5 Problem Identification 23 There are heater,cooler,condenser and reboiler in production of Ammonium Nitrate from reaction between ammonia and nitric acid. In this chapter, we only show the detail calculation for designing a cooler which is E-106. The function of E-106 is to exchange the temperature between the stream 18 and stream 19 from 204.3ºC to 100ºC. 5.3.1.6 Selection of Basic Heat Exchanger Type Shell and tube heat exchangers are the most widely used equipments in chemical industry. They are mostly used as heat transfer equipments but in few cases they are also used as Reactors and Falling Film Absorbers. However, in the case of recovering Sulphur from Sour Water, shell and tube heat exchangers was chosen as the heat exchanger. Sizes of various part of shell and tube of heat exchangers like shell, tubes, tie-rods are standardized. Standards developed by Tubular Exchanger Manufacturers Association, USA (TEMA) are universally used for design of shell and tube heat exchangers. 5.3.1.6.1 Shell Shell is the costliest part of heat exchanger. The cost sensitively changes with change in the diameter of shell. Using TEMA standard, shell size ranges from 6 in (152 mm) to 60 in (1520 mm). Standard pipes are available up to 24 in (608 mm). Thus, shell greater than 24 in (608 mm) is fabricated by a rolling plate. Shell diameter depends on tube bundle diameter. For fixed tube sheet shell and tube heat exchanger, the minimum gap between shell and tube is ranging from 10 mm to 20 mm. For maximum pull through floating head heat exchanger the range is from 90 mm to 100 mm. 5.3.1.6.2 Tube Tube size ranges from ¼ in (6.35 mm) to 2.5 in (63.5 mm) in shell and tube heat exchanger. Data for the standard tubes are given in TEMA standard. For the standard tubes, its size is equal to outer diameter of tube. Thickness of standard tubes is expressed in BWG (Birmingham Wire Gauge). Increase in BWG means decrease in thickness of tube. For no phase change heat exchangers and for condensers, ¾ in (19.05 mm) OD tube is widely used in practice. Tubes are available in standard lengths like 6 ft (1.83 m), 8 ft (2.44 m), 12 ft (3.66 m), 16 ft (4.88 m) and 19.67 ft (6.00 m). 24 Tube side passes are provided to decrease the tube side flow area and to decrease tube side velocity thereby to improve the tube side heat transfer coefficient, at the expense of pressure drop. This is true only if there is no phase change on tube side. Hence, more number of tube side passes are recommended only if there is no change in phase of the tube side fluid. 5.3.1.7 Fluid Allocation Factor According to the following factors, cooling water as cooler at tube side while organic solvent on shell side only if there is no phase change in shell side and tube side. Fluid allocation depends on following factors: Table 5.4: Fluid allocation factor Factor Fluid Allocation Corrosion Corrosive fluid will be in tube side Fouling Fluid that has the greatest tendency to foul on heat transfer surface will be allocated in the tube side. Fluid Very hot or very cold liquid is place in tube side to avoid the temperatur use of costlier material for shell. At moderate temperature, e hotter fluid is better passed through tubes. If placed on shell side more insulation required not only to reduce the heat loss but also the safety purpose. Operating High pressure stream should be place on tube side as high pressure pressure tube is cheaper than high pressure shell. Fluid flow Fluid which provides very low value of Reynolds number rate and viscosity should be placed on shell side as the dependency of shell side heat transfer rate coefficient on Reynolds number is less as compared to the same of tube side heat transfer coefficient. (Sinnott R. K, 2005) 5.3.2 Chemical Design The design procedure is as shown below: 1. Define the duty: heat transfer rate, fluids flow rate and temperature 25 2. Collect the fluid physical properties required: density, viscosity and thermal conductivity 3. Decide the type of heat exchanger to be used 4. Select the trial value for the overall coefficient, Uo 5. Calculate the mean temperature different, ∆Tlm and heat transfer area required 6. Decide the heat exchanger layout and calculate the individual coefficient 7. Calculate the overall coefficient and compare with the trial value. If the calculated value differs from the estimated value, substitute the calculated for the estimated value and return to step 4 8. Calculate the heat exchanger pressure drop; if unsatisfactory return to the step above. 5.3.2.1 Process Condition The process condition for designing tube heat exchanger as follows. Table 5.5: Process condition of the tube side Inlet Properties 0 Temperature ( C) Pressure (kPa) Specific heat (kJ/kg0C) Thermal conductivity, Kf (W/m0C) Density (kg/m3) Viscosity (N sm-2) (stream 18) 204.3 520 3.157 65 1138 0.7013 Mean 152.15 525 2.9737 70.74 1239 0.8414 Outlet (stream 19) 100 530 2.79 76.48 1340 0.9815 Table 5.6 was shown the process condition consideration in designing shell for heat exchanger. Table 5.6: Process condition of shell fluid (cooling water) Properties Temperature (0C) Pressure (kPa) Specific heat (kJ/kg0C) Thermal conductivity, Kf (W/m0C) Density (kg/m3) Viscosity (N sm-2) Inlet 25 101.3 4.2 166.89 997 0.8904 5.3.2.2 Design of Shell and Tube Heat Exchanger Mean 49.5 101.3 4.2 174.335 986 0.6349 Outlet 74 101.3 4.2 181.78 975 0.3793 26 As mention before, we have decided to choose type of shell and tube as our heat exchanger. Table 5.7 was shown the design criteria for shell and tube heat exchanger. Table 5.7: Criteria design of Shell and Tube Heat Exchanger Heat exchanger type Design type Heat exchanger orientation Tube inlet direction Tube side inlet stream Tube side outlet stream Tube side Shell side inlet stream Shell side outlet stream Shell side Heat duty (kW) 5.3.2.3 Heat Load One shell and two tubes Fixed and Tube Horizontal Horizontal 1 1 Mixture of Ammonium Nitrate, Ammonia, Nitric Acid and water 1 1 Cooling water 1109.6 For Tube side, Q = Wmixture Cp ∆t = (12880/3600) x (2.9737) x (204.3-100) = 1109.6 kJ/s For Shell side, Wwater = Q/ (Cp x ∆T) = [1109.6/ (4.2 x (74-25))] = 5.39 kg/s = 19409.91 kg/hr 5.3.2.4 Overall Coefficient From Table 12.1 ( R Sinnott, 2009), using typical overal coefficient for shell and tube heat exchanger. Hot fluid = Organic solvent Cold fluid = water Therefore the overall coefficient = 250 – 750 W/m2oC. We take the maximum value of the overall coefficient of the heat exchanger, Uo = 750 W/m2oC. 5.3.2.5 Heat Transfer Area 27 The flow arrangement for the heat exchanger is counterflow and the type of construction is one shell pass and two tube passes. Table 5.8: Temperature for shell and tube Tube or Shell side Tin (C) Tout (C) Tlm Logarithmic mean temperature, Tube 204.3 100 Shell 25 74 (T1 t2 ) (T2 t1 ) (T t ) In 1 2 (T2 t1 ) Equation 5.12 Where, T1 = Inlet shell side fluid temperature T2 = Outlet shell side fluid temperature t1 = Inlet tube side fluid temperature t2 = Outlet tube side fluid temperature From equation, ∆Tlm = [(25-100)-(74-204.3)] / ln [(25-100)/(74-204.3)] = 100.12 oC True temperature difference is given by; ∆Tm = Ft x ∆Tlm Where, Ft = Temperature correction factor R = (T1-T2)/(t2-t1) = (25-74)/(100-204.3) = 0.47 S = (t2-t1)/(T1-t1) = (100-204.3)/(25-204.3) = 0.58 From Figure 12.19 ( R Sinnott, 2009) the value of Ft = 0.91. ∆Tm = Ft x ∆Tlm = 0.91 x 100.12 = 91.11 oC The heat transfer area of heat exchanger, Ao = Q/(Uo x ∆Tm) = (1109.6 x 1000) / (750 x 91.11) = 16.24 m2 28 5.3.2.6 Tube Dimensions From Table A.5-2 (C.J Geankoplis, 2003), we take standard pipe of: Table 5.9: Tube dimension Particulars Tube length, L Outer diameter, OD Inside diameter, ID Birmingham wire gage (BWG) Area of the tube, At = LπOD Tube Dimension 5m 0.01905m 0.01483m 14 = 5π(0.01905) = 0.2992 m2 Using the area needed from the duty and the area for each tube, we can calculate the number of tube needed, Nt. Nt = Ao/At = (16.24/0.2992) = 54.28 Therefore, Nt = 58 Tube cross sectional area, Atc = π (OD/2)2 = π (0.01905/2)2 = 2.85 x 10-4 