Environmental Engineering Test 2

Concept

 Fixed dissolved Solids = Total solids - Suspended solids

Calculation

Total Solids = 100 + 200 + 168 + 120 + 111

 = 699 mg/l

Fixed dissolved solids

= Total Solids – suspended solids

= 699 – 100 – 200

Concept

\(\% \;Removal\;\left( {\% R} \right)\; = \;\frac{{{N_o} - {N_t}}}{{{N_o}}} \times 100\)

Log removal (log R) = log₁₀(N_o) – log₁₀(N_t)

Where, N_o = Initial no. of organisms

N_t = Number of organisms at time t.

 Calculations

 Given, N_o = 10⁶, N_t = 102

\(∴ \;\% \;R\; = \;\frac{{{{10}^6} - {{10}^2}}}{{{{10}^6}}} \times 100\; = \;99.99\)

∴ % R = 99.99

log R = log₁₀(10⁶) – log₁₀(10²) = 4

 ∴ log R = 4 

Concept:

C/N ratio is an important factor for bacterial activity to be continued as bacteria uses nitrogen for building their cell structure (as proteins) and use carbon as source of energy.

For proper development of anaerobic digestion C/N ratio of the digestive material should be kept in between 30-50 for optimum digestion. This limit is based on the fact that microorganism utilizes carbon 30 to 50 times faster than Nitrogen.

Case 1:

When C/N ratio is higher than optimum, then Nitrogen will be exhausted prior to carbon i.e. nitrogen will be used up and carbon will be left over thereby leaving the digestion of organic matter incomplete.

Case 2:

When there is too much nitrogen i.e. C/N ratio less than the optimum value, carbon will get exhausted prior to nitrogen i.e. carbon will be used up and nitrogen will left over, the left over nitrogen combine with the Hydrogen and leads to formation of ammonia gas which can kill microorganism and act as disaster for methane producers.

Therefore, anaerobic decomposition will require an optimum C/N ration of about 30 – 50.

Location A witness the liberation of Ammonia gas, implies that it pertains to case 2: i.e. X < 45

Location B witnessed the incomplete digestion of Organic Matter implies that it pertains to case 1: i.e.Y > 45.

Concept

Large G values with short times tend to produce small, dense flocs,

While low G values and long times produce large, lighter flocs. Since large, dense flocs are more easily removed in the settling basin, it may be advantageous to vary the G values over the length of the flocculation basin.

The small dense flocs produced at high G values subsequently combine into larger flocs at the lower G values.

Reduction in G values by a factor of 2 from the influent end to the effluent end of the flocculator has been shown to be effective.

Concept:

Sludge volume index is the volume occupied in mL by one gm of solids in the mixed liquor after settling for 30 minutes.

\(SVI = \frac{{{V_{ob}}}}{{{X_{ob}}}}\) (in ml/gm)

Where V_ob = settled sludge volume (in ml/lit)

X_ob = Concentration of suspended solids in the mixed liquor in gm/lit.

Calculation:

V_ob=200 ml/lit

X_ob = 2 gm/lit

SVI = 200/2 = 100 

Concept:

In general, water sample consist of three type of bacteria’s:

1. Aerobic bacteria which require oxygen for their survival.

2. Anaerobic bacteria which flourish and thrive in the absence of free oxygen.

3. Facultative bacteria which can survive with or without of free oxygen.

Coliforms are the harmless aerobic microorganisms whose presence or absence indicated the presence or absence of pathogens in waste water.

The Water sample X has been collected from sewage influents, so it contains very large number of pathogens and sample Y is collected after giving some treatment to waste water, so it will contain less amount of bacterial concentration than that of sample X.

