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Question
With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m2 and specific gravity 0.9? Density of air is 1.29 kg/m3.
Solution
Given:
d = 0.4mm,
η = 0.1Pa·s,
ρL = 0.9 × 103 kg/m3 = 900 kg/m3,
ρair = 1.29 kg/m3,
g = 9.8 m/s2
To find:
Terminal velocity of the air bubble.
Solution:
At terminal velocity, the downward viscous force is equal in magnitude to the net upward force.
Viscous force = buoyant force - the gravitational force
∴ 6πηrvt = `4/3 pi"r"^3 (rho_"air" - rho_"L")`g
∴ `"v"_"t" = (2"r"^2"g"(rho_"air" - rho_"L"))/(9η)`
`= (2(2 xx 10^-4)^2 (9.8 )(1.29 - 900))/(9(0.1))`
`= (2 xx 4 xx 9.8 xx -898.7 xx 10^-8)/0.9`
`= -7.829 xx 10^4 xx 10^-8`
`= -7.829 xx 10^-4` m/s ... [The negative sign indicates that the bubble rises up]
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