1. Determine temperature at where the cloud ceiling forms. · OlympiadBench #1179

Physics · thermodynamics

Problem

1. Determine temperature T1T_{1} at M1M_{1} where the cloud ceiling forms. Given constants: Moist air is streaming adiabatically across a mountain range as indicated in the figure. Equal atmospheric pressures of 100kPa100 \mathrm{kPa} are measured at meteorological stations M0\mathrm{M}_{0} and M3\mathrm{M}_{3} and a pressure of 70kPa70 \mathrm{kPa} at station M2\mathrm{M}_{2}. The temperature of the air at M0\mathrm{M}_{0} is 20∘C20^{\circ} \mathrm{C}. As the air is ascending, cloud formation sets in at 84.5kPa84.5 \mathrm{kPa}. Consider a quantity of moist air ascending the mountain with a mass of 2000 kg2000 \mathrm{~kg} over each square meter. This moist air reaches the mountain ridge (station M2\mathrm{M}_{2} ) after 1500 seconds. During that rise an amount of 2.45 g2.45 \mathrm{~g} of water per kilogram of air is precipitated as rain. <image_1> 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 K1 \mathrm{~K}, the height of the cloud ceiling to an accuracy of 10 m10 \mathrm{~m} and the precipitation level to an accuracy of 1 mm1 \mathrm{~mm}. Specific heat capacity of the atmosphere in the pertaining temperature range: cp=1005 J⋅kg−1⋅K−1\mathrm{c}_{\mathrm{p}}=1005 \mathrm{~J} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1} Atmospheric density for p0\mathrm{p}_{0} and T0\mathrm{T}_{0} at station M0:ρ0=1.189 kg⋅m−3\mathrm{M}_{0}: \quad \rho_{0}=1.189 \mathrm{~kg} \cdot \mathrm{m}^{-3} Specific latent heat of vaporisation of the water within the volume of the cloud: Lv=2500 kJ⋅kg−1\mathrm{L}_{\mathrm{v}}=2500 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} cpcv=χ=1.4\frac{\mathrm{c}_{\mathrm{p}}}{\mathrm{c}_{\mathrm{v}}}=\chi=1.4 and g=9.81 m⋅s−2\mathrm{g}=9.81 \mathrm{~m} \cdot \mathrm{s}^{-2}

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