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PHYSICAL CHEMISTRY
M. V. VELICHKO, L. V. RADUSHKEVICH
PROPERTIES OF A HYDRODYNAMIC FLOW PAST VERY SMALL STATIONARY OBSTACLES, AND THE EFFICIENCY OF CAPTURE OF AEROSOL PARTICLES
(Presented by Academician M. M. Dubinin, 8 VIII 1963)
The possibility of applying the Navier—Stokes hydrodynamic equations to solving problems of flow past obstacles whose characteristic size \(R\) is comparable with the mean free path \(\lambda\) of gas molecules depends on the value \(M(\lambda/R)\), where \(M\) is the Mach number, equal to the ratio of the flow velocity to the speed of sound. As shown in \((^1)\), p. 310, ordinary hydrodynamics remains valid for \(M \ll 1\), even if \(\lambda/R \cong 1\). In this case, however, it is necessary to introduce special boundary conditions. Slightly modifying the method proposed by Fuchs \((^2)\), we have calculated the slip velocity for the cases of gas flow past a sphere and a cylinder whose radii are close to the value of \(\lambda\). The solutions obtained have been used by us to calculate the efficiency of capture of aerosol particles from a flow on a stationary cylinder.
We shall assume that ordinary hydrodynamics is applicable at distances \(r\) from the obstacle greater than \(a = R + \beta\lambda\), with \(\beta \cong 1\), while at distances in the interval \(R \ll r \ll a\) the process is carried out by free molecular flow. Matching the differential hydrodynamic and molecular flows on a surface element \(dS\), we determine the parameters entering the expressions for the flow field of a viscous fluid.
In the general case, the friction force applied to a surface element \(dS\) with radius of curvature \(a\) is directed along the flow and is equal to
\[ dF_a = (p_{rr}\cos\theta - p_{r\theta}\sin\theta)\,dS, \tag{1} \]
where the components of the hydrodynamic momentum are expressed in polar coordinates by the known relations
\[ p_{rr} = -p + 2\mu \frac{\partial v_r}{\partial r}; \qquad p_{r\theta} = \mu\left(\frac{1}{r}\frac{\partial v_r}{\partial \theta} + \frac{\partial v_\theta}{\partial r} - \frac{v_\theta}{r}\right); \tag{2} \]
here \(v_r\) and \(v_\theta\) are the velocity components, \(\mu\) is the gas viscosity, and \(p\) is the total pressure.
Assuming a Maxwellian distribution of gas-molecule velocities, let us write the expression for the force acting on a surface element \(dS'\) of radius \(R\), from molecules incident upon it and diffusely reflected with accommodation under free molecular flow. In doing so we use Epstein’s results \(((^3); (^1), p. 283)\); then
\[ dF_{\mathrm{m}} = \frac{1}{2}Nmc\,(\overline{v_r}\delta\cos\theta - \frac{1}{2}v_\theta\sin\theta)\,dS'. \tag{3} \]
Here \(\delta = 1 + \pi/4\). Taking into account that \(\mu = 0.499\,Nmc\lambda \cong \frac{1}{2}Nmc\lambda\), and equating (1) and (3), we obtain a series of identities with respect to functions of \(\theta\), from which we find the constants of the hydrodynamic equations for the flow.
