Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Energy, economic and environmental issues of power plants

305 views

Published on

Energy, economic and environmental issues of power plants

Published in: Education
  • DOWNLOAD FULL BOOKS INTO AVAILABLE FORMAT ......................................................................................................................... ......................................................................................................................... 1.DOWNLOAD FULL PDF EBOOK here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL EPUB Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL doc Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL PDF EBOOK here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL EPUB Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL doc Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... .............. Browse by Genre Available eBooks ......................................................................................................................... Art, Biography, Business, Chick Lit, Children's, Christian, Classics, Comics, Contemporary, Cookbooks, Crime, Ebooks, Fantasy, Fiction, Graphic Novels, Historical Fiction, History, Horror, Humor And Comedy, Manga, Memoir, Music, Mystery, Non Fiction, Paranormal, Philosophy, Poetry, Psychology, Religion, Romance, Science, Science Fiction, Self Help, Suspense, Spirituality, Sports, Thriller, Travel, Young Adult,
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
  • DOWNLOAD FULL BOOKS INTO AVAILABLE FORMAT ......................................................................................................................... ......................................................................................................................... 1.DOWNLOAD FULL PDF EBOOK here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL EPUB Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL doc Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL PDF EBOOK here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL EPUB Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... 1.DOWNLOAD FULL doc Ebook here { https://tinyurl.com/y8nn3gmc } ......................................................................................................................... ......................................................................................................................... ......................................................................................................................... .............. Browse by Genre Available eBooks ......................................................................................................................... Art, Biography, Business, Chick Lit, Children's, Christian, Classics, Comics, Contemporary, Cookbooks, Crime, Ebooks, Fantasy, Fiction, Graphic Novels, Historical Fiction, History, Horror, Humor And Comedy, Manga, Memoir, Music, Mystery, Non Fiction, Paranormal, Philosophy, Poetry, Psychology, Religion, Romance, Science, Science Fiction, Self Help, Suspense, Spirituality, Sports, Thriller, Travel, Young Adult,
       Reply 
    Are you sure you want to  Yes  No
    Your message goes here
  • Be the first to like this

Energy, economic and environmental issues of power plants

  1. 1. POWER PLANT ENGINEERING S.BALAMURUGAN - M.E ASSISTANT PROFESSOR MECHANICAL ENGINEERING AAA COLLEGE OF ENGINEERING & TECHNOLOGY UNIT 5 โ€“ ENERGY, ECONOMIC & ENVIRONMENTAL ISSUES OF POWER PLANTS