The first stage efficiency of trickling filter, η_I = \(\frac{{100}}{{1 + 0.0044 \times \sqrt {\frac{{{W_1}}}{{{V_1} \times {F_1}}}} }}\)

Where W₁ = Total organic loading in kg/day

V₁ = Filter volume in a hectare. m

F₁ = Recirculation factor 

The first stage efficiency of trickling filter, η_II = \(\frac{{100}}{{1 + \frac{{0.0044}}{{1 - {\eta _I}}} \times \sqrt {\frac{{{W_2}}}{{{V_2} \times {F_2}}}} }}\)

W₂ = Total organic loading in kg/day = (1 - η_I)

V₂ = Filter volume in a hectare-m

F₂ = Recirculation factor

Overall Efficiency = η_I + (1-η_I) × η_II

Calculation:

The overall efficiency = 0.8 + (1 – 0.8) × 0.7

η = 0.95

Concept:

 1) Modified Winkler’s method for Dissolved oxygen:

The modified Winkler test for estimation of Dissolved Oxygen (DO) in water is used on-site, as delays between sample collections and testing may result in an alteration in oxygen content.

For this test, Manganese Sulphate (MnSO4­)   and mixture of (NAOH + KI) is added in water sample.

Now, if DO is absent in water it leads to formation of white colored precipitate of Manganese Hydroxide (MnOH₂).

If DO is present, then Red- Brown precipitate of MnO2 is formed, which is further titrated using Sodium thiosulphate reagent using soluble starch as an indicator.

(In absence of oxygen)

Mn²⁺ + 2OH^- → Mn(OH)₂  (White ppt.)

(In presence of Oxygen)

Mn²⁺ + 2OH^- + ½ O₂ → Mn(OH)₂ ↓ (red – brown ppt)

 2) Dichromate method for COD:

The COD test is widely used as an alternative to BOD test to estimate the strength of domestic and industrial wastes as COD test can give results in few hours as compared to BOD test which takes normally 5 days.

The COD test uses potassium dichromate (K₂Cr₂O₇) in presence of concentrated sulfuric acid (H₂SO₄) solution that oxidizes both organic (predominate) and inorganic substances in a waste water sample.

The reagents for COD test are followings:

a) Standard potassium dichromate solution (0.25N)

b) Concentrated sulfuric acid

c) Standard ferrous ammonium sulphate titrant (0.1N)

d) Ferroin indicator solution.

 3) EDTA titrimetric method: Hardness is normally expressed in terms of CaCO₃. The most widely used method its determination is by using EDTA as titrant. The titration is done by adding a small amount of buffer (pH 10) followed by the addition of Eriochrome black T (EBT) as indicator.

 4) Mohr method of chlorides:

The presence of chloride content in water in high concentration indicates its pollution due to industrial waste and sewage.

Chloride content in water sample is determined by Mohr’s Method i.e. by titrating the water sample with Standard Silver Nitrate (AgNO₃) solution with Potassium Chromate (K₂CrO₄) as an indicator..

Concept

When chlorine is added to water, it forms hypochlorous acid (HOCl) and hypochlorite ions (OCl^‑).

However, the action of hypochlorous acid is about 80 times more effective than hypochlorite ions.

The proportion of hypochlorous acid (HOCl) and hypochlorite ions (OCl^‑) in the water sample depends upon the pH value (alkalinity or acidity) of the water sample.

∴ As pH increases, the concentration of hypochlorite ions increases and disinfection efficiency decreases. 

Concept:

Efficiency (η) of particle removal/sedimentation is given by:

η = Settling velocity (V_s)/Overflowing velocity (V_o)

Overall efficiency (η_o) \(= \frac{{{{\rm{\eta }}_1} \times {{\rm{P}}_1} + {{\rm{\eta }}_2}{{\rm{P}}_2} + \ldots + {{\rm{\eta }}_{\rm{n}}}{{\rm{P}}_{\rm{n}}}}}{{{{\rm{P}}_1} + {{\rm{P}}_2} + {{\rm{P}}_3} + \ldots + {{\rm{P}}_{\rm{n}}}}}\) 

Calculation:

V_o = 32.6 m/day = 1.358 m/hour

\({{\rm{\eta }}_1} = \frac{{{{\rm{v}}_{{{\rm{s}}_1}}}}}{{{{\rm{v}}_{\rm{o}}}}} = \frac{{0.5}}{{1.358}} = 0.368,{{\rm{P}}_1} = 0.20\)

Similarly

\({{\rm{\eta }}_{\rm{o}}} = \frac{{\frac{{0.5}}{{1.358}} \times 0.2 + \frac{{0.4}}{{1.358}} \times 0.3 + \frac{{0.6}}{{1.358}} \times 0.25 + \frac{{0.3}}{{1.358}} \times 0.35}}{{0.2 + 0.3 + 0.25 + 0.25}}\)

η_o = 0.3276

Hardness is expressed in terms of degree of hardness as per Clarke’s scale.

