8.2: Turbulence

If molecular viscosity is important only over distances of a few millimeters, and if it is not important for most oceanic flows, unless of course you are a zooplankton trying to swim in the ocean, how then is the influence of a boundary transferred into the interior of the flow? The answer is: through turbulence.

Turbulence arises from the non-linear terms in the momentum equation (\(u \ \partial u/\partial x\), etc.). The importance of these terms is given by a non-dimensional number, the Reynolds Number \(Re\), which is the ratio of the non-linear terms to the viscous terms: \[Re = \frac{\text{Non-linear Terms}}{\text{Viscous Terms}} = \frac{\left(u \ \dfrac{\partial u}{\partial x}\right)}{\left(\nu \dfrac{\partial^{2} u}{\partial x^{2}}\right)} \approx \frac{U \dfrac{U}{L}}{\nu \dfrac{U}{L^{2}}} = \frac{UL}{\nu} \nonumber \]

where \(U\) is a typical velocity of the flow and \(L\) is a typical length describing the flow. You are free to pick whatever \(U\), \(L\) might be typical of the flow. For example, \(L\) can be either a typical cross-stream distance, or an along-stream distance. Typical values in the open ocean are \(U = 0.1 \ \text{m/s}\) and \(L = 1 \ \text{megameter}\), so \(Re = 10^{11}\). Because non-linear terms are important if \(Re > 10 – 1000\), they are certainly important in the ocean. The ocean is turbulent.

The Reynolds number is named after Osborne Reynolds (1842–1912) who conducted experiments in the late 19th century to understand turbulence. In one famous experiment (Reynolds 1883), he injected dye into water flowing at various speeds through a tube (figure \(\PageIndex{1}\)). If the speed was small, the flow was smooth. This is called laminar flow. At higher speeds, the flow became irregular and turbulent. The transition occurred at \(Re = V \ \frac{D}{\nu} \approx 2000\), where \(V\) is the average speed in the pipe, and \(D\) is the diameter of the pipe.

Figure \(\PageIndex{1}\): Reynolds apparatus for investigating the transition to turbulence in pipe flow, with photographs of near-laminar flow (left) and turbulent flow (right) in a clear pipe much like the one used by Reynolds. After Binder (1949: 88-89).

As Reynolds number increases above some critical value, the flow becomes more and more turbulent. Note that flow pattern is a function of Reynold’s number. All flows with the same geometry and the same Reynolds number have the same flow pattern. Thus flow around all circular cylinders, whether 1 mm or 1 m in diameter, look the same as the flow at the top of figure \(\PageIndex{2}\) if the Reynolds number is 20. Furthermore, the boundary layer is confined to a very thin layer close to the cylinder, in a layer too thin to show in the figure.

Figure \(\PageIndex{2}\): Flow past a circular cylinder as a function of Reynolds number between one and a million. From Richardson (1961). The appropriate flows are: A—a toothpick moving at 1 mm/s; B—finger moving at 2 cm/s; F—hand out a car window at 60 mph. All flow at the same Reynolds number has the same streamlines. Flow past a 10 cm diameter cylinder at 1 cm/s looks the same as 10 cm/s flow past a cylinder 1 cm in diameter because in both cases \(Re = 1000\).

Turbulent Stresses: The Reynolds Stress

Prandtl, Karman and others who studied fluid mechanics in the early 20th century hypothesized that parcels of fluid in a turbulent flow played the same role in transferring momentum within the flow that molecules played in laminar flow. The work led to the idea of turbulent stresses. To see how these stresses might arise, consider the momentum equation for a flow with mean \((U, V, W)\) and turbulent \((u', v', w')\) components: \[u = U + u'; \quad v = V + v'; \quad w = W + w'; \quad p = P + p' \nonumber \]

where the mean value \(U\) is calculated from a time or space average: \[U = \langle u \rangle = \frac{1}{T} \int_{0}^{T} u(t) \ dt \quad \text{or} \quad U = \langle u \rangle = \frac{1}{X} \int_{0}^{X} u(x) \ dx \nonumber \]

The non-linear terms in the momentum equation can be written \[\begin{align} \biggl \langle (U + u') \frac{\partial (U + u')}{\partial x} \biggr \rangle &= \biggl \langle U \frac{\partial U}{\partial x} \biggr \rangle + \biggl \langle U \frac{\partial u'}{\partial x} \biggr \rangle + \biggl \langle u' \frac{\partial U}{\partial x} \biggr \rangle + \biggl \langle u' \frac{\partial u'}{\partial x} \biggr \rangle \nonumber \\ \biggl \langle (U + u') \frac{\partial (U + u')}{\partial x} \biggr \rangle &= \biggl \langle U \frac{\partial U}{\partial x} \biggr \rangle + \biggl \langle u' \frac{\partial u'}{\partial x} \biggr \rangle \end{align} \nonumber \]

The second equation follows from the first since \(\langle U \ \partial u' / \partial x \rangle = 0\) and \(\langle u' \ \partial U / \partial x \rangle = 0\), which follow from the definition of \(U\): \(\langle U \ \partial u' / \partial x \rangle = U \partial \langle u' \rangle / \partial x = 0\).

Using \((\PageIndex{2})\) in \((7.7.3)\) gives: \[\frac{\partial U}{\partial x} + \frac{\partial V}{\partial y} + \frac{\partial W}{\partial z} + \frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y} + \frac{\partial w'}{\partial z} = 0 \nonumber \]

Subtracting the mean of \((\PageIndex{5})\) from \((\PageIndex{5})\) splits the continuity equation into two equations: \[\begin{align} \frac{\partial U}{\partial x} + \frac{\partial V}{\partial y} + \frac{\partial W}{\partial z} = 0 \\ \frac{\partial u'}{\partial x} + \frac{\partial v'}{\partial y} + \frac{\partial w'}{\partial z} = 0 \end{align} \nonumber \]

Using \((\PageIndex{2})\) in \((8.1.1)\) taking the mean value of the resulting equation, then simplifying using \((\PageIndex{4})\), the \(x\)-component of the momentum equation for the mean flow becomes: \[\frac{Du}{Dt} = -\frac{1}{\rho} \frac{\partial P}{\partial x} + 2 \Omega \ V \sin \varphi + \frac{\partial}{\partial x} \left[\nu \frac{\partial U}{\partial x} - \langle u'u' \rangle\right] + \frac{\partial}{\partial y} \left[\nu \frac{\partial U}{\partial y} - \langle u'v' \rangle\right] + \frac{\partial}{\partial x} \left[\nu \frac{\partial U}{\partial z} - \langle u'w' \rangle\right] \nonumber \]

The derivation is not as simple as it seems. See Hinze (1975: 22) for details. Thus the additional force per unit mass due to the turbulence is: \[F_{x} = - \frac{\partial}{\partial x} \langle u'u' \rangle - \frac{\partial}{\partial y} \langle u'v' \rangle - \frac{\partial}{\partial z} \langle u'w' \rangle \nonumber \]

The terms \(\rho \langle u'u' \rangle\), \(\rho \langle u'v' \rangle\), and \(\rho \langle u'w' \rangle\) transfer eastward momentum \((\rho u')\) in the \(x\), \(y\), and \(z\) directions. For example, the term \(\rho \langle u' w' \rangle\) gives the downward transport of eastward momentum across a horizontal plane. Because they transfer momentum, and because they were first derived by Osborne Reynolds, they are called Reynolds Stresses.