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APhO 2010, Теория — Задача 3. Electron and Gas Bubbles in Liquids

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Theoretical Competition, 25 April 2010 — Theoretical Question 3: Electron and Gas Bubbles in Liquids · 25 април 2010 г. · 10 т.

Условие

This question deals with physics of two bubble-in-liquid systems. It has two parts:

Part A. An electron bubble in liquid helium

Part B. Single gas bubble in liquid

Part A. An Electron Bubble in Liquid Helium

When an electron is planted inside liquid helium, it can repel atoms of liquid helium and form what is called an electron bubble. The bubble contains nothing but the electron itself. We shall be interested mainly in its size and stability.

We use $\Delta f$ to denote the uncertainty of a quantity $f$ . The components of an electron’s position vector $\vec{q} = (x, y, z)$ and momentum vector $\vec{p} = (p_x, p_y, p_z)$ must obey Heisenberg’s uncertainty relations $\Delta q_\alpha \Delta p_\alpha \geq \hbar/2$ , where $\hbar$ is the Planck constant divided by $2\pi$ and $\alpha = x, y, z$ .

We shall assume the electron bubble to be isotropic and its interface with liquid helium is a sharp spherical surface. The liquid is kept at a constant temperature very close to $0\ \mathrm{K}$ with its surface tension $\sigma$ given by $3.75 \times 10^{-4}\ \mathrm{N \cdot m^{-1}}$ and its electrostatic responses to the electron bubble may be neglected.

Consider an electron bubble in liquid helium with an equilibrium radius $R$ . The electron, of mass $m$ , moves freely inside the bubble with kinetic energy $E_{\mathrm{k}}$ and exerts pressure $P_{\mathrm{e}}$ on the inner side of the bubble-liquid interface. The pressure exerted by liquid helium on the outer side of the interface is $P_{\mathrm{He}}$ .

(a) Find a relation between $P_{\mathrm{He}}$ , $P_{\mathrm{e}}$ , and $\sigma$ . [0.4 point]

Find a relation between $E_{\mathrm{k}}$ and $P_{\mathrm{e}}$ . [1.0 point] [1,4 т.]

(b) Denote by $E_0$ the smallest possible value of $E_{\mathrm{k}}$ consistent with Heisenberg’s uncertainty relations when the electron is inside the bubble of radius $R$ . Estimate $E_0$ as a function of $R$ . [0,8 т.]

(c) Let $R_{\mathrm{e}}$ be the equilibrium radius of the bubble when $E_{\mathrm{k}} = E_0$ and $P_{\mathrm{He}} = 0$ .

Obtain an expression for $R_{\mathrm{e}}$ and calculate its value. [0,6 т.]

(d) Find a condition that $R$ and $P_{\mathrm{He}}$ must satisfy if the equilibrium at radius $R$ is to be locally stable under constant $P_{\mathrm{He}}$ . Note that $P_{\mathrm{He}}$ can be negative. [0,6 т.]

(e) There exists a threshold pressure $P_{\mathrm{th}}$ such that equilibrium is not possible for the electron bubble when $P_{\mathrm{He}}$ is less than $P_{\mathrm{th}}$ . Find an expression for $P_{\mathrm{th}}$ . [0,6 т.]

Part B. Single Gas Bubble in Liquid — Collapsing and Radiation

In this part of the problem, we consider a normal liquid, such as water.

When a gas bubble in a liquid is driven by an oscillating pressure, it can show dramatic responses. For example, following a large expansion, it can collapse rapidly to a small radius and, near the end of the collapse, emit light almost instantly. In this phenomenon, called single-bubble sonoluminescence, the gas bubble undergoes cyclic motions which typically consist of three stages: expansion, collapse, and multiple after-bounces. In the following we shall focus mainly on the collapsing stage.

We assume that, at all times, the bubble considered is spherical and its center remains stationary in the liquid. See Fig 1. The pressure, temperature, and density are always uniform inside the bubble as its size diminishes. The liquid containing the bubble is assumed to be isotropic, nonviscous, incompressible, and very much larger in extent than the bubble. All effects due to gravity and surface tension are neglected so that pressures on both sides of the bubble-liquid interface are always equal.