  2. 2. SITE SELECTION FOR POWER PLANTS ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  3. 3. SITE SELECTION FOR POWER PLANTS ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  4. 4. TERMS AND DEFINITON OF LOAD CONNECTED LOAD โ€“ It is the sum of rating in kilowatts of each equipment's installed by the consumer. If consumer connects four tube lights of 40 watts, 2 fan of 60 watts. Then total connected load = (4 ร— 40) + (2 ร— 60) = 280 watts MAXIMUM LOAD โ€“ It is the maximum possible load which a customer uses at any time. DEMAND FACTOR = ๐‘€๐ด๐‘‹๐ผ๐‘€๐‘ˆ๐‘€ ๐ฟ๐‘‚๐ด๐ท ๐ถ๐‘‚๐‘๐‘๐ธ๐ถ๐‘‡๐ธ๐ท ๐ฟ๐‘‚๐ด๐ท , less than one, Connected load always high Example โ€“ Maximum demand 85 MW, Connected load 100 MW, Demand factor = 85/100 = 0.85. It is vital in determining the capacity of the plant equipment. ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  5. 5. TERMS AND DEFINITON OF LOAD LOAD FACTOR = ๐ด๐‘‰๐ธ๐‘…๐ด๐บ๐ธ ๐ฟ๐‘‚๐ด๐ท ๐‘ƒ๐ธ๐ด๐พ ๐ฟ๐‘‚๐ด๐ท = ๐ธ๐‘๐ธ๐‘…๐บ๐‘Œ ๐ถ๐‘‚๐‘๐‘†๐‘ˆ๐‘€๐ธ๐ท ๐ผ๐‘ ๐ด ๐บ๐ผ๐‘‰๐ธ๐‘ ๐‘ƒ๐ธ๐‘…๐ผ๐‘‚๐ท ๐‘‚๐น ๐‘‡๐ผ๐‘€๐ธ ๐‘ƒ๐ธ๐ด๐พ ๐ฟ๐‘‚๐ด๐ท ร— ๐ป๐‘‚๐‘ˆ๐‘…๐‘† ๐‘‚๐น ๐‘‚๐‘ƒ๐ธ๐‘…๐ด๐‘‡๐ผ๐‘‚๐‘ ๐ผ๐‘ ๐‘‡๐ป๐ธ ๐บ๐ผ๐‘‰๐ธ๐‘ ๐‘ƒ๐ธ๐‘…๐ผ๐‘‚๐ท It is always less than one. It plays key role in determining the per unit generation cost. Higher load factor of the power station, lesser will be the cost per unit generated. DIVERSITY FACTOR = ๐‘†๐‘ˆ๐‘€ ๐‘‚๐น ๐ผ๐‘๐ท๐ผ๐‘‰๐ผ๐ท๐‘ˆ๐ด๐ฟ ๐‘€๐ด๐‘‹๐ผ๐‘€๐‘ˆ๐‘€ ๐ท๐ธ๐‘€๐ด๐‘๐ท๐‘† ๐‘€๐ด๐‘‹๐ผ๐‘€๐‘ˆ๐‘€ ๐ท๐ธ๐‘€๐ด๐‘๐ท ๐‘‚๐น ๐‘‡๐ป๐ธ ๐‘Š๐ป๐‘‚๐ฟ๐ธ Example โ€“ Four individual feeder circuits with connected loads 250 kVA, 200 kVA, 150 kVA, 400 kVA 7 its demand factors of 90%, 80%, 75% & 85%. Use diversity factor of 1.5. Maximum demand of customer 1 = 250 kVA ร— 90% = 225 kVA Maximum demand of customer 2 = 200 kVA ร— 80% = 160 kVA Maximum demand of customer 3 = 150 kVA ร— 75% = 112.5 kVA Maximum demand of customer 4 = 400 kVA ร— 85% = 340 kVA Sum of individual maximum demands = 225 + 160 + 112.5 + 340 = 837.5 kVA Diversity factor = 1.5, maximum demand on the feeder circuits = 837.5 / 1.5 = 558 kVA This situation requires 600 kVA Transformer. If the diversity factor 1 means, it requires 850 kVA Transformer. Higher the diversity factor of their loads, the smaller will be the capacity of the power plant required & leads to reduce the fixed charges due to capital investment. ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  6. 6. TERMS AND DEFINITON OF LOAD PLANT CAPACITY FACTOR = ๐ธ๐‘›๐‘’๐‘Ÿ๐‘”๐‘ฆ ๐‘๐‘Ÿ๐‘œ๐‘‘๐‘ข๐‘๐‘’๐‘‘ ๐‘–๐‘› ๐‘Ž ๐‘”๐‘–๐‘ฃ๐‘’๐‘› ๐‘๐‘’๐‘Ÿ๐‘–๐‘œ๐‘‘ ๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ก๐‘•๐‘’ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ร— ๐‘‡๐‘œ๐‘ก๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘•๐‘œ๐‘ข๐‘Ÿ๐‘  ๐‘–๐‘› ๐‘”๐‘–๐‘ฃ๐‘’๐‘› ๐‘๐‘’๐‘Ÿ๐‘–๐‘œ๐‘‘ A power station is so designed that it has some reserve capacity for meeting the increased load demand in future. Therefore, the installed capacity of the plant is always greater than the maximum demand of the plant. Reserve Capacity = Plant capacity โ€“ Maximum Demand PLANT USE FACTOR = ๐ธ๐‘›๐‘’๐‘Ÿ๐‘”๐‘ฆ ๐‘๐‘Ÿ๐‘œ๐‘‘๐‘ข๐‘๐‘’๐‘‘ ๐‘–๐‘› ๐‘Ž ๐‘”๐‘–๐‘ฃ๐‘’๐‘› ๐‘๐‘’๐‘Ÿ๐‘–๐‘œ๐‘‘ ๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ก๐‘•๐‘’ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ร—๐ด๐‘๐‘ก๐‘ข๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘•๐‘œ๐‘ข๐‘Ÿ๐‘  ๐‘ก๐‘•๐‘’ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ๐‘–๐‘› ๐‘œ๐‘๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ๐‘› A plant having installed capacity of 20 MW produces annual output of 7.8 ร— 106 kWh & remains in operation for 2190 hours in a year. Then Plant use factor = 7.8 ร— 106 / (20 ร— 106 ร— 2190) = 0.178 = 17.8 % UTILIZATION FACTOR = ๐‘€๐‘Ž๐‘ฅ๐‘–๐‘š๐‘ข๐‘š ๐‘‘๐‘’๐‘š๐‘Ž๐‘›๐‘‘ ๐‘œ๐‘› ๐‘ก๐‘•๐‘’ ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ก๐‘•๐‘’ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  7. 