section 3-2.pptx
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The document provides information to design a cylindrical storage bin for 450 tons of soda ash. It includes the material properties and design requirements. The summary is: The dimensions of the bin are estimated as a diameter of 5m, conical bottom height of 6.5m, and cylindrical height of 11.5m. The minimum bottom opening diameter is 141.5mm. With a bottom diameter twice the minimum, the solid flow rate out of the bin would be 146 kg/s. The vertical pressure on the bottom is 29.3 kPa and equivalent liquid pressure is 211 kPa.
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The document discusses common mechanical operations involving particulate matter including calculating the shape factor of different particle shapes, performing a sieve analysis to determine properties like specific surface area and average particle diameter, and using a correlation equation between cumulative fraction and particle size to calculate constants and the average surface volume diameter. Sieve analysis data is provided for a problem involving calculating values like specific surface area, average particle diameter, median diameter, and average surface volume particle diameter.
section 4.pptx
This document discusses belt conveyor design and calculations. It provides guidelines for belt thickness and idler spacing based on material density being conveyed. It also outlines the equations needed to calculate the mass of the belt, idlers, and material per unit length, which are required for power calculations. These include factors like the number of idlers per side, mass of individual idlers, belt thickness and density, and flow rate and velocity of the material being conveyed. An example problem is provided to demonstrate how to apply these calculations.
section 9.pptx
This document discusses three questions related to mechanical operations and flow through porous media:
1. It calculates the pressure drop across a catalyst bed packed with cylindrical pellets for an air stream, and how this would change with larger pellet size.
2. It estimates the inlet pressure of a catalyst tower given the outlet pressure, dimensions, flow properties, and contact time.
3. It calculates the porosity of a bed packed with Raschig rings given the pressure drop and flow properties of an air stream.
section 6.pptx
This document discusses common mechanical operations used to transport materials, including bucket elevators, screw conveyors, and pneumatic transportation. It provides examples of how to calculate the power required for a bucket elevator to lift crushed alumina, the design of a screw conveyor to transport fly ash horizontally, and the conveying capacity of a pneumatic transportation system moving ground rock vertically and horizontally through pipes and elbows. Key parameters discussed include bulk density, spacing, pitch, filling ratio, pressure change, flow rate, and power.
material science 1 .pptx
This document provides notes on material science concepts. It discusses key points about assignments, quizzes, reports and presentations. It also covers classifications of materials, basic concepts like deformation, stress, strain and mechanical properties including elasticity, ductility, brittleness, hardness and toughness. Graphs of stress-strain curves are presented and terms like elastic limit, yield point and ultimate tensile strength are defined. Students are instructed to submit assignments on time and follow the instructor's directions.
material science 2 .pptx
This document discusses material science concepts including stress, strain, Young's modulus, shear modulus, and bulk modulus. It provides definitions and formulas for these terms and includes examples of calculating stress, strain, elongation, shear stress, and changes in volume using given material properties and applied forces. Several practice problems are worked through applying these concepts to situations like stretching a cylinder, cutting a sheet, and determining changes to a bowling ball underwater.
Material 5.pptx
This document discusses phase diagrams for binary systems with different types of solubility between liquid and solid phases. It provides examples of phase diagrams for systems with: (1) complete liquid and solid solubility, (2) complete liquid but zero solid solubility, and (3) complete liquid and limited solid solubility. It also works through examples of determining phase compositions and amounts for steel alloys based on their placement on an iron-carbon phase diagram.
material 4.pptx
This document discusses phase diagrams and examples of using lever rule calculations to determine phase percentages and compositions. It provides an example phase diagram of the lead-tin system and asks questions about:
1) Solubility of tin in lead and lead in tin from the diagram
2) Using the Pb-10% Sn alloy and diagram to find phase percentages that form upon cooling
3) Masses of tin and lead in each phase for a Pb-10% Sn alloy cooled to 0°C
Section 3 .pptx
The document discusses various crystal structures including simple cubic, body centered cubic, and face centered cubic structures. It then discusses atomic packing factors for these different structures. Next, it covers alloys and phase diagrams. It defines key concepts for phase diagrams including solubility, liquidus and solidus curves, and how to determine composition using tie lines. Finally, it provides an example problem of determining the composition of phases in a Cu-40% Ni alloy at different temperatures using the phase diagram.
