Environmental Engineering Test 1

Complete removal of hardness cannot be accomplished by chemical precipitation. Precipitation of the supersaturated solution of CaCO₃ will continue slowly, however, resulting in deposits in water lines and storage facility. It is, therefore, necessary to stabilize the water concerting supersaturated CaCO₃ back to soluble form. This process is generally known as re-carbonation.

The most common practice is

\(\begin{array}{l} CaC{O_3} + C{O_2} + {H_2}O \to C{a^{2 + }} + 2\left( {HCO_3^ - } \right)\\ Mg{\left( {OH} \right)_2} + 2C{O_2} \to M{g^{2 + }} + 2\left( {HCO_3^ - } \right) \end{array}\)

Concept:

The total microbial mass in the aeration system (M) is computed by multiplying the average concentration of solids in the mixed liquor of the aeration tank, called Mixed Liquor Suspended Solids with the volume of the tank (V).

∴ M = MLSS × V

Calculation:

The concentration of mixed liquor suspended solids,

MLSS = 1000 mg/l ⇒ \(\frac{{1000 \times {{10}^{ - 6}}}}{{{{10}^{ - 3}}}}\;kg/{m^3} = \;1\;kg/{m^3}\)

For a volume of 400 m³,

M = 1 × 400 = 400 kg

Concept:

The self-cleansing velocity is the minimum velocity through which the sewage should flow through sewer such that there’s no silt deposition along its length.

\({{\rm{v}}_{\rm{s}}} = \sqrt {\frac{{8 \times k \times g \times d \times \left( {G - 1} \right)}}{f}}\)

where

k = sediment coefficient

f = frictional factor

g = acceleration due to gravity

G = specific gravity

d = mean diameter of the silt

Calculation:

From the above formula, \({{\rm{v}}_{\rm{s}}} \propto \frac{1}{{\sqrt f }}\)

\(\therefore \frac{{{{\rm{v}}_{{\rm{s}}1}}}}{{{v_{s2}}}} = \frac{{\sqrt {{f_2}} }}{{\sqrt {{f_{\;1}}} }} \Rightarrow {v_{s2}} = \frac{1}{{\sqrt 2 }} \times {v_{s1}}\)

Concept:

Carbon-nitrogen ratio:

• C/N of the input material is an important factor for the bacterial activity to continue, since the bacteria use nitrogen for building their cell structures and carbon for food.
• The anaerobic bacteria, use up carbon about 30-50 times faster than they use up nitrogen. For proper growth of anaerobic digestion, C/N ratio of the digestive material should be between 30 to 50 for optimum digestion.
• When there is too much of carbon, i.e. C/N ratio is higher than the optimum, then nitrogen will be used up and carbon left over, thereby leaving the digestion of organic matter incomplete.
• When there is too much of nitrogen, i.e C/N ratio is lower than the optimum, then the carbon get exhausted and fermentation stop, leaving nitrogen in the digestor which combines with hydrogen to form ammonia (NH₃). This can inhibit growth of bacteria mostly the methane formers.
• The anaerobic digestion requires an optimum C/N ratio of 30-50.

Concept:

Hardness: It is defined as the concentration of multivalent metallic cations in solution. Multivalent metallic ions most abundant in natural water are Calcium & Magnesium. Other ions that lead to hardness are Fe²⁺, Mn²⁺, Strontium (Sr²⁺) and Aluminium (Al³⁺). For all practical purposes, hardness may be represented by the sum of Ca²⁺ and Mg²⁺, ions as other cations are present in negligible amount.

Hardness can be divided into two parts i.e. carbonate hardness and non-carbonate hardness.

