数理解析研究所講究録
1075 巻 1999 年 37-52

37

蒸発凝縮を伴う希薄気体の Couette 流: 流れの逆転と分岐

Cylindrical Couette flows of a rarefied gas with evaporation andcondensation: Reversal and bifurcation of flows
京大工航空宇宙

曾根良夫, 杉元宏, 青木 –生

Yoshio Sone, Hiroshi Sugimot , and Kazuo Aoki$\mathit{0}$Department of Aeronautics and Astronautics, Graduate School of
Kyoto 606-8501, Japan$En_{\mathit{9}^{in}\mathit{9}}nee\dot{n}$, Kyoto University,

A rarefied gas between two coaxial circular cylinders made of the condensed phase of the gas is considered, where eachcylinder is kept at a uniform temperature and is rotating at a constant angular velocity around its axis (cylindrical
Couette flows of a rarefied gas with evaporation or condensation on the cylinders). The steady behavior of the gas,with special interest in bifurcation of a flow, is studied on the basis of kinetic theory from the continuum to the
Knudsen limit. The solution shows profound variety: reversal of direction of evaporation-condensation with variationof the speed of rotation of the cylinders; contrary to the conventional cylindrical Couette flow without evaporationand condensation, bifurcation of a flow in a simple case where the state of the gas is circumferentially and axially
uniform.

1

Introduction, respectively; let the corresponding quantities of theouter cylinder be, and. The saturation gas
pressure at temperature $T$ (or) is denoted by(or). The Knudsen number Kn of the system is, where is the mean free path of thedefined bygas molecules in the equilibrium state at rest with pressure Ps 1 and temperature . We will investigate thesteady behavior of the gas under the following assumptions: i) The behavior of the gas is described by the
Boltzmann-Krook-Welander equation (BKW or BGK
Notes on the generalization of the result for the standard Boltzmann equation will be given
in several places. ii) The velocity distribution of thegas molecules leaving the inner (or outer) cylinder isthe corresponding part of the Maxwellian distribution
with pressure(or), circumferential velocity(or), and temperature(or). (This is called
the complete condensation condition on the interface
of a gas and its condensed phase.18-20) iii) The behavior of the gas is axially and circumferentially uniform.
Then the axial velocity vanishes.
The Boltzmann-Krook-Welander equation for a steadyflow with axial and circumferential uniformity is givenin the cylindrical coordinate system $(r, \theta, z)$ , with theaxis of the cylinder as the axis (Fig. 1), as follows:$V_{\theta 1}$

Bifurcation of flows attracts many scientists and engineers, and a lot of works have been done in the framework of classical gas dynamics, especially in relation toturbulence.1-9 B\’enard and-Couette flows are its
most well-known examples. The study of these flows fora rarefied gas is also being done in recent years.
In this paper we will present a new example of bifurcation of a simple flow, which is found in a rarefied gas
as well as in a gas in the continuum limit. That is, weconsider a gas between two coaxial circular cylinders
made of the condensed phase of the gas, on the surfaceof which evaporation or condensation thus may takeplace. Each cylinder is kept at a uniform temperatureand is rotating at a constant angular velocity around
its axis. We investigate the behavior of the gas (cylindrical Couette flows of a rarefied gas with evaporationor condensation on the cylinders) under the condition
that the state of the gas is axially and circumferenuniform on the basis of kinetic theory from thecontinuum to the Knudsen limit. The centrifugal force
induced in the gas by rotation of the cylinders affects
evaporation and condensation on the cylinders, and thisintroduces profound variety of the behavior of the gas.
In the following analysis, we find reversal of direction of-condensation with variation of the sPeed ofrotation of the cylinders and bifurcation of a flow in thesimple case where the state of the gas is circumferentially and axially uniform, contrary to the conventional
cylindrical Couette flow without evaporation and condensation.
$r_{\mathrm{R}\mathrm{y}\mathrm{l}\mathrm{o}\mathrm{r}}$$10^{-}14$$\mathrm{t}\mathrm{i}\mathrm{a}\mathrm{U}\mathrm{y}$$L_{2},$$T_{2}$$V_{\theta 2}$$T_{1},$$p_{s1},$$T_{2}$$p_{s}$$p_{2}\epsilon$$l_{1}/L_{1}$$l_{1}$$T_{1}$$\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{i}_{0}\mathrm{n})^{1}.5-18$$p_{s1}$$V_{\theta 1}$$p_{s2}$$T_{1}$$V_{\theta 2}$$T_{2}$$z$$\mathrm{e}\mathrm{v}\mathrm{a}\mathrm{p}_{0}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$,$\xi_{r}\frac{\partial f}{\partial r}+\frac{\xi_{\theta}^{2}}{r}\frac{\partial f}{\partial\xi_{r}}-\frac{\xi_{r}\xi_{\theta}}{r}\frac{\partial f}{\partial\xi_{\theta}}=A_{Col}\rho(fe-f)$(1)$f_{\mathrm{e}}= \frac{\rho}{(2\pi RT)^{3/2}}\exp(-\frac{(\xi_{r}-u_{r})^{2}+(\xi_{\theta}-u_{\theta})2\xi_{z}+2}{2RT})(2)$ ’

2Problem and basic equation
Consider a rarefied gas between two coaxial circularcylinders made of the condensed phase of the gas, where
each cylinder is kept at a uniform temperature and isrotating at a constant angular velocity around its axis.
Let the radius, temperature, and circumferential velocity of the surface of the inner cylinder be, and$L_{1},$$T_{1}$,(3a)$\rho$$=$$\int\int\int fd\xi_{r}d\xi_{\theta}d\xi_{z}$$\rho u_{r}$$=$$\int\int\int\xi_{r}fd\xi_{r}d\xi_{\theta}d\xi_{z}$

,

(3b)

$\rho u_{\theta}$$=$$\int\int\int\xi_{\theta}fd\xi rd\xi_{\theta}d\xi_{z}$

,

(3c)

38

$3R\rho T$$=$$\int\int\int[(\xi_{r}-ur)^{2}+(\xi\theta-u_{\theta})2\xi^{2}+z]f$

$p$

$=$

$R\rho T$’$d\xi_{r}d\xi_{\theta}d.\xi z$

,

(3d)
(3e)

where
, and are, respectively, the radial $(r)$ , az, and axial $(z)$ components of the molecularimuthalvelocity;is the velocity distribution function of thegas molecules, which is a function of, and ;
is the density of the gas;andare, respectively,
the radial and circumferential components of the flow
velocity, which are its nonvanishing components;
isthe temperature of the gas; is the pressure of the gas;
is a constant; $R$ is the specific gas constant (orthe Boltzmann constant divided by the mass of a gas
molecule). The integrations in Eqs.
, and inwhat follows unless otherwise stated, are carried out
over the range $(-\infty<\xi_{r}<\infty,$ $-\infty<\xi_{\theta}<\infty,$ $-\infty<$. The
is the collision frequency of a gas
molecule, which is independent of molecular velocity
for the BKW equation.
The complete condensation boundary conditions on
the cylinders are given as follows: on the inner cylinder$\xi_{r},$$\xi_{\theta}$$\xi_{z}$$(\theta)$

$f$

$r,$

$\rho$$u_{r}$$\xi_{r},$$\xi_{\theta}$$\xi_{z}$$u_{\theta}$

$T$

$p$

$A_{co\mathrm{t}}$$(3\mathrm{a})-(3\mathrm{d})$$\xi_{z}<\infty)$$(r=L_{1})$$A_{\mathrm{c}\circ l\beta}$,

Fig. 1. Coordinate systems.

where$D= \zeta\cos\theta_{\zeta}\frac{\partial}{\partial r\wedge}-\frac{\zeta\sin\theta_{\zeta}}{r\wedge}\frac{\partial}{\partial\theta_{\zeta}}$$f= \frac{\beta s1}{(2\pi R\tau_{1})^{3/}2}.\exp(-\frac{\xi_{f}^{2}+(\xi\theta-V_{\theta 1})^{2}+\xi_{z}^{2}}{2RT_{1}})$$(\xi_{r}>0)$,(4)$\ =\frac{l_{1}}{L_{1}}$,$l_{1}= \frac{(8R\tau_{1}/\pi)1/2}{A_{c\circ\iota}\rho S1}$

,

(10a)

,

(10b)

where
$\rho_{s1}=p_{s1}/RT_{1}$,(5)and on the outer cylinder $(r=L_{2})$ ,$= \frac{\rho\wedge}{\pi}$xexp$f= \frac{\rho_{s2}}{(2\pi RT_{2})^{3/}2}\exp(-\frac{\xi_{f}2+(\xi\theta-V_{\theta 2})^{2}+\xi z2}{2RT_{2}})$$(\xi_{r}<0)$