material 9.pptx
This document discusses diffusion and Fick's first law of diffusion. It provides the Arrhenius equation for modeling diffusion rate as an exponential function of temperature. It also presents Fick's first law, which defines flux as the rate of diffusion. The document then provides an example problem calculating the number of nickel ions diffusing through magnesium oxide per second given the diffusion coefficient and concentrations. Another example calculates the thickness of an iron membrane needed to limit nitrogen and hydrogen diffusion rates. Finally, it presents a bonus problem using experimental diffusion rates to calculate activation energy.
material Science 7.pptx
This document discusses diffusion and Fick's laws of diffusion. It defines diffusion as the net flux of particles down a concentration gradient, with flux depending on concentration gradient and temperature. Diffusion processes like carburization can increase surface hardness by introducing carbon atoms. Diffusing atoms require activation energy to move into new sites. The Arrhenius equation relates diffusion rate to temperature and activation energy. The document also defines Fick's first law of diffusion, relating flux to the diffusion coefficient and concentration gradient. It provides examples calculating diffusion rates, activation energies, and concentration gradients.
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section 3-2.pptx
- 1. Common Mechanical Operation Section 3 Eng. Kareem H. Mokhtar
- 2. Rules • Volume of cone= pi/12 * D2 * H • Mass flow rate= C* (bulk density) * 𝑔 ∗ 𝐷5/2 • C= (pi/6) * (1−𝐶𝑜𝑠1.5(ℎ𝑎𝑙𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒) sin2.5(ℎ𝑎𝑙𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒) • Pv= 𝐷∗𝑔∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 4 (tan 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∗ 𝑘 ∗ 𝑒−4∗tan 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∗𝑘∗ 𝐻 𝐷 • Pl = k *Pv • K= 1−sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ) 1+sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 )
- 3. Rules
- 4. Sheet 2 • It is required to design a cylindrical bin to store 450 tons of soda ash of bulk density 1600 kg/m3, a particle density of 2500 kg/m3 and an angle of repose = 35°, angle of friction on steel = 25o and angle of internal friction = 38o. • Estimate the dimensions of the bin as well as its minimum bottom opening diameter. Assume mass flow to occur. • What would be the solid flow rate when the bottom opening diameter is twice the minimum? • The particle size distribution follows the expression: • 𝜑𝑐 = 0.0192𝐷𝑝 2 − 0.682𝐷𝑝 + 1.254
- 5. Sheet 2 • 450 tons of soda ash • bulk density 1600 kg/m3 • particle density of 2500 kg/m3 • angle of repose = 35° • angle of friction on steel = 25o • angle of internal friction = 38o • The particle size distribution follows the expression: • 𝜑𝑐 = 0.0192𝐷𝑝 2 − 0.682𝐷𝑝 + 1.254 • Estimate the dimensions of the bin as well as its minimum bottom opening diameter. Assume mass flow to occur. • What would be the solid flow rate when the bottom opening diameter is twice the minimum?