● Radial motion of the bubble-liquid interface

As the bubble’s radius $R = R(t)$ changes with time $t$ , the bubble-liquid interface will move with radial velocity $\dot{R} \equiv dR/dt$ . It follows from the equation of continuity of incompressible fluids that the liquid’s radial velocity $\dot{r} \equiv dr/dt$ at distance $r$ from the center of the bubble is related to the rate of change of the bubble’s volume $V$ by

\frac{dV}{dt} = 4\pi R^2\dot{R} = 4\pi r^2\dot{r}. \qquad (1)

This implies that the total kinetic energy $E_{\mathrm{k}}$ of the liquid with mass density $\rho_0$ is

E_{\mathrm{k}} = \frac{1}{2}\int_R^{r_0} \rho_0\,(4\pi r^2 dr)\dot{r}^2 = 2\pi\rho_0 R^4\dot{R}^2\int_R^{r_0}\frac{1}{r^2}dr = 2\pi\rho_0 R^4\dot{R}^2\left(\frac{1}{R} - \frac{1}{r_0}\right) \qquad (2)

where $r_0$ is the radius of the outer surface of the liquid.

A spherical gas bubble of radius R at the centre of a large shaded liquid sphere of radius r_0. Inside the bubble: pressure P and temperature T, with a small arrow on the bubble surface. The liquid is labelled with density ρ_0 and temperature T_0; a line from the centre marks a distance r in the liquid; an arrow from outside labelled P_0 presses on the outer surface. — Fig. 1

(f) Assume the ambient pressure $P_0$ acting on the outer surface $r = r_0$ of the liquid is constant.

Let $P = P(R)$ be the gas pressure when the radius of the bubble is $R$ .

Find the amount of work $dW$ done on the liquid when the radius of the bubble changes from $R$ to $R + dR$ . Use $P_0$ and $P$ to express $dW$ . [0.4 point]

The work $dW$ must be equal to the corresponding change in the total kinetic energy of the liquid. In the limit $r_0 \to \infty$ , it follows that we have Bernoulli’s equation in the form

\tfrac{1}{2}\rho_0\, d\left(R^{\mathrm{m}}\dot{R}^2\right) = (P - P_0)R^{\mathrm{n}}dR. \qquad (3)

Find the exponents m and n in Eq. (3). Use dimensional arguments if necessary. [0.4 point] [0,8 т.]

● Collapsing of the gas bubble

From here on, we consider only the collapsing stage of the bubble. The mass density of the liquid is $\rho_0 = 1.0 \times 10^3\ \mathrm{kg \cdot m^{-3}}$ , the temperature $T_0$ of the liquid is $300\ \mathrm{K}$ and the ambient pressure $P_0$ is $1.01 \times 10^5\ \mathrm{Pa}$ . We assume that $\rho_0$ , $T_0$ , and $P_0$ remain constant at all times and the bubble collapses adiabatically without any exchange of mass across the bubble-liquid interface.

The bubble considered is filled with an ideal gas. The ratio of specific heat at constant pressure to that at constant volume for the gas is $\gamma = 5/3$ . When under temperature $T_0$ and pressure $P_0$ , the equilibrium radius of the bubble is $R_0 = 5.00\ \mathrm{\mu m}$ .

Now, this bubble begins its collapsing stage at time $t = 0$ with $R(0) = R_{\mathrm{i}} = 7R_0$ , $\dot{R}(0) = 0$ , and the gas temperature $T_{\mathrm{i}} = T_0$ . Note that, because of the bubble’s expansion in the preceding stage, $R_{\mathrm{i}}$ is considerably larger than $R_0$ and this is necessary if sonoluminescence is to occur.

(g) Express the pressure $P \equiv P(R)$ and temperature $T \equiv T(R)$ of the ideal gas in the bubble as a function of $R$ during the collapsing stage, assuming quasi-equilibrium conditions hold. [0,6 т.]

