1 / 63

NBS-M016 Contemporary Issues in Climate Change and Energy 2010

N.K. Tovey ( 杜伟贤 ) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science Director C Red Project HSBC Director of Low Carbon Innovation. NBS-M016 Contemporary Issues in Climate Change and Energy 2010. Introduction to Thermodynamics Combined Cycle Gas Turbines

joan-craft
Download Presentation

NBS-M016 Contemporary Issues in Climate Change and Energy 2010

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. N.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science DirectorCRedProject HSBC Director of Low Carbon Innovation NBS-M016 Contemporary Issues in Climate Change and Energy 2010 • Introduction to Thermodynamics • Combined Cycle Gas Turbines • Combined Heat and Power • Heat Pumps Lecture 2 Lecture 3 Lecture 1 1

  2. 10. Elementary Thermodynamics - History. 1) Boil Water> Steam Problem: Cylinder continually is cooled and heated. 2) Open steam valve pushes piston up (and pumping rod down) 3) At end of stroke, close steam valueopen injection valve 4) Water sprays in condenses steam in cylinder creating a vacuum and sucks piston down - and pumping rod up Newcomen Engine 2

  3. 10. Elementary Thermodynamics - Watt Engine. 1) Cylinder is always warm 2) cold water is injected into condenser 3) vacuum is maintained in condenser so “suck” out exhaust steam. 4) steam pushes piston down pulling up pumping rod. Higher pressure steam used in pumping part of cycle. • Watt Engine 3

  4. 10. Elementary Thermodynamics. •  Thermodynamics is a subject involving logical reasoning. • Much of it was developed by intuitive reasoning. • 1825 - 2nd Law of Thermodynamics - Carnot • 1849 - 1st Law of Thermodynamics - Joule • Zeroth Law - more fundamental - a statement • about measurement of temperature • Third Law - of limited relevance for this Course 4

  5. 10. Elementary Thermodynamics. Carnot’s reasoning Water at top has potential energy Water at bottom has lost potential energy but gained kinetic energy 5

  6. H1 H2 10. Elementary Thermodynamics. Carnot’s reasoning Water looses potential energy Part converted into rotational energy of wheel Potential Energy = mgh • Theoretical Energy Available = m g (H1 - H2) • Practically we can achieve 85 - 90% of this 6

  7. 10. Elementary Thermodynamics. Carnot’s reasoning • Temperature was analogous to Head of Water • Energy  Temperature Difference • Energy  (T1 - T2) • T1 is inlet temperature • T2 is outlet temperature • Just as amount of water flowing in = water flowing out. • Heat flowing in = heat flowing out. • In this respect Carnot was wrong • However, in his day the difference was < 1% 7

  8. 10. Elementary Thermodynamics. • Joule 1849 • Identified that “Lost” Heat = energy out as Work • Use a paddle wheel to stir water - the water will heat up • Mechanical Equivalent of Heat • Berlin Demonstration • Symbols • W - work Q - heat • Over a complete cycle • Q = W • Heat in +ve Heat out -ve • Work in -ve Work out +ve • FIRST LAW: “You can’t get something for nothing” 8

  9. Heat In Q1 Work Out W Heat Out Q2 Schematic Representation of a Power Unit 10. Elementary Thermodynamics. First Law: W = Q1 - Q2 so efficiency Heat Engine But Carnot saw that Heat  Temperature • What do we mean by temperature? Rankine, Kelvin? Reamur, Fahrenheit, Celcius, • Which should we use? 9

  10. 10. Elementary Thermodynamics. Is this a sensible definition of efficiency? If T1 = 527oC ( = 527 + 273 = 800K) and T2 = 27oC ( = 300K) Note: This is a theoretical MAXIMUM efficiency 10

  11. 10. Elementary Thermodynamics. • Second Law is more restrictive than First • “It is impossible to construct a device operating in a cycle which exchanges heat with a SINGLE reservoir and does an equal amount of work on the surrounds” • This means Heat must always be rejected • Second Law cannot be proved • - fail to disprove the Law • If heat is rejected at 87oC (360K) By keeping T2 at a potentially useful temperature, efficiency has fallen from 62.5% 11

