Attached Eddies and High-Order Statistics

Abstract

The attached eddy hypothesis of Townsend [16] is the basis of a model of the logarithmic region in wall-bounded turbulent flows, in which the inertially dominated part of the flow is described by a hierarchy of self-similar eddying motions that extend to the wall. The hypothesis has gained considerable support from high Reynolds number experiments and recently from DNS Sillero et al., Phys. Fluids 25:105102, 2013, [14]. Meneveau and Marusic, J. Fluid Mech., 719:R1, 2013, [9] also recently used the attached eddy hypothesis, together with the central limit theorem, to deduce that all even-ordered moments of the streamwise velocity will exhibit a logarithmic dependence on the distance from the wall. This was also further supported by experimental evidence.

Appendix

Since Campbell’s theorem is usually applied to systems in which events are randomly placed in time, rather than space as the eddies are here, it is worthwhile to extend the existing proofs to cover spacial rather than temporal averages. This is an extension of a proof of Campbell’s theorem by Rice [13], and applies to the mean velocity and the Reynolds stresses.

Central to the following proof are two physical assumptions we have made about the eddies. The first is the inherently reasonable assumption that there is a limited region of space over which the velocity field corresponding to a single eddy is non-negligible. Mathematically, this manifests as the fact that any integral over the x-y plane of the velocity field corresponding to a single eddy will be equivalent.

We illustrate here in one dimension: For any function f(x),

$$\begin{aligned} \int \limits _{-\infty }^{ \infty } f(x-a ) \, \text{ d }a = \int \limits _{-\infty }^{\infty } f(x) \, \text{ d }x. \end{aligned}$$

(18)

In fact, it is not necessary for the integral to be taken over the entire real line. It is sufficient that the value of f(x) should be zero (or negligible) outside the bounds of integration. In this work, we will always take integrals over the entire x-y plane when using the above equation.

The second assumption we make use of is more controversial, namely that the locations of each eddy are independent of each other. As has been stated in Sect. 1, the angled brackets refer to ensemble averages in this work. Again we demonstrate with a one-dimensional example: If F(x) represents the sum of K copies of the function f(x), each of which is randomly located, so that

$$\begin{aligned} F(x)=\sum _{k=1}^K f(x-a_k ), \end{aligned}$$

(19)

then the ensemble average of F(x) will be given by

$$\begin{aligned} \langle F(x) \rangle =\int \limits _{-\infty }^{ \infty } p_1 (a_1) \int \limits _{-\infty }^{ \infty } p_2 (a_2) \dots \int \limits _{-\infty }^{ \infty } p_K (a_K) \sum _{k=1}^K f(x-a_k ) \, \text{ d }a_1 \, \text{ d }a_2 \dots \text{ d }a_K, \end{aligned}$$

(20)

where the $p_k(a_k)$ represent the probability that the kth variable has the value $a_k$. For each element k of the sum above, it is only the integral over $a_k$ that will be non-zero. This is because, as we have assumed, the location of each eddy is independent of the location of every other eddy. The implication is that the above equation can be simplified to

$$\begin{aligned} \langle F(x) \rangle = \sum _{k=1}^K \int \limits _{-\infty }^{ \infty } p_k (a_k) f(x-a_k ) \, \text{ d }a_k. \end{aligned}$$

(21)

We will make use of (18) and (21) subsequently. We now explicitly consider eddies in three-dimensional space. Imagine there are N eddies randomly placed on a two-dimensional space of area $L^2$. (We will ultimately consider the limit as $L\rightarrow \infty$.) The velocity field that corresponds to these N eddies is given by

$$\begin{aligned} \mathbf {U}_N(\mathbf {x}) = \sum _{k=1}^{N} \mathbf {Q}\left( \frac{\mathbf {x}-\mathbf {x}_{e_k}}{h_k}\right) . \end{aligned}$$

(22)

When we take the ensemble average of $\mathbf {U}_{N}(\mathbf {x})$, we are averaging over infinitely many realisations in which N eddies are randomly placed on a plane of area $L^2$. Because the eddies are perfectly randomly placed, $p(\mathbf {x}_e)$, the probability density function for the location of an eddy is given by

$$\begin{aligned} p(\mathbf {x}_e) = \frac{1}{L^2}. \end{aligned}$$

(23)