(h) Let $\beta \equiv R/R_{\mathrm{i}}$ and $\dot{\beta} = d\beta/dt$ . Eq. (3) implies a conservation law which takes the following form

\tfrac{1}{2}\rho_0\,\dot{\beta}^2 + U(\beta) = 0. \qquad (4)

Let $P_{\mathrm{i}} \equiv P(R_{\mathrm{i}})$ be the gas pressure of the bubble when $R = R_{\mathrm{i}}$ . If we introduce the ratio $Q \equiv P_{\mathrm{i}}/[(\gamma - 1)P_0]$ , the function $U(\beta)$ may be expressed as

U(\beta) = \mu\beta^{-5}[Q(1 - \beta^2) - \beta^2(1 - \beta^3)]. \qquad (5)

Find the coefficient $\mu$ in terms of $R_{\mathrm{i}}$ and $P_0$ . [0,6 т.]

(i) Let $R_{\mathrm{m}}$ be the minimum radius of the bubble during the collapsing stage and define $\beta_{\mathrm{m}} \equiv R_{\mathrm{m}}/R_{\mathrm{i}}$ . For $Q \ll 1$ , we have $\beta_{\mathrm{m}} \approx C_{\mathrm{m}}\sqrt{Q}$ .

Find the constant $C_{\mathrm{m}}$ . [0.4 point]

Evaluate $R_{\mathrm{m}}$ for $R_{\mathrm{i}} = 7R_0$ . [0.3 point]

Evaluate the temperature $T_{\mathrm{m}}$ of the gas at $\beta = \beta_{\mathrm{m}}$ . [0.3 point] [1 т.]

(j) Assume $R_{\mathrm{i}} = 7R_0$ . Let $\beta_u$ be the value of $\beta$ at which the dimensionless radial speed $u \equiv |\dot{\beta}|$ reaches its maximum value. The gas temperature rises rapidly for values of $\beta$ near $\beta_u$ .

Give an expression and then estimate the value of $\beta_u$ . [0.6 point]

Let $\bar{u}$ be the value of $u$ at $\beta = \bar{\beta} \equiv (\beta_{\mathrm{m}} + \beta_u)/2$ . Evaluate $\bar{u}$ . [0.4 point]

Give an expression and then estimate the duration $\Delta t_{\mathrm{m}}$ of time needed for $\beta$ to diminish from $\beta_u$ to the minimum value $\beta_{\mathrm{m}}$ . [0.6 point] [1,6 т.]

● Sonoluminescence of the collapsing bubble

Consider the bubble to be a surface black-body radiator of constant emissivity $a$ so that the effective Stefan-Boltzmann’s constant $\sigma_{\mathrm{eff}} = a\sigma_{\mathrm{SB}}$ . If the collapsing stage is to be approximated as adiabatic, the emissivity must be small enough so that the power radiated by the bubble at $\beta = \bar{\beta}$ is no more than a fraction, say 20 %, of the power $\dot{E}$ supplied to it by the driving liquid pressure.

(k) Find the power $\dot{E}$ supplied to the bubble as a function of $\beta$ . [0.6 point]

Give an expression and then estimate the value for an upper bound of $a$ . [0.8 point] [1,4 т.]

Appendix

$\dfrac{d}{dx}x^n = nx^{n-1}$
Electron mass $m = 9.11 \times 10^{-31}\ \mathrm{kg}$
Planck constant $h = 2\pi\,\hbar = 2\pi \times 1.055 \times 10^{-34}\ \mathrm{J \cdot s}$
Stefan-Boltzmann’s constant $\sigma_{\mathrm{SB}} = 5.67 \times 10^{-8}\ \mathrm{W \cdot m^{-2} \cdot K^{-4}}$

END

Отговори

Решение

Part A. An Electron Bubble in Liquid Helium

(a) Consider a half of the spherical interface (see Fig. A1 below). The condition for its static equilibrium implies that the total force acting on it must be zero. This implies

\pi R^2(P_{\mathrm{e}} - P_{\mathrm{He}}) = 2\pi R\sigma \qquad (a-1)

which leads to

A half of the spherical interface: a semicircular arc over a dashed diameter labelled 2R. At the top a downward arrow labelled P_He pushes inward and an upward arrow from inside labelled P_e pushes outward; at both ends of the diameter downward arrows labelled σ represent the surface tension. — Fig. A1

P_{\mathrm{e}} = P_{\mathrm{He}} + \frac{2\sigma}{R} \qquad ((a-2)*)

An equation marked with an asterisk gives key answers to the problem.