  12. 10. Elementary Thermodynamics. The Practical efficiency will always be less than the Theoretical Carnot Efficiency. To obtain the "real" efficiency we define the term Isentropic Efficiency as follows:- Thus "real" efficiency = carnot x isen Typical values of isen are in range 75 - 80% Hence in a normal turbine, actual efficiency = 48% A power station involves several energy conversions. The overall efficiency is obtained from the product of the efficiencies of the respective stages. 12

  13. 10. Elementary Thermodynamics. • EXAMPLE: • In a large coal fired power station like DRAX (4000MW), the steam inlet temperature is 566oC and the exhaust temperature to the condenser is around 30oC. • The combustion efficiency is around 90%, while the generator efficiency is 95% and the isentropic efficiency is 75%. • If 6% of the electricity generated is used on the station itself, and transmission losses amount to 5% and the primary energy ratio is 1.02, how much primary energy must be extracted to deliver 1 unit of electricity to the consumer? 13

  14. 10. Elementary Thermodynamics. • (566 + 273) - (30 + 273) • Carnot efficiency = ------------------------------ = 63.9% • 566 + 273 • so overall efficiency in power station:- = 0.9 x | combustion loss 0.639 x | Carnot efficiency 0.75 x | Isentropic efficiency 0.94 | Station use 0.95 x | Generator efficiency = 0.385 14

  15. 10. Elementary Thermodynamics. Transmission Loss ~ 91.5% efficient Primary Energy Ratio for Coal ~ 1.02 Overall efficiency 1 x 0.385 x 0.915 = -------------------------- = 0.345 units of delivered energy 1.02 i.e. 1 / 0.345 = 2.90 units of primary energy are needed to deliver 1 unit of electricity. 15

  16. 10. Elementary Thermodynamics. • How can we improve Carnot Efficiency? • Increase T1 or decrease T2 • If T2 ~ 0 the efficiency approaches 100% • T2 cannot be lower than around 0 - 30oC i.e. 273 - 300 K • T1 can be increased, but properties of steam limit maximum temperature to around 600oC, (873K) • We can improve matters by the use of combined cycle gas turbine stations CCGTs. 16

  17. 10. Elementary Thermodynamics. In this part of the lecture we shall explore ways to improve efficiency Most require us to ensure we work with thermodynamics rather than against it The most important equation:

  18. 11. Applications of Thermodynamics. Other modes of Electricity Generation: Open Circuit Gas turbinesc Efficiency is low - exhaust temperature is high --- (T1 - T2)/T1 - similar to an aircraft engine 18

  19. 11. Applications of Thermodynamics. Combined Cycle Gas Turbines Waste Heat Combined Cycle Gas Turbine Practical Efficiencies:- Gas Turbine alone 20 - 25% Steam Turbine alone 35 - 38% CCGT 47 - 52% 19

  20. 11. Applications of Thermodynamics. Combined Cycle Gas Turbines: Multiple Shaft Example 0.22 0.23 1.0 Generator Gas Turbine 0.77 0.01 0.15 Waste Heat Boiler 0.02 0.28 0.62 0.30 Generator Steam Turbine 0.32 Condenser Gas turbine T1 = 950oC = 1223 K T2 = 500oC = 823K Electricity Steam turbine T1 = 500oC = 773 K T2 = 30oC = 303K Electricity Isentropic efficiency ~ 80% • Output from Gas Turbine: 0.23 units of power to generator and 0.77 units to WHB • Generator is ~ 95% efficient so output ~ 0.22 units • Waste Heat boiler is ~ 80% efficient so there will be ~ 0.15 units lost with 0.8*0.77=0.62 • units effective for raising steam. • Shaft power from Steam turbine = 0.62 * 0.486 = 0.30 units with 0.32 units to condenser • Total electrical output = 0.22 + 0.28 = 0.50units of which 0.03 units are used on station • Overall efficiency = 47% 20

