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17.7: Mountain Waves

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  • Screen Shot 2020-03-30 at 10.20.22 PM.png
    Figure 17.29 Mountain-wave characteristics.

    17.7.1. Natural Wavelength

    When statically stable air flows with speed M over a hill or ridge, it is set into oscillation at the Brunt-Väisälä frequency, NBV. The natural wavelength λ is

    \(\ \begin{align} \lambda=\frac{2 \pi \cdot M}{N_{B V}}\tag{17.30}\end{align}\)

    Longer wavelengths occur in stronger winds, or weaker static stabilities.

    These waves are known as mountain waves, gravity waves, buoyancy waves, or lee waves. They can cause damaging winds, and interesting clouds (see the Clouds chapter).

    Friction and turbulence damp the oscillations with time (Fig. 17.29). The resulting path of air is a damped wave:

    \(\ \begin{align} z=z_{1} \cdot \exp \left(\frac{-x}{b \cdot \lambda}\right) \cdot \cos \left(\frac{2 \pi \cdot x}{\lambda}\right)\tag{17.31}\end{align}\)

    where z is the height of the air above its starting equilibrium height, z1 is the initial amplitude of the wave (based on height of the mountain), x is distance downwind of the mountain crest, and b is a damping factor. Wave amplitude reduces to 1/e at a downwind distance of b wavelengths (that is, b·λ is the e-folding distance).

    Sample Application (§)

    Find and plot the path of air over a mountain, given: z1 = 500 m, M = 30 m s–1, b = 3, ∆T/∆z = –0.005 K m–1, T = 10°C, and Td = 8°C for the streamline sketched in Fig. 17.29.

    Find the Answer

    Given: (see above). Thus T = 283 K

    Find: NBV = ? s–1, λ = ? m, and plot z vs. x

    From the Stability chapter:

    \(\ N_{B V}=\left[\frac{9.8 \mathrm{m} \cdot \mathrm{s}^{-2}}{283 \mathrm{K}} \cdot(-0.005+0.0098)\right]^{1 / 2}= 0.0129 s^{-1} \)

    Use eq. (17.30)

    \(\ \lambda=\frac{2 \pi \cdot(30 \mathrm{m} / \mathrm{s})}{0.0129 \mathrm{s}^{-1}}=14.62 \mathrm{km}\)

    Solve eq. (17.31) on a spreadsheet to get the answer:

    Screen Shot 2020-03-30 at 10.36.11 PM.png

    Check: Units OK. Physics OK. Sketch OK.

    Exposition: Glider pilots can soar in the updraft portions of the wave, highlighted with white boxes.

    17.7.2. Lenticular Clouds

    In the updraft portions of mountain waves, the rising air cools adiabatically. If sufficient moisture is present, clouds can form, called lenticular clouds. The first cloud, which forms over the mountain crest, is usually called a cap cloud (see Clouds chapter).

    The droplet sizes in these clouds are often quite uniform, because of the common residence times of air in the clouds. This creates interesting optical phenomena such as corona and iridescence when the sun or moon shines through them (see the Atmospheric Optics chapter).

    Knowing the temperature and dew point of air at the starting altitude before blowing over the mountain, a lifting condensation level (LCL) can be calculated using equations from the Water Vapor chapter. Clouds will form in the crests of those waves for which z > zLCL.

    Sample Application (§)

    Replot the results from the previous Sample Application, indicating which waves have lenticular clouds.

    Find the Answer

    Given: (see previous Sample Application).

    Find: zLCL = ? m.

    From the Moisture chapter: zLCL = a · (T – Td) .

    zLCL = (125m °C–1)·(10°C – 8°C) = 250 m

    above the reference streamline altitude. From the sketch below, we find 1 cap cloud and 2 lenticular clouds.

    Screen Shot 2020-03-30 at 10.45.02 PM.png

    Check: Units OK. Physics OK. Sketch OK.

    Exposition: Most clouds are blown with the wind, but standing-lenticular clouds are stationary while the wind blows through them!

    17.7.3. Froude Number - Part 2

    For individual hills not part of a continuous ridge, some air can flow around the hill rather than over the top. When less air flows over the top, shallower waves form.

    The third variety of Froude Number Fr3 is a measure of the ability of waves to form over hills. It is given by

    \(\ \begin{align} F r_{3}=\frac{\lambda}{2 \cdot W}\tag{17.32}\end{align}\)

    where W is the hill width, and λ is the natural wavelength. Fr3 is dimensionless.

