(a) A dust particle with a diameter of 0.0012 in and a specific gravity of 3.6 falls in the air (standard atmosphere at sea level, ρ = 0.0023768 slugs/ft^3 , μ= 0.37372 × 10^−6 lb-sec/ft^2 , γ = 0.076472 lb/ft^3 ), as illustrated in Figure EP 10.8a, and settles to the ground after a windstorm.

Question

(a) A dust particle with a diameter of 0.0012 in and a specific gravity of 3.6 falls in the air (standard atmosphere at sea level, [latex] ho = 0.0023768 slugs/ft^{3} [/latex], [latex]\mu = 0.37372 imes 10^{-6} lb-sec/ft^{2} [/latex], [latex]\gamma = 0.076472 lb/ft^{3} [/latex]), as illustrated in Figure EP 10.8a, and settles to the ground after a windstorm. Determine the terminal (free-fall) velocity for the dust particle as it falls to the ground. (b) Mathematically illustrate how Stokes law is applied in the use of a falling sphere viscometer to determine the viscosity of a fluid.

Accepted Answer

(a) In order to determine the drag force, Equation 10.19 [latex] \sum\limits_{z}{F_{z}} = W - F_{B} - F_{D} = ma_{z} =0 [/latex] (Newton’s second law of motion) is applied in the vertical direction as follows:

[latex] \sum\limits_{z}{F_{z}} = W - F_{B} - F_{D} = ma_{z} =0 [/latex]
[latex] D_{dust}: = 0.0012 in = 1 imes 10^{-4} ft [/latex] [latex] V_{dust}: = \frac{\pi . D_{dust}^{3} }{6} = 5.236 imes 10^{-13} ft^{3} [/latex]

[latex] s_{dust}: = 3.6 [/latex] [latex] \gamma _{w}: = 62.417 \frac{lb}{ft^{3}} [/latex] [latex] \gamma _{dust}: = s_{dust}. \gamma _{w} = 224.701 \frac{lb}{ft^{3}} [/latex]
[latex] W_{dust}: = \gamma _{dust}. V_{dust} = 1.177 imes 10^{-10} lb [/latex]

[latex] \gamma _{air}: = 0.076472 \frac{lb}{ft^{3}} [/latex] [latex] F_{B}: = \gamma _{air} V_{dust}= 4.004 imes 10^{-14} lb [/latex]

Guess value: [latex] F_{D}: =1 lb [/latex]

Given

[latex] W_{dust} - F_{B} - F_{D} = 0 [/latex]
[latex] F_{D}: = Find (F_{D}) = 1.176 imes 10^{-10}lb[/latex]

In order to determine the terminal (free-fall) velocity for the dust particle as it falls to the ground, the drag force equation, Equation 10.12 [latex] F_{D} = C_{D} \frac{1}{2} ho v^{2}A [/latex], is applied. Since the dust particle is small in diameter, assume creeping flow, R ≤ 1. Furthermore, since the dust particle is assumed to be a sphere, [latex] C_{D} [/latex] = 24/R, this result for [latex] C_{D} [/latex] from applying Stokes law is applied as follows:

[latex] slug: = 1 lb \frac{sec^{2}}{ft} [/latex] [latex] ho _{air} : = 0.0023768 \frac{slug}{ft^{3}} [/latex] [latex]\mu _{air} := 0.37372 imes 10^{-6} lb \frac{sec}{ft^{2}} [/latex]

[latex] A_{front}: = \frac{\pi .D_{dust}^{2}}{4} = 7.854 imes 10^{-9} ft^{2} [/latex]

Guess value: [latex] R:= 0.1 [/latex] [latex] C_{D}: = 1 [/latex] [latex] V_{dust}: =0.5 \frac{ft}{sec} [/latex]

Given

[latex] F_{D} = C_{D} \frac{1}{2} ho _{air}.V_{dust}^{2} . A_{front } [/latex] [latex] R = \frac{ ho _{air} .V_{dust} .D_{dust}}{\mu _{dust}} [/latex] [latex] C_{D} = \frac{24}{R} [/latex]
[latex] \left ( \begin{matrix} R \ C_{D} \ V_{dust} \end{matrix} ight ) : = Find (R , C_{D}, V_{dust}) [/latex]

[latex] R= 0.212 [/latex] [latex] C_{D} = 113.012 [/latex] [latex] V_{dust} =0.334 \frac{ft}{s} [/latex]

Since R = 0.212, this confirms the assumption that R ≤1. Finally, it is important to note that when solving a numerical solve block (as opposed to an analytical solve block) in Mathcad, especially those problems that have more than one unknown to solve for, the solution procedure is highly sensitive to the guessed values. Thus, one may have to continue to adjust the guess values (for R, [latex] C_{D} [/latex] , and [latex] v_{dust} [/latex] above) until the numerical procedure converges to a solution. As such, application of the drag force equation and Newton’s second law of motion serve as check, respectively, as follows:

[latex] F_{D} := C_{D} \frac{1}{2} ho _{air}.V_{dust}^{2} . A_{front } = 1.176 imes 10^{-10} lb [/latex]
[latex] W_{dust} - F_{B} - F_{D} = 0 lb [/latex]

(b) Stokes law is applied in the use of a falling sphere viscometer to determine the viscosity of a fluid as follows. Assume the fluid with an unknown viscosity, μ is placed in a very tall clear cylinder, as illustrated in Figure EP 10.8b. A small sphere ([latex] \gamma _{s} [/latex] ) with a given weight, [latex] W = \gamma _{s} (\pi D^{3}/6)[/latex] and a given diameter, D is dropped in the cylinder of fluid. In order to determine the unknown fluid ([latex] \gamma _{f} [/latex]) viscosity, μ, first, Equation10.19 [latex] \sum\limits_{z}{F_{z}} = W - F_{B} - F_{D} = ma_{z} =0 [/latex](Newton’s second law of motion) is applied in the vertical direction as follows:

[latex] \sum\limits_{z}{F_{z}} = W - F_{B} - F_{D} = ma_{z} =0 [/latex]

Then, assuming creeping flow at R ≤ 1, Stokes law (Equation 10.17 [latex] F_{D} = 3 \pi \mu vD [/latex]) [latex] F_{D} [/latex] = 3πμvD is substituted in Equation 10.19 [latex] \sum\limits_{z}{F_{z}} = W - F_{B} - F_{D} = ma_{z} =0 [/latex] as follows

[latex] \sum\limits_{z}{F_{z}} = \gamma _{s} \frac{\pi D^{3}}{6} - \gamma _{f} \frac{\pi D^{3}}{6} - 3\pi \mu vD =0 [/latex]

Isolating the fluid viscosity, μ and simplifying the equation yields the following:

[latex] \mu = \frac{D^{2} (\gamma _{s} - \gamma _{f} )}{18v} [/latex]

Furthermore, if the viscosity, μ of the fluid is known, then the expression for the terminal (free-fall) velocity is given as follows:

[latex] v = \frac{D^{2} (\gamma _{s} - \gamma _{f} )}{18\mu } [/latex]

Question 10.8: (a) A dust particle with a diameter of 0.0012 in and a speci...

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