According to the de Broglie relation $p = h/\lambda \propto 1/R$ , the non-relativistic kinetic energy $E_{\mathrm{k}}$ is inversely proportional to $R^2$ , i.e.

E_{\mathrm{k}} = \frac{p^2}{2m} = \frac{\mathrm{const.}}{R^2}. \qquad (a-3)

By the work- energy theorem, we have

-P_{\mathrm{e}}dV = dE_{\mathrm{k}} = (-2)\frac{\mathrm{const.}}{R^3}dR = -\frac{2}{R}E_{\mathrm{k}}dR \qquad (a-4)

Thus,

-P_e(4\pi R^2 dR) = -\frac{2}{R}E_{\mathrm{k}}\,dR \qquad (a-5)

P_{\mathrm{e}} = \frac{1}{2\pi R^3}E_{\mathrm{k}} \qquad ((a-6)*)

[Alternative]

The state of an electron confined in the bubble corresponds to standing waves which vanish on the interface. According to Part B of Question 1, these are equivalent to the superposition of two travelling waves moving in opposite directions and continually being reflected at the interface. They give rise to pressure on the interface and the relation between the non-relativistic kinetic energy $E_{\mathrm{k}}$ and the pressure $P_{\mathrm{e}}$ for an electron inside a bubble is similar to that obtained from the kinetic theory of gases. Thus we have

P_{\mathrm{e}} = \frac{2}{3}\frac{E_{\mathrm{k}}}{V} = \frac{2}{4\pi R^3}E_{\mathrm{k}} = \frac{1}{2\pi R^3}E_{\mathrm{k}} \qquad ((a-7)*)

(b) Let $\hbar = h/(2\pi)$ . From the uncertainty relations, we have

\Delta x\,\Delta p_x \geq \tfrac{1}{2}\hbar, \qquad \Delta y\,\Delta p_y \geq \tfrac{1}{2}\hbar, \qquad \Delta z\,\Delta p_z \geq \tfrac{1}{2}\hbar. \qquad (b-1)

From symmetry considerations implied by isotropy, we have

\bar{x} = \bar{y} = \bar{z} = 0, \quad \overline{p_x} = \overline{p_y} = \overline{p_z} = 0 \qquad (b-2)

(\Delta x)^2 = \overline{x^2} - \bar{x}^2 = \overline{x^2} = (\Delta y)^2 = (\Delta z)^2, \quad (\Delta p_x)^2 = \overline{p_x{}^2} = (\Delta p_y)^2 = (\Delta p_z)^2. \qquad (b-3)

where $\bar{f}$ denotes the mean value of the quantity $f$ . Therefore, we have

3(\Delta \mathrm{x})^2 = (\Delta \mathrm{x})^2 + (\Delta \mathrm{y})^2 + (\Delta \mathrm{z})^2 = \overline{\mathrm{x}^2} + \overline{\mathrm{y}^2} + \overline{\mathrm{z}^2} = \overline{\mathrm{r}^2}. \qquad (b-4)

3(\Delta p_x)^2 = (\Delta p_x)^2 + (\Delta p_y)^2 + (\Delta p_z)^2 = \overline{p_x^2} + \overline{p_y^2} + \overline{p_z^2} = \overline{p^2}. \qquad (b-5)

Thus we obtain (cf. 18 $^{\mathrm{th}}$ IPhO)

\overline{r^2}\,\overline{p^2} = 9(\Delta x)^2(\Delta p_x)^2 \geq \tfrac{9}{4}\hbar^2 \qquad (b-6)

and the kinetic energy must satisfy the following inequality:

E_{\mathrm{k}} = \frac{\overline{p^2}}{2m} \geq \frac{1}{2m}\left(\frac{9\hbar^2}{4}\right)\frac{1}{\overline{r^2}} \qquad (b-7)