m2 Area per pass = 58 tubes x 2.85 x 10-4 = 0.0157 m2 29 Volumetric flow, Vt = Wmixture/ρmixture = (12880/3600) / 1239 = 0.003 m3/s Tube side velocity, Ut = Vt/Area per pass = 0.003/0.0157 = 0.19 m/s Reynolds number, Re = (ρUtID)/viscosity = (1239 x 0.19 x 0.01483)/ (0.8414/1000) = 4149.19 Prandtl number, Pr = (Cpx viscosity)/Kf = [(2.9737 x 1000) x (0.8414/1000)]/ 70.74 = 0.04 L/ID = 5/ 0.01483 = 337.15 From Figure 12.23 (R Sinnott, 2009), heat transfer factor, jh = 0.032 30 Nusselt number, Nu = jh x Re x (Pr)0.33 = 0.032 x 4149.19 x 0.04 = 5.31 Tube heat transfer coefficient, ht = Nu x (Kf/ID) = 5.31 x (70.74/0.01483) = 25329.02 W/m2oC From Figure 12.24 ( R Sinnott, 2009), jf = 0.06 Pressure drop on tube side, ∆Pt = 2 x 8 x jh (L/ID) x (ρUt2/µ) = 2 x 8 x 0.06 (337.15) x (1239 x 0.192/0.8414) = 17205.6 N/m2 5.3.2.7 Bundle and Shell Diameter Calculation The tubes in a heat exchanger are usually arranged in an equilateral triangular, square or rotated square pattern. The triangular and rotated square patterns five higher heat transfer rates. Here we use the triangular pattern. Tube pitch is the distance between tube centers and formulated as, Pt = 1.25 x OD 31 = 0.0238 m From Table 12.4 (R Sinnott,2009), Triangular pitch, K1 = 0.249 (4 passes), n = 2.207 Pt Figure 5.10: Triangular pitch The bundle diameter, Db = OD (Nt / K1) 1/n = 0.01905 (55/0.249)1/2.207 = 0.2198 m = 219.8 mm From Figure 12.12 ( R Sinnott, 2009), for bundle diameter of o.2198m, bundle clearence is 10.05mm. Shell diameter, Ds = 0.2198 + 0.01005 = 0.23m Baffle spacing, IB = 5 Ds/5 = 0.23/5 = 0.046m Shell cross flow area, As =[( Pt – ODtube) x Ds x (Ds/5)] / Pt = [(0.0238 – 0.01905) x 0.23 x 0.046] / 0.0238 = 0.0021m2 Equivalent triangular pitch arrangement diameter, De = (1.10/OD) x (Pt2 – 0.917 x OD2) = (1.10/0.01905) x (0.02382 – 0.917 x 0.019052) = 0.0135m Volumetric flow rate on the shell side, 32 Vt = Wwater/ρwater = (19409.91/3600)/988.15 = 0.0055 m3/s Shell side velocity, Us = Vt/As = 0.0055/0.0021 = 2.62 m/s Reynolds number, Re = (ρ x Us x De)/µ = (988.15 x 2.62 x 0.0135)/(0.6349/1000) = 55049 Prandtl number, Pr = (Cp x µ)/Kf = [( 4.2 x 1000) x (6.349 x 10-4)]/174.335 = 0.0153 Use segmental baffles with 25% cut. From figure 12.29 (R Sinnott, 2009), the value of heat transfer factor, jh is 0.0027 by neglecting the viscosity correction. The heat transfer coefficient on shell side, hs = (Kf/De) x jh x Re x Pr0.33 = (174.335/0.0135) x 0.0027 x 55049 x 0.01530.33 = 483183 W/m2oC From figure 12.24 (R Sinnott, 2009), the value of heat transfer factor, jf is 0.0056. Pressure drop on shell side, ∆Ps = 8 x jf x (Ds/De) x (L/IB) x (ρUt2/µ) x (µ/µw)0.14 = 18025.4 N/m2 The relationship between overall coefficient and individual coefficients is given by: 1 1 1 ODln(OD/ID ) 1 = + + + U o h s h od 2k w h id OD ID + 1 OD ( ) h t ID Equation 5.13 Where, Uo = the overall coefficient based on the outside area of the tube, W/m2. °C hs = tube heat transfer coefficient W/m2. °C ht = shell heat transfer coefficient W/m2. °C 33 hid = inside dirt coefficient W/ m2 . oC hod = outside dirt coefficient W/ m2. oC kw = thermal conductivity of the tube wall material W/m2.°C OD =tube inside diameter, m OD = tube outside diameter, m Table 5.10: Heat transfer coefficient hs (W/m2.oC) 483183 Ht (W/m2.oC) 25329.02 hod (W/m2.oC) 5000 hid (W/m2.oC) 5000 kw (W/m.oC) 16 ID (m) 0.01483 OD (m) 0.01905 Overall heat transfer coefficient can be calculated by using the formula using Equation 5.13 abbove: 1 1 1 ODln(OD/ID ) 1 = + + + U o h s h od 2k w h id OD ID + 1 OD ( ) h t ID = 0.00162 Uo = 617.28 W/m2.oC Error pencentage = Uassumed – Ucalculated x 100 Ucalculated = 750 – 617.28 x 100 617.28 = 21.5 % Hence, the calculation accepted because the percentage is less than 30%. 5.3.2.8 Summary of Chemical Design and Sizing for Heat Exchanger Table 5.11 shown the summary for design and sizing of shell and tube heat exchanger. Table 5.11: Chemical Design and Sizing for Heat Exchanger Parameter Item No Function Heat Exchanger Specification Cooler, E-106 To cool down process line from 34 204.3 0C (stream 18) to 100 0C Heat Duty, Q (kW) Hot Fluid Properties Flowrates (kg/hr) Inlet temperature (°C) Outlet temperature (°C) (stream 19) 1109.6 kW Process Stream from 18 to 27 12880.00 204.3 100.0 Utility (water) Cold Fluid Properties Flowrates (kg/hr) Inlet temperature (°C) Outlet temperature (°C) Heat transfer area, A (m2) Number of tubes, Nt Tube inside diameter, di (m) Tube outside diameter, do (m) Length of tube, L (m) Bundle diameter, Db (m) Shell diameter, Ds (m) Tube side heat transfer 19409.91 25.0 74.0 16.24 58 0.01483 0.01905 5.0 219.8 0.23 25329.02 coefficient, (W/ m2°C) Shell side heat transfer 483183 2 coefficient, (W/ m °C) Overall Coefficient, 2 617.28 o Uo (W/ m . C) % error Tube Pressure Drop (N/m2) Shell Pressure Drop (N/m2) 5.4 AMMONIA RECOVERY COLUMN 5.4.1 introduction 21.5 17205.6 18025.4 Distillation separates two or more liquid components in a mixture using the principle of relative volatility or boiling points. The greater the difference in relative volatility the greater the non-linearity and the easier it is to separate the mixture using distillation process. The process involves production of vapor by boiling the liquid mixture in a distillation column and removal of the vapor from the distillation column by condensation. Due to differences in relative volatility or boiling points, the vapor is rich in light components and the liquid is rich in heavy components. 35 Often a part of the condensate is returned (reflux) back to the still and is mixed with the outgoing vapor. This allows further transfer of lighter components to the vapor phase from the liquid phase and transfer of heavier components to the liquid phase from the vapor phase. Consequently, the vapor stream becomes richer in light components and the liquid stream becomes richer in heavy components. Different types of devices called plates, trays or packing are used to bring the vapor and liquid phases into intimate contact to enhance the mass transfer. Depending on the relative volatility and the separation task (i.e. purity of the separated components) more trays (or more packing materials) are stacked one above the other in a cylindrical shell to form a column. The distillation process can be carried out in continuous, batch or in semi-batch (or semi-continuous) mode. 5.4.1.1 Continuous Distillation Ammonia Recovery Column (ARC) T-101 is a distillation column. ARC operates in continuous mode. Continuous distillation is an ongoing separation in which a mixture is continuously (without interruption) fed into the process and separated fractions are removed continuously as output streams. The feed of ARC contains three components; ammonia, ammonium nitrate, and water which are more than two components; called multi-component column. In the distillation process, volatile vapor phase and a liquid phase that vaporizes are involved. The liquid mixture (feed), which is to be separated into its components, is fed into the column at one point along the column. Liquid runs down the column due to gravity while the vapor runs up the column. The vapor is produced by partial vaporization of the liquid reaching the bottom of the column. The remaining liquid is withdrawn from the column as bottom product rich in heavy components. The vapor reaching the top of the column is partially or fully condensed. Part of the condensed liquid is refluxed into the column while the remainder is withdrawn as the distillate product. The column section above the feed tray rectifies the vapor stream with light components and therefore is termed as rectifying section. The column section below the feed tray strips heavy components from the vapor stream to the liquid stream and is termed as stripping section. 