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APhO 2014, Теория — Задача 1. Question 1

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15TH ASIAN PHYSICS OLYMPIAD, 11 - 18 MAY 2014, SINGAPORE — Question 1 · 11 май 2014 г.

Условие

The schematic below shows the Hadley circulation in the Earth’s tropical atmosphere around the spring equinox. Air rises from the equator and moves poleward in both hemispheres before descending in the subtropics at latitudes $\pm\varphi_d$ (where positive and negative latitudes refer to the northern and southern hemisphere respectively). The angular momentum about the Earth’s spin axis is conserved for the upper branches of the circulation (enclosed by the dashed oval). Note that the schematic is not drawn to scale.

Schematic of the Hadley circulation at the spring equinox: two closed circulation loops with arrows rising near the equator (point X on the central dashed line) and descending at the dashed lines labelled +φd (left) and −φd (right); points Y at the top of each loop where the dashed oval encloses the upper branches; a shaded Earth segment labelled Equator, an arrow labelled northward pointing left, and a sun symbol at the top.

(a) (2 points) Assume that there is no wind velocity in the east-west direction around the point X. What is the expression for the east-west wind velocity $u_Y$ at the points Y? Convention: positive velocities point from west to east.

(The angular velocity of the Earth about its spin axis is $\Omega$ , the radius of the Earth is $a$ , and the thickness of the atmosphere is much smaller than $a$ .)

(b) (1 point) Which of the following explains ultimately why angular momentum is not conserved along the lower branches of the Hadley circulation?

Tick the correct answer(s). There can be more than one correct answer.

(I) There is friction from the Earth's surface.

(II) There is turbulence in the lower atmosphere, where different layers of air are mixed

(III) The air is denser lower down and so inertia slows down the motion around the spin axis of the Earth.

(IV) The air is moist at the lower levels causing retardation to the wind velocity.

Around the northern winter solstice, the rising branch of the Hadley circulation is located at the latitude $\varphi_r$ and the descending branches are located at $\varphi_n$ and $\varphi_s$ as shown in the schematic below. Refer to this diagram for parts (c), (d) and (e).

Schematic of the Hadley circulation around the northern winter solstice: a large loop on the left rising at the dashed line φr and descending at φn, with points P (left) and Q (top centre) on the upper branch, and a small loop on the right descending at φs with point R; point Z at the rising branch; the dashed oval encloses the upper branches; shaded Earth segment labelled Equator, arrow labelled northward pointing left, sun symbol at top right; labels φn, Equator, φr, φs along the bottom.

(c) (2 points) Assume that there is no east-west wind velocity around the point Z. Given that $\varphi_r = -8^{\circ}$ , $\varphi_n = 28^{\circ}$ and $\varphi_s = -20^{\circ}$ , what are the east-west wind velocities $u_P$ , $u_Q$ and $u_R$ respectively at the points P, Q and R?

(The radius of the Earth is $a = 6370\ \mathrm{km}$ .)

Hence, which hemisphere below has a stronger atmospheric jet stream?

(I) Winter Hemisphere

(II) Summer Hemisphere

(III) Both hemispheres have equally strong jet streams.

(d) (1 point) The near-surface branch of the Hadley circulation blows southward across the equator. Mark by arrows on the figure below the direction of the east-west component of the Coriolis force acting on the tropical air mass

(A) north of the equator;

(B) south of the equator.

A horizontal line labelled Equator crossed by a broad hollow arrow pointing downward (southward), labelled near-surface branch of Hadley; a small arrow labelled N pointing up on the left.

(e) (1 point) From your answer to part (d) and the fact that surface friction nearly balances the Coriolis forces in the east-west direction, sketch the near-surface wind pattern in the tropics near the equator during northern winter solstice.

Suppose the Hadley circulation can be simplified as a heat engine shown in the schematic below. Focusing on the Hadley circulation reaching into the winter hemisphere as shown below, the physical transformation of the air mass from A to B and from D to E are adiabatic, while that from B to C, C to D and from E to A are isothermal. Air gains heat by contact with the Earth's surface and by condensation of water from the atmosphere, while air loses heat by radiation into space.