,

(6)

where

.
(7)
For the convenience of analysis, we introduce the following nondimensional variables (Fig. 1) and marginal
velocity distribution functions:$\rho_{s2}=_{Ps2}/RT_{2}$$(- \frac{\zeta^{2}+u_{r}^{2}+u\theta\wedge\wedge 2-2ur\zeta\wedge\theta\zeta-2u\theta\zeta\cos \mathrm{s}\mathrm{i}\wedge \mathrm{n}\theta_{\zeta}}{\hat{T}}).(10\mathrm{c})$The is the mean fiiee path of the gas molecules in theequilibrium state at rest with pressure Ps1 and temperatureand Kn is the Knudsen number at the state.
By the introduction of the marginal velocity distributions and , the variable
is eliminated, and in the
new independent variables, the equations forand contain only the derivatives with respect toand .
The macroscopic variables, and are expressed by and as$l_{1}$$T_{1_{)}}$

,

(8a)

$\zeta_{z}$$(r\wedge, \zeta, \theta_{\zeta})$

$h$

$g$

$r=L_{1^{\wedge}}r$

$h$

$g$

$r\wedge$$\theta_{\zeta}$$\rho\wedge,$$\xi_{r}=(2R\tau_{1})^{1}/2\zeta\cos\theta\zeta$,$\xi_{\theta}=(2R\tau_{1})^{1}/2\zeta\sin\theta_{\zeta}$,$\xi_{z}=(2RT1)^{1/2}\zeta z$(8b)$\rho\wedge=$$u_{r}\wedge$’$u_{\theta}\wedge$,$\rho=\rho_{s1\rho}\wedge$$\hat{T}$$u_{\theta}\wedge$$\int\int\zeta gd\zeta d\theta_{\zeta}$

,

(lla)

$=$

$\frac{1}{\rho\wedge}\int\int\zeta^{2}\cos\theta\zeta gd\zeta d\theta\zeta$$=$$\frac{1}{\rho\wedge}\int\int\zeta^{2}\sin\theta_{\zeta}gd\zeta d\theta_{\zeta}$

,

(llb)

,

(11C)

$u_{r}=(2R\tau_{1})^{1/}2u_{r_{)}}\wedge$(8c)$u_{\theta}=(2RT_{1})1/2\wedge u_{\theta}$$u_{r}\wedge,$

$h$

$g$

,

,

$p=ps1p\wedge$$T=T_{1}\hat{T}$$\frac{3}{2}\hat{T}=$,$\frac{1}{\rho\wedge}\int\int\zeta(\zeta^{2}g+h)d\zeta d\theta\zeta-u-u_{\theta}^{2}.(\wedge 11\mathrm{d}\wedge r2)$

The two-fold integrations with respect to andin Eqs.
, and in what follows unless otherwise stated, are carried out over the domain $(0\leq\zeta<$$\zeta$$= \frac{2RT_{1}}{\rho_{s1}}\int_{-\infty}^{\infty}fd\xi_{z}$

.

(8d)

With these new variables, the BKW equation (1) is
reduced to$D= \frac{2}{\sqrt{\pi}\mathrm{R}}\rho\wedge(-)$,(9)$(1\mathrm{l}\mathrm{a})-(1\mathrm{l}\mathrm{d})$$\infty,$$-\pi<\theta_{\zeta}\leq\pi)$.

The boundary conditions are, at$r\wedge=1$,.

$\theta_{\zeta}$

39$= \frac{1}{\pi}\exp(-\zeta^{2}-\frac{V_{\theta 1}^{2}}{2RT_{1}}+\frac{2V_{\theta 1}\zeta\sin\theta_{\zeta}}{(2R\tau_{1})^{1/2}})$and at(12):$(|\theta_{\zeta}|<\pi/2),$,$\wedge r=L_{2}/L_{1}$$= \frac{1}{\pi}\frac{p_{s2}T_{1}}{p_{s1}T_{2}}$$\cross\exp(-\frac{T_{1}}{T_{2}}(\zeta^{2}+\frac{V_{\theta 2}^{2}}{2RT_{1}}-\frac{2V_{\theta 2}\zeta\sin\theta_{\zeta}}{(2R\tau_{1})^{1/2}}))$$(\pi/2<|\theta_{\zeta}|\leq\pi)$

.

(13)

The problem is determined by the six parameters
Kn,
, Ps
, and
. We will analyze this boundary-value problem
mainly numerically for various sets of values of theseparameters. The saturation gas pressure
is generally
a rapidly increasing function of temperature . For
many gases, $p_{s2}/p_{s1}=1.2$ , for example, corresponds to$1<T_{2}/T_{1}<1.02$ ; thus, it is sufficient to take $T_{2}/T_{1}=$in illustrative examples.
$L_{2}/L_{1_{)}}V_{\theta 1}/(2R\tau_{1})^{1}/2,$ $V_{\theta 2}/(2R\tau_{1})^{1/2}$$2/p_{s1}$$T_{2}/T_{1}$$p_{s}$

$T$

$1$

3

Outline of method of numerical solution

The numerical analysis of the boundary-value problem (9), (12), and (13) will be carried out by a finitedifference method. We studied a cylindrical problem in
Refs. 21 and 22, where steady evaporation of a rarefied
gas from a stationary cylindrical condensed phase into
an infinite expanse of the gas with various pressures orinto a vacuum was considered. We can make use of thefinite-difference method developed in Ref. 21. Thus, weonly outline the method here.
(i) We consider the problem in a finite domain $(1\leq$, $0\leq\zeta\leq\zeta_{D}$ ,,inspace, whereis chosen properly depending on
the situations. IYom our numerical tests, and areseen to decay rapidly with , and therefore accuratecomputation of the problem can be carried out with. The discrete solution
reasonable size ofof $(g, h)$ at the lattice points inspace is constructed as the limit of the sequenceobtained as follows. The initial solutionof the$\wedge r\leq L_{2}/L_{1}$$-\pi\leq\theta_{\zeta}\leq\pi)$$\theta_{(})$$(r\wedge,$

Fig. 2. Diagram of the process of computation.

$\zeta$$\zeta_{D}$

$h$

$g$

$\zeta$$(\mathit{9}\#, h_{\#})$$\zeta_{D}$$(\hat{r}, \zeta, \theta_{\zeta})$$(g_{\#}^{(n)}, h_{\#}^{(n)})$$(g_{\#’\#}^{(0)}h^{(0}))$$\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{a}(g_{\#’\#}^{(n})\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}h^{(}n\mathrm{P})\mathrm{b}\mathrm{e}\mathrm{k}\mathrm{n}\mathrm{o}\mathrm{w}\mathrm{n}.\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{s}\mathrm{u}\mathrm{t}\mathrm{r}\mathrm{o}\mathrm{c}\mathrm{e}\mathrm{S}\mathrm{s}\mathrm{i}_{\mathrm{S}\mathrm{c}}\mathrm{h}\mathrm{o}\mathrm{S}\mathrm{e}\mathrm{n}\mathrm{p}_{\mathrm{O}}\mathrm{r}_{1\mathrm{i}}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}1\mathrm{y}.\mathrm{L}\mathrm{e}\mathrm{t}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{S}\mathrm{o}1\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}(g\#’ h^{()}(n+1)\#\mathcal{R}+1)\mathrm{i}\mathrm{n}\mathrm{o}\mathrm{n}$and $(\pi/2<\theta_{\zeta}\leq\pi)$ is computed
successively from(the solution intodomains
and I2 in Fig. 2) and then $(g_{\#}^{(n+1)}, h(n+1))\#$
in $(-\pi/2<\theta_{\zeta}<\pi/2)$ fromto $r\wedge=L_{2}/L_{1}$ (thesolution in domains
and II2 in Fig. 2) with the aidof a finite-difference equation for Eq. (9). In the com(or I2),
putation ofin the domain),(orthe computation along the line
which can be done independently of the surrounding, is done first, and then thedata because of(or I2) is done with the aid of thecomputation in(ordata on $r\wedge=L_{2}/L_{1}$ andand. The computation ofin the(or) is done with the aid of the data on
domain(or $r\wedge=1$ and). Theand$(-\pi\leq\theta_{\zeta}<-\pi/2)$$\wedge r=L_{2}/L_{1}$$\wedge r=1$$\mathrm{I}_{1}$$\wedge r=1$

, which is the common boundary ofdata on, are computed independently of the surroundanding data because of.
(ii) On the surface of the cylinders (),
andthe marginal velocity distribution functions and are(or $\xi_{r}=0$ ), because thediscontinuous atnature of the velocity distribution function of the incoming molecules and that of the outgoing molecules
are different. The discontinuity propagates into the gas
along the characteristic of Eq. (9) ffom the inner cylinder, but not from the outer cylinder, since the characteristic does not enter the gas
there.23 Therefore,
the discontinuity of $(g, h)$ lies on the surface$\theta_{\zeta}=0$$\mathrm{I}\mathrm{I}_{1}$$\mathrm{I}\mathrm{I}_{2}$$\sin\theta_{\zeta}=0$$r\wedge=1$$L_{2}/L_{1}$