It is defined as an ability of the water to cause precipitation of insoluble calcium and magnesium salts of higher fatty acids from soap.

As per Clarke’s one grains of CaCO₃ dissolved in one gallon of water will produce 1^o of hardness.

Now,

One grain = 64.8 mg and one gallon = 4.546 litres

∴ 1^o of hardness will be equivalent to 64.8/4.546 = 14.254 mg of CaCO₃.

⇒ 1^o of hardness will be equivalent to 14.3 mg of calcium carbonate in a liter of water.

 Calculation

The energy content of the waste (as a discarded basis) is given by:

\( = \frac{{\sum \left( {percent\;by\;mass \times energy\;\left( {\frac{{kJ}}{{kg}}} \right)} \right)}}{{100}}\)

\( = \frac{{1137500}}{{100}} = 11375\;kJ/kg\)

The energy content on dry waste basis is given by:

\( = Energy\;\left( {as\;discarded} \right) \times \frac{{100}}{{100 - moisture\;content}}\)

\( = 11375 \times \frac{{100}}{{\left( {100 - 25} \right)}}\)

= 15166.67 kJ/kg

The energy content on ash-free dry basis is given by

\( = Energy\;\left( {as\;discarded} \right) \times \frac{{100}}{{100 - mc - ac}}\)

Where,

mc = moisture content

ac = ash content

\( = 11375 \times \frac{{100}}{{100 - 25 - 6}} = 16485.5\;\left( {\frac{{kJ}}{{kg}}} \right)\)

The difference between energy content on the ash-free dry basis and on a dry waste basis

= (16485.5 – 15166.67) kJ/kg

= 1318.83 kJ/kg

Concept

10 % dilution means out of 100 ml, 10 ml is waste water and 90 ml is tap water.

\(D{0_{initial}} = \frac{{D{0_{waste\;water}} \times 10\;ml + D{O_{water}} \times 90}}{{100}}\)

5 day BOD is given by

\(BO{D_5} = \left( {Initail\;DO - Final\;DO} \right) \times Dilution\;factor\)

Ultimate BOD is given by,

\({L_5} = {L_0} \times \left( {1 - {{10}^{ - \left( {K \times t} \right)}}} \right)\)

Where,

L₀ is ultimate B.O.D, L₅ is 5- Day BOD

K is BOD decay constant, t is time

Calculation

Given,

10% dilution

∴ Dilution factor = 100/10 = 10

Final DO = 0.5 mg/lit, K = 0.12 Day^-1

DO in waste water sample = 1.2 mg/lit

DO in water = 6 mg/lit

\(D{0_{initial}} = \frac{{D{0_{waste\;water}} \times 10\;ml + D{O_{water}} \times 90}}{{100}}\)

\(D{0_{initial}} = \frac{{1.2 \times 10\;ml + 6 \times 90}}{{100}} = 5.52\;mg/lit\)

i) 5 Day BOD

\(BO{D_5} = \left( {Initail\;DO - Final\;DO} \right) \times Dilution\;factor\)

= (5.52- 0.50) × 10

= 50.2 mg/lit

∴ 5 Day BOD = 50.2 mg/lit

ii) Ultimate BOD

\({L_0} = \frac{{{L_5}}}{{1 - {{10}^{ - \left( {K \times t} \right)}}}}\)

\({L_0} = \frac{{50.2}}{{1 - {{10}^{ - \left( {0.12 \times 5} \right)}}}} = 67.04\;mg/lit\)

∴ Ultimate BOD = 67.04 mg/lit

Calculation

i) Dry weather flow (D. W. F):

