[A]
(a) The buoyant and drag forces on the sphere:
$
F_B = \rho \frac{\pi}{6} D^3 g
\quad \text{and} \quad
F_D = \frac{1}{2} \rho u_z^2 \frac{\pi}{4} D^2 C_D
= \frac{1}{2} \rho u_z^2 \frac{\pi}{4} D^2
\left( \tfrac{1}{2} + \tfrac{24}{Re_D} \right)
$
The approximate equation for the sphere's velocity:
$
m \frac{du_z}{dt}
= mg - \rho \frac{\pi}{6} D^3 g
- \frac{1}{2} \rho u_z^2 \frac{\pi}{4} D^2
\left( \tfrac{1}{2} + \tfrac{24}{Re_D} \right)
$
Use
$u_z = \mu Re_D / \rho D$ to eliminate $u_z$:
$
m \frac{du_z}{dt}
= m \frac{\mu}{\rho D} \frac{dRe_D}{dt}
= mg - \rho \frac{\pi}{6} D^3 g
- \frac{1}{2} \rho \left( \frac{\mu Re_D}{\rho D} \right)^2 \frac{\pi}{4} D^2
\left( \tfrac{1}{2} + \tfrac{24}{Re_D} \right) $
$\Rightarrow \frac{m \mu}{\rho D} \frac{dRe_D}{dt}
= \Bigg( m - \rho \frac{\pi}{6} D^3 \Bigg) g
- \frac{\pi}{8\rho} \mu^2 Re_D^2
\left( \tfrac{1}{2} + \tfrac{24}{Re_D} \right)
$
For the dimensionless time $t^*$ using the chain rule for differentiation and $t^* = \rho g t D / \mu$
$
\frac{m \mu}{\rho D} \frac{dRe_D}{dt}
= \frac{m \mu}{\rho D} \left( \frac{dRe_D}{dt^*} \frac{dt^*}{dt} \right)
= \frac{m \mu}{\rho D} \left( \frac{dRe_D}{dt^*} \frac{\rho g D}{\mu} \right)
= mg \frac{dRe_D}{dt^*}
= \Bigg( m - \rho \frac{\pi}{6} D^3 \Bigg) g
- \frac{\pi}{8\rho} \mu^2 Re_D^2
\left( \tfrac{1}{2} + \tfrac{24}{Re_D} \right)
$
Divide both sides of the final equality by $mg$:
$
\frac{dRe_D}{dt^*}
= - \frac{\pi}{16} \left( \frac{\mu^2}{\rho m g} \right) Re_D^2
- 3\pi \left( \frac{\mu^2}{\rho m g} \right) Re_D
+ \left( 1 - \frac{\pi}{6} \frac{\rho D^3}{m} \right)
$
\(\therefore A = - \frac{\pi}{16} \left( \frac{\mu^2}{\rho m g} \right),
\quad B = -3\pi \left( \frac{\mu^2}{\rho m g} \right),
\quad C = 1 - \frac{\pi}{6} \frac{\rho D^3}{m}\)
(b) Assuming $C$ is positive, intergrate the scaled equation:
$
\int \frac{dRe_D}{A Re_D^2 + B Re_D + C}
= \int dt^*,
t^* + \text{const.} = -\frac{2}{\sqrt{B^2 - 4AC}}
\tanh^{-1} \left( \frac{2A Re_D + B}{\sqrt{B^2 - 4AC}} \right)
$
$\Rightarrow Re_D = \frac{1}{2A} \Bigg\{ -B + \sqrt{B^2 - 4AC} \, \tanh \Bigg[ -\tfrac{1}{2} \sqrt{B^2 - 4AC} \, (t^* + \text{const.}) \Bigg] \Bigg\}
$
With the initial condition $Re_D = 0$ at $t = 0$:
$
Re_D = \frac{1}{2A} \Bigg\{ -B + \sqrt{B^2 - 4AC} \, \tanh \Bigg[
-\frac{\sqrt{B^2 - 4AC}}{2} t^* + \tanh^{-1} \Bigg( \frac{B}{\sqrt{B^2 - 4AC}} \Bigg) \Bigg] \Bigg\}
$
(c) Terminal velocity will occur when $\tfrac{dRe_D}{dt^*} = 0$, $Re_T$ is the Reynolds number at terminal velocity: \(0 = -\frac{\pi}{16} \left( \frac{\mu^2}{\rho m g} \right) Re_T^2
- 3\pi \left( \frac{\mu^2}{\rho m g} \right) Re_T
+ \left( 1 - \frac{\pi}{6} \frac{\rho D^3}{m} \right)
\Rightarrow 0 = Re_T^2 + 48 Re_T - \frac{16}{\pi} \left( \frac{\rho m g}{\mu^2} \right)
\left( 1 - \frac{\pi}{6} \frac{\rho D^3}{m} \right)\)
\(\because\) $Re_T > 0$: \(Re_T = \frac{1}{2} \Bigg( -48 + \sqrt{48^2 - 4(1) \left(-\frac{16}{\pi} \left(\frac{\rho m g}{\mu^2}\right)
\left(1 - \frac{\pi}{6} \frac{\rho D^3}{m}\right)\right)} \Bigg)\)
\(\therefore (u_z)_T = \frac{\mu}{\rho D} \left( \frac{-B}{2A} + \sqrt{\left(\frac{B}{2A}\right)^2 - \frac{C}{A}} \right)
= \frac{\mu}{\rho D} \left( -24 + \sqrt{24^2 + \frac{16}{\pi} \left(\frac{\rho m g}{\mu^2}\right)
\left(1 - \frac{\pi}{6} \frac{\rho D^3}{m}\right)} \right)\)