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Free Engineering essays Cp1090K = a + bT + cT2 + dT3. = 29.0277 ‘ (2.1865 x 10-3 x 1090) + (11.6437 x 10-6 x 10902) + (4.7065 x 10-9 x 10903) = 34.383 KJ/Kmol K = 8.214 Kcal/Kmol K. Cp Value for H2 : Cp1058K = a + bT + cT2 + dT3 = 28.6105 + (1.0194 x 10-3 x 1058) + (-01476 x 10-6 x 10582) + (0.769 x 10-9 x 10583) = 30.4345 KJ/Kmol K buy essay online cheap link between health and wealth 7.271 Kcal/Kmol K. Cp1090K = a + bT + cT2 + dT3 = 28.6105 + (1.0194 x 10-3 x 1090) + (-01476 x 10-6 x 10902) = (0.769 x 10-9 x 10903) 30.5422 KJ/Kmol K = 7.296 Kcal/Kmol K. Cp Value for H2O : Cp1058K = a + bT + cT2 + dT3 = 32.4921 + (0.0796 x 10-3 x 1058) + (13.2107 x 10-6 x 10582) + (-4.5474 x 10-9 x 10583) = 41.9785 KJ/Kmol K = 10.029 Kcal/Kmol K Cp1090K = a + bT + cT2 + dT3 = 32.4921 + (0.0796 x 10-3 x 1090) + (13.2107 x 10-6 x 10902) + (-4.5474 x 10-9 x 10903) = 42.3855 KJ/Kmol K = 10.126 Kcal/Kmol K. Component Cp 1058 Cp1090 Cpmolal CH4 17.77 18.17 1.11 CH2 13.15 13.22 0.299 N2 7.89 7.94 0.283 CO 8.16 8.21 0.291 H2 7.27 7.30 3.64 H2O 10.03 10.13 0.557. Since, the Cp Values at the inlet and outlet are mostly the same, taking Cp values at the inlet only. Cpmixture dry: = (0.125 (17.77) + (0.1251) (13.15) + (0.0083) (7.89) + (0.08706) (7.89) + (0.0806) (8.16) + (0.6549) (7.27) + (0.6549) (7.27) Heat load For Dry Gas : Hd = 9.351 (1090 ‘ 1058) x 1979.26 = 592257.3779 Kcal/hr. H H20 = 10.03 (1090 ‘ 1058) x 6916.67 = 2219973.33 Kcal/hr. Total Heat load = Hd + H H20 = 592257.3779 + 2219973.33 = 2812230.708 Kcal/hr. ‘ Total Head load in the Primary Reformer : = 431034714.70 + 2812230.708. 6.5) Heat load in Secondary Reformer : T1 = Inlet temp. = 797’C = 1070 ??K T2 = Outlet temp = 961’C = 1234 ??K. TABLE: 6.7 Inlet and outlet values. Component Inlet Kmol/hr Enthalpy Outlet Enthalpy 797’C (Kmol/hr) 961’C Kcal/Kgmol Kcal/Kgmol N2 0.15 5686 1761.51 6996 CO 2190.36 -206.74 2380.00 -19341 CH4 276.80 -7505 0.831 -4466 CO2 61.95 -85145 116.88 -82951 H2 6332.19 5452 6427.05 6658 H2O 4726.29 -50862 5183.36 -49180 Air introduced to the Secondary Reformer is 2229.57 Kmol/hr. ‘ OxygenContent ‘ =468.209(21%) Kmol/hr. ‘ NitrogenContent =1761.36(79%) Kmol/hr. Absolute Enthalpy at this temperature, for N2 = 5130Kcal/Kgmol for O2 = 5250 Kcal/Kgmol Heat of Reaction (Hr) = (Absolute Enthalpies of Product) ‘ (Absolute Enthalpy of Reactants) Flow rate of outlet gas:4659.997 kgmol/hr. So, enthalpy of outlet gas=2966802.5 *4659.997/100 = 1.82268*106 KJ/hr. 1) Around High temp. S.C. : Inlet temp = 365’C. Taking reference temp, as 25??C Enthalpy of reactants. = 14508.75 [28.1337(365-25) + 1972*10-3 (3652-252) + 0.909*10-6(3653-253) ‘ 0.528*10-9 (3652-252) Now for outlet flow rate, amix = 27.491 bmix = 10.574*10-3 cmix = -3.905*10-6 dmix = .8878*10-9. For reaction at 25??C Hp = 14508.75 [27.491 (422-25) + 10.574*10-3 (4222-252) + 3.905*10-6(3653-253) ‘ 0.8878*10-9 (4224-254) 3 = 170643064.3 Kcal/hr. Co (g) + H2O (g) = CO2(g) + H2(g) Heat of reaction at 25??C. Heat of Reaction (Hr) = (Heat of formation of Product) ‘ (Heat of formation of reactants) = (-94081 + 0) ‘ (-26415 ‘ 57797) Total heat of reaction = 9839 (249.145 ‘ 1142.25) = -8787260.095 Kcal/hr. Desired heat of reaction, Hr = 40755448.8 ‘ 8787260.095 ‘ 143745266.5 = -111777077.7 Kcal/hr. By heat balance, heat required to remove by indirect heat exchanges= enthalpy of gas in +heat of reaction ‘ enthalpy of gas receiving. =14374526.6 + 14207493 ‘ 34569897 = 72.084049*106 KJ/hr. Reaction (1) CO2 + 4H2 = CH4 + 2H2O + Heat (2) CO + 3H2 = CH4 + H2O + Heat. Inlet temp = 315 C = 588 K Outlet temp = 360 C = 633 33K. amix = 7.9495 bmix = .9933*10-3 cmix = 3.46*10-6 dmix = -1.3792*10-9. H = -m[a + bT + cT2 + dT3] dt. = 8266.08 [7.9495 (588-298) – .9933*10-3 (5882-2982) + 3.46*10-6 (5883-2983) 1.3792*10-9 ‘ (5884-2984) Now for outlet flow rate, amix = 28.7416 bmix = .3944*10-3. cmix = 3.178*10-6 dmix = .796*10-9. Hp = 9630 [28.741 (633-298) – 0.3944*10-3 (6332-2982) 2. + 3.178*10-6 (6333-2983) – 0796*10-9 (6334-2984)] Heat of Reaction on No.1 (Hr) = (Heat of formation of Product)-(Heat of formation of reactants) = (-17888 + 2(-57797)) ‘ ((-94051+410)) Moles of CO2 at inlet flow rate of .01 kmol/hr. Heat of Reaction on No.2 (Hr) = (Heat of formation of Product) ‘(Heat of formation of reactants) Moles of CO at inlet flow rate of 7.33 kmol/hr Hr2 = -49270 * 7.33. By energy balance, loss= heat with in + heat f reaction ‘ heat with out. = 25.441471*106 + 380470.29 *106 ‘ 60.51862 * 10 6. Flow outlet of outlet stream=2768.393 kmol/hr Q= * 2768.393. = 60.51862 * 10 6 kmol/hr. 6.8) Around Ammonia Convertor: Inlet written they get my papers back in a timely manner = 135 C = 408 K Outlet temp = 320 C = 593 K. amix = 27.294 bmix = 3.1164*10-3 cmix = 3.664*10-6 dmix = -1.265*10-9. H = -m[a + bT + cT2 + dT3] dt. = 70814.73 [27.294 (408-298) + 3.1164*10-3 (4082-2982) + 3.664*10-6 (4083-2983) ‘ 1.265*10-9 (4084-2984) = 53590516.24 Kcal/hr amix = 145.66 bmix = 7.891*10-3 cmix = 4.636*10-6 dmix = -1.6705*10-9. HP= 62757.96 [145.66 (593-298) + 7.891*10-3 (5932-2982) + 4.636*10-6 (5933-2983) ‘ 1.6705*10-9 (5934-2984)] = 275477364.8 Kcal/hr Reaction : N2 + 3H2 = 2NH3 + Heat Heat of this reaction at 25??C Hr = 2(-11039) = -22078 Kcal/kmol My childhood essay writing Windermere Preparatory School Heat of reaction = -22708(9432.52 ‘ 1430.45) = -166371485.7 Kcal/hr. This heat is taken on by the input stream to attain the temp. of 135??C. By energy balance, loss= heat input + heat of reaction ‘ heat out. = 1.501479*108 + 65.63 * 106 – 1.52577468*108 =63.201321*106 kj/hr. ENERGY BALANCE DATASHEET: For diameter : The flow chart for steps of calculating the diameter is shown below: CALCULATION; Assume specific velocity is 100 feet / min. ?? mi yipressure 20 bart = 250 c. Total flow rate = 4833.73 kg/hr. Total mol flow rate = 323.63 kmol/ hr. Q = 20*8.73/0.08250*250 m3 / hr = 8.66 m3/hr. vol. flow rate = 4833.73 / 6.903. assume NTU = 15HTU = 0.6. Area = vm / specific velocity = 0.3828 m2 Now di = 0.7 mheight = 9 m. ”internal design pressure = (20 bar ‘ 1 atm) * unt thesis boot camp th = p ri / f j ‘ 0.6p + C.A. MOC of material is M.SSA516 GRADE 70. allowable stress f = 6 * 107 pa = 600.16698 = 600 bar j= 0.85 C.A =1.5 By internal design pressure p = 21.9. ”For shell = P ri / fj ‘ 0.6 p + C.A = 21.9 *0.35 /(600 * 0.85) ‘(0.6 * 21.9) +0.015 = 0.03042 m = 30 mm. Weight of shell = ??/4 (D02 ‘ Di2) * L *?? D0 = 0.76 m Weight = 5000 kg (approx.) ”For top head th = p Di v / 2fj ‘ 0.2 p + C.A = 21.9 * 0.7 * 1 / (2*600*0.45)-(0.2*21.9) + 0.015 = 30 mm. now weight of top head = ??/4 * (B.D)2 * th * ?? dia = 0.7m = 70 cm = 28 inch. B.D = 36 inch = 91.44 mm = 0.91 m. Weight of top head = 650 kg. So total weight of top and bottom head = 2* 650 = 1300 kg. ”total weight = 5000+1300 = 6300 kg. CHAPTER 8 PLANT LOCATION AND LAYOUT. After the process flow diagrams are completed and before detailed piping, structural, and electrical design can begin, the layout of the process units in a plant and equipment within these process units must be planned. This layout can play an important part in determining construction and manufacturing costs, and thus must be planned carefully with attention being given to future problems that may