Average per capita demand = 250 lpcd

\(\therefore {\rm{Water\;supplied}} = \frac{{250 \times {{10}^{ - 3}} \times 90000}}{{24 \times 60 \times 60}} = 0.26{{\rm{m}}^3}/{\rm{sec}}\)

∴ Sewer discharge = 0.75 × 0.26 = 0.195 m³/sec

∴ Maximum design sewer discharge = 3 × Q_average = 3 × 0.195 = 0.586 m³/sec

ii) Storm water:

Time of concentration = 3 + 17 = 20 minutes

∴ Time of duration of storm for more discharge = 20 min

\({\rm{R}} = \frac{{25.4 \times {\rm{a}}}}{{6 + {\rm{b}}}} = \frac{{25.4 \times 30}}{{20 + 10}} = 25.4{\rm{\;mm\;per\;hour}}\)

Where

a = 30, b = 10 for t ≤ 20 min and a = 40, b = 20 for t > 20 min

\(\therefore {\rm{Q}} = \frac{{{\rm{AIR}}}}{{360}}\)

Where

Q = discharge in m³/sec, A = area in hectares, I = impermeability factor and R = rainfall intensity in mm/hour.

\(\therefore {\rm{Q}} = \frac{{100 \times 0.50 \times 25.4}}{{360}} = 3.52{{\rm{m}}^3}/{\rm{sec}}\)

∴ Total discharge = 3.52 + 0.586 = 4.11m²/sec

Area (A) = Q (Discharge)/V (Velocity)

\(\therefore \frac{{\rm{\pi }}}{4} \times {{\rm{d}}^2} = \frac{{4.1137}}{3}\)

Concept

Permissible BOD of the stream,

\({y_{mix}} = \frac{{{Q_w}.{y_w} + {Q_s}.{y_s}}}{{{Q_W} + {Q_s}}}\)

Where, y_w = BOD of treated waste.

 y_s = BOD of stream water on U/S side

 Calculation

 Given,

 Population = 40000

 BOD contribution per capita = 0.07 kg/day

 BOD of the stream on the U/S side = 3 mg/l

 Permissible max. BOD of the stream on D/S side = 5 mg/l

 Waste water flow, Qw = 0.081 m3/s

 Minimum flow of stream, Qs = 0.13 m3/s.

\({y_{mix}} = \frac{{{Q_w}.{y_w} + {Q_s}.{y_s}}}{{{Q_W} + {Q_s}}}\)

 \(5 = \frac{{0.081 \times {y_w} + 0.13 \times 3}}{{0.081 + 0.13}}\)

 y_w = 8.21 mg/l

 let, y_i = Influent BOD

 ∴ Total BOD produced/day, Q_wy_i = 40000 × 0.07

 Q_w.y_i = 2800 kg/day

\({y_i} = \frac{{2800 \times {{10}^6}}}{{0.081 \times {{10}^3} \times \left( {24 \times 3600} \right)}} = 400\;mg/l\)

∴ % of purification required \(= \frac{{{y_i} - {y_w}}}{{{y_i}}} \times 100\)

Concept

2% of solids (by weight) + 98% of water (by weight) = 100 % of sludge (by weight)

As density ρ = m/V and Specific gravity (G) =\({G_s} = \frac{{{\rho _{solid}}}}{{{\rho _{water}}}}\)

Equating volume 

\(\frac{{100}}{x} = \frac{{98}}{{1 \times {\rho _w}}} + \frac{2}{{{G_s} \times {\rho _w}}}\)

Calculation

Given,

Solids content = 2 % by weight

Specific gravity of Sludge solid = 2.2

i) 2 % of solids means there is 98% of water in sludge.

