Question

An ideal wind-turbine, 12 m in diameter, operates at a theoretical efficiency (th) (equation given below) of 50% in u. = 14 m/s wind. If the air density is 1.2 kg/m and (a +u,)(x - 4) Пh 2u 1 23 U2 = U3 = (uj + u4)/2 P1 = P4 = Patm Determine, a) The axial force on the windmill. (assume uniform flow at positions) b) The mean (gage) pressures immediately in front of (P-Patm) and behind (P3-Patm) the disc, c) The shaft power developed by the turbine.

Answer 1

for wind turbine, D=12m, nth= (14+44) (19?-12) 203 s 0.5 la 14m is S Or 5 given (14+04) (14 24143 2744 746 - (14 + U4)114348) (W4+14) (196-02) - 2744 U4 = 8.65 M Cubic s polynomial [solved .: Now ų = = Uitua by Calculates 2 ų =ų = 11-325m

i] Axial force (F) = Pa-p13 A by bernauli Equation, Pa-Piz (vik-V42) 1.2 [142-8.652] € 72.70 N/m² (72-70) 6= ( 72-70) à 2122 Nlo e APE F = APA - = 8222.91 N F=8.22 KN.

meon gauge pressuse b] finding front side (P2-Patm) = P2-P, = p [vi- v 12[14?-1.3254] 2 40.64 Pa P2-Patm = 40.64 Pa. behind side, (Po-Palm) = [VA-ves] = 12 (868+1.34] Pz- PA Poz-Patm = -31.99 Pa. c] Shaft Power a Axial force & Velouty (F) (42) (8.22) (11.325) 43.0915kW Р .