Technical Analysis of Design Tanks - Assignment Example

TANKS DESIGN REPORT Prepared by Showb Inc April, 2012 Student Name Student Number 1 Introduction This report will present technical analysis of design section 1, which is dealing with the size of each tank. This design section will present the dimensions of all tanks that will be used in the project, volume of the tanks, and maximum mass of material held by each tank. The report will highlight how this section of the design will/may impact other sections and conclude with key findings and equations that will be used in the system model. 2 Technical Analysis Design section 1 The size of each tank (A to F) – multiple processing system may be considered (Dimensions, Volume (m3), maximum mass of material held (kg)). Dimensions Dimensions of the tanks will vary because they are not meant to hold the same capacity Tank B, C, D and F dimensions will be equal while Tanks A and E dimensions will be twice the dimensions of Tank B, C, D and F since they are holding a bigger capacity than either of them. Shapes of the tanks will be in rectangular prism H W L Fig 1: Rectangular prism shape Where L = length (m), W = width (m) and H = height (m) Dimensions of tanks B, C, D and F will therefore be L (m), W (m) and H (m) Dimensions of tanks A and E will be 2L (m), 2W (m) and 2H (m) Volume (m3) Let V (m3) represents the volume of the tanks Tanks B, C, D and F will have equal volumes as indicated by the following equation V (m3) = L (m) * W (m) * H (m) (Equation 1a) = LWH (m3) Volumes of tank A and E will be V (m3) = 2L (m) * 2W (m) * 2H (m) V (m3) = 8LWH (m3) (Equation 1b) Maximum mass of material held (kg) From the fundamental equation D = M/V where D = density (kg/m3), M = mass (kg), V = volume (m3)[Ste11]. Therefore mass (M) is given by: M (kg) = D (kg/m3) * V (m3) (Equation 2a) Table 1: Ingredients Specifications (copied from Client brief, 2012) Material Variable (letter) Value or equation Canola oil – density ρc 920kg/m3 Methanol – density ρm 791 kg/m3 Sodium Hydroxide – density ρs 2130 kg/m3 Glycerin – density ρg 1260 kg/m3 Oil/Methanol/Sodium Hydroxide mix -density (average) -specific heat capacity ρc с 911kg/m3 1.80 kJ/kg/C Maximum mass of Canola oil held in tank A = density of canola oil * tank volume M (kg) = 920 kg/m3 * 8 LWH (m3) (Equation 2b) Maximum mass of Methanol held in tank B = density of Methanol * tank volume M (kg) = 791 kg/m3 * LWH (m3) (Equation 2c) Maximum mass of Sodium Hydroxide held in tank C = density of Sodium hydroxide * tank volume M (kg) = 2130 kg/m3 * LWH (m3) (Equation 2d) Maximum mass of Sodium hydroxide/Methanol mixture held in tank D = density of the mixture * tank volume From the client brief 1 litre of Methanol will be mixed with 7g of Sodium hydroxide. Therefore, the ratio of the mixture is 1:0.007 From this ratio the average density of the mixture will be 799 kg/m3 Therefore, M (kg) = 799 kg/m3 * LWH (m3) (Equation 2e) Maximum mass of Oil, Methanol, and Sodium Hydroxide mixture in tank E = density of the mixture * volume of the tank M (kg) = 911 kg/m3 * 8LWH (m3) (Equation 2f) Maximum mass of Glycerol held in F = density of Glycerin * volume of the tank M (kg) = 1260 (kg/m3) * LWH (m3) (Equation 2g) Cost of material Cost of the material which will be used to make the tanks can be determined by the following formula: Cost (C) = Tank surface area (m2) * Cost per unit ($) (Equation 3) Cost will vary depending on the material used, that is, Stainless steel or Polypropylene From the fundamental formula, surface area of a rectangular prism = 2 (lw * wh * lh) where l = length, w = width, h = height[Jea07]. Surface areas (A) of tanks B, C, D and F are equal and can be determined from the following formula: A (m2) = 2 (LW * WH * LH) (Equation 4a) Surface areas of tank A and E are equal and can be determine using the following formula: A (m2) = 128 (LW * WH * LH) (Equation 4b) Table 3: Tank material specifications (copied from Client brief, 2012) Material Convective Heat transfer coefficient of the tank surface (see assumptions) (W/m2/K) Cost per unit ($/m2 of tank surface area) Stainless steel – fabricated (any size) 32 100 Polypropylene tanks – fabricated (Max size 3000L) 2 40 Cost of the tanks depending on the material used can be determined using the following formulas: Cost of tanks B, C, D and F using stainless steel is C = 2 (LW * WH * LH) * $100 (Equation 5a) = $200 (LW * WH * LH) Cost of tanks B, C, D and F using Polypropylene C = 2 (LW * WH * LH) * $40 (Equation 5b) = $80 (LW * WH * LH) Cost of tanks A and E using stainless steel is C = 128 (LW * WH * LH) * $100 (Equation 5c) = $12800 (LW * WH * LH) Cost of tanks A and E using Polypropylene C = 128 (LW * WH * LH) * $40 (Equation 5d) = $5120 (LW * WH * LH) NB: Stainless steel cannot be used to make tanks A and E because the value has already exceeded the maximum capital cost of $12000. 3 Conclusion The technical analysis of design section 1 shows that it is possible to design the tanks using a rectangular prism shape in which the two bigger tanks dimensions are twice the smaller ones. It is possible to design all the tanks while using Polypropylene material. Using Stainless steel to design the tanks will be a problem because the cost of designing the two bigger tanks exceed the maximum capital cost, thus, making the whole project not feasible. Therefore, it is recommended that the tanks will be designed using the cheapest material. The key equations which will be used in modeling the system include: V (m3) = L (m) * W (m) * H (m) M (kg) = D (kg/m3) * V (m3) , and Cost (C) = Tank surface area (m2) * Cost per unit ($) 4 References Lists Ste11: , (Holzner, 2011), Jea07: , (Bell, 2007), Read More