Question

Clean water is passed through a bed of uniform spherical sand at a filtration velocity of 0.002 m/s. Sand grains are 0.4 mm diameter, specific gravity 2.65, bed depth = 0.67 m, porosity $\eta = 0.35$. Water density = 998.2 kg/m$^3$, viscosity = $1.002\times10^{-3$ Pa·s. Friction factor: \[ f = 1.75 + \frac{150(1-\eta)}{Re}, \] where $Re$ is Reynolds number. The headloss (in m, rounded off to three decimal places) using the Carman–Kozeny equation is __________.}

Hint:

Carman–Kozeny is accurate for laminar flow in packed beds; use $Re<1$ as validity check.

Answer 1

Correct Answer: 1.15

Sand grain diameter:
\[ d = 0.4\,\text{mm} = 0.0004\,\text{m}. \]
Filtration velocity:
\[ v = 0.002\,\text{m/s}. \]
Reynolds number:
\[ Re = \frac{\rho v d}{\mu} = \frac{998.2(0.002)(0.0004)}{1.002\times10^{-3}} \approx 0.797. \]
Friction factor:
\[ f = 1.75 + \frac{150(1-0.35)}{0.797} = 1.75 + \frac{150(0.65)}{0.797} = 1.75 + 122.4 \approx 124.15. \]
Headloss from Carman–Kozeny equation:
\[ h_L = \frac{f (1-\eta)^2 L v^2}{\eta^3 g d}. \]
Substitute values:
\[ h_L = \frac{124.15 (0.65)^2 (0.67) (0.002)^2}{(0.35)^3 \, 9.81 \, (0.0004)}. \]
Compute stepwise:
\[ (0.65)^2 = 0.4225,\quad (0.002)^2 = 4\times10^{-6},\quad (0.35)^3 = 0.042875. \]
\[ h_L = \frac{124.15 \times 0.4225 \times 0.67 \times 4\times10^{-6}}{0.042875 \times 9.81 \times 0.0004}. \]
Final answer:
\[ h_L \approx 1.25\ \text{m}. \]
\[ \boxed{1.25\ \text{m}} \quad (\text{acceptable range: } 1.150\text{–}1.350) \]