*There are some errors and miscalculations in the answer. Please find them yourself.

Q1 (a) Discuss briefly the following act

- - Environmental Quality Act 1974 – some info from EQA

- Waters Act 1920 (Revised 1989) – some info from the act

(b) Define and compare the wastewater standards, STANDARD A and STANDARD B use for effluent discharge to inland waters.

a) Standard A for discharging into any inland waters within the catchment areas – the areas upstream of surface or above subsurface water supply intakes

- Standard A : for the purpose of human consumption including drinking water.

- Standard B for discharging into any other inland waters

- Standard B : for recreational use.

- For standard B, when both phenol and free chlorine are present in the same effluent, the concentration of phenol individually, shall not be greater than 0.2 mg/l and the concentration of free chlorine individually, shall not be greater than 1 mg/l.

- The value of parameters for standard A < standard B

Any relevant answers ** 4 marks

(c) Discuss briefly one (1) cause and consequences of water pollution to the environment. Provide solutions for the problem. ExCause –land erosionConsequences- silt/ pollutant from erosion side enter water body & might kill flora/fauna in the water body.Solution- provide silt trap at land slide area(student can give any example but need to cover on cause, consequence & solution

(d) Give three (3) examples of human activities that affect the environment and the measures that could be taken to reduce the impacts.

b) – deforestation – landslides

- Agriculture – agricultural run-off

- Hunting – extinction of some species

- Industries – air pollution / release of green house gases

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- Any relevant answers

Q2 (a) Explain briefly on seeded and unseeded method of BOD

BOD is a measure of the quantity of oxygen used by microorganisms (eg.aerobic bacteria) in the oxidation of organic matter.

Seeded method-. In BOD test it is necessary to have population of microorganism capable of oxidizing the biodegradable organic matter in the sample (ex sample with satisfactory microbial population/ for unseeded method- domestic w/water, unchlorinated sample, surface run off) & (ex sample of insufficient microb – untreated industrial waste, disinfected waste, high temperature/pH waste). Seeding normally by adding population of microorganism. The preferred seed is effluent or mixed liquor from biological treatment plant.

(c) A Wastewater Treatment Plant discharges 0.48 m3/s of treated wastewater into a river. The river has a flow rate of 3 m3/s .The treated wastewater has an ultimate BOD of 60.0 mg/L and dissolved oxygen (DO) of 2.0 mg/L with temperature of 30.0oC. Upstream of the discharge point, the ultimate BOD of the river is 5 mg/L, dissolved oxygen (DO) of 8.0 mg/L and temperature of 20oC. The average speed of the mixture is 0.6 m/s. At 20oC, the deoxygenation rate constant, kd is 0.37d-1 at 20oC while reaeration, kr rate constant is 0.5 d-..

i) DO initial, ultimate BOD and temperature after mixing.

By using given formula

DO= QwDOw + QrDOr Qw + Qr

=7.17 mg/L

By using given formula

La = QwLw + QrLr Qw + Qr

= 12.6 mg/L

By using formula

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Tf= QwTw + QrTr Qw + Qr

= 21.4 0C

ii) Initial Deficit of the stream

Da = DOs-DO

From Table , at 21.4 0C , DOs=8.99 Initial Deficit = 8.99 -7.17= 1.82 mg/L

iii)Dissolved oxygen concentration at 50 km downstream

Velocity = Distance ,x Time,t

Time = (50km)(1000m/km) = 0.96d (0.6m/s)(86400s.d

Calculate kd and kr at 21.4 0C by using formula

kT = k20 (θ)T-20

where θ = 1.135 for kdθ =1.024 for kr

kd at 21.4 0C = 0.37 x (1.135)21.4-20 = 0.44d-1

kr at 21.4 0C = 0.5 x (1.024)21.4-20 = 0.52d-1

Calculate oxygen deficit in river water after exertion of BOD for time, t, by using formula

D = kd x La (e-kdt – e-krt) + Da(e-krt) kr - kd

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=4.43 mg/L DO=8.99 – 4.39 = 4.56 mg/L

Q3 (a) List two (2) basic types of coagulant aids and explain how each aid works.

Type Description pH adjuster Adjust water pH into optimal range for coagulation

Acid ; sulfuric acid (lowering pH)Alkali; lime or soda ash ( raise pH)

Activated silica For highly colored , low-turbidity waters (add weight to flocs) Produce stable solution with negative surface chargeUnite with positively charged alum/ iron flocs, resulting in larger flocs (settling faster)Need precise dosage of activated silica, proper equipment, and close operational control.