36 Figure 5.11: Continuous distillation column 5.4.1.2 Distillation in Packed Column The type of internal used in ARC is packing. Packed column is classified where trays of various designs are used to hold up the liquid to provide better contract between vapour and liquid, hence better separation. Packed columns are often used for distillation when the separation is relatively easy and the required column diameter is not very large. They are generally less expensive than plate columns when the column diameter is relatively snall and have a lower pressure drop. 5.4.1.2.1 Comparing Trays and Packings Almost every separation can be performed either with trays or with packings. These factors only represent economic pros and cons, and each may be overridden. A couple of trays-versus-packing comparisons have appeared in the recent literature. These comparisons emphasize capacity, efficiency, and costs. Factors favoring packed-column: a) Small-diameter columns When column diameter is less than 3ft, it is difficult to access the column from inside in order to install and maintain the trays, “Cartridge” trays are often 37 installed, or an oversized diameter is used. Either option is expensive. Cartridge trays also run into leakage and hold-down problems. Packing is normally a cheaper and more desirable alternative. b) Corrosive systems The range of packing materials is wider than that commonly available for trays. Ceramic and plastic packings are cheap and effective. Trays can be manufactured in nonmetals, but packing is usually a cheaper and more desirable alternative. The main disadvantage is the difficulty in getting good liquid distribution, particularly for large diameter columns or very tall columns. Even if liquid is spread evenly over the packing at the top of the column, liquid tends to move toward the wall and to flow through the packing in preferred channels. The following factors generally factor favored trays compared to structured packings, but usually not compared to random packings. a) Material of Construction Due to the thin sheets used in structured packings, their materials of construction need to have better resistance for oxidation or corrosion. For a service in which carbon steel is usually satisfactory with trays, stainless steel may be required with structured packings. b) Column Wall Inspection With structured packings, it is often difficult to inspect the column wall without damaging the structured packings. Due to their snug fit, structured packings are easily damaged during removal. c) Washing and Purging Thorough removal of residual liquid, wash water, air, or process gas trapped in structured packings at startup and shutdown is more difficult than with trays. Inadequate removal of these fluids may be hazardous. 5.4.1.3 Distillation Column Design Steps The design of a distillation column can be divided into the following steps: 1. 2. 3. 4. Specify the degree of separation required: set product specifications. Select the operating conditions: batch or continuous; operating pressure. Select the type of contacting device: plates or packing. Determine the stage and reflux requirements: the number of equilibrium stages. 38 5. Size the column: diameter, number of real stages. 6. Design the column internals: plates, distributors, packing supports. 7. Mechanical design: vessel and internal fittings. Distillation consumes huge amounts of energy, both in terms of cooling and heating requirements. The best way to reduce operating costs of existing units is to improve the efficiency and operation via process optimization and control. 5.4.2 Chemical Design Of Ammonia Recovery Column 5.4.2.1 Design Data The Ammonia Recovery Column is shown in Figure 5.12 below: Figure 5.12: T-101 Ammonia Recovery Column The design data for ARC presented in Table 4.1 for feed, distillate, and bottom stream. Table 5.12 Flow rate inlet and outlet ARC Feed Component Ammonia (NH3) Product Feed inlet Distillate Bottom Stream 15 Stream 16 Stream 18 Molar flow rate (kgmol/hr) Molar flow rate (kgmol/hr) Molar flow rate (kgmol/hr) 75.0530 0.0075 145.4499 39 Ammonium Nitrate (NH4NO3) 145.4500 74.9850 0.0553 Water (H2O) 68.4480 0.0075 68.4448 Total 288.9510 75.0000 213.9500 5.4.2.2 Determination of Plate Number The main components in the feed to the ARC are ammonia, ammonium nitrate, and water as listed in the previous table. Since the feed is a mixture of those three components, the ARC is then considered as a multicomponent distillation column. The separation between the top and bottom products of multicomponent system is usually specified by setting limits on two key components, between which it is desired to make the separation. The light key (KLK) will be the component that it is desired to keep out of the bottom product (high volatility), and the heavy key (KHK) is the component to be kept out of the top product (low volatility). For ARC, the key components chose are: a) b) KHK = Water (H2O) KLK = Ammonia (NH3) The minimum number of stages is calculated by using Fenske Equation (Fenske, 1932) since McCabe-Thiele method is only applicable for two components system, not applicable in determining minimum number of stages for multicomponent system. 5.4.2.2.1 Determination of Bubble and Dew Point Temperature A saturated liquid is at its bubble point where any rise in temperature will cause a bubble of vapour to form whereas a saturated vapour is at its dew point where any drop in temperature will cause a drop of liquid to form. Therefore, the determination of bubble point and dew point temperature is crucial in order to calculate the minimum number of stages, as well as the condenser and reboiler temperatures. Dew points and bubble points can be calculated from the vapour-liquid equilibrium for the system. In terms of equilibrium constants, bubble and dew point are defined by equations 40 yi Ki xi 1.0 Bubble point : xi Dew point yi 1.0 Ki : For multicomponent mixtures, the temperature that satisfies these equation, at a given system pressure, must be found by iteration. The vapour pressure can be determined by using the Antoine’s equation as follows: ln P* a b d ln T eT T c f Equation 5.14 where a, b, c, d, e and f are the Antoine constant taken from HYSYS simulation, P in kPa and T in Kelvin. With related at equilibrium, constant K, Ki P* P Equation 5.15 Table 5.13: Antoine constant Component NH4NO3 NH3 H2O a) a 219.862 59.655 65.9278 b -23155.8 -4261.5 -7227.53 c 0.00 0.00 0.00 d -27.1074 -6.9048 -7.17695 e 1.70 x 10-17 1.00 x 10-5 4.03 x 10-6 f 6.00 2.00 2.00 Dew point temperature (top column) By using goal seek in the excel program, the value of the dew point temperature is 369.6 K with the constant operating pressure is 450 kPa. Operating P = 450 kPa Dew point T = 369.6 K 41 Table 5.14: Dew point temperature data Component NH4NO3 NH3 H2O Total b) Y 0.0001 0.9998 0.0001 1.0000 P* (kPa) 0.0488 5815.6343 88.9981 K 0.0001 12.9236 0.1978 y/K 0.9227 0.0774 0.0005 1.0006 Bubble point temperature (bottom column) By using goal seek in the excel program, the value of the bubble point temperature is 470.8 K with the constant operating pressure is 520 kPa. Operating P = 150 kPa Bubble point T = 470.8 K Table 5.15: Bubble point temperature Component NH4NO3 NH3 H2O Total 5.4.2.2.2 X 0.6798 0.0003 0.3199 1.0000 P* (kPa) 56.4429 30603.3974 1480.5590 K 0.1085 58.8527 2.8472 xK 0.0738 0.0152 0.9109 0.9999 Determination of Relative Volatility, α The determination of relative volatility, α of the components can be determined as the ratio between K values of light key component to heavy key component. K LK K HK Equation 5.16 where KLK = KHK = Light key component, NH3 Heavy key component, H2O The equilibrium vaporization constant K is defined for a compound by: Ki yi xi Equation 5.17 42 Where yi = mole fraction of component i in the vapour phase xi = mole fraction of component i in the liquid phase Table 5.16: Relative volatily for each component Component K α Distillate Bottom Distillate Bottom Average NH4NO3 0.0001 0.1085 0.0005 0.0381 0.0046 NH3 12.9236 58.8527 65.3456 20.6702 36.7519 H2O 0.1978 2.8472 1.0000 1.0000 1.0000 5.4.2.2.3 Determination of Number of Stages Using Fenske Equation The Fenske equation (Fenske, 1932) can be used to estimate the minimum stages required at total reflux. This equation is applicable for multicomponent systems. The minimum number of stages can be obtained by equation: x LK x HK x HK d x LK log LK log N min b Equation 5.18 where αLK = average relative volatility of the light key with respect to the heavy key xLK = light key concentrations xHK = heavy key concentrations N min 0.9998 0.3199 log 0.0001 0.0003 log 36.7519 N min 4.5 N min 5 stages Then, the theoretical number of stage is given by equation: 43 N T 2N min Equation 5.19 N T 2(5) N T 10 stages; including reboiler. 