Question

5. (4x5pts = 20pts) An anemometer mounted 10 m above a surface with crops, hedges, and shrubs, shows a wind speed of 5 m/s. Assuming 15°C and 1 atm pressure, determine the following for a wind turbine with hub height 80 m and rotor diameter of 80 m: V top - - 80 m 120 m V bottom - 80 m 5 m/s 40 m 10 m Crops, shrubs, hedges FIGURE P7.2 a. Estimate the wind speed and the specific power in the wind (W/m) at the highest point that the rotor blade reaches. Assume no air density change over these heights. b. Repeat (a) at the lowest point at which the blade falls. c. Compare the ratio of wind power at the two elevations using results of (a) and (b) and compare that with the ratio obtained using Equa- tion 7.20. d. What would the power density be at the highest tip of the blade if we include the impact of elevation on air density? Assume the temperature is still 15 C. Does air density change seem worth considering in the above analysis?

Answer 1

Given: Anemometer mounted at height-lom luind speed = 5 m/s temp. = 15°c at Lampressure Hub height=80m Rotor diameter=80m (@) = 1.225 kq @ 15°C Latm pressure Nows ( Vood - Chole where dccoops, hedges, shouba) : Vuze 5 mg ( 120102 Vi20 = 8-219 ņ =0.2 10.2 P120 = specific power = t (812) (vist = £*(1-225)(8-219)=340 wom? (b) At lowest point d (crops, hedges, Shoubs) = -2 . Mto) - ( 40 )°*?V40 = 5(43°26-597ms! Now; Pato = specific pavert (940) (V40)* =4 (1-225) (6.59733 c) We have been wat om t - (128032 2. Po to (eo) 02922 1-933

288 d) At 120m; ajo domuity change g (K9/m) = 353.200342 % ...3120 = 353.4 2 0 0 0 = 1.208 kgm? Now; specific power = Pizo = 4 (1120) (V120)3 | - (1-2) (3290-235-4k ..% Change = 340-335-4 x 100% = 1-35% It means; air density change û worth considering 340
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