Heat-engine schematic of the winter Hadley cell: a closed loop A → B (rising, through a cloud) → C (top) → D (descending) → E → A along the surface; wavy arrows labelled blackbody radiation leave the top branch near C, wavy arrows labelled surface heat transfer enter between E and A from the shaded Earth; the cloud is labelled cloud; arrow labelled northward pointing left; sun symbol at top right.

(f) (2 points) Given that atmospheric pressure at a vertical level owes its origin to the weight of the air above that level, order the pressures $p_A$ , $p_B$ , $p_C$ , $p_D$ , $p_E$ , respectively at the points A, B, C, D, E by a series of inequalities.

(Given that $p_A = 1000\ \mathrm{hPa}$ and $p_D = 225\ \mathrm{hPa}$ . Note that 1 hPa is 100 Pa.)

(g) (2 points) Let the temperature next to the surface and at the top of the atmosphere be $T_H$ and $T_C$ respectively. Given that the pressure difference between points A and E is 20 hPa, calculate $T_C$ for $T_H = 300\ \mathrm{K}$ .

Note that the ratio of molar gas constant ( $R$ ) to molar heat capacity at constant pressure ( $c_p$ ) for air, $\kappa$ , is 2/7.

(h) (2 points) Calculate the pressure $p_B$ .

(i) For an air mass moving once around the winter Hadley circulation, using the molar gas constant, $R$ , and the quantities defined above, obtain expressions for

(A) (2 points) the net work done per unit mole $W_{net}$ ignoring surface friction;

(B) (1 point) the heat loss per unit mole $Q_{loss}$ at the top of the atmosphere. [3 т.]

(j) (1 point) What is the value of the ideal thermodynamic efficiency $\varepsilon_i$ for the winter Hadley circulation?

(k) (2 points) Prove that the actual thermodynamic efficiency $\varepsilon$ for the winter Hadley circulation is always smaller than $\varepsilon_i$ , showing all mathematical steps.

(l) (1 point) Which of the following statements best explains why $\varepsilon$ is less than the ideal value? Tick the correct answer(s). There can be more than one correct answer.

(I) We have ignored work done against surface friction.

(II) Condensation occurs at a temperature lower than the temperature of the heat source.

(III) There is irreversible evaporation of water at the surface.

(IV) The ideal efficiency is applicable only when there is no phase change of water.

Отговори

Решение

(a) Solution:

As the problem is symmetric about the equator, we need only consider the northern hemisphere as shown below.

Cross-section of the northern hemisphere: a quarter circle of radius a from Earth's centre, with the equator at the top and the north pole on the left; a rotation arrow around the spin axis at the north pole; a dashed line from the centre at angle φd from the equator direction; velocity markers (⊗) labelled Ωa at the equator and Ωacos φd + uY at latitude φd; labels equator, a, north pole, Earth's centre.

Conservation of angular momentum about the Earth's spin axis implies that:

\Omega a^2 = (\Omega a\cos\varphi_d + u_Y)a\cos\varphi_d \qquad \textbf{(1.5 point)}

u_Y = \Omega a\left(\frac{1}{\cos\varphi_d} - \cos\varphi_d\right) \qquad \textbf{(0.5 point)}

(b) Solution: (I) & (II) (0.5 point each)

To discourage guessing, minus 0.5 point for each wrong answer.

The minimum points to be awarded in this part is 0.

