Problem 1 Calculate the terminal velocity ... [FREE SOLUTION]

91影视

Rain or Shine

Tyson Ochsner, Eric Howerton, Braiden Ellis

Combined Science

2019 Edition

Chapter 2: Problem 1

Calculate the terminal velocity in water at \(20^{\circ} \mathrm{C}\) for each of the following. Assume a particle density of \(2650 \mathrm{~kg} \mathrm{~m}^{-3}\), a water density of \(1.00 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{-3}\), and a viscosity of \(1.00 \times 10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}\) \(\mathrm{s}^{-1} .\) 1\. 100. \mum diameter sand particle 2\. \(10.0 \mu \mathrm{m}\) diameter silt particle 3\. \(1.00 \mu \mathrm{m}\) diameter clay particle

Short Answer

Expert verified

Question: Calculate the terminal velocities in water for sand, silt, and clay particles. Answer: The terminal velocity of the sand particle in water is approximately 0.0389 m/s, for the silt particle is approximately 3.89 x 10^-5 m/s, and for the clay particle is approximately 3.89 x 10^-8 m/s.

Step by step solution

1. Calculating Reynolds number for the sand particle

First, we need to find the Reynolds number for the sand particle with a diameter of \(100 \,\mu m\). We need to convert the diameter to meters, which gives us \(d = 100 \times 10^{-6} \, \mathrm{m}\). To find the Reynolds number, we will initially assume a terminal velocity of \(1 \, \mathrm{m}\mathrm{~s}^{-1}\) and calculate the Reynolds number as: \(Re = \frac{vd\rho}{\mu} = \frac{1 \times 100 \times 10^{-6} \times 1.00 \times 10^{3}}{1.00 \times 10^{-3}} = 100\) Since the Reynolds number is less than 500, we can use the Stokes' law to calculate the terminal velocity.

2. Calculating terminal velocity for the sand particle

Using Stokes' law, the formula for the terminal velocity is given by \(v = \frac{2}{9} \times \frac{(\rho_p - \rho)gd^2}{\mu}\), where \(\rho_p\) is the particle density, \(\rho\) is the fluid density, \(g\) is the acceleration due to gravity, and \(d\) is the diameter of the particle. We will use the following values: \(\rho_p = 2650 \mathrm{~kg}\mathrm{~m}^{-3}\), \(\rho = 1.00 \times 10^{3}\mathrm{~kg}\mathrm{~m}^{-3}\), \(g = 9.81 \, \mathrm{~m}\mathrm{~s}^{-2}\), and \(d = 100 \times 10^{-6} \, \mathrm{m}\). \(v = \frac{2}{9} \times \frac{(2650 - 1.00 \times 10^{3})(9.81)(100 \times 10^{-6})^2}{1.00 \times 10^{-3}} = 0.0389 \, \mathrm{~m}\mathrm{~s}^{-1}\) The terminal velocity of the sand particle in water is approximately \(0.0389 \, \mathrm{~m}\mathrm{~s}^{-1}\).

3. Calculating terminal velocities for silt and clay particles

We will repeat steps 1 and 2 for the silt (\(d = 10.0 \times 10^{-6} \, \mathrm{m}\)) and clay (\(d = 1.00 \times 10^{-6} \, \mathrm{m}\)) particles. For the silt particle: Reynolds number \(Re = \frac{vd\rho}{\mu} = \frac{1 \times 10.0 \times 10^{-6} \times 1.00 \times 10^{3}}{1.00 \times 10^{-3}} = 10\), Stokes' law can be used. \(v = \frac{2}{9} \times \frac{(2650 - 1.00 \times 10^{3})(9.81)(10.0 \times 10^{-6})^2}{1.00 \times 10^{-3}} = 3.89 \times 10^{-5} \, \mathrm{~m}\mathrm{~s}^{-1}\) For the clay particle: Reynolds number \(Re = \frac{vd\rho}{\mu} = \frac{1 \times 1.00 \times 10^{-6} \times 1.00 \times 10^{3}}{1.00 \times 10^{-3}} = 1\), Stokes' law can be used. \(v = \frac{2}{9} \times \frac{(2650 - 1.00 \times 10^{3})(9.81)(1.00 \times 10^{-6})^2}{1.00 \times 10^{-3}} = 3.89 \times 10^{-8} \, \mathrm{~m}\mathrm{~s}^{-1}\) The terminal velocity of the silt particle in water is approximately \(3.89 \times 10^{-5} \, \mathrm{~m}\mathrm{~s}^{-1}\) and for the clay particle is approximately \(3.89 \times 10^{-8} \, \mathrm{~m}\mathrm{~s}^{-1}\).