Section 9.pptx
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The document discusses optimum design parameters and provides examples of determining optimum values that minimize costs. Question 1 finds the optimum insulation thickness. Question 2 determines optimum temperature and pressure values that minimize total cost. Question 3 calculates the daily profit at the production rate giving minimum cost per unit and maximum daily profit. Question 4 involves determining the batch cycle time that gives the minimum total annual cost of producing 1 million kg of chemical product.
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This document provides design calculations for the aeration system of a conventional activated sludge plant treating 10 MLD of sewage. Key details include:
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In this project the steam generator is water tube boiler fired with rice husk. The steam is transferred to the tyre/tube moulds where tyres/tubes are cured while the heat is rejected to the tyres the condensate forms and this condensate is put back to the boiler. While doing so the steam is also stopped back to boiler without rejecting complete heat to the product. This gets flashed into atmosphere at feed water tank. The science of separation of condensate from steam saves energy. Better the separation more the fuel conservation.
In the steam generator the fuel is burnt to heat the water and form steam. This fuel burnt flue gas carries lot of energy, out through chimney. Prior to exhausting through the heat left in flue need to be recovered, through heat recovery mechanisms’. In this project an air-preheater condensate heat recovery unit is the major energy consuming station.
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1. Size reduction involves breaking down large particles into smaller pieces for easier handling and processing. It is commonly used in industries like mining, cement, and food processing.
2. The main methods of size reduction are compression, impact, attrition, and cutting. An ideal crusher would have a large capacity, require little power, produce a uniform product size, and have low costs.
3. The power required for size reduction depends on factors like feed rate, particle sizes, material properties, and the specific reduction technique used according to models like Rittinger's law, Bond's law, or Kick's law. Efficiency is also important to minimize energy costs, which make up a major portion of size reduction
Process costing
This document discusses principles and examples of process costing. Some key points covered include:
- Production is classified by processes and each process account tracks direct materials, labor, expenses and overheads. Cost per unit is calculated at the end of each process.
- Normal loss is credited to the process account at scrap value while abnormal loss is credited at full cost.
- Examples show setting up process accounts to track costs of different products as they move through multiple processes, and calculating unit costs and profits.
- Process loss and wastage distinguishes between normal and abnormal loss.
- Joint products and by-products examples show allocating joint costs between products.
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Standard costing involves setting predetermined expected costs for cost components like direct materials, direct labor, and factory overhead. Variances are calculated as the difference between actual and standard costs. This includes direct material, direct labor, and factory overhead variances. The direct material variance has a price and usage component. The factory overhead variance separates variable from fixed costs, with the controllable variance measuring variable cost efficiency and volume variance measuring fixed cost utilization.
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The document provides information on calculating mass storage in stockpiles and storage facilities using bulk density, angle of repose, and dimensional measurements. It includes examples of:
1) Calculating the mass of gravel that can be stored in a 6x6m area using the given bulk density of 1700kg/m3 and angle of repose of 45°, finding the answer is 48 tons.
2) Finding the minimum dimension H of 8.5m for a storage compartment with a 5x5m base to store 360 tons of bauxite ore over 3 days at a furnace rate of 5 tons/hr, using the given bulk density of 2000kg/m3 and angle of repose of
section 5.pptx
This document provides rules and examples for calculating mechanical operations related to bulk solids storage and transport. It includes formulas for calculating volume and mass flow rate of cones, as well as vertical and lateral pressures at the bottom of a bin. An example calculates these pressures for a given bin geometry. It also discusses belt conveyor design, including belt and idler specifications and formulas for power calculations. The final example uses these formulas to calculate the motor power and storage bin dimensions needed for a given solid transportation rate and production period.
section7.pptx
1. The document discusses various rules and equations related to mechanical operations like crushing, grinding and milling including Rittinger's law, Kick's law, and equations for calculating ball mill dimensions, rotation speeds, and charge volumes.
2. It provides an example calculation for estimating the theoretical power required to grind a one ton batch in a roller mill over 30 minutes given the feed and product screen analyses.
3. Questions and examples are given for calculating reasonable ball mill speeds, ball charges, and rock feed charges given mill dimensions and properties of the balls and feed material.
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 8.pptx
This document provides information on mechanical operations like crushing and grinding. It includes:
- Rules for operations like Rittinger's law, Kick's law, and Bond work index.
- Parameters for equipment like ball mill rotation speed, ball size, and charge mass.
- Worked examples calculating time required to achieve a specific surface increase in crushing and grinding, based on parameters like feed size, product size, power, and work index.
- An example calculating the time required to grind one ton of material in a batch rod mill, given the motor power, grinding efficiency, feed and product sizes, and work index.
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.
section 3-2.pptx
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.
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.
