Discussion
Air flows over a spherical storage vessel of diameter 4 m at a speed of 1 m/s. To find the drag force on the vessel, a test run is to be carried out in water using a sphere of diameter 100 mm. The density and dynamic viscosity of air are 1.2 kg/m3 and \(1.8 \times 10^{-5}\) Pa.s, respectively. The density and dynamic viscosity of water are 1000 kg/m3 and \(10^{-3}\) Pa.s, respectively. The drag force on the model is 4 N under dynamically similar conditions. The drag force (in N) on the prototype is approximately
Answer: C
The drag on the prototype is about 1.08 N.
Dynamic similarity for a submerged body means equal Reynolds numbers. That fixes the model speed:
\[ \begin{aligned} \frac{1.2(1)(4)}{1.8 \times 10^{-5}} &= \frac{1000\,V_{m}(0.1)}{10^{-3}} \\ 2.667 \times 10^{5} &= 10^{5}V_{m} \Rightarrow V_{m} = 2.667\ \text{m/s} \end{aligned} \]
With equal drag coefficients, drag scales as \(\rho V^{2}D^{2}\):
\[ \begin{aligned} \frac{F_{p}}{F_{m}} &= \frac{1.2(1)^{2}(4)^{2}}{1000(2.667)^{2}(0.1)^{2}} = \frac{19.2}{71.1} \\ F_{p} &= 4 \times 0.27 \approx 1.08\ \text{N} \end{aligned} \]
4 N assumes the drag is unchanged between model and prototype, and the other values come from scaling with the wrong variables.
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