  21. 11. Applications of Thermodynamics. Combined Cycle Gas Turbines: • Early CCGTs had multiple shafts with separate generators attached to gas turbines • Some had two or more gas turbines providing heat to waste heat boilers which powered a single steam turbine • Modern CCGT’s tend to have a common shaft with a gas turbine and steam turbine turning a single generator. • Advantages of single shaft machines: • tend to have lower capital cost • Tend to have higher overall efficiencies up to 55/56% - e.g. Great Yarmouth • Disavantages: • No option to run gas turbine by itself • Gas Turbines can reach full output in a matter of minutes. • Steam turbines take 12 hours or more • Gas Turbines tend to have higher NOx emissions and special provision is needed to reduce these levels – e.g. injecting steam into gas turbine.

  22. 12. Applications of Thermodynamics. Combined Heat and Power (2) • Heat is normally rejected at ~ 30oC • Too low a temperature for useful space heating • Reject heat at 100oC • i.e. Less electricity is generated, but heat is now useful • Typically there are boiler and other losses before steam is raised • Assume only 80% of energy available in coal is available. • And technical (isentropic efficiency) is 75% • Then for 1 unit of coal - electricity generated • case 1 = 0.8*0.75*0.639 = 0.38 units • case 2 = 0.8*0.75*0.555 = 0.33 units + up to 0.47 units of heat • or up to 0.8 units in total. Typically 10% of heat is lost so 0.73 units available

  23. The first Law of Thermodynamics states that we can neither create or destroy energy ie Work out = Heat in – Heat Out   Second Law states we must always reject Heat and efficiency = If we could utilise all of rejected heat The 1947 Act stated Electricity must be generated as efficiently as possible i.e. Work/Electricity (not energy) was King 12. Applications of Thermodynamics. Combined Heat and Power (1)

  24. 12. Applications of Thermodynamics. Combined Heat and Power (3) Boiler Heat Exchanger To District Heat Main ~ 90oC GT Heat Exchanger Gas in To District Heat Main ~ 90oC Boiler Heat Exchanger Normal Condenser To District Heat Main ~ 90oC Back Pressure Steam Turbine ITOC or Pass out Steam Turbine Problem: For most CHP plant, electrical output will be limited if there is no requirement for heat. ITOC provides greater flexibility Gas Turbine with CHP also Diesel/gas engine with CHP

  25. 12. Applications of Thermodynamics. Combined Heat and Power (4) Process Integrated Electricity Generation, Process Heat, Space Heat and Air compression at ICI Wallerscote Plant in late 1970s

  26. 12. Applications of Thermodynamics. CCGT with CHP (1) Heat Lost 24 MW Fuel in 239 MW Steam Turbine GT Temp 1127oC Generator Generator Electricity 62 MW Electricity 55 MW Gas Turbine Useful Heat 98 MW

  27. Engine Generator 9. Applications of Thermodynamics. Combined Heat and Power

  28. 12. Applications of Thermodynamics. Example of Small Scale Scheme (4) In most cases, CHP plant is based on an approximate summer time heat load with supplementary heating provided by normal boilers in coldest months of year

  29. Electricity generation in summer is restricted and import is highest when demand is least 12. Applications of Thermodynamics. Example of Small Scale Scheme (5)

  30. 12. Applications of Thermodynamics. Example of Small Scale CHP Scheme 6000 kWe (1) Hot water and process heat demand is constant over the year at 4000 kW Heat loss rate for buildings is 1000 kW oC-1 Existing Heating provided by gas (80% efficiency). Mean space heat demand in January = (15.5 – 1.9) * 1000 = 13 600 kW This is the balance temperature – we shall discuss this in 2 weeks time. In such schemes approximately 1.4 kW heat is rejected for every 1 kW electricity generated. In this case 8400kW

