Extended Navier–Stokes equations, ultrasonic absorption and shock structure

Abstract

The Navier–Stokes (N–S) equations for a steady or unsteady, compressible, continuum flow are modified. The extension is based on a Stokesian fluid with a single nonlinear term in an isotropic stress, rate-of-deformation relation. This is the simplest possible nonlinear extension that also satisfies the second law of thermodynamics. The transport coefficient of this term is referred to as the third viscosity coefficient. In the extended version, the momentum and energy equations each contain a nonlinear term that is proportional to this new coefficient. These terms are significant only when the velocity gradient is extremely large. They are inconsequential, e.g., in a laminar boundary layer. Nevertheless, there are flows where the extended version of the N–S equations is relevant. The first of these is an ultrasonic, unsteady, one-dimensional flow, which is used for evaluating the bulk viscosity. In this case, the linearized N–S equations become singular as the ultrasonic frequency increases toward infinity. When the frequency is sufficiently large, nonlinear terms in the extended N–S equations need to be retained. The terms that are proportional to the third viscosity coefficient increase in importance, relative to linear terms, as the fourth power of the frequency. A second example is shock wave structure. A model is established and numerically solved for the normalized density derivative. Results are compared with corresponding measurements for argon when the upstream Mach number is 1.058 and 1.23. Good agreement between the extended N–S predictions and measurements is obtained for both Mach numbers with a single, but extremely small, value for the third viscosity coefficient. An important difference between conventional and extended N–S shock structure solutions is that the extended-model solution depends on the upstream pressure, whereas the conventional solution does not.

Appendix A: shock wave structure formulation

The non-dimensional governing equations for a perfect gas can be written as:

$$\begin{aligned}&\rho u = M_1 \end{aligned}$$

(7.1)

$$\begin{aligned}&p = M_1 \frac{T}{u} \end{aligned}$$

(7.2)

$$\begin{aligned}&\rho u\frac{{\mathrm{d}}u}{\mathrm{d}x} \!=\! -\frac{1}{\gamma }\frac{{\mathrm{d}}p}{\mathrm{d}x}\!+\!\frac{1}{Re}\frac{\mathrm{d}}{\mathrm{d}x}\left( {\hat{{\mu }}\frac{{\mathrm{d}}u}{{\mathrm{d}}x}} \right) \!+\!E\frac{\mathrm{d}}{{\mathrm{d}}x}\left[ {\zeta \left( {\frac{{\mathrm{d}}u}{{\mathrm{d}}x}} \right) ^{3}} \right] \quad \end{aligned}$$

(7.3)

$$\begin{aligned}&\rho u\frac{{\mathrm{d}}T}{{\mathrm{d}}x} \!=\! \frac{\gamma \!-\!1}{\gamma }u\frac{{\mathrm{d}}p}{{\mathrm{d}}x}\!+\!\frac{\gamma \!-\!1}{Re}\hat{{\mu }}\left( {\frac{{\mathrm{d}}u}{{\mathrm{d}}x}} \right) ^{2}\!+\!\left( {\gamma \!-\!1} \right) E\zeta \left( {\frac{{\mathrm{d}}u}{{\mathrm{d}}x}} \right) ^{4} \nonumber \\&\qquad \qquad +\frac{1}{\left( {\frac{4}{3}+\alpha } \right) PrRe}\frac{\mathrm{d}}{{\mathrm{d}}x}\left( {\kappa \frac{{\mathrm{d}}T}{{\mathrm{d}}x}} \right) , \end{aligned}$$

(7.4)

where

$$\begin{aligned} \mu =\left( {\frac{4}{3}+\alpha } \right) T^{\omega },\quad \kappa =T^{\omega },\quad \quad \,\zeta =T^{\omega _{\zeta }} \end{aligned}$$

(7.5)

The upstream and downstream conditions are

$$\begin{aligned} u_1&= M_1,\quad T_1 =1 \end{aligned}$$

(7.6)

$$\begin{aligned} u_2&= \frac{2}{\gamma +1}\frac{1+\frac{\gamma -1}{2}M_1^2 }{M_1}, \nonumber \\ T_2&= \left( {\frac{2}{\gamma +1}}\right) ^{2}\frac{\left( {1+\frac{\gamma -1}{2}M_1^2 } \right) \left( {\gamma M_1^2 -\frac{\gamma -1}{2}} \right) }{M_1^2 } \end{aligned}$$

(7.7)