The difference between the various realisations of this system will simply be the different placements of each of the eddies. Averaging over the ensemble therefore entails averaging over the possible locations of each of the eddies. Therefore,

$$\begin{aligned} \langle \mathbf {U}_{N}(\mathbf {x}) \rangle = \iint \limits _{-L/2}^{ L/2} p(\mathbf {x}_{e_1})\dots \iint \limits _{-L/2}^{ L/2} p(\mathbf {x}_{e_N})\sum _{k=1}^{N} \int \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} \mathbf {Q}\left( \frac{\mathbf {x}-\mathbf {x}_{e_k}}{h}\right) P(h ) \, \text{ d }h \, \text{ d }\mathbf {x}_{e_N} \dots \text{ d }\mathbf {x}_{e_1}. \end{aligned}$$

(24)

(Note that because the wall-normal component of $\mathbf {x}_e$ is universally zero, it has been taken to be a two-dimensional vector in the above integrals. Throughout this work, any integral over $\mathbf {x}_e$ should be assumed to be over the x and y planes only.) By substituting (23) into the above, we get

$$\begin{aligned} \langle \mathbf {U}_{N}(\mathbf {x}) \rangle = \iint \limits _{-L/2}^{ L/2} \frac{1}{L^2} \dots \iint \limits _{-L/2}^{ L/2} \frac{1}{L^2} \sum _{k=1}^{N} \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} \mathbf {Q}\left( \frac{\mathbf {x}-\mathbf {x}_{e_k}}{h}\right) P(h ) \, \text{ d }h \, \text{ d }\mathbf {x}_{e_N} \dots \text{ d }\mathbf {x}_{e_1}.\nonumber \\ \end{aligned}$$

(25)

We can now make use of the fact that the locations of each eddy are independent, in the same manner as Eq. (21), to simplify this to

$$\begin{aligned} \langle \mathbf {U}_{N}(\mathbf {x}) \rangle = \frac{1}{L^2} \sum _{k=1}^N \int \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} P(h) \iint \limits _{-L/2}^{ L/2} \mathbf {Q}\left( \frac{\mathbf {x}-\mathbf {x}_{e_k}}{h}\right) \text{ d }\mathbf {x}_{e_k} \, \text{ d }h. \end{aligned}$$

(26)

It is here that we make use of (18) and the fact that any integrals of a certain function over the entire x-y plane will be equivalent. The above equation therefore simplifies to

$$\begin{aligned} \langle \mathbf {U}_{N}(\mathbf {x}) \rangle = \frac{N}{L^2} \int \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} P(h) \iint \limits _{-L/2}^{ L/2}\mathbf {Q}\left( \frac{\mathbf {x}}{h}\right) \, \text{ d }x \, \text{ d }y \, \text{ d }h. \end{aligned}$$

(27)

We can now average $\mathbf {U}$ over all possible values of N. The overall velocity will therefore be expected to be

$$\begin{aligned} \langle \mathbf {U}(\mathbf {x}) \rangle = \sum _{N=0}^\infty \mathscr {P}(N) \langle \mathbf {U}_{N}(\mathbf {x}) \rangle , \end{aligned}$$

(28)

where $\mathscr {P} (N)$ represents the probability that there are exactly N eddies. According to Poisson’s law of small probabilities, the probability that there will be exactly N eddies on a plane of area $L^2$ will be

$$\begin{aligned} \mathscr {P}(N) = \frac{(\beta L^2 )^N}{N!} e^{-\beta L^2}. \end{aligned}$$

(29)

Using this value of $\mathscr {P}(N)$, it is easy to verify that

$$\begin{aligned} \sum _{n=0}^\infty N \mathscr {P}(N) = L^2 \beta . \end{aligned}$$

(30)

Using (27) and (30) we can evaluate the sum in (28). In doing so we find that

$$\begin{aligned} \langle \mathbf {U}(\mathbf {x}) \rangle = \ \beta \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h ) \iint \limits _{-L/2}^{ L/2}\mathbf {Q}\left( \frac{\mathbf {x}}{h}\right) \, \text{ d }x \, \text{ d }y \, \text{ d }h. \end{aligned}$$

