Friday, April 3, 2015

Derivation and Nondimensionalization of the Orr-Sommerfeld Equation

The Orr-Sommerfeld equation is a famous equation that can give some insight into the stability of the velocity profile of a fluid flow. This is one of two parts on the derivation and stability analysis of the Orr-Sommerfeld equation. In this post, I derive the Orr-Sommerfeld equation starting from the 2-D Navier-Stokes equations. I then show how it can be nondimensionalized. It may look like a lot of math at first glance, but it is all relatively simple.

Derivation

The 2-D Navier-Stokes equations are given as follows:
$$\begin{align}
\nabla\vec{V}&=0\\
\rho\frac{D\vec{V}}{Dt}&=-\nabla p+\mu\nabla^{2}\vec{V}
\end{align}$$
Letting \(V_{x}=U+u'\), \(V_{y}=V+v'\), and \(p=P+p'\) and performing a small disturbance analysis gives the small perturbation version of the Navier-Stokes Equations
$$\begin{align}\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}&=0\\
\frac{\partial u'}{\partial t}+U\frac{\partial u'}{\partial x}+V\frac{\partial u'}{\partial y}+u'\frac{\partial U}{\partial x}+v'\frac{\partial U}{\partial y}&=-\frac{1}{\rho}\frac{\partial p'}{\partial x}+\frac{\mu}{\rho}\left(\frac{\partial^{2}u'}{\partial x^{2}}+\frac{\partial^{2}u'}{\partial y^{2}}\right)\\
\frac{\partial v'}{\partial t}+U\frac{\partial v'}{\partial x}+V\frac{\partial v'}{\partial y}+u'\frac{\partial V}{\partial x}+v'\frac{\partial V}{\partial y}&=-\frac{1}{\rho}\frac{\partial p'}{\partial y}+\frac{\mu}{\rho}\left(\frac{\partial^{2}v'}{\partial x^{2}}+\frac{\partial^{2}v'}{\partial y^{2}}\right)
\end{align}\tag{1}$$
Assuming parallel flow, where \(U\approx U(y)\) and \(V\approx0\), we can simplify this to the the following form of the Navier-Stokes equations.
$$\begin{align}\frac{\partial u'}{\partial x}+\frac{\partial v'}{\partial y}&=0\\\frac{\partial u'}{\partial t}+U\frac{\partial u'}{\partial x}+v'\frac{\partial U}{\partial y}&=-\frac{1}{\rho}\frac{\partial p'}{\partial x}+\frac{\mu}{\rho}\left(\frac{\partial^{2}u'}{\partial x^{2}}+\frac{\partial^{2}u'}{\partial y^{2}}\right)\\\frac{\partial v'}{\partial t}+U\frac{\partial v'}{\partial x}&=-\frac{1}{\rho}\frac{\partial p'}{\partial y}+\frac{\mu}{\rho}\left(\frac{\partial^{2}v'}{\partial x^{2}}+\frac{\partial^{2}v'}{\partial y^{2}}\right)\end{align}\label{simp_NS}\tag{2}$$
In this analysis, disturbances are assumed to be Tollmien-Schlichting waves, with the general form as follows.
$$\begin{align}\psi&=\phi(y)e^{i(\alpha x-\omega t)}\\u'&=\frac{\partial\psi}{\partial y}=\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\v'&=-\frac{\partial\psi}{\partial x}=-i\alpha\phi e^{i(\alpha x-\omega t)}\end{align}\tag{3}$$
The temporal and spatial derivatives are then calculated as follows.