The smallest possible kinetic energy $E_0$ of the electron consistent with the uncertainty relations is thus obtained if the mean-squared-radius $\overline{r^2}$ is set equal to its largest possible value of $R^2$ . This gives

E_{\mathrm{k}} \geq E_0 = \frac{9\hbar^2}{8mR^2} \qquad ((b-8)*)

P_{\mathrm{e}} = \frac{1}{2\pi R^3}E_0 = \frac{9\hbar^2}{16m\pi R^5} = \frac{2\sigma}{R} + P_{\mathrm{He}} \qquad (c-1)

For $P_{\mathrm{He}} = 0$ , this gives the following equilibrium radius of the electron bubble:

\begin{aligned} R_{\mathrm{e}} &= \left(\frac{9\hbar^2}{32\pi m\sigma}\right)^{\frac{1}{4}} = \left(\frac{9 \times (1.055 \times 10^{-34})^2}{32\pi \times 9.11 \times 10^{-31} \times 3.75 \times 10^{-4}}\right)^{\frac{1}{4}} \\ &= (2.91674 \times 10^{-36})^{1/4}\ \mathrm{m} \cong 1.31\ \mathrm{nm} \end{aligned} \qquad ((c-2)*)

It might be of some interest to note that, from Eq. (c-1), the corresponding minimum kinetic energy is

E_0 = 4\pi R_0^2\sigma = 3h\left(\frac{\sigma}{2\pi m}\right)^{1/2} = 0.100\ \mathrm{eV} \qquad (c-3)

(d) The condition for stable local equilibrium of the electron bubble at radius $R$ is that when $R$ is increased by a small amount $dR > 0$ , the inward force pushing on the interface must be greater than the outward force so as to decrease the radius. Thus, from Eq. (c-1), we obtain

\frac{2\sigma}{(R + dR)} + P_{\mathrm{He}} > \frac{9\hbar^2}{16m\pi(R + dR)^5} \qquad (d-1)

By keeping only terms linear in $dR$ after both sides of the inequality are expanded as a power series and making use of Eq. (c-1) to eliminate $P_{\mathrm{He}}$ , we obtain

(-1)\frac{2\sigma}{R^2} > (-5)\frac{9\hbar^2}{16m\pi R^6} \qquad (d-2)

Note that the same inequality is obtained if we consider a small change $dR < 0$ . Using Eq. (c-1), we may express Eq. (d-2) in terms of $P_{\mathrm{He}}$ as

\frac{2\sigma}{R} < 5\left(P_{\mathrm{He}} + \frac{2\sigma}{R}\right) \qquad (d-3)

or equivalently,

P_{\mathrm{He}} > -\left(\frac{8\sigma}{5R}\right) \qquad ((d-4)*)

(e) From Eqs. (a-2), (a-6), and (b-8), we have

\frac{2\sigma}{R} + P_{\mathrm{He}} = P_{\mathrm{e}} = \frac{E_{\mathrm{k}}}{2\pi R^3} \geq \frac{E_0}{2\pi R^3} = \frac{9\hbar^2}{16m\pi R^5} \qquad (e-1)

or equivalently,

P_{\mathrm{He}} \geq \frac{9\hbar^2}{16m\pi R^5} - \frac{2\sigma}{R} \qquad (e-2)

The minimum of the right-hand side of the inequality occurs when its derivative vanishes, i.e.

\frac{-45\hbar^2}{16m\pi R^6} + \frac{2\sigma}{R^2} = 0 \qquad (e-3)

R = R_{\mathrm{th}} = \left(\frac{45\hbar^2}{32m\pi\sigma}\right)^{1/4} \qquad (e-4)

Substituting the last result back into Eq. (e-2), we obtain

P_{\mathrm{He}} \geq P_{\mathrm{th}} \equiv \frac{9\hbar^2}{16m\pi R_{\mathrm{th}}^5} - \frac{2\sigma}{R_{\mathrm{th}}} = \left(\frac{1}{5} - 1\right)\frac{2\sigma}{R_{\mathrm{th}}} = \frac{-8\sigma}{5R_{\mathrm{th}}} = \frac{-16\sigma}{5}\left(\frac{2m\pi\sigma}{45\hbar^2}\right)^{1/4} \qquad ((e-5)*)

For $\mathrm{P_{He}} < \mathrm{P_{th}}$ , no equilibrium is possible for the electron bubble.

