Rising bubbles · RMPH 2021 #E

Physics · fluid-mechanics

Problem

All situations take place in an enclosure thermostated at 70°C with air and saturated water vapor. The pressure is constant at p0=1.0×105 N m−2p_0 = 1.0 \times 10^5 \, \text{N m}^{-2}. Physical constants: g=9.8 m s−2g = 9.8 \, \text{m s}^{-2}, μ0=4×10−7 H m−1\mu_0 = 4 \times 10^{-7} \, \text{H m}^{-1}, ε0=8.9×10−12 F m−1\varepsilon_0 = 8.9 \times 10^{-12} \, \text{F m}^{-1} Water properties: Density ρ=1.0×103 kg⋅m−3\rho = 1.0 \times 10^3 \, \text{kg} \cdot \text{m}^{-3}, Saturated vapor pressure at 70°C ps=3.1×104 N⋅m−2p_s = 3.1 \times 10^4 \, \text{N} \cdot \text{m}^{-2}, Surface tension coefficient at 70°C σ=6.5×10−2 N⋅m−1\sigma = 6.5 \times 10^{-2} \, \text{N} \cdot \text{m}^{-1}, Magnetic susceptibility χ=−9.0×10−6\chi = -9.0 \times 10^{-6} The gas bubbles that appear during boiling rise to the free surface of the water. Consider that there are only water-saturated vapors in the bubbles and that the radius of a bubble remains constant as it rises. Consider a vapor bubble with a constant radius Rbubble=1.0×10−3 mR_{\text{bubble}} = 1.0 \times 10^{-3} \, \text{m}, which rises on the distance h0=1.0×10−1 mh_0 = 1.0 \times 10^{-1} \, \text{m}, with a constant speed vbubblev_{\text{bubble}}. E.1. Determine the expression of the mechanical work LbubbleL_{\text{bubble}} done by the vapor bubble on the water in the vessel, during its lifting. Express the result as function of RbubbleR_{\text{bubble}}, h0h_0 and vbubblev_{\text{bubble}}. E.2. Determine the value of time interval tupt_{\text{up}}, in which the bubble reaches the free surface of the water in the vessel.

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