$$\begin{align}\frac{\partial u'}{\partial t}&=-i\omega\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\\frac{\partial u'}{\partial x}&=i\alpha\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\\frac{\partial^{2}u'}{\partial x^{2}}&=-\alpha^{2}\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\\frac{\partial u'}{\partial y}&=\frac{\partial^{2}\phi}{\partial y^{2}}e^{i(\alpha x-\omega t)}\\\frac{\partial^{2}u'}{\partial y^{2}}&=\frac{\partial^{3}\phi}{\partial y^{3}}e^{i(\alpha x-\omega t)}\\\frac{\partial v'}{\partial t}&=-\alpha\omega\phi e^{i(\alpha x-\omega t)}\\\frac{\partial v'}{\partial x}&=\alpha^{2}\phi e^{i(\alpha x-\omega t)}\\\frac{\partial^{2}v'}{\partial x^{2}}&=i\alpha^{3}\phi e^{i(\alpha x-\omega t)}\\\frac{\partial v'}{\partial y}&=-i\alpha\frac{\partial\phi}{\partial y}e^{i(\alpha x-\omega t)}\\\frac{\partial^{2}v'}{\partial y^{2}}&=-i\alpha\frac{\partial^{2}\phi}{\partial y^{2}}e^{i(\alpha x-\omega t)}\end{align}\tag{4}$$
We can then substitute each of these derivatives into Equation \(\ref{simp_NS}\) and we get the following relations.
$$\begin{align}e^{i(\alpha x-\omega t)}\left[i\alpha\frac{\partial\phi}{\partial y}-i\alpha\frac{\partial\phi}{\partial y}\right]&=0\\-\rho e^{i(\alpha x-\omega t)}\left[-i\omega\frac{\partial\phi}{\partial y}+i\alpha U\frac{\partial\phi}{\partial y}+-i\alpha\phi\frac{\partial U}{\partial y}-\frac{\mu}{\rho}\left(-\alpha^{2}\frac{\partial\phi}{\partial y}+\frac{\partial^{3}\phi}{\partial y^{3}}\right)\right]&=\frac{\partial p'}{\partial x}\\-\rho e^{i(\alpha x-\omega t)}\left[-\alpha\omega\phi+U\alpha^{2}\phi-\frac{\mu}{\rho}\left(i\alpha^{3}\phi-i\alpha\frac{\partial^{2}\phi}{\partial y^{2}}\right)\right]&=\frac{\partial p'}{\partial y}\end{align}$$
To eliminate the pressure fluctuation term, differentiate the x- and y-momentum equations by \(y\) and \(x\), respectively.
$$\begin{align}\frac{1}{-\rho e^{i(\alpha x-\omega t)}}\frac{\partial^{2}p'}{\partial x\partial y}&=-i\omega\frac{\partial^{2}\phi}{\partial y^{2}}+i\alpha\frac{\partial U}{\partial y}\frac{\partial\phi}{\partial y}+i\alpha U\frac{\partial^{2}\phi}{\partial y^{2}}-i\alpha\frac{\partial\phi}{\partial y}\frac{\partial U}{\partial y}\\&\quad-i\alpha\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{\mu}{\rho}\left(\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}-\frac{\partial^{4}\phi}{\partial y^{4}}\right)\\\frac{1}{-i\alpha\rho e^{i(\alpha x-\omega t)}}\frac{\partial^{2}p'}{\partial x\partial y}&=-\alpha\omega\phi+U\alpha^{2}\phi+\frac{\mu}{\rho}\left(-i\alpha^{3}\phi+i\alpha\frac{\partial^{2}\phi}{\partial y^{2}}\right)\end{align}\tag{5}$$
Equating the two momentum equations gives
$$-i\omega\frac{\partial^{2}\phi}{\partial y^{2}}+i\alpha U\frac{\partial^{2}\phi}{\partial y^{2}}-i\alpha\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{\mu}{\rho}\left(2\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}-\frac{\partial^{4}\phi}{\partial y^{4}}-\alpha^{4}\phi\right)+i\alpha^{2}\omega\phi-iU\alpha^{3}\phi=0$$
This simplifies to the Orr-Sommerfeld Equation.
$$\left(U-\frac{\omega}{\alpha}\right)\left(\frac{\partial^{2}\phi}{\partial y^{2}}-\alpha^{2}\phi\right)-\phi\frac{\partial^{2}U}{\partial y^{2}}+\frac{i\nu}{\alpha}\left(\frac{\partial^{4}\phi}{\partial y^{4}}-2\alpha^{2}\frac{\partial^{2}\phi}{\partial y^{2}}+\alpha^{4}\phi\right)=0\label{orrsommerfeld}\tag{6}$$