Case 1: If Total hardness is greater than Total alkalinity

Carbonate hardness = Total alkalinity and Non-carbonate hardness = Total hardness - Carbonate Hardness

Case 2: If Total hardness is less than Total alkalinity

Carbonate hardness = Total alkalinity and Non-carbonate hardness = 0 (zero)

Calculation:

TH = 320 mg/L, Total alkalinity = 280 mg/L

∵ TH > TA

∴ Carbonate Hardness = Minimum of TH and TA = 280 mg/L

∴ Non Carbonate Hardness = Total hardness – CH = 320 – 280 = 40 mg/L

Concept:

When more than three dilutions of samples are used (like 10 ml, 1 ml, 0.1 ml, 0.01 ml, 0.001 ml, 5-different dilutions), following procedure is adopted to obtain MPN from MPN chart:

(a) Only three consecutive dilutions are used to determine MPN

(b) If all dilutions tested yield positive result, select the three highest dilutions.

(c) If one or more dilutions have all tubes positive, select the highest dilutions (i.e. smallest sample quantity like from 10 ml, 1 ml chose 0.1 ml) with positive result and next two higher dilutions.

(d) When none of the dilution yield all tubes positive, select the three lowest dilutions for which middle dilution contains positive result.

(e) If a positive result occurs in a higher unselected dilution, add the no. of positive tubes in the dilution to the result of the highest dilution of the three selected.

Calculation:

Choosing the three lowest dilution because none of the dilutions yield all positive results,

MPN will be corresponding to 4 – 3 – 2 of 10 mL, 1 mL and 0.1 mL respectively.

∴ MPN = 26

There are mainly two methods (Indore and Bangalore method) adopted in India for decomposition of Solid wastes generated. The main difference between Indore Method and Bangalore Method for decomposition of MSW is given below:

The most effective chemical to be used as coagulants are alum, ferrous sulphate, ferric chloride, ferric chloride, and ferric sulphate.

a) Use of Alum (Al₂ (SO₄)₃.18H₂O) as coagulant:

Alum or filter alum has proved to be a very effective coagulant, and it is now extensively used throughout the world. It is quite cheap, forms an excellent stable floc, and does not require any skilled supervision for handling.

Al₂(SO4)₃.18H₂O + 3Ca(HCO₃)₂ → 3CaSO₄ + 2Al(OH)₃ + 6 CO₂

The only drawback in the use of alum is that the effective pH range for its use is small, i.e. 6.5 to 8.3, and in many cases it requires the addition of external alkali salts, thereby rendering it costlier.

b) Use of Copperas (FeSO₄.7H₂O) as coagulant:

Copperas (FeSO₄.7H₂O) is extensively used as a coagulant for raw waters that are not coloured. It is generally cheaper than alum and functions effectively in the pH range of 8.5 and above.

\({\bf{FeS}}{{\bf{O}}_4}.7{{\bf{H}}_2}{\bf{O}} + {\bf{Ca}}{\left( {{\bf{OH}}} \right)_2} \to {\bf{CaS}}{{\bf{O}}_4} + {\bf{Fe}}{\left( {{\bf{OH}}} \right)_2} + 7{{\bf{H}}_2}{\bf{O}}\)

it is not used for coloured raw waters. The quantity of copperas required is almost the same as that of alum.

c) Use of Chlorinated copperas as coagulant:

The resultant combination of ferric sulphate and ferric chloride is known as chlorinated copperas, and is a valuable coagulant of removing colours, especially where the raw water has a low pH value.

6(FeSO₄.7H₂O) + 3Cl₂ → 2Fe₂(SO₄)₃ + 2FeCl₃ + 42H₂O

Ferric sulphate is quite effective in the pH range of 4 to 7 and above 9, whereas Ferric chloride is quite effective in the pH range of 3.5 to 6.5 and above 8.5. The combination has, therefore, proved to be a very effective coagulant for treating low pH waters.

d) Use of sodium aluminate (N_a2 Al₂ O₄) as coagulant:

\({\rm{N}}{{\rm{a}}_2}{\rm{A}}{{\rm{l}}_2}{{\rm{O}}_4} + {\rm{Ca}}{\left( {{\rm{HC}}{{\rm{O}}_3}} \right)_2} \to {\rm{CaA}}{{\rm{l}}_2}{{\rm{O}}_4} + {\rm{N}}{{\rm{a}}_2}{\rm{C}}{{\rm{O}}_3} + {\rm{C}}{{\rm{O}}_2} \uparrow + {{\rm{H}}_2}{\rm{O}}\)

This coagulant is about 1½ times costlier than alum, and is, therefore, generally avoided for treating ordinary public supplies, but, however, it is very useful for treating waters which do not have the natural desired alkalinity, and thus cannot be treated with pure alum.

