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Information about ESwindgeothermal07

Published on January 17, 2008

Author: Marcell


Slide1:  Incident power in the wind Incident power P given by (mass/sec)(KE/mass) P = (dm/dt)½u2 = (ruA)½u2 P = ½ rAu3 Note strong dependence on wind speed Example: r = 1 kg m-3 , A = 1 m2 , u = 12 m s-1 (rated wind speed) P = 864 W ~ 1 kW Typically turbine efficiency ~ 40% so power output 300-400 W Why are wind turbines not more efficient? Wind Power Slide2:  Theoretical limit to Efficiency of a Wind Turbine ‘Betz Limit’ Consider streamtube of air passing through turbine The turbine extracts energy from the air so the air speed decreases across the turbine and the cross-sectional area of the streamtube increases u0 A0 = u1 A1 = u2 A2 Force on turbine = rate of loss of momentum of air F = (dm/dt)(u0 - u2) Power extracted P = Fu1 = (dm/dt)(u0 - u2)u1 (rate of work done) Also P = ½(dm/dt)(u02 - u22 ) (rate of loss of KE) Slide3:  P = (dm/dt)(u0 - u2)u1 = ½(dm/dt)(u02 - u22 ) (u0 - u2)u1 = ½(u0 - u2) (u0 + u2 ) u1 = ½(u0 + u2 ) or u2 = 2 u1 - u0 now dm/dt = ruA = ru1A1 so P = ru1A1(u0 - u2)u1 or P = 2ru12 A1(u0 - u1) Let u1 = a u0 then P = ½ru03 A1{4a2(1 - a)} or P = P(wind)F(a) where P(wind) = ½ru03 A1 and F(a) = 4a2(1 - a) Slide4:  P max/min when d{4a2(1 - a)}/da = 0 8a - 12a2 = 0 so a = 0 (min) or 2/3 (max) At max F(a) = 16/27 so Maximum efficiency is 59% (Betz criterion) Power coefficient CP = P/ {½ru03 A1} ‘Lost power’ is due to fact that air needs KE to go downstream Thrust dF = dLcosf Power dP = dLsinfv dP = dFtanfv tanf = u1/v so dP = dFu1 Slide5:  Materials absorb radiation differently so temperature gradients arise causing convection and pressure changes which result in winds. A simple example being the off-shore night-time wind often found on coasts, caused by the sea retaining the heat from the sun better than the land. Slide6:  Simplified representation of world wind circulation Slide7:  Persian Windmill Windmills thought to have been in existence for about 4000 years Slide8:  Some examples of Wind Turbines Slide9:  Flow around an Aerofoil Slide10:  Forces on an Aerofoil Lift Force: ½CLrAu2 Drag Force: ½CDrAu2 CL, CD functions of non-dimensional parameters, ie Reynolds number Re = r u l/h , where l is a characteristic length Shape of the aerofoil Angle of attack a Slide11:  Blade speed v = wr so angle of attack a depends on radius r Twist of blade changes with radius to optimise a. Betz condition u1 =2u/3 cot f = v /u1 = 3rl/2R Stall regulated- as u increases a increases and blade stalls Tip-speed vtip = wR Wind speed u Tip-speed ratio l = vtip/u Horizontal Axis Wind Turbine (HAWT) Angular velocity w Effect of Drag on CP:  Effect of Drag on CP As a result of drag rotational force becomes L sin f - D cos f  L sin f (1-g cot f) where g  CD/CL . cot f = v /u1 = 3rl/2R , so reduction decreases with r Typical r = 2R/3 so cot f  l and CPmax  (1-gl)CP(Betz) g ~ 1/40 and l ~ 10 so CPmax ~ 45% Modern wind turbine CP – l curve:  Modern wind turbine CP – l curve Turbine designed to have maximum efficiency at l ~ 10 Slide14:  Rotor efficiencies of some wind turbines l The width W and angle of attack a are for a particular l. If the wind speed alters, then angle f of wind to motion of blade and therefore the lift L changes. This changes the thrust from it optimal value and CP decreases. Slide15:  Thrust on a Wind Turbine p0 u0 p1 u1 p2 u2 A0 A1 A2 upstream turbine downstream p0 /r + ½ u02 = p1 /r + ½ u12 p1* /r + ½ u1*2 = p2 /r + ½ u22 Conservation of mass u1= u1* and p0 = p2 = atmospheric. So (p1 - p1*) /r = ½ (u02 - u 22) Fthrust = ½ r (u02 - u22)A1 p1* u1* Consider streamtubes of air before and after turbine, not across turbine because flow unsteady and not streamlined. Thrust is maximum when u2 is minimum – this corresponds to maximum power extraction for which u2 equals u0 / 3. Therefore Fthrust = ½ r u02A1×8/9 Similar to a circular disc of area A1 which has a drag force FD = ½ CD r u02A1 and CD ~1 Slide16:  Probability distribution for wind speed at North Ronaldsay, Orkney Probability distribution F(u) can often be approximated by Rayleigh distribution: F(u) = (2u/c2) exp[-(u/c)2] where c = 