5.4.2.2.4 Calculation for Minimum Reflux Ratio, Rmin Colburn (1941) and Underwood (1948) have derived equations for estimating the minimum reflux ratio for multicomponent distillations. As the Underwood equation is more widely used it is presented and the equation can be stated in the form: i xi,d i R min 1 Equation 5.20 where, αi = the relative volatility of component i with respect to some reference component, usually heavy key Rmin = the minimum reflux ratio xi, d = concentration of component i in the distillate at minimum reflux θ = the root of the equation, from the following equation: i xi, f i 1 q Equation 5.21 where, xi,f = concentration of component i in the feed at minimum reflux 44 q = depends on the condition of the feed The value of θ must satisfy the relation α HK < θ < αLK. By assuming the feed is enter at its boiling where q = 1, the value of θ is determined using goal seek in excel application. i xi, f i 0 By using goal seek, θ = 1.8694 when Table 5.16: Relative volatility at feed Component Α xf αxf αxf / α – θ NH4NO3 0.0046 0.5034 0.0023 -0.0012 NH3 36.7519 0.2597 9.5445 0.2736 H2O 1.0000 0.2369 0.2369 -0.2725 Σ [αxf / (α – θ)] -0.0001 Table 5.17: Relative volatility at distillate Component Α Xd αxd αxd / α – θ NH4NO3 0.0046 0.0001 0.0000 0.0000 NH3 36.7519 0.9998 36.7446 1.0534 H2O 1.0000 0.0001 0.0001 -0.0001 Σ [αxd / (α – θ)] From the previous equation, R min i x i,d i 1 1.0533 45 R min 1.0533 1 R min 0.05 The minimum reflux ratio can also be determined by using the Underwood equation, R min 1 1 x d , LK x d , HK x f , HK x f , LK Equation 5.22 where xd,LK = light key component at top flow xd,HK = heavy key component at top flow xf,LK = light key component at feed flow xf,HK = heavy key component at feed flow R min 0.9998 1 0.0001 36.7519 36.7519 1 0.2597 0.2369 R min 0.11 Since the value of Rmin calculated using Underwood equation is bigger than the previous calculation, therefore the value of R min = 0.11 will be used for the next calculations. For many systems, the optimum reflux ratio will lies between 1.2 to 1.5 times the minimum reflux ratio. Optimum reflux ratio, R = 0.11 x 1.5 = 0.16. 5.4.2.2.5 Plate Efficiency The O’Connell correlation shown in Figure 4.3 can be used to estimate the overall column efficiency. The equation correlated with the product of the relative volatility of the light key and the molar average viscosity of the feed with estimated at the average column temperature. Eduljee (1958) has expressed the correlation in the form of equation: 46 E o 51 32.5 log( a a ) Equation 5.23 where μa = the molar average liquid viscosity, mNs/m2 αa = average relative volatility of the light key Figure 5.13: Distillation column efficiencies (O’Connell, 1946). 5.4.2.2.6 Viscosity at Average Temperature The average viscosity can be found by using equation below where T is the average temperature. 1 1 T VISB log VISA Equation 5.24 where T = Tavg = (T at top + T at bottom)/2 = (1.192+204.3)/2 = 375.9 K Table 5.18: Viscosity for each component 47 Component VIS A VIS B NH4NO3 NH3 H2O 0.7109 -0.1391 -0.8678 0.4628 -0.2470 -1.1067 , mN m2 0.0292 0.2732 0.1635 Mole fraction 0.5034 0.2597 0.2369 μ(avg)= Molar viscosity 0.0147 0.0709 0.0387 0.0415 From feed composition: (0.0292 0.5034) (0.2732 0.2597) (0.1635 0.2369) 3 Molar average viscosity = αaμa Eo 5.4.2.2.7 = 0.0415 mN/m2 = 36.7519 x 0.0415 = 1.5240 = 51 – 32.5 log (1.5240) = 45.05 % The Actual Number of Stages The actual number of stages, N is the number is stages required for the column. This stage can be determined by: N NT Eo Equation 5.25 N 10 1 0.4505 N 19.98 N 20 5.4.2.2.8 stages Determination of Feed-Point Location 48 Feed point location is calculated using equation by Kirkbride (1944): N log r Ns B x f , HK 0.206 log D x f , LK x b , LK x d , HK 2 Equation 5.26 where Nr = number of stages above the feed, including any partial condenser Ns = number of stages below feed, including the reboiler 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 Therefore, 214 0.2369 0.0003 2 Nr log 0.206 log Ns 75 0.2597 0.0001 Nr 1.80 Ns N r 1.80 N s From the previous calculation, the number of stages excluding reboiler is 20. Hence, N r N s 20 N s 20 N r N s 20 1.8 N s N s 7.14 Ns 8 ; feed enters at stage number 8. 49 5.4.2.3 Equipment Sizing of Ammonia Recovery Column The following table shows the mole fraction, molecular weight, and liquid density for each component. Table 5.19: Mole fraction, molecular weight, and liquid density for each component Feed Componen t Distillate F Bottom D (kgmol/hr xf xd ) 145.4500 0.5034 0.0075 NH3 75.0530 0.2597 74.9850 H2O 68.4480 0.2369 0.0075 Total 288.9510 1.0000 75.0000 (kgmol/hr xb 0.000 1 0.999 8 0.000 1 1.000 0 145.4499 0.0553 68.4448 213.9500 0.679 80.0 1720.0 8 0.000 5 17.0 0 3 0.319 3 18.0 9 1.000 0 0 Table 5.20: RMM and vapor density for each component NH4NO3 NH3 H2O RMM Top 0.0080 17.0266 0.0018 Bottom 54.4205 0.0044 5.7584 ρV(kg/m3) Top Bottom 0.0016 7.2258 3.4057 0.0006 0.0004 0.7646 Sample of calculation at distillate: The relative molecular mass (RMM) in distillate; ( x MW ) (0.0001 80.05) (0.9998 17.03) (0.0001 18.00) 17.0364kg / mol (kg/m3) ) NH4NO3 Component ρL B (kgmol/hr ) MW 616.07 997.99 50 Liquid density at distillate; x (1720 0.0001) (616.07 0.9998) (997.99 0.0001) 616.22kg / m 3 Vapor density at distillate; RMM T STP POP V STP TOP PSTP 0.008 273K 4.5bar 17.0266 273K 4.5bar 3 3 274.192 K 1bar 22.4kg / m 274.192 K 1bar 22.4kg / m 0.0018 273K 4.5bar 3 22.4kg / m 274.192 K 1bar 3.41kg / m 3 All calculated data is tabulate in Table 5.21. Table 5.21: Liquid and vapor density at distillate and bottom Reflux ratio, R = 0.16 Feed, F = 3.93 kg/s Distillate, D = 0.35 kg/s Bottom, B = 3.58 kg/s ρL above feed = 616.22 kg/m3 ρV above feed = 3.41 kg/m3 ρL below feed = 1488.74 kg/m3 ρV below feed = 7.99 kg/m3 Mass balance on top distillation column Liquid mass flow rate, Lm = R x D = 0.16 x 0.35 = 0.06 kg/s Vapor mass flow rate, Vm = D + Lm = 0.35 + 0.06 = 0.41 kg/s Mass balance on bottom distillation column Liquid mass flow rate, Ln = (R x D) + F = 0.06 + 3.93 51 = 3.99 kg/s Vapor mass flow rate, Vn = Ln - B = 3.99 – 3.58 = 0.41 kg/s 5.4.2.3.1 Calculation for Column Diameter a) First trial: sieve tray column internal Try tray spacing = 0.45 m From Chemical Engineering, Volume 6; V L Lm Vm Top liquid-vapor flow factor, FLV = 0.06 0.41 3.41 616.22 = = 0.01 Ln Vn Bottom liquid-vapor flow factor, FLV V L = 3.99 7.99 0.41 1488.74 = = 0.719 From Chemical Engineering, Volume 6, Figure 4.4 below: Top K1 = 0.078 Bottom K1 = 0.037 52 Figure 5.14: Flooding velocity, sieve plate Correction for surface tension: From HYSYS simulation; Surface tension, σtop = 0.032 N/m Surface tension, σbottom = 0.700 N/m 2 Top K (correction) = 0.02 K1 2 = = 0.032 0.02 0.01 0.086 2 Bottom K (correction) = = = 0.02 K1 0 .7 0.02 2 0.719 0.075 From Chemical Engineering Design, Volume 6 53 K1 Top flooding velocity, uf L V V = 0.086 616.22 3.41 3.41 = = 1.15 m/s K1 Bottom flooding velocity, uf L V V = 0.075 1488.74 7.99 7.99 = = 1.03 m/s Design for 80% flooding at maximum flow rate: Top ûv Bottom ûv = 0.80 x uf = 0.80 x 1.15 = 0.92 m/s = 0.80 x uf = 0.80 x 1.03 = 0.82 m/s Maximum volumetric flow rate: Top Bottom Net area required = Vm / ρV = 0.41 / 3.41 = 0.119 m3/s = Vn / ρV = 0.41 / 7.99 = 0.051 m3/s = Maximum volumetric flow rate uv 54 Top Bottom = 0.119 / 0.92 = 0.13 m2 = 0.051 / 0.82 = 0.062 m2 As first trial take downcomer as 19% of total: Column cross-sectional area Top Bottom = Net area / 0.81 = 0.13 / 0.81 = 0.16 m2 = Net area / 0.81 = 0.062 / 