ε = 0.33

ρ = 62.3 lb/ft³

ΔL = 1 ft

g = 32.2 ft/sec²

Applying Bernoulli's equation from the top surface of the fluid to the outlet of the packed bed and ignoring the kinetic-energy term and the pressure drop through the support screen, which are both small, we find

g(ΔL) = h_f

h_f = Δp/ρ

For laminar flow, (Blake-Kozeny Equation)

\(\displaystyle \frac{\Delta p}{\rho} = 150\frac{V_s\mu(1-\varepsilon)^2\Delta L}{D_p^2\varepsilon^3\rho} \)

Therefore, V_s = 32.2 x 1.25 x (0.05/12)² x 0.33³ x 62.3 / ( 150 x (1 x 6.72x10^-4) x (1 - 0.33)² x 1)

= 0.035 ft/sec = 0.011 m/sec.

Q = AV_s = (2/12)² x (π/4) x 0.035 = 0.00075 ft³/sec = 21 cm³/sec

Before accepting this as the correct solution, we check the NRe_pm.

NRe_pm= (0.05/12) x 0.035 x 62.3 / (1 x 6.72x10^-4 x (1 - 0.33) ) = 20.2

This is slightly above the value of 10 (up to which the Blake-Kozeny Equation can be used), for which we can safely use without appreciable error.