(c) Solution:

The angular velocity of the Earth about its spin axis is:

\Omega = \frac{2\pi}{24\times 60\times 60s} = 7.27\times 10^{-5}\ s^{-1}

so we have:

\Omega \mathrm{a} = (7.27\times 10^{-5}s^{-1})(6.37\times 10^{6}m) = 463\ ms^{-1}

Conservation of angular momentum about the Earth's spin axis implies that the wind velocity $u$ at latitude $\varphi$ is:

\Omega a^2\cos^2\varphi_r = (\Omega a\cos\varphi + u)a\cos\varphi

u = \Omega a\left(\frac{\cos^2\varphi_r}{\cos\varphi} - \cos\varphi\right) \qquad \textbf{(0.5 point)}

The required east-west wind velocities are:

(0.5 point for each correct answer, but capped at 1 point maximum)

u_p = \Omega a\left(\frac{\cos^2\varphi_r}{\cos\varphi_n} - \cos\varphi_n\right) = 463ms^{-1}\times\left(\frac{\cos^2 8^{\circ}}{\cos 28^{\circ}} - \cos 28^{\circ}\right) = 105\ ms^{-1}

u_Q = \Omega a\left(\frac{\cos^2\varphi_r}{\cos 0^{\circ}} - \cos 0^{\circ}\right) = 463ms^{-1}\times\left(\frac{\cos^2 8^{\circ}}{1} - 1\right) = -8.97\ ms^{-1}

u_R = \Omega a\left(\frac{\cos^2\varphi_r}{\cos\varphi_s} - \cos\varphi_s\right) = 463ms^{-1}\times\left(\frac{\cos^2 8^{\circ}}{\cos 20^{\circ}} - \cos 20^{\circ}\right) = 48.1\ ms^{-1}

Thus, the winter hemisphere (I) has a stronger atmospheric jet stream. (0.5 point)

(d) Solution: (0.5 point for each correct arrow)

Solution to (d): the equator line crossed by the hollow southward arrow labelled near-surface branch of Hadley circulation; a solid arrow pointing left north of the equator labelled westward Coriolis force and a solid arrow pointing right south of the equator labelled eastward Coriolis force; N arrow on the left.

(e) Solution:

As surface friction nearly balances the Coriolis forces in the east-west direction, the east-west component of surface friction must act eastward and westward north and south of the equator respectively. Since friction always opposes motion, the east-west wind velocity near the surface must be westward and eastward north and south of the equator respectively. So the resultant near-surface wind pattern looks like below.

Solution to (e): three grey curved arrows labelled near-surface wind pattern, coming from the north-east, crossing the equator line and turning to the south-east; the line labelled Equator; N arrow on the left.

(0.5 point for consistency with part (d), even if part (d) was wrong)

(0.5 point for correct answer)

(f) Solution:

Since there is less and less air above as one climbs upward in the atmosphere, atmospheric pressure must decrease upwards.

So,

p_A > p_B > p_C \quad and \quad p_E > p_D > p_C \ \textbf{(0.5 point)}

The process EA represents an isothermal expansion as heat is gained from the surface. So,

p_E > p_A \quad \textbf{(0.5 point)}

Since the total heat gain must equal the total heat loss, more heat must be lost in the isothermal compression CD than in the isothermal expansion BC. So net heat loss occurs from B to D and hence

p_D > p_B \quad \textbf{(0.5 point)}

So with the values of the pressure at A and D, we deduce that:

p_A > p_D \quad \textbf{(0.5 point)}

Collecting all inequalities together,

p_E > p_A > p_D > p_B > p_C

(g) Solution:

Since $p_E > p_A$ and $p_A = 1000\ \mathrm{hPa}$ , we have $p_E = 1020\ \mathrm{hPa}$ .

From the adiabatic compression from D to E, we have:

p_E^{-\kappa}T_H = p_D^{-\kappa}T_C \qquad \textbf{(1 point)}

T_C = \left(\frac{p_D}{p_E}\right)^{\kappa}\times T_H = \left(\frac{225}{1020}\right)^{2/7}\times 300K = 195K \qquad \textbf{(1 point)}

(h) Solution:

From the adiabatic expansion AB and adiabatic compression DE,

\left.\begin{aligned} p_A^{-\kappa}T_H &= p_B^{-\kappa}T_C \\ p_E^{-\kappa}T_H &= p_D^{-\kappa}T_C \end{aligned}\right\}\ \frac{p_A}{p_E} = \frac{p_B}{p_D} \quad \cdots\cdots\cdots(*) \quad \textbf{(1 point)}