(31)

The mean velocity can now be related to the eddy contribution functions, $I_{k,l,m}(Z)$, which have been defined in (8). For the streamwise velocity in the limit as $L\rightarrow \infty$,

$$\begin{aligned} \langle U\rangle = \beta \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} I_{1,0,0}(Z) h^2 P(h ) \, \text{ d }h , \end{aligned}$$

(32)

and similarly for $\langle V\rangle$ and $\langle W\rangle$. This result could be achieved through other means, since it essentially states that the velocity field corresponding to many geometrically identical eddies will simply be the sum of the velocity fields corresponding to each individual eddy. However, by extending this methodology to the averages of higher powers of the velocity, we see its utility.

We demonstrate this now by deriving the Reynolds shear stress. It would be trivial to modify the following derivation so that it instead derives $\langle u^2\rangle$, $\langle v^2\rangle$ or $\langle w^2\rangle$.

$$\begin{aligned} \langle uw\rangle =&\left\langle ( U - \langle U\rangle )( W - \langle W\rangle ) \right\rangle \nonumber \\ =&\langle UW\rangle -\langle U\rangle \langle W\rangle . \end{aligned}$$

(33)

The definition of $\mathbf {U}_{N}$ given in (22) can be extended to

$$\begin{aligned} \textit{UW}_N (\mathbf {x}) = \sum _{k=1}^{N} \sum _{m=1}^{N} Q_x \left( \frac{\mathbf {x}-\mathbf {x}_{e_k}}{h_k}\right) Q_z \left( \frac{\mathbf {x}-\mathbf {x}_{e_m}}{h_m}\right) . \end{aligned}$$

(34)

Once again, the ensemble average is found by averaging over all of the possible heights and locations of the eddies, as it was in (24). This leads to

$$\begin{aligned} \langle U&W_{N} \rangle = \sum _{k=1}^N \sum _{m=1}^N \iint \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} P(h_k )P(h_m ) \nonumber \\&\ \times \iint \limits _{-L/2}^{ L/2} p(x_{e_1}) \dots \iint \limits _{-L/2}^{ L/2}p(x_{e_N}) Q_x \left( \frac{\mathbf {x}-\mathbf {x}_{e_k}}{h_k}\right) Q_z\left( \frac{\mathbf {x}-\mathbf {x}_{e_m}}{h_m}\right) \, \text{ d }h_k \, \text{ d }h_m \, \text{ d }\mathbf {x}_{e_N} \dots \text{ d }\mathbf {x}_{e_1}. \end{aligned}$$

(35)

Naturally, there will be N cases in which $Q_x$ and $Q_z$ refer to the same eddy (i.e. where $k=m$). In these cases, the multiple integrals over $h_k$, $h_m$ and the various $\mathbf {x}_e$ simplify to

$$\begin{aligned} \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h_k ) \iint \limits _{-L/2}^{ L/2} p(x_{e_k}) Q_x \left( \frac{\mathbf {x}-\mathbf {x}_{e_k}}{h_k}\right)&Q_z\left( \frac{\mathbf {x}-\mathbf {x}_{e_k}}{h_k}\right) \text{ d }\mathbf {x}_{e_k} \, \text{ d }h_k = \nonumber \\&\int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h ) \iint \limits _{-L/2}^{ L/2} Q_x\left( \frac{\mathbf {x}}{h}\right) Q_z \left( \frac{\mathbf {x}}{h}\right) \frac{\text{ d }\mathbf {x}}{L^2} \, \text{ d }h. \end{aligned}$$

(36)

We have again made use of (18), and so the right-hand side above reflects the fact that we are taking the limit as $L\rightarrow \infty$, and in this limit, any integral over the entire x-y plane will be equivalent. It has also been made use of (23) on the right-hand side above.