2uaverage /(p)1/2 P = ½rA<u3> ≈ rA<u>3 as <u>3 ≈ 2<u3> Wind speed increases with height z : uz ≈ u10(z/10)0.14 where z is in metres Slide17:  Wind Turbine chosen to have output capacity ~ 3 times average power output to take advantage of high wind speed periods. Cut-out value to protect turbine installation. Typical spacing of turbines on a wind farm is 4D(crosswind) x 7D(downwind), where D is the diameter of the turbine. Wind Farm P ≈ 0.2D2<u>3, D diameter of turbine Slide18:  Engineering Designs : Pros/Cons Vertical axis Advantages: a) No Yaw necessary b) Direct coupling to electrical generator Disadvantages: a) Many natural resonances leading to vibration and fatigue b) Variable torque leading to uneven output c) Less cost-effective than HAWTs Horizontal axis Advantages: Low solidity machines (few blades) a) low moment of inertia hence fast b) high frequency good for power generation High solidity machines (many blades) a) high moment of inertia hence slow b) low frequency good for battery charging or water lifting Slide19:  Horizontal axis Disadvantages: a) Upwind blades need Yaw (fan-tail for alignment) b) Downwind blades self-orientate but tower blocks some wind Fatigue Many revolutions gives rise to fatigue which gives rise to cracks Wind turbines ~108 cycles - very demanding on materials Environmental impact Appearance: matter of opinion Noise: gearbox, electrical generator, aero noise (swish) eg Denmark has requirement that wind turbines are located >150 m from houses and that noise level < 45dB noise  u5  low rotational speeds E.M. Interference: reflection of em waves/TV signals from metal blades Wildlife: wind farms are a hazard for birds Slide20:  Relative Noise Levels I(dB) = 10 log10(I/I0), where I0 is the threshold of hearing (at 1000 Hz I0 = 10-12 Wm-2) Slide21:  Economics Capital cost £600-1000/ kW California (Reagan) : tax breaks (most imported from Denmark) Cost of production goes down as demand goes up Best sites competitive with fossil fuels Applications Battery chargers: 105 in World (China mostly) Wind pumps: >106 worldwide (fast growth in developing world) Electricity generators: increasing worldwide. Low carbon so very important as alternative to fossil fuel Future Potential Could reach 10-20% of electricity needs of World ~2020-2050 Higher % needs increasing more ‘spinning reserve’ unless good energy storage developed (eg Fuel Cells) due to wind variability Electrical transmission from windy sites to main population centres also required Slide22:  Annual incremental installed capacity (GW) Slide23:  Worlds’ land-based wind energy resources estimated as 53,000 TWh per year World electricity demand by 2020 estimated as 26,000 TWh per year (equivalent to 3 TW continuous cf ~20 TW continuous for estimated total energy demand) Land-based wind energy resources in TWh per year Slide24:  Potential UK offshore wind generation resource Current UK electrical energy demand is ~350 TWh per year (1.3  1018 J per year or ~40 GW continuous power) Slide25:  Estimates of renewable-energy resources for 2025 in the UK Slide26:  Theoretical or Gross potential Estimate of total annual energy that could be produced Technical potential Maximum annual energy that could be extracted taking into account practical, environmental, and social constraints (estimates of 4% by WEC, and 10% of the land area with suitable winds have been made) Practicable or accessible potential The amount of the technical potential that can be utilized by a particular time Economic potential Amount of the technical potential that is economically viable Depends on the cost of alternative supplies, on the cost of borrowing, and on policies such as a carbon tax (nb definitions of potential differ) Slide27:  Guardian Tuesday May 3rd 2005 Slide28:  Geothermal Energy Origin: Heat from a) cooling of Core (loss of heat of formation) b) decay of radioactive isotopes, 232Th, 238U and 40K Total geothermal power ~ 1021 J/yr cf total solar power ~5.4 1024 J/yr Mantle (depth > 30 km), temperature ~ 1000 oC Convective heat flow > ~ 100 km depth Outer shell (depth ~ 30 km) fissures (volcanoes/geysers) Thermal conduction k ~ 2 W m-1K-1, no convection Heat flux q = - k dT/dr ~ - k (Tc- Ts)/d q ~ 6.10-2 W/m2 Total geothermal power = q 4pR2 ~ 1021 J/yr Slide29:  Forms of exploitation Warm water springs Spa towns (eg Bath) New Zealand (Maoris) Geysers Eg Italy, New Zealand, Iceland, USA(California) All situated in geological fault regions Total output ~ 6 GW Aquifers Porous layer sandwiched between non-porous rock Hot dry rock mining Like aquifers but water pumped through natural fissures (cracks) in rocks Slide30:  Map of the Earth’s plates Movement generally 1-10cm per annum Slide31:  Aquifer Properties Porous medium (e.g. sand, gravel) Define porosity = (Volume of cavities)/(Total volume) Pressure difference, DP = rgH H = ‘Head’ of water kw= Hydraulic conductivity % Porosity () kw(m/day) Clay 50 < 10-2 Silt 40 10-2 – 1 Sand 30 1 – 500 Gravel 30 103 – 104 Slide32:  Heat Extraction from Aquifer Water volume V flows per second through narrow porous layer area A and thickness h and is heated by surrounding rock at temperature T+T1 Heat lost by rock plus water = Heat gained by water -[(1-f)rrcr+ frwcw]Ah d T = VrwcwT dt dT/dt = -T/t T = Toexp(-t/t) where t = C/Vrwcw and C = [(1-f)rrcr+ frwcw]Ah ~3km Slide33:  Lifetime of an Aquifer Heat extracted is stored energy- time to replace is longer than lifetime if used as a power source- so not renewable Example: A=1 km2, h= 0.5 km, f=5%, rr= 2700 kg/m3, cr= 840 J/kg/K rw= 1000 kg/m3, cw= 4200 J/kg/K, V = 100 l/s, T = 100 C Lowest useful temperature Tl = 40 C C = [(1-f)rrcr+ frwcw]Ah = 1.2 1015 J/K t = (1.2 1015)/(4.2 105) = 90 years Energy stored, Es = C(T - Tl) E(t) = Esexp(-t/t) so dE/dt = -(Es/t)exp(-t/t) Substituting: Initial Power = 25 MW Slide34:  Volcanic Geothermal System Slide35:  Hot Dry Rock Mining USA, UK (Canborne, Cornwall), Germany, Japan Look for high temperature gradients (so less drilling) Granite good – higher than average radioactivity Depths 3 - 6 km Temperatures 200 - 300 C Resource in Cornwall approximately equal to UK coal reserves, but currently too expensive to exploit Commercial Exploitation Southampton (Hampshire Geological Basin) 1980/81 Depth 1.7 km Temp 74 C Vol/s 12 l/s Lifetime ~20 years District heating: Civic Centre, Swimming Baths, Department Stores Cost 1p/KW Output 1 MW Capital Cost: Government + EC as a demonstration project Slide36:  Hot Dry Rocks in the UK a) Predicted temperature-depth curves in parts of the UK b) Projected temperature contours in centigrade at 6 km depth in the SE Slide37:  Extraction Techniques a) Hot dry rock system b) Hyperthermal power station (temperature gradient > 80 C/km- tectonic plate boundaries) Geothermal Power Plants:  Geothermal Power Plants >150 C ~ 100-150 C SO2 (lbs/MW-hr) CO2 (lbs/MW-hr) Slide39:  Potential UK geothermal energy resource at different temperatures Current UK electrical energy demand is ~ 350 TWh per year (1.3  1018 J per year or ~ 40 GW continuous power) Slide40:  Economics Drilling costs increase exponentially with depth increasing which results in deep-mined geothermal energy not economic (but drilling technology is getting cheaper) Main potential in exploiting surface or near-surface fissures Global potential ~12 GW by 2005 Environmental factors Drilling noisy Waste disposal of ‘spoil’, water loss down cracks H2S (bad eggs) from geysers Geothermal ‘brine’ corrosive/toxic _ secondary heat exchangers Safety Ok if properly managed Future Development Restricted to geologically unstable regions, particularly developing World and Pacific Rim. BUT USA evaluating hot dry rock drilling Slide41:  MegaWatts Geothermal Heat Pumps:  Geothermal Heat Pumps Take advantage of relatively constant temperature below ground- typically 100-400 ft Heat pump either extracts or transfers heat Q using a compressor (as in a refrigerator) that requires work W . The ratio Q/W is the coefficient of performance, COP. For an ideal pump heating a building COP = T1/(T1 - T2) Eg for DT  (T1 - T2) =31 C and a ground temperature T2 = 6 C = 279 K, then COP = 10 Actual COPs are typically 3 - 4.5 Over 40% of CO2 emissions in the US are from space heating and cooling- geothermal heat pumps powered by ‘green electricity’ are an important source of very-low carbon energy

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