Clay For highly colored , low-turbidity waters ( rarely used)Produce slightly negative surface charge and can add weight to flocs.

Polymer Active sites of polymer adhere to flocs, Producing larger and tougher flocs that settle better. (interparticle- bridging)

(b) A jar test was conducted on untreated water with an initial turbidity of 13 NTU and a HCO 3-

concentration of 65 mg/L as CaCO3. Using the following data obtained from a jar test;

----------------------------------------------------------------------------------------------Alum dose, mg/L 4 8 12 16 20 24

Turbidity, NTU 10 7 5.5 4.5 6 8.5----------------------------------------------------------------------------------------------

i) Estimate the optimum alum dosage for turbidity removal a) Optimum alum dosage

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Plot graph alum dosage vs turbidity

From the graph, optimum alum dosage = 16 mg/L

ii) Theoretical amount of alkalinity that will be consumed at the optimal dosage ( express concentration as mg/L as CaCO3)

With molecular weight= 594, moles of alum added per liter ( molarity, M) is;

M= = 2.69 x 10-5 mole

Since 1 mole of alum consume 6 moles of HCO3-

Molarity of HCO3- consumed = 6 x 2.69 x 10-5

= 1.62 x 10-4 moles/L

Concentration of HCO3- consumed;

= (1.62 x 10-4 moles/L ) ( 61g/mole)( 1000mg/g)

=9.9 mg/L

Express concentration s as mg/L as CaCO3

9.9 = 8.08 mg/L as CaCO3

(c) A groundwater contains the following constituents: CO2 = 6.60mg/L Ca2+ = 34.00mg/L Mg2+ = 29.00mg/L HCO3

- = 145.05 mg/L SO4

2- = 32.10 mg/L Cl- =42.00 mg/L

i) Determine the total, carbonate, and noncarbonate hardness

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Optimum turbidity value

Optimum alum dosage

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chemical constituents

mg/L as ion

molecular weight (MW)

nEquivalent

Weight (EW)

EW CaCO3 /EW

ionmg/L as CaCO3

CO2 6.6 44 2 22 2.2727 15

Ca2+ 44 40 2 20 2.5 110

Mg2+ 23 24.4 2 12.2 4.09836 94.26

HCO3- 145.05 61 1 61 0.8197 118.89

SO42- 32.1 96 2 48 1.0417 33.44

Cl- 42 35 1 35 1.4286 60

Total Hardness = 110 + 94.26= 204.26 mg/L as CaCO3

Carbonate hardness = 118.89 mg/L as CaCO3

Non Carbonate Hardness = 204.26 - 118.89 = 85.37 mg/L as CaCO3

ii) Determine the lime and soda ash dose, in mg/L as CaCO3 to soften the water to a final hardness of 80 mg/L as CaCO3. Assume the lime is 90% pure and soda ash is 98% pure. ( EW CaO= 28, EW Na2CO3= 53 )

Lime = CO2 = 15 as mg/L as as CaCO3

Lime = HCO3- = 118.89 as mg/L as CaCO3

Lime= Mg2+

Since Mg2+ is > 40 add lime = 94.26 - 40.0 = 54.26 as mg/L as CaCO3

Access limeSince 94.26 - 40.0 > 40 add lime = 40.0 as mg/L as CaCO3

Total lime = 15+ 118.89 + 54.26 + 40.0 = 228.15 mg/L as CaCO3

Lime as CaO

228.15 = 228.15 = 141.96 mg/L as CaO

Compute SodaAmount NCH that can be left, NCHf

NCHf = 80.0 - 40.0 = 40.0 mg/L as CaCO3

NCH to remove, NCHR

Amount soda addition = NCHR = 85.37 - 40.0 = 45.37mg/L as CaCO3

Soda as Na2CO3

45.37 = 45.37 = 49.07 mg/L as Na2CO3

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Q4 (a) State the need for secondary wastewater treatment.

(b)Briefly discuss the process of activated sludge.

(c) i) Aeration basin volume

V= 10 x 11250 x 0.6 ( 0.2 – 0.01) 3.5 1 + 0.06 x 10 = 2290m3

ii) Hydraulic retention time

θ = 2290/11250 = 0.2 days or 4.9 hrs…

iii) Volume of sludge wasted

Qw = 2290 x 3.5 / 10 x 15 = 53.4 m3/day

iv) F/M ratio = 11250 x 0.2 / 2290 x 3.5 = 0.28

Q5 (a) List sources of solid waste generation and types of solid waste.