$g$

$h$

$\theta_{\zeta}=\pm\pi/2$$\mathrm{h}\mathrm{o}\mathrm{m}$$\mathrm{I}\mathrm{I}_{1}$$(g_{\#}^{(n+1)}, h_{*}^{(n+)})1$$\mathrm{I}_{1}$$\theta_{\zeta}=\pi$$\theta_{\zeta}=-\pi$$\sin\theta_{\zeta}=0$$\mathrm{I}_{1}$$\theta_{\zeta}=\pi$$\theta_{\zeta}=-\pi)$$\mathrm{I}\mathrm{I}_{1}$$\wedge r=1$$\wedge r=L_{2}/L_{1}$$(g_{\#}^{(1}n+),$$h_{\#}^{(n+1)})$$\mathrm{I}\mathrm{I}_{2}$$\theta_{\zeta}=\pi/2$$\theta_{\zeta}=-\pi/2$$\wedge r\sin\theta_{\zeta}=1$$(-\pi/2\leq\theta_{\zeta}\leq\pi/2)$

.

(14)

The position of the discontinuity is independent of themolecular speed . As the distance increases, thediscontinuity decays owing to molecular collisions. For
small Knudsen numbers, the discontinuity from the inner cylinder is practically negligible on the outer cylinbut for intermediate or large Knudsen numbers, itremains appreciable there.
When we discretize Eq. (9), which has the derivatives, we should not apply finite-difference
and$\zeta$$\mathrm{d}\mathrm{e}\mathrm{r}_{\rangle}$$\partial/\partial r\wedge$$\partial/\partial\theta_{\zeta}$$\wedge r$

40
formulas for differentiation that contain the data acrossthe discontinuity. Therefore, we divide the domain)(or ,) into two regions
(or
and
and
(Fig. 2) by discontinuity surface (14) and apply standard finite-difference approximations in each region. In
on thethis scheme, the limiting values of
discontinuity surface from both sides are required as
the boundary condition. They are obtained separately
by integration of Eq. (9) along the characteristic (14).
In the region where the discontinuity has decayed sufficiently, we use a standard finite-difference scheme without dividing the domain for efficiency.
$\mathrm{I}\mathrm{I}_{1}$$\mathrm{I}\mathrm{I}_{1b}$$\mathrm{I}\mathrm{I}_{1a}$$\mathrm{I}\mathrm{I}_{2}$$\mathrm{I}\mathrm{I}_{2b}$$\mathrm{I}\mathrm{I}_{2a}$$(g_{\#}^{(n)}, h_{\#}^{(n)})$Free molecular and continuum solutions
In this section we explain the general behavior of thesolutions of two extreme cases: the free molecular so-

The mass flow rate is independent of the circumferof the inner cylinder, and increases
ential velocity
with the speedof the outer cylinder.
4.2 Continuum solution) can beThe solution of the continuum limit (Ikn
obtained by the asymptotic theory developed in Refs.
24 and 25. When evaporation or condensation is taking place on the cylinders, the behavior of the gas is described by the following system of equations and boundary conditions. The equations are the Euler set of equations:$J$$V_{\theta 1}$$|V_{\theta 2}|$$arrow 0$$\frac{d}{dr}(\rho u_{r}r)=0$

,

(18)

4

$u_{r} \frac{du_{\theta}r}{dr}=0$lution and the continuum solution.
4.1 Free molecular solution
The free molecular solution is easily obtained by theas follows:
standard$p=R_{\beta}T$

In terms of the Mach number
radial velocity component :

(19)

,

(20)

$u_{r} \frac{d}{dr}(u_{r}^{2}+u_{\theta}^{2}+5RT)=0$$\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{d}^{18}’ 19$$f= \frac{\rho_{s1}}{(2\pi RT_{1})^{3/2}}\exp(\frac{V_{\theta 1}^{2}}{2RT_{1}}[(\frac{r}{L_{1}})^{2}-1])$,$\rho(u_{r}\frac{du_{r}}{dr}-\frac{u_{\theta}^{2}}{r})+\frac{dp}{dr}=0$

,

(21)

.

(22)
$M_{r}$, with sign, of the$u_{r}$$\mathrm{x}\exp(-\frac{\xi_{r}^{2}+(\xi_{\theta}-V\theta 1r/L1)2+\xi_{z}^{2}}{2RT_{1}}.)$

and,$M_{r}=u_{r}/\sqrt{5RT}/3$$[\xi_{r}>0$$-L_{1}/r<\xi_{\theta}/(\xi_{r}^{2}+\xi_{\theta}^{2})^{1/2}<L_{1}/r],$$(15\mathrm{a})$(23)

the boundary conditions on the cylinders are given as
follows: at $r=L_{1}$$u_{\theta}=V_{\theta 1},$$= \frac{\rho_{s2}}{(2\pi RT_{2})^{3/2}}\exp(-\frac{V_{\theta 2}^{2}}{2RT_{2}}[1-(\frac{r}{L_{2}})^{2}])$,$p=p_{s1}h_{1}(M_{r}),$$T=T_{1}h_{2}(M_{r})$$(0<M_{r}\leq 1)$$\cross\exp(-\frac{\xi_{r}^{2}+(\xi_{\theta}-V\theta 2r/L_{2})^{2}+\xi_{z}^{2}}{2RT_{2}})$[for the other values of$\xi_{r}$and$\xi_{\theta}$

,

,

(24a)

$p=p_{s1}F_{S}(-M_{r)}|(u_{\theta}-V_{\theta 1})/\sqrt{5RT}/3|, T/T_{1})$$(-1<M_{r}<0)$ ,

(15b)

].

,

(24b)

$p>p_{s1}F_{b(}-Mr’|(u_{\theta}-V_{\theta 1})/\sqrt{5RT}/3|,$ $T/T_{1})$The contributions of the two cylinders are independent.
The macroscopic variables, which are obtained by moments of the distribution function, are expressed by thesum of two contributions.
and the energy flow $W$ from the
The mass flow
inner cylinder to the outer per unit time and unit lengthof the cylinders are
$J$

$J$

$2\pi L_{1}(m_{I}-m_{\circ)}$$=$$(M_{r}\leq-1)$,

and at $r=L_{2}$$u_{\theta 2}=V_{\theta 2},$$p=p_{s2}h_{1}(-M_{r}),$$T=T_{2}h_{2}(-M_{r})$$(-1\leq M_{r}<0)$ ,$p=p_{s2}F_{S}(M_{r}, |(u_{\theta}-V_{\theta 2})/\sqrt{5RT}/3|,T/T_{2})$

,

(16)

$=$

$\frac{p_{s1}}{(2\pi R\tau_{1})^{1/2}}$$m_{O}$$=$$\frac{p_{s2}}{(2\pi R\tau_{2})^{1/2}}\exp(-\frac{V_{\theta 2}^{2}}{2RT_{2}}[1-(\frac{L_{1}}{L_{2}})^{2}])$,$(M_{r}\geq 1)$

,

,

$W$

$=$

$2\pi L_{1}(_{W_{I}}-wo)$$w_{I}$$=$$p_{s1}( \frac{RT_{1}}{2\pi})^{1/2}(2+\frac{V_{\theta 1}^{2}}{2RT_{1}})$$w_{O}$

$=$

(

Ps 2

,

and , obtained from the sowhere the functions
lution of the half-space problem of evaporation, areand ,listed in Refs. 24 and 26, and the function
obtained from the half-space problem of condensation
with a parallel flow, are shown in Ref. 27. In the flow
a discontinuity (or a shock wave) is allowed, where the
Rankine-Hugoniot relation should be satisfied.28
The general solution of the set of equations (18)(22) (an isentropic circulating source or sink flow)29,30
is expressed by the following parametric representationin terms of the Mach number $M$$h_{1}$$h_{2}$$\frac{RT_{2}}{2\pi})^{1/2}[2+\frac{V_{\theta 2}^{2}}{2RT_{2}}(\frac{L_{1}}{L_{2}})^{2}]$,$M=\sqrt{\frac{3(u_{r}^{2}+u_{\theta})2}{5RT}}$andare the contributions of the inner
where
and
are those of the outer cylinder.
cylinder and$m_{I}$$w_{I}$$m_{O}$$w_{O}$,$F_{b}$-(17)$\cross\exp(-\frac{V_{\theta 2}^{2}}{2RT_{2}}[1-(\frac{L_{1}}{L_{2}})^{2}])$

,

(25c)

$F_{s}$

and

,

(25a)

(25b)

$p>p_{s2}F_{b}(M_{r}, |(u_{\theta}-V_{\theta 2})/\sqrt{5RT}/3|,T/T_{2})$$m_{I}$,$(24_{\mathrm{C}})$$(0<M_{r}<1)$ ,

,

,

.