Concept:

Estimation of solids:

(1) Total solid (Suspended + Dissolved)

a) Determined by gravimetry test

b) Raw water when heated at 104 - 105°C for 4-5 hours resulted in total solid.

(2) Suspended solid

Raw water when passed through Whattman filter paper no. 9, the filter paper collects suspended solids which when dried under hot air oven gives total suspended solid.

(3) Dissolved solid

The filtered water when heated at 104° - 105°C for 4 to 5 hours gives dissolved solids, which when subjected to mettle furnace (700°C - 1000°C for 4 to 5 hours) results in nonvolatile solids and hence the difference obtained is volatile solid.

Calculation:

From above:

Total dissolved solids: 150 milligrams

Nonvolatile solids = 100 milligrams

Volatile solids = 50 milligrams

Population after n years

P_n  = P₀ + n

\(\begin{array}{l} \Rightarrow \;\bar x = \frac{{{p_n} - {p_0}}}{n}\\\therefore \bar x = \frac{{325\;000 - 200\;000}}{{10}}\end{array}\)

\(\bar x = 12500\)

Discharge required to reach design capacity

= 75000 – 50000

= 25000 m³/d

Present average water consumption per head per day

\(= \frac{{50000}}{{200000}} = \frac{1}{4}{m^3} = 0.25{m^3}\)

Increase in water consumption each year

= 0.25 × 12500

= 3125 m³/ day.

Number of years required to reach design capacity \(= \frac{{25000}}{{3125}}\;years\)

Concept:

Sewers are never designed for the full flow condition because the fluctuation in discharge can be higher and in wastewater some gases are also released, and these gases occupies the empty space.

These are always designed for the maximum hourly discharge and for branch sewer maximum hourly discharge is three times the average daily discharge.

For half flow condition, discharge is also half of full flow discharge. Since flow in sewers is open channel flow so manning equation is valid here,

Discharge in open channel by manning’s equation:

\({\rm{Q}} = \frac{1}{{\rm{n}}} \times {\rm{A}} \times {{\rm{R}}^{\frac{2}{3}}} \times {{\rm{S}}^{\frac{1}{2}}}\)

Where,

Q = discharge, n = manning’s roughness coefficient, A = area of flow, R = hydraulic radius, and S = slope of channel

Calculation:

Population = 10000, Water supply rate = 200 lpcd, and Wastewater generation rate = 80%

Design discharge= 3 × 10000 × 200 × 0.80 = 3 × 1.6 × 10⁶ L/day = \(\frac{{3 \times 1.6 \times {{10}^6} \times {{10}^{ - 3}}}}{{24 \times 3600}} = 0.05555{\rm{\;}}{{\rm{m}}^3}/{\rm{s}}\)

Discharge in half flow sewer is half of full flow sewer:

∴ Discharge in full flow sewer = 2 × 0.0555 = 0.1111 m³/s.

