## Chapter 14

## Q. 14.4

A modern sculpture includes a wind gage in the form of a circular cylinder suspended from two ﬁne wires as shown in Figure 14.21. If the presence of the wires is assumed to have a negligible inﬂuence on the ﬂow ﬁeld, at what angle will the cylinder hang in a wind of 25 km/h? The cylinder weighs 6 N and its dimensions are D = 10 cm and L = 1 m.

## Step-by-Step

## Verified Solution

Since the wind applies a lift and drag to the cylinder, after neglecting the tiny buoyancy force on the cylinder, and writing a force balance on the cylinder in the x and y directions we obtain

(F_{D} cos θ + F_{L} sin θ) − W sin θ = 0 and 2T − W cos θ + (F_{L} cos θ − F_{D} sin θ) = 0

Noting that the x component of force of the wind (F_{D} cos θ + F_{L} sin θ) = C_D\frac{1}{2} ρ_{air} (U \cos θ)^2A, the force balance in the x direction shows that

\sin \theta =\frac{\left(C_D\frac{1}{2} ρ_{air} U^2A\right) ( \cos θ)^2}{W} (A)

We expect that the angle will be larger for a lighter cylinder at any given wind speed but can never exceed 90°. For small angles, cos²θ = 1, and since A = DL, we obtain the approximate result

\theta =\sin ^{-1}\left(\frac{C_D \rho_{air} U^2L}{2W} \right) (B)

Assuming 20°C air for which ρ = 1.2 kg/m^{3} and \nu = 1.51 × 10^{−5} m^{2}/s, the Reynolds number is

Re=\frac{UD}{\nu}=\frac{(25\ km/h)(1000\ m/km)(h/3600\ s)(0.1\ m )}{1.51×10^{−5}\ m^2/s}=4.6×10^4

From Figure 14.18, at this Reynolds number C_{D} = 1.2. Inserting the data into (A) we must iterate or use a symbolic code to solve

The result is θ = 27.2º. It is easy to conﬁrm that (B) does not deliver good accuracy in this case. Depending on the range of wind speeds expected at the sculpture site, it may be best to employ a lighter cylinder to obtain a larger angle of deﬂection. We should also be aware that the aspect ratio of the cylinder affects the drag coefficient so it may be best to validate a design based on (A) or (B) by careful calibration experiments.