Question 10.2: A wind anemometer has two 10 cm diameter cups with centre–ce......
A wind anemometer has two 10 cm diameter cups with centre–centre distance of 30 cm. The pivot is stuck in a vertical plane as a result of some malfunction and the cups stop rotating. For a wind speed of 20 m/s and air density of 1.23 kg/m³, determine the maximum torque this anemometer exerts on the pivot.
Diameter of the cup = 10 cm = 0.10 m
Wind speed, u = 20 m/s
\rho_{air} = 1.23 kg/m³
Projected area =\;\frac{1}{4}\times\pi d^{2}=\frac{1}{4}\times\pi\times0.10^{2}=7.85\times10^{-3}\;\mathrm{m^{2}}
Lever arm distance, r_{1}=r_{2}=0.5\times30\ \mathrm{cm}=0.15\ \mathrm{m}
Torque by Cup-1, {T_{1}}={F_{D1}}\times{r_{1}} \\ \begin{array}{l}{{=\,C_{D1}\times0.5\times\rho\times U^{2}\times\text{area}\times r_{1}}}\\ {{=\,0.4\times0.5\times1.23\times20^2\times7.85\times10^{-3}\times0.15}}\\ {{=\,0.772\mathrm{~Nm}}}\end{array}
Torque due to Cup-2, {T_{2}}={F_{D2}}\times{r_{2}} \\ \begin{array}{l}{{=\,C_{D2}\times0.5\,\times\,\rho\,\times\,U^{2}\times\,\text{area}\times\,r_{2}}}\\ {{=\,1.4\,\times\,0.5\,\times\,1.23\times\,20^{2}\times7.85\times\,10^{-3}\times\,0.15}}\\ {{=\,2.70\ \mathrm{Nm}}}\end{array}
Maximum net torque on the stuck pivot = T_{2}-T_{1} = 2.70 – 0.772 = 1.928 N.m