![Solution(contd.)
Now theoretical air can be found out by
Theoretical Air
100/23[(8/3 C+8H2+S)-O2]
𝑂2 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑓𝑜𝑟 𝐶𝑂2 𝑖𝑠 ,
C + O2 = CO2
1kg + 8/3kg = 11/3
𝐶 =
8
3
× 0.656 = 1.7493 𝑘𝑔](https://image.slidesharecdn.com/thermodynamicspblpresentationgroupa1-200812064402/85/Analysis-Procedures-of-Coal-21-320.jpg)
Analysis Procedures of Coal
- 2. PBL Case Study Presentation in Group for the Mid-Semester Examination of Thermodynamics-II Teacher:- Prof. Dr. Abdul Fatah Abbasi Group:- A Subgroup:- A1 Dated: 27-07-2020 Group Members 18ME01 Muhammad Taha 18ME02 Usama Kaleem 18ME05 Umama Qureshi 18ME06 Awais Ali Memon 18ME07 Saad-ud-din 18ME08 Wahid Dino 18ME09 Syed Muhammad Hassan
- 3. Problem Statement In coal fired power plants coal is a main fuel for combustion purpose. Before use of coal different tests are to be carried out to analysis the constituent elements and some undesirable contamination in the coal. Discuss the analysis procedures of the coal. The analysis of coal is as follows C=82%, H=6%,O2=4% and remaining is ash. Determine the amount of theoretical air required for complete combustion. If the actual air supplied is 40% in excess and 80% of given carbon is burnt to CO2 and remaining is CO. Conduct the volumetric analysis of dry products of combustion.
- 4. What is a Fuel? Any material that can be burned and give us thermal energy for our useful work those materials is known as FUEL. Mostly organic compounds are being used as best thermal energy producers ,so we often choose hydrocarbon ( carbon and hydrogen subsuming compounds) as fuel. Types of Fuels • There are three types of fuels . 1. Solid ( mostly coal) 2. Liquid ( mostly gasoline) 3. Gas (mostly sui gas) Now , actually we are discussing about coal as mention in our given question.
- 6. Coal Coal is an organic rock (as opposed to most other rocks in the earth's crust, such as clays and sandstone, which are inorganic); it contains mostly carbon (C), but it also has hydrogen (H), oxygen (O), sulfur (S) and nitrogen (N), as well as some inorganic constituents (minerals) and water (H2O)
- 7. Properties of Coal Because of wide variations in the composition and properties of coals, a classification system is needed to describe the different kinds available for use in homes and power plants
- 8. Need For Analysis Worldwide, coals are used for various purposes such as generation of electricity, iron and steel making, cement production, paper manufacturing and chemical and pharmaceutical productions. However, the application of coals in manufacturing industries depend on its types, grades and quality, which are function of temperature and pressure and the length of time of its formation . There are two classes of coals: low rank coals (the lignite and subbituminous) and hard coals (the bituminous and anthracite).
- 9. Need For Analysis Rank refers to steps in a slow, natural process called “coalification,” during which buried plant matter changes into an ever denser, drier, more carbon rich, and harder material. Low rank coals are characterized with high moisture content and hard coals with high carbon/energy content. Both low rank and hard coals can be used in railway transportation, power generation and cement manufacturing because high quality coals is not pre-requisite. Thus, little or no laboratory analysis may be required after exploration and prior to the usage. However, in iron and steel making industries, smokeless fuel manufacturing industries and town gas supply, high quality coal is needed This implies that there is a need for coal analysis to meet the desire objectives.
- 10. Types of Analysis of Coal. Analysis of Coal Ultimate Analysis Proximate Analysis
- 11. Ultimate Analysis The determination of the percentage of constituent elements of a chemical substance. In case of coal and coke, the determination of carbon and hydrogen in the material as found in the gaseous products of its complete combustion the determinations of Sulphur, nitrogen and ash in the material as a whole, and the calculation of oxygen by difference The principal reason for the ultimate analysis of coal is the classifications of coals according to rank, although it is often used for commercial and industrial purposes when it is most desirable to known the Sulphur content of coal also known as total analysis of coal.
- 12. Ultimate Analysis (contd.) 1) Carbon and Hydrogen : 𝐶 + 𝑂2 → 𝐶𝑂2 −−− −𝐾𝑂𝐻 𝐻2 + 1 2 𝑂2 → 𝐻2 𝑂 −−− −𝐶𝑎𝐶𝑙2 % 𝑜𝑓 𝐶𝑎𝑟𝑏𝑜𝑛 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓 𝐾𝑂𝐻 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑜𝑎𝑙 × 12 44 × 100 % 𝑜𝑓 𝐻𝑦𝑑𝑟𝑜𝑔𝑒𝑛 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑜𝑓 𝐶𝑎𝐶𝑙2 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑜𝑎𝑙 × 2 18 × 100
- 13. Ultimate Analysis(contd.) 2) % of Nitrogen 𝐶𝑜𝑛𝑐 𝐻2 𝑆𝑂4 𝐾2 𝑆𝑂4 𝑁𝐻4 2 𝑆𝑂4 2𝑁 + 3𝐻 + 𝐻2 𝑆𝑂4 → 𝑁𝐻4 2 𝑆𝑂4 % 𝑜𝑓 𝑛𝑖𝑡𝑟𝑜𝑔𝑒𝑛 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑐𝑖𝑑 𝑐𝑜𝑚𝑝𝑢𝑛𝑑 ×𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑜𝑎𝑙 × 14 × 100
- 14. Significance of Coal Carbon and Hydrogen: They increases the calorific values. Nitrogen: Less amount of nitrogen is good fuel it does no have calorific values. Sulphur: It increases the calorific values, but more mount of Sulphur leads to smoky flame. Ash: It decreases the calorific values. Oxygen: It increases the calorific values.
