A rapid sand filter bed of depth 0.8 m has 40 % porosity during service cycle. It is recommended that during backwash operation, the expanded filter bed should have 70 % porosity. The uniform expanded depth (in m) of the filter bed during backwash is

The correct option is 1.6.

To find the uniform expanded depth of the filter bed during the backwash operation, we can use the principle of conservation of mass for the solid sand particles. During backwashing, the volume of sand grains (solid particles) remains constant, even though the total volume of the bed increases due to expansion (which increases the porosity/void volume).

Let:
_L1 = initial depth of the filter bed during the service cycle = 0.8 m
_n1 = porosity of the filter bed during the service cycle = 40% = 0.40
_L2 = expanded depth of the filter bed during backwashing
_n2 = porosity of the expanded filter bed during backwashing = 70% = 0.70
_A = cross-sectional area of the filter bed (which remains constant)

The volume of the solid sand particles (_Vs) can be expressed in terms of the total bed volume (_V) and the porosity (_n) as follows:
_Vs = _V × (1 - _n)

Since the cross-sectional area _A is constant, the total volume of the bed is _V = _A × _L. Therefore, the volume of the solid particles is:
_Vs = _A × _L × (1 - _n)

Equating the volume of solid particles before expansion (during the service cycle) and after expansion (during the backwash operation):
_A × _L1 × (1 - _n1) = _A × _L2 × (1 - _n2)

Since the area _A is the same on both sides, it cancels out, leaving the relation:
_L1 × (1 - _n1) = _L2 × (1 - _n2)

Now, we substitute the given values into the equation to solve for the expanded depth _L2:

$$0.8\times (1-0.40)=L_2$$

Simplify the terms inside the parentheses:

$$0.8\times 0.60=L_2$$

Calculate the product on the left side:

$$0.48=L_2\times 0.30$$

Solve for _L2:

$$L_2=\frac{0.48}{0.30}$$

$$L_2=1.6\text{m}$$

Thus, the uniform expanded depth of the filter bed during the backwash operation is 1.6 m.