Evaporator Mass and Energy Balance, and Mechanical Design - Essay Example

Evaporator mass and energy balance, and mechanical design Evaporator mass and energy balance and mechanical design The main purposeof evaporator in a system is to carry out evaporation process. In the evaporation the concentration of the feed (product) is usually increased by boiling out (evaporating out) solvent which is usually water (GEA 2012). The concentrate (evaporator end product) normally has the optimum solid content (concentration) that has the desired economics of operation and quality (Wacharawichanant 2012). As in this case the manufacture of the paracetamol was carried out using p-nitrochlorobenzene (PNCB) rout which involves five main stages, namely: i. Neutralization: Neutralization reaction sodium hydroxide (caustic soda) and p-nitrochlorobenzene to form water and sodium p-nitrophenol. ii. Nitration: This reaction involves the nitration of the produced sodium p-nitrophenol using nitric acid and sulphuric acid to form p-nitrophenol. iii. Reduction: The produced p-nitrophenol is reduced by iron (powered) to produce p-aminophenol. iv. Acetylation: Acetylation of the produced p-aminophenol using Acetic Anhydride to form Acetic Acid and Paracetamol. v. This Acetic Acid stream contains water and some traces of Paracetamol. It is fed into a fractionating column (evaporator) to remove acetic acid (as vapour at 89.12 degrees celcius) and produce paracetamol concentrate and water. The acetic acid which is the vapour is then cooled to obtain its distillate. From the cooler the acetic acid stream containing paracetamol and water (product) enters evaporator for concentration (increasing its concentration) by removal of certain portions of water from the product. Consider the flow diagram below; Figure 1: Simplified flow diagram of Evaporator of pharmaceutical plant (concentration of the paracetamol) There are two types of balances in evaporator operations; material mass balance and energy balance. Mass balance According to mass balance principle, mass of materials entering a system is equal to the mass of materials leaving the system assuming that there are neither losses within the system nor losses associated with the processes under which the materials entering the system undergo (Wacharawichanant 2012). This can be expressed mathematically as shown below, As of the case of processing paracetamol, the rate of mass of products (paracetamol) entering the evaporator is equal to rate of mass of mass of concentrate leaving the system plus rate of mass of vapour leaving the system. Mass balance can be expressed mathematically as shown below, (consider Figure 1) Let, i. The rate of mass of the product (dilute paracetamol) entering the evaporator be F. ii. The rate of mass of the concentrate (concentrated paracetamol) leaving the evaporator be C. iii. The rate of mass of the Vapour leaving the evaporator be V. Therefore, according to mass balance principle, F = Products entering the Evaporator include: Acetic acid, Water, and paracetamol. C = Concentrate: concentrated paracetamol and acetic acid. V = water vapour. Given the following (obtained from the report 2) The mass of the products from cooler, (the mass of Acetic acid + Water + paracetamol) was 2319 Kg per hour (Kg/hr) (From L-37) of the Engineering line diagram (report 2). After evaporation the mass of the concentrate obtained per hour was 1380.85 Kg per hour. Therefore, the mass of the vapour (acetic acid vapur) assuming that no loss occurred as a result of evaporation process can be calculated as follows. According to mass balance Given that F = 2319 Kg/hr, and C = 1380.85Kg/hr Therefore, the rate at which acetic vapour is removed from the evaporator (fractional distillatory) is 938 Kilogram per hour. In order that required quantity of paracetamol per year is obtained, and assuming other factors are constant, the product from the cooler should be fed at rate of 2319 Kilograms per hour or 55656 Kilograms per day (assuming that the plant will work 24 hours per day). Concentration Considering the concentrates alone Let the concentration of Feed (product) be CA, and concentration of concentrate be CB Therefore, evaporation ratio (e) can be calculated as follows For this case the evaporation ratio can calculated as follows, Therefore, evaporation ratio (e) for case is 1.68 A measure of concentration process that takes place in the evaporator is what is known as evaporation ratio, and it is calculated as shown above. When concentrating the solution F (dilute paracetamol) at a steady rate, the concentration rate starts slowly but increases tremendously to a maximum value(GEA 2012). At this maximum value, none of the solvent (Vapour, V) will be still in the solution (concentrate, C) (APV 2011). When the initial concentration of the feed (product) is low, the concentration curve would be steep. It can therefore be concluded that: “the lower the CA (Product concentration) the steeper the concentration curve”. This relationship (the relationship between initial concentration (product concentration) and gradient of the concentration curve) is amongst the most essential components in the controlling of the evaporating plants (GEA 2012). It is important to note that while calculating the evaporation ratio the mass flow rates rather than volume flow rates are used. The diagram below show mass typical evaporation ratios and concentrations of feeds (products) (CA) and concentrates (CB). Figure 2: Relationship between evaporation ratio (e) and concentrations of Feed (Products) and concentrate (GEA 2012). Energy