Environmental Engineering Test 1

Concept:

The intensity of taste and odour is measured by Threshold odour number (TON).

It represents the dilution ratio at which odour is hardly detectable.

The formula for TON \(= \frac{{A\; + \;B}}{A}\),  where A is the volume of odorous water in mL and B is the volume of odour free water required to produce a mixture in which odour is hardly detectable.

Calculation:

The volume of odorous water (A) = 50 ml

Total volume of mix = 500 ml

The volume of odour free water (B) = 450 ml

Threshold odour number \(= \frac{{A\; +\; B}}{A} = \frac{{500}}{{50}} = 10\)

Important Point:

TON allowed is 1 – 3. Hence the given sample is not acceptable as per GOI recommendations.

Hardness: Hardness in water is determined by Versonate Method. It is also known as EDTA (Ethylene diamine tetracetate acid) as EDTA is added to the wine red colored solution (mixture water sample and Eriochrome Black T) to estimate hardness.

Chlorine: Chlorine in water can be detected by three tests: i) Starch iodide test, ii) Ortho-toluidine test and iii) Ortho- toluidine arsenate test. The first two test determines total form of chlorine in water and the later detects only free form of chlorine in water sample.

Dissolved Oxygen: Dissolved oxygen in a water sample is determined by Wrinkle’s Method. Main compounds for conducting this test are Azide Iodide, Starch solution and Sodium thiosulphate.

Chloride: Chloride in water is determined by Mohr’s Test. It involves addition of potassium chromate and silver nitrate sequentially to the water sample. Excess of chlorides may lead to cardiac problems and kidney diseases.

Institutional and Commercial Water Demand:

The individual requirements would be as follows:

(i) Schools/Colleges: 45 to 135 lpcd

(ii) Offices: 45 lpcd

(iii) Restaurants: 70 lpcd

(iv) Cinema and theatre: 15 lpcd

(v) Hotels: 180 lpcd

(vi) Hospitals: 340 lpcd (when is less than 100) & 450 (beds exceeding 100).

Industrial Water Demand:

Water Demand of certain important industries (As per GOI MANUAL):

Name of Industry	Unit of Production	Approximate quantity of water required per unit of production
1. Automobiles	Vehicle	40
2. Fertilizers	Tonne	80 – 200
3. Leather	Tonne (or 100 kg)	40 4
4. Paper	Tonne	200 – 400
5. Petroleum Refinery	Tonne (crude)	1 – 2

Concept:

(i) Maximum Daily Consumption:

Maximum daily consumption = 1.8 × Avg. daily consumption = 1.8 × q

(ii) Maximum Hourly Consumption:

Maximum hourly consumption of the maximum day i.e. Peak demand

= 1.5 × Avg hourly consumption of the maximum day \(= 1.5 \times \left[ {1.8 \times \frac{q}{{24}}} \right] = 2.7\left[ {\frac{q}{{24}}} \right]\)

Coincident Draft: The maximum daily demand (i.e. 1.8 × average daily demand) when added to the fire demand is known as the coincident draft.

i.e Coincident Draft = Maximum daily demand + Fire Demand

Calculation:

Average daily demand (q) = 140 lpcd

∴ Maximum daily demand = 1.8 × q = 1.8 × 140 = 252 lpcd

∴ Coincident draft = Maximum daily demand + Fire demand= 252 + 10 = 262 lpcd

Important Point:-

The total draft is taken as the sum of maximum daily demand and fire demand or the maximum hourly demand, whichever is more.

Concept:

pH:

It is a measure of acidity or alkalinity of a solution. It measures the presence of hydrogen ion concentration relative to that of a standard solution. Mathematically, pH is the negative logarithm of the hydrogen ion concentration [H+] present in the solution expressed in molarity.

pH = -log10 [H+]

Molarity: It is defined as the number of moles of solute per litre of solution.

Calculation:

For solution A:

pH = 6 ⇒ - log10 [H+] = 6 ⇒ [H+] = 10-6 mol/ℓ

For solution B:

pH = 8 ⇒ -log10 [H+] = 8 ⇒ [H+] = 10-8 mol/litre

When A and B are mixed, total volume of solution = 2ℓ

Total hydrogen ion = (10-6 + 10-8) moles in 2ℓ of solution.

∴ Hydrogen ion concentration \(= \frac{{\left( {{{10}^{ - 6}} + {{10}^{ - 8}}} \right)}}{2}\;mol/litre\)

∴ pH of the mixture \(= - {\log _{10}}\left[ {\frac{{{{10}^{ - 6}} + {{10}^{ - 8}}}}{2}} \right]\) ≈ 6.3

Mistake point:

The hydrogen ion concentration of the solution is given as mole/litre in the pH calculation.

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 which 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

It is measured by using spectrophotometric techniques.

