Answer
given,
hydraulic loading rate = 6 m/h
total Volumetric flow rate = 3 MGD
Effective filtration rate = 5.8 m/h
Production efficiency = 97%
filter cycle duration = 48 h
3 MGD = 3 x 0.04381 = 0.131 m³/s = 471.6 m³/hr
[tex]Q = V_a A_f[/tex]
Q = 6 x A_f
471.6 = 6 x A_f
A_f = 78.6
x² = 78.6
x = 8.86 m
b) Production efficiency
[tex]\eta_f = \dfrac{V_f-V_b-V_r}{V_f}[/tex]
[tex]0.97 = \dfrac{V_f-V_b-V_r}{V_f}[/tex]..........(1)
eff. filtration rate
[tex]5.8 = \dfrac{V_f-(V_b+V_r)}{A_ft_c}[/tex]
[tex]0.97 V_f = 5.8 \times A_f \times t_c[/tex]
[tex]V_f = 22559\ m^3[/tex]
from equation (1)
volume of water required for back washing = V_b + V_r
= 676.77 m³
