IPhO 1987 · Theory problem 1

Ascending moist air

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Moist air is streaming adiabatically across a mountain range as indicated in the figure.

Equal atmospheric pressures of 100 kPa are measured at meteorological stations \(M_0\) and \(M_3\) and a pressure of 70 kPa at station \(M_2\). The temperature of the air at \(M_0\) is \(20^\circ\) C.

As the air is ascending, cloud formation sets in at 84.5 kPa.

Consider a quantity of moist air ascending the mountain with a mass of 2000 kg over each square meter. This moist air reaches the mountain ridge (station \(M_2\)) after 1500 seconds. During that rise an amount of 2.45 g of water per kilogram of air is precipitated as rain.

Figure 1

  1. Determine temperature \(T_1\) at \(M_1\) where the cloud ceiling forms.

  2. What is the height \(h_1\) (at \(M_1\)) above station \(M_0\) of the cloud ceiling assuming a linear decrease of atmospheric density?

  3. What temperature \(T_2\) is measured at the ridge of the mountain range?

  4. Determine the height of the water column (precipitation level) precipitated by the air stream in 3 hours, assuming a homogeneous rainfall between points \(M_1\) and \(M_2\).

  5. What temperature \(T_3\) is measured in the back of the mountain range at station \(M_3\)? Discuss the state of the atmosphere at station \(M_3\) in comparison with that at station \(M_0\).

Hints and Data

The atmosphere is to be dealt with as an ideal gas. Influences of the water vapour on the specific heat capacity and the atmospheric density are to be neglected; the same applies to the temperature dependence of the specific latent heat of vaporisation. The temperatures are to be determined to an accuracy of 1 K, the height of the cloud ceiling to an accuracy of 10 m and the precipitation level to an accuracy of 1 mm.

Specific heat capacity of the atmosphere in the pertaining temperature range:

\[ c_p=1005\ \mathrm{J\cdot kg^{-1}\cdot K^{-1}} \]

Atmospheric density for \(p_0\) and \(T_0\) at station \(M_0\): \[ \rho_0=1.189\ \mathrm{kg\cdot m^{-3}} \]

Specific latent heat of vaporisation of the water within the volume of the cloud:

\[ L_v=2500\ \mathrm{kJ\cdot kg^{-1}} \]

\[ \frac{c_p}{c_v}=\chi=1.4 \qquad\text{and}\qquad g=9.81\ \mathrm{m\cdot s^{-2}} \]

Official English text of the problem.

Answers

  1. \(T_1\approx279\ \mathrm K\), or \(6^\circ\mathrm C\).
  2. \(h_1\approx1410\ \mathrm m\).
  3. \(T_2\approx271\ \mathrm K\), or approximately \(-2^\circ\mathrm C\), to the requested accuracy.
  4. The precipitation level in 3 hours is approximately \(35\ \mathrm{mm}\).
  5. \(T_3\approx300\ \mathrm K\), or approximately \(27^\circ\mathrm C\). The air is about \(7\ \mathrm K\) warmer than at \(M_0\) and is drier: the foehn effect.

Solution

All temperatures in thermodynamic relations are absolute temperatures. Set \[ T_0=293.15\ \mathrm K,\qquad \kappa=\frac{\chi-1}{\chi}=\frac27. \] The specific gas constant of air satisfies \[ R=c_p-c_v,\qquad \frac{R}{c_p}=\kappa. \]

We use the given approximation that water does not appreciably alter the air’s heat capacities or density. Condensed water is treated as being removed as rain.

1. Temperature at the onset of cloud formation

Before reaching \(M_1\), no condensation occurs. The rising air therefore follows the dry adiabatic relation, even though it contains water vapour.

For an ideal gas, the first law for an adiabatic change gives \[ c_p\,dT=\frac{RT}{p}\,dp. \] Consequently, \[ \frac{dT}{T}=\kappa\frac{dp}{p}, \qquad T p^{-\kappa}=\text{constant}. \]

Thus \[ T_1=T_0\left(\frac{p_1}{p_0}\right)^\kappa =293.15\left(\frac{84.5}{100}\right)^{2/7} =279.38\ \mathrm K. \]

Therefore, \[ \boxed{T_1\approx279\ \mathrm K\approx6^\circ\mathrm C.} \]

2. Height of the cloud formation level

First determine the density at \(M_1\). From \(p=\rho RT\), \[ \frac{\rho_1}{\rho_0} =\frac{p_1}{p_0}\frac{T_0}{T_1}. \] Hence \[ \rho_1 =1.189\left(\frac{84.5}{100}\right) \left(\frac{293.15}{279.38}\right) =1.0542\ \mathrm{kg\,m^{-3}}. \]

Hydrostatic equilibrium requires \[ \frac{dp}{dh}=-\rho g. \] With density decreasing linearly between \(M_0\) and \(M_1\), its average over this height interval is \[ \overline{\rho}=\frac{\rho_0+\rho_1}{2}. \] Integrating the hydrostatic equation therefore gives \[ p_0-p_1=g h_1\frac{\rho_0+\rho_1}{2}. \]

Thus \[ h_1 =\frac{2(p_0-p_1)}{g(\rho_0+\rho_1)} =\frac{2(100000-84500)} {9.81(1.189+1.0542)} =1408.7\ \mathrm m. \]

To the requested accuracy of 10 m, \[ \boxed{h_1\approx1410\ \mathrm m.} \]

3. Temperature at the mountain ridge

During the cloudy ascent, condensation releases latent heat. Although the parcel exchanges no heat with its surroundings, this internal conversion of latent energy reduces its cooling.