  31. 12. Applications of Thermodynamics. Example of Small Scale Scheme (2) Col [7] is actual amount of heat that can be usefully used. i.e if col [6] is greater than demand then the useful amount = demand Column 3 values = (15.5 – col [2])* 1000 15.5oC is the balance or neutral temperature at which no heating is required. Incidental gains from appliance heat and body heat increase temperature to comfort level. Column [5] is electricity demand from Previous Sheet Column [6] indicates the potential amount of heat which would be available. Typically it is around 1.4 times the electricity generation so Col [6] = 1.4 * col [5] subject to a maximum electricity generation of 6000 kW i.e. when electrical demand > 6000kW, only 6000 * 1.4 = 8400 kW will be available for heat. Column [4] values = col[3] + 4000 The 4000 is hot water and process heat requirement. Maximum Electricity generation = 6000 kW electrical 8400kW heat

  32. 12. Applications of Thermodynamics. Example of Small Scale Scheme (3) Column [9] is actual electricity that can be generated. If the heat demand is greater than 8400, then units can be run at full output – i.e. 6000 kW. If heat requirement is less than 8400kW, then output of generators will be restricted to a maximum of Col [7] / 1.4 Column [8] is supplementary heat required from back up boilers Col [8] = col [4] – col [7] Column [10] is additional electricity needed. Note: highest import occurs when electricity demand is least. The totals represent the total amount of heat or electricity generated or required over the year. Using 30 day months the totals in each column will be: mean values * 24 * 30

  33. 12. Applications of Thermodynamics. Example of Small Scale Scheme (4) In most cases, CHP plant is based on an approximate summer time heat load with supplementary heating provided by normal boilers in coldest months of year 33

  34. Electricity generation in summer is restricted and import is highest when demand is least 12. Applications of Thermodynamics. Example of Small Scale Scheme (5) 34

  35. 12. Applications of Thermodynamics. CCGT with CHP (1) large scale 1.0 Gas Turbine Waste Heat Boiler Generator Steam Turbine Station Electricity Use Heat Out Condenser Heat Losses from Mains Generator Electricity Out Irrecoverable Losses Useful Heat

  36. 12. Applications of Thermodynamics. CCGT with CHP (2) large scale Gas Turbine 0.2375 Electricity 1.0 0.0125 Waste Heat Boiler Generator 0.25 0.75 0.125 Irrecoverable Losses 0.625 Gas turbine efficiency Electricity generated: 0.25 * 0.95 = 0.2375 Energy to Steam Turbine = 0.75 – 0.125 = 0.625

  37. 12. Applications of Thermodynamics. CCGT with CHP (3) large scale Gas Turbine 0.2375 0.25 1.0 0.0125 0.75 Waste Heat Boiler 0.125 Irrecoverable Losses 0.0133 0.625 0.2523 0.2656 Generator Steam Turbine Condenser Generator 0.3594 Mechanical power to generator = 0.425 * 0.625 = 0.2656 Heat to Condenser = 0.625 – 0.2656 = 0.3594 Electricity out = 0.95 * 0.2656 = 0.2523 steam turbine efficiency

  38. 12. Applications of Thermodynamics. CCGT with CHP (4) large scale Gas Turbine 0.2375 0.25 1.0 Electricity Out 0.0125 0.75 Waste Heat Boiler 0.125 0.470 Irrecoverable Losses 0.0133 0.625 0.2523 0.2656 Generator Steam Turbine Station Electricity Use 0.3594 Heat Out 0.0196 Useful Heat 0.3594 Condenser 0.3048 Heat Losses from Mains 0.0546 Generator Station use of electricity = (0.2375 + 0.25230) * 0.04 = 0.196 Useful Heat = 0.3594 * (1 – 0.152) = 0.3048

  39. 12. Applications of Thermodynamics. CCGT with CHP (5) Summary of Scheme For each unit of fuel • Electricity available = 0.470 units • Heat sent out = 0.3594 units • Station efficiency = 0.470 + 0.3594 = 82.9% • But heat is lost form mains so only 0.3048 is actually useful • Overall system efficiency = 0.47 + 0.3048 = 77.5%