(26)

41
That is,it follows that $p(r=L_{2})<p_{s2}$ and $p(r=L_{2})>Ps1$ ;the two relations contradict when $Ps1>p_{s2}$ .
, the situation
On the other hand, whenis complicate: evaporation on either of the cylinders is
possible and bifurcation of flow is expected. For example, when both the cylinders are at rest, the gas evaporates obviously on the outer cylinder and proceeds radially to the inner cylinder without circulating motion]. If the inner cylinder[a flow of type (i) with
begins to rotate from this state, the flow remains radialfrom Eq. (20) or (27d) and the boundary condition (25a) on the outer cylinder, which is at rest, but
the mass flow rate decreases. The boundary condition(24b) on the inner cylinder, where condensation is tak$p_{s1}<p_{\epsilon 2}$$\frac{p}{\rho*}=(\frac{4}{3+M^{2}})^{3/2}$$\frac{\rho}{\rho*}=(\frac{4}{3+M^{2}})^{3/2},$$\frac{p}{p_{*}}=(\frac{4}{3+M^{2}})^{\epsilon}/2(27\mathrm{a})$$\frac{T}{T_{*}}=\frac{4}{3+M^{2}},$$\rho*=\frac{p*}{RT_{*}}$,$u_{\theta}=0$$M_{r}=M( \frac{3+M^{2}}{4})^{3/2}$$(27\mathrm{b})$$\cross[\sin^{2}\alpha_{*}+(\frac{3+M^{2}}{4})^{3}\cos^{2]\mathrm{c}}\alpha_{*}-1/2\mathrm{o}\mathrm{s}\alpha_{*}$$(u_{\theta}=0)$.

$\frac{r}{r_{*}}=\frac{1}{M}(\frac{3+M^{2}}{4})^{1/2}$(27C)$\cross[\sin^{2}\alpha_{*}+(\frac{3+M^{2}}{4})^{3}\cos^{2}\alpha_{*}]1/2$$u_{r}=( \frac{\rho_{*}r_{*}}{\rho r})(\frac{5RT}{3}*)^{1/2}\cos\alpha_{*}$

,
(27d)

$u_{\theta}=( \frac{r}{r}*)(\frac{5RT}{3}*)^{1/2}\sin\alpha_{*}$

,

, and
arbitrary constants. The
where
are, respectively, the, andconstants
density, the pressure, the temperature, the angle of deflection of the flow velocity from the radial direction,and the radial coordinate at the hypothetical point with
$M=1$ (sonic point). In Fig. 3, the relations
vs$M$ andare shown for various . The
takes the minimum value at $|M_{r}|=1$ . Thus the solution is classified into four types: $M_{r}\leq-1$ , $-1\leq$$M_{r}<0,0<M_{r}\leq 1$ , and $M_{r}\geq 1$ . The
naturallyvaries along the flow but never exceeds the above limits
in the continuous flow. A shock wave (discontinuity) isallowed in a flow with $|M_{r}|>1$ , where the flow can be
decelerated ffom $|M_{r}|>1$ to $|M_{r}|<1$ .
With the aid of the general solution, let us overview
the feature of the evaporation-condensation problem.
, the flow cannot
andFrom the relation between
be accelerated or decelerated across $|M_{r}|=1$ (the sonic), and from Eqs. (24a) andspeed with respect tois(25a) the evaporating sPeed on the condensed
limited by $|M_{r}|\leq 1$ . Thus, only three types of flow are
possible in the present evaporation-condensation problem: (i) When evaporation is taking place on the utercylinder $(M_{r}<0)$ , the flow is subsonic with respecttoand accelerating (circulating sink flow). Whenthe evaporation is taking place on the inner cylinder$(M_{r}>0),$
the flow is subsonic with respect toand decelerating (circulating source flow) or (iii) theflow is an accelerating flow evaporating ffom the innercylinder with $M_{r}=1$ and may be accompanied by ashock wave in the middle region.
of the inner
When the saturation gas pressureof the outer cylinder
cylinder is larger than that
(ps $1>p_{s2}$ ), evaporation on the outer cylinder [or a flow
is impossible. In fact, in the flow of typeof type(i) the pressure decreases along the flow $[p(r=L_{1})<$$p(r=L_{2})]$ , but from the boundary conditions (24b) and(25a) with the inequalities $h_{1}(|M_{r}|)\leq 1$ and $F_{s}\geq 1$ ,$p_{*},$ $T_{*},$$r_{*}$$\rho_{*},$$p_{*},$ $T_{*},$$\alpha_{*}\mathrm{a}r\mathrm{e}$$r_{*}$$\alpha_{*}$$|M_{r}|$$r/r_{*}\mathrm{v}\mathrm{s}M_{r}$$\alpha_{*}$$r/r_{*}$$M_{r}$$M_{r}$$r/r_{*}$$M_{r}$$\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{S}\mathrm{e}$$\mathit{0}$$M_{r}$$M_{r}$$(\mathrm{i}\mathrm{i})$$p_{s1}$$p_{s2}$$(\mathrm{i})]$vs $M$ ;Fig. 3. Solution in the continuum limit. (a). The arrows on the curves in (b) indicate(b)
is a monotonic
the direction of flow. It is found thattakes the minimum
increasing function of $M$ and that.
value at$|M_{r}|$$r/r_{*}\mathrm{v}\mathrm{s}M_{r}$$|M_{r}|$$r/r_{*}$$|M_{\mathrm{r}}|=1$

42
ing place, binds the pressure but not the circumferential velocity there. Next let us imagine that the system
is making a solid rotation with a large angular speed.
Then the pressure of the gas increases rapidly with radial distance because of the centrifugal force, and thepressure on the outer cylinder would be larger thanand that on the inner cylinder would be smaller thanPs 1. Thus evaporation takes place on the inner cylinderand condensation on the outer (even if Ps $1<p_{s2}$ ). Thenthe speed of rotation of the outer cylinder is decreased
to rest. Similar flow (a circulating source flow) would
be maintained, because the circumferential motion isdetermined by the condition of the inner cylinder andthe boundary condition of the outer cylinder does not
restrict the circumferential velocity . This consideration suggests the possibility of bifurcation of flow when
the inner cylinder is rotating fast and the outer is at
rest.
When the evaporation or condensation happens tovanish $(u_{r}=0)$ , the set of Eqs. (18) –(22) degenerates to an undetermined system, and the higher-orderanalysis of the asymptotic expansion of the Boltzmann
equation is required to obtain the determined system,
which is derived in Ref. 25 together with its boundary
condition on the cylinders (see also Ref. 31). That is,
the set of equations is$p_{s2}$$u_{\theta}$$\frac{pu_{\theta}^{2}}{r}-\frac{dp}{dr}=0$A,

Fig. 4.
speed

The mass-flow rateversus the circumferentialof the outer cylinder $(L_{2}/L_{1}=2,$ $p_{s2}/p_{S}1=1.1$ ,$T_{2}/T_{1}=1$ , and). The numerical solutions of the
BKW equation for various Kn are shown by the symbols:
0.005; 2: 0.01; $\otimes:0.02;\square :0.1;\mathrm{O}$ : 1;
. The solid line –: the solution atby theasymptotic analysis in Sec. 4.2. The dashed line—- : thefree molecular solution.
$J$$|V_{\theta 2}|$

$V_{\theta 1}=0$$\blacksquare:\mathrm{K}=0.002;\bullet_{:}$$\mathrm{O}:10$$\mathrm{R}=0_{+}$(28)$( \frac{du_{\theta}}{dr}+\frac{u_{\theta}}{r})=\frac{\gamma_{1}}{r}\frac{d}{dr}[r^{2}\tau^{m}(\frac{du_{\theta}}{dr}-\frac{u_{\theta}}{r})]$, . (29)$\Lambda\frac{d}{dr}(\frac{u_{\theta}^{2}}{2}+\frac{5RT}{2})$algebraic equation and one first-order and two secondorder differential equations for four functions
with an undetermined constant (A) is to be obtained
under six boundary conditions. It is noted that the behavior of the gas at
is determined together
with a quantity of the order of Kn (or). That is,
it is not the conventional cylindrical Couette flow [theaxially and circumferentially uniform solution of theNavier-Stokes equation under the nonslip boundary condition without evaporation-condensation $(u_{r}=0)$ on
the boundary] except when $\Lambda=0$ .
In the next section we will discuss the flow in thewhole range of the Knudsen number (including the more
explicit description of the two limiting solutions) for$p_{s2}>p_{s1}$ , where bifurcation and reversal of a flow are$(u_{\theta}, \rho,p,T)$$\mathrm{R}=0_{+}$(30)$= \gamma_{1}\frac{d}{dr}[rT^{m}u_{\theta}(\frac{du_{\theta}}{dr}-\frac{u_{\theta}}{r})]$$+ \frac{5\gamma_{2}R}{2}\frac{d}{dr}[rT^{m_{\frac{dT}{dr}]}}$

