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GPhO 2022, Теория — Задача 2. Global warming

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Всички материали от GPhO 2022

The 4th Gulf Physics Olympiad — Theoretical Competition · 14 март 2022 г. · 10 т.

Указания

The examination lasts for 5 hours. There are 3 problems worth in total 30 points.
You must not open the envelope with the problems before the signal of the beginning of competition.
You are not allowed to leave your working place without permission. If you need any assistance (broken calculator, need to visit a restroom, etc), please raise your hand until an organizer arrives.
Use only the front side of the sheets of paper.
For each problem, there are dedicated Solution Sheets (see header for the number and pictogram). Write your solutions onto the appropriate Solution Sheets. For each Problem, the Solution Sheets are numbered; use the sheets according to the enumeration. Always mark which Problem Part and Question you are dealing with. Copy the final answers into the appropriate boxes of the Answer Sheets. There are also Draft papers; use these for writing things which you don’t want to be graded. If you have written something that you don’t want to be graded onto the Solution Sheets (such as initial and incorrect solutions), cross these out.
If you need more paper for a certain problem, please raise your hand and tell an organizer the problem number; you are given two Solution sheets (you can do this more than once).
You should use as little text as possible: try to explain your solution mainly with equations, numbers, symbols and diagrams. Though in some places textual explanation may be unavoidable.
After the signal signifying the end of examination you must stop writing immediately. Put all the papers into the envelope at your desk. You are not allowed to take any sheet of paper out of the room. If you have finished solving before the final sound signal, please raise your hand.

Условие

The average temperature of planet Earth ( $15\ {}^{\circ}\mathrm{C}$ ) is the result of a combination of two factors: (a) Earth’s surface absorbs a fraction $\alpha \approx 0.7$ of the incident solar radiation; (b) The surface of the Earth radiates according to Stefan-Boltzamann’s law: the power radiated per unit area equals to $\sigma\varepsilon T^4$ , where $\sigma = 5.67\times 10^{-8}\ \mathrm{W/m^2K^4}$ , and $T$ denotes the surface temperature in Kelvin, and $\varepsilon$ — the emissivity, the ratio of the energy radiated from a material’s surface to that radiated from an ideal black body. Due to the second law of thermodynamics, the emission factor must be equal to the absorption coefficient $\alpha$ , but only if the radiation and absorption occur at the same wavelengths. The heat radiation from the Sun is mostly in the optical range for which the atmosphere is fully transparent (the above value $\alpha \approx 0.7$ corresponds to the optical wavelengths). The Earth emits predominantly infrared radiation, and in this wavelength range the Earth’s surface (mostly water) has an average emissivity $\epsilon_0 \approx 0.9$ . At the same time, the Earth’s heat balance is affected by greenhouse gases, which partially reflect Earth’s heat radiation back to Earth, and thus reducing the emissivity to a new effective value $\epsilon_{\mathrm{eff}}$ .

Below, you can use the following data: The emissivity of the Sun is 1.00; Distance between the Sun and Earth $D = 1.47\times 10^{8}\ \mathrm{km}$ ; Radius of the Sun $R_{\odot} = 6.96\times 10^{5}\ \mathrm{km}$ ; the surface temperature of the Sun $T_{\odot} = 5780\ \mathrm{K}$ ; Average surface temperature of Earth $T_0 = 15\ {}^{\circ}\mathrm{C}$ ; zero degrees celsius in Kelvin is $273.15\ \mathrm{K}$ ; Earth’s radius $R_0 = 6370\ \mathrm{km}$ ; total energy produced by mankind in one year: $A = 8.8\times 10^{20}\ \mathrm{J}$ (assume that this energy ends up mostly being dissipated as heat into the environment).

1. What is the ratio of the total power consumed by mankind to the total solar radiation power absorbed by the Earth? A year has 365.24 days. [2 т.]

2. How large of a surface area on the Earth needs to be covered by solar panels to supply enough electrical energy to the humankind assuming a solar panel efficiency of $\eta = 20\%$ ? Compare this with the surface area of the Empty Quarter desert of about $650\,000\ \mathrm{km^2}$ . [1 т.]

3. What is the average effective emissivity $\epsilon_{\mathrm{eff}}$ of the Earth’s surface at the wavelengths corresponding to the heat radiation of the Earth? [2 т.]

4. By how many degrees would the average temperature of the Earth’s surface decrease if, at some point of time, humanity stopped producing any energy, and assuming that the value of $\epsilon_{\mathrm{eff}}$ would remain unchanged? [2 т.]

5. The average surface temperature of the Earth has increased by $\Delta T = 0.9\ \mathrm{K}$ over the last 50 years and this is mainly explained by the increase in greenhouse gas concentrations in Earth’s atmosphere. In a simplified manner, this effect can be modelled as the greenhouse gases absorbing a fraction $k$ of the heat radiation emitted by Earth and then radiating the absorbed energy thermally so half of the radiated energy goes towards the surface of the Earth, and the other half goes into space. By how much has the factor $k$ changed over the last 50 years? [1,5 т.]