2% of solids + 98% of water = 100 % of sludge

Let ‘x’ be the density of sludge

Equating total volume 

 \(\frac{{100}}{x} = \frac{{98}}{{1 \times {\rho _w}}} + \frac{2}{{{G_s} \times {\rho _w}}}\)

\(\frac{{100}}{x} = \frac{{98}}{{1 \times 1000}} + \frac{2}{{2.2 \times 1000}}\)

X = 1011 kg/m³

∴ Density of sludge = 1011 kg/m³

ii) Let V₁ = volume of sludge (Initial)

 = Volume of water + Volume of solids

\( = \frac{{98}}{{1 \times 1000}} + \frac{2}{{2.2 \times 1000}}\)

= 0.0989 m³

∴ Total volume initially = 0.0989 m³

After sludge thickening (Volume of water changes)

New volume of sludge (V₂) = V₁/2

V₂ = 0.04945m³

∴ 0.04945 = Vol. of solids + volume of water 

\(0.04945 = \frac{2}{{2.2 \times 1000}} + \frac{x}{{1000}}\)

 x = 48.545 % 

∴ Water content by weight = \(\frac{{48.545}}{{48.545\; + 2\;}} \times 100\)

= 96.04 % (by weight)

Calculation

At STP, 64 gm of SO₂ at 0°C occupies 22.4 liters

\(1\;ppm = \frac{{1{m^3}of\;S{O_2}}}{{{{10}^6}{m^3}of\;air}}\)

1 m³ of air has 80 mg of SO₂

10⁶m³ of air has 80 × 10⁶μg of SO = 80 gm of SO₂

Number of a mole of SO₂ in 10⁶ m³ of air

 \( = \frac{{given\;wt.}}{{Molecular\;wt.}} = \frac{{80}}{{64}}\) = 1.25 moles

Using ideal gas law : -

P x V = n x R x T

\(V = \frac{n}{{\left( {\frac{P}{{RT}}} \right)}} = \frac{{1.25}}{{41.6}} = 0.03\;{m^3} = 30\;L\)

∴ 30 litres of SO₂ has wt. of 64 gm

∴ (1m³ = 10³ litres) of SO₂ has wt. = \(\frac{{64}}{{30}} \times {10^3}\)

= 2133.33 gm

∴ 1 ppm of \(S{O_2} = \frac{{1{m^3}of\;s{o_2}}}{{{{10}^6}{m^3}\;of\;air}} = \frac{{2133.33\;gm}}{{{{10}^6}{m^3}\;of\;air}}\)

\(\begin{array}{l} 1ppm = 2133.33\;\frac{{ugm\;of\;S{O_2}}}{{1{m^3}of\;air}}\\ \Rightarrow \frac{{1ugm}}{{{m^3}}} = \frac{1}{{2133.33}}ppm\\ \Rightarrow \frac{{80ugm}}{{{m^3}}} = \frac{{80}}{{2133.33}}ppm\\ \Rightarrow \frac{{80ugm}}{{{m^3}}} = 0.0375\;ppm \end{array}\)

Concept

Rapid and filter is designed for maximum daily demand = 1.8 × average daily demand.

As 4 % water is required for washing filter

Total filtered water required per day = Max daily demand/0.96

As there is 1 Hour lost in cleaning,

Filtered Water required per hour = Q/23

Calculation

Given,

Population = 75,000, Avg Rate supply = 150 lit/capita/day.

Rate of filtration = 100 litre/min/sq.m

= 6000 lit/hr/sq.m

Average daily demand = 75,000 × 150 = 11.25 × 10⁶ lit/day

= 11.25 MLD

Maximum Daily demand = 1.8 × 11.25

= 20.25 MLD

Total filtered water required per day (Q) = 20.25/0.96

= 21.09 MLD

As there is 1 Hour lost in cleaning,

Filtered Water required per hour = Q/23

= 21.09/23

= 0.917 MLH

i) Area of filter is

\(Area\;of\;filter = \frac{{Filtered\;water\;required\;per\;hour}}{{Rate\;of\;filteration}} = \frac{{0.917 \times {{10}^6}lit/hr}}{{6000\;lit/hr/sqm}}\)

= 152.833 m²

∴ Area of filter bed is 152.833 m²

ii) Number of filter beds required roughly given by

Morell and Wallace as

N = 1.22 × √Q

Where, N is number of filter beds,

Q is plant capacity in MLD 

N = 1.22 × √21.09

= 5.6

≈ 6 beds

∴ 6 Beds are provided.