1. Stationary sphere. The velocity components for a viscous flow past a sphere of radius \(R\) at small Reynolds numbers are expressed—
are of the form
\[ v_r=u_0\cos\theta\left(1-\frac{A}{r}+\frac{B}{r^3}\right), \]
\[ v_\theta=-u_0\sin\theta\left(1-\frac{A}{2r}+\frac{B}{2r^3}\right). \tag{4} \]
The coefficients \(A\) and \(B\) are determined by the boundary conditions. In ordinary hydrodynamics, for the motion of a sphere in a viscous liquid, it is known that
\[ A={}^{3}/_{2}R;\qquad B={}^{1}/_{2}R^3. \tag{4′} \]
In the problem under consideration, with the use of (1), (2), and (3), \(A\) and \(B\) take other values, which we find as follows. Since it is known from (4) that \(p=-\mu\dfrac{u_0 A}{r^2}\cos\vartheta\), from (2) and (4) we obtain
\[ p_{rr}=3\mu u_0\cos\theta\left(\frac{A}{r^2}-2\frac{B}{r^4}\right);\qquad p_{r\theta}=-3\mu u_0\sin\theta\cdot\frac{B}{r^4}. \]
Substituting these expressions into (1) and taking into account that \(dS=2\pi a^2\sin\theta\,d\theta\), we have
\[ dF_a=3\mu u_0\left[\left(A-2\frac{B}{a^2}\right)\cos^2\theta+\frac{B}{a^2}\sin^2\theta\right]2\pi\sin\theta\,d\theta. \]
Introducing \(v_r\) and \(v_\theta\) from (4) into relation (3), we find, for \(dS'=2\pi R^2\sin\theta\,d\theta\),
\[ dF_{\mathrm{M}}={}^{1}/_{2}Nmcu_0\left[\delta\left(1-\frac{A}{r}+\frac{B}{r^3}\right)\cos^2\theta+\right. \]
\[ \left.+{}^{1}/_{2}\left(1-\frac{A}{2r}+\frac{B}{2r^3}\right)\sin^2\theta\right]2\pi R^2\sin\theta\,d\theta. \]
Equating \(dF_a\) and \(dF_{\mathrm{M}}\), we obtain identities with respect to \(\theta\), from which, taking into account that \(\mu\cong{}^{1}/_{2}Nm c\lambda\), it follows that
\[ A=\frac{1}{G}\,3(1+\beta\xi)\,[\delta+4\xi(1+\beta\xi)(\delta+1)]\,R; \tag{5} \]
\[ B=\frac{1}{G}(1+\beta\xi)^3[\delta+6\xi(1+\beta\xi)]\,R^3, \tag{6} \]
where
\[ G=36\xi^2(1+\beta\xi)^2+12\xi\delta(1+\beta\xi)+9\xi(1+\beta\xi)+2\delta, \tag{7} \]
with \(\xi=\lambda/R\). Relations (5), (6), and (7) give the solution of the problem. In the limiting case, for a large sphere radius \(R\) and small \(\lambda\), we have \(\xi\to0\), and then (5) and (6) give the classical values of \(A\) and \(B\) from (4′), and \(G=2\delta\). Integrating the expression \(dF_a\) over the sphere of radius \(a\), we obtain the formula for the resistance
\[ F=4\pi\mu u_0 A. \]
Let us compare this relation with Millikan’s formula for a small sphere:
\[ F=6\pi\mu Ru_0\,[1+P\lambda/R+Q(\lambda/R)e^{-bR/\lambda}]^{-1}. \]
Here \(P=1.246;\ Q=0.42;\ b=0.87;\ \lambda=0.635\cdot10^{-5}\ \mathrm{cm}\). Put \(R=\lambda\). Then the coefficient \(A\) is transformed, for \(\xi=1\), into the form
\[ A=\frac{3R(1+\beta)[\delta+4(1+\beta)(\delta+1)]}{36(1+\beta)^2+12\delta(1+\beta)+9(1+\beta)+2\delta}. \]
From this we obtain the value \(\beta\cong0.42\), i.e., the two flows merge at a distance \(\beta\lambda=2.69\cdot10^{-6}\ \mathrm{cm}\) from a sphere of radius \(R=0.653\cdot10^{-5}\ \mathrm{cm}\).