\therefore p_B = \frac{p_A}{p_E}p_D = \frac{1000}{1020}225hPa = 220hPa \quad \textbf{(1 point)}

(i) Solution:

(A) Work done per mole in an isothermal process is generally given by

W = \int p\,dV = \int p\,d\left(\frac{RT}{p}\right) = -RT\int p^{-1}dp = -RT\ln p + const. \ \textbf{(1 point)}

Work done per mole in processes EA and BCD are respectively,

W_{EA} = -RT_H\ln p_A + RT_H\ln p_E = RT_H\ln\left(\frac{p_E}{p_A}\right)

W_{BCD} = RT_C\ln\left(\frac{p_B}{p_D}\right)

Work done in an adiabatic process is used entirely to raise the internal energy of the air mass. Since the decrease in internal energy in process AB exactly cancels the increase in internal energy in process DE because the respective decrease and increase in temperature cancel, no net work is done in the adiabatic processes.

So the net work done per mole on the air mass is:

\begin{aligned} W_{net} &= W_{EA} + W_{BCD} \\ &= RT_H\ln\left(\frac{p_E}{p_A}\right) + RT_C\ln\left(\frac{p_B}{p_D}\right) \\ &= R(T_H - T_C)\ln\left(\frac{p_E}{p_A}\right) + RT_C\ln\left(\frac{p_B}{p_D}\frac{p_E}{p_A}\right) \\ &= R(T_H - T_C)\ln\left(\frac{p_E}{p_A}\right) \quad or \quad R(T_H - T_C)\ln\left(\frac{p_D}{p_B}\right) \end{aligned}

using equation $(*)$ in part (h)

(1 point)

(B) The heat loss per mole at the top of the atmosphere is the same as the work done per mole on the air mass because there is no change in internal energy for an isothermal process.

\begin{aligned} Q_{loss} &= W_{CD} & \textbf{(0.5 point)} \\ &= RT_C\ln\left(\frac{p_D}{p_C}\right) & \textbf{(0.5 point)} \end{aligned}

(j) Solution:

\begin{aligned} \varepsilon_i &= 1 - \frac{T_C}{T_H} & \textbf{(0.5 point)} \\ &= 1 - \frac{195}{300} = 0.35 & \textbf{(0.5 point)} \end{aligned}

(k) Solution:

\varepsilon = \frac{W_{net}}{Q_{loss} + W_{net}}

\begin{aligned} \frac{1}{\varepsilon} - 1 = \frac{Q_{loss}}{W_{net}} &= \frac{RT_C\ln\left(\frac{p_D}{p_C}\right)}{R(T_H - T_C)\ln\left(\frac{p_E}{p_A}\right)} \\ &= \frac{T_C\ln\left(\frac{p_D}{p_B}\times\frac{p_B}{p_C}\right)}{(T_H - T_C)\ln\left(\frac{p_E}{p_A}\right)} \\ &> \frac{T_C\ln\left(\frac{p_D}{p_B}\right)}{(T_H - T_C)\ln\left(\frac{p_E}{p_A}\right)} \quad \text{as } \frac{p_B}{p_C} > 1 \quad \textbf{(1 point)} \\ &= \frac{T_C}{T_H - T_C} \qquad \text{using equation } (*) \text{ in part (h)} \end{aligned}

\frac{1}{\varepsilon} > 1 + \frac{T_C}{T_H - T_C} = \frac{T_H}{T_H - T_C}

\varepsilon < \frac{T_H - T_C}{T_H} = \varepsilon_i \qquad \textbf{(1 point)}

(l) Solution: (II) & (III) (0.5 point each)

To discourage guessing, minus 0.5 point for each wrong answer.

The minimum points to be awarded in this part is 0.

Оригинал в Архива: 1.pdf · официални решения: 1_sol.pdf