Chapter 10, Problem 125P

To determine

The terminal velocity of the particle.

Answer to Problem 125P

The terminal velocity of the particle is $0.19357\text{m}/\text{s}$.

Explanation of Solution

Given information:

The density of air is $0.8588\text{kg}/{\text{m}}^{\text{3}}$, the temperature of air is $-50°\text{C}$ the pressure of air is $55\text{kPa}$, the density of the particle is $1240\text{kg}/{\text{m}}^{\text{3}}$, and the viscosity of air is $1.474\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}$.

Write the expression for the force acting on the particle in the downward direction.

  $F_{down}=\pi \frac{D^3}{6}{\rho }_{P}g$  ...... (I)

Here, the mass of the particle is $m$ the density of the particle is ${\rho }_P$ the diameter is $D$ and the acceleration due to gravity is $g$.

Write the expression for the drag force.

  $F_D=3\pi \mu VD$  ...... (II)

Here, the viscosity is $\mu$, the terminal velocity is $V$ and the diameter is $D$.

Write the expression for the buoyancy force.

  $F_{Buoyancy}=\pi \frac{D^3}{6}{\rho }_{P}g$  ...... (III)

Here, the mass of the particle is $m$ the density of the particle is ${\rho }_P$ the diameter is $D$ and the acceleration due to gravity is $g$.

Write the expression for the upward force.

  $F_{up}=F_D+F_{Buoyancy}$  ...... (IV)

Substitute $3\pi \mu VD$ for $F_D$, and $\pi \frac{D^3}{6}{\rho }_{P}g$ for $F_{Buoyancy}$ in Equation (IV).

  $F_{up}=3\pi \mu VD+\pi \frac{D^3}{6}{\rho }_{P}g$  ...... (V)

Write the expression for the force Equilibrium.

  $F_{up}=F_{down}$  ...... (VI)

Calculation:

Substitute $65\mu \text{m}$ for $D$, $1240\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_P$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$.

  $\begin{array}{c}{F}_{down}=\pi \frac{{\left(65\mu \text{m}\right)}^3}{6}\left(1240\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =\pi \frac{{\left(65\mu \text{m}\left(\frac{1\times {10}^{-6}\text{m}}{1\mu \text{m}}\right)\right)}^3}{6}\left(1240\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =5.56774\pi \times {10}^{-9}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\\ =1.749{10}^{-9}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\end{array}$

Substitute $65\text{μm}$ for $D$, $0.8588\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_P$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.475\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (VI).

  $\begin{array}{c}{F}_{up}=\left[\begin{array}{l}3\pi \left(1.474\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}\right)V\left(65\mu \text{m}\right)+\\ \pi \frac{{\left(65\mu \text{m}\right)}^3}{6}\left(0.8588\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\end{array}\right]\\ =\left[\begin{array}{l}3\pi \left(1.474\times {10}^{-5}\text{kg}/\text{m}\cdot \text{s}\right)V\left(65\mu \text{m}\left(\frac{1\times {10}^{-6}\text{m}}{1\mu \text{m}}\right)\right)+\\ \pi \frac{{\left(65\mu \text{m}\left(\frac{1\times {10}^{-6}\text{m}}{1\mu \text{m}}\right)\right)}^3}{6}\left(0.8588\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\end{array}\right]\\ =9.029\times {10}^{-9}V\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}+3.856\times {10}^{-13}\pi \text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\\ =9.029\times {10}^{-9}V\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}+1.2114\times {10}^{-12}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\end{array}$

Substitute $9.029\times {10}^{-9}V\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}+1.2114\times {10}^{-12}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}$ for $F_{up}$, $1.749{10}^{-9}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}$ for $F_{down}$ in Equation (VI).

  $\begin{array}{l}9.029\times {10}^{-9}V\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}+1.2114\times {10}^{-12}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}=1.749\times {10}^{-9}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\\ 9.029\times {10}^{-9}V\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}=1.749\times {10}^{-9}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}-1.2114\times {10}^{-12}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\\ V=\frac{1.749\times {10}^{-9}\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{9.029\times {10}^{-9}\text{kg}/\text{s}}\\ V=0.1937\text{m}/\text{s}\end{array}$

Conclusion:

The terminal velocity of the particle is $0.19357\text{m}/\text{s}$.