0.81 = 0.076 m2 Column diameter Top = = = Bottom = = = √ Column cross section area× 4 π 0.160 4 0.452 m √ Column cross section area× 4 π 0.076 4 0.312 m Since the column diameter is smaller than 0.6m, packing should be considered because plates would be difficult to install and expensive. For the design of packed distillation columns, it is simpler to treat the separation as a staged process and use the concept of the height of an equivalent equilibrium stage to convert the number of ideal 55 stages required to a height of packing. The method for estimating the number of ideal stages from the previous calculation can then be applied to packed column. b) Second trial: 1T Intalox sheet metal structured packing column internal Figure 5.15: Intalox Structured Packing (Koch-Glitsch®) Intalox structured packing is a unique family of patented distillation devices that offer better efficiency and capacity than other metal-sheet structured packings. Intalox structured packing delivers superior performance in these and other applications: i. ii. iii. iv. v. vi. vii. viii. FCC main fractionators Aromatic distillations Alcohol distillations Amine scrubbers and regenerators Amine distillations Glycol distillations Flavor/fragrance fractionators Refrigerant distillations The characteristic of Intalox Structured Packing is as follows: Table 5.22: Characteristic of Intalox Structured Packing Surface area (ft2/ft3) Voidage, ε (%) Packing factor, Pf (ft-1) 96 95 20 56 Since the data available for packing is in English unit, thus the packing distillation column is calculated by using English unit, instead of SI unit. The properties from calculation and HYSYS simulation are as follows: Table 5.23: Properties from calculation and HYSYS L (lb/s) G (lb/s) L/G ѵ (cSt) μ (cP) σ (dyne/cm) ρL (lb/ft3) ρG(lb/ft3) Vl (ft3/s) Vg (ft3/s) Top 0.12 0.90 0.1379 0.2592 0.1635 31.658 38.47 0.21 0.0032 4.2075 Bottom 8.79 0.90 9.8177 0.6159 0.7014 699.93 92.94 0.50 0.0946 1.7942 Flood prediction by GPDC interpolation and Kister & Gill correlation Table 5.24: Flow parameter and capacity parameter at flooding Top Bottom X or Flv 0.010 0.719 Co 1.89 0.65 Cs,fl (ft/s) 0.452 0.149 Flow parameter, Flv is taken from the previous calculation; calculated using equation; L G FLV G L Equation 5.27 Pressure drop at flooding is calculated using equation; PFl 0.115F p 0.7 PFl 0.115( 20) 0.7 PFl 0.9363in.H 2 O / ft Capacity parameter, Co is determined from Figure 5.16: 57 FLOW PARAMETER Figure 5.16: GPDC for structured packing only The capacity parameter is then inserted into the following equation to calculate Cs; C o C s F p0.5 0.05 Equation 5.28 Therefore, Cs Co F 0.05 0.5 p Packed column are usually design to 70-80 % margin from flooding velocity. This practice provides sufficient margin to allow the uncertainties associated with the flood point concept and prediction to avoid the flood point. For the calculation purpose, the column will be designed for 75% of flood. Cs design = 0.75Cs,fl. The derating factor is 0.85. Table 5.25: Diameter calculation due to flooding Cs nonderated derating factor Top 0.339 0.85 Bottom 0.112 0.85 58 Cs derated ρG/(ρL-ρG) sqrt [ρG/(ρL-ρG)] Us (ft/s) CFS (vapor flow rate, (ft3/s) At (ft2) Dt (ft) Dt (m) 0.288 0.006 0.075 3.865 4.208 1.089 1.177 0.359 0.095 0.005 0.073 1.292 1.794 1.388 1.330 0.405 Velocity, us, area, At, and diameter, Dt are calculated using; G L G Cs us Equation 5.29 AT CFS us Equation 5.30 Diameter calculation due to maximum pressure drop criterion; DT 4 AT Equation 5.31 Maximum pressure drops, ΔP (in. H2O/ft) recommended for packed column for atmospheric and low-to-medium pressure is between 0.5 and 1.0. Table 5.26: Diameter calculation at maximum pressure drop -1 Fp (ft ) ѵ (cSt) ΔPmax (in H2O/ft) Co Cs, max ρG/(ρL-ρG) sqrt [ρG/(ρL-ρG)] Us, max (ft/s) Top 20 0.2592 1.000 1.94 0.464 0.006 0.075 6.224 Bottom 20 0.6159 1.000 0.66 0.151 0.005 0.073 2.058 59 CFS (ft3/s) 4.208 2 0.676 At (ft ) 0.928 Dt (ft) 0.283 Dt (m) The diameter is taken from the largest one, which 1.794 0.872 1.054 0.321 is 1.177 ft, and then round off to the nearest standard pipe size 18 in (1.5 ft). 5.4.2.3.2 Determination of Packed Height Table 5.27: Determination of packed height dp(in) Dt (ft) 12 Dt/dp Area (ft2) HETP (ft) n (stages) Total packed height (ft) Height of equivalent theoretical plate Top 1.00 1.50 18 1.77 1.50 12 18 Bottom 1.00 1.50 18 1.77 1.50 8 12 (HETP) is determined from www.cheresoures.com with refers to the packing size used in the tower; tabulate in Table 4.18. For distillation process, HETP is equal to 1.5 ft (0.46 m) for packing size of 1.0 in. The total packed height is 30 ft; equivalent to 9.144 m. Height to diameter (H/D) ratio is 20 which is satisfactory the range given that is H/D ratio must be kept below 20 to 30. Table 5.28: HETP with respect to process and packing size SETUP Method Distillation Vacuum Distillation Absorption/Stripping HETP expressed Packing as ft (meters) Size (in) 1.0 1.5 2.0 1.5 (0.46) 2.2 (0.67) 3.0 (0.91) 1.0 2.0 (0.67) 1.5 2.0 All Sizes 2.7 (0.82) 3.5 (1.06) 6.0 (1.83) Table 5.29: Chemical Design Specification Sheet 60 CHEMICAL DESIGN SPECIFICATION SHEET Identification: Item: Distillation Column (Recovery Column) Item Code: T-101 No. Required: 1 By: Siti Hajar binti Abd Kadir Function: To separate ammonia to be recycled back to the feed stream Operating Condition 0 Operating Temperature : 204.3 C Operating Pressure: 503.33 kPa Operation: Continuous Column Specification Design type Vertical Feed location Stage 8 Material of Stainless steel HETP 0.4572 m construction 1T Intalox Column Diameter 0.4572 m Packing Structured Column Height 9.144 m Packing Column Cross Sectional 1.77 m2 20 Area 0.16 Top: 12 Reflux Ratio Top: 0.12 m3/s Bottom: 8 CFS (vapor flow rate) Bottom: 0.05 m3/s H/D ratio 20 Number of Stages Stream Flow rate (kg/s) Feed flowrate 3.93 Distillate flowrate 0.35 Bottom flowrate 3.58 1T Intalox Structured Packing Design Surface area 315 m2/m3 Voidage 95 % Size 2.54 cm Packing factor 66 m-1 Flooding criteria Flooding % 75 Pressure drop 78.3 mm H2O/m Derating factor 0.85 Superficial velocity Top: 1.18 m/s Bottom: 0.39 m/s Maximum pressure drop criteria 61 Pressure drop 83.3 mm H2O/m Superficial velocity Top: 1.90 m/s Bottom: 0.63 m/s 5.5 DISTILLATION COLUMN 2, T-102 5.5.1 Introduction Generally, the function of distillation is to separate components based on difference of its boiling point by vaporization, a liquid mixture of miscible and volatile substances into individual components or some into groups of components. Distillation employs heat to generate vapors and cooling to effect partial or total condensation as needed. In reboiled absorbers, partial stripping of the lighter components is performed in the lower part of the equipment. The feasibility of separation of mixtures by distillation, absorption, or stripping depends on the fact that the compositions of vapor and liquid phases are different from each other at equilibrium (Ulrich, 1984). When a liquid mixture of two volatile materials is heated, the vapor that comes off will have a higher concentration of the more volatile (i.e., lower boiling point) material than the liquid from which it was evolved. The greater relative volatility, the mixture is easier to separate. Vapour flows up the column and liquid counter-currently down the column. In other words, the more volatile component is discharge at the top in vapour phase and the less volatile is release at the bottom as liquid (Duncan and Reimer, 1998). This means that the top product has low boiling point compared to the bottom product. The purpose of this distillation column is to separate the components mixture to get the desired product of ammonium nitrate. A distillation column design can be divided into chemical and mechanical designs. 5.5.2 Chemical Design 62 There are several steps that can be followed in designing the distillation column, (Sinnot and Towler, 2009): 1. 2. 3. 4. 5. 6. 