SOLUTION The power generation and loading of a wind turbine are to be analyzed. The efficiency and the force exerted on the mast are to be determined, and the effects of doubling the wind velocity are to be investigated.
Assumptions 1 The wind flow is steady and incompressible. 2 The efficiency of the turbine–generator is independent of wind speed. 3 The frictional effects are negligible, and thus none of the incoming kinetic energy is converted to thermal energy. 4 The average velocity of air through the wind turbine is the same as the wind velocity (actually, it is considerably less—see Chap. 14). 5 The wind flow is nearly uniform upstream and downstream of the wind turbine and thus the momentum-flux correction factor is 𝛽 = 𝛽1 = 𝛽2 ≅ 1.
Properties The density of air is given to be 0.076 lbm/ft³.
Analysis Kinetic energy is a mechanical form of energy, and thus it can be converted to work entirely. Therefore, the power potential of the wind is proportional to its kinetic energy, which is V ²/2 per unit mass, and thus the maximum power is .m V ²/2 for a given mass flow rate:
\begin{aligned}V_{1} &=(7 mph )\left(\frac{1.4667 ft / s }{1 mph }\right)=10.27 ft / s \\\dot{m} &=\rho_{1} V_{1} A_{1}=\rho_{1} V_{1} \frac{\pi D^{2}}{4}=\left(0.076 lbm / ft ^{3}\right)(10.27 ft / s ) \frac{\pi(30 ft )^{2}}{4}=551.7 lbm / s \\\dot{W}_{\max } &=\dot{m} ke _{1}=\dot{m} \frac{V_{1}^{2}}{2} \\&=(551.7 lbm / s ) \frac{(10.27 ft / s )^{2}}{2}\left(\frac{1 lbf }{32.2 lbm \cdot ft / s ^{2}}\right)\left(\frac{1 kW }{737.56 lbf \cdot ft / s }\right) \\&=1.225 kW\end{aligned}
Therefore, the available power to the wind turbine is 1.225 kW at the wind velocity of 7 mph. Then the turbine–generator efficiency becomes
\eta_{\text {wind turbine }}=\frac{\dot{W}_{\text {act }}}{\dot{W}_{\max }}=\frac{0.4 kW }{1.225 kW }= 0 . 3 2 7
(b) The frictional effects are assumed to be negligible, and thus the portion of incoming kinetic energy not converted to electric power leaves the wind turbine as outgoing kinetic energy. Noting that the mass flow rate remains constant, the exit velocity is determined to be
\dot{m} ke _{2}=\dot{m} ke _{1}\left(1-\eta_{\text {wind turbine }}\right) \rightarrow \dot{m} \frac{V_{2}^{2}}{2}=\dot{m} \frac{V_{1}^{2}}{2}\left(1-\eta_{\text {wind turbine }}\right) (1)
or
V_{2}=V_{1} \sqrt{1-\eta_{\text {wind turbine }}}=(10.27 ft / s ) \sqrt{1-0.327}=8.43 ft / s
To determine the force on the mast (Fig. 6–25), we draw a control volume around the wind turbine such that the wind is normal to the control surface at the inlet and the outlet and the entire control surface is at atmospheric pressure (Fig. 6–23). The momentum equation for steady flow is given as
\sum \vec{F}=\sum_{\text {out }} \beta \dot{m} \vec{V}-\sum_{\text {in }} \beta \dot{m} \vec{V} (2)
Writing Eq. 2 along the x-direction and noting that \beta=1, V_{1, x }=V_{1},, and V_{2, x }=V_{2} \text { give }
F_{R}=\dot{m} V_{2}-\dot{m} V_{1}=\dot{m}\left(V_{2}-V_{1}\right) (3)
Substituting the known values into Eq. 3 gives
\begin{aligned}F_{R} &=\dot{m}\left(V_{2}-V_{1}\right)=(551.7 lbm / s )(8.43-10.27 ft / s )\left(\frac{1 lbf }{32.2 lbm \cdot ft / s ^{2}}\right) \\&=-31.5 lbf\end{aligned}
The negative sign indicates that the reaction force acts in the negative x-direction, as expected. Then the force exerted by the wind on the mast becomes F_{\text {mast }}=-F_{R}= 31.5 lbf.
The power generated is proportional to V ³ since the mass flow rate is proportional to V and the kinetic energy to V ². Therefore, doubling the wind velocity to 14 mph will increase the power generation by a factor of 2³ = 8 to 0.4 × 8 = 3.2 kW. The force exerted by the wind on the support mast is proportional to V ². Therefore, doubling the wind velocity to 14 mph will increase the wind force by a factor of 2² = 4 to 31.5 × 4 = 126 lbf.
Discussion Wind turbines are treated in more detail in Chap. 14.
