The Production of Low-Cost Bio-Diesel - Case Study Example

Title of Report Prepared by Name of Company Month, Year Student Name Student Number 1 Executive Summary The company uses latest technology to provide an effective bio diesel plant that is economical and effective. The proposal design adheres to the requirements and constraints that affect the project. The proposal is below the $12000 budget using Heater model (H55), Pump type (P2) and the material for the tank is stainless fabricated. The design will produce a total of 12000 Litres of bio diesel within 8hrs. The report contains: Introduction – gives a background information goals and constraints of the project. Technical analysis – it has five derived equations and calculations of volumes and mass of the reactants and raw material used in the oil production. It also gives the total cost for putting up the plant. Design solution – all equations are derived in a spreadsheet with actual numbers. The equations in the spreadsheet are based on volume, mass and density. Recommendation - lastly the report will recommend that the pump to be used in the plant should be P2 and the heater model should be H22. In addition the material for the tanks should be stainless fabricated. Table of Contents 1 Executive Summary 2 2 Technical Analysis 3 2.1 Size and volume of the tanks 5 2.2 Number and model of electric heater 8 2.3 Type of Pump and rating 10 2.4 Processing system 11 2.5 Cost of tanks and heater 11 3 Design Solutions 12 4 Recommendation 13 References 14 Introduction Bio-neutral production facility presented by design is an economical and accurate in the production process. The plant has a capability to produce up to 12000 Litres per 8hr shift. The plant utilises readily available raw materials and straightforward procedures through out the process. The oil goes through four main stages of filtration mixing reaction with chemicals and purification. The presentation displayed all these sections, and step by step procedure carried out by the oil processing plant. The process is well monitored. All the raw materials are quantified and mixed within the specified amounts for a complete reaction and accurate results. Some of the reactants and raw material used include; methanol, sodium hydroxide, and fresh canola oil. This report keenly looks into the processing system and analyses how the plant meets its production target, as well as the total cost for putting up this plant. 2 Technical Analysis The technical design of the whole project indicates that the plant minimises on cost completely. At the same time, the plant is able to meet its target of producing 12000 litres within 8 hours. The plant contains fully enclosed tanks and provides space for the required volume. This enables the tank to cut down on energy loss and consistent pressure in the pipes and pumps. The report will work out the volume of the tank the type of pump and model and model and power input of the electric heater. All these will be determined by the cost. The design portfolio suggests that the resources and raw material used to construct the processing plant are adequate to make the plant succeed. The budget for acquiring all the resources and raw material is $12000. The plant has six tanks, (tank A, B, C, D, E and F). Tank A is used for storing filtered canola oil which is directed to tank E. Tank B is used for storing methanol which is directed to tank D. Tank C holds sodium hydroxide which is also directed to tank D. Tank D is used for mixing the sodium hydroxide and methanol. Tank E is the reaction tank where the mixture of sodium hydroxide and methanol reacts together with canola oil. At the completion of the reaction glycerol and bio diesel are formed. Glycerol is directed to tank f and bio diesel is circulated via the piping to remove impurities. Figure 1: Sample of biodiesel production plant The table below shows the specifications of the ingredients used in the process. Table 1: Ingredients Specification Material Variable (letter) Value or equation Canola oil- density ρo 920Kg/m3 Methanol- density ρm 791 Kg/m3 Sodium hydroxide- density ρs 2130 Kg/m3 Glycerine- density ρg 1260 Kg/m3 Oil/Methanol/Sodium Hydroxidemix-density(average) ρc 911Kg/m3 The volume of raw materials and reactants used are determined in the calculations below. 