Part B. Single Gas Bubble in Liquid — Collapsing and Radiation

(f) When the bubble’s radius $R$ changes by $dR$ , the volume of the liquid displaced by the interface is $dV = 4\pi R^2 dR$ . But the total volume of the incompressible liquid cannot change, so the change of the volume at the outer surface of the liquid must also be $dV$ . Thus the amount of work done on the liquid is

dW = PdV - P_0 dV = (P - P_0)4\pi R^2 dR \qquad ((f-1)*)

From Eq.(2), the change in total kinetic energy of the liquid is, in the limit $r_0 \to \infty$ ,

dE_{\mathrm{k}} = d\left[2\pi\rho_0 R^4\dot{R}^2\left(\frac{1}{R} - \frac{1}{r_0}\right)\right] = 2\pi\rho_0 d\left(R^3\dot{R}^2\right). \qquad (f-2)

Since $dE_{\mathrm{k}} = dW$ , we obtain

\frac{1}{2}\rho_0 d\left(R^{\mathrm{m}}\dot{R}^2\right) = (P - P_0)R^{\mathrm{n}}dR. \qquad (f-3)

with

\mathrm{m} = 3, \qquad \mathrm{n} = 2. \qquad ((f-4)*)

(g) The initial gas temperature is $T_0$ . According to the ideal gas law, the initial gas pressure $P_{\mathrm{i}} = P(R_{\mathrm{i}})$ is thus given by

P_{\mathrm{i}}R_{\mathrm{i}}^3 = P_0 R_0^3. \qquad ((g-1)*)

Since the process is adiabatic, the radial dependence of the gas pressure $P$ is

P \equiv P(R) = \left(\frac{R_{\mathrm{i}}^3}{R^3}\right)^{\gamma} P_{\mathrm{i}} = \left(\frac{R_{\mathrm{i}}}{R}\right)^5 P_{\mathrm{i}} = \left(\frac{R_{\mathrm{i}}}{R}\right)^5 P_0\left(\frac{R_0}{R_{\mathrm{i}}}\right)^3. \qquad ((g-2)*)

and the temperature $T$ corresponding to the radius $R$ is given by

T \equiv T(R) = \left(\frac{R_{\mathrm{i}}^3}{R^3}\right)^{(\gamma - 1)} T_0 = \left(\frac{R_{\mathrm{i}}}{R}\right)^2 T_0. \qquad ((g-3)*)

(h) From Eqs. (3) and (g-2), we have

\frac{1}{2R^2}\frac{d}{dR}\left(R^3\dot{R}^2\right) = \frac{P - P_0}{\rho_0} = \frac{P_0}{\rho_0}\left[\frac{P_i}{P_0}\left(\frac{R_i}{R}\right)^{3\gamma} - 1\right] \qquad (h-1)

In terms of $\beta = R/R_{\mathrm{i}}$ and $\dot{\beta} = \dot{R}/R_{\mathrm{i}}$ , the last equation may be rewritten as

\frac{1}{2\beta^2}\frac{d}{d\beta}\left(\beta^3\dot{\beta}^2\right) = \frac{P_0}{\rho_0 R_{\mathrm{i}}^2}\left(\frac{P_{\mathrm{i}}}{P_0}\beta^{-3\gamma} - 1\right). \qquad (h-2)

This may be integrated to give

\begin{aligned} \frac{1}{2}\beta^3\dot{\beta}^2 &= \frac{P_0}{\rho_0}\int_1^{\beta}\left(\frac{P_{\mathrm{i}}}{P_0}y^{2 - 3\gamma} - y^2\right)dy = \frac{P_0}{\rho_0}\left[\left(\frac{P_{\mathrm{i}}}{P_0}\right)\frac{\beta^{3 - 3\gamma} - 1}{3(1 - \gamma)} - \frac{\beta^3 - 1}{3}\right] \\ &= \frac{P_0}{3\rho_0\beta^2}\left[-\left(\frac{P_{\mathrm{i}}}{P_0}\right)\frac{1}{(\gamma - 1)}(1 - \beta^2) + \beta^2(1 - \beta^3)\right]. \end{aligned} \qquad (h-3)

