Question

An ideal wind-turbine, 12 m in diameter, operates at a theoretical efficiency (9th) (equation given below) of 50% in ui = 14 m/s wind. If the air density is 1.2 kg/m² and 4 (4 + )(x - 1) nih 2u 1 2 3 = U2 = U3 (ui + 04)/2 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 (P2-Patm) and behind (P3-Patm) the disc, c) The shaft power developed by the turbine.

Answer 1

men: 0.5 u, = 14 m/s S 1.2 kg/m3 mth Uit UM D x (418-44²) 2 U13 0.5 = (14+ UH) x ( 142 - U42] ax(14) 3 on solving we get Geo U4 = 8.65247 mls. acial force T SxAX U₂3 (ui-uh) A = I x D2 4 De diameter of teu bine = 12m 2 А = π x 122 113.09733 m F = 1.2x 113.09733x/14+8.65247 (14-8-65247 (14 786547) x 2. F 8 220 N

422 UltUH 14+ 8.6524 7 Il 68241 = 11:326235 arial force 8220 N P2 - Patm apply beenoulig the onem a cross 1 - 2 Pq + 1 + I 8 U23 = Pit ₂ Su,2 (P,= Patm Р. -Patm = S(U,3 - U₂ 2 23 4x12x [Brokastis tortas (11-32623) 142 P₂ - Patm 2 P₂ - Patm 40.6299 N/m2 G P3 Patm = - P₂ - Patm. Pa - Py = $ 30,2 – 432) 11.32 6 235 m/s Ug = Uz :

P3-Patm 5 x1.2 * ( 8.6524 7? - 11.92 6235 52) Pa - Patm = 32.05 N/m2 9 Shaft power Ps 19 a uz Ps - Axial force X Fuz Ps Ps 8220 X 11:326235 Ps 93101.65 watts Ps 93-10165 Kwatts