Съдържание

IOM 2017, Теория — Задача 1. CARBONATED WATER

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IOM 2017

The 2nd International Olympiad of Metropolises — physics, theoretical part, 2017 · 25 т.

Внимание: Бележка към темата

Three whole native tasks with ten connected requests each. The archived radiography key contains two full proposed solutions with different flash/wavelength/angle values and scores35/20; both are retained, and question2/paper total points are unspecified. Printed equations, mass ranges, signs and notation defects remain literal with separate archive source notes.

Native source header and key scope

The 2nd International Olympiad of Metropolisesphysics — theoretical part — 2017.

The native question edition contains three problems: CARBONATED WATER, RADIOGRAPHY and Neutrino, each with ten connected numbered requests. Neutrino has three native titled parts. The original header has the concatenated word “Metropolisesphysics”. No examination date, time limit or whole-paper score is printed in this question packet.

The retained matching archived key repeats the questions, supplies full proposed solutions and grading tables, and contains two distinct complete radiography key variants. Carbonated water is graded25, radiography35 in the first table and20 in the second, and Neutrino40. Neither radiography table is designated as the sole scoring authority in these retained sources. Key pages16 and27 are blank. All38 original pages remain retained; the keys are labelled as archive solutions.

Условие

Suppose you have a regular 1-liter factory sealed bottle of carbonated water. You have turned the bottle cap slightly to unscrew it (so a hissing sound was heard) and screw the cap tightly again. Now you would observe bubbles of carbon dioxide (CO2) rising upwards – large, at first, and then smaller ones. Let us study the process of bubble surfacing.

One can easily see that the shape of a small bubble is much closer to spherical than that of a bigger one.

1) Estimate the maximal size of an immobile bubble such that the bubble shape approximates sphere with an accuracy of 10% or better. Water density is $\rho\approx1\ \mathrm{g/cm^3}$ , the surface tension $\sigma\approx0{,}07\ \mathrm{N/m}$ , and the free fall acceleration is taken to be $g\approx9{,}8\ \mathrm{m/s^2}$ . The numerical answer should be in mm.

Consider a bubble so small that it can be regarded as almost spherical. For instance, let the bubble initial diameter near the bottle bottom be $d_0=0{,}3\ \mathrm{mm}$ . [1 т.]

2) Figure out the bubble acceleration right after it has detached from the bottom. The CO2 density inside the bubble at the given water temperature is $\rho_v\approx0{,}002\ \mathrm{g/cm^3}$ . The numerical answer should be in m/s².

Drag force exerted on a bubble, when it is moving in water, is linearly proportional to its cross-sectional area, water density, and the bubble velocity squared. We assume the proportionality factor for this problem to be $\beta=0{,}2$ . [2 т.]

3) Suppose that the bubble volume remains constant. What is the terminal velocity the bubble can reach? Write down the equation and evaluate the numerical value (in m/s). [2 т.]

4) Estimate the time it takes the bubble to reach the terminal velocity after detachment, i.e. when its acceleration becomes much less than $g$ . Write down the equation and evaluate the numerical value (in seconds).

Actually, the density of CO2 molecules dissolved in a liquid is much higher than that in a gas bubble. Therefore, the dissolved gas diffuses into the bubble and its radius grows. It is reasonable to assume that the growth rate of the bubble volume due to the diffusion is proportional to the bubble surface area and to an excess of the dissolved gas density and inversely proportional to the thickness of the liquid layer through which the gas diffuses (called the «depletion layer»). The faster bubble is moving the thinner is the effective depletion layer (due to circulation of water surrounding the bubble). A kinetic theory calculation gives for the depletion layer thickness: $\delta=\mathrm{const}\cdot\sqrt{d/v}$ , where $v$ is the bubble velocity and $d$ is its diameter. [2 т.]

5) Suppose it takes time $T$ for a bubble to surface. Figure out the bubble diameter as a function of time $t$ , providing it has increased by a factor $k=3^{4/5}\approx2{,}4$ during the surfacing. The answer should be an equation for $d(t)$ in terms of $T$ , $t$ , $d_0$ , and $k=3^{4/5}$ . Neglect a change in the density of CO2 dissolved in water during the bubble surfacing. [4 т.]

6) Determine time dependence $v(t)$ of the bubble velocity (the answer should include a formula for $v(t)$ expressed in terms of the parameters listed above and the terminal velocity $v_0$ ). [2 т.]

7) Determine the law of motion of a bubble, i.e. time dependence of its elevation $h$ above the bottom (the answer should include a formula for $h(t)$ in terms of $T$ , $t$ , and $v_0$ ; take the same value $k=3^{4/5}$ ). [3 т.]

8) Suppose the height of water column, which the bubble traverses on its way upward, equals $H=30\ \mathrm{cm}$ . Evaluate the time of bubble surfacing (the answer should be the explicit formula including the parameters given in the problem and calculate the numerical value in seconds).

