Ex.Mats 5 New

Engineering SECOND YEAR Part IB Paper 8: Elective (2) Wind Turbines Examples Paper 2 Solutions

Vestas/Platts Guest lectures 1. (a) Extreme wind, turbine must survive. Lean out of the way? Change shape? Or just resist the loads. Must have a good idea of the likely magnitude of extreme events. Failure occurs when the “expected” extreme event is exceeded. Lightning, blades may conduct (if carbon fibre or aluminium). Need to design for the high currents. Blades can explode if heat softens the composite matrix when loads are high. Instrumentation can be destroyed by lightning. Cold causes ice formation. Ice changes shape of blade causing drop in lift and lower efficiency. Ice can fly off and is a hazard. Extreme heat causes gearbox and generator, bearings to overheat and fail unless cooling system is appropriately designed. Also composite blades are reinforced polymers whose strength reduces with temperature. (b) Challenges:.•Foundations – serious precision required on a large scale. Millimetre positioning and alignment is critical on a tens-of-metres scale. Huge quantities of material (concrete). •Tower is typically a huge rolled steel fabrication. 50mm thick – massive rolling task. Transport to site is a big job. Site assembly of smaller bits is a nightmare, especially offshore. •Blades are very long. Manufacture takes time and needs to have high quality control especially of inclusions and of shape. •Hub contains the pitch control mechanism. Very complex, hydraulically controlled with a clever bit of mechanics to get the linkage right. May contain over-speed protection triggered by centrifugal force or wind pressure. •Nacelle contains all the real mechanical engineering. •Gearbox reduces speed by, say, 150:1. Transmits huge power (say 6MW) and at 99% efficiency dissipates 60kW, so gets very hot. Need oil cooling. •Bearings take huge thrust (order of 10MN), as do the gearbox mounts which need to be strong and stiff. But gearbox vibration transmits through mounts so they need to be resilient. A real compromise. •Emergency brake best sited on generator side of gearbox where torque is lower, parking “hand-brake” best on hub-side of gearbox to allow for gearbox maintenance. •Drive shaft has to take huge torques – very clever engineering for long life. Modern shafts are high-stiffness carbon-fibre composites. •Generator options, number of poles, synchronous vs asynchronous, permanent magnet vs electromagnet desing. •Control system is very sophisticated to balance load to wind conditions and to deal with variable wind direction. •Electrical systems of great complexity to link 30 (say) turbines into the grid. Think of it as a power station. (c) Gearbox needed for high rotation speed at the generator. Gearboxes are notoriously unreliable, maintenance-hungry and sources of noise. Huge advantages of avoiding gearbox but need v. strong permanent magnets at a large radius to generate power at low shaft speed.

1 2. (a) (extended answer) The problem with placing several tonnes of material into a mould, quickly and accurately, when all the material comes in small pieces, is that there is no time for any trimmimg or fitting. Everything has to be already to size and be supplied in a way that enables it to be placed quickly and accurately. This is known as "kitting". Each layer of glass cloth - there is one outside of the bamboo which goes into the mound first and then a second layer which is inside the bamboo layer and so goes into the mould last - needs to be pre-cut exactly to shape and made into a roll so that it can be unrolled up the mould. The bamboo planks come in sections about 2m long, with chamfered ends which taper fit with the chamfered beginnings of tthe next planks. If each 2m long set is made into a bundle by stapling two tapes across, they can be made into a roll which is unrolled across the blade and then slid along to mesh with the previously placed bundle. All the steel studs can be prefitted into the semicircular end plate which forms the end of the mould, thus all being placed simultaneously. These sorts of kitting operations, which take place at the supplier's factory or at a preparation bay in the blade factory, enable operations in the mould itself to be simple, quick and reliable. The problem with resin infusion and resin curing is that the resin viscosity reduces (and so the resin flows more easily) as the temperature rises, but also the resin/hardener reaction proceeds more quickly and so there is less time available before the resin begins to harden and will no longer flow. The "flow and cure" process thus has to be very carefully managed by not only having the resin at the right temperature but also the moulds and also all the dry material which is put into the mould and around which the resin flows. Not only the factory space but also the storage space needs to be temperature controlled. In parts of Inner Mongolia the external temperature can reach +40C in summer and -40C in winter. There is then a second problem. As the resin cures it is an exothermic reation and this can generate a lot of heat. It is like having a 50kW heater in a 40m blade. If there are pockets of resin concentrated in parts of the infusion, this can cause local overheating which can damage the material properties. There is thus a need to be able to cool the mould, and also to be able to warm the mould, to control it at a steady temperature. This can be done with water pipes embedded in the mould, fed from a heater/cooler unit such as is used for heating and cooling plastic injection moulding tools. (b) Wood is a natural composite. It has high specific strength-to-weight and strength-to- cost rations, it has excellent fracture toughness and it is renewable. But if the right trees are not available nearby then other options (CFRP etc) may make better options. This clealy depends on where you are. Bamboo, of woods, is terrific – but possibly not for the UK. (c) High wind generally gives more output, but no large benefit if turbine is already at maximum output. Hence (a) below is only marginally better even though the wind is a lot faster. This is because no benefit is gained above 10m/s. In (b) the turbine runs at 5kW in the high winds so more time at lower winds means loss of power.

