## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## WriteWise AI Model by Holooly Genius

Your Ultimate AI Essay Writer & Assistant.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 12.4

Consider the autocorrelation function $R_{11}(τ)=\overline{u^{2}_{1}}/(1+ω^{2}_{o}\tau^{2})$ for the random variable $u_{1}(t)$. What are the integral $(Λ_{t})$ and Taylor $(λ_{t})$ length scales, and the spectrum $(S_{e})$ of $u_{1}(t)$?

## Verified Solution

The integral scale can be found directly by substituting the given autocorrelation function into (12.18) and evaluating the integral using an integration variable $\beta=\omega_{o}\tau$:

$\Lambda_{t}\equiv\overset{\infty}{\underset{0}{\int}}r_{11}(\tau)d\tau=(1/R_{11}(0))\overset{\infty}{\underset{0}{\int}}R_{11}(\tau)d\tau$,                     (12.18)

$\Lambda_{t}=\frac{1}{R_{11}(0)}\overset{+\infty}{\underset{0}{\int}}R_{11}(\tau)d\tau=\frac{1}{\overline{u^{2}_{1}}}\overset{+\infty}{\underset{0}{\int}}\frac{\overline{u^{2}_{1}}}{1+\omega^{2}_{o}\tau^{2}}d\tau=\frac{1}{\omega_{o}}\overset{+\infty}{\underset{0}{\int}}\frac{d\beta}{1+\beta^{2}}=\frac{1}{\omega_{o}}[\tan^{-1}\beta]^{+\infty}_{0}=\frac{\pi}{2\omega_{o}}$.

The Taylor scale can be found from (12.19). Here, $r_{11}(\tau)=1/(1+\omega^{2}_{o}\tau^{2})$, which can be expanded around τ = 0 to find: $r_{11}(\tau)=1-\omega^{2}_{o}\tau^{2}+\cdot\cdot\cdot$. Thus, $[d^{2}r_{11}/d\tau^{2}]_{\tau=0}=-2\omega^{2}_{o}$, so (12.19) implies

$\lambda^{2}_{t}\equiv-2/[d^{2}r_{11}/d\tau^{2}]_{\tau=0}$       (12.19)

$\lambda_{t}\equiv(-2/[d^{2}r_{11}/d\tau^{2}]_{\tau=0})^{1/2}=\frac{1}{\omega_{o}}$.

As expected, the integral length scale is larger than the Taylor length scale. In high-Reynolds number turbulence, the ratio $\Lambda_{t}/\lambda_{t}$ can be much greater than that found here. The spectrum $S_{e}(\omega)$ is obtained from (12.20):

$S_{e}(\omega)\equiv\frac{1}{2\pi}\overset{+\infty}{\underset{-\infty}{\int}}R_{11}(\tau)\exp\left\{-i\omega\tau\right\}d\tau$.        (12.20)

$S_{e}(\omega)\equiv\frac{\overline{u^{2}_{1}}}{2\pi}\overset{+\infty}{\underset{-\infty}{\int}}\frac{\exp\left\{-i\omega\tau\right\}}{1+\omega^{2}_{o}\tau^{2}}d\tau=\frac{\overline{u^{2}_{1}}}{2\omega_{o}}\exp\left\{-\frac{|\omega|}{\omega_{o}}\right\}$,

where the integral is readily evaluated using complex contour integration techniques.