During bubble surfacing some heat is being released (the drag force does a work by increasing the kinetic energy of turbulent flow which eventually dissipates as heat) and at the same time some heat is being absorbed (due to CO2 evaporation from water into a bubble). [4 т.]

9) Assume that all the heat released during the surfacing is converted into heating the «column» of water which cross-section equals the average cross-section of a rising bubble. Using this assumption estimate (by the order of magnitude) the temperature increment of water in the «column». The specific heat capacity of water is $c_w\approx4200\ \mathrm{J/(kg\cdot K)}$ . The answer should be given in Kelvin. [2 т.]

10) Evaporation into bubbles of 1 mole of CO2 dissolved in water requires approximately 20 kJ of energy. Estimate (by the order of magnitude) the cooling of the water «column» (see 9)) due to this effect during the bubble rising and compare to the heating effect (see 9)). The answer should be given in Kelvin. What will the net result be? The answer should be either «+» (the temperature increases) or «–» (the temperature decreases). Assume that the pressure and temperature in the bottle change slightly and remain close to $p_0\approx120\ \mathrm{kPa}$ and $T_0\approx290\ \mathrm K$ . [3 т.]

Решение

PROPOSED SOLUTION AND ANSWERS

Bubble «oblateness» is caused by the difference of hydrostatic pressure above and below the bubble, while surface tension tries to keep the bubble shape spherical. Therefore, one obtains a reasonable estimate by assuming that the bubble shape remains spherical at the given accuracy if the force of surface tension exceeds the net force of hydrostatic pressure (i.e. buoyancy force for the bubble at rest) by one order of magnitude (i.e. approximately 10 times):

\sigma\cdot\pi d\ge10\rho g\frac{\pi d^3}{6}.

Hence $d\le\sqrt{3\sigma/(5\rho g)}\approx2\ \mathrm{mm}$ , so $d_{\max}\approx2\ \mathrm{mm}$ .

There is no drag force exerted on the bubble at zero velocity. Only the gravity and buoyancy forces act on it. However, using a «common» expression for the buoyancy force gives the absurd result: the acceleration of gas bubble must be about $500g$ ! Actually, at a non-zero acceleration the net force of hydrostatic pressure differs from the «static» value of buoyancy force. The simplest way to understand this is to take into account that bubble rising is equivalent to going down of the displaced amount of water due to gravity force. Obviously, the corresponding acceleration cannot exceed $g$ even if there is no drag. At the start, the bubble acceleration is close to the maximum, i.e. $a_0\approx g$ .

Note: an accurate evaluation of the acceleration of a body of density $\rho_v$ , which is surfacing in a liquid of density $\rho$ , yields: $a=\frac{\rho-\rho_v}{\rho+\rho_v}g$ . This indeed tends to $g$ when $\rho\gg\rho_v$ .

At the stationary regime, when the bubble velocity equals to its terminal value $v_0$ , the drag balances buoyancy force (which at zero acceleration equals its «static» value), and the gravity force exerted on the bubble can be neglected. Therefore,

\rho g\frac{\pi d_0^3}{6}=\beta\rho\frac{\pi d_0^2}{4}v_0^2,\qquad v_0=\sqrt{\frac{2gd_0}{3\beta}}\approx0{,}1\ \mathrm{m/s}.

The bubble acceleration decreases from $g$ to zero while its speed grows from zero to $v_0$ . Therefore, it is reasonable to estimate that the bubble reaches the terminal velocity in a time

\tau\approx\frac{v_0}{g/2}\approx2\sqrt{\frac{2d_0}{3g\beta}}\approx0{,}02\ \mathrm s.

Notice that the height of water in the bottle significantly exceeds the distance traveled during this time (about 1 mm), so practically during the whole of surfacing the bubble is going up at the terminal velocity.

According to the previous estimate the bubble is going at the terminal velocity almost all the time during the surfacing, i.e. its velocity and the diameter are related as $v(y)=\sqrt{2gy/(3\beta)}$ (where $y$ is the bubble diameter at a particular time). According to the problem statement, the rate of the bubble volume change is

\frac{d}{dt}\left(\frac{\pi y^3}{6}\right)=A\cdot\pi y^2\cdot\frac{\Delta n}{\delta}

(where $\Delta n$ is the difference between CO2 molecules density in water and inside the bubble and $A$ is a numerical constant). Differentiating this equation one obtains: $dy/dt=2A\Delta n/\delta$ . Now, using the equation given in 5) for the thickness of depletion layer and recalling that approximately $\Delta n=\mathrm{const}$ , one obtains:

\frac{dy}{dt}=B\sqrt{\frac vy}=C\cdot y^{-1/4}

(where $B$ and $C$ are new constants). Then

\int_{d_0}^{d(t)}y^{1/4}\,dy=C\int_0^t dt=Ct,

which gives after integration

\frac45\big[(d(t))^{5/4}-d_0^{5/4}\big]=Ct\quad\Rightarrow\quad d(t)=\left\{d_0^{5/4}+\frac{5C}{4}t\right\}^{4/5}.

Since $d(T)=D=k\cdot d_0$ one can determine the constant:

d(t)=\left\{d_0^{5/4}+[D^{5/4}-d_0^{5/4}]\frac tT\right\}^{4/5}=d_0\cdot\left\{1+[k^{5/4}-1]\frac tT\right\}^{4/5}.

Substituting $k=3^{4/5}$ , one gets finally: $d(t)=d_0\cdot\{1+2t/T\}^{4/5}$ .

Using the relation between velocity and diameter one obtains: $v(t)=\sqrt{2gd(t)/(3\beta)}$ or

v(t)\approx v_0\cdot\left\{1+[k^{5/4}-1]\frac tT\right\}^{2/5}=v_0\cdot\left\{1+2\frac tT\right\}^{2/5}.

To determine the law of motion one should integrate the velocity: the elevation above the bottom is $h(t)=\int_0^t v(t)dt$ . Hence,

h(t)\approx\frac{5v_0T}{14}\cdot\left\{\left[1+2\frac tT\right]^{7/5}-1\right\}.

Now we substitute $t=T$ in the derived equation and obtain $H\approx5(3^{7/5}-1)v_0T/14$ . Therefore

T\approx\frac7{5(3^{7/5}-1)}H\sqrt{\frac{6\beta}{gd_0}}\approx2{,}3\ \mathrm s.

The experimental value is a bit greater because the bubble at the end of surfacing is bigger and therefore more oblate, which leads to increasing drag force.

The heat released is equal (up to the sign) to the work done by drag force, so it can be estimated as $Q_+\approx\beta\cdot\rho S_{\mathrm{av}}\cdot v_{\mathrm{av}}^2\cdot H$ (the subscript «av» stands for the average value of a physical quantity). The mass of the water «column» described in the problem is $m\approx\rho S_{\mathrm{av}}H$ . Then the temperature increment of the «column» is estimated as

\Delta T_+\approx\frac{\beta v_{\mathrm{av}}^2}{c_w}\sim\frac{2gd_{\mathrm{av}}}{c_w}\sim10^{-6}\ \mathrm K.

The absorbed heat is proportional to the amount of evaporated gas $\nu$ : $Q_-\approx r\cdot\nu$ , where $r\approx2\cdot10^4\ \mathrm{J/mole}$ . Since

\nu\approx\frac{p_0}{RT_0}\frac{4\pi}{3}(k^3-1)d_0^3\approx7\cdot10^{-8}\ \mathrm{mole},\qquad\Delta T_-\approx-\frac{r\nu}{c_wm}\approx-\frac{r\nu}{c_w\rho S_{\mathrm{av}}H}\sim-10^{-3}\ \mathrm K.

Therefore, the net effect is, obviously, cooling. Notice that surfacing of a single bubble has almost no effect on the water temperature in the bottle.

TABLE OF ANSWERS

№	Answer	Maximum score
1	$d_{\max}\approx2\ \mathrm{mm}$ . An answer deviating from this value by no more than 1 mm is accepted.	1
2	A numerical answer must agree with the free fall acceleration by order of magnitude. $a_0\approx g=9{,}8\ \mathrm{m/s^2}$ .	$1+1=2$
3	$v_0=\sqrt{2gd_0/(3\beta)}\approx0{,}1\ \mathrm{m/s}$ .	1 (equation) + 1 (number) = 2
4	$\tau\approx v_0/(g/2)\approx2\sqrt{2d_0/(3g\beta)}\approx0{,}02\ \mathrm s$ .	1 (equation) + 1 (number) = 2
5	$d(t)=d_0\cdot\{1+2t/T\}^{4/5}$ .	4
6	$v(t)\approx v_0\cdot\{1+2t/T\}^{2/5}$ , where $v_0=\sqrt{2gd_0/(3\beta)}$ .	2
7	$h(t)\approx(5v_0T/14)\cdot\{[1+2t/T]^{7/5}-1\}$ .	3
8	$T\approx\frac7{5(3^{7/5}-1)}H\sqrt{6\beta/(gd_0)}\approx2{,}3\ \mathrm s$ .	3 (equation) + 1 (number) = 4
9	$\Delta T_+\sim2gd_{\mathrm{av}}/c_w\sim10^{-6}\ \mathrm K$ .	2
10	$\Delta T_-\sim10^{-3}\ \mathrm K$ ; «–» (the net effect is cooling).	$2+1=3$
Total		25

Archive source notes: The proposed key calls the acceleration expression in 2) “accurate”; that claim and its formula are retained as printed. In 10), the native key uses $4\pi d_0^3/3$ although $d_0$ is defined as a diameter. Its worked cooling estimate has a minus sign, while its table gives the positive magnitude. The order estimate in 9) omits the factor $1/3$ from the preceding velocity formula. These native expressions have not been silently corrected. The key's repeated question 10 shortens the primary question's water-column description; the original question edition governs the statement above.

Оригинал в Архива: theoretical.pdf · решение от архива: theoretical-solutions.pdf