Since $Q \equiv P_{\mathrm{i}}/[(\gamma - 1)P_0]$ and $\gamma = 5/3$ , the last equation leads to

\begin{aligned} \frac{1}{2}\rho_0\dot{\beta}^2 = -U(\beta) &\equiv \frac{-P_0}{3R_{\mathrm{i}}^2\beta^5}[Q(1 - \beta^2) - \beta^2(1 - \beta^3)] \\ &= \frac{-P_0(1 - \beta^2)}{3R_{\mathrm{i}}^2\beta^5}\left[Q - \frac{\beta^2(1 - \beta^3)}{(1 - \beta^2)}\right]. \end{aligned} \qquad (h-4)

Thus we obtain

\mu = \frac{P_0}{3R_{\mathrm{i}}^2}. \qquad ((h-5)*)

(i) The radius of the bubble reaches its minimum value when $\dot{R} = R_{\mathrm{i}}\dot{\beta} = 0$ . Thus, from Eq. (h-4), we obtain

Q = \frac{\beta_{\mathrm{m}}^2}{1 - \beta_{\mathrm{m}}^2}(1 - \beta_{\mathrm{m}}^3) = \beta_{\mathrm{m}}^2\left(1 + \frac{\beta_{\mathrm{m}}^2}{1 + \beta_{\mathrm{m}}}\right). \qquad (i-1)

The last equality shows that $\beta_{\mathrm{m}}$ must be very small in order that $Q \ll 1$ . Thus

Q \approx \beta_{\mathrm{m}}^2, \text{ or } \beta_{\mathrm{m}} \approx \sqrt{Q}, \text{ i.e. } C_{\mathrm{m}} = 1 \qquad ((i-2)*)

For $R_{\mathrm{i}} = 7R_0 = 35.0\ \mathrm{\mu m}$ , we have, from Eq. (g-1),

Q = \frac{P_{\mathrm{i}}}{P_0(\gamma - 1)} = \frac{1}{(\gamma - 1)}\left(\frac{R_0}{R_{\mathrm{i}}}\right)^3 = \frac{3}{2}\left(\frac{1}{7}\right)^3 = 0.00437. \qquad (i-3)

Therefore

\beta_{\mathrm{m}} = \sqrt{Q} = 0.0661, \qquad (i-4)

R_{\mathrm{m}} = \beta_{\mathrm{m}}R_i = \sqrt{Q}R_i = 0.0661 \times 35\ \mathrm{\mu m} = 2.31\ \mathrm{\mu m}, \qquad ((i-5)*)

and from Eq. (g-3), the corresponding temperature $T_{\mathrm{m}}$ is

T_{\mathrm{m}} = \left(\frac{1}{\beta_{\mathrm{m}}}\right)^2 T_0 = \left(\frac{1}{0.0661}\right)^2 \times 300\ \mathrm{K} = 6.86 \times 10^4\ \mathrm{K}. \qquad ((i-6)*)

(j) From Eq. (h-4), the maximum value of the radial speed $u \equiv |\dot{\beta}|$ occurs at $\beta = \beta_u$ where $-U(\beta)$ is also at its maximum, i.e. the derivative of $U(\beta)$ with respect to $\beta$ must vanish at $\beta = \beta_u$ . Since

U(\beta) = \frac{P_0}{3R_{\mathrm{i}}^2}\left[Q\left(\frac{1}{\beta^5} - \frac{1}{\beta^3}\right) - \left(\frac{1}{\beta^3} - 1\right)\right], \qquad (j-1)

we have

\left.\frac{dU}{d\beta}\right|_{\beta = \beta_u} = \frac{-P_0}{3R_{\mathrm{i}}^2\beta_u}\left[Q\left(\frac{5}{\beta_u^5} - \frac{3}{\beta_u^3}\right) - \frac{3}{\beta_u^3}\right] = 0. \qquad (j-2)