Balance Assuming no is heat lost through convection and/or radiation, Energy balance principle states that total heat energy entering a system is equal to the total energy leaving the system. Amount heat in any system can be calculated by using the formula below. Q=mCp∆t Where, Q = Energy of Component (Kilo joules, kJ) m = Mass of Component (Kilogram, kg) Cp = Specific Heat Capacity of Component (Kilo joules, kJ/kgK) ∆t = Change temperature between (Kelvin, K) The total energy in a process stream is then given by: Q =∑mCp×∆t Therefore, according to energy balance, total heat energy entering the evaporator heat energy associated with product (Acetic acid + Water + paracetamol) is equal to the total heat energy leaving the system (heat energy of acetic acid vapour or distillate + heat energy of concentrate). Let heat energy associated with product (Acetic acid + Water + paracetamol) be Q1, and that associated with materials leaving the system be Q2. Therefore according to energy balance; Q1 = Q2 Q2 = Qac + Qc Where, Qac = Heat energy associated with acetic acid vapour Qc = Heat energy associated with concentrate mfCpf∆tf = Q2 = Heat energy associated with materials leaving the system macCpac∆tac = Heat energy associated with acetic acid vapour mcCpc∆tc = Heat energy associated with the concentrate (APV 2011) Assumptions made while carrying out Energy balance calculations i. The ambient plant temperature is 25⁰C. ii. There no heat loss through the plant’s vessels, pipelines and other equipment. iii. The process of manufacturing the paracetamol is steady; therefore the sum of the total energy of the system (accumulation) is zero. iv. Energy balance basis time one day (that is, any amount of calculated heat quantity is on daily basis). v. The system’s datum temperature is 0 ⁰C. Material entering the system (Evaporator) Cp = specific Heat capacities Cp for Carbon (C) at 25 degrees = 9.75 KJ/KgK Cp for Hydrogen (H) at 25 degrees = 28.7 KJ/KgK Cp for Oxygen (O) at 25 degrees = 25.9 KJ/KgK Cp for Nitrogen (N) at 25 degrees = 28.45 KJ/KgK Using Koops Rule the specif heat capacities for the various products can be calculated as follows; Cp for Paracetamol (C8H9NO2) at 25 degrees (298 K) = (8 * 9.75 + 9 * 14.2 + 14.2 + 13) = 233KJ/KMolK Cp for Water (H2O) at 25 degrees (298 K) = 75.37KJ/KMol/K Paracetamol (C8H9NO2) Cp for Acetic acid (C2H4O2) at 25 degrees (298 K) = (2 * 9.75) + (4 * 14.4) + (2 * 14.05) = 19.5 + 57.6 + 28.0 = 105.1 KJ/KMol/K Amount of energy available in the Feed = 233 + 75.37 + 105.1 = 413.47KJ/KMol/K Amount of energy that is possessed by the evaporating steam Assumptions The initial temperature of steam was 25 degrees (298 K) Steam temperature = 100 (398 K) Temperature difference = Initial temperature – Final temperature Using energy balance Steam energy entering the evaporator = Concentrate energy + Condensate energy + Vapour energy Concentrate Energy + Vapour energy = 650882 Condensate energy = 538813 KJ/day. Steam energy entering the evaporator per day = 650882 + 538813 = 874593 KJ/day Evaporator’s mechanical Design The following assumptions are made in the Evaporator’s heat exchanger. i. Tubing in which steam is flowing are made copper. ii. Transfer of energy occurs from steam via the copper tubing walls in water (feed, Acetic acid + Paracetamol + water). iii. The heat transfer coefficient of copper is 1160W/m2. iv. Steam temperature is 100 degrees Celsius. v. Feed (product) temperature is 25 degrees Celsius. Therefore, the evaporator’s heating surface required can be calculated as follows. Where A = Evaporators Heating Surface K = Heat transfer coefficient Q = Heat transferred across the surface in KW T = Temperature difference between the two exchanging liquids (Wacharawichanant 2012) Heat Transferred across the surface per day is given by 650882 KJ/day = 650882 X 1000 = 650882000 J/day Therefore heat transferred across the evaporator in J/s is calculated as shown below; Therefore, Area of the Evaporator’s heating surface is calculated as follows Therefore the heating surface of the evaporator should be 0.087m2 Typically, the distance between the tube plates in evaporator are normally taken to be 2 meters. Therefore, in the Evaporator too this distance will be approximately 2 meters. Tube take down diameters can calculated using the formula shown below Therefore the take down diameter of the evaporator should be 0.0058m Feed pump power consumption The feed pump (product) must be able to pump the feed at 2319 Kilograms per hour (0.644Kg per second). Density of acetic acid = 1049 Kg/m3, Density of paracetamol = 1029 Kg/m3, and density of water = 1000 Kg/m3 Assuming that feed has density of 1049 kg/ Kg/m3 (density of acetic acid) Therefore pump rate (m3/second) can be calculated as follows; Assuming pump head of 3m Power in horse power is calculated as follows, Q = flow rate in gallons per min, H = Total head in feet, WHP = horsepower Q = 0.000614m^3/s = 9.732gal/min, H = 3 m = 9.84252feet Therefore, An 18 watt electric pump will be enough to maintain the flow rate. In order that efficient heat exchange takes place, a similar pump should be installed pump steam in a counter direction with the evaporator. Process Flow Diagram for the process Figure 3: The Process Flow diagram Engineering Flow Diagram Figure 4: Engineering flow diagram for the evaporator (GEA 2012) References APV., 2011. Evaporation Handbook. New York: An Invensys company. GEA., 2012. Evaporation Technology. Columbia: GEA Process Engineering Inc. Wacharawichanant, S., 2012. Heat Transfer Operations: Evaporation. Bangkok: Silpakorn University. Read More