,

,
(31)
where $m=1/2,$
1.270042,
1.922284 for a
hard-sphere molecular gas, $m=1,$ $\gamma_{1}=1,$ $\gamma_{2}=1$
for the BKW equation, and A is a constant to be determined with the solution. The physicalmeaning of A is as follows: If the rarefaction of the gas(or the effect of the mean free path of the gas molecules)
were taken into account, the radial velocity
would
not be zero. Then,$p=R\rho T$$\gamma_{1}=$$\gamma_{2}=$$(u_{\theta}, \rho,p, \tau)$$u_{r}$$\Lambda=c0u_{r}r\tau m(RT)^{-1}/2l-1$

,

(32)

where is the local mean free path of the gas molecules
and is a nondimensional constant [$c_{0}=2\sqrt{2}/\pi$ (hardsphere and BKW)]. The set of equations (28) $-(32)$ is
equivalent to the set of the axially and circumferentially uniform Navier-Stokes equations where the radial
velocity of the order of the Knudsen number is retained.
The boundary conditions for the set of equations on thetwo cylinders are$l$$c_{0}$$u_{\theta}=V_{\theta 1}$$u_{\theta}=V_{\theta 2}$,,$T=T_{1}$$T=T_{2}$,,$p=p_{s1}$$p=ps2$(at $r=L_{1}$ ),(at $r=L_{2}$ ).

(33)

The system of equations (28) $-(31)$ and boundary conditions (33) is a determined system, where the set of one$u_{r}$

expected.

5

Flows in the whole range of the Knudsen
number and discussions

5.1 Reversal of flow
First, we consider the case where the inner cylinder is
at rest $[V_{\theta 1}/(2R\tau_{1})^{1/2}=0]$ . Taking $p_{s2}/p_{s1}=1.1$ ,$T_{2}/T_{1}=1$ , and $L_{2}/L_{1}=2$ , we investigate the behaviorof the gas numerically forious values of the parameters Kn and. The temperature ratio is
put equal to unity by the reason explained in the lastparagraph of Sec. 2. The results of computation together with those for the free molecular and continuum
flows are given in Figs. 4-9, on the basis of which wewill discuss the feature of the flow.
In order to grasp the overall feature of the flow in thedomain of the parameters Kn and, wefirst discuss the mass-flow rate between the cylinders.
$\mathrm{v}\mathrm{a}r$$V_{\theta 2}/(2R\tau_{1})^{1/2}$$V_{\theta 2}/(2R\tau_{1})^{1/2}$

43

Fig. 5. The profiles of the flow velocity, pressure, and temperature $T$ of the gas for various circumferential
velocitiesof the outer cylinder when the inner cylinder is
at rest 1: (Kn $=0.1,$ $p_{s2}/p_{s1}=1.1,$ $T_{2}/T_{1}=1,$ $L_{2}/L_{1}=2$ ).
The symbol
indicates the position corresponding to, Ps 1, or(, or ).
$(u_{r}, u_{\theta})$$p$$V_{\theta 2}$$\nu(\sqrt)$$V_{\theta 1}$$T_{1}$$V_{\theta 2,p_{s}2}$$T_{2}$The mass-flow rate ffom the inner to outer cylinder
versus the speedof the outer cylinder is shown forvarious Knudsen numbers fromtoin Fig. 4.
In the ffee molecular flow, as explained in Sec. 4.1,
the effects of the inner and outer cylinders are independent of each other. All the molecules leaving theinner cylinder reach the outer cylinder. On the other$J$$V_{\theta 2}$$\mathrm{I}\mathrm{f}\mathrm{f}\mathrm{i}=0$$\infty$Fig. 6. The profiles of the flow velocity, pressure, and temperature $T$ of the gas for various circumferential
velocitiesof the outer cylinder when the inner cylinder is
at rest 11: (Kn $=10,$ $p_{s2}/p_{s1}=1.1,$$L_{2}/L_{1}=2$ ).
The symbol($) indicates the position corresponding to, Ps 1, or(, or ).
$(u_{r}, u_{\theta})$$p$$V_{\theta 2}$$T_{2}/T_{1}-rightarrow 1,$$\rangle$$V_{\theta 1}$$T_{1}$$V_{\theta 2,p_{s}2}$$T_{2}$

hand, some molecules leaving the outer cylinder do notreach the inner cylinder and return directly to someother point on the outer cylinder. This is determinedby the direction of the velocity of a molecule leaving
the inner cylinder. As the speed of the outer cylinder
increases, the number of molecules that reach the inner cylinder decreases and finally no molecules reach it,

44$\sim\overline{\mathrm{h}}\mathrm{h}$, pressure ,Fig. 7. The profiles of the flow velocity
and temperature $T$ of the gas for various Knudsen numbers
when the inner cylinder is at rest 1: $(V_{\theta 2}/(2RT_{1})^{1/2}=$$0.15,$ $p_{S}2/p_{s1}=1.1,$ $T_{2}/T_{1}=1,$ $L_{2}/L_{1}=2)$ . The dotted. indicates the Knudsen layer flattened on the cylinline. The symbol ’ ($) indicates the position
ders at, or ).
, Ps 1, or(
corresponding to, pressure ,Fig. 8. The profiles of the flow velocity
and temperature $T$ of the gas for various Knudsen numbers
Kn when the inner cylinder is at rest 1I: $[V_{\theta 2}/(2RT_{1})^{1/2}=$$0.75,$ $p_{s2}/ps1=1.1,$ $T_{2}/T_{1}=1,$ $L_{2}/L_{1}=2]$ . The symbol, Ps 1, or($) indicates the position corresponding to, or ).
(

since the high speed motion of the outer cylinder imparts a large circumferential motion to the molecules
leaving the outer cylinder. Thus, the net mass-flowrate (ffom the inner cylinder to the uter) increases
from $2\pi L_{1}(2\pi R\tau_{1})-1/2[Ps1-p_{s2}(\tau_{1}/T_{2})^{1/}2]$ to(or a negative to a positive value). Thatis, evaporation takes place on the outer cylinder andcondensation on the inner when both the cylinders areat rest. As the speed of the outer cylinder increases,
the rate of evaporation on the outer cylinder decreases
and vanishes at some speed, and then reversal of theflow or evaporation on the inner cylinder occurs.

$(u_{r}, u_{\theta})$$p$$\mathrm{M}$$\cdots\cdot$$\mathrm{M}=0_{+}$$V_{\theta 1}$$T_{1}$$V_{\theta 2,p_{s}2}$

$p$

$\nu$

$V_{\theta 1}$$V_{\theta 2,Ps}2$$T_{1}$$T_{2}$$T_{2}$$\mathit{0}$$2\pi L_{1Ps1}$$(2\pi R\tau_{1})^{-}1/2$$(u_{r}, u_{\theta})$45
In the continuum flow (Kn $=0_{+}$ ), as in the free molecular flow, evaporation takes place on the outer cylinderand condensation on the inner when both the cylinders
are at rest, and as the speed of the outer cylinder increases, the rate of evaporation on the outer cylinder
decreases and vanishes at the speed$|V_{\theta 2}|$$=$$\frac{(5RT_{2})^{1}/21^{1}-(p_{\epsilon}2/p_{s}1)-2/5]^{1/2}}{[(L_{2}/L1)^{2}-1]1/2}$$=$$0.1766(2RT_{1})^{1/2}$

.