In general 1 standby bed is provided.

∴ Number of beds = 6 + 1 = 7 beds 

Concept:

As per NRC equations:

Equation for efficiency of trickling filter (η) is given by:

\({\rm{\eta }} = \frac{{100}}{{1 + 0.0044\sqrt {\frac{{\rm{Y}}}{{{\rm{VF}}}}} }}\)

Where

Y = Amount of BOD in kg/day

V = Volume of tank in ha-m

F = Recirculation factor

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

R = Recirculation ratio

Calculation:

Given,

 BOD concentration = 150 mg/l,Sewage Qty = 10 MLD

 Volume = 100 ha-m  

∴ Y = 150 × 10 = 1500 kg/day

Note: mg/l × MLD = kg/day

\(\because{\rm{F}} = \frac{{1 + {\rm{R}}}}{{{{\left( {1 + 0.1{\rm{R}}} \right)}^2}}} = \frac{{1 + 1.15}}{{{{\left( {1 + 0.1 × 1.15} \right)}^2}}} = 1.729\)

Now,

\({\rm{\eta }}\left( {\rm{\% }} \right) = \frac{{100}}{{1 + 0.0044\sqrt {\frac{{\rm{\gamma }}}{{{\rm{V}} × {\rm{F}}}}} }}\)

\({\rm{\eta \;}}\left( {\rm{\% }} \right) = \frac{{100}}{{1 + 0.0044\sqrt {\frac{{1500}}{{100 × 1.729}}} }} = 98.72\% \)

Concept:

Water softening: It is the process of hardness removal from the water.

Hardness is caused by multivalent cation and affects the water quality. Lime soda method is a water softening method in which lime and soda ash is added to the water, which cause the precipitation of multivalent cation as CaCO₃.

Precipitation of CaCO₃ occurs only when the pH of water is greater than 9, so in case of less pH alkalinity is added to the water. In this process small amount of Ca²⁺ and Mg²⁺ precipitates very late, which will create incrustation in the pipe, so to avoid this recarbonation is done to dissolve back this small amount of cation.

Due to this, the method does not give zero hardness.

Total lime required based on total hardness is given by:

Line as CaO required (mg/l) = TH × 56/100

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

Soda required (mg/l) = NCH × 106/100

∴ The role of soda ash is to remove the non-carbonate hardness of the calcium.

Concept

Let, for the steady flow to a fully penetrating well in a confined aquifer, the drawdown at radial distance r₁ and r₂ from the well be s₁ and s₂ respectively.

Therefore, the pumping rate can be written as

\(Q = \frac{{2\pi kD\left( {{s_2} - {s_1}} \right)}}{{\ln \left( {{r_1}/{r_2}} \right)}}\)

Where,k = coefficient of permeability

D = height of the confined aquifer

Calculation:

Given,

Radius of well, r = 15 cm = 0.15 m

Radius of influence, R = 3000 m

Drawdown at face of the well = 3 m

Drawdown at (x = R) = 0

\(\therefore Q = \frac{{2\pi KD \times \left( {3 - 0} \right)}}{{\ln \left( {\frac{{3000}}{{0.15}}} \right)}}\)

\(\frac{Q}{{KD}} = \frac{{2\pi \times 3}}{{\ln \left( {\frac{{3000}}{{0.15}}} \right)}}\)

\(\frac{Q}{{KD}} = 1.9033\)

Now, the drawdown at 300 m away from the center of the well

\(Q = \frac{{2\pi KD\left( {3 - {S_1}} \right)}}{{\ln \left( {\frac{{300}}{{0.15}}} \right)}}\)

\(\frac{Q}{{KD}} = \frac{{2\pi \left( {3 - {S_1}} \right)}}{{\ln \left( {\frac{{300}}{{0.15}}} \right)}}\)

\(\Rightarrow \frac{{1.9033 \times \ln \left( {\frac{{300}}{{0.15}}} \right)}}{{2\pi }} = 3 - {S_1}\)

⇒ 3 – S₁ = 2.303

⇒ S₁ = 3 – 2.303