Question

For a dual-media filter with 0.5 m of anthracite coal and 0.3 of sand. Other data for anthracite: specific gravity is 1.67, porosity is 0.5, d₉₀ grain size is 1.5 mm. Other data for sand: specific gravity is 2.65, porosity is 0.4, d₉₀ grain size is 0.9 mm. The water temperature is 20C, so the mass density of water (ρ) is 998.2 kg/m³ , density of sand is 1650 kg/m³ , density of anthracite coal is 1506 kg/m^3, the absolute viscosity is 1.002 x 10^-3 kg/m-s. Assuming a 30% margin of safety for backwash, calculate the following:

a) the recommended backwash rate for the filter (gpm/sf)

b) the pressure drop through the filter during fluidization (psi)

Note: multiply (L/m²-s) by 1.473 to get gpm/sf. Multiply (N/m²) by 0.000145 to get psi.

Answer 1

Ans a) We know,

V_b = [(1135.69 + 0.0408 G_n)^0.50 / d₉₀ ] - (33.7/ d₉₀)  

where, = viscosity

   = density of water

   G_n = Galileo number

   d₉₀ = sieve size that passes 90% by weight

Also, G_n = ( s - )g d₉₀³ / ²

where, _s = density of sand

Putting values,

G_n = 998.2 ( 1650 - 998.2) (9.81) (0.0015)³ / (0.001002)²

=> G_n = 21455.53

    Hence,

V_b = [0.001002((1135.69 + 0.0408(21455.53))^0.50 / (998.2 x 0.0015) ] - (33.7 x 0.001002/ 998.2 x 0.0015)

= 0.030 - 0.0225

= 0.0075 m³ / m² s

or 7.5 L /m² s

Design backwash rate = 1.3 x V_b

= 9.75 L/m²s

=> 9.75 x 1.47 = 14.5 gpm/sf

b) Pressure drop through filter = L g ( 1 - n) (_s - ) ,where n = porosity

= 0.8 x 9.81 ( 1 - 0.5) (1650- 998.2)

= 2557.67 N/m²

or 0.37 psi