Question

An ideal wind-turbine, 12 m in diameter, operates at a theoretical efficiency (9th) (equation given below) of 50% in uj = 14 m/s wind. If the air density is 1.2 kg/m² and (u + ua)(u? – uz) 2u 2 3 nin U2 = u3 = (uj + u4)/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

Tax 122 meth = 0.5 ui 14 m/s 8 = 1.2 kg/m3 mth Uit Uly х (412-Un ² 2 U13 0.5 = (14+Uh ) x ( 142 - U4²] 2x(14) 3 on solving we get 8 GB Un = 8.65247 m/s. a arral foace F = fx ax UqX (4,-44) A IX DR 4 Da diameter of teubine 12m 2 113.097 33 m F = 1.2x 113.09733x/14+8.65247 19.04783 x(* ** *5241) 865244) (4-865247 2 - 8 220 N

U2= UltUH 2 147 8.6524 2 11:326235 m/s anal force 8220N ره P2 - Patm = apply beenoulis the onem across 1-2 P2 + 1 2 3 U₂3 = Pit % 8 4,2 Patm Pq Patm ₂ 2 3[4,? - Ug2 142 (1-32623 P₂ - Patm |X12X [ Boost 2. P₂ - Patm = 40.6299 N/m2 P₂ - Patm . + 3(U.,? _ Ug?) P₂ - Patm. P3-Py = 2 Uz = U2 = 11:326235 m/s

P3 – Patm = x1-2 * ( 8-65247? - 11.32 6235? 352) eot Po - Patm = - 32.05 N/m² Shaft power Pgio u a Ps= Axial force X Ps Fuz Pg 8220 X 10.326235 Ps 93101.65 watts Ps E 93-10165 Knatts