2.1 Size and volume of the tanks Formula for calculating volume = mass/ density (Equation 1) Calculation of the total volume of canola oil, Methanol and sodium hydroxide Canola oil = V litres Methanol = 0.2V litres Sodium hydroxide = 0.007kg * 0.2V Mixture in tank D = Methanol + Sodium Hydroxide {0.2Vlitres + (0.007kg /2130Kg/m3)0.2V} = 0.2V litres + [(3.286385*10-6)1000litres] 0.2V The reactants in tank E = V + 0.2V+ (0.003286385*0.2V) = 1.2V + 0.000657277V Litres In order to determine V we first determine the volume of glycerol. Glycerol is 8% of the content in the reaction tank i.e. 0.08 * 1.200657277V = 0.09605258V Calculation of V Total Volume of the raw materials = Total volume out of the products 1.200657277V = 0.09605258V + 6000 1.200657277V - 0.09605258V = 6000 1.104604697V = 6000 => V = 5431.80742966 L Volume in the reaction tank = 1.200657277 * 5431.80742966 L = 6521.7391177 L Tank E = 6.5217391177 m3 Therefore; volume for Tank A = V/1000 which is 5431.80742966 L converted to volume 5431.80742966 /1000 = 5.43180742966 m3 Volume for tank B = 0.2V/1000 = > (0.2 * 5431.80742966) / 1000 => 1086.3614859/1000 = 1.0863614859 m3 Tank B = 1.0863614859 m3 Volume of Tank C = volume of sodium hydroxide used Sodium hydroxide = 0.000657277V => 0.000657277 * 5.43180742966 = 0.0035702 m3 Tank C = 0.0035702 m3 Tank D = tank B + tank C => 1.0863614859 m3 + 0.0035702 m3 = 1.0899316859 m3 Tank D = 1.0899316859 m3 Tank F = 8% of Tank E => 0.08 * 6.5217391177 m3 = 0.521739129416 m3 Tank F = 0.521739129416 m3 2.1.1 Mass of raw material used Canola oil From the equation; Density = Mass/Volume, Therefore, Mass = Density * Volume (Equation 2) The density of Canola from the table above is 920Kg/m3 Mass of Canola is, therefore, given by Density of Canola * Volume of tank A => 920Kgm-3 * 5.43180742966 m3 = 4997.2628352 Kg Methanol Mass of Methanol needed for 1 process (Mass = Density * Volume) Density of Methanol is 791Kg/m3 therefore, mass will be => 791 * 1.0863614859 = 859.3119353 Kg Sodium hydroxide Mass of Sodium hydroxide (Mass = Density * Volume) Density of sodium hydroxide is 2130Kg/m3 therefore, mass will be => Mass = 2130 * 0.0035702 = 7.604526 Kg Methoxide Mass of methoxide = (mass of methanol + mass of sodium hydroxide) => 7.604526 + 859.3119353 = 866.9164613 Kg Mixture in tank E (Oil, Methanol and Sodium hydroxide combined) Mass of the mixture in tank E = Density * Volume (Density of the mixture is 911Kg/m3) = 911 * 6.5217391177 = 5941.3043362247 Kg Mass of glycerine (Mass = Density * Volume) 1260 * 0.521739129416 = 657.3913031 kg The volumes of the tanks and the mass they hold are summarised in the table below; Table 2: Volume of Tank and Mass of Raw Material they contain Tank Volume of tank Raw material and reactants Mass of raw material A 5.4318074 m3 Canola oil 4997.2628352 Kg B 1.0863615 m3 Methanol 859.3119353 Kg C 0.0035702 m3 Sodium hydroxide 7.604526 Kg D 1.0899317 m3 Methoxide 866.9164613 Kg E 6.5222731 m3 Oil, Methanol and Sodium hydroxide combined 5941.3043362 Kg F 0.5217391 m3 glycerine 657.3913031 kg 2.2 Number and model of electric heater From the portfolio the amount of heat required to raise 1kg of the total mixture in tank E is 267380.586kj This was calculated as below; The formula for calculating quantity of heat is Q= cmΔT (Equation 3) Where: Q = Quantity of heat added in Kilo Joules (KJ) c = Specific heat capacity of the liquid in KJ/Kg/C (1.8KJ/KG/C) ΔT = change in temperature (50-25=25) m = Mass of the liquid in tank E (as calculated above) = 5941.3043362 Kg Assumption given: All energy from the heater is transferred to the liquid (100% efficiency). =>Heat supplied by heater = Heat absorbed by liquid Heat absorbed by liquid = cmΔT 1.8*5941.7908*25 = 267358.695129 KJ =267358695.129 J Converting joules to kilowatt per hour (Power is energy per second) 1 Watt is 1 Joule per second 1Kwh = 1000*3600seconds = 1Kwh = 3.6 x 106 J Conversion constant 3.6 x 106 J = 1Kwh => 267358695.129 J = (267358695.129 J *1)/ 3.6 x 106 J = 74.2663042025 Kwh Electric heater rating is given in kW in the client brief The process in the reaction tank takes 1.75hrs The heater used must provide 74.2663042025 kWh in 1.75 hrs Rating of the heater is therefore given by =>74.2663042025/1.75 = 42.4378881157 kW =42.44kW We refer to the table of heater specifications to identify the exact model and rating. Table 3: Heater model Specification Heater Model Input Power rating Ph (kW) Cost Ch ($) H11 11 1000 H22 22 2000 H33 33 2700 H55 55 4000 H80 80 6500 The H55 is much appropriate for the plant. It produces 55kW/h which gives allowance of up to 12.56kW to be lost in the air every hour. The tanks used in the plant are cylindrical in shape since they are more efficient and economical. 