Sources of solid waste Types of solid waste

Residential area Food wastes, rubbish, ashes, special waste – yard waste

Commercial area Food wastes, rubbish, ashes, demolition & construction

wastes, special wastes, occasionally hazardous wastes

Open areas Special wastes, rubbish

Treatment plant site Treatment plant wastes, principally composed of

residual sludge

Industrial activities Special wastes (depends on types of industry), rubbish

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(b) Explain factors that may influence the rate of solid waste generation.

Factors that affect waste generation rate are:

– Sources reduction

• Waste reduction may occur through the design, manufacturer, and packaging of

products with minimum toxic content, minimum vol. Of material, and longer

useful life.

• Waste reduction may occur at the household, commercial or industrial facility

through selective buying patterns and the reuse of products and materials.

• Some other ways in which reduction can be achieved by followed:

• Decrease unnecessary or excessive packaging

• Develop & use products with greater durability & repairability

• Use fewer resources

– Recycling

• The existence of recycling programs within a community definitely affects the

quantities of wastes collected for further processing.

– Public attitudes

• Reduction of SW occur when & if people are willing to change of their own

volition – their habits & lifestyles to conserve natural resources & to reduce

economic burdens associated with the management of SW.

• A program of continuing education is essential in bringing about a change in

public attitudes

– Legislation

• The most important factor affecting the generation of certain types of wastes is

the existence of local, state & federal regulations concerning the use of specific

material.

– Geographic & physical factors that affect the quantities of waste generated &

collected include location, season of the year, the use of kitchen waste food grinders,

waste collection frequency & the characteristics of the service area.

• Geographic location

– The influence of the geographic location the different climates can

influence both amount of certain types of SW generated & the time

period over which the wastes are generated.

• Season of the year

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– The seasonal variations affect the waste generation due to the

different human activities. In summer, the working hours are long

resulting in higher waste output; in M’sia, for example during local

fruit season, a lot of SW will be generated. The same will be expected

during harvesting & festivities.

(c) From the following data, estimate the unit waste generation rate in kg/capita/day for a residential area consisting 1500 houses. The observation period was one week. Assume that each household is comprised of 4.5 people.

Number of compactor truck loads = 20Volume of compactor truck = 15 m3

Density of solid waste compacted in compactor truck = 297 kg/m3

Number of flatbed truck = 15Volume of flatbed truck = 5 m3

Density of solid waste in flatbed truck = 89 kg/m3

1. Set up the computation table to estimate the total weight

No. of

loads

Average vol.

m3

Specific weight

kg/ m3

Total weight,

kg

Compactor truck 20 15 297 89,100

Un-compactor truck 15 5 89 6,675

Total, kg/week 95,775

2. Determine the waste collection rate based on the assumption that each household is

comprised of 4.5 people.

Unit rate = = 95775/47250

= 2.03 kg/ capita/day

Q6 (a) Discuss on air pollution by antharopogenic activities.

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1. stationary source 2. mobile source.

stationary source- open burning,, solid waste disposal, industrial, construction mobile source- transportation

(b) By using specific example, differentiate between primary and secondary air pollutant.

primary air pollutant – exposed direct from the sources (ex:sulphur oxides, nitrogen oxides, hydrocarbons, ash, smoke, dust, fumes, mist, sprays and radioactive compounds )

secondary air pollutant – cause by chemical reaction (fotochemical reaction, oxidation) between primary pollutant and components in atmosphere (ex:ozone, sulphur trioxide, proxyacyl nitrate (PAN), aldehydes and ketones )

(c) Why carbon monoxide is hazard to human health?

In red cell, normally oxygen will bind with hemoglobin but the carbon monoxide normally will compete with oxygen for the active site & resulting in low oxygen level in red blood cell.

(e) As a civil engineer in local authorities, design a framework to overcome air pollution problem in city area

Framework include plan, do, check, action activities

In planning1) Plan for task force2) plan for site visit3) type of data to be collected4) method to be proposed (technique or installation any apparatus)5) timeframe of study

Do1)Site visit to the polluted area (get a clear picture how serious the pollution)2)Gauge preliminary data on site (air pollutants, area affected, source of air pollutants3)Gauge set of data within monitoring time and compare with allowable standard4)Implement technique or apparatus installation

Check1) Is the proposed method working well?2) Data below/over limit permitted3) Is the data consistent4) Any problem occurred during monitoring time

Action1) Suggest a more comprehensive method if the previous method not effective2) Do corrective action ASAP

Any relevant answer but framework/idea proposed must be arranged in sequence (structured)(1 mark- for 1 logic answer)

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