6. Solar radiation is the main driving force of the atmospheric motion; on a global scale, the atmospheric circulation can be seen as a heat engine. To explain this, consider the so-called Hadley circulation: the Sun heats the ground, which in turn warms the air; the warm air initially travels along the ground, from the middle latitudes towards the equator, and at a certain point, when it has warmed up sufficiently, rises adiabatically and cools down. In the high atmospheric layers, at an altitude of $\sim 10\ \mathrm{km}$ , the cooler air travels back to the middle latitudes and slowly radiates heat into space. During its travel, air becomes cooler and the density increasingly higher, until at a certain point, it becomes so dense that it descends down back to the ground level. After that, the process starts repeating. Based on the model of ideal heat engine and the data provided and calculated above, estimate an upper limit for the total power of producing the wind energy on Earth, assuming that the temperature in the atmosphere drops by about one degree celsius per every 100 meters increase in the altitude. How many times does it differ from the total energy produced by humanity? [1,5 т.]

Отговори

Решение

1. (2 pts) The total radiated power of the Sun is given by Stefan-Boltzmann’s law:

L_{\odot} = 4\pi R_{\odot}^2\sigma T_{\odot}^4.

Here we used that the emissivity of the sun is 1.00.

Using Stefan-Boltzmann’s law correctly for the Sun with its values for radius, emissivity, and temperature (0.5 pts)

As the radiation reaches Earth, it assumes an intensity given by the ratio of $L_{\odot}$ and the area of a sphere of radius $D$ :

I_0 = \frac{L_{\odot}}{4\pi D^2} = \frac{R_{\odot}^2\sigma T_{\odot}^4}{D^2} = 1419\ \mathrm{W}.

Arriving at the solar intensity or flux at Earth’s distance from the Sun using the surface area, and getting the correct value within 25% with units (0.2 + 0.2 + 0.1) (0.5 pts)

The cross-sectional area of Earth is $\pi R_{\odot}^2$ , hence the total solar power reaching Earth’s surface before being absorbed is

P_0 = I_0\pi R_{\odot}^2 = \frac{R_{\odot}^2\sigma T_{\odot}^4}{D^2}R_0^2\pi = 1.81\times 10^{17}\ \mathrm{W}.

Obtaining the solar power at Earth while using $I_0$ and Earth’s cross sectional area with correct value within 25% and units (0.2 + 0.2 + 0.1) (0.5 pts)

The absorbed amount is $P_0\alpha$ . The duration of a year is $\Delta t = 365.24\cdot 24\cdot 60\cdot 60\ \mathrm{s} = 3.15\times 10^{7}\ \mathrm{s}$ . Hence, the ratio of mankind’s power consumption and the solar radiation is

\frac{A}{P_0\alpha\Delta t} = 2.20\times 10^{-4}.

Combining $P_0$ correctly with $\alpha = 0.7$ for 1 year in correct units with $A$ to get the ratio (0.2 + 0.3). Full mark for a value within 25% (0.5 pts)

2. (1 pt) The average amount of solar radiation arriving on a unit area of Earth before being absorbed is $I_1 = P_0/(4\pi R_{\odot}^2) = I_0/4 = 355\ \mathrm{W}$ . (0.3 pts)

Solar panels convert to power an intensity of $I_1\eta$ . Thus, the total area needed for the solar panels is

A_S = \frac{A}{\Delta tI_1\eta} = 393\,000\ \mathrm{km^2},

which is around 1.5 times bigger than the surface area of the Persian Gulf.

Combining $I_1$ with the efficiency, $A$ , and time (0.3 + 0.2 + 0.2). Full mark for a value within 25% (0.7 pts)

3. (2 pts) Because the Earth is in thermal equilibrium, it must release the same amount of heat as it absorbs. The absorbed heat flux has two contributions, one is the absorbed flux $I_1\alpha$ from the Sun (0.5 pts) , and the other is the flux produced by Mankind $I_h = A/(\Delta t4\pi R_{\odot}^2) = 0.0547\ \mathrm{W}$ . (0.5 pts)

The total absorbed and released flux are therefore $I_{\mathrm{tot}} = I_1\alpha + I_h$ . (0.5 pts)

Humanity’s contribution to the total flux is small, but we keep track of this for the sake of the next subtask. The ideal black body flux from the surface is $\sigma T_0^4$ (note that we’re missing $\varepsilon_0$ because calculating the effective emissivity requires comparing the actual flux with ideal black body radiation) so the effective emissivity is given by

\varepsilon_{\mathrm{eff}} = \frac{I_1\alpha + I_h}{\sigma T_0^4} = 0.635. \qquad \mathbf{(0.5\,pts)}