According manning’s equation:

\({\rm{Discharge\;}}\left( {\rm{Q}} \right) = \frac{1}{{\rm{n}}} \times {\rm{A}} \times {{\rm{R}}^{\frac{2}{3}}} \times {{\rm{S}}^{\frac{1}{2}}}\)

For circular section,

\(\therefore {\rm{Q}} = \frac{1}{{\rm{n}}} \times \frac{{{\rm{\pi }}{{\rm{D}}^2}}}{4} \times {\left( {\frac{{\frac{{{\rm{\pi }}{{\rm{D}}^2}}}{4}}}{{{\rm{\pi D}}}}} \right)^{\frac{2}{3}}} \times {{\rm{S}}^{\frac{1}{2}}} = \frac{1}{{\rm{n}}} \times \frac{{{\rm{\pi }}{{\rm{D}}^2}}}{4} \times {\left( {\frac{{\rm{D}}}{4}} \right)^{\frac{2}{3}}} \times {{\rm{S}}^{\frac{1}{2}}} = \frac{{{\rm{\pi }} \times {{\rm{D}}^{8/3}} \times {{\rm{S}}^{1/2}}}}{{10.08{\rm{\;n}}}}\)

Concept:

Chlorine Demand: The difference between applied chlorine and residual chlorine at or beyond break-point is called chlorine demand of water.

Calculation:

Total water disinfected = 50000 × 50 = 2.5 × 10⁶ L/Day

Chlorine Demand = 2 mg/L

∴ Total amount of chlorine required = (2.5 × 10⁶) × 2 × 10^-6 kg/day= 5 kg/day

Since bleaching powder contains only 25% chlorine

∴ Total Bleaching powder required \(= \frac{5}{{0.25}} = 20\;kg\)

Important Point:

The dose of chlorine should be such that 0.2 mg/L is left in the water after a contact period of 30 min  to prevent against future contamination.

The survey conducted on the waste is as follows:

The average density of the solid waste (ρ_sw) is given by:

\(\frac{{100}}{{{{\rm{\rho }}_{{\rm{sw}}}}}} = \frac{{50}}{{300}} + \frac{{30}}{{500}} + \frac{{20}}{{100}}\)

ρ_sw = 234.375 kg/m³

Solid waste produced annually = 6000 × 3 × 365 = 65700000 kg

∴ Volume of landfill = \(\frac{{65700000{\rm{\;}}}}{{234.375}} = 280320{\rm{\;}}{{\rm{m}}^3}\) 

∴ Area of landfill \(= \frac{{280320}}{5} \times {10^{ - 4}} = 5.60{\rm{\;hectares\;}}\)

Given:

Ultimate BOD(L₀) = 2000 mg/L

BOD₅ = L₀ (1 – e^-kt)

BOD₅ = 2000(1 – e^-0.23×5)

BOD₅ = 1366.73 mg/L

∴ S_o = BOD₅ = 1366.73 mg/L

\({S_o} = \frac{{1366.73 \times {{10}^{ - 6}}}}{{{{10}^3}}} = 1.367\;kg/{m^3}\)

Also, S = 15% of S_o

Therefore, \(S = \frac{{15}}{{100}} \times 1.367\)

S = 0.205 kg/m³

Flow, Q = 0.1 × 10⁶ × 10^-3 m³/day

Q = 100 m³/day

Therefore, sludge age can be written as

\(\frac{1}{{{\theta _c}}} = \frac{{\left( {S - {S_o}} \right)YQ}}{{VX}} - {k_d}\)

\(\frac{1}{{10}} = \frac{{\left( {1.367 - 0.205} \right) \times 0.1 \times 100}}{{VX}} - 0.01\)

\(\Rightarrow VX = \frac{{1.162 \times 10}}{{0.09}}\)

⇒ VX = 105.63 kg

Hence, sludge wasted \(= {Q_W}{X_u} = \frac{{VX}}{{{\theta _C}}}\)

\(\Rightarrow {Q_w}{X_u} = \frac{{105.63\;}}{{10}}\) kg/day

⇒ Q_wX_u = 10.56 kg/day

Concept:

the efficiency of trickling filter is given by:

\({\rm{\eta }} = \frac{{100}}{{1 + 0.44\sqrt {\frac{{\rm{W}}}{{{\rm{V}} \times {\rm{F}}}}} }}\)

Where,

W = BOD of wastewater entering in filter unit (kg/day)

V = Volume of filter (in m³)