(34)

(This result is independent of molecular models. The,
first formula applies to bitrary values of) For the higher speed of the outer cylinder,andin contrast to the free molecular flow, no reversal offlow takes place, and neither evaporation nor condensation occurs on the cylinders, because evaporation on
the inner cylinder at rest does not supply circumferential motion. The solution without evaporation andcondensation is obtained by the system (28) $-(31)$ and(33). According to , determined by these equations,
the Kn-order term of the nondimensional mass-flow rate
ps 1
is negative for0.4697 and positive for $|V_{\theta 1}|/(2R\tau_{1})^{1/2}>0.4697$ . (Thisvalue is 0.4693 for a haxd-sphere molecular gas.) Corredifspondingly, the profiles of the flow field at Kn
ferent from that of the conventional Couette flow (thesolution of the Navier-Stokes equation with $u_{r}=0$ )excePt at $|V_{\theta 1}|/(2R\tau_{1})^{1/2}=0.4697$. This is studied in
the incomdetail in Refs. 25 as an example
pleteness of the classical gas dynamics.
The mass-flow rate in the intermediate Knudsen numof thebers increases monotonically with the speed
outer cylinder. The variation of the mass-flow rate withisthe Knudsen number is not monotonic when
. This is discussed in relasmaller than abouttion to the profile of flow field later.
The profiles of the velocity, pressure, and temperature at Kn $=0.1$ and 10 are shown for various speeds ofthe outer cylinder in Figs. 5 and 6, where we can see
theiation of the flow field with the speed of the outercylinder. The circumferential motion of the outer cylinder is transmitted to the whole region of the gas, inducing pressure variation owing to the centrifugal force. As
the speed of the outer cylinder increases, the pressure
and the pressurenear the outer cylinder exceeds
near the inner cylinder goes down below Ps 1. This results in the reversal of the flow (from $u_{r}<0$ to $u_{r}>0$ ).
The qualitative feature is the same for both the Knudsen numbers. The temperature field at &=0.1 clearlyshows the combination of the heating effect of gas motion and temperature jump (or drop) on the condensing (evaporating) boundary. The heating is obviously
stronger for faster motion, and the jump (or drop) isbigger for stronger condensation (or evaporation). The.
latter contribution is not so evident at
Figures 7 and 8 show the profiles for various Knudsen
numbers at $|V_{\theta 2}|/(2R\tau_{1})^{1/2}=0.15$ and 0.75. In thesefigures we can see the variation of the flow field with
the Knudsen number.
In the case of $|V_{\theta 2}|/(2R\tau_{1})^{1/2}=0.15$ (fairly small
Fig. 7), the circumferential velocity at Ikn $=0_{+}$ ,$p_{s2}/p_{s1},$$\mathrm{a}x$$\tau_{2}/T_{1}$$L_{2}/L_{1}.$$\Lambda$$J/2\pi L_{1}$$|V_{\theta 1}|/(2R\tau_{1})^{1/2}<$$(2R\tau_{1})^{1/2}$$=0_{+}\mathrm{i}_{\mathrm{S}}$$\mathrm{s}\mathrm{h}\mathrm{o}\mathrm{w}\mathrm{i}32\mathrm{n}\mathrm{g}$$|V_{\theta 2}|$$|V_{\theta 2}|$$\sqrt{RT_{1}}/2$$\mathrm{v}\mathrm{a}r$$p_{s2}$$\mathrm{M}=10$$|V_{\theta 2}|)$

Fig. 9. The type of solutions classified by the direction) for $L_{2}/L_{1}=2$ ,of flow in the plane (Kn,$\mathrm{P}s2/p_{s1}=1.1,$ $T_{2}/T_{1}=1$ , and $V_{\theta 2}=0$ .
The type of a(point) andsolution is indicated by the symbols:(range): a solution with $u_{r}>0;\blacksquare$ (point) and(range): $u_{r}<0$ ; EEEB at Kn $=0_{+}:$
at Kn $=0_{+}:$ $u_{r}=O(l\mathrm{h})<0$ .
$V_{\theta 2}/(2RT_{1})^{1/2}$$\square$$\mathrm{R}$–$u_{r}=\mathrm{O}(\mathrm{f}\mathrm{f}\mathrm{i})>0;\mathrm{E}\mathrm{E}\coprod$

where the flow is isentropic and evaporation is taking
place on the outer cylinder, is determined by the velocityof the outer cylinder and is not affected by thevelocityof the inner cylinder. As the Knudsen number increases, the effect of the stationary inner cylinder
is transmitted into the gas and the circumferential motion of the gas is retarded. This reduces the pressurenear the outer cylinder and increases the pressure near
the inner cylinder. Thus the rate of evaporation on theouter cylinder increases. This increase ceases at about$\mathrm{M}=0.1$ ; then it decreases to the value at Kn(seealso Fig. 4). This maximum is the result of different dependence on the Knudsen number for large and small, as in the Knudsen minimum in the Poiseuille flowof a rarefied gas.19,33 This behavior for large Knudsen
numbers can be understood in the following way. At Kn, all the molecules from the inner cylinder reach theouter directly. At a ge but not infinite Kn, molecular collisions block some of the molecules to reach theouter cylinder; in return the molecules from the outercylinder are subject to collision effect, but only a partof them was to reach the inner cylinder without collisions. Thus the latter blocking effect is smaller than
the former. The number of molecules from the outer tothe inner cylinder, therefore, is bigger with molecularcollision. This nonmonotonic dependence on Kn is seenfor $0<|V_{\theta 2}|/(2R\tau_{1})^{1/2}\leq 0.45$ .
In the case $|V_{\theta 2}|/(2R\tau_{1})^{1/2}=0.75$ (Fig. 8), neither
evaporation nor condensation occurs on the cylinders$V_{\theta 2}$$V_{\theta 1}$$=\infty$$\mathrm{R}$$=\infty$$1\mathrm{a}r$46

Fig. 10. The mass-flow rateversus the circumferential speedof the inner cylinder $(L_{2}/L_{1}=2,$ $p_{S}2/p_{s1}=1.2$ ,, and). The numerical solutions of the BKW equation for various Kn are shown by the symbols:
Kn
0.005; 2: 0.01; $\otimes:0.02;\ominus:0.05;\square :0.1_{1}\nabla:0.2;\triangle:0.5;\mathrm{O}$: 1. The solid line –: the solution atby theasymptotic analysis in Sec. 4.2. The dashed line–: the free molecular solution.
$J$$T_{2}/T_{1}=1$$|V_{\theta 1}|$

$V_{\theta 2}=0$$\bullet_{:}$$=$$\mathrm{R}=0_{+}$–. As already mentioned, the profiles, obat
tained by Eqs. (28) $-(31)$ and (33), are different fromthose of the conventional cylindrical Couette flow. At
this value of, A is positive, that is, theradial velocityof the orderis positive. As theKnudsen number increases, the effect of the circumferential motion of the outer cylinder is further transmitted into the gas for small Kn and induces higher
pressure near the outer cylinder owing to the additional
centrifugal force, or the blocking effect of molecular collisions, discussed in the case of $|V_{\theta 2}|/(2RT_{1})^{1/2}=0.15$ ,decreases for large Kn. Thus, in contrast to the case of$|V_{\theta 2}|/(2R\tau_{1})^{1/2}=0.15$ ,the flow to the outer cylinder
increases monotonically with R.
As the summary of this subsection, the type of solutions classified by the direction of flow ( $u_{r}>0$ or $u_{r}<$is shown on the parameter plane (Kn,)
in Fig. 9. At, the evaporation-condensationceases for0.1766; the sign ofofthe order of Kn in this range of, whichis obtained by asymptotic analysis of the BKW equation, is also shown. Only one solution exits for a set of(Kn,), and bifurcation of flow does notoccur. Except for the continuum limit, reversal of flow
occurs at some speed of the outer cylinder.
$\mathrm{M}=0_{+}$$|V_{\theta 2}|/(2RT_{1})^{1/2}$$u_{r}$$0)$$\mathrm{M}$$|V_{\theta 2}|/(2RT_{1})^{1/2}$$\mathrm{R}=0_{+}$$|V_{\theta 2}|/(2R\tau_{1})^{1/2}\geq$$u_{r}$$|V_{\theta 2}|/(2RT_{1})^{1/2}$$|V_{\theta 2}|/(2R\tau_{1})^{1}/2$-