7. LOAD CURVE & LOAD DURATION CURVE Load curve โ€“ A graphical record showing the power demands for every instant during a certain time interval. Load Duration Curve โ€“ It represents re-arrangements of all the load elements of chronological load curve in order of descending magnitude. Significance of load curve โ€ข It shows the variations of load on the power station during different hours of the day โ€ข The area under the daily load curve gives the number of units generated in the day โ€ข The highest point on the daily load curve represents the maximum demand on the station on that day โ€ข It helps in selecting the size & the number of generating units โ€ข It helps in preparing the operation schedule of the station โ€ข The area under the curve divided by the total number of hours gives the average load. AVERAGE LOAD = ๐‘‡๐‘‚๐‘‡๐ด๐ฟ ๐ธ๐‘๐ธ๐‘…๐บ๐‘Œ ๐‘‡๐ผ๐‘€๐ธ ๐‘ƒ๐ธ๐‘…๐ผ๐‘‚๐ท = ๐ธ๐‘๐ธ๐‘…๐บ๐‘Œ ๐ถ๐‘‚๐‘๐‘†๐‘ˆ๐‘€๐ธ๐ท ๐ผ๐‘ 24 ๐ป๐‘‚๐‘ˆ๐‘…๐‘† 24 ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  8. 8. POWER FACTOR The power factor plays a major role in the plant economics. The low power factor increases the load current which increases the losses in the system. Thus, the regulation becomes poor. For improving the power factor, the power factor correction equipment is installed at the generating station. Thus, the cost of the generation increases. ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  9. 9. POWER TARIFF TYPESโ€ข The actual tariffs that the customer pay depends on the consumption of the electricity. โ€ข The consumer bill varies according to their requirements. โ€ข The industrial consumers pay more tariffs because they use more power for long times than the domestic consumers. The electricity tariffs depends on the following factors โ€ข Type of load โ€ข Time at which load is required. โ€ข The power factor of the load. โ€ข The amount of energy used. The total bill of the consumer has three parts, namely, fixed charge C, semi-fixed charge Ax and running charge By. Z = A X + B Y + C where, Z โ€“ total charge for a period x โ€“ maximum demand during the period (kW), A โ€“ cost per kW y โ€“ Total energy consumed during the period (kW), B โ€“ cost per kWh of energy consumed, C โ€“ fixed charge during each billing period Power Factor is a measure of how effectively incoming power is used in your electrical system and is defined as the ratio of Real (working) power to Apparent (total) power ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  10. 10. FLAT DEMAND TARIFF TOTAL AMOUNT Z = A X โ€ข In this type of tariff, the bill of the power consumption depends only on the maximum demand of the load. โ€ข The generation of the bill is independent of the normal energy consumption. โ€ข This type of tariff is used on the street light, sign lighting, irrigation, etc., where the working hours of the equipment are known. โ€ข The metering system is not used for calculating such type of tariffs. STRAIGHT LINE METER RATE TOTAL AMOUNT Z = B Y โ€ข The generation of the bills depends on the energy consumption of the load. โ€ข The customer need not pay anymore even though the plant incurred some expenditure to provide readiness to serve the customers โ€ข It doesnโ€™t encourage the customers to use electricity POWER TARIFF TYPES ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  11. 11. BLOCK METER RATE TOTAL AMOUNT Z = B1 Y1 + B2 Y2 + B3 Y3 โ€ข In this type of tariff, a particular amount per unit is charged for the particular block of each unit โ€ข The reduced price per unit are charged for the succeeding block of units. โ€ข This type of tariff reduces the unit energy charges with increasing consumptions. โ€ข Demerit โ€“ Lacks a measure of the customer demand. TWO PART TARIFF TOTAL AMOUNT Z = A X + B Y โ€ข The generation of the bills depends on maximum demand & the energy consumption of the load. โ€ข It requires two meters to record maximum demand & energy consumption of the consumer. โ€ข Used for industrial customers. POWER TARIFF TYPES ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  12. 