arise. Since each plant differs in many ways and no two plant sites are exactly alike, there is no one ideal plant layout. However, proper layout in each case will include arrangement of processing areas, storage areas, and handling areas in efficient coordination and with regard to such factors as: ”New site development or addition to previously developed site. ”Type and quantity of products to be produced. ”Possible future expansion. ”Operational convenience and scholarship applications with essays of process and product control. ”Economic distribution of utilities and services. ”Type of process and product control. ”Economic distribution of utilities and services. ”Health and safety considerations. ”Space available and space required. Preparation of the layout: Scale drawings, complete with elevation indications, can be used for determining the best location for equipment and facilities. Elementary layouts are essay writing about environment Berlin Brandenburg International School first. These show the fundamental relationships between storage space and operating equipment. The next step requires consideration of the operational sequence and gives a primary layout based on flow of materials, unit operations, storage, and future expansion. Finally, by analyzing all the factors that are involved in plant layout, a detailed recommendation can be presented, and drawings and elevations, including isometric drawings of the piping systems, can be prepared. Three-dimensional models are often made. Plant location is however one of the most important part of final planning. Selection of suitable location of the plant is very important because success of the plant is based on this point. The geographical location of the final plant can have a strong influence on the success of an industrial venture. Much care must be exercised in choosing the plant site, and many different factors must be considered. Primarily, the plant should be located where the minimum cost of production and distribution can be obtained, but other factors, such as room for expansion and general living conditions, are also important. An approximate idea as to the plant location should be obtained before a design project reaches the detailed-estimate stage, and a firm location should be established upon completion of the detailed-estimate design. The choice of the final site should first be based on a complete survey of the advantages and disadvantages of various geographical area and, ultimately, on the advantages and disadvantages of available real estate. The following factors should be considered in choosing a plant site: ”Raw materials ”Markets ”Power and fuel ”Climate Sinario ”Transportation facilities ”Water supply ”Waste disposal ”Labor supply ”Taxation and legal restrictions ”Site characteristics ”Flood and fire protection ”Community factors. CHAPTER 9 SAFETY AND ENVIRONMENT. 9.1) PRINCIPLE OF PROTECTION & PREVENTION: Industrial accidents are caused by negligence of employer, the worker or the both. Employers’ efforts to reduce the accidents are generally motivated by four considerations. ‘ To lessen human suffering ‘ To prevent damage to plant and machinery ‘ To reduce the amount of time lost as a result ‘ To hold the expenses of workman’s compensation to a minimum. The basic reasons for preventing industrial accidents are human and economic. The most important of these should be to avoid human suffering. Pain, suffering and wrecked lives are not to be the byproducts of any industry. FIRE FIGHTING APPLIANCES : Type of Class A Class B Class C extinguisher Carbon dioxide Suitable for surface Suitable. Does