Axial-momentum analysis led to a power coefficient C_P = 4a(1 − a)² for a HAWT eq. (30.7)

Compute the power coefficient another way: start from the tangential force per unit length f T in the simple model that led to eq. (30.12)

implement the axial momentum theory constraint (30.15)

find the torque per unit length dτ/dr R and compute the power

Show that the result is the same as the result of the axial-momentum analysis (30.7).

Transcribed Image Text:

v4 = (1 – 2a)v1 v2 = (1 – a)v1, AT 1- 2a |A1 = (1 – a)AT, A4 %3| P = ;PoviAT4a(1 – a)² . Sr=-K Bu°ci sinø 2 pOK Bu²(1 – a)²c , 2. 2 sin o FN=PoK Bw?cj cos o âf = poK Bv²(1 – a)²cj cos ê, 2 sin? o