Съдържание

NBPhO 2023, Теория и експеримент — Задача 2. Nitrogen explosion

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от NBPhO 2023

Nordic-Baltic Physics Olympiad 2023 · 8 т.

Условие

— Päivo Simson.

A half-sphere of radius $r = 0.1\ \mathrm{m}$ is filled with liquid nitrogen at the boiling point temperature of $T_1 = 77.4\ \mathrm{K}$ ( $-195.8\ ^{\circ}\mathrm{C}$ ). The other half is then firmly sealed onto the first, creating a sphere containing liquid nitrogen and nitrogen gas, each occupying one-half of the volume. The sphere is immediately thrown into $T_w = 20\ ^{\circ}\mathrm{C}$ temperature water, where it floats exactly as shown in the figure below. After some time, it explodes.

A sphere floating half submerged: the upper half, labelled gas (dotted), is above the water line in air; the lower half, labelled liquid, is below the dashed water surface; labels air, water and 20 °C.

The sphere is made of PCTFE plastic of density $\rho_p = 2130\ \mathrm{kg\,m^{-3}}$ , maximum tensile strength $\sigma = 3.4 \times 10^7\ \mathrm{N\,m^{-2}}$ (above this stress, the plastic will break) and thermal conductivity $k = 0.84\ \mathrm{W\,m^{-1}\,K^{-1}}$ . For liquid nitrogen, under the conditions considered here, the latent heat of vaporization is $\lambda = 2.0 \times 10^5\ \mathrm{J\,kg^{-1}}$ , specific heat $c_v = 2000\ \mathrm{J\,kg^{-1}\,K^{-1}}$ and density $\rho_n = 808\ \mathrm{kg\,m^{-3}}$ . Molar mass $\mathrm{M(N_2)} = 28\ \mathrm{g\,mol^{-1}}$ . The ideal gas constant $R = 8.31\ \mathrm{J\,K^{-1}\,mol^{-1}}$ . Temperature dependence of saturated vapor pressure of nitrogen is shown below.

Graph of the saturated vapour pressure of nitrogen p (MPa), from 0 to 2.5, against temperature T (K), from 70 to 120; a rising convex curve from about 0.03 MPa at 70 K to about 2.5 MPa at 120 K, on a fine grid.

i) What is the wall thickness $d$ of the sphere? [1,5 т.]

ii) What is the pressure $p_2$ inside the sphere right before it explodes? The outside pressure is $p_a = 1.0 \times 10^5\ \mathrm{Pa}$ . [1,5 т.]

iii) What is the temperature $T_2$ of liquid nitrogen right before the explosion? [1,5 т.]

iv) Calculate the mass of nitrogen that evaporates inside the sphere before the explosion. [1,5 т.]

v) Estimate the time it will take for the sphere to explode. The heat capacity of the plastic and the heat flux through the upper half of the sphere can be neglected. [2 т.]

Решение

— Solution by Päivo Simson, grading schemes by Päivo Simson and ....

i) (1,5 т.)

As shown in the first figure of the problem, the sphere floats so that exactly half of it is submerged in water. When studying buoyancy, the mass of nitrogen gas can be considered negligibly small compared to the masses of liquid nitrogen and the plastic sphere. According to Archimedes’ principle, we have

\rho_n V_n + \rho_p V_p = \rho_w V_w,

where $V_n$ , $V_p$ , and $V_w$ are the volumes of liquid nitrogen, plastic, and displaced water, respectively. We have

\frac{1}{2}\cdot\frac{4\pi r^3\rho_n}{3} + d\cdot 4\pi r^2\rho_p = \frac{1}{2}\cdot\frac{4\pi r^3\rho_w}{3},

and solving for $d$ we get

d = \frac{r}{6}\cdot\frac{\rho_w - \rho_n}{\rho_p} = 1.5\ \mathrm{mm}.

Grading:

Archimedes’ principle (0.3 pts)
Correct volumes of liquid nitrogen, plastic, and displaced water (0.6 pts)
Correctly solving for $d$ (0.3 pts)
Correct answer $d = 1.5\ \mathrm{mm}$ (0.3 pts)

ii) (1,5 т.)

The sphere will explode when the pressure difference $p_2 - p_a$ is such that the maximum tensile strength of the plastic is reached. Let’s look at only one-half of the sphere and study the balance of the forces.

A hemisphere of the plastic shell cut through its equator, with small arrows along the cut rim labelled elastic forces and a large arrow into the flat face labelled resultant of the pressure forces; the other hemisphere is drawn dashed, separated, at the upper right.

