Filtration MCQ Quiz - Objective Question with Answer for Filtration - Download Free PDF

Explanation:

Rapid Gravity Type Filters

• Used for large-scale municipal water treatment to remove suspended solids after coagulation and flocculation.

• Water passes rapidly through a sand bed under gravity (rate: 4,000–12,000 liters/m²/hr).

• Coagulation is essential before filtration to form flocs for efficient removal.

• Requires periodic backwashing (usually every 1–2 days) to clean clogged filter media.

• Needs skilled supervision for operation, backwash management, and chemical handling.

• Initial and operational costs are high compared to slow sand filters.

• Removes turbidity, color, and microorganisms effectively when used with disinfection.

• Typically consists of a filter bed (sand), underdrain system, and wash water troughs.

Additional InformationSlow Sand Filters:

• Operate at a slow filtration rate (100–200 liters/m²/hr), much slower than rapid filters.

• Mainly used in small-scale or rural water treatment plants with low turbidity water.

• Filtration is done through a fine sand bed, forming a biologically active layer called schmutzdecke on top.

• No need for coagulation or chemical treatment before filtration.

• Require large land area due to slow filtration rate and longer filter runs.

• Cleaning is done by scraping the top sand layer, not backwashing.

• Low operational cost and does not require highly skilled supervision.

• Effective in removing bacteria, turbidity, and pathogens through biological and physical processes.

Explanation:

Concepts:

Filters are classified into three types as follows, according to the types of media used:  

Single-media filters: These filters have only one type of media and either sand or crushed anthracite coal is used as filter media.

Dual-media filters: These filters have two types of media and usually anthracite coal and sand are used as filter media.

Multi-media filters: These filters have three types of media and usually crushed anthracite coal, sand, and garnet are used as filter media.

The ideal filter is one in which pore size is increasing from bottom to top. This is done so that maximum pore volume is available at the top of the bed and gradually decreases to a minimum at the bottom. Also, the media have a gradation which is from coarse at the top to fine at the bottom.

 Based on the above two facts different filter media are arranged in the following way:

For a dual-media filter: The correct order filter material from bottom to top is sand and anthracite coal.

For Multi-Media filters: The correct order filter material from bottom to top is garnet, sand, and anthracite coal.  Garnet is placed at the bottom because it is very dense.

Explanation:

Slow Sand Filters

Slow sand filters are effective in treating raw water by removing turbidity, but their efficiency is limited to a certain threshold. This threshold is determined by the filter's ability to handle suspended solids without clogging and maintaining a proper filtration rate. The maximum turbidity level that slow sand filters can handle is discussed below:

• Key Design Considerations: Slow sand filters are designed for low turbidity levels, making them suitable for water with limited suspended solids.

• Maximum Turbidity: The raw water turbidity that can be effectively treated by slow sand filters is up to 60 mg/litre.

• Clogging Issues: Higher turbidity levels (above 60 mg/litre) can lead to clogging of the filter bed, reducing efficiency.

Analyzing the Given Options

1. Option 1: 60 mg/litre (Correct Answer)

   • Slow sand filters are designed to handle a turbidity of up to 60 mg/litre efficiently without clogging.

   • This is the recommended limit for ensuring optimal performance.

2. Option 2: 75 mg/litre (Incorrect)

   • While slow sand filters might function at 75 mg/litre, their efficiency drastically decreases.

   • This turbidity level can cause frequent clogging, requiring more maintenance.

3. Option 3: 100 mg/litre (Incorrect)

   • 100 mg/litre is beyond the capacity of slow sand filters to operate effectively.

   • This level of turbidity would require pre-treatment before filtration.

4. Option 4: 150 mg/litre (Incorrect)

   • 150 mg/litre is far beyond the design capacity of slow sand filters.

   • Such high turbidity levels would lead to immediate clogging and failure of the filter bed.