Nondimensionalization

The Orr-Sommerfeld equation is nondimensionalized using the following nondimensional parameters,
$$\bar{U}=\frac{U}{U_{\infty}}\quad\xi=\frac{y}{\delta}\quad\bar{\phi}=\frac{\phi}{U_{\infty}\delta}\quad\bar{c}=\frac{c}{U_{\infty}}\quad\bar{\alpha}=\alpha\delta\quad Re_{\delta}=\frac{U_{\infty}\delta}{\nu}\tag{7}$$
where \(c=\frac{\omega}{\alpha}\) and \(\delta\) is the boundary layer thickness. Substituting these into Equation \(\ref{orrsommerfeld}\) gives
$$\begin{align}0&=\left(\bar{U}U_{\infty}-\bar{c}U_{\infty}\right)\left(\frac{1}{\delta^{2}}\frac{\partial^{2}}{\partial\xi^{2}}\left(\bar{\phi}U_{\infty}\delta\right)-\left(\frac{\bar{\alpha}}{\delta}\right)^{2}\left(\bar{\phi}U_{\infty}\delta\right)\right)-\frac{\bar{\phi}U_{\infty}\delta}{\delta^{2}}\frac{\partial^{2}}{\partial\xi^{2}}\left(\bar{U}U_{\infty}\right)\\&\quad+\frac{i\nu\delta}{\bar{\alpha}}\left(\frac{1}{\delta^{4}}\frac{\partial^{4}}{\partial\xi^{4}}\left(\bar{\phi}U_{\infty}\delta\right)-\frac{2\bar{\alpha}^{2}}{\delta^{4}}\frac{\partial^{2}}{\partial\xi^{2}}\left(\bar{\phi}U_{\infty}\delta\right)+\frac{\bar{\alpha}^{4}}{\delta^{4}}\left(\bar{\phi}U_{\infty}\delta\right)\right)\end{align}\tag{8}$$
Applying the chain rule for each partial derivative gives
$$\begin{align}0&=U_{\infty}\left(\bar{U}-\bar{c}\right)\left(\frac{U_{\infty}}{\delta}\frac{\partial^{2}\bar{\phi}}{\partial\xi^{2}}-\frac{\bar{\alpha}^{2}U_{\infty}}{\delta}\bar{\phi}\right)-\frac{U_{\infty}^{2}}{\delta}\frac{\partial^{2}\bar{U}}{\partial\xi^{2}}\bar{\phi}\\&\quad+\frac{i\nu\delta}{\bar{\alpha}}\left(\frac{U_{\infty}}{\delta^{3}}\frac{\partial^{4}\bar{\phi}}{\partial\xi^{4}}-\frac{2\bar{\alpha}^{2}U_{\infty}}{\delta^{3}}\frac{\partial^{2}\bar{\phi}}{\partial\xi^{2}}+\frac{\bar{\alpha}^{4}U_{\infty}}{\delta^{3}}\bar{\phi}\right)\end{align}\tag{9}$$
Finally, factor out \(\frac{U_{\infty}^{2}}{\delta}\) and substitute for \(Re_{\delta}\) to get
$$\left(\bar{U}-\bar{c}\right)\left(\frac{\partial^{2}\bar{\phi}}{\partial\xi^{2}}-\bar{\alpha}\bar{\phi}\right)-\frac{\partial^{2}\bar{U}}{\partial\xi^{2}}\bar{\phi}+\frac{i}{\bar{\alpha}Re_{\delta}}\left(\frac{\partial^{4}\bar{\phi}}{\partial\xi^{4}}-2\bar{\alpha}^{2}\frac{\partial^{2}\bar{\phi}}{\partial\xi^{2}}+\bar{\alpha}^{4}\bar{\phi}\right)=0\tag{10}$$
For convenience, derivatives with respect to the station coordinate \(\xi\) are hereafter denoted with prime notation. This gives the final nondimensional form of the Orr-Sommerfeld equation:
$$\left(\bar{U}-\bar{c}\right)\left(\bar{\phi}''-\bar{\alpha}\bar{\phi}\right)-\bar{U}''\bar{\phi}+\frac{i}{\bar{\alpha}Re_{\delta}}\left(\bar{\phi}''''-2\bar{\alpha}^{2}\bar{\phi}''+\bar{\alpha}^{4}\bar{\phi}\right)=0\tag{11}$$

15 comments:

  1. Hello Christopher, thanks a lot. This will help me a lot in my exam.

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  2. Thanks a ton, Christopher.

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  3. Hi Christopher. I really don't know how u^\prime.\dfrac{\partial u^\prime}{\parital x} and v^\prime.\dfrac{\partial u^\prime}{\parital x} are gone at the 2nd equation, in (1)? I'll very happy if you explain it to me. Thank you so much.

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    Replies
    1. it is due to small disturbance assumption, meaning that the problem is linearized, so any products of primes are neglected (small term squared).

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  4. Really thank you Christopher from Japan!

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  5. Hello Christopher, I'd like to ask whether this is for 3D or 2D? And where does the gravity term ( gravity x density) from the navier stokes equation disappear? This is really easy to understand, thank you!

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    Replies
    1. it's 2D and the body force (gravity) is neglected, such as in flow along straight pipe without any elevation change.

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  6. Thanks a lot this help for examination because complete calculation available but I follow viscous fluid book by white where we observe many difficulties about calculation

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  7. This is a good one Christopher. Pls how do we derive the fluctuating x and y momentum equation?

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  8. Why we eliminate pressure term???

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  9. Thanks Chris, amazing work. Only thing: I noticed a typo in equations 10 and 11, where your alpha_bar term in the second parenthesis goes from a power to 2 to a power of 1.

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  10. Hi Christopher, I'd like to mention one issue I have when skimming your derivation, I found in eq.(9), at he RHS inside the first parenthesis, the second term has the "alpha^2" as the factor, however, in eq.(10) I noticed it has changed to the "alpha". why?

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  11. Christopher,

    Great work. However, as others have mentioned, the power of non-dimensionalized numerical wavenumber should be 2 in equations 10 and 11.

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