5.2 Bifurcation of flow
In this section we consider the case where the outercylinder is at rest $[V_{\theta 2}/(2R\tau_{1})^{1/2}=0]$ .
$1.2,$ $T_{2}/T_{1}=1$ , and $L_{2}/L_{1}=2$ , we investigate the behavior of the gas numerically for various values of theparameters Kn and. The temperature ratio is put equal to unity by the reason explained in thelast paragraph of Sec. 2. The results of computation together with those for the hee molecular and continuum
flows are given in Figs. 10-16, on the basis of which wewill discuss the feature of the flow.
The mass-flow rate from the inner to outer cylinder
versus the speedof the inner cylinder is shown forvarious Knudsen numbers
toin Fig. 10.
In the free molecular flow, where all the molecules
leaving the inner cylinder reach the outer cylinder, thenumber of molecules leaving the inner cylinder is independent of the velocity of the inner cylinder. Thus the
mass-flow rate isirrespective of.
In the continuum flow (Kn $=0_{+}$ ), on the other hand,the situation is complicate. When both the cylinders
are at rest, evaporation takes place on the outer cylinder. It decreases gradually with the speed of the innercylinder, but never vanishes until the speed becomes
infinite. Besides this solution, there are two other solutions when theof the inner cylinder exceeds the$r_{\mathrm{B}\mathrm{k}\mathrm{i}\mathrm{n}\mathrm{g}pS2}/p_{s1}=$$V_{\theta 1}/(2R\tau_{1})^{1/2}$$J$$|V_{\theta 1}|$$\mathrm{h}\mathrm{o}\mathrm{m}\mathrm{M}=0$$\infty$$2\pi L_{1}p_{s1}(2\pi R\tau_{1})-1/2[1-(\mathrm{p}_{\epsilon 2}/p_{s1})$$(T_{1}/T_{2})^{1/}2]$$V_{\theta 1}$$\mathrm{s}_{\mathrm{P}^{\mathrm{e}\mathrm{e}\mathrm{d}}}$47$\overline{\mathrm{b}}$$\sim \mathrm{h}$$1^{\epsilon \mathrm{t}/}$$(\cup)$Fig. 11. The profiles of the flow velocity, pressure , and temperature $T$ of the gas for various circumferential
velocitiesof the inner cylinder when the outer cylinder is at rest 1: (Kn $=0.005,$ $p_{s2}/p_{s1}=1.2,$ $\tau_{2}/\tau_{1}=1,$ $L_{2}/L_{1}=2$ ).
(a) The family of solutions with $J<0;(\mathrm{b})$ The family of solutions with $J>0$ . The symbol
indicates the position
corresponding to, or, or ).
($(u_{r}, u_{\theta})$$p$$V_{\theta 1}$$\rangle$$V_{\theta 1,Ps1}$$T_{1}$$V_{\theta 2,p_{s}2}$

value

, and
) One of them is a flow with
evaporation on the inner cylinder. The rate of evaporation increases from zero to $0.2360(2\pi L1)\rho_{s}1(2RT1)^{1/2}$
as the speedof the inner cylinder increases from$0.5022(2R\tau 1)^{1/2}$ to $1.373(2R\tau 1)^{1/2}$ ,and it remains constant for the larger. As
increases, the speedof evaporation (or) on the inner cylinder increases
and reaches $M_{r}=1$ . Thus for the larger, the rate$p_{s2}/p_{s1},$ $T_{2}/T_{1}$$|V_{\theta 1}|$$(\sqrt)$$T_{2}$$=$$\frac{(5RT_{1})^{1}/2[(p_{2}S/p_{s}1)^{2}/\mathrm{s}-1]1/2}{1^{1-}(L_{1}/L2)^{2}]^{1}/2}$$=$$0.5022(2RT_{1})^{1/2}$$L_{2}/L_{1}.$$J$$|V_{\theta 1}|$

.

(35)

$|V_{\theta 1}|$

(The bifurcation of flow occurs at this speed of the inner cylinder, which is independent of molecular models. The first formula applies to arbitrary values of$|V_{\theta 1}|$

$M_{r}$$|V_{\theta 1}|$

48
term of the nondimensional mass-flow rate $J/2\pi L_{1}$is negative for $|V_{\theta 1}|/(2R\tau_{1})^{1/2}<$ 0.6672
and positive for $|V_{\theta 1}|/(2R\tau_{1})^{1/2}>0.6672$. (This value
is 0.6593 for a hard-sphere molecular gas.) Correspondingly, the profiles of the flow field at Kn $=0_{+}$ is different from that of the conventional Couette flow except
at $|V_{\theta 1}|/(2R\tau_{1})^{1/2}=$ 0.6672. This is also an example
showing the incompleteness of the classical gas dynamics discussed in Refs. 25 and 32.
The mass flow rate increases monotonically with the
speedof the inner cylinder when the Knudsen
number Kn is larger than about 0.02. For smaller Knudsen numbers the situation is quite different. There are
two branches of solutions for a given Knudsen number:
one is a family of flows with evaporation on the outer, the ratecylinder existing in the range, andof evaporation of which slowly decreases withthe other is a family of flows with condensation on the, the rateouter cylinder existing in the range. Here,of condensation of which increases withdepend on Kn, and $V_{A}>V_{B}$ . There are two
and(bifurcation ofsolutions in the rangeflow). There should be the third family of solutions,corresponding to the solution with $u_{r}=0$ at Kn $=0_{+}$ ,, but ittothat connects the two families fromcould not be obtained numerically, probably because of
instability of the solution.
The profiles of the velocity, pressure, and temperature at $\Re=$ 0.005 and 0.05 are shown for various
speeds of the inner cylinder in Figs. 11 and 12, where
we can see the variation of the flow field with the speedof the inner cylinder.
At Kn $=0.005$ , there are two families of proffies: one]; theexists in $0\leq|V_{\theta 1}|/(2R\tau_{1})^{1/2}\sim<1.068$ [Fig. 11], in eachother in $|V_{\theta 1}|/(2R\tau_{1})^{1/2}\sim>0.6255$ [Fig. 11. Twoof which the profiles vary smoothly withfiles are possible for $0.6255\sim<|V_{\theta 1}|/(2R\tau_{1})^{1/2}\leq 1.068$ .
At $|V_{\theta 1}|=0$ , the flow is radial from the outer cylinincreases, the circumferender to the inner. As
tial motion of the inner cylinder is transmitted into thegas, but its penetration is strongly blocked by the convection radial flow and reaches only half way from the(a little higher
inner cylinder even atthan sonic speed). Owing to this limited angular motion resulting in only small pressure rise by centrifugal
force, the pressure near the outer cylinder is lower than
, and the gas remains evaporating from the outercylinder. For $|V_{\theta 1}|/(2R\tau_{1})^{1/2}\sim>1.075$ , this type of flow
no longer exists, and completely different flows appear.
The angular motion comparable to that of the innercylinder prevails over the whole field, except in a thinlayer near the outer cylinder, inducing a pressure higher
thannear the outer cylinder, and therefore evaporation occurs on the inner cylinder. This type of flow
1.075 but also
exists not only forfor smaller
down to about 0.6255. It
was impossible to obtain a family of solutions that connects the two branches smoothly, probably because of
instability of the flow.
iation is smooth for all
At Kn $=0.05$ (Fig. 12), the; only one profile is possible for a given$\rho_{s1}(2R\tau 1)^{1/2}$$|V_{\theta 1}|$$0\leq|V_{\theta 1}|\leq V_{A}$$|V_{\theta 1}|$$|V_{\theta 1}|\geq V_{B}$$V_{A}$$|V_{\theta 1}|$

$V_{B}$$V_{B}\leq|V_{\theta 1}|\leq V_{A}$$V_{B}$$V_{A}$$(\mathrm{a})$$(\mathrm{b})$$|V_{\theta 1}|$$|V_{\theta 1}|$$|V_{\theta 1}|=(2R\tau_{1})^{1/2}$$p_{s2}$, pressure
Fig. 12. The profiles of the flow velocity, and temperature $T$ of the gas for various circumferentialof the inner cylinder when the outer cylinder isvelocities
at rest 1I: (Kn $=0.05,$ $p_{\theta}2/p_{s1}=1.2,$ $T_{2}/T_{1}=1,$ $L_{2}/L_{1}=2$ ).
indicates the position corresponding to
The symbol, Ps 1, or(or ).
$(u_{r}, u_{\theta})$$p$$V_{\theta 1}$$\nu(\sqrt)$$T_{1}$$V_{\theta 1}$$V_{\theta 2,p_{s2)}}$$T_{2}$$p_{s2}$of evaporation remains constant, since evaporation with$M_{r}>1$ is impossible [see Eq.
]. The flow is an accelerating supersonic flow accompanied by a shock wave
(Fig. 14). The other one is a flow without evaporationand condensation on the cylinders, the solution of which
is obtained by the system (28) $-(31)$ and (33). According to , determined by these equations, the Kn-order$(24\mathrm{a})$$|V_{\theta 1}|/(2R\tau_{1})^{1/2}\sim>$$|V_{\theta 1}|/(2R\tau_{1})^{1/2}$$\mathrm{v}\mathrm{a}r$$|V_{\theta 1}|/(2R\tau_{1})^{1}/2$$\Lambda$$|V_{\theta 1}|/(2R\tau_{1})^{1/2}$.

