### In this article, we will discuss filter numerical questions.

** 1. Formula **

Surface Area Required = Q/Rate of filtration |

Geometrical Method P_{n} = P_{o}(1+r/100)^{n} |

Q = A/V |

**2. Filter Numerical **

**1. Determine the amount of bleaching powder required annually in a water treatment plant treating 10 MLD of water if 0.3 ppm of chlorine dosage is required. Available bleaching powder contains 27% of chlorine.**

**Solution:**

Flow Q = 10 x 10^{6} lit/day = 10000 m^{3}/day

Dosage= 0.3 mg/lit

Chlorine required= Flow x Dosage

= 0.3 x 10^{-6} kg/lit x 10 x 10^{6} lit/day

= 3 kg/day

Bleaching powder required= 3/0.27

= 11.11 kg/day

Bleaching powder required = 4.05 tones

**2. Chlorine usage in the treatment of 25 MLD of water is 9 kg/day. The residual chlorine after 10 minutes of contact is 0.2 mg/lit. Calculate the dosage in mg/lit and chlorine demand of water.**

Solution:

Flow of water (Q) = 25 x 10^{6} lit/day = 25000 m^{3}/day

Quantity of chlorine used = 9 kg/day = 9×10^{6} mg/day

Dosage or chlorine used= 9×10^{6} / 25×10^{6} = 0.36 mg/lit

Chlorine demand = Dosage – Residual chlorine

= (0.36-0.2) mg/lit

= 0.16 mg/lit

**3. Design slow sand filter beds from the following data:**

**Population to be served= 60,000 nos**

**Average rate of demand= 160 lpcd**

**Rate of filtration= 150 liters/hr/m ^{2}**

**L:B = 2:1**

**Assume design discharge as twice the average flow and;**

**Assume that one unit will be kept on standby.**

**Solution:**

The average quantity of water to be treated = 60,000 x 160 lpcd

= 96,00,000 lit/day

Design discharge (Q) = 2x 96,00,000 lit/day

= 19.2 x 10^{6} lit/day

Rate of filtration= 150 liters/hr/m^{2}

= 24 x 150 = 3600 liters/day/m^{2}

**Surface area required= Q/ Rate of filtration**

= 19.2×10^{6}/3600

=5333.34 m^{2}

Assuming 5 filter units;

Area for each filter bed(a_{s}) = 5333.34/5 = 1066.67 m^{2}

As = L x B = 2B^{2} = 1066.67

B= √1066.66/2

B= 23.09 m

L= 46.02 m

Breadth(B) = 23.09 m

Length (L) = 46.02m

Providinf freeboard = 0.5 m, water depth=1 m, depth filter media = 1m, depth of base material= 0.6, depth of under drain pipe= 0.2 m, therefore overall depth of filter= 3.30

Providing one filter unit standby, required filter units= 6 with a dimension of 46.02 m x 23.10 m x 3.30 m.

Read More: Hardness and Alkalinity Numerical |

**4. Average water consumption rate is 150 lpcd in an urban area. Design a slow sand filter for a community having a population of 10000 at the base year 2068.**

Solution:

Assume annual population growth rate 1.7% and design period= 15 years

Using a geometrical method for forecasting population of the design year 2083;

**Geometrical Method P _{n} = P_{o}(1+r/100)^{n}**

The average quantity of water to be treated,

= 12877 nos x 150 lpcd

= 1931550 lit/day

Let design discharge is equal to the average quantity of water (Q)

= 1931550 lit/day

Assuming rate of filtration = 150 liters/hr/m^{2}

= 24 x 150 = 3600 liters/hr/m^{2}

Surface area required= Q/Rate of filtration

= 1931550/3600

=536.54 m^{2}

Assuming 3 filter units;

Area for each filter bed (a_{s}) = 536.54/3

= 178.84 m^{2}

As= Lx B = 2B2 = 178.84 m^{2}

B= √178.84/2

B= 9.5 m

L= 19 m

Breadth (B)= 9.5 m, Length (L) = 19m

Providing free board=0.5 m, water depth= 1m, depth filter media= 1m, depth of base material=0.6m, depth of under drain pipe= 0.2m. Therefore overall depth of filter= 3.30m

Providing one filter unit standby, required filter units= 4 with a dimension of 19 m x 9.5 m x 3.30 m.

Read Also: Sedimentation Tank Numerical |

**5. Calculate the dimension of a set of rapid sand filters for treating water required for a population of 0.1 million with an average rate of demand of 200 lpcd.**

Solution:

Average quantity of water to be treated= 0.1 x 10^{6} x 200 lpcd

= 20 x10^{6} lit/day

Assuming that 3 % of filtered water is required for filter backwashing and time required for backwashing= 30 minutes

Discharge (Q) = 20000000 lit/day

= 20×10^{6} lit/day

= 833333.34 lit/hr

Design **discharge** (Q) = 833333.34/(1-0.03)x(24-0.5)

= 36557.72 lit/hr

Assume rate of filtration = 4000 liters/hr/m^{2}

= 24 x 4000

= 96000 liters/day/m^{2}

Surface area required = Q/ Rate of filtration

= 877385.39/96000

=9.13 m^{2}

Minnimum Surface area (As) = 10 m^{2}

Assuming,

L:B=1.3:1

As= LxB

B= 2.8 m

L= 3.65 m

Breadth (B) = 2.8 m, Length (L) = 3.65 m

Provide, freeboard=0.5 m, water depth= 1.6 m, Filter media = 0.6 , Base material = 0.6m, Under drain pipe depth= 0.2m. Therefore, overall depth = 3.5 m

Providing one filter unit standby, required filter units= 2 with the dimension of 3.65m x 2.8m x 3.5m.