7. Specify the degree of separation required: set product specification. Select the operating conditions: batch or continuous; operating pressure. Select the type of contacting device: plate or packing. Determine the stage and reflux requirements. Size the column: diameter, number of real stages. Design the column internals: plates, packing supports. Mechanical design: vessel and internal fittings. Figure 5.17: Schematic Diagram for Distillation Column For this design, the base column is heated by using condensing steam. The operation of partial condensation of rising vapour and partial vaporization of the reflux is repeated on each tray. On ideal tray the vapour rising from it will be in equilibrium with the liquid leaving, though in practice a smaller degree of enrichment will occur. Continuous fractionating column will be used in this design. The purpose of this distillation column is to separate the components mixture (ammonium nitrate, water, ammonia and nitric acid) to get the desired product of ammonium nitrate in the bottom stream. The complete composition of the inlet and outlet streams for the distillation column is shown in Table 5.30. 63 Table 5.30: Summary of inlet and outlet composition Component Feed Molar Flow Mole Rate Fraction (Kmol/hr) NH4NO2 HNO3 NH3 H2O Total 145.4513 0.0008 0.0485 68.4480 213.9486 Assumption: Product 0.6798 0.0000 0.0002 0.3199 1.0000 Top Molar Flow Bottom Molar Flow Mole Mole Rate Fraction Rate Fraction (Kmol/hr) 0.0085 0.0008 0.0485 41.5193 41.5771 0.0002 0.0000 0.0012 0.9986 1.0000 (Kmol/hr) 145.4428 0.0000 0.0000 26.9287 172.3715 0.8438 0.0000 0.0000 0.1562 1.0000 If the presence of the other components does not significantly affect the volatility of the key components, the keys can be treated as a pseudo-binary. Thus, for this distillation column, pseudo-binary system will be used. Since the nitric acid and ammonia content are too small which is less than 1%, all calculation will be based on binary system calculations. Since the calculation is based on pseudo-binary system, the mole faction will have to be recalculated. The new mole fraction is shown in table below: Table 5.31: Binary mol fraction and mol flow rate of inlet and outlet composition Feed Componen t Product Molar Mole Flow Fraction Rate 145.4513 68.4480 213.8993 Molar Mole Flow Fraction Rate (Kmol/hr) NH4NO2 H2O Total Top 0.6800 0.3200 1.0000 (Kmol/hr) 0.0085 41.5193 41.5278 Bottom Molar Mole Flow Fraction Rate 0.0002 0.9998 1.0000 (Kmol/hr) 145.4428 26.9287 172.3715 0.8438 0.1562 1.0000 Besides that, there are a few other assumptions can be made for this distillation column design (Couper, 2005): i. There are no chemical reactions in the distillation column. ii. A cylindrical vertical vessel is suitable for this separation because of large production and will be the cheapest shape. iii. The position of the interface can be controlled with or without the use of instruments, by use of a siphon take off for the heavy liquid iv. Neglecting friction loss in the pipes. 64 Table 5.32: Temperature and pressure at feed, top and bottom distillation column Feed 100 530 Temperature (˚c) Pressure (kPa) Top 141.1 400 Bottom 227.7 520 5.5.2.1 Bubble and Dew Point Temperature To estimate the stage and condenser and reboiler temperature, procedures are required for calculating dew and bubble points (Ulrich, 1984). By definition, a saturated liquid is at its bubble points (any rise in temperature will cause a bubble of vapour to form), and a saturated vapour is at its dew point (any drop in temperature will causes a drop in liquid form). It can be calculated in term of equilibrium constant, K. Table below show the constant of Antoine equation for each component (HYSYS). Bubble point : Σyi = Σkixi = 1.0 Dew point Σxi = Σyi/ki = 1.0 : Using Antoine equation to find vapour pressure (P0) (Hysys) ln P0= A− B + D × ln ( T ) + E ×T F T +C Equation 5.32 K= P0 PT Equation 5.33 65 Compone A C D E F 0 -2.711E+1 1.697E-17 6 0 6.593E+1 -7.228E+3 -7.177 0 1.632E+2 -1.008E+4 -2.277E+1 0 5.966E+1 -4.262E+3 -6.905 Table 5.33: Antoine Constant 4.031E-6 2.729E-5 1.002E-5 2 2 2 nt NH4NO2 H2O HNO3 NH3 B 2.199E+2 -2.316E+4 Table 5.34: Bubble point Calculation, Bottom (PT = 530 kPa) T Component NH4NO2 H2O Xb 0.8438 0.1562 221.8597438 ˚C ln P P(kPa) 5.14315E+00 171.2549828 7.78488E+00 2403.984301 495.0097438 ˚K K K*Xb 0.329336505 0.277894143 4.623046733 0.7221199 TOTAL 1.000014043 Table 5.35: Dew Point Calculation, Top (PT= 400 kPa) T Component NH4NO2 H2O Yb 0.0002 0.9998 143.6425413 ˚C ln P P(kPa) 8.66005E-01 1 5.99168E+00 400.0867244 416.7925413 ˚K K K*Yb 0.0025 0.0000005 1.000216811 1.000016768 TOTAL 1.000017268 Table 5.36: Bubble-Point calculation, feed (liquid feed) (PT=530 kPa) T Component NH4NO2 H2O 199.1414003 ˚C X ln P 0.68 4.10523E+00 0.32 7.33114E+00 P(kPa) 60.65658132 1527.125937 472.2914003 ˚K K K*X 0.11444638 0.077823538 2.881369692 0.922038301 TOTAL 0.99986184 Temperature dew point and bubble point are calculated by using goal and seek (Microsoft Excel) 5.5.2.2 Relative Volatility 66 It is in need to identify the light and heavy key to determine average relative volatility. Water is needed to recover much at distillate; therefore it is a light key Average relative volatility, αa can be determined by knowing top and bottom column temperature. α= K LK K HK Equation 5.34 Where; KLK = light key component KHK = heavy key component LK = water, H2O HK = ammonium nitrate, NH4NO2 And For this case, the light component is water and heavy component is ammonium nitrate. Table 5.37: The relative volatility for component Feed Top Bottom Component NH4NO2 H2O NH4NO2 H2O NH4NO2 H2O X 0.68 0.32 0.0002 0.9998 0.8438 0.1562 K 0.11444638 2.88136969 0.0025 1.00021681 0.32933651 4.62304673 Sample calculation: At the top KHK = 0.0025 KLK = 1.00021681 α,T = At the bottom 1.00021681 0.0025 = 400.086724 α 1 25.176591 1 400.086724 1 14.0374561 67 KHK = 0.32933651 KLK = 4.62304673 α,B = 4.62304673 0.32933651 = 14.0374561 5.5.2.3 Average volatility α LK , AVG =√ α T × α B Equation 5.35 α LK , AVG =√ 400.086724 ×14.0374561 = 74.9413092 5.5.2.4 Number of Stages Number of stages to this distillation column is calculate using Frenske equation in continuous fractional distillation is an equation used for calculating the minimum number of theoretical plates required for the separation of a binary feed stream by a fractionation column that is being operated at total reflux (Duncan and Reimer, 1998). x HK ¿ x LK b ¿ x LK ¿ ¿ x HK d log ¿ N m=¿ Equation 5.36 0.9998 0.8438 ][ ] 0.0002 0.1562 log 74.9413092 log [ N m= N m=2.3638 stages ≈ 3 stages Normally after using Frenske’s equation, the value of Nm is given by the equation below to get the number of theoretical stages NT. N T =2(N m ) Equation 5.37 = 2(3) =6 5.5.2.5 Plate Efficiency Feed composition, mole fraction: Ammonium nitrate : 0.68 Water : 0.32 The overall column efficiency is correlated with the product of the relative volatility of the light key component and the molar average viscosity of the feed, estimated at the average column temperatures. The Cornell’s correlation is mainly based on data obtained with hydrocarbon system, but also includes some values for water-alcohol mixtures as well. The method only considered the viscosity and volatility of each component and does not take into account the plate design parameters. αμ ¿0.25 ¿ 0.5 Eo = ¿ Equation 5.38 Where, µ = the average liquid viscosity, mNs/m2 α = average volatility of the light key Average Liquid Viscosity From HYSYS Simulation and calculation, 1. For distillate, µ = 0.1445 mNs/m2 2. For bottom, µ = 0.7783 mNs/m2 Average liquid viscosity = 0.1445 + 0.7783 2 = 0.4614 mNs/m2 From the previous calculation, relative volatility of light key is: α a=¿ 74.9410392 Therefore, overall column efficiency, 74.9413092× 0.4614 ¿0.25 ¿ 0.5 E O= ¿ = 20.62 % = 0.2062 To get the real number of stages, the efficiency of the process must be considered and the efficiency is calculated based on the equation by O’Connell’s (J. Douglous, 1988) N= NT EO ¿ 6 0.2062 = 29.09 ≈ 30 stages 5.5.2.6 Feed Point Estimation of the feed point location can be calculated using Kirkbride