2. A stationary cylinder. Carrying out an analogous derivation for a cylinder of radius \(R\) with unit length, we use Lamb’s solution \(({}^4)\) for small \(\mathrm{Re}\) in the form
\[ v_r=\frac{A_0}{r}-\frac{A_1\cos\theta}{r^2}+u_0\cos\theta-C_0\left(\frac{1}{2kr}+\frac{1}{2}\cos\theta-\frac{1}{2}\cos\theta\ln z\right), \]
\[ v_\theta=-\sin\theta\left(u_0+\frac{A_1}{r^2}+\frac{C_0}{2}\ln z\right), \tag{8} \]
where \(z = \frac{1}{2}\gamma k r\). In ordinary hydrodynamics the values of the coefficients of these equations are expressed as
\[ C_0=\frac{4u_0}{1-2\ln z};\qquad A_0=\frac{4\nu}{1-2\ln z};\qquad A_1=-\frac{u_0}{1-2\ln z}R^2, \]
where \(\nu\) is the kinematic viscosity. It is known that
\[ p=-\rho u_0\frac{A_0}{r}\cos\theta-\rho u_0\frac{A_1}{r^2}\cos 2\theta+\ldots \]
Therefore, according to (2), we find:
\[ p_{rr}=2\mu\left[\cos\theta\left(\frac{\rho u_0}{2\mu}\frac{A_0}{r} +\frac{2A_1}{r^3}+\frac{C_0}{2r}\right) +\left(\frac{C_0}{2kr^2}-\frac{A_0}{r^2}\right)\right] -\rho u_0A_1\frac{\cos 2\theta}{r^2}; \]
\[ p_{r\theta}=\mu\frac{4A_1}{r^3}\sin\theta. \]
Here \(k=u_0/2\nu\).
Introducing these expressions into formula (1) and taking into account that \(dS=a\,d\theta\), we find \(dF_a\). Substituting again the expressions for \(v_r\) and \(v_\theta\) for the cylinder into formula (3), taking into account that \(dS'=R\,d\theta\), we obtain \(dF_m\). As before, equating \(dF_a\) and \(dF_m\), and from the resulting identity finding the coefficients with the adopted boundary conditions, we then have
\[ C_0=\frac{4u_0}{\varkappa-2\ln z};\qquad A_0=\frac{4\nu}{\varkappa-2\ln z};\qquad A_1=-\frac{u_0}{\varkappa-2\ln z}\omega a^2; \]
\[ \omega=\frac{2\lambda/R+\delta/2}{28\lambda/R+\delta/2+\lambda/R};\qquad \varkappa=(8\lambda/R+1)\omega. \]
Introducing these expressions into formulas (8); then putting \(C=u_0(\varkappa-2\ln z)^{-1}\), we finally find
\[ \begin{aligned} v_r&=C\cos\theta\left(\omega a^2/r^2+\varkappa-2+2\ln r/a\right),\\ v_\theta&=C\sin\theta\left(\omega a^2/r^2-\varkappa-2\ln r/a\right). \end{aligned} \tag{9} \]
These relations are valid for distances from the cylinder greater than \(\beta\lambda\). In the limiting case, when \(\lambda/R\to 0\), we have \(\varkappa=1\), \(\omega=1\), and then the expressions for \(C_0, A_0\), and \(A_1\) pass into the usual relations written above.
3. Efficiency of deposition of aerosol particles on a cylindrical obstacle
Consider the deposition of aerosol particles from a flow onto a cylinder arranged perpendicular to the flow, the particle radius \(r_{\mathrm{ch}}\) being smaller than the cylinder radius \(R\), but under the condition that \(R\) is close to \(\lambda\). Let the particle move along a streamline. From formulas (9) we easily find the stream function
\[ \Psi=\frac{u_0\sin\theta}{\varkappa-2\ln z} \left[(\varkappa-2)r+\omega\frac{a^2}{r}+2r\ln\frac{r}{a}\right]. \tag{10} \]
Following Langmuir\({}^{(5)}\), we approximately find the efficiency \(\eta\) of deposition due to contact and diffusion. Let \(x_0\) be the thickness of the layer at \(\theta=\pi/2\), from which the particles are removed from the flow as a result of one of the two mechanisms. In the case of contact \(x_0=r_{\mathrm{ch}}\). Then, by (6), from (9) and (10) we have, per unit length of cylinder,
\[ \eta=\frac{2}{u_0\cdot 2R} \int_a^{a+r_{\mathrm{ch}}} \left.\frac{d\Psi}{dr}\right|_{\theta=\pi/2}dr = -\frac{1}{u_0R} \int_a^{a+r_{\mathrm{ch}}} \left.v_\theta\right|_{\theta=\pi/2}dr = \]
\[ =\frac{1+\beta\lambda/R}{\varkappa-2\ln z} \left[ (\varkappa-2)q+\frac{\omega}{q}+2q\ln q-(\varkappa+\omega-2) \right], \tag{11} \]
where \(q=1+r_{\mathrm{ch}}/a\).