  40. N.K. Tovey (杜伟贤) M.A, PhD, CEng, MICE, CEnv Н.К.Тови М.А., д-р технических наук Energy Science DirectorCRedProject HSBC Director of Low Carbon Innovation NBS-M016 Contemporary Issues in Climate Change and Energy 2010 13. Heat Recovery : Heat Pumps 40

  41. 13. Heat Recovery Systems and Heat Pumps • Parallel Plate Heat Exchanger Hot Fluid In Cold Fluid In

  42. 13. Heat Recovery Systems and Heat Pumps Parallel Flow Shell and Tube Exchanger Inefficient: maximum temperature achieved is ~ 50% of temperature difference Hot Fluid In Cold Fluid In Hot Fluid Temperature Cold Fluid Distance

  43. 13. Heat Recovery Systems and Heat Pumps Contra Flow Shell and Tube Exchanger Inefficient: maximum temperature achieved is ~ 50% of temperature difference Hot Fluid In Cold Fluid In Hot Fluid Temperature Cold Fluid Distance

  44. Operation of Regenerative Heat Exchangers Fresh Air Fresh Air Stale Air Stale Air B Stale air passes through Exchanger A and heats it up before exhausting to atmosphere Fresh Air is heated by exchanger B before going into building A After ~ 90 seconds the flaps switch over B Stale air passes through Exchanger B and heats it up before exhausting to atmosphere Fresh Air is heated by exchanger A before going into building A

  45. Work IN W Heat In Q2 Schematic Representation of a Heat Pump 13. Applications of Thermodynamics. Heat Pumps A Heat Pump is a reversed Heat Engine: NOT a reversed Refrigerator Heat Out Q1 Heat Pump If T1 = 323K (50oC) and T2 = 273K (0oC) Schematic Representation of a Heat Pump. IT IS NOT A REVERSED REFRIGERATOR.

  46. Responding to the Challenge: Technical Solutions The Heat Pump Compressor Heat supplied to house High Temperature High Pressure Condenser • Any low grade source of heat may be used • Typically coils buried in garden • Bore holes • Example of roof solar panel (Look East: Tuesday) Throttle Valve Evaporator Low Temperature Low Pressure A heat pump delivers 3, 4, or even 5 times as much heat as electricity put in. We are working with thermodynamics not against it. Heat extracted from outside 46

  47. 13. Applications of Thermodynamics. Heat Pumps • Performance is measured by Coefficient of Performance • (COP) • Theoretical Performance of 6.46 • Practical COP in excess of 3. • i.e. Three times as much heat is obtained as work put in. • Remaining heat comes from the environment • The closer the temperature difference, the better the COP • Can be used for efficient heat recovery • Can recover the energy lost in electricity generation • Will out perform even a gas condensing boiler • Working with Thermodynamics - NOT against it

  48. Condenser Throttle Valve Compressor Evaporator 13. Applications of Thermodynamics. Heat Pumps and Refrigerators A heat pump refrigerator consists of four parts:- 1) an evaporator (operating under low pressure and temperature) 2) a compressor to raise the pressure of the working fluid 3) a condenser (operating under high pressure and temperature) 4) a throttle value to reduce the pressure from high to low.

  49. The Norwich Heat Pump Original Paper by John Sumner Proc. Institution of Mechanical Engineers (1947): Vol 156 p 338

  50. The History of the Site • The building was unique - the very first heat pump in the UK. • Installed during in early 1940s during the War. • Built from individual components which were not ideal. • Compressor was second hand built in early 1920’s ! for Ice making. • The evaporator and condenser had to be built specifically on site. • Refrigerant choice was limited during War - only sulphur dioxide was possible. • A COP of 3.45 was obtained - as measured over 2 years. • Even in 1940s, the heat pump was shown to perform as well as, if not better than older coal fired boiler.

More Related