The resultant force due to pressure is $(p_2 - p_a)\pi r^2$ . This must always be balanced by the elastic forces in the plastic. Right before the explosion, we have the equality

(p_2 - p_a)\pi r^2 = \sigma 2\pi r d,

where $2\pi r d$ is the cross-sectional area of the plastic. Now solving for $p_2$ we get

p_2 = p_a + \frac{2\sigma d}{r} = 1.1\cdot 10^6\ \mathrm{Pa}.

Grading:

Qualitative understanding of the conditions for the sphere to explode - resultant pressure force equals the total tensile strength (0.4 pts)
The idea of breaking the sphere in half and analysing the forces acting on only one side of the sphere (0.4 pts)
Correct condition for the sphere to explode: $(p_2 - p_a)\pi r^2 = \sigma 2\pi r d$ (0.4 pts)
Correct final expression and answer (0.3 pts)

iii) (1,5 т.)

During boiling at constant pressure, the temperature of the liquid does not change, even though the liquid is gaining energy all the time. If however the pressure changes, as we have in our problem, then so does the boiling point, following exactly the phase transition line on the phase diagram. Since we know the final pressure $p_2$ , we can simply read the corresponding temperature of the diagram:

p_2 = 1.1\cdot 10^6\ \mathrm{Pa} \implies T_2 = 106\ \mathrm{K}.

Grading:

Correctly stating that the process $1 - 2$ follows the phase transition line. (1.0 pts)
Reading the correct temperature from the graph (0.5 pts)

iv) (1,5 т.)

If $p$ , $V$ , and $T$ are known, the mass of the nitrogen gas can be calculated using the ideal gas equation:

pV = \frac{m}{M}RT \implies m = \frac{pVM}{RT}.

Since we know the initial and final states, the mass $\Delta m$ of nitrogen that evaporated between these states is

\begin{aligned} \Delta m &= m_2 - m_1 = \\ &= \frac{2\pi r^3 M}{3R}\left(\frac{p_2}{T_2} - \frac{p_1}{T_1}\right) = 0.065\ \mathrm{kg}. \end{aligned}

Grading:

The idea of calculating the mass of nitrogen gas in the initial and final state (0.3 pts)
Realizing that ideal gas law can be used to calculate the mass of nitrogen gas (0.3 pts)
Correct ideal gas law (0.3 pts)
Correctly using the molar mass (0.3 pts)
Correct calculation and correct final answer (0.3 pts)

v) (2 т.)

Neglecting the heat capacity of the plastic and the heat flux through the upper half of the sphere, we only need to consider the heat of evaporation:

Q_e = \lambda\Delta m = 1{,}31\cdot 10^4\ \mathrm{J},

and the heat for raising the temperature of the liquid nitrogen:

\begin{aligned} Q_l &= c_v m_l(T_2 - T_1) = \\ &= \frac{2}{3}\pi r^3\rho_n c_v(T_2 - T_1) = 9{,}7\cdot 10^4\ \mathrm{J}. \end{aligned}

The heat for rising the temperature of the nitrogen gas can be neglected, as the mass of the gas is very small compared to the mass of liquid nitrogen ( $\approx 1.7\ \mathrm{kg}$ ). The total heat $Q_e + Q_l$ is taken from the water through the plastic. The average temperature of nitrogen is $T_n = (T_1 + T_2)/2 = 91.7\ \mathrm{K}$ , and the average temperature difference between water and nitrogen is $T_w - T_n = 201.5\ \mathrm{K}$ . The heat flux through the plastic is

q = k\cdot\frac{T_w - T_n}{d}.

The total amount of heat carried through the lower half of the sphere during time $\Delta t$ is

\Delta Q = qA\Delta t = Q_e + Q_l,

where $A = 2\pi r^2$ is half of the surface area of the sphere. Finally, solving for $\Delta t$ , we have the estimated time it takes for the sphere to explode:

\Delta t = \frac{d\cdot(Q_e + Q_l)}{2\pi r^2 k\,(T_w - T_n)} = 15.5\ \mathrm{s}.

Grading:

Understanding that $Q_e + Q_l$ must be equal to the total heat received from the water (0.3 pts)
Correct heat of evaporation (0.2 pts)
Correct heat for rising the temperature of liquid nitrogen (0.2 pts)
Correctly using the average temperature of nitrogen (0.2 pts)
Correctly using the average temperature difference (0.2 pts)
Correct expression for the heat flux $q$ though the plastic (0.3 pts)
Correct expression for the total heat $\Delta Q$ received through the plastic (0.3 pts)
Correct final expression for calculating the time till explosion (0.3 pts)

Give full points for this part if the final expression and the answer are correct.

Оригинал в Архива: NBPhO_2023_problems.pdf · официални решения: NBPhO_2023_solutions.pdf