4. (2 pts) If we remove humanity’s contribution, then using similar thermodynamic equilibrium arguments, the total released flux must be equal to $I_1\alpha$ . If Earth’s new temperature is $T_1$ , then $I_1\alpha = \sigma\varepsilon_{\mathrm{eff}}T_1^4$ . (0.5 pts)

(0.5 pts)

0.5 for LHS and RHS each

Plugging in $\varepsilon_{\mathrm{eff}}$ from the previous part, we get

1 + \frac{I_h}{I_1\alpha} = \frac{T_0^4}{T_1^4} \qquad \mathbf{(0.5\,pts)}

so the change in Earth’s temperature is

\Delta T_1 = T_0 - T_1 = T_0\left(1 - \left(1 + \frac{I_h}{I_1\alpha}\right)^{-1/4}\right) = 0.016\ \mathrm{K}. \qquad \mathbf{(0.5\,pts)}

Full mark for correct values within 25%

5. (1.5 pts) Our system consists of Earth’s surface and the atmospheric layers with greenhouse gases. Both must be in thermodynamic equilibrium. This is in contrast with the previous part where we could look at the total resulting system and its aggregated properties. If we denote the black body radiation being released from the surface with $I_s$ , then a fraction of $I_sk$ is absorbed in the greenhouse gases and $I_sk/2$ (0.5 pts) is being directed back to the surface of the Earth (and the other $I_sk/2$ to space). (0.5 pts)

Hence, the heat flux balance of the surface reads $I_1\alpha + I_sk/2 = I_s$ (0.5 pts)

(we have neglected humanity’s thermal contribution, as that leads to a negligible change in temperature $\Delta T_1 \ll \Delta T$ ). Therefore, $k = 2(1 - I_1\alpha/I_s)$ . (0.5 pts)

Additionally, we know that the emissivity of Earth’s surface is $\varepsilon_0$ so $I_s = \varepsilon_0\sigma T^4$ , where $T$ is Earth’s temperature. This gives us an analytic expression for $k$ :

k = 2\left(1 - \frac{I_1\alpha}{\varepsilon_0\sigma T^4}\right).

For $T = T_0$ we calculate $k_0 = 0.5886$ and for $T = T_0 - \Delta T$ , $k_1 = 0.5708$ . Therefore, the absorption factor of the greenhouse gases has changed by

\Delta k = k_0 - k_1 = 0.0178, \qquad \mathbf{(0.5\,pts)}

which on a global scale is very significant!

Full mark for correct values within 25%

6. (1.5 pts) The Hadley circulation forms a big heat engine. The circulating air receives heat from Earth’s surface at temperature $T_h = T_0 = 15\ {}^{\circ}\mathrm{C}$ , (0.2 pts)

and releases heat in the form of radiation in the upper atmospheric layers at an effective temperature $T_c \approx T_h - 10\ \mathrm{km}/(1\ {}^{\circ}\mathrm{C}/100\mathrm{m}) = 188\ \mathrm{K}$ . (0.3 pts)

The released heat amount is constrained by the total heat radiated into space by the atmospheric gases found in the previous subtask, i.e. $P_{\mathrm{out}} \le I_sk_02\pi R_0^2 = \varepsilon_0\sigma T^4k_02\pi R_0^2$ . . (0.3 pts)

On the other hand, the maximal efficiency of the heat engine corresponds to a Carnot cycle

\eta \le \frac{T_h - T_c}{T_h}. \qquad \mathbf{(0.2\,pts)}

If the power going into the cycle is $P_{\mathrm{in}}$ , then from conservation of energy we have $P_{\mathrm{in}} = P_{\mathrm{work}} + P_{\mathrm{out}}$ , where $P_{\mathrm{work}}$ is the amount of wind energy produced on Earth. Further, $P_{\mathrm{work}} = P_{\mathrm{in}}\eta = (P_{\mathrm{work}} + P_{\mathrm{out}})\eta$ so

\begin{aligned} P_{\mathrm{work}} &= P_{\mathrm{out}}\frac{\eta}{1-\eta} \le P_{\mathrm{out}}\frac{\frac{T_h-T_c}{T_h}}{1 - \frac{T_h-T_c}{T_h}} \\ &= P_{\mathrm{out}}\left(\frac{T_h}{T_c} - 1\right) \le \varepsilon_0\sigma T_0^4k_02\pi R_0^2\left(\frac{T_h}{T_c} - 1\right) \\ &= 2.8\times 10^{16}\ \mathrm{W}. \end{aligned} \qquad \mathbf{(0.3\,pts)}

This is $P_{\mathrm{work}}\Delta t/A = 1000$ times bigger than humanity’s total production. (0.2 pts)

Full mark for correct values within 25%

Оригинал в Архива: th.pdf · официални решения: th_sol.pdf