equation (1944): [( )( )( ) ] [ ] N B log r =0.206 log Ns D x f , HK x f , LK x b , LK x d ,HK 2 Equation 5.39 Where, Nr = Number of stages above the feed, including any partial Ns xf,HK xf,LK xd,HK xb,LK B D = = = = = = = condenser Number of stages below the feed, including the reboiler Concentration of the heavy key in the feed Concentration of the light key in the feed Concentration of the heavy key in the top product Concentration of the light key in the bottom product Molar flow of bottom product Molar flow of top product [( )( )( ) ] [ ] N B log r =0.206 log Ns D log x f , HK x f , LK [ ] x b , LK x d ,HK [( Nr 172.3715 =0.206 log Ns 41.5278 2 Equation 5.40 0.68 0.1562 0.32 0.0002 )( )( )] 2 [ ] Nr =24.3524 Ns Where actual number of stages is 30 Nr + Ns = 30 24.3524 N + N s s Ns Nr Feed point = 30 = = = 1.1833 28.8267 2 from bottom 5.5.2.7 Physical properties 5.5.2.7.1 Calculation for Density and Relative Molar Mass (RMM) Table 5.38: Properties component Molecular Componen t Ammonium Nitrate Water 5.5.2.7.2 NH4NO2 H2O Total Weight (MW) 80.052 18.015 98.067 Liquid Feed,x Top, xD xW F 0.6800 0.3200 1.0000 0.0002 0.9998 1.0000 Calculation for Relative Molar Mass (RMM) RMM = ΣComponent mole fraction × MW RMM feed = 0.6800(80.052) + 0.3200(18.015) = 60.2002 kg/kmol = = 0.0002(80.052) + 0.9998(18.015) 18.0274 kg/kmol RMM Bottom Product = 0.8438(80.052) + 0.1562(18.015) RMM Top Product Bottom, 0.8438 0.1562 1.0000 Density (kg/m3) 1725 1000 = 5.5.2.7.3 70.3618 kg/kmol Pressure Drop The pressure drop over the plate is an important design consideration. There are two main sources of pressure loss: 1. Vapour flow through the holes (an orifice loss) 2. Static head of liquid on the plate It is convenient to express the pressure drops in term of milimetres of liquid. In pressure units: ∆ Pt =9.81 ×10−3 ht ρL Equation 5.41 ∆ Pt = Total pressure drop, Pa, (N/m2) ht = Total plate pressure drop, mm liquid ∆ Pt =9.81 ×10−3 ht ρL Top pressure, 5.132 atm = 520 kPa Bottom pressure, 3.9477 atm = 400 kPa 5.5.2.7.4 Calculation for density a. Bottom product Liquid density, ρL,w = ΣxB,i ρi = 0.8438(1725) + 0.1562(1000) = 1611.755 kg/m3 T Vapour density, ρV,w P = ∑ VMW × TSTP × P OP = 70.3618 kg /kmol 273.15 5.132 × K× atm 3 500.85 1 22.4 m /lmol STP OP STP = 8.7916 kg/m3 = ΣxB,i ρi = 0.0002(1725) + 0.9998(1000) = 1000.145 kg/m3 b. Top product Liquid density, ρL,T T Vapour density, ρV,T ∑ VMW × TSTP × P OP = 18.0274 kg/kmol 273.15 3.9477 × K× atm 3 414.25 1 22.4 m /lmol STP OP STP 2.0949 kg/m3 = 5.5.2.7.5 P = Summary Table 5.39: Summary vapour and liquid flow Vapour and Liquid Flow F 213.9486 kmol/hr D 41.5771 kmol/hr W 172.3715 kmol/hr Enriching Section Ln 192.5537 kmol/hr Vn 234.1308 kmol/hr Stripping Section Lm(Lw) 406.5023 kmol/hr Vm(Vw) 234.1308 kmol/hr Table 5.40: Summary of density Density (kg/m3) Top Bottom Liquid 1000.145 1611.755 Vapour 2.0949 2.0949 5.5.2.8 Column Sizing 5.5.2.8.1 Column Diameter The principal factor that determines the column diameter is the vapour flow rate. The vapour velocity must be below that which caused excessive liquid entrainment or a high-pressure drop. The equation given below, which is based on the well-known Souder’s and Brown equation (Duncan and Reimer, 1998), can be used to estimate the maximum allowable superficial vapour velocity, and hence the column area and diameter. 2 t uv =( −0.171l +0.27 l t −0.047 ) [ ( ρ L −ρV ) ρV 1 /2 ] Equation 5.42 uv = Maximum allowable vapour velocity, based on the gross (total) column cross-sectional area, m/s l t = Plate spacing (range 0.5 -1.0) Where: In order to use the given equation, the plate spacing has to be specified. Plate spacing 0.55 m is chosen. Thus, 2 t uv =( −0.171l +0.27 l t −0.047 ) [ ( ρ L −ρV ) ρV 1 /2 ] [ ( 1611.755−2.0949 ) uv =( −0.171(0.55) +0.27 (0.55)−0.047 ) 2.0949 2 = 0.0497725 × 27.7195 = 1.37966 m/s With the value gained, the column diameter can be calculated using: Dc = √ 4V w πρv uv Where Vw is the vapour flow rate (kg/s). So, V w= ¿ 234.1308 kmol /hr × RMM (kg/kmol) 3600 s 234.1308 kmol / hr ×70.3618(kg /kmol) 3600 s = 4.576075 kg/s Therefore, Dc= √ 4 × 4.576075 kg/ s π ×2.0949 kg /m3 ×1.37966 m/ s Dc = 1.419814 m 1 /2 ] Approximate to 1.5 m. 5.5.2.8.2 Column Area The column area can be calculated from the calculation of internal diameter A c= πD 2c 4 1.5 ¿ ¿ ¿2 π¿ A c =¿ A c =1.767146 m2 5.5.2.8.3 Column Diameter Flooding velocity can be estimated from the correlation by Fair (1961): U f =Ki √ ρ L −ρV ρV Equation 5.43 Where, Uf Ki = = Flooding vapour velocity, m/s, based on the net column cross-sectional area. A constant obtain from (Coulson and Richardson, 2009) The liquid-vapour flow factor, Fv (Coulson and Richardson, 2009) is given by: F LV = LW VW LW VW √ ρV ρL = Liquid mass flow rate, kg/s = Vapour mass flow rate, kg/s Equation 5.44 Where: F LV , bottom= √ 406.5023 8.7916 234.1308 1611.755 = 0.1282 F LV , bottom= √ 192.5537 2.0949 234.1308 1000.145 = 0.03764 Take plate spacing as 0.55 m. (Plate spacing is the important for determined the overall height of column. Plates spacing from 0.15 m to 1 m are normally used. The spacing chosen depends on the column diameter and operating conditions. For columns above 1 m diameter, plate spacing 0.3 to 0.6 m will normally be used (Sinnot and Towler, 2009). This will be revised as necessary. Therefore, the selected plate spacing is satisfactory since the column diameter is 1.5 m. From Figure 11.34: Flooding Velocity, Sieve Plate (Coulson and Ricardson, 2009) Bottom Ki = 0.0625 Top Ki = 1.1× 10-1 Hence: U f =Ki √ ρ L −ρV ρV Equation 5.45 Bottom U f =0.0625 √ 1611.755−8.7916 8.7916 = 0.8439 m/s Top U f =1.1 ×10−1 √ 1000.145−2.0949 2.0949 = 0.4009 m/s For design, a value of 80 to 85 percent of the flooding velocity should be used. Bottom Uf = 0.8439 m/s (0.85) = 0.7173 m/s Top 5.5.2.8.4 Uf = 2.4009 m/s (0.85) = 2.0408 m/s Maximum Volumetric Flow Rate Maximum volumetric flow rate Bottom Top 234.1308 (18.0274) 1 × 2.0949 3600 0.5597 m3/s Net Area, An Net area required Bottom 0.5205 m3/s = = 5.5.2.8.5 234.1308 (70.3618) 1 × 8.7916 3600 = = = volumetric flow rate flooding velocity = m (¿¿ 3/s ) 0.52051 0.7173( m/ s) ¿ = Top = V m(¿ RMM ) 1 × ρv 3600 ¿ = = 0.7257 m2 m 0.5597(¿¿ 3/ s) 2.0408(m/s) ¿ 0.2743 m2 As first trial, take down corner area as 12% of total. Column cross-sectional area: Bottom Top = 0.7257 0.88 = 0.8247 m2 = 0.2743 0.88 = 0.3117 m2 Column Diameter Bottom = Top √ = 0.9612 m √ = = 4 ×0.7257 π 4 ×0.2743 π 0.59097 m 5.5.2.9 Liquid Flow Pattern Maximum volumetric liquid rate = = = L w × RMM bottom 3600 × ρ L 406.5023 kg/ s ×70.3618 kg /kmol 3 3600 s × 1611.755 kg /m 4.9294 × 10-3 m3/s The choice of plate type will depend on the liquid flow rate and column diameter. Table 5.41: Advantages and disadvantages of different type of plates (Sinnot and Towler, 2009) Type of plate Advantages The cheapest among the Disadvantages plate by vapour flow others Sieve Plate (Perforated Simplest type through the holes in the Highest capacity rating plate Low flow rate of liquid Good operating range Plate) Liquid is returned on the will ‘weep’ through the holes, reducing the plate efficiency Bubble Cap Plate Traditional and oldest Expensive Risers ensures that level Highest pressure drop of liquid maintained on Lowest capacity rating tray at all vapour flow rate Operate efficiently at low vapour rates Valve plate (Floating Cap Plate) Operates efficiently at lower Expensive than sieve flow rate than sieve plates Valve closing at low vapour rates plates Pressure drop higher than sieve plates In this project, a sieve plate has been selected due to various factors such as cost, capacity, operating range, efficiency and pressure drop. Figure 5.18: Normal operation of sieve plate (Sinnot and Towler, 2009) 5.5.2.10 Provisional Plate Design Column diameter, Dc Column area, Ac Down corner area, Ad Net area, An Active area, Aa Hole area, Ah 5.5.2.11 = = = = = = = = = = = 1.5 m 1.767146 m2 0.12 × 1.767146 m2 (at 12%) 0.2121 m2 Ac- Ad 1.767146 m2 – 0.2121 m2 1.5551 m2 Ac- 2Ad 1.767146 m2 – 2(0.2121 m2) 1.342946 m2 0.08 m2 (take 6% of Aa as second trial after 10%) Weir Length Ad 0.2121 × 100 = ×100 =12 Ac 1.767146 From figure 11.39: Relation Between Downcomer Area And Weir Length (Coulson and Richardson, 2009) Iw =¿ Dc 0.76 Iw = 0.76 × 1.5 m = Weir height, hw Hole diameter, dh Plate thickness 1.14 m = = = 50mm 5mm 5mm Take Area of 1 hole, Alh = π d 2h 4 = 0.005 ¿ ¿ ¿2 π¿ ¿ Equation 5.46 = 1.963 ×10−5 m 2 Number of holes per plate, N h= Ah Alh Equation 5.47 ¿ 0.08 −5 1.963 ×10 ¿ 4075 holes 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. It is determined based on the condition inside the column. 