# 17.4: Open-channel Hydraulics

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Sometimes a dense cold-air layer lies under a lessdense warmer layer, with a relatively sharp temperature discontinuity (∆T = ∆θ) between the two layers (Fig. 17.15a and b). This temperature jump marks the **density interface** between the two layers. Examples of such a** two-layer system** include arctic air advancing behind a cold front and sliding under warmer air, cold gust fronts from thunderstorms, and cool marine air moving inland under warmer continental air.

These two-layer systems behave similarly to water in an **open channel** — a two layer system of dense water under less-dense air. Hence, you can apply **hydraulics** (applications of liquid flow based on its mechanical properties) to the atmosphere, for cases where air compressibility is not significant.

Sometimes the cold air can be stably stratified (Figs. 17.15c & d) as idealized here with constant lapse rate, where ∆θ/∆z = ∆T/∆z + Γ_{d} , using the dry adiabatic lapse rate Γ_{d} = 9.8 °C km^{–1}. You can use modified hydraulic theory for these cases.

Hydraulic theory depends on the speed of waves on the interface between cold and warm air.

These two-layer systems behave similarly to water in an **open channel** — a two layer system of dense water under less-dense air. Hence, you can apply **hydraulics** (applications of liquid flow based on its mechanical properties) to the atmosphere, for cases where air compressibility is not significant.

Sometimes the cold air can be stably stratified (Figs. 17.15c & d) as idealized here with constant lapse rate, where ∆θ/∆z = ∆T/∆z + Γ_{d} , using the dry adiabatic lapse rate Γ_{d} = 9.8 °C km^{–1}. You can use modified hydraulic theory for these cases.

Hydraulic theory depends on the speed of waves on the interface between cold and warm air.

# 17.4.1. Wave Speed

**Waves** (vertical oscillations that propagate horizontally on the density interface) can exist in air (Fig. 17.16), and behave similarly to water waves. For hydraulics, if water in a channel is shallow, then long-wavelength waves on the __water__ surface travel at the intrinsic “shallow-water” **phase speed** c_{o} of:

\(\ \begin{align} c_{o}=\sqrt{|g| \cdot h}\tag{17.14}\end{align}\)

where |g| = 9.8 m s^{–2} is gravitational acceleration magnitude, and h is average water depth. **Intrinsic phase speed** is the speed of propagation of any wave crest __relative__ to the mean fluid motion.

For a two-layer __air__ system, the effect of gravity is reduced because it depends on the relative buoyancy between the warm and cold air layers. Define a **reduced gravity** as |g’|= |g|·∆θ_{v}/T_{v}. For a shallow bottom layer, the resulting intrinsic phase speed of **surface waves** on the __interface between the coldand warm-air layers__ is:

\(\ \begin{align} c_{o}=\sqrt{\left|g^{\prime}\right| \cdot h}=\left(|g| \cdot \frac{\Delta \theta_{v}}{T_{v}} \cdot h\right)^{1 / 2}\tag{17.15}\end{align}\)

where ∆θ_{v} is the virtual potential temperature jump between the two air layers, T_{v} is an average absolute virtual temperature (in Kelvin), h is the depth of the cold layer of air, and |g| = 9.8 m·s^{–2} is gravitational acceleration magnitude.

The speed that wave __energy__ travels through a fluid is the **group speed** c_{g}. Group speed is the speed that hydraulic information can travel relative to the mean flow velocity, and it determines how the upstream flow reacts to downstream flow changes. For a two-layer system with bottom-layer depth less than 1/20 the wavelength, the group speed equals the phase speed

\(\ \begin{align} c_{g}=c_{o}\tag{17.16}\end{align}\)

**Sample Application**

For a two-layer __air__ system with 5°C virtual potential temperature difference across the interface and a bottom-layer depth of 20 m, find the intrinsic group speed, and compare it to the speed of a __water__ wave.

**Find the Answer**

Given: ∆θ_{v} = 5°C = 5K, h = 20 m. Assume T_{v} = 283 K

Find: c_{g} = ? m s^{–1}, for air and for water

For a 2-layer air system, use eqs. (17.15 & 17.16):

c_{g} = c_{o} = [(9.8 m·s^{–2})·(5K)·(20m)/(283K)]^{1/2} = __1.86 m s__^{–1}

For water under air, use eq. (17.14 & 17.16)):

c_{g} = c_{o} = [(9.8 m·s^{–2})·(20m)]^{1/2} = __14 m s__^{–1}

**Check:** Units OK. Magnitude OK.

**Exposition:** Atmospheric waves travel much slower than channel or ocean waves, and have much longer wavelengths. This is because of the reduced gravity |g’| for air, compared to the full gravity |g| for water.

For a statically-stable __atmospheric__ system with constant lapse rate, there is no surface (no interface) on which the waves can ride. Instead, **internal waves** can exist that propagate both horizontally and vertically inside the statically stable region. Internal waves reflect from solid surfaces such as the ground, and from statically neutral layers.

For internal waves, the horizontal component of group velocity u_{g} depends on both vertical and horizontal wavelength λ. To simplify this complicated situation, focus on infinitely-long waves in the horizontal (which propagate the fastest in the horizontal), and focus on a wave for which the vertical wavelength is proportional to the depth h of the statically stable layer of air. Thus:

\(\ \begin{align}u_{g}=N_{B V} \cdot h=\left(\frac{|g|}{T_{v}} \cdot \frac{\Delta \theta_{v}}{\Delta z}\right)^{1 / 2} \cdot h\tag{17.17}\end{align}\)

where N_{BV} is the Brunt-Väisälä frequency, and ∆θ_{v}/ ∆z is the vertical gradient of virtual potential temperature (a measure of static-stability strength).