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Section 9.pptx
- 1. OPTIMUM DESIGN ENG. KAREEM HOSSAM
- 2. WHAT ARE WE GOING TO DEAL WITH • 1- one variable • 2- two variables • 3- optimum production rates
- 3. HOW TO KNOW THE OPTIMUM DESIGN PARAMETERS
- 4. HOW TO KNOW THE OPTIMUM DESIGN PARAMETERS
- 5. QUESTION 1 • Find the optimum thickness to achieve minimum total cost knowing that the fixed cost of the insulation is 3x+20 egp where x is the insulation in cm, and the cost of heat lost is 20/x +50. • Solution Total Cost = fixed + variable = 3x+20 + 20/x +50. dcost/ dx = 3-20/x2 x= 2.58 cm D2cost/dx2= 40/x3 at x = 2.58 the value will be positive
- 6. QUESTION 2 • if you know that the temperature and pressure are two variables affecting the industry, on the other hand both of them contribute in the total cost if you know that the effect of temperature is 2.33T + (3000/PT) +3 and the effect of pressure is 5P+ (5000/PT) +5 • 1- Determine the optimum values for P and T which gives you the minimum cost knowing that T in C and P in Psi. • 2- Determine the total cost at the optimum conditions
- 7. QUESTION 2 • Total cost = 2.33T + 3000/PT +3 + 5P+ 5000/PT +5 =2.33T+8000/PT +5P+ 8 • Derivative w.r.t T = 2.33 – 8000/PT2 = 0 • Derivative w.r.t P = 5 – 8000/TP2 = 0 • Solving both equations • P = 9 psi and T = 19.5 C • Second derivatives are: • 2*8000/PT3 and 2*8000/TP3 and both of them has + values after substitution • The total cost at optimum conditions is 144 EGP
- 8. QUESTION 3 • A plant produces at a rate of P units per day. The variable costs per unit was 30 + 0.1P, The total daily fixed charges are 1000, and all other expenses are constant at 7000 per month. If the selling price per unit is 100 determine: • (a)The daily profit at production giving the minimum cost per unit. • (b)The daily profit at production giving the maximum daily profit.
- 9. QUESTION 3 – CONTINUE (A) a) Total Cost per unit = 30+0.1P + 1000/P + (7000/30)/P = 30 + 0.1 P + 1233.33/P d(cost)/P = 0.1 -1233.33/P2 P= 111 Unit • Second derivative of the previous equation is 2* 1233.33/P3 (+ve at all cases) • Daily profit is = selling revenue per day – total cost per day • Total cost per unit at p =111 = 52.2 $ • Total cost per day = 52.2 * 111 = 5794.2$ • Selling revenue per day = 100*111 = 11100$ • Daily profit is 5305.8 $
- 10. QUESTION 3 – CONTINUE (B) • B) daily profit = selling price per unit * no. of units per day - Total cost per unit * no. of units per day • daily profit = 100P – [30 + 0.1 P + 1233.33/P]P • d profit/dP = 100-30 - 0.2P P = 350 unit • Second derivative = - 0.2 always negative which indicates a maximum value • Daily profit at a production giving the maximum daily profit = 11016.67 $
- 11. QUESTION 4 • An organic chemical is being produced by a batch operation. Each cycle consists of the operating time necessary to complete the reaction plus a discharging time of 0.7 hour and charging time of 0.3 hour. The operating time per cycle is equal to 2P0.5 h, where P is the kilograms of product produced per batch. The operating costs during the operating period are $20 per hour, and the costs during the discharge-charge period are $10 per hour. The annual fixed costs for the equipment vary with the size of the batch as the fixed cost is equal to 300P1.2 dollars per year. the plant can be operated 24 h per day for 300 days per year. The annual production is 1 million kg of product. At this capacity, the other costs rather than those already mentioned is $200,000 per year. Determine the cycle time for conditions of minimum total cost per year.
- 12. QUESTION 4 SOLUTION • Givens: • P production per cycle (kg) • discharging time of 0.7 hour/cycle charging time of 0.3 hour/cycle operating time per cycle = 2P0.5 h • operating costs during the operating= $20/h during the discharge-charge are $10/h • fixed cost= 300P1.2 $/year • Operation time 24 h per day for 300 days per year • annual production is 106 kg of product • other costs= 200000$/year • Determine the cycle time for conditions of minimum total cost per year. • Total cost/year = fixed cost/year + variable cost/year • Cycle time= operating + charge + discharge = 2P0.5+1 • Total cost/year = 300P1.2 + 200000 + variable cost/year • Variable cost /year = 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡 𝑏𝑎𝑡𝑐ℎ * 𝑁𝑜. 𝑜𝑓𝑏𝑎𝑡𝑐ℎ 𝑦𝑒𝑎𝑟 • 𝑁𝑜. 𝑜𝑓 𝐵𝑎𝑡𝑐ℎ 𝑦𝑒𝑎𝑟 = 𝑎𝑛𝑛𝑢𝑎𝑙 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒 = 106 𝑃 • 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡 𝑏𝑎𝑡𝑐ℎ = cost of operation + cost of charge and discharge • = 20 ℎ𝑜𝑢𝑟 * time of operation + 10 $ ℎ𝑜𝑢𝑟 * time of charge and Discharge
- 13. QUESTION 4 SOLUTION • Givens: • P production per cycle (kg) • discharging time of 0.7 hour/cycle charging time of 0.3 hour/cycle operating time per cycle = 2P0.5 h • operating costs during the operating= $20/h during the discharge-charge are $10/h • fixed cost= 300P1.2 $/year • Operation time 24 h per day for 300 days per year • annual production is 106 kg of product • other costs= 200000$/year • 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡 𝑏𝑎𝑡𝑐ℎ = cost of operation + cost of charge and discharge • = 20 ℎ𝑜𝑢𝑟 * time of operation + 10 $ ℎ𝑜𝑢𝑟 * time of charge and Discharge • = 20 ℎ𝑜𝑢𝑟 * 2P0.5+ 10 $ ℎ𝑜𝑢𝑟 * (0.7+0.3) • Total cost/year = 300P1.2 + 200000 +[ 20 ℎ𝑜𝑢𝑟 * 2P0.5+ 10 $ ℎ𝑜𝑢𝑟 * (0.7+0.3)]* 106 𝑃 • From dTotal annual Cost/d(p)= zero get = 625 kg/bath • Substitute in Cycle time= 2P0.5+1 get the minimum time
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