i) If the ground water contains iron and magnesium, then it is to be removed by Aeration process.

Fe²⁺ + O₂ + H₂O → Fe (OH)₃ ↓ + H⁺

Mn²⁺+ O₂ + H₂O → MnO₂ ↓ + H⁺

ii) If the groundwater also contains Ca²⁺ and Mg²⁺ then its softening is required as both calcium and magnesium add hardness to the water.

iii) Disinfection is the last treatment given to any water.

iv) Aeration is always performed before coagulation and flocculation, sedimentation, softening, and filtration.

Chlorine Demand × Amount of sample = Amount of Bleaching powder × 100/ (% chlorine in bleaching powder)

$0.2\times 1000=AmountofBleachingPowder\times \frac{30}{100}$

Amount of Bleaching powder$=\frac{0.2\times 1000\times 100}{30}=666.67mg=0.67g$

Concept:

Starch Iodide Test (Indo-metric Test):

It is used when the presence of nitrite, N₂, manganic compounds which makes orthotolidene test unsuitable. This test is more precise than orthotolidene test particularly when residual chlorine is more than 1 ppm.

The indicators are starch and potassium iodide and the titrant is sodium thiosulphate.

1 litre water + 10 ml (kI) + 5 ml (starch) → Blue colour is formed titrated with $\frac{N}{100}N{a}_2{S}_2{O}_3$.

Important Point:

Different tests for Chlorine Residue:

(a) Orthotolidene test

(b) DPD test

(c) Chlorotex test

(d) Starch iodide test

Concept:

Weir Loading Rate (WLR):- Discharge per unit length of weir is known as weir loading rate.

$\mathrm{W}\mathrm{L}\mathrm{R}=\frac{\mathrm{D}\mathrm{i}\mathrm{s}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}{\mathrm{L}\mathrm{e}\mathrm{n}\mathrm{g}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{f}\mathrm{w}\mathrm{e}\mathrm{i}\mathrm{r}}=\frac{\mathrm{Q}}{\mathrm{L}}$

It is expressed in m3/day/m

Surface Loading Rate (SLR) or Surface Overflow Rate (SOR):- Volume of water applied in unit time per unit surface area of settling tank is known as Surface Loading Rate. It is also denoted by Vo.

$\mathrm{S}\mathrm{O}\mathrm{R}\mathrm{o}\mathrm{r}{\mathrm{V}}_{\mathrm{o}}=\frac{\frac{\mathrm{V}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}}{\mathrm{T}\mathrm{i}\mathrm{m}\mathrm{e}}}{\mathrm{S}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}}=\frac{\mathrm{D}\mathrm{i}\mathrm{s}\mathrm{c}\mathrm{h}\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{e}}{\mathrm{S}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}}$

It is expressed in m3/d/m2

Calculation:

Surface overflow rate $=\frac{flowrate}{surfacearea}$

$40m/day=\frac{Q}{B\times L}$

⇒ Q = 40 m/day × (50 × 5) m²

⇒ Q = 10000 m³/day

Weir length provided is 40m

Therefore, weir loading rate

$=\frac{Q}{L}=\frac{10000}{40}=250m^3/day/m$

Concept:

Temporal mean velocity gradient (G):-  It is a measure of the relative velocity of two particles of fluid and distance between them. The unit of G is sec^-1.

$G=\frac{V_2-V_1}{x}$

Turbulence and the resultant intensity of mixing is based on the rate of power input to the water,

$G=\sqrt{\frac{P}{\mu V}}$

Where, P is power in watts and μ is dynamic viscosity in $\frac{NS}{m^2}$

Calculation:

G = 300/sec, ν  = 1.3 × 10^-6 m²/s, P = 1 kW

$\mu =\nu \times \rho =1.31\times {10}^{-6}\times {10}^3=1.31\times {10}^{-3}$

$\therefore 300=\sqrt{\frac{1\times {10}^3}{1.31\times {10}^{-3}\times V}}\Rightarrow V=8.48m^3$

Mistake point:

The dynamic viscosity is used in the calculation of the Temporal mean velocity gradient. Note that kinematic viscosity is given in the Qs and hence we need to obtain the dynamic viscosity.

Concept:

The slow sand filter is designed for maximum daily demand.

Maximum Daily Demand = 1.8 × Average daily demand

$\frac{Q_{design}}{Rateoffiltration}=Planarea$

Calculation:

Q_avg = 40000 × 150 = 6 × 10⁶ litres per day

∴ Q_design = 1.8 × Q_avg = 1.8 × 6 × 10⁶ = 10.8 × 10⁶ litres per day

Total plan area = $\frac{Q_{design}}{Rateoffiltration}$ ⇒ Total plan area = $\frac{10.8\times {10}^6}{150\times 24}$ = 3000 m²

∴ Number of filters required $=\frac{3000}{500}=6$

Assuming 1 filter on standby, the total number of filters required = 7

Concept:

Total lime required based on total hardness is given by:

Lime as CaO required (mg/l) = TH × $\frac{56}{100}$

Total soda required based on non-carbonate hardness is given by

Soda required (mg/l) = NCH × $\frac{106}{100}$

Calculation:

Total hardness (TH) = 200 mg/l

Total alkalinity (TA) = 150 mg/l

∴ Non carbonate hardness (NCH) = 50 mg/l

∴ Lime as CaO required (mg/l) $=200\times \frac{56}{100}=112$ mg/l

∴ Total lime required per day = 112 × 1 = 112 kg

Total soda required $=50\times \frac{106}{100}=53$ mg/l

∴ Total soda required = 53 kg per day

Concept:

Chicks Watson Law:

Chemical inactivation of a specific species of microorganism is a function of disinfection concentration and contact time. Other important factors are the kind of disinfectant, temperature, pH, and the presence of suspended organic matter.

The rate equation for the inactivation of microorganism is given by Chicks-Watson Law

$\frac{dN}{dt}=-K{C}^{n}N$

K = Rate constant which depends on the type of microorganism, the type of disinfectant and temperature

n = Constant for a particular microorganism and type of disinfectant

C = Disinfectant concentration (weight/volume)

N = Number of microorganism present at any time ‘t’

For the constant concentration of disinfectant,

$\int \frac{dN}{N}=-K{C}^n\int dt$

⇒ In N = -KCnt + C1

At t = 0, N = N0 ⇒ C1 = In N0

$\Rightarrow ln\frac{N}{N_0}=-K{C}^{n}t\Rightarrow N=N_0{e}^{-k{C}^{n}t}$

Calculation:

Q = 2760 m³/day

The volume of the tank = 50 m³

∴ Detention time (t_d) $=\frac{Volumeofthetank}{Discharge}=\frac{50}{2760}$ day = 26.09 min

Disinfection model, N_t = N₀ e^{-0.15 t}

∴ Micro-organisms present after detention time (t_d),

$N_{t_d}=N_0{e}^{-0.15\times 26.09}\approx 0.02N_0$

∴ % Removal of microorganisms = $\frac{N_0-N_{t_d}}{N_0}\times 100=\frac{N_0-0.02N_0}{N_0}\times 100$ = 98%

Concept:

The surface overflow rate is defined as the volume of water flow per unit of time per unit surface area. It can also be thought of as settling velocity of that particle which if introduced at the topmost point at the inlet will reach the bottom-most point at the outlet of the tank.

A particle having setting velocity less than overflow rate will not get 100% removed.

Percentage removal of the particle having setting velocity $V_s^{\prime }=\frac{V_s^{\prime }}{V_s}\times 10$

Where, V_s = Overflow rate

Efficiency of tank $=\frac{WeightofsuspendedsolidsRemoved}{WeightofsuspendedsolidsPresent}$

Weight Removed = Quantity Present × % Removed

= (1000 × 0.1) × 0.667 + (1000 × 0.2) × 0.834 + (1000 × 0.15) + (1000 × 0.05) + (1000 × 0.3) + (1000 × 0.2) = 933.50 gm

∴ Efficiency of tank $=\frac{WeightofsuspendedsolidsRemoved}{WeightofsuspendedsolidsPresent}=\frac{933.50}{1000}\times 100=93.35\mathrm{\%}$

Concept:

Hydraulic Head Loss and Expansion of the Filter bed during Backwash:

Backwashing is accomplished by reversing the flow in filter media to force clean water to move upward through the filter media. In order to clean the filter bed, it is required to expand the bed, so that granular media filters are no longer in contact with each other.

To hydraulically expand a porous bed, the head loss (h_Le) must be at least equal to the buoyant weight of the filter bed. Even when the bed gets expanded to depth D_e, the head loss through the expanded bed essentially remains unchanged, because the total buoyant weight of the bed is constant.

Head loss, h_Le = D (1 – n) (G – 1)       ----(i)

where D = undisturbed depth of filter bed in m.

n = porosity

G = specific gravity

Also, h_Le = D_e (1 - n­_e)(G – 1)        ----(ii)

where D_e = the depth of expanded bed in m.

n_e = porosity of expanded bed

∴ from equation (i) and (ii) we get:

$\frac{D_e}{D}=\frac{1-n}{1-n_e}$

Calculation:

$\frac{D_e}{D}=\frac{1-n}{1-n_e}$

$\frac{0.75}{0.675}=\frac{\left(1-0.42\right)}{1-n_e}$

⇒ n_e = 0.478

Therefore,

h_Le = D_e (1 – n_e) (G – 1)

h_Le = 0.75 (1 – 0.478) (2.72 – 1)

h_Le = 0.67338 m

∴ h_Le = 673.4 mm

Alternate Solution:

Head loss in both the condition will be same i.e.

h_Le = D (1 – n) (G – 1) = De (1 – n_e)(G – 1)

h_Le = 0.675 (1 – 0.42) (2.72 – 1)

h_Le = 0.67338 m

h_Le = 673.38 mm