w ind speed w ind speed (a) (b) Line 2

2 Materials

3. (a) Need to consider manufacture, structures, materials, operational factors. For example the blade needs to have relatively low weight to keep the dynamic loads down, while its aerodynamic design dictates a complex shape with a good surface finish. Moreover the blades have significant fatigue strength and stiffness constraints. These factors make composites attractive, and in particular GFRP with its lower cost. Finally the good corrosion resistance of GFRP is advantageous.

The tower, however, is static, so weight is not so much of an issue. Steel or concrete are obvious choices, the former easier to make into a slender tower while concrete might benefit from better corrosion properties offshore. It seems that lattice structures are not so preferred for aesthetic reasons, these would tend to dictate a metallic construction.

(b) Growth exponent n (Ln) Cross-sectional area A ~ L2 2 Mass ~ AL 3 Second moment of area I ~Ad2 4* Aerodynamic load ~ pL2 2 (where p is pressure) Aerodynamic root moment ~ pL2×L 3 (where depends on the shape of the load) Self-weight root moment ~ weight × L 4 (where depends on the weight distribution) Aerodynamic stress ~ dM/I ~L×L3/L4 0 Self weight stress ~ dM/I ~ L×L4/L4 1

*assuming t << d, (although this will scale with L4 even if t isn’t much less that d )

Dimensional analysis Aerodynamic stress  (Nm–2) = fn(L, p, and shape e.g. d, c etc. ): /p = constant Self weight stress  (Nm–2) = fn(L, g, and shape e.g. d, c etc. ): /(gL) = constant

3 L L x A L 4. Mass m   Adx  A dx  0   0 L 2 0 0

3 WL  x  Moment M x    3  L 

3 2 2  x  Second moment of area Ix  Ad  A0d0   assuming t << d  L  3 Md WL  x  d WL  x  Peak stress  x      2   I 3  L  Ad 3A0d0  L  d 2 y M WL dy WL     x  L 2 EI 2 dx 2 dx 3EA0d0 3EA0d0 WL  L2  WL3   x  0     2  2  2 3EA0d0   6EA0d0

A L  WL4 Mass m  0   2 2 12Ed0 

For m/m greater than one, stiffness is critical, while the converse is true for m/m < 1. Here, with E = 45 GPa, f = 150 MPa,  = 5m and d0 = 0.05 L

2 m  L 0.1E  1 f so that the cross-over occurs when L   150m   m 2 0.05 EL f

5. (a) See notes

(b) 2-3, 3-3a, 4-5. 5-5a, 6-7, 7-7a, 1-8, 8-13, 9-10, 10-12b, 11-12, 12-12a, 13-14

4 6. For each entry in the table identify the average stress range value S and average mean stress value Sm . Obtain equivalent values of stress range  0 using Goodman’s rule:

S =  Sm     0 1    ts  with ts = 200 MPa. M 9  S   S  Number to failure using N fi        S0   400  N Miner’s rule, find lifetime used up in one month block:  i   i N fi Total life is 1/, with a factor 1000 to account for numbers of cycles being quoted in thousands, and a factor of 12 to convert to years. See spreadsheet below for calculation.

300 300 200 Number of cycles ('ooo per month) 200 300 200 200 200 100

10 20 30 Mean stress 10 20 30 10 20 30

20 20 20 Alternating stress 30 30 30 40 40 40

21.1 22.2 23.5 Effective stress range - Goodman's rule 31.6 33.3 35.3 42.1 44.4 47.1

3.227E+11 1.984E+11 1.186E+11 Number of cycles to failure - Nfi 8.394E+09 5.160E+09 3.085E+09 6.302E+08 3.874E+08 2.316E+08

9.297E-07 1.512E-06 1.687E-06 Miner's rule - lifetime used up - N/Nfi 2.383E-05 5.814E-05 6.484E-05 3.173E-04 5.162E-04 4.317E-04

Sum N/Nfi Life (months) Life (years) 1.416E-03 706.1 58.8

5 2 2   S   6 (b)(S)  exp    S  S        M  N S 2 2  S  2N 2 1  Tot dS  N exp S S    dS  Tot S M exp S S  dS  N S  Tot S   S  M    0 f 0  0  S0 S 0

2 2SdS  S  dt  Make the substitution t    , 2  S  S M  M 1 2   2NTot M 1 2 S dt  S   M 1 1  S t exp t  NTot   M  2    2  S0 S 0  S0   using z  t z1etdt 0 Here, with M =9, we can use the results that (z)=(z-1)! for z integer to give M  S  1 N   0  =555×106 cycles = 46.2 years Tot  S  M 1 2!  

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