12. THREE PART TARIFF TOTAL AMOUNT Z = A X + B Y + C โ€ข In this type of tariff, the total charge to be made from the consumer is split into three parts. โ€ข By adding fixed charges to two part tariff, it becomes three part tariff. โ€ข This type of tariff is applied to big consumers. POWER TARIFF TYPES ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  13. 13. Time hours A power station supplies the following loads are necessary to the customers Find 1) Draw the load curve & Estimate the load factor of the plant 2) What is the load factor of a standby equipment of 30 MW capacity if it takes up all loads above 70 MW 3) What is its Use factor ? Solution Energy Generated = Area under the curve = 30 ร— 6 + 70 ร— 4 + 90 ร— 2 + 60 ร— 4 + 100 ร— 4 + 80 ร— 2 + 60 ร— 2 = 1560 MWh Load factor = ๐‘จ๐‘ฝ๐‘ฌ๐‘น๐‘จ๐‘ฎ๐‘ฌ ๐‘ณ๐‘ถ๐‘จ๐‘ซ ๐‘ท๐‘ฌ๐‘จ๐‘ฒ ๐‘ณ๐‘ถ๐‘จ๐‘ซ , Average Load = ๐Ÿ๐Ÿ“๐Ÿ”๐ŸŽ ๐Ÿ๐Ÿ’ = 65 MW Maximum Demand = 100 MW, Load factor = ๐Ÿ”๐Ÿ“ ๐Ÿ๐ŸŽ๐ŸŽ = 0.65 Load Factor of a Standby Equipment The standby equipment's supplies the power at three situations (more than 70 MW) 90 โ€“ 70 = 20 MW for 2 hours, 100 โ€“ 70 = 30 MW for 4 hours, 80-70 = 10 MW for 2 hours Energy Generated = Area under the curve = 20 ร— 2 + 30 ร— 4 + 10 ร— 2 = 180 MWh Total Operation hours for standby equipment = 2 + 4 + 2 = 8 hours Average Load = ๐Ÿ๐Ÿ–๐ŸŽ ๐Ÿ– = 22.5 MW, Load factor = ๐Ÿ๐Ÿ.๐Ÿ“ ๐Ÿ‘๐ŸŽ = 0.75 Use Factor = ๐ธ๐‘›๐‘’๐‘Ÿ๐‘”๐‘ฆ ๐‘๐‘Ÿ๐‘œ๐‘‘๐‘ข๐‘๐‘’๐‘‘ ๐‘–๐‘› ๐‘Ž ๐‘”๐‘–๐‘ฃ๐‘’๐‘› ๐‘๐‘’๐‘Ÿ๐‘–๐‘œ๐‘‘ ๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ก๐‘•๐‘’ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ร—๐ด๐‘๐‘ก๐‘ข๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘•๐‘œ๐‘ข๐‘Ÿ๐‘  ๐‘ก๐‘•๐‘’ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ๐‘–๐‘› ๐‘œ๐‘๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ๐‘› = ๐Ÿ๐Ÿ–๐ŸŽ ๐Ÿ‘๐ŸŽ ร— ๐Ÿ– = 0.75
  14. 14. The maximum demand of a power station is 96000 kW & daily load curve is described as follows (i) Determine the load factor of power station (ii) What is the load factor of standby equipment rated at 30 MW that takes up all load in excess of 72 MW? Also calculate its Use factor. Solution Energy Generated = Area under the curve = 48 ร— 6 + 60 ร— 2 + 72 ร— 4 + 60 ร— 2 + 84 ร— 4 + 96 ร— 4 + 48 ร— 2 = 1632 MWh Load factor = ๐‘จ๐‘ฝ๐‘ฌ๐‘น๐‘จ๐‘ฎ๐‘ฌ ๐‘ณ๐‘ถ๐‘จ๐‘ซ ๐‘ท๐‘ฌ๐‘จ๐‘ฒ ๐‘ณ๐‘ถ๐‘จ๐‘ซ , Average Load = ๐Ÿ๐Ÿ”๐Ÿ‘๐Ÿ ๐Ÿ๐Ÿ’ = 68 MW Maximum Demand = 96000 kW = 96 MW, Load factor = ๐Ÿ”๐Ÿ– ๐Ÿ—๐Ÿ” = 0.71 Load Factor of a Standby Equipment The standby equipment's supplies the power at two situations (more than 72 MW) 84 โ€“ 72 = 12 MW for 4 hours, 96 โ€“ 72 = 24 MW for 4 hours Energy Generated = Area under the curve = 12 ร— 4 + 24 ร— 4 = 144 MWh Total Operation hours for standby equipment = 4 + 4 = 8 hours Average Load = ๐Ÿ๐Ÿ’๐Ÿ’ ๐Ÿ– = 18 MW, Load factor = ๐Ÿ๐Ÿ– ๐Ÿ๐Ÿ’ = 0.75 Use Factor = ๐ธ๐‘›๐‘’๐‘Ÿ๐‘”๐‘ฆ ๐‘๐‘Ÿ๐‘œ๐‘‘๐‘ข๐‘๐‘’๐‘‘ ๐‘–๐‘› ๐‘Ž ๐‘”๐‘–๐‘ฃ๐‘’๐‘› ๐‘๐‘’๐‘Ÿ๐‘–๐‘œ๐‘‘ ๐ถ๐‘Ž๐‘๐‘Ž๐‘๐‘–๐‘ก๐‘ฆ ๐‘œ๐‘“ ๐‘ก๐‘•๐‘’ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ร—๐ด๐‘๐‘ก๐‘ข๐‘Ž๐‘™ ๐‘›๐‘ข๐‘š๐‘๐‘’๐‘Ÿ ๐‘œ๐‘“ ๐‘•๐‘œ๐‘ข๐‘Ÿ๐‘  ๐‘ก๐‘•๐‘’ ๐‘๐‘™๐‘Ž๐‘›๐‘ก ๐‘–๐‘› ๐‘œ๐‘๐‘’๐‘Ÿ๐‘Ž๐‘ก๐‘–๐‘œ๐‘› = ๐Ÿ๐Ÿ’๐Ÿ’ ๐Ÿ‘๐ŸŽ ร— ๐Ÿ– = 0.6 Time hours ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  15. 