5.5.2.12 Column Height Height = (no.stages-1)(tray spacing)+(tray spacing×2)+(no.stages-1)(thickness of plate) Height = (30 – 1)(0.55 m)+(0.55 m × 2)+(30 – 1)(0.005 m) = 17.195 m = 18.9145 m (including 10% safety) Approximate = 19 m Table 5.42: Summary Item Column Diameter Dc Value 1.5 m No. of Plates 30 Plate Spacing 0.55 m No. of stages below feed 2 Plate thickness 5 mm Total Column Height, Ht 19 m Down Corner Area, Ad Column Area, Ac Net Area, An Active Area, Aa Hole Area, Ah Number of Holes 0.2121 m2 1.767146 m2 1.5551 m2 1.342946 m2 0.08 m2 4074 Weir Length 1.14 m Weir Height 0.05 m(standard) 5.5.2.13 Weir Liquid Crest Check weeping (Enough vapour to prevent liquid flow through hole) LW (RMM ) 3600 S Maximum liquid rate = 406.5023 (70.3618 ) 3600 S = Minimum liquid rate = 7.9451 kg/s = 0.7 ×7.9451 kg/s (at 70% turn-down ratio) = 5.5615 kg/s The height of the liquid crest over the weir can be estimated using the Francis weir formula. For a segmental down corner this can be written as: how =750 Lw ρl (l w ) 2 /3 ( ) Where, lw Lw = Weir length = Liquid flow rate, kg/s ρl = Liquid density Maximum ,h ow =750 ( 7.9451 kg /s 1611.75 kg /m3 ×1.14 m 2 /3 ) =19.906 mm liquid 5.5615 kg /s Minimum , how =750 1611.75 kg /m3 ×1.14 m ( = 15.694 mm liquid At minimum rate, clear liquid depth, 2/ 3 ) how +hw =( 15.694+50 ) mm liquid = 65.694 mm liquid From the figure 11.37: Weep-point correlation (Eduljee, 1959), we can find K 2 (a constant, dependent on the depth of clear liquid on the plate) When how +hw =65.694 mm liquid K2 = 30.4 5.5.2.14 Weeping Point Liquid flow through sieve-plate perforations occurs when the gas pressure drop through the perforations is not sufficient to create bubble surface and support the static head of froth above the perforations. Weeping can be harmful in that liquid tends to short-circuit the primary contacting zones. Minimum vapour velocity through the holes based on the holes area can be calculated using Eduljee (1959) equation: ρ ¿ ¿ ¿ K 2−0.9(25.4−d h) U h (min)= ¿ Uh dh K2 = = = Minimum vapour velocity through the holes based on the holes area Hole diameter, mm A constant, dependent on the depth of clear liquid on the plate. Obtain from figure 11.37: Weep-point Correlation (Coulson & Richardson, 2005) ρ ¿ ¿ ¿ K 2−0.9(25.4−d h) U h (min)= ¿ Where, 1 /2 8.7916 ¿ ¿ 30.4−0.9 (25.4−5) ¿ ¿ ¿ 4.06757 m/s Actual minimum vapour velocity = minimum vapour rate Ah 3 ¿ 0.7 (0.52051 m / s) 0.08 = Actual minimum vapour velocity 4.5545 m/s ¿Uh So, minimum operating rate will be below the weep point. 5.5.2.15 Plate Pressure Drop 5.5.2.15.1 Dry Plate Drop The pressure drop through the dry plate can be estimated using expression derived for flow through orifices: 2 [ ] U hd =51 h Co ρv ρL Equation 5.48 Uh Co Uh = = = Velocity through the holes Orifice coefficient Where, volumetric flowrate hole area , A h Uh = 0.52051m 3 /s 2 0.08 m = 6.506375 m/s When: plate thickness 5 mm = =1 hole diameter 5 mm And: 2 Ah 0.08 m = ×100 A a 1.342946 m2 =6% So, orifice coefficient value can be obtained from figure 11.42: Discharge Coefficient Sieve Plates (Coulson & Richardson, 2009) Co = 0.81 Therefore, dry plate drop: 2 [ ] 6.506375 8.7916 hd =51 0.812 1611.755 =17.86097 mm liquid 5.5.2.15.2 Residual Head Residual head can be calculated using Hunt et al (1955) equation: hr = 12.5 × 103 ρL hr = 12.5 × 10 1611.755 Equation 5.49 3 = 7.75552 mm liquid 5.5.2.15.3 Total Plate Pressure Drop ht =hd + ( h w + how ) + hr ht =17.86097+ ( 50+15.694 )+ 7.75552 ht =91.31049 mm liquid 5.5.2.15.4 Downcomer Liquid Back-up Downcomer pressure loss: hap=hw −10 mm hap=50 mm−10 mm = 40 mm Where hap is the height of the bottom edge of the apron above the plate. Area under apron, A ap=hap (I w ) = 40×10-3 m (1.14 m) = 0.0456 m2 ≈ 0.05 m2 Where A ap is the clearance area under downcomer. As this is less that Ad= 0.965 m2, this equation can be used to calculate the head loss in downcomer. hdc =166 2 [ ] Lw ρL Am Equation 5.50 Where, hd = Head loss in down comer, mm c Lw Am = Liquid flow rate in down, kg/s = Either the down comer area Ad or the clearance area under A ap :whichever is the smaller down comer, hdc =166 [ 7.9451 1611.755 ×0.05 2 ] = 1.6135 = 2 mm Back-up in the down comer, (hb), hbc =h w + how + hht +h dc Equation 5.51 hbc =50+15.694+ 91.31049+2 = 159.00449 mm liquid or 0.159 m For safe design the clear liquid back-up should not exceed half the plate spacing it, to avoid flooding. Therefore, hb < 1/2 (plate spacing + weir height) 0.159m < 1/2 (plate spacing + weir height) 0.159m < 1/2 (0.55 + 0.05) m 0.159m < 0.3 m So, tray spacing is acceptable (to avoid flooding) 5.5.2.15.5 Downcomer Residence Time Sufficient residence time must be allowed in the downcomer for the entrained vapour to disengange from the liquid stream; to prevent heavily ‘aerated’ liquid being carried under the downcomer. At time of at least 3 seconds is recommended. (Couper, 2005) Check residence time t r= A d × hb × ρ L Lw Equation 5.52 ¿ 0.2121 ×0.159 × 1611.755 7.9451 = 6.84128 s ¿ 3 s (satisfactory) 5.5.2.15.6 Check Entrainment Entrainment can be estimated from the correlation given by Fair (1961). The percentage flooding is given by: flooding= U n actual velocity (based on net area) Uf Equation 5.53 U n= Uf An ¿ 0.5205 1.5551 ¿ 0.3347 m/s flooding= 0.3347 ×100 0.7173 = 46.66 % From figure 11.36: Entrainment Correlation for Sieve Plates (Coulson & Richardson’s, Chemical Engineering Design, 2009) Fractional entrainment, φ φ = 0.03 value is below 0.1, so column diameter which is proposed earlier is acceptable. 5.5.2.15.7 Perforated Area From (Coulson & Richardson’s, 2009) At= ¿ Iw Dc Equation 5.54 1.14 m 1.5 m = 0.76 Therefore, θc = 99 ˚ Angle subtended at plate edge by unperforated strips, 5.5.2.15.8 (180 – 99) = 81˚ Mean Length, Unperforated Edge Strips ¿( Dc −0.05) × π × 81 180 ¿(1.5−0.05)× π × 81 180 = 1.1745m 5.5.2.15.9 Area of Unperforated Edge Strips = (50×10-3) m × (1.1745) m = 0.058725 m2 5.5.2.15.10 Area of Calming Zones = 2 × 0.05 (Iw – 2 × 50 × 10-3) m2 = 2 × 0.05 (1.41 – 2 × 50 × 10-3) m2 = 0.113 m2 5.5.2.15.11 Ap = = = = Total Area Available for Perforation, Ap Active area – (Area of unperforated edge + Area of calming zones) 1.342946 m2 – (0.058725 + 0.113)m2 1.171221 m2 Ah 0.08 m2 = A p 1.171221 m2 = 0.0683 From figure 11.41: Relation between hole area and pitch (Coulson & Richardson’s, Chemical Engineering Design, 2009,) satisfactory , range normally within 2.5 Ip =3.34( ¿4.0) dh 5.5.2.15.12 Trial Layout 50 mm 1.14 m θc 50 mm Figure 5.19: Trial layout 1.5 m 5.5.3 Summary of Chemical Design of T-102 Table 5.43: Summary of chemical design Item Value Column diameter, Dc 1.5 m Plate spacing, ts 0.55 m Downcomer area, Ad 0.2121 m2 Downcomer Material Stainless steel, type 304 Plate material Stainless steel, type 304 Plate Thickness tτ 0.005 m Column Area, Ac 1.7671 m2 Net Area, An 1.5551 m2 Perforated plate area, Ap 1.1712 m2 Active area, Aa 1.3429 m2 Total hole area, Ah 0.08 m2 Weir length 1.14 m Weir height 0.005 m Hole diameter 0.005 m Number of holes 4075 Flooding % 47 % REFERENCES (On-line) http://www.chemeng.queensu.ca/courses/CHEE218/projects/HeatExchanger (25 January 2011) (On-line) http://www.engineeringtoolbox.com ( 17 February 2011) (On-line) http://www.homebuildingremodeling.com ( 2 March 2011) (On-line) http://www.scribd.com/doc/38723673/Heat-Exchanger (21 February 2011) (On-line) www.aksteel.com ( 27 February 2011) A. 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