    Figure 17.30 Mountain wave behavior vs. Froude number, Fr3.
    Screen Shot 2020-03-30 at 10.51.04 PM.pngScreen Shot 2020-03-30 at 10.52.08 PM.png

    For strong static stabilities or weak winds, Fr3 << 1. The natural wavelength of air is much shorter than the width of the mountain, resulting in only a little air flowing over the top of the hill, with small waves (Fig. 17.30a). If H is the height of the hill (Fig. 17.29), then wave amplitude z1 < H/2 for this case. Most of the air is blocked in front of the ridge, or flows around the sides for an isolated hill.

    For moderate stabilities where the natural wavelength is nearly equal to twice the hill width, Fr3 ≈ 1. The air resonates with the terrain, causing very intense waves (Fig. 17.30b). These waves have the greatest chance of forming lenticular clouds, and pose the threat of violent turbulence to aircraft. Extremely fast near-surface winds on the downwind (lee) side of the mountains cause downslope wind storms that can blow the roofs off of buildings. Wave amplitude roughly equals half the hill height: z1 ≈ H/2. Sometimes rotor circulations and rotor clouds will form near the ground under the wave crests (Fig. 17.30b; also see the Clouds chapter).

    For weak static stability and strong winds, the natural wavelength is much greater than the hill width, Fr3 >> 1. Wave amplitude is weak, z1 < H/2. A turbulent wake will form downwind of the mountain, sometimes with a cavity of reverse flow near the ground (Fig. 17.30c). The cavity and rotor circulations are driven by wind shear like a bike chain turning a gear.

    For statically neutral conditions, Fr = ∞. A large turbulent wake occurs (Fig. 17.30d). These wakes are hazardous to aircraft.

    Screen Shot 2020-03-30 at 11.00.51 PM.png
    Figure 17.31 Vertical wave propagation, tilting crests, and wave drag.

    Sample Application

    For a natural wavelength of 10 km and a hill width of 15 km, describe the type of waves.

    Find the Answer

    Given: λ = 10 km, W = 15 km.

    Find: Fr = ? (dimensionless)

    Use eq. (17.32): Fr3 = (10km)/[2· (15km)] = 0.333

    Check: Units OK. Magnitude OK.

    Exposition: Waves as in Fig. 17.30a form off the top of the hill, because Fr3 < 1. Some air also flows around sides of hill.

    17.7.4. Mountain-wave Drag

    For Fr3 < 1, wave crests tilt upwind with increasing altitude (Fig. 17.31). The angle α of tilt relative to vertical is

    \(\ \begin{align} \cos (\alpha)=F r_{3}\tag{17.33}\end{align}\)

    For this situation, slightly lower pressure develops on the lee side of the hill, and higher pressure on the windward side. This pressure gradient opposes the mean wind, and is called wave drag. Wave drag adds to the surface friction (skin drag). The wave drag (WD) force per unit mass near the ground is:

    \(\ \begin{align} \frac{F_{x W D}}{m}=-\frac{H^{2} \cdot N_{B V}^{2}}{8 \cdot h_{w}} \cdot F r_{3} \cdot\left[1-F r_{3}^{2}\right]^{1 / 2}\tag{17.34}\end{align}\)

    where hw is the depth of air containing waves.

    Not surprisingly, higher hills cause greater wave drag. The whole layer of air containing these waves feels the wave drag, not just the bottom of this layer that touches the mountain.

    Sample Application

    For Fr3 = 0.8, find the angle of the wave crests and the wave drag force over a hill of height 800 m. The Brunt-Väisälä frequency is 0.01 s–1, and the waves fill the 11 km thick troposphere.

    Find the Answer

    Given: Fr3 = 0.8 , H = 800 m, NBV = 0.01 s–1.

    Find: α = ?°, Fx WD/m = ? m·s–2

    Use eq. (17.33): α = cos–1(0.8) = 36.9°

    Use eq. (17.34):

    \(\ \frac{F_{x W D}}{m} =-\frac{\left[(800 \mathrm{m}) \cdot\left(0.01 \mathrm{s}^{-1}\right)\right]^{2}}{8 \cdot(11,000 \mathrm{m})} \cdot 0.8 \cdot\left[1-(0.8)^{2}\right]^{1 / 2}
    =\underline{\bf{3.5 \times 10^{-4} \mathrm{m} \cdot \mathrm{s}^{-1}}} \)

    Check: Units OK. Physics OK.

    Exposition: This is of the same order of magnitude as the other forces in the equations of motion.