not Suitable. Non- fires only leave residue or conductor and affect equipment or does not damage food stuff. equipment. Dry chemical Suitable for small Suitable. Chemical Suitable. fire releases smothering Chemical is non- gas and shields conductor or dry operator from heat. chemical shields operator from heat. Foam Suitable. Has both Suitable. Unsuitable. Foam smothering effect Smothering blanket being a conductor and wetting action. does not dissipate, Should not be floats on top of Used on live spilled liquid. equipment. Water Suitable. Water Unsuitable. Water Unsuitable. Water saturates material a will spread and not being conductor. 9.2) SAFETY CONSIDARATION: Mild concentrations of product will cause conjunctivitis. Contact with higher concentrations of product will cause swelling of the eyes and lesions with a possible loss of vision. Mild concentrations of product will cause dermatitis or conjunctivitis. Contact with higher european language institute gamesgames of product will cause caustic-like dermal burns and inflammation. Toxic #1 Memoir Ghostwriter exposure may cause skin lesions resulting in early necrosis and scarring. INGESTION EFFECTS: Since product is a gas at room temperature, ingestion is unlikely. Corrosive and irritating to the upper respiratory system and all mucous type tissue. Depending on the concentration inhaled, it may cause burning sensations, coughing, wheezing, shortness of breath, headache, nausea, with eventual collapse. Inhalation of excessive amounts affects the upper airway (larynx and bronchi) by causing caustic-like burning resulting in edema and chemical pneumonitis. If it enters the deep lung, pulmonary edema will result. 9.3) FIRST AID MEASURES:- Flush contaminated eye(s) with copious quantities of water. Part eyelids to assure complete flushing. Continue for a minimum buy essay online cheap link between health and wealth 15 minutes. persons with potential exposure to ammonia should not wear contact lenses. Remove contaminated clothing as rapidly as possible. Flush affected area with copious quantities of water. In cases of frostbite or cryogenic “burns” flush area with lukewarm water. DO NOT USE HOT WATER. A physician should see the patient promptly if the cryogenic “burn” has resulted in blistering of the dermal surface or deep tissue freezing. prompt medical attention is mandatory in all cases of overexposure. rescue personnel should be equipped with self-contained breathing apparatus. 9.4) HAZARD: 1)What happens to ammonia when it enters the environment? Because ammonia occurs naturally, it is found throughout the environment in soil, air, and water. Ammonia is recycled naturally in the environment as part of the nitrogen cycle. It does not last very long in the environment. Plants and bacteria rapidly take up ammonia from soil and water. Some ammonia in water and soil is changed to nitrate and nitrite by bacteria. Ammonia released to air is rapidly removed by rain or snow or by reactions with other chemicals. Ammonia does not build up in the food chain, but serves as a nutrient source for plants and bacteria. 2) How might I be exposed to ammonia? Everybody is regularly exposed to low levels of ammonia in air, food, soil, and water. Ammonia has a strong irritating odor that people can easily smell before it may cause harm. If you use ammonia cleaning products at home, you will be exposed to ammonia released to the air and through contact with your skin. If you apply ammonia fertilizers or live near farms where these fertilizers have been applied, you can breathe ammonia released to the air. You may be exposed to ammonia from leaks and spills from production plants, storage facilities, pipelines, tank trucks, and rail cars. You may be exposed to higher levels if you apply ammonia fertilizers or live near farms where these fertilizers have been applied. You may be exposed to high levels if you go into enclosed buildings that contains lots of animals (such as on farms). 