- 6. Sheet 2 • 450 tons of soda ash • bulk density 1600 kg/m3 • particle density of 2500 kg/m3 • angle of repose = 35° • angle of friction on steel = 25o • angle of internal friction = 38o • The particle size distribution follows the expression: • 𝜑𝑐 = 0.0192𝐷𝑝 2 − 0.682𝐷𝑝 + 1.254 • Estimate the dimensions of the bin as well as its minimum bottom opening diameter. Assume mass flow to occur. V total = 450000/1600=281.25 m3 Assume cylindrical volume: 281.25 = (pi/4) * D2 * H Let H=3D Then D= 4.9 m around 5 m Top Section: Tan ( alpha r) = 2 h /D h = 1.75 Volume at top= 11.45 m3 Bottom section (Mass flow)
- 7. Sheet 2 • 450 tons of soda ash • bulk density 1600 kg/m3 • particle density of 2500 kg/m3 • angle of repose = 35° • angle of friction on steel = 25o • angle of internal friction = 38o • The particle size distribution follows the expression: • 𝜑𝑐 = 0.0192𝐷𝑝 2 − 0.682𝐷𝑝 + 1.254 • Estimate the dimensions of the bin as well as its minimum bottom opening diameter. Assume mass flow to occur.
- 8. Sheet 2 • 450 tons of soda ash • bulk density 1600 kg/m3 • particle density of 2500 kg/m3 • angle of repose = 35° • angle of friction on steel = 25o • angle of internal friction = 38o • The particle size distribution follows the expression: • 𝜑𝑐 = 0.0192𝐷𝑝 2 − 0.682𝐷𝑝 + 1.254 • Estimate the dimensions of the bin as well as its minimum bottom opening diameter. Assume mass flow to occur. • Hlaf apex angle = 20 • Tan(half apex angle) = D/2hb • Hb= 6.5 m • Volume at bottom = pi/12 * 52 * 6.5 = 42.5 m3 • Cylindrical section • Vol= 281.25-11.45-42.5=227.3 m3 • 227.3= (pi/4) * 52 * H • H= 11.57m • FinalDimensions: • Dia= 5m, the conical bottom height is 6.5m and the cylindrical part is 13.5m
- 9. Sheet 2 • 450 tons of soda ash • bulk density 1600 kg/m3 • particle density of 2500 kg/m3 • angle of repose = 35° • angle of friction on steel = 25o • angle of internal friction = 38o • The particle size distribution follows the expression: • 𝜑𝑐 = 0.0192𝐷𝑝 2 − 0.682𝐷𝑝 + 1.254 • Estimate the dimensions of the bin as well as its minimum bottom opening diameter. Assume mass flow to occur. • Dmin = 2.45 (80+Dp max) tan (alpha r) • Using the equation and substituting 𝜑𝑐 = zero • Dp max = 1.95 mm • Dmin= 141.5 mm
- 10. Sheet 2 • 450 tons of soda ash • bulk density 1600 kg/m3 • particle density of 2500 kg/m3 • angle of repose = 35° • angle of friction on steel = 25o • angle of internal friction = 38o • The particle size distribution follows the expression: • 𝜑𝑐 = 0.0192𝐷𝑝 2 − 0.682𝐷𝑝 + 1.254 • What would be the solid flow rate when the bottom opening diameter is twice the minimum • D = 2 Dmin = 283 mm • Mass flow rate= C* (bulk density) * 𝑔 ∗ 𝐷5/2 • C= (pi/6) * (1−𝐶𝑜𝑠1.5(ℎ𝑎𝑙𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒) sin2.5(ℎ𝑎𝑙𝑓 𝑎𝑝𝑒𝑥 𝑎𝑛𝑔𝑙𝑒) = 0.682 Mass flow = 0.682*1600* 9.81*(0.283)2.5 = 146 kg/s
- 11. Sheet 2 • In the above problem, calculate the vertical and the lateral pressure on the bottom of the cylindrical part of the bin. Also calculate the equivalent vertical pressure exerted by a liquid having the same (bulk) density of the solid. • Pv= 𝐷∗𝑔∗𝑏𝑢𝑙𝑘 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 4 (tan 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∗ 𝑘 ∗ 𝑒−4∗tan(𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛)∗𝑘∗𝐻/𝐷 • Pl = k *Pv • K= 1−sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ) 1+sin(𝑎𝑙𝑝ℎ𝑎 𝑖𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ) K= 0.238 Pv= kPa Pl= 29.3 kPa Pliq = 211 kPa