Equivalent weight of CaCO₃ = (40 + 12 + 16 × 3)/2 = 50

Calculation:

Hardness in the sample given is caused by Ca²⁺, Mg²⁺ and Al³⁺

Total meq of hardness causing cations = 30 + 12 + 13 = 55 meq/L

∴ Total hardness in mg/ℓ as CaCO₃ = (Number of gm. eq × Equivalent weight of CaCO₃)

= (55 × 10^-3) × 50 gm/L as CaCO₃ = 2.75 gm/L = 2.75 × 10³ mg/L = 2750 mg/L as CaCO₃ .

Mistake Point:

Aluminium ion (Al³⁺) is also a multivalent metallic cation and hence it will cause hardness in the sample.

Concept:

Geometric Increase Method:

In this method, the per decade percentage increase or percentage growth rate (r) is assumed to be constant, and the increase is compounded over the existing population every decade. The forecasted population (P_n) after n decades is given by,

\({P_n} = {P_0}{\left( {1 + \frac{r}{{100}}} \right)^n}\)

Where,

P₀ = Population at the end of last known census.

r = Assumed growth rate (%)

 The growth rate is calculated as :

\(r = {\left( {{r_1} \times {r_2} \times {r_3} \ldots \times {r_n}} \right)^{1/n}}\)

Calculation:

Year	Population	Increasing population	% Increase in population
1960	50000
1970	60000	10000	\(\frac{{10000}}{{50000}} \times 100 = 20\% \)
1980	75000	15000	\(\frac{{15000}}{{60000}} \times 100 = 25\%\)
1990	92000	17000	\(\frac{{17000}}{{75000}} \times 100 = 22.67\%\)

 

The geometric mean of past growth rate (r) \(= \sqrt[3]{{\left( {20 \times 25 \times 22.67} \right)}} = 22.46\%\)

∴ Population in 2010 i.e. after 2 decades,

\({P_n} = {P_0}{\left( {1 + \frac{r}{{100}}} \right)^n} = 92000 \times {\left( {1 + \frac{{22.46}}{{100}}} \right)^2}\) ≈ 137968

Important Point:

The geometric progression method gives the highest value of the forecasted population and it is suitable for new town expanding rapidly.

Concept:

The Thomas equation is used when the MPN table is not available. According to this equation:

MPN/100ml \(= \frac{{Number\;of\;positive\;tubes\; \times \;100}}{{\sqrt {\left( {ml\;of\;sample\;in\;negative\;tube} \right)\; \times \;\left( {ml\;of\;sample\;in\;all\;tube} \right)} }}\)

Calculation:

Total number of positive tubes = 4 + 3 + 3 = 10

ml of sample in negative tube = 1 × 1 + 2 × 0.1 + 2 × 0.01 + 5 × 0.001 = 1.225

ml of sample in all tube = 1 × 5 + 0.1 × 5 + 0.01 × 5 + 5 × 0.001 = 5.555

∴ MPN as per Thomas equation \(= \frac{{10\; \times \;100}}{{\sqrt {\left( {1.225} \right)\; \times \;\left( {5.555} \right)} }} = 383.34\)

Mistake Point:

The count of positive tubes should begin with the highest dilution in which at least one negative result has occurred.

Concept:

There are different ways to represent the concentration of a solution:

Molarity: It is concentration of solution expressed as no. of moles per liter solution.

\({\rm{Molarity}} = \frac{{{\rm{No}}.{\rm{\;of\;moles}}}}{{{\rm{volume\;of\;solution\;in\;liter}}}}\)

Where, \({\rm{No}}.{\rm{\;of\;moles}} = \frac{{{\rm{Given\;weight}}}}{{{\rm{Molecular\;weight}}}}\)

Normality: It is concentration of solution expressed as no. of gram equivalents per liter solution.

\({\rm{Normality}} = \frac{{{\rm{Gram\;equivalents}}}}{{{\rm{volume\;of\;solution\;in\;liter}}}}\)

Where, \({\rm{No}}.{\rm{\;of\;gram\;equivalents}} = \frac{{{\rm{Given\;weight}}}}{{{\rm{Equivalent\;weight}}}}\)

\({\rm{And\;equivalent\;weight}} = \frac{{{\rm{Molecular\;weight}}}}{{{\rm{Valency}}}}\)

Calculation:

Molecular weight of H₂SO₄ = 2 × 1 + 32 + 16 × 4 = 98 grams

No. of moles present in the original solution = 0.1 × 0.8 = 0.08 moles

Weight of 0.15 moles of H₂SO₄ = 0.08 × 98 = 7.84 grams

Equivalent weight of H­₂SO₄ = 98/2 = 49 grams

No. of gram equivalent of H₂SO₄ = \(\frac{{7.84}}{{49}}\) = 0.16 gram-equivalents

Final volume = 800 + 200 = 1000 ml

Normality of new solution = 0.16/1 = 0.16 N