- 15. Proximate Analysis of Coal. It subsumes quantitative determination of :- 1. % Moisture • Higher percentage of moisture , lower the caloric value in outcome. • Increase the cost of transport. • Considerable amount of heat is lost in evaporation. 2. % Volatile Matter • Higher value causes smoke ,long flame and decrease the caloric value. 3. % Ash • High ash content decease the calorific value. 4. % Carbon • High carbon percentage gives good quality coal, increase its calorific value.
- 16. Determination of moisture:- • About 1g of finely powdered coal is weighed in silica crucible. • The crucible is placed without lid an electric hot-air-oven , maintained at 105C for an hour . The crucible is then taken out , cooled in a desiccator and weighed for loss in weight. % moisture = 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑎𝑙 𝑠𝑎𝑚𝑝𝑙𝑒 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑎𝑙 𝑡𝑎𝑘𝑒𝑛 ×100
- 17. Determination volatile matter • The coal in the crucible in in covered with a lid and placed in a muffle furnace maintained at 925 ± 20 C for 7 minutes. • The crucible is cooled in air, then in desiccator and weighed again. % of volatile matter = 𝑙𝑜𝑠𝑠 𝑖𝑛 𝑤𝑒𝑢𝑔ℎ𝑡 𝑎𝑡 925 ±20𝐶 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑎𝑙 𝑡𝑎𝑘𝑒𝑛 ×100
- 18. Determination of Ash • The residual coal in the crucible in is then heated without lid in muffle furnace at 700 ± 50 C for half-an-hour. • The crucible is then taken out , cooled first in air , then in desiccator and weighed. • The residual is reported as ash on percentage basis. • % of Ash = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 𝑙𝑒𝑓𝑡 𝑖𝑛 𝑐𝑟𝑢𝑐𝑖𝑏𝑙𝑒 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑐𝑜𝑎𝑙 𝑡𝑎𝑘𝑒𝑛 Determination of Carbon • Percentage of fixed carbon = 100 -% of ( moisture + volatile matter + ash)
- 19. Problem The analysis of coal is as follows C=82%, H=6%,O2=4% and remaining is ash. Determine the amount of theoretical air required for complete combustion. If the actual air supplied is 40% in excess and 80% of given carbon is burnt to CO2 and remaining is CO. Conduct the volumetric analysis of dry products of combustion. Data • C = 82% • H = 6% • 𝑂2= 4% • Ash = 8% • Excess air = 40% • Theoretical Air= ??
- 20. Solution As we know 80% of carbon is burnt into 𝐶𝑂2 and remaining into 𝐶𝑂, So for Carbon dioxide, C = 0.82 x 0.8 = 0.656 And for Carbon monoxide C = 0.82 - 0.656 = 0.164
- 21. Solution(contd.) Now theoretical air can be found out by Theoretical Air 100/23[(8/3 C+8H2+S)-O2] 𝑂2 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑓𝑜𝑟 𝐶𝑂2 𝑖𝑠 , C + O2 = CO2 1kg + 8/3kg = 11/3 𝐶 = 8 3 × 0.656 = 1.7493 𝑘𝑔
- 22. Solution(contd.) 𝑂2 required for 𝐻2 𝑂 is 2H2 + O2 = 2H2O 1kg + 8 kg = 9 kg H = 8 x 0.06 = 0.48 kg So total oxygen required will be O2 = 1.7493 + 0.48 − 0.04 = 2.1893kg
- 23. Solution(contd.) So , theoretical air will, Theoretical air = 100 23 2.1893 = 9.5186 𝑘𝑔 𝑘𝑔 𝑐𝑜𝑎𝑙 Since, 𝐸𝑥𝑐𝑒𝑠𝑠 𝑎𝑖𝑟 % = 𝐴𝑖𝑟 𝐹𝑢𝑒𝑙 𝑎𝑐𝑡𝑢𝑎𝑙 − 𝐴𝑖𝑟 𝐹𝑢𝑒𝑙 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝐴𝑖𝑟 𝐹𝑢𝑒𝑙 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝐴𝑖𝑟 𝐹𝑢𝑒𝑙 𝑎𝑐𝑡𝑢𝑎𝑙 = 0.4 + 1 × 9.5186 = 13.326
- 24. Solution(contd.) • Thus, mass of 𝑂2 supplied with excess air will • 𝑂2 = 0.23 × 13.326 − 2.1893 = 0.87568 • Masses of the product will • 𝐶𝑂2 = 11 3 × 0.656 = 2.40553 𝑘𝑔 • 𝐶𝑂 = 7 3 × 0.164 = 0.382667 kg • 𝐻2 𝑂 = 9 × 0.06 = 0.54 kg
- 25. Solution(contd.) Products Mass (kg) Molecular Mass No of Mole 𝒏 = 𝒈𝒊𝒗𝒆𝒏 𝒎𝒂𝒔𝒔 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒂𝒓 𝒎𝒂𝒔𝒔 %Volume 𝑛𝑖 Σ𝑛𝑖 × 100 𝐶𝑂2 2.40533 44 0.05466 47.575% 𝐶𝑂 0.382667 28 0.013667 11.895% 𝐻2 𝑂 0.54 28 0.0192 16.711% 𝑂2 0.87568 32 0.02736 23.814% Total Σ𝑛𝑖 = 0.11489 100% Volumetric Analysis
- 26. Thank You