This leaves the $N^2 -N$ cases in which $Q_x$ and $Q_z$ refer to different eddies (i.e. $k\not = m$). In these cases, the multiple integral in (35) becomes

$$\begin{aligned} \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h ) \iint \limits _{-L/2}^{ L/2} Q_x \left( \frac{\mathbf {x}}{h_k} \right) \frac{\text{ d }\mathbf {x}}{L^2} \, \text{ d }h \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h ) \iint \limits _{-L/2}^{ L/2} Q_z \left( \frac{\mathbf {x}}{h_m} \right) \frac{\text{ d }\mathbf {x}}{L^2} \, \text{ d }h. \end{aligned}$$

(37)

By substituting these two into (35), we arrive at

$$\begin{aligned}&\langle \textit{UW}_{N} \rangle = \frac{N}{L^2} \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h ) \iint \limits _{-L/2}^{ L/2} Q_x\left( \frac{\mathbf {x}}{h}\right) Q_z \left( \frac{\mathbf {x}}{h}\right) \, \text{ d }x \, \text{ d }y \, \text{ d }h \nonumber \\&+ \frac{N^2 -N}{L^4} \int \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} P(h ) \iint \limits _{-L/2}^{ L/2} Q_x(\mathbf {x})\, \text{ d }x \, \text{ d }y \, \text{ d }h \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h )\iint \limits _{-L/2}^{ L/2} Q_z(\mathbf {x}) \, \text{ d }x \, \text{ d }y \, \text{ d }h. \end{aligned}$$

(38)

To determine $\langle \textit{UW}_{N} \rangle$, we must now sum over all possible values of N, as we did in (31). This gives

$$\begin{aligned} \langle \textit{UW} \rangle = \sum _{N=0}^\infty \mathscr {P}(N) \langle \textit{UW}_{N} \rangle . \end{aligned}$$

(39)

By using (29) we can easily show that

$$\begin{aligned} \sum _{n=0}^\infty (N^2 -N) \mathscr {P}(N) = L^4 \beta ^2. \end{aligned}$$

(40)

After we substitute (30), (38), (40) and the above into (39), we find that the sum over N reduces to

$$\begin{aligned} \langle \textit{UW}&\rangle = \beta \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h ) \iint \limits _{-L/2}^{ L/2} Q_x\left( \frac{\mathbf {x}}{h}\right) Q_z \left( \frac{\mathbf {x}}{h}\right) \, \text{ d }x \, \text{ d }y \, \text{ d }h \nonumber \\&\ + \beta \int \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} P(h ) \iint \limits _{-L/2}^{ L/2} Q_x(\mathbf {x})\, \text{ d }x \, \text{ d }y \, \text{ d }h \, .\, \beta \int \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} P(h )\iint \limits _{-L/2}^{ L/2} Q_z(\mathbf {x}) \, \text{ d }x \, \text{ d }y \, \text{ d }h. \end{aligned}$$

(41)

If we take the limit as $L\rightarrow \infty$ then, after taking (9) into account, the above becomes

$$\begin{aligned} \langle \textit{UW} \rangle = \beta \int \limits _{h_{\textit{min}}}^{h_{\textit{max}}} P(h ) \iint \limits _{-\infty }^{ \infty } Q_x\left( \frac{\mathbf {x}}{h}\right) Q_z \left( \frac{\mathbf {x}}{h}\right) \, \text{ d }x \, \text{ d }y \, \text{ d }h + \langle U\rangle \langle W\rangle . \end{aligned}$$

(42)

By substituting the above into (33), we obtain

$$\begin{aligned} \langle \textit{uw}\rangle = \beta \int \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} P(h ) \iint \limits _{-\infty }^{ \infty } Q_x\left( \frac{\mathbf {x}}{h}\right) Q_z \left( \frac{\mathbf {x}}{h}\right) \, \text{ d }x \, \text{ d }y \, \text{ d }h. \end{aligned}$$

(43)

By comparing the above to (8), we can relate the Reynolds shear stress above to the eddy contribution function. This results in

$$\begin{aligned} \langle \textit{uw}\rangle = \beta \int \limits _{h_{\textit{min}}}^{ h_{\textit{max}}} I_{1,0,1}(Z) h^2 P(h ) \, \text{ d }h. \end{aligned}$$

(44)