15. A peak load on the thermal power plant is 75MW. The loads having maximum demands of 35MW, 20MW, 15MW & 18MW are connected to the power plant. The capacity of the plant is 90MW and annual load factor is 0.53. Calculate the average load on power plant, energy supplied per year, demand factor and Diversity factor. Given Max. demand or Peak load = 75 MW, Capacity of the plant = 90 MW, Annual Load factor = 0.53 Solution Load factor = ๐‘จ๐‘ฝ๐‘ฌ๐‘น๐‘จ๐‘ฎ๐‘ฌ ๐‘ณ๐‘ถ๐‘จ๐‘ซ ๐‘ท๐‘ฌ๐‘จ๐‘ฒ ๐‘ณ๐‘ถ๐‘จ๐‘ซ , 0.53 = ๐‘จ๐‘ฝ๐‘ฌ๐‘น๐‘จ๐‘ฎ๐‘ฌ ๐‘ณ๐‘ถ๐‘จ๐‘ซ ๐Ÿ•๐Ÿ“ , Average Load = 39.75 MW Energy Supplied per Year = Average load ร— Number of hours in one year = 39.75 ร— 365 ร— 24 = 348210 MWh Demand Factor = ๐‘€๐ด๐‘‹๐ผ๐‘€๐‘ˆ๐‘€ ๐ฟ๐‘‚๐ด๐ท ๐ถ๐‘‚๐‘๐‘๐ธ๐ถ๐‘‡๐ธ๐ท ๐ฟ๐‘‚๐ด๐ท = 75 35+20+15+18 = 0.85 Diversity Factor = ๐‘†๐‘ˆ๐‘€ ๐‘‚๐น ๐ผ๐‘๐ท๐ผ๐‘‰๐ผ๐ท๐‘ˆ๐ด๐ฟ ๐‘€๐ด๐‘‹๐ผ๐‘€๐‘ˆ๐‘€ ๐ท๐ธ๐‘€๐ด๐‘๐ท๐‘† ๐‘€๐ด๐‘‹๐ผ๐‘€๐‘ˆ๐‘€ ๐ท๐ธ๐‘€๐ด๐‘๐ท ๐‘‚๐น ๐‘‡๐ป๐ธ ๐‘Š๐ป๐‘‚๐ฟ๐ธ = 35+20+15+18 75 = 1.17 ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  16. 16. ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  17. 17. The following data for a 2200 kW diesel power station is given. The peak load on the plant is 1600kW and its load factor is 45%. Capacity cost / kW installed = Rs.15000, Annual costs = 15% of capital, Annual Maintenance cost = Fixed Rs. 1,00,000 and Variable cost Rs.2,00,000, annual operating cost = Rs.6,00,000. , Cost of Fuel = Rs.0.8 per kg, Cost of lubricating oil = Rs. 40 per kg, C.V of Fuel = 40,000 kJ/kg, Consumption of fuel = 0.5 kg/kwh, Consumption of Lubricating oil = 1/400 kg/kwh. Determine (a) the annual energy produced (b)cost of generation per kwh (c) Efficiency Solution Capital Cost = Rs. 15,000 / kW Capital cost of 2200 kW plant = 15,000 ร— 2200 = Rs. 33 ร— 106 Annual cost = 15 % of Capital = 15 100 ร— 33 ร— 106 = Rs. 4.95 ร— 106 Load Factor = ๐‘จ๐‘ฝ๐‘ฌ๐‘น๐‘จ๐‘ฎ๐‘ฌ ๐‘ณ๐‘ถ๐‘จ๐‘ซ ๐‘ท๐‘ฌ๐‘จ๐‘ฒ ๐‘ณ๐‘ถ๐‘จ๐‘ซ , 0.45 = ๐‘จ๐‘ฝ๐‘ฌ๐‘น๐‘จ๐‘ฎ๐‘ฌ ๐‘ณ๐‘ถ๐‘จ๐‘ซ ๐Ÿ๐Ÿ”๐ŸŽ๐ŸŽ , Average Load = 720 kW Average Load = ๐‘‡๐‘‚๐‘‡๐ด๐ฟ ๐ธ๐‘๐ธ๐‘…๐บ๐‘Œ ๐‘‡๐ผ๐‘€๐ธ ๐‘ƒ๐ธ๐‘…๐ผ๐‘‚๐ท , Total energy produced = 720 ร— 365 ร— 24 = 6.31 ร— 106 kWh Consumption of fuel = 0.5 kg/kwh Fuel cost = Rs.0.8 per kg For 6.31 ร— 106 kWh, Mass of Fuel = 0.5 ร— 6.31 ร— 106 = 3.154 ร— 106 kg Fuel cost = 0.8 ร— 3.154 ร— 106 = Rs. 2.523 ร— 106 Consumption of Lubricating oil = 1/400 kg/kwh Cost of lubricating oil = Rs. 40 per kg For 6.31 ร— 106 kWh, Mass of Lubricating oil = (1/400) ร— 6.31 ร— 106 = 15775 kg Lubri. oil cost = 0.40 ร— 15775 = Rs. 0.631 ร— 106 ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  18. 18. The following data for a 2200 kW diesel power station is given. The peak load on the plant is 1600kW and its load factor is 45%. Capacity cost / kW installed = Rs.15000, Annual costs = 15% of capital, Annual Maintenance cost = Fixed Rs. 1,00,000 and Variable cost Rs.2,00,000, annual operating cost = Rs.6,00,000. , Cost of Fuel = Rs.0.8 per kg, Cost of lubricating oil = Rs. 40 per kg, C.V of Fuel = 40,000 kJ/kg, Consumption of fuel = 0.5 kg/kwh, Consumption of Lubricating oil = 1/400 kg/kwh. Determine (a) the annual energy produced (b)cost of generation per kwh (c) Efficiency. Solution Continued from previous slide Fixed cost = Annual cost + Annual maintenan. fixed cost = 4.95 ร— 106 + 1,00,000 = Rs. 5.05 ร— 106 Variable cost = 6,00,000 + 2,00,000 + 2.523 ร— 106 + 0.631 ร— 106 = Rs. 3.954 ร— 106 Total Cost = Fixed cost + Variable cost = 5.05 ร— 106 + 3.954 ร— 106 = Rs. 9.004 ร— 106 Cost of Power = ๐‘ป๐’๐’•๐’‚๐’ ๐‘ช๐’๐’”๐’• ๐‘ฌ๐’๐’†๐’“๐’ˆ๐’š ๐‘ท๐’“๐’๐’…๐’–๐’„๐’†๐’… = 9.004 ร— 106 6.31 ร— 106 = Rs. 1.43 Efficiency = ๐‘ถ๐’–๐’•๐’‘๐’–๐’• ๐‘ฐ๐’๐’‘๐’–๐’• Output power = E = 6.31 ร— 106 kWh = 6.31 ร— 106 ร— 3600 = 2.2716 ร— 1010 kW-sec = kJ Input Power = mโ€™ ร— CV = 3.154 ร— 106 ร— 40,000 = 1.2626 ร— 1011 kJ ฮท = 2.2716 ร— 1010 1.2626 ร— 1011 = 0.18 = 18% kW.sec = k( ๐‘ฑ ๐‘บ ).S = kJ ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  19. 19. POLLUTION CONTROL TECHNOLOGY โ€“ COAL BASED POWER PLANT โ€ข Environment pollution by thermal power plant is more than other power plant โ€ข Air pollution is caused by thermal power plant is very high by burning fuels like coal. โ€ข Combustible elements of fuel are converted into Gaseous products โ€ข Non-Combustible elements to Ash โ€ข Normally the thermal power plant has high sulphur content coal & it must be removed before burnt. โ€ข Some sulphur oxides & the fly ash will be removed by coal washing. ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  20. 20. EMISSION GASEOUS EMISSION SULHUR DIOXIDE OXIDES OF NITROGEN HYDROGEN SULPHIDE CARBON MONOXIDE PARTICULATE EMISSION DUST (DIA 1 MICRON) PARTICLES (SMOKE, FUMES, FLY-ASH, CINDERS) (DIA 10 MICRON) SOLID WASTE EMISSION ASH, CALCIUM & MAGNESIUM SALTS THERMAL POLLUTION DISCHARGE OF THERMAL ENERGY INTO WATER ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  21. 21. REMOVAL OF SULPHUR DIOXIDE (SO2) WET SCRUBBER โ€“ Polluted gas is sent into the scrubber tangentially. โ€“ Water spray absorbs these gases & particulate matters โ€“ It is collected on the surface of the scrubber are washed down by water โ€“ This water again treated, filtered & reused โ€“ Used to remove the particulate matters. โ€“ Here the gases are cooled, they must be reheated before being sent to the stack โ€“ The used water contain sulphuric & sulphurous acid which corrode the pipeline & scrubber ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  22. 22. POLLUTION CONTROL TECHNOLOGY โ€“ COAL BASED POWER PLANT EMISSION OF NOx โ€ข The combustion of fossil fuels in air is accomplished by the formation of nitric oxide which is subsequently partly oxidized to Nitrogen Oxide. โ€ข NOx present in stack gases from coal, oil & gas furnaces METHODS TO REDUCE THE EMISSION OF NOx โ€ข Reduction of residence period in combustion zone โ€ข Reduction of temperature in combustion zone PARTICULATE EMISSION & ITS CONTROL โ€ข Dust โ€“ Dia โ€“ 1micron โ€“ which do not settle down โ€ข Particles โ€“ Dia โ€“ 10 microns โ€“ which settle down to the ground EMISSION OF PARTICLES โ€ข Smoke โ€“ Stable suspension of particles โ€“ Visible โ€“ Dia < 10 microns โ€ข Fumes โ€“ Very small particles which is obtained from chemical reaction. It is composed of metals & metallic oxides โ€ข Fly-Ash โ€“ Diameter less than or equal to 100 microns โ€ข Cinders โ€“ Diameter of particles are more than or equal to 100 microns ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  23. 23. POLLUTION CONTROL TECHNOLOGY โ€“ COAL BASED POWER PLANT CINDER CATCHERS โ€ข The performance parameters for any particulate remover is called the collector efficiency. Collector efficiency = ๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘‘๐‘ข๐‘ ๐‘ก ๐‘Ÿ๐‘’๐‘š๐‘œ๐‘ฃ๐‘’๐‘‘ ๐‘€๐‘Ž๐‘ ๐‘  ๐‘œ๐‘“ ๐‘‘๐‘ข๐‘ ๐‘ก ๐‘๐‘Ÿ๐‘’๐‘ ๐‘’๐‘›๐‘ก ร— 100 Collector efficiency varies from 50 โ€“ 99 % Electrostatic precipitator > 90 % 50 โ€“ 75 % Stroker & small cyclone furnace(Crushed coal) Centrifugal force 50 โ€“ 75 % ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  24. 24. POLLUTION CONTROL TECHNOLOGY โ€“ COAL BASED POWER PLANT BAGHOUSE FILTERS โ€ข Used to remove the particulate matters where low sulphur coal is used. โ€ข Filter is cleaned by applying a forced air in reversed direction โ€ข Large filter area required โ€“ Cost high โ€ข Life period โ€“ 3 years โ€ข Coal firing systems ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  25. 25. ELECTRO STATIC PRECIPITATOR โ€ข The medium between the electrodes is air, and due to the high negativity of negative electrodes, there may be a corona discharge surround the negative electrode rods or wire mesh. The air molecules in the field between the electrodes become ionized, and hence there will be plenty of free electrons and ions in the space โ€ข The flue gases enter into the electrostatic precipitator, dust particles in the gases collide with the free electrons available in the medium between the electrodes and the free electrons will be attached to the dust particles. ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  26. 