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Chapter 9: Problem 78

Compare the rise velocity of an \(\frac{1}{8}\) -in.-diameter air bubble in water to the fall velocity of an \(\frac{1}{8}\) -in.- -diameter water drop in air. Assume each to behave as a solid sphere.

Short Answer

Expert verified
The air bubble in water has a higher terminal velocity than the water droplet in air, hence the air bubble rises faster.

Step by step solution

01

Set up the parameters

Firstly, determine the values of the radius, r, the gravitational acceleration, g, the density of the particle, \(ρ_p\), the density of the fluid, \(ρ_f\), and the drag coefficient, \(C_d\). We know that g is approximately 9.8 m/s², the diameter is 1/8 inch which gives a radius of 1/16 inch. We also know that the density of air is approximately 1.2 kg/m³, the density of water is approximately 1000 kg/m³, and we assume that \(C_d\) for both fluid is 24/Re.
02

Calculate the Reynolds number

The Reynolds number, Re, is a dimensionless quantity that is used to predict fluid flow patterns. It is calculated using the equation: Re = \(ρ_f * v_t * 2r / μ\), where μ is the dynamic viscosity of the fluid. For air, μ is 0.000018 Pa.s and for water, it's 0.001002 Pa.s. We have to substitute \(v_t\) in terms of other parameters to calculate the value of \(C_d\).
03

Calculate the terminal velocities

With all parameters defined, insert them into the terminal velocity equation to solve for the terminal velocities. One should notice that \(ρ_p - ρ_f\) would be negative for air bubble in water and positive for water droplet in air, this indicates the direction of the motion.
04

Compare the velocities

Finally, check which terminal velocity is higher. This would tell you whether the air bubble rises faster or the water droplet falls faster.

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