Equation (7.3) can be integrated once, with the result

$$\begin{aligned} p=1+\gamma M_1^2 -\gamma M_1 u+\frac{\gamma \hat{{\mu }}}{Re}\frac{{\mathrm{d}}u}{{\mathrm{d}}x}+\gamma E\zeta \left( {\frac{{\mathrm{d}}u}{{\mathrm{d}}x}} \right) ^{3}, \end{aligned}$$

(7.8)

where the constant of integration is evaluated at state 1. Eliminate $p$ in (7.4), and integrate once, to obtain

$$\begin{aligned} \frac{{\mathrm{d}}T}{{\mathrm{d}}x}&= \frac{\left( {\frac{4}{3}+\alpha } \right) PrRe}{\kappa }\left[ -M_1 \left( {1+\frac{\gamma -1}{2}M_1^2} \right) \right. \nonumber \\&\left. +\frac{\gamma -1}{\gamma }\left( {1+\gamma M_1^2 } \right) u-\frac{\gamma -1}{2}M_1 u^{2}+\frac{M_1 }{\gamma }T \right] \nonumber \\ \end{aligned}$$

(7.9)

This result is (6.5a) for both the N–S and the extended N–S equations. By setting all derivatives equal to zero, it can be shown that the above two equations are consistent with (7.6, 7.7).

By replacing $p$ in (7.8) with $M_{1}T/u$, a cubic equation for ${\mathrm{d}}u/{\mathrm{d}}x$ is obtained.

$$\begin{aligned} E\zeta \left( {\frac{{\mathrm{d}}u}{{\mathrm{d}}x}} \right) ^{3}\!+\!\frac{\hat{{\mu }}}{Re}\frac{{\mathrm{d}}u}{{\mathrm{d}}x}\!+\!\left[ {\frac{1+\gamma M_1^2 }{\gamma }-M_1 \left( {u+\frac{T}{\gamma u}} \right) } \right] =0\nonumber \\ \end{aligned}$$

(7.10)

The conventional N–S result is obtained by setting $E$ equal to zero; the result has the form of (6.5b). An exact solution of the above cubic equation can be written as

$$\begin{aligned}&a = \frac{\hat{{\mu }}}{E\zeta Re},\quad b=\frac{1}{E\zeta }\left[ {\frac{1+\gamma M_1^2 }{\gamma }-M_1 \left( {u+\frac{T}{\gamma u}} \right) } \right] \nonumber \\ \end{aligned}$$

(7.11)

$$\begin{aligned}&A = \left[ {-\frac{b}{2}+\left( {\frac{b^{2}}{4}+\frac{a^{3}}{27}} \right) ^{1/2}} \right] ^{1/3},\nonumber \\&B=\left[ {\frac{b}{2}+\left( {\frac{b^{2}}{4}+\frac{a^{3}}{27}} \right) ^{1/2}} \right] ^{1/3} \end{aligned}$$

(7.12)

$$\begin{aligned}&\frac{{\mathrm{d}}u}{{\mathrm{d}}x} = A-B \end{aligned}$$

(7.13)

which also has the form of (6.5b).

At the upstream state, set

$$\begin{aligned} u=M_1 -\epsilon _{u},\quad T=1+\epsilon _T \end{aligned}$$

(7.14)

To first order, $a$ and $b$ are

$$\begin{aligned} a=\frac{1}{ERe},\quad b=\frac{1}{\gamma E}\left( {\frac{\gamma M_1^2 -1}{M_1 }\epsilon _u -\epsilon _T } \right) \end{aligned}$$

(7.15)

and $A$ and $B$ become

$$\begin{aligned} A&= \left( {\frac{a}{3}} \right) ^{1/2}-\frac{1}{\gamma Ea}\left( {\frac{\gamma M_1^2 -1}{M_1 }\epsilon _u -\epsilon _T } \right) \end{aligned}$$

(7.16)

$$\begin{aligned} B&= \left( {\frac{a}{3}} \right) ^{1/2}+\frac{1}{\gamma Ea}\left( {\frac{\gamma M_1^2 -1}{M_1 }\epsilon _{u} -\epsilon _{T}} \right) \end{aligned}$$

(7.17)

The final result is

$$\begin{aligned} \frac{\mathrm{d}\epsilon _u }{{\mathrm{d}}x}&= \frac{2Re}{\gamma }\left( {\frac{\gamma M_1^2 -1}{M_1 }\epsilon _u -\epsilon _T } \right) \end{aligned}$$

(7.18)

$$\begin{aligned} \frac{\mathrm{d}\epsilon _T }{{\mathrm{d}}x}&= \frac{\left( {\frac{4}{3}+\alpha } \right) PrRe}{\gamma }\left[ {M_1 \epsilon _T -\left( {\gamma -1} \right) \epsilon _{u}} \right] \end{aligned}$$

(7.19)

By taking the ratio of these two quantities, a phase-plane analysis [23] yields a nondegenerate nodal point.