The condensed mass per kilogram of air is \[ \Delta r=2.45\times10^{-3}\ \mathrm{kg\,kg^{-1}}, \] so the latent energy released per kilogram of air is \[ Q_\ell=L_v\Delta r =(2.500\times10^6)(2.45\times10^{-3}) =6125\ \mathrm{J\,kg^{-1}}. \]

Without condensation, the ridge temperature would be \[ T_{2,\mathrm{dry}} =T_0\left(\frac{p_2}{p_0}\right)^\kappa =293.15\left(\frac{70}{100}\right)^{2/7} =264.75\ \mathrm K. \]

The latent energy corresponds to a temperature increase \[ \frac{Q_\ell}{c_p} =\frac{6125}{1005} =6.095\ \mathrm K. \] Adding this increase at the final pressure gives \[ T_2\simeq T_{2,\mathrm{dry}}+\frac{Q_\ell}{c_p} =270.84\ \mathrm K. \]

To check that this approximation meets the specified accuracy, let \(dr\) be an increment of condensed water per kilogram of air. The first law during condensation is \[ c_p\,dT-\frac{RT}{p}\,dp=L_v\,dr. \] Multiplying by the integrating factor \(p^{-\kappa}\) and integrating gives \[ T_2 =T_1\left(\frac{p_2}{p_1}\right)^\kappa +\frac{L_v}{c_p} \int_0^{\Delta r} \left(\frac{p_2}{p(r)}\right)^\kappa dr. \]

Condensation occurs at pressures between \(p_1\) and \(p_2\), so \[ \left(\frac{p_2}{p_1}\right)^\kappa \leq \left(\frac{p_2}{p(r)}\right)^\kappa \leq1. \] Therefore, \[ 270.52\ \mathrm K\leq T_2\leq270.84\ \mathrm K. \]

The simple latent-heat correction overestimates \(T_2\) by at most \(0.32\ \mathrm K\), well within the required accuracy. Thus \[ \boxed{T_2\approx271\ \mathrm K \quad\text{(approximately }-2^\circ\mathrm C\text{).}} \]

4. Precipitation level in 3 hours

The air mass associated with each square metre is \[ \Sigma=2000\ \mathrm{kg\,m^{-2}}. \] During one stated 1500 s interval, it supplies a rainfall mass per unit area of \[ \Sigma\Delta r =2000(2.45\times10^{-3}) =4.90\ \mathrm{kg\,m^{-2}}. \]

For the steady stream, the rainfall accumulated in \(3\ \mathrm h=10800\ \mathrm s\) is therefore \[ \frac{m_{\mathrm w}}{A} =4.90\frac{10800}{1500} =35.28\ \mathrm{kg\,m^{-2}}. \]

Using the usual liquid-water density \[ \rho_{\mathrm w}\simeq1000\ \mathrm{kg\,m^{-3}}, \] the corresponding water-column height is \[ H=\frac{m_{\mathrm w}/A}{\rho_{\mathrm w}} =\frac{35.28}{1000} =0.03528\ \mathrm m. \]

Homogeneous rainfall gives this same precipitation level throughout the specified rain region: \[ \boxed{H\approx35\ \mathrm{mm}\text{ in 3 hours}.} \]

5. Temperature and atmospheric conditions at \(M_3\)

With the condensed water removed as rain, the descending air becomes unsaturated as compression raises its temperature. Its descent is therefore dry adiabatic: \[ T_3=T_2\left(\frac{p_3}{p_2}\right)^\kappa. \]

Using the approximation from part 3, \[ T_3 \simeq270.84\left(\frac{100}{70}\right)^{2/7} =299.90\ \mathrm K. \] The more precise bounds from part 3 imply \[ 299.54\ \mathrm K\leq T_3\leq299.90\ \mathrm K, \] so the uncertainty from the condensation history remains below \(1\ \mathrm K\).

Hence \[ \boxed{T_3\approx300\ \mathrm K\approx27^\circ\mathrm C.} \]

Compared with \(M_0\), the air at \(M_3\):

  • is approximately \(7\ \mathrm K\) warmer;
  • has lost \(2.45\ \mathrm g\) of water per kilogram of air;
  • has lower relative humidity, owing both to the water loss and to its higher temperature.

Latent heat released during ascent reduces the cooling, while the removed rain cannot produce corresponding evaporative cooling during descent. The resulting warm, dry descending wind is the foehn effect.

The solution was checked against the official solution.