F = Recirculation factor \(= \frac{{1 + R}}{{{{\left( {1 + 0.1R} \right)}^2}}}\)

R = Recirculation ratio

Calculation:

Total sewage discharge \(= \frac{{30000 \times 120}}{{{{10}^6}}} = 3.6{\rm{MLD}}\)

BOD of wastewater initially = 250 mg/l

BOD of water after primary clarifier = 250 × 0.7 = 175 mg/l

BOD of desired effluent = 30 mg/l

\({\rm{Efficiency\;}}\left( {\rm{\eta }} \right) = \frac{{175 - 30}}{{175}} \times 100 = 82.86{\rm{\% }}\)

Now,

W = 3.6 × 175 = 630 kg/day

\({\rm{F}} = \frac{{1 + 1.5}}{{{{\left( {1 + 0.1 \times 1.5} \right)}^2}}} = 1.89\)

\(\therefore 82.86 = \frac{{100}}{{1 + 0.44\sqrt {\frac{{650}}{{{\rm{V}} \times 1.89}}} }}\)

Concept:

The average value of the various recorded sound pressure levels (L_p) over a given period cannot be computed by simple averaging due to log scale involved in their values.

Average pressure level

\({\bar L_p} = 20\log \frac{1}{N}\mathop \sum \limits_{n = 1}^{n = N} {\left( {10} \right)^{{L_{n/20}}}}\)

Where,

L̅_p = Average sound pressure level in dB

N = Number of measurement readings

L_n = nth sound pressure level in dB

n = 1, 2, 3, …. N

Calculation:

L₁ = 60 dB, L₂ = 80 dB, L₃ = 100 dB, N = 3

∴ Average sound pressure level (L_avg)

\(= 20{\log _{10}}\frac{1}{N}\mathop \sum \limits_{n = 1}^{n = N} {\left( {10} \right)^{{L_{n/20}}}}\)

\( = 20{\log _{10}} \times \frac{1}{3}\left[ {{{10}^{\frac{{60}}{{20}}}} + {{10}^{\frac{{80}}{{20}}}} + {{10}^{\frac{{100}}{{20}}}}} \right]\) = 91.36 dB

Concept:

Rapid sand filters are designed for the maximum daily demand and backwash water required.

Maximum Daily Demand = 1.8 × Average daily demand

Calculation:

Q_avg = 100000 × 160 = 16 × 10⁶ litre per day

Assuming 5% as backwash and 30 minutes of back-washing period,

\({Q_{design}}\;\left( {l/d} \right) = \left[ {\frac{{1.8 \times {Q_{avg}} + 0.05 \times {Q_{avg}}}}{{23.5}}} \right] \times 24\) \(= \left[ {\frac{{1.8 \times 16 \times {{10}^6} + 0.05 \times 16 \times {{10}^6}}}{{23.5}}} \right] \times 24\) = 30.23 × 10⁶ ℓ/d

Rate of filtration = 4000 ℓ/hr/m²  = 4000 × 24 = 96000 ℓ/day/m²

∴ Total plan area of filters required \(= \frac{{{Q_{design}}}}{{Rate\;of\;filtration}} = \frac{{30.23 \times {{10}^6}}}{{96000}} = 314.90\;{m^2}\)

∴ Number of filters required \(= \frac{{Total\;plan\;area}}{{Plan\;area\;of\;1\;filter}} = \frac{{314.90}}{{50}} = 6.3\)

Thus adopting 7 filters and 1 on stand by, total number of filters required = 8

Important Point:

The number of Rapid sand filters required if the rate of filtration is not known is given as

\(n = 1.22\;\sqrt Q\),  Where, Q = Design Discharge in MLD

Concept:

Surface overflow rate: It can be thought of as the settling velocity of that particle if introduced at the topmost point at the inlet of the tank gets settled at the bottom-most point just before the outlet of the tank.

Particles having sizes greater than the particles for which setting velocity is equal to overflow rate gets 100% removed.