Thus

Q = \frac{3\beta_u^2}{(5 - 3\beta_u^2)}, \text{ or } \beta_u^2 = \frac{5}{3}\left(\frac{Q}{1 + Q}\right). \qquad (j-3)

which implies

\beta_u = \sqrt{\frac{5}{3}\left(\frac{Q}{1 + Q}\right)} = 0.0852. \qquad ((j-4)*)

The radius midway between $\beta_u$ (corresponding to maximum speed) and $\beta_{\mathrm{m}}$ (corresponding to zero speed) is given by

\bar{\beta} \equiv \frac{1}{2}(\beta_{\mathrm{m}} + \beta_u) \cong \frac{1}{2}(0.0661 + 0.0852) = 0.0757 \qquad (j-5)

From Eq. (h-4), the dimensionless radial speed at radius $\bar{\beta}$ is

\begin{aligned} \bar{u} = -\dot{\beta}(\bar{\beta}) &= \sqrt{\frac{-2}{\rho_0}U(\bar{\beta})} \\ &= \sqrt{\frac{2P_0(1 - \bar{\beta}^2)}{3\rho_0 R_{\mathrm{i}}^2\bar{\beta}^3}\left[1 - \frac{Q}{\bar{\beta}^2} + \frac{\bar{\beta}^2}{(1 + \bar{\beta})}\right]} = 5.52 \times 10^6. \end{aligned} \qquad ((j-6)*)

Thus an estimate of the duration $\Delta t_{\mathrm{m}}$ for the radius of the bubble to diminish from $\beta_u$ to the minimum value $\beta_{\mathrm{m}}$ is

\Delta t_{\mathrm{m}} = \frac{(\beta_u - \beta_{\mathrm{m}})}{\bar{u}} = \frac{(0.0852 - 0.0661)}{5.52 \times 10^6} = 3.45 \times 10^{-9}\ \mathrm{s}. \qquad ((j-7)*)

(k) Suppose the bubble is a surface radiator with emissivity $a$ . By making use of Eq. (g-3), the radiant power $W_{\mathrm{r}}$ of the bubble at temperature $T$ can be written as a function of $\beta$ , i.e.

W_{\mathrm{r}} = a(\sigma_{\mathrm{SB}}T^4)4\pi R^2 = 4\pi R_{\mathrm{i}}^2 a\sigma_{\mathrm{SB}}T_0^4\frac{1}{\beta^6}, \qquad (k-1)

where $\sigma_{\mathrm{SB}}$ is the Stefan-Boltzmann constant. The power supplied to the bubble is

\dot{E} = -P\frac{dV}{dt} = -P_{\mathrm{i}}\left(\frac{V_{\mathrm{i}}}{V}\right)^{\gamma}\frac{dV}{dt} = -4\pi R_{\mathrm{i}}^3 P_{\mathrm{i}}\frac{\dot{\beta}}{\beta^3} \qquad ((k-2)*)

The assumption of an adiabatic collapsing of the bubble is deemed reasonable when the radiant power is less than 20 % of the power supplied to the bubble at $\beta = \bar{\beta}$ . Thus we have

4\pi R_{\mathrm{i}}^2 a\sigma_{\mathrm{SB}}T_0^4\frac{1}{\bar{\beta}^6} \leq 4\pi R_{\mathrm{i}}^3 P_{\mathrm{i}}\frac{\bar{u}}{\bar{\beta}^3} \times 20\ \% \qquad (k-3)

a \leq \frac{P_{\mathrm{i}}R_{\mathrm{i}}}{5\sigma_{\mathrm{SB}}T_0^4}\bar{\beta}^3\bar{u} = \frac{P_0 R_{\mathrm{i}}}{5\sigma_{\mathrm{SB}}T_0^4}\left(\frac{R_0}{R_{\mathrm{i}}}\right)^3\bar{\beta}^3\bar{u} = 0.0107 \qquad ((k-4)*)

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf