Question

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

• A. 1.0
• B. 1.2
• C. 1.4
• D. 1.6

Hint:

The volume of sand grains stays constant during backwash, so equate (1 - porosity) times depth before and after expansion.

Answer 1

The Correct Option is D

A cleaner way to see this problem is through the bed expansion ratio, which relates the expanded depth directly to the original depth using only the two porosity values, without writing out the solids volume balance explicitly each time.

Since the mass and volume of sand grains stay fixed while only the void space grows during backwash, the ratio of expanded depth to original depth equals the ratio of the solid fractions (before over after):

\[ \frac{L_2}{L_1} = \frac{1 - e_1}{1 - e_2} \]

Here $e_1 = 0.40$ is the service porosity and $e_2 = 0.70$ is the expanded porosity, so:

\[ 1 - e_1 = 0.60, \qquad 1 - e_2 = 0.30 \] \[ \frac{L_2}{L_1} = \frac{0.60}{0.30} = 2.0 \]

This says the bed must expand to exactly twice its original depth to raise the porosity from 40% to 70%, since the packed solids get spread over double the depth while the void fraction rises. With $L_1 = 0.8$ m:

\[ L_2 = 2.0 \times 0.8 = 1.6 \text{ m} \]

This confirms option (D). It also makes physical sense: going from 40% voids to 70% voids is a big jump in looseness, so a doubling of depth during the vigorous backwash agitation is reasonable.

\[ \boxed{L_2 = 1.6 \text{ m}} \]