Question 5 of 7: Anemometry

Question 5: Anemometry (20 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.

a) A Pitot tube has an opening facing directly into the flow (the stagnation/impact port) alongside a port sensing the surrounding static pressure. The flow is brought to rest at the stagnation port, converting all of its kinetic energy into pressure (Bernoulli's equation), so the measured pressure there is the total (stagnation) pressure, while the side port reads only the static pressure of the moving stream. The instrument outputs the difference between the two, the dynamic pressure $\Delta P=P_{total}-P_{static}=\tfrac{1}{2}\rho V^2$, from which the air velocity is recovered as $$V=\sqrt{\dfrac{2\,\Delta P}{\rho}}.$$

b) The ideal relation above assumes perfectly inviscid, incompressible flow that is brought to rest with no losses and that the tube is perfectly aligned with the flow. In practice the coefficient (typically written $V=C_p\sqrt{2\Delta P/\rho}$) absorbs the real, non-ideal effects: viscous/friction losses and boundary-layer effects near the probe tip, sensitivity to the tube's angular misalignment with the true flow direction (yaw/pitch angle), and any compressibility effects at higher velocities where the flow is no longer well approximated as incompressible — the coefficient is determined experimentally for a given probe geometry to correct for all of these together.

c) A thermal anemometer holds a small heated element (a fine wire or film) at an elevated temperature above the surrounding air, usually with active electronic control (constant-temperature or constant-current operation). Moving air convectively carries heat away from the element; the rate of forced-convection heat loss increases with air velocity, following a relation of the form (King's Law) $P=(A+B\sqrt{V})\Delta T$. By measuring either the electrical power needed to hold the element at constant temperature, or the temperature drop for constant heating power, the instrument infers $V$ from this known heat-transfer relationship — effectively using the element itself as both the heater and the temperature/heat-loss sensor.

d) Key design factors: the cups must have a strongly asymmetric drag coefficient between their concave (facing) and convex (back) sides so that the assembly reliably rotates in one direction regardless of instantaneous wind direction, and that drag asymmetry (together with the arm radius) sets the calibration constant relating spin rate to wind speed. Low-friction bearings and low rotational inertia are essential so the cups start spinning at low wind speeds (a low starting threshold) and track rapid gusts with minimal dynamic lag, while the assembly must still be robust/durable to withstand sustained high winds. The device must also respond essentially the same way to wind from any horizontal direction (rotational symmetry about the vertical spin axis), which is what makes it direction-independent for speed measurement (a separate wind vane is needed for direction).

e) Given. A ball of mass $m$ and frontal (projected) area $A$ hangs from a thread of length $L$; wind of velocity $V$ blows horizontally, deflecting the thread to angle $X$ from vertical (Fig. 3).

Find. An equation for $V$ in terms of the measured angle $X$.

Approach. Write the horizontal and vertical force balance on the ball (string tension, weight, aerodynamic drag) and eliminate the unknown tension $T$.

1. Resolve the string tension. The ball is in static equilibrium under three forces: weight $mg$ (down), horizontal drag $F_d$ (in the wind direction), and string tension $T$ (along the thread, toward the pivot). Resolving $T$ along the vertical and horizontal directions, $$T\cos X=mg,\qquad T\sin X=F_d.$$
2. Eliminate $T$. Dividing the two equations cancels $T$ entirely: $$\tan X=\dfrac{F_d}{mg}\quad\Longrightarrow\quad F_d=mg\tan X.$$
3. Substitute the drag-force relation. Standard aerodynamic drag on a bluff body is $F_d=\tfrac{1}{2}\rho V^2 C_d A$ (air density $\rho$, drag coefficient $C_d$, frontal area $A$). Setting this equal to the result of Step 2 and solving for $V$: $$\tfrac{1}{2}\rho V^2 C_d A=mg\tan X$$ $$\boxed{V=\sqrt{\dfrac{2\,mg\tan X}{\rho\,C_d\,A}}.}$$

The equation shows this simple sensor's calibration is entirely geometric once $m$, $A$ and $C_d$ are fixed — $V$ grows only as $\sqrt{\tan X}$, so the instrument becomes progressively less sensitive (larger $V$ change needed for the same change in $X$) as $X$ approaches 90°, a practical limit on its usable range.