Concept:

To calculate the area of a rapid sand filter, use:

\( \text{Area} = \frac{\text{Total Water Demand (litres/hour)}}{\text{Filtration Rate}} \)

• Population = 100,000
• Demand per person = 160 litres/day
• Filtration rate = 4500 litres/hour/m²

Total hourly demand = \( \frac{100000 \times 160}{24} = 666666.67 \) litres/hour

Area = \( \frac{666666.67}{4500} \approx 148.15 \, \text{m}^2 \)

Explanation:

For a slow sand filter

The rate of filtration is 100 - 200 lit/hour/sqm

Explanation:

Given Data:

Village population = 200

The average rate of water demand = 100 litres per capita per day

average filtration rate = 100 liter/hour/m² = 100 x 24 liter/day/m²

Ratio of maximum demand to average demand = 1.5

Design dicharge = 1.5 x Village population x average rate of water demand

= 1.5 x 200 x 100 = 30000 liter/day/m2

Area of rapid sand filter = \(\frac{Design \ discharge}{Average \ filtration \ rate}\)

Area of rapid sand filter = \(\frac{30000}{100\ \times 24}\)

Area of rapid sand filter = 12.5 m2

Additional InformationFor Rapid Sand Filter

• Numbers of filters (N) = 1.22 \(\sqrt (\)Q), where Q is in MLD
• Total cross-sectional area of perforations = 0.2% of filtered area
• Cross-sectional area of on lateral is 2 or 4 times the total cross-sectional area of perforation.

Concept:

Area of filtration = Capacity/Loading rate to the filters

Number of filters = Area of filtration/Area of one filter

Given:

Capacity of plant = 1m³/s = 3600 × 24 = 86400 m³/day

Loading rate to the filters = 120 m³/day/m²

Calculation:

Area of filtration = 86400/120 = 720 m²

Number of filters required = 720/(6 × 10) = 12

When two filters are out of service for backwashing then number of filters available = 12 - 2 = 10

Area of filtration available = 60 × 10 = 600 m²

Explanation:

Explanation:

Slow sand/gravity filter and rapid gravity/sand filter:

Hence, coagulation is an essential pre-treatment for rapid gravity-type filters.

Additional Information

Pressure filter

The difference between the rapid sand filters and pressure filters:

• The flow of water is not under gravity. This implies that water inside the closed chamber is under pressure.
• The diameter of the tank is 1.5 m - 3 m and the height or length is 3.5 to 8 m.
• Operation is the same as rapid gravity filter.
• Rate of filtration is 6000 - 15000 l/m2/hr.

Explanation:

Explanation:

Comparison of slow and rapid sand filters is shown below:

Explanation:

Slow sand filter:

(i) The first filter to be used on a widespread basis for water purification were slow sand filters. These filters were constructed of fine sand with an effective size of about 0.2 mm.

Efficiency of Slow Sand Filters:

1. Bacterial Load:

(i) The slow sand filters are highly efficient in the removal of bacterial load from water. It is expected that they may remove about 98 to 99% of bacterial load from raw water. However, for complete removal of bacteria, disinfection is essential.

2. Colour:

(i) The slow sand filters are less efficient in the removal of colour of raw water. It is estimated that they remove about 20 to 25% colour of raw water.

3. Turbidity:

(i) The slow sand filters can remove turbidity to the extent of about 50 ppm. For water having greater turbidity than 60 ppm, it is necessary to give preliminary treatment.

Concept:

Rapid Sand Filter:

• Rapid sand filters use relatively coarse sand and other granular media to remove particles and impurities that have been trapped in a flocculation process.
• The unfiltered water flows through the filter medium under gravity or under pumped pressure and the material is trapped in the sand matrix surfaces.
• Rapid sand filters must be cleaned frequently by backwashing, which involves reversing the direction of the water and adding compressed air.

For rapid sand filter, Surface area of filter required = Flow rate/ rate of filtration.

Calculations:

Given, Flow rate = 10 MLD = 10 × 10⁶ litres/day and rate of filtration = 100,000 litres/day per m²

∴ Surface area of filter required \({\rm{}} = \frac{{10{\rm{\;}} \times {\rm{\;}}{{10}^6}{\rm{\;litres}}/{\rm{day\;}}}}{{100,000{\rm{\;litres}}/{\rm{day\;per\;}}{{\rm{m}}^2}}} = 100{\rm{\;}}{{\rm{m}}^2}\)

Explanation:

Hence, the most appropriate answer is option 1.