Particles having settling velocity less than overflow rate will not get 100% removed.

The percentage removal of particles having settling velocity V_s^' is given as \(=\frac{{{V}_{s}}'}{{{V}_{s}}}\times 100\)

Where, VS’ = Settling velocity of smaller size particles.

Settling velocity is calculated by Stoke’s law for discrete particle settling. It is given as-

\(V_{s}^{'}=\frac{\left( {{G}_{s}}-1 \right) {g{d}^{2}}}{18~\nu }\)

Where Gs = Specific gravity of particles.

d = Particle size

ν = Kinematic Viscosity of fluid in which particle is settling.

Calculation:

Plan area = 800 m², Q = 32000 m³/day

∴ Overflow rate \(({V_s}) = \frac{Q}{{Plan\;area}} = \frac{{32000}}{{800}}\) = 40 m/day = 0.463 mm/s

For sand particles:

G_s = 2.70, d_p = 0.018 mm, ν_water = 1.01 × 10^-2 cm²/s

As per stoke’s law, settling velocity of the sand particles,

\(V_s' = \frac{{\left( {{G_s} - 1} \right) \times g \times d_p^2}}{{18\;\nu }} = \frac{{\left( {2.7 - 1} \right) \times 9.81 \times {{\left( {0.018 \times {{10}^{ - 3}}} \right)}^2}}}{{18 \times \left( {1.01 \times {{10}^{ - 2}} \times {{10}^{ - 4}}} \right)}}\) = 2.97 × 10^-4 m/s = 0.297 mm/sec

∴ Percentage removal \(= \frac{{{V_s}'}}{{{V_s}}} \times 100 = \frac{{0.297}}{{0.463}} \times 100\) = 64.15%

Important point:

• Discrete Particle settling is assumed in stokes law and applicable for laminar flow for Reynolds number < 1.
• The efficiency of the sedimentation tank is independent of the depth of the tank and depends only on the plan area of the tank.

Concept:

Fractional collection efficiency of Electrostatic precipitator is given by:

\({\rm{\eta }} = 1 - {{\rm{e}}^{ - \frac{{{\rm{WA}}}}{{\rm{Q}}}}}\)

Where

W = terminal drift velocity (m/s), A = Total collection area (m²), and Q = Volumetric air flow rate (m³/s)

Calculation:

Let,

η₁ = Exiting fractional efficiency of electrostatic precipitator, η₂ = New fractional efficiency of electrostatic precipitator, A₁ = Existing collection area, and A₂ = New collection area

\({{\rm{\eta }}_1} = 1 - {{\rm{e}}^{\left( { - \frac{{{\rm{W}}{{\rm{A}}_1}}}{{\rm{Q}}}} \right)}}\)

\( ⇒ 0.92 = 1 - {{\rm{e}}^{\left( { - \frac{{{\rm{W}}{{\rm{A}}_1}}}{{\rm{Q}}}} \right)}}\\ ⇒ 0.08 = {{\rm{e}}^{\left( { - \frac{{{\rm{W}}{{\rm{A}}_1}}}{{\rm{Q}}}} \right)}}\)

⇒ 2.5257 = WA₁/Q

\({{\rm{\eta }}_2} = 1 - {{\rm{e}}^{\left( { - \frac{{{\rm{W}}{{\rm{A}}_2}}}{{\rm{Q}}}} \right)}}\)

\(⇒ 0.98 = 1 - {{\rm{e}}^{\left( { - \frac{{{\rm{W}}{{\rm{A}}_2}}}{{\rm{Q}}}} \right)}}\\⇒ 0.028 = {{\rm{e}}^{\left( { - \frac{{{\rm{W}}{{\rm{A}}_2}}}{{\rm{Q}}}} \right)}}\)

⇒ 3.5756 = WA₂/Q

Solving both equations,

A₂ = 1.548 A₁

So, collecting area must be 1.548 times of existing area