26. โ€ข As a result, the dust particles become negatively charged. Then these negatively charged particles will be attracted due to electrostatic force of the positive plates. โ€ข Consequently, the charged dust particles move towards the positive plates and deposited on positive plates. Here, the extra electron from the dust particles will be removed on positive plates, and the particles then fall due to gravitational force. โ€ข We call the positive plates as collecting plates. The flue gases after travelling through the electrostatic precipitator become almost free from ash particles and ultimately get discharged to the atmosphere through the chimney ELECTRO STATIC PRECIPITATOR ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  27. 27. โ€ข Large amount of ash is formed by burning the coal in the furnace โ€ข This ash is removed as bottom ash from furnace โ€ข From discharged solid wastes, calcium & magnesium can be generated by absorption of SO2 & SO3 by reactant like lime stone. POLLUTION CONTROL TECHNOLOGY โ€“ COAL BASED POWER PLANT SOLID WASTE DISPOSAL ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  28. 28. โ€ข Discharge of thermal energy into water is called thermal pollution โ€ข Water is the main medium to condense the steam. If this heated water is discharged into lakes or rivers, the water temperature goes up. โ€ข At about 35ยฐC, the dissolved oxygen will be low that the aquatic life will die. โ€ข As per regulation - 1ยฐC more then the atmospheric temperature acceptable โ€ข In very cold regions, discharging the hot water into lakes helps in increasing fish growth. THERMAL DISCHARGE INDEX โ€“ Estimates the thermal efficiency of plant TDI = ๐‘‡๐‘•๐‘’๐‘Ÿ๐‘š๐‘Ž๐‘™ ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ ๐‘‘๐‘–๐‘ ๐‘๐‘•๐‘Ž๐‘Ÿ๐‘”๐‘’๐‘‘ ๐‘ก๐‘œ ๐‘’๐‘›๐‘ฃ๐‘–๐‘Ÿ๐‘œ๐‘›๐‘š๐‘’๐‘›๐‘ก ๐‘–๐‘› ๐‘€๐‘Š ๐‘ก๐‘•๐‘’๐‘Ÿ๐‘š๐‘Ž๐‘™ ๐ธ๐‘™๐‘’๐‘๐‘ก๐‘Ÿ๐‘–๐‘๐‘Ž๐‘™ ๐‘๐‘œ๐‘ค๐‘’๐‘Ÿ ๐‘œ๐‘ข๐‘ก๐‘๐‘ข๐‘ก ๐‘–๐‘› ๐‘€๐‘Š ๐ธ๐‘™๐‘’๐‘๐‘ก๐‘Ÿ๐‘–๐‘๐‘Ž๐‘™ POLLUTION CONTROL TECHNOLOGY โ€“ COAL BASED POWER PLANT THERMAL POLLUTION ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  29. 29. METHODS TO REDUCE THERMAL POLLUTION โ€“ Construction of a separate lake โ€“ once through cooling the condenser is adopted. If natural cooling of water from the lake is not sufficient means, floating spray pumps can be employed. Expensive โ€“ Cooling pond - A cooling pond with continuously operating fountains can be adopted. Aesthetic look in the power plant โ€“ Cooling tower โ€“ In order to throw heat to atmosphere, most power stations use cooling tower. POLLUTION CONTROL TECHNOLOGY โ€“ COAL BASED POWER PLANT THERMAL POLLUTION ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  30. 30. Mining โ€“ Depending on the depth and concentration of the uranium source, and the conditions of the surrounding rock, mining companies will extract uranium ore in one of three ways: open pit mining, underground mining or in-situ recovery. Milling โ€“ To extract the uranium, the ore is crushed in a mill and ground to a fine slurry. The slurry is then leached in sulfuric acid, which produces a solution of uranium oxide (U3O8). The concentrate of this solution is called yellowcake. Refining โ€“ A series of chemical processes separate the uranium from impurities, producing high-purity uranium trioxide (UO3). Conversion โ€“ UO3 is converted to uranium dioxide (UO2) for use in heavy water reactors, UO3 to uranium hexafluoride (UF6) for enrichment, before it can be used in light water reactors. Enrichment (Gas Centrifugation)โ€“ Uranium-235 is the uranium isotope that can be used in fission, but it makes up only 0.7% of naturally occurring uranium, which is not concentrated enough for light water reactors. So, enrichment processes increase the concentration of U-235 to about 3% โ€“ 5%. After undergoing enrichment, the UF6 is chemically transformed back into UO2 powder. NUCLEAR FUEL CYCLE ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  31. 