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

Step 1: Understanding the Question:
A filter bed has a fixed volume of solid sand grains. During service, this sand occupies a depth of 0.8 m with 40% porosity (voids). During backwash, the upward water flow lifts and expands the bed, spreading the same solid grains over a larger depth, so the porosity increases to 70%. We need the new (expanded) depth.

Step 2: Key Formula or Approach:
The total volume of solid sand grains does not change between the service condition and the expanded (backwash) condition, only the bed depth and the void space change. For a bed of depth \(L\) and porosity \(e\), the solids occupy a fraction \((1-e)\) of the bed volume. Since the cross-sectional area of the filter is the same before and after expansion, the volume of solids per unit area is \((1-e)L\), and this stays constant:
\[ (1 - e_1) L_1 = (1 - e_2) L_2 \]
where subscript 1 is the service condition and subscript 2 is the expanded (backwash) condition.

Step 3: Detailed Explanation:
Given: \(L_1 = 0.8\) m, \(e_1 = 0.40\), \(e_2 = 0.70\).
Solid fraction during service: \(1 - e_1 = 1 - 0.40 = 0.60\).
Solid fraction during backwash: \(1 - e_2 = 1 - 0.70 = 0.30\).
Apply the constant-solids relation:
\[ 0.60 \times 0.8 = 0.30 \times L_2 \]
\[ 0.48 = 0.30 \times L_2 \]
\[ L_2 = \frac{0.48}{0.30} = 1.6 \text{ m} \]

Step 4: Final Answer:
The expanded depth of the filter bed during backwash is 1.6 m. \[ \boxed{L_2 = 1.6 \text{ m}} \]