Question 6 of 13: Thrust on a Wind Turbine (Actuator-Disk Theory)

Question 6: Thrust on a Wind Turbine (Actuator-Disk Theory) (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The thrust (axial force) the wind exerts on the turbine.

Approach. Apply mass conservation along the streamtube to get the velocity at the disk and far downstream, then apply the linear-momentum theorem over the whole tube (pressure is atmospheric at both open ends, so only the momentum flux terms survive).

1. Continuity along the streamtube (same mass flow at every station): $$A_1V_1=A_{disk}V_{disk}=A_2V_2$$
2. Velocity at the disk: $$V_{disk}=V_1\frac{A_1}{A_{disk}}=V_1\left(\frac{D_1}{D_{disk}}\right)^2=10\left(\frac{3}{4}\right)^2=\boxed{5.625\ \text{m/s}}$$
3. Velocity far downstream: $$V_2=V_1\left(\frac{D_1}{D_2}\right)^2=10\left(\frac{3}{4.5}\right)^2=4.44\ \text{m/s}$$
4. Mass flow rate through the disk (this is the mass flow for the whole tube): $$\dot m=\rho A_{disk}V_{disk}=1.2\left(\frac{\pi}{4}(4)^2\right)(5.625)=84.82\ \text{kg/s}$$
5. Momentum theorem on the streamtube control volume (atmospheric pressure cancels at the two open ends; the turbine reaction force equals the momentum lost by the air): $$F=\dot m (V_1-V_2)=84.82(10-4.44)=\boxed{471\ \text{N}}$$