31. Fuel manufacturing โ€“ Natural or enriched UO2 powder is pressed into small cylindrical pellets, which are then baked at high temperatures, and finished to precise dimensions. Electricity generation โ€“ Fuel is loaded into a reactor, and nuclear fission generates electricity. After fuel is consumed, it is removed from the reactor and stored onsite for a number of years while its radioactivity and heat subside. Optional chemical reprocessing โ€“ After a period of storage, residual uranium or by-product plutonium, both of which are still useful sources of energy, are recovered from the spent fuel elements and reprocessed. Disposal โ€“ Depending on the design of the disposal facility, the nuclear fuel may be recovered if needed again, or remain permanently stored. At some point in the future the spent fuel will be encapsulated in sturdy, leach-resistant containers and permanently placed deep underground where it originated. NUCLEAR FUEL CYCLE ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  32. 32. TYPES OF NUCLEAR WASTE โ€ข High Level Waste (3%) โ€“ Spent fuel containing 95% of radioactivity in the nuclear waste โ€ข Intermediate Waste (7%) โ€“ Used filters, steel components from within the reactor & some effluents from reprocessing containing 4% of radioactivity in the nuclear waste โ€ข Low level waste (90%) โ€“ Lightly contaminated items like tools & clothing containing only 1% of radioactivity in the nuclear waste. โ€ข Both LLW and ILW from a nuclear power station will be isolated from the environment for typically 300 years so that their radioactivity will reduce with time to natural levels. โ€ข HLW is highly radioactive. After being extracted from the spent fuel, it typically needs a period of 20 to 50 years to cool down. โ€ข Technology has been developed to pack the waste in glass in a process known as vitrification. โ€ข The packaged vitrified waste can then be stored underground in a stable geological formation to isolate it and prevent its movement for over thousands of years. Furthermore, the radioactivity will fall over time and after several thousand years it will fall to natural levelsME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  33. 33. โ€ข Spent fuel is processed at facilities in Trombay near Mumbai, at Tarapur on the west coast north of Mumbai, and at Kalpakkam on the southeast coast of India. โ€ข Plutonium will be used in a fast breeder reactor (under construction) to produce more fuel, and other waste vitrified at Tarapur and Trombay. โ€ข Interim storage for 30 years is expected, with eventual disposal in a deep geological repository in crystalline rock near Kalpakkam. NUCLEAR WASTE PROCESSING IN INDIA ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  34. 34. NUCLEAR WASTE - PROCESSING โ€ข Highly reactive materials created during production of nuclear power. โ€“ Core of the nuclear reactor or nuclear weapon โ€“ Uranium, plutonium, high reactive elements from fission โ€“ Long lives> 100000years Management of high level waste in India following three stages Immobilization- Vitrified borosilicate glasses Engineered interim storage of vitrified waste for passive cooling & surveillance over a period of time Ultimate storage / disposal of the vitrified waste a deep geological depository deep geological depository requirements โ€“ remoteness from environment โ€“ absence of circulating ground water ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  35. 35. ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  36. 36. ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET
  37. 37. Joule Heated Ceramic Melter (JHCM) ME 6701 POWER PLANT ENGG. S.BALAMURUGAN AP/MECH AAACET

ร—