$\mathrm{p}\mathrm{r}\not\in$49$(^{\mathrm{U}}$$\mathrm{t}^{\mathrm{d}})$’, pressure , and temperature $T$ of the gas for various Knudsen numbers
Fig. 13. The profiles of the flow velocity
Kn when the outer cylinder is at rest 1: $[V_{\theta 1}/(2RT_{1})^{1/2}=0.7, ps2/ps1=1.2, T_{2}/T_{1}=1, L_{2}/L_{1}=2]$.
The profiles for. indicates the Knudsen layer flattened on the
0–0.0125.; (b) The profiles for0.0135–00. The dotted line
, or
, Ps 2, or ).
(cylinders at $\Re=0_{+}$ . The symbol($) indicates the position corresponding to$(u_{r}, u_{\theta})$$p$$(\mathrm{a})$$\mathrm{R}=$$\mathrm{R}=$$\rangle$

Figures 13-15 show the profiles for various Knudsen
numbers at $|V_{\theta 1}|/(2R\tau_{1})^{1/2}=0.7,2.5$ , and 0.3. In
these figures, we can see the variation of the flow fieldwith the Knudsen number.
At $|V_{\theta 1}|/(2R\tau_{1})^{1/2}=0.7$ (Fig. 13), there are two
families of the profiles, one with $u_{r}<0$ and the otherwith $u_{r}>0$ , for small Knudsen numbers (Kn is smaller
than about 0.0125). (There is another solution with$u_{r}=0$ at Kn $=0_{+}.$ ) This nonuniqueness can be under-$\cdots\cdot$$V_{\theta 1,\mathrm{P}s1}$$T_{1}$$V_{\theta 2}$$T_{2}$

stood in the following way. Owing to the convection of
the radial gas flow, the effect of circumferential motionof a condensing cylinder (the inner or outer cylinder
depending on $u_{r}<0$ or $u_{r}>0$ ) on the circumferential
motion of the gas hardly penetrates deep into the gas
when the Knudsen number is small. Let $u_{r}<0$ . The
flow is radial, and the circumferential motion is confined in a narrow region near the inner cylinder; thus
the pressure is nearly uniform outside the region and50$\backslash -\aleph$$\propto \mathrm{o}\mathrm{e}\overline{\mathrm{h}}$$\backslash \vee$$\mathrm{a}\approx$.

$\backslash \approx\Omega^{n}-$$\dot{\mathrm{h}}$$\backslash \mathrm{h}$Fig. 14. The profiles of the flow velocity, pressure ,and temperature $T$ of the gas for various Knudsen numbers
Kn when the outer cylinder is at rest 11: $[V_{\theta 1}/(2RT_{1})^{1/2}=$$2.5,$ $Ps2/p_{S}1=1.2,$ $T_{2}/T_{1}=1,$ $L_{2}/L_{1}=2]$ .
The dotted. indicates the Knudsen layer flattened on the cylinline. The symbol $>($ indicates the position
ders at, or, or ).
(
corresponding to, pressure ,Fig. 15. The profiles of the flow velocity
and temperature $T$ of the gas for various Knudsen numbers
Kn when the outer cylinder is at rest 111:
$0.3,$ $ps\mathit{2}/ps1=1.2,$ $T_{2}/T_{1}=1,$ $L_{2}/L_{1}=2]$ . The dotted. indicates the Knudsen layer flattened on the cylinline. The symbol ($) indicates the position
ders at, or, Ps 2, or ).
(corresponding to

the pressure near the outer cylinder is lower than.
Thus in conformity with the assumption, evaporation
takes place on the outer cylinder. Let $u_{r}>0$ . The flow
is circulating over the whole field except near the outercylinder, and a considerable pressure variation due tocentrifugal force is induced and the pressure near theouter cylinder is higher than. Thus, in conformity

with the assumption, condensation (evaporation) takes
place on the outer (inner) cylinder. As the Knudsen
number increases, the effect of the condensing cylinder
on the circumferential motion penetrates deeper into
the gas, and two solutions approach. One of the solutions $(J<0)$ , however, disappears at about Kn$(u_{r}, u_{\theta})$$p$$\cdots\cdot$$\mathrm{M}=0_{+}$$V_{\theta 1,p_{s1}}$$T_{1}$$V_{\theta 2,p_{s2}}$$T_{2}$$p_{s\mathit{2}}$$p_{s2}$$(u_{r}, u_{\theta})$$p$$[V_{\theta 1}/(2RT_{1})^{1/\mathit{2}}=$$\ldots.$$\mathrm{R}=0_{+}$$\rangle$$V_{\theta 1,p_{s}1}$$T_{1}$$V_{\theta 2}$$T_{2}$$=$

51
0.0135, and the other solution loses its radial motion
and finally reverses its direction.
At larger speedof the inner cylinder (Fig. 14), nonunique situation of the solution is similar to that in the previous example. The
larger circumferential velocity induces the larger pressure variation to reverse the direction of flow even when
the penetration of circumferential motion is not deep,
and the first type flow (a flow evaporating on the outercylinder) occurs for only smaller Knudsen numbers.
Among Knudsen numbers for which the computation
is carried out, only the case $\Re=0_{+}$ is the first typeof flow. Along the second type flow, the flow is firstaccelerated and then decelerated. The deceleration becomes sharper as the Knudsen number decreases; andfinally at Kn $=0_{+}$ , the decelerated region is reduced
to a discontinuity (a shock wave) accompanied by amoderately decelerating region.
At small speedof the innercylinder (Fig. 15), where the solution with $u_{r}=0$ at
Kn $=0_{+}$ does not exist, the pressure variation inducedby circumferential motion is too small to reverse thedirection of flow even if the circumferential motion is
induced over the whole field, and thus evaporation takes
place on the outer cylinder for all Knudsen numbers.
As the summary of this subsection, the type of solutions classified by the direction of flow $(u_{r}>0,$ $u_{r}=$, or $u_{r}<0$ ) is shown on the parameter plane (Kn,in Fig. 16. There exist three typesof solution at Kn $=0_{+}$ when $|V_{\theta 1}|/(2R\tau_{1})^{1/2}\geq\sqrt{5/2}$$[(p_{s2}/p_{S}1)^{2/5}-1]^{1/2}[1-(L_{1}/L_{2})^{2}]-1/2=0.5022$ . Two
solutions are obtained in some range of small Knudsen numbers. For the larger Knudsen numbers, there
is only one solution. Bifurcation as well as reversal of
a flow is seen in the second example, where the innercylinder is rotating.
$[|V_{\theta 1}|/(2R\tau_{1})^{1/\mathit{2}}=2.5]$$\backslash -\mathrm{N}\wedge \mathrm{t}_{\mathrm{t}}\circ\overline{\mathrm{h}}$$\underline{\underline{\vee\backslash }\triangleright^{\dot{\Phi}}}$$[|V_{\theta 1}|/(2R\tau_{1})^{1/\mathit{2}}=0.3]$$0$$|V_{\theta 1}|/(2R\tau_{1})^{1}/\mathit{2})$6Concluding remarks

In the present paper, a rarefied gas between two coaxial
circular cylinders made of the condensed phase of thegas was considered, where each cylinder is kept at a uniform temperature and is rotating at a constant angularvelocity around its axis (cylindrical Couette flows of ararefied gas with evaporation or condensation on thecylinders). The steady behavior of the gas was studied
on theis of kinetic theory from the continuum to the
Knudsen limit. The solution showed profound variety,especially bifurcation of flow was seen even in the simple case where the state of the gas is circumferentiallyand axially uniform. In view of the Taylor-Couette instability of a flow, the stability of the bifurcated flows
in the absence of the constraint of circumferential oraxial uniformity is a problem to be studied.
Bifurcation and reversal of flow, typical features ofthe problem, are not special for the BKW equation. At
least for the continuum limit, these are found for the
standard Boltzmann equation. The fact that asymptotic analysis for small Knudsen numbers goes parallel
for these equations and the physical discussion about
the behavior of the gas, which is not special for the
BKW model, support this.
$\mathrm{b}\mathrm{a}s$Fig. 16. The type of solution classified by the directionof flow in the plane (Kn,) for $L_{2}/L_{1}=2$ ,$p_{s2}/p_{s1}=1.2,$ $T_{2}/T_{1}=1$ , and $V_{\theta 2}=0$ . The symbol(point) and(range): a solution with $u_{r}>0;\blacksquare$(point) and(range): $u_{r}<0$ ;: two solutions$(u_{r}<0, u_{r}>0)$ exist;
at: three solutions$[u_{r}<0, u_{r}=0(u_{r}=O(\mathrm{R})<0), u_{r}>0]$ exist; $\mapsto$ atthree solutions [ $u_{r}<0,$ $u_{r}=0$ ( $u_{r}=$ O(Kn) $>0$ ),$V_{\theta 2}/(2RT_{1})^{1/2}$$\square$$\subset\supset$–$\mathrm{D}$$\ovalbox{\tt\small REJECT}$$\mathrm{R}=0_{+}$$\mathrm{R}=0_{+}:$$u_{r}>0]$

exist.

The numerical computation was carried out on several workstations such as
and VT-Alpha(CPU: Alpha533), and took several years of the CPU time.
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