For the downstream singularity, set

$$\begin{aligned} u=u_2 +\epsilon _u,\quad T=T_2 -\epsilon _T \end{aligned}$$

(7.20)

To first order, $a$ and $b$ are

$$\begin{aligned} a=\frac{T_2^{\omega -\omega _{\zeta }} }{ERe},\quad b=\frac{M_1}{\gamma Eu_2^2 T_2^{\omega _{\zeta }} }\left[ {\left( {T_2 -\gamma u_2^2} \right) \epsilon _u +u_2 \epsilon _T } \right] \nonumber \\ \end{aligned}$$

(7.21)

This yields

$$\begin{aligned} \frac{\mathrm{d}\epsilon _u }{{\mathrm{d}}x}&= \frac{M_1 Re}{\gamma u_2^2 T_2^\omega }\left[ {\left( {T_2 -\gamma u_2^2 } \right) \epsilon _u +u_2 \epsilon _T} \right] \end{aligned}$$

(7.22a)

$$\begin{aligned} \frac{\mathrm{d}\epsilon _T }{{\mathrm{d}}x}&= \frac{\gamma -1}{\gamma T_2^\omega }\left( {\frac{4}{3}+\alpha } \right) PrRe\left\{ \left[ {\left( {1+M_1^2 } \right) -\gamma M_1 u_2} \right] \epsilon _u \right. \nonumber \\&\left. -\frac{M_1}{\gamma -1}\epsilon _{T} \right\} \end{aligned}$$

(7.22b)

The $\epsilon \hbox {s}$ are linearly transformed by [23]

$$\begin{aligned} \epsilon _u&= \left( {\frac{\lambda _+ -d}{c}} \right) \hat{{\epsilon }}_{_u } +\left( {\frac{\lambda _- -d}{c}} \right) \hat{{\epsilon }}_T \end{aligned}$$

(7.23a)

$$\begin{aligned} \epsilon _T&= \hat{{\epsilon }}_u +\hat{{\epsilon }}_T \end{aligned}$$

(7.23b)

with the result

$$\begin{aligned} \frac{\mathrm{d}\hat{{\epsilon }}_u }{{\mathrm{d}}x}=\lambda _+ \hat{{\epsilon }}_u,\quad \frac{\mathrm{d}\hat{{\epsilon }}_T }{{\mathrm{d}}x}=\lambda _- \hat{{\epsilon }}_T \end{aligned}$$

(7.24)

The $\lambda _\pm$ are the eigenvalues, and are given by

$$\begin{aligned} a&= M_1 Re\left( {\frac{T_2 }{\mu u_2^2 }-1} \right) \end{aligned}$$

(7.25a)

$$\begin{aligned} b&= \frac{M_1 Re}{\gamma u_2 } \end{aligned}$$

(7.25b)

$$\begin{aligned} c&= \frac{2}{\gamma }\left( {\frac{\gamma -1}{\gamma +1}} \right) \left( {\gamma M_1^2 -\frac{\gamma -1}{2}} \right) \left( {\frac{4}{3}+\alpha } \right) PrRe \nonumber \\\end{aligned}$$

(7.25c)

$$\begin{aligned} d&= -\frac{M_1 }{\gamma }\left( {\frac{4}{3}+\alpha } \right) PrRe \end{aligned}$$

(7.25d)

$$\begin{aligned} \lambda _\pm =\frac{1}{2}\left\{ {a+d\pm \left[ {\left( {a-d} \right) ^{2}+4bc} \right] ^{1/2}} \right\} \end{aligned}$$

(7.26)

With the foregoing relations, initial conditions at the downstream state are

$$\begin{aligned} u\left( 0 \right) =u_2 +\epsilon _u,\quad T\left( 0 \right) =T_2 -\frac{c}{\lambda _+ -d}\epsilon _u, \end{aligned}$$

(7.27)

where $\epsilon _u$ is set equal to $10^{-4}$. The integration proceeds smoothly from $\bar{{x}} = 0$, where $\bar{{x}} = -x$.