We find the deposition efficiency in diffusion by introducing the absolute displacement \(\bar{x}+\beta\lambda\) of a particle in Brownian motion over the time \(t\):
\[ \beta\lambda+\bar{x}=\left(\frac{4}{\pi}Dt\right)^{1/2}. \tag{12} \]
We determine this time by considering the motion of a particle near the cylinder between the points \(\theta_1=\frac{1}{6}\pi\) and \(\theta_2=\frac{5}{6}\pi\) along the streamline passing at \(\theta_0=\pi/2\) at a distance \(x_0+\beta\lambda\) from the surface of the cylinder. Integrating the expression for \(v_\theta\), we find:
\[ t=\frac{\bar{r}}{v_{\theta_0}}\ln\left(\operatorname{tg}\frac{\pi}{12}\bigg/\operatorname{tg}\frac{5}{12}\pi\right), \tag{13} \]
where \(\bar{r}=\bar{x}+a\). We define the quantity \(\bar{x}\) as the mean distance of the particle from the cylinder of radius \(a\) as it moves along the streamline from \(\theta_1\) to \(\theta_2\), if at \(\theta_0\) we set \(x=x_0\) and \(r_0=x_0+a\). Averaging \(\Psi\) over \(\theta\), we find the relation between \(x_0\) and \(\bar{x}\) in the form
\[ \frac{(\chi-2)r_0+\omega a^2/r_0+2r_0\ln r_0/a} {(\chi-2)\bar{r}+\omega a^2/\bar{r}+2\bar{r}\ln \bar{r}/a} =\overline{\sin\theta}, \tag{14} \]
we eliminate \(\bar{x}\) and \(t\) from equations (12), (13), and (14) and compute \(x_0\) for the specified values of \(R\), \(r_{\mathrm{ч}}\), \(u_0\), and \(\beta\).
As an example we take: \(R=2\cdot10^{-5}\) cm; \(r_{\mathrm{ч}}=2\cdot10^{-6}\) cm; \(D=3.8\cdot10^{-5}\) cm\(^2\)/sec and \(u_0=2\) cm/sec. Assuming, on the basis of the preceding, \(\beta=0.42\), we find \(\bar{x}=1.18x_0\); \(x_0=3.39(R+\beta\lambda)\); \(\chi-2\ln z=18\); \(\chi=2.35\); \(\omega=0.63\). The deposition efficiency due to diffusion is obtained in this case as \(\eta_{\mathrm{д}}=0.75\); for the contact mechanism we find \(\eta_{\mathrm{к}}=0.35\).
Institute of Physical Chemistry
Academy of Sciences of the USSR
Received
27 VII 1963
REFERENCES
- H. S. Tsien, J. Aeron. Sci., 13, No. 12 (1946); collection Gas Dynamics, IL, 1950.
- Н. А. Fuchs, Zs. phys. Chem., A, 171, 199 (1934).
- P. S. Epstein, Phys. Rev., 23, 710 (1924); collection Gas Dynamics, IL, 1950.
- G. Lamb, Hydrodynamics, Ch. XI, §§ 335, 343, 1947.
- I. Langmuir, The Collected Works, 10, 1961, p. 354.