3)How can ammonia affect my health? No health effects have found in humans exposed to typical environmental concentrations of ammonia. Exposure to high resume access database of animal farm of ammonia in air may be irritating to your skin, eyes, throat, and lungs and cause coughing and burns. Lung damage and death may occur after exposure to very high concentrations of ammonia. Some people with asthma may be more sensitive to breathing ammonia than others. Swallowing concentrated solutions of ammonia can cause burns in your mouth, throat, and stomach. Splashing ammonia into your eyes can cause burns and even blindness. 4)Has the federal government made recommendations to protect human health? The Occupational Safety and Health Administration (OSHA) has set an acceptable eight-hour exposure limit at 25 parts of ammonia per one million parts of air (ppm) and a short-term (15 minutes) exposure level at 35 ppm. CHAPTER. From this project we envisage the practical work going in the chemical plants.some of the practical espects of the chemical engineering we have putted it nicely in this project prior to which we don’t have any practical experience. Our practical view about the chemical plant is align somewhat by doing the project. By perfuming the calculation of material balanceenergy balance,in which how we Put into practice is cleared. how the chemical plants operation runs on the basis of calculation is totally envisage now. CHAPTER 11 BIBLOGRAPHY. 1) Perry, R.H. and Green D.Perry’s Chemical Engineer’s Handbook, McGraw-Hill book company, ed. 8th, pg no. 16-61 to 16-64, 20-57 to 20-62. 2) George T. Austin, Shreve’s Chemical Process Industries, Fifth Edition, Nitrogen Industries. 3) K.A Gavhane,introduction to process calculation,18th edition,nirali prakasan. 4) B.I BHATT ,S.M VORA,stoichiometry 4th edition tata mcgraw-hill publishers. 5).John J. Mcketta and William Aaron Cunningham, Encyclopedia of Chemical Processing And Design, Volume-2, Adsorption design. 6). IFFCO ‘ Indian Farmers Fertilisers Cooperative Limited, Ammonia Plant Manual. 7)Prof. M.N. Vyas, Safety and Hazards Management in Chemical Industries, Atlantic publishers and distributors ltd. 8).S.B. Thakore, B.I.Bhatt, Introduction to Process Engineering and Design, Tata McGraw Hill Education Pvt. Ltd.(design of adsorber) Search our thousands of essays: If this essay isn't quite what you're looking for, why not order your own custom Engineering essay, dissertation or piece of coursework that answers your exact question? There are UK writers just responsibility cookie policy discount policy me on hand, waiting to help you. Each of us is qualified to a high level in our area of expertise, and we can write you a fully researched, fully referenced complete original answer to your essay question. Just complete our simple order form and you could have your customised Engineering work in your email box, in as little as 3 hours. This Engineering essay was submitted to us by a student in order to help you with your studies. This page has approximately words. If you use part of this page in your own work, you need to provide a citation, as follows: Essay UK, Report: ‘SYNTHESIS MANUFACTURING OF AMMONIA’. Available from: [05-09-18]. 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