2.3 Type of Pump and rating The energy required to raise the mixture. Potential energy = m*g*h (Equation 4) m = mass of the liquid g = gravitational constant (9.80665 m/s2) h = height (m) 1m (height of tank E) If power = energy/time (joules/s), from equation 4 we can say that power = mgh / t Since mass (m) = Volume (V)* density (Ρ) Power = [Volume (V)* density (Ρ) *g*height (h)]/time (t) = kgm2s-3 Tank E contains 6522.2731 litres of liquid collected after the reaction. 8% of this is glycerol (8% * 6522.2731 = 521.78185). The glycerol is drained off leaving 6000.49125 litres of bio diesel. To ensure efficiency 25% of the volume is circulated every minute. i.e. 6000.49125 Litres * 25% = 1500.1228125 litres/minute. The volume flow rate per second will be translated to 1500.1228125 litres/60 = 25.002046875 litres = 0.025002046875 m3 Power input for the pump = [Volume (V)* density (Ρ) *g*height (h)]/time (t) = kgm2s-3 [0.025002046875 m3 * 911 Kg/m3]* 9.80665 m/s2 = 223.36474024090078125 W 223.36474024090078125 W /1000 = 0.22347402409 kW Power input by the pump = 0.2234 kW. From the pump specification table we select the appropriate pump. Table 4: Pump Specification Pump Type Input power rating Pp (kW) Efficiency ηp Cost Cp ($) P1 0.15 0. 90 500 P2 0.25 0. 90 900 P3 0.50 0. 90 1250 P4 1.00 0. 90 1500 The pump type P2 is much appropriate for the plant. It produces input of 0.25 kW and gives an allowance of 0.02664 kW to be lost in the air. 2.4 Processing system With the analysed volumes of the tanks used in the bio diesel production process, the plant will be able to produce 6000L of bio-diesel in one process. For one complete process is 4 hrs. i.e. 15 minutes of mixing methanol and sodium hydroxide in Tank D, 1.75hrs of mixing sodium hydroxide oil and methanol in the reaction Tank E, 1hr for allowing the reaction to settle and 1hr for dry washing process. (15/60 + 1.75 + 1 + 1 = 4 hrs.). This implies that the plant will run two complete processes to meet the target of 12000L. The design parameters described above shows that the plant is economical and efficient. 2.5 Cost of tanks and heater Heater specified for the plant will be H55 while the pump that suits the plant is P2. From the design portfolio the tank should be made of stainless fabricated the features are shown below; Table 5: Tank material Specification Material Convective heat transfer coefficient of the tank surface (W/m2/K) Cost per unit ($/m2 of tank surface area) Stainless steel – fabricated (any size) 32 100 Polypropylene tanks - fabricated (max size 3000L) 2 40 Table 6: Total cost of the plant Cost Heater model (H55) $ 4000 Pump type (P2) $ 900 Tank material (total surface of all the tanks used = 14.264m2) 14.264*100) $ 1426.4 The total cost of the plant is $6326.4 3 Design Solutions In order to identify design solution and selection criteria a comparison of elements that may influence selection criteria are looked into thoroughly. The major consideration is to ensure the design meets production target as well as the constraints. The decision for selecting an appropriate heater model widely depends on the amount of heat input required. In the graph below it shows the comparison of heat input by the heaters with relevant cost. Cost is not a strong determinant in the selection of heater model. The cost of the heater increases with the increase in heat input produced by the heater. Figure 2: comparison between cost and heat input of the heater models Similarly selection of pump will depend on the appropriate heat input required to pump the liquid. Figure 3: comparison between cost and energy input of the pump The graph shows increase in cost with the increase in heat input as efficiency remains constant. Therefore having an appropriate pump will be determined by the amount of heat input required. 4 Recommendation The cost of the material used plays a crucial role in determining the commercial viability of the whole plant. The tank should be made of stainless fabricated material. The heater model H55 is much appropriate for the plant since it produces adequate heat in the reaction tank. The pump type P2 is also appropriate for the plant. The balance that remains from the budget is $5673.6. This remaining cash will be utilised in the purchase of reactants and raw material. References ‘Client Brief.’ University Publication. University Publishers. ‘Indian inventor develops a process technology for the production of low-cost bio-diesel’2010, Indian Patents News, pp. 13. Iliopoulos, C & Rozakis, S 2010, ‘Environmental cost-effectiveness of bio diesel production in greece: Current policies and alternative scenarios’, Energy Policy, vol. 38, no.2, pp. 1067. Rossi, C C, Cardozo-filho, L & Guirardello, R 2012, ‘Parameter estimation and thermodynamic model fitting for components in mixtures for bio-diesel production’. Read More