Sanitary engineering revision questions and answers -1.pdf

SANITARY ENGINEERING REVISION QUESTIONS & ANSWERS By Albert Kenyani Inima 16 April 2015

QUESTION ONE [24 marks]

a. A factory factory is powered using a diesel power generator. generator. The generator generator is placed at the bottom of a corner of a room measuring 12 m [long], 6 m [wide] and 5 m [high]. The noise absorption coefficients for the room are 0.9 for the ceiling, 0.2 for the floor and 0.5 for the walls. The noise absorption of the door and window is negligible. The Noise Power Level [SWL] of the generator is 100 decibels [A-scale]. List of formulae:

Lp = Lw + 10 log( Q/ ( 4 π r 2 ) + 4 / R c) Rc = ST α/ (1- αm) (i) Explain the meaning of the noise directivity factor Q [2 marks] ====Model answer=== Q = Area of a sphere/ Actual Area of Noise Radiation (ii) What is the value of the Q factor for the generator , assuming it can be regarded to

be a point noise source [ 2 marks] ====Model answer ==== Q= 8 because the area of noise radiation is 1/8 of a sphere at the room corner. (iii) Calculate the total room surface area S T [ 1 mark] ====Model answer ==== Calculation of total surface area,

ST:

= 6 x12 [floor] + 6 x12 [ceiling] + 2 x [5 x12 + 6 x5] [walls] = 72 [floor] + 72 [ceiling] + 180 [walls] = 324 m2

(iv) Calculate the average room noise absorption coefficient ====Model answer ====

ST x α = 72 x 0.2 +72 x 0.9 + 180 X 0.5 324 x α =14.4 + 64.8 + 90 α = 169.2 / 324 = 0.52 v) Calculate the room constant R c [4 marks] ====Model answer ====

Rc = ST αm/ (1- αm) = [324 x 0.52] / [1 -0.52]

αm [4 marks].

=

351 Sabines [m2]

(Vi) Calculate the noise level experienced at the floor of the far corner [opposite corner] from the generator [4 marks] ===Model answer===

Distance to far corner = r = SQRT [ 6 2 + 122] = SQRT [180] = 13.4 m

Lp = Lw + 10 log( Q/ ( 4 π r 2 ) + 4 / R c) Lp = 100 + 10 log( 8/ ( 4 π X 13.4 2 ) + 4 / 351 ) = 100 + 10 log( 8/ ( 4 π X 13.4 2 ) + 4 / 351) =100 + 10 log {0.00354 + 0.01140} = 100 - 1.83 =98.17 dB b. Two power generator machines have a noise power rating [SWL] of 90 dB(A) and 100 dB(A). What would be the total noise level if both machines are placed side-by-side? [3 marks] Note the threshold noise power level is 1.0 x 10 ===Model answer=== SWL = 10 log [ W/W 0] W = W0 10 SWL/10

-12

W/M2 and SWL = 10 log [ W/W0]

Total power = W 1 + W2 = W 0 10 SWL1/10 + W0 10 SWL2/10 = 10-12 [109 +1010] = 0.011 Watts Convert back to decibels, SWLT = 10 log [0.011/10 -12] = 100.4 dB

c. The noise pressure level experienced 20m away from a factory alarm siren is 90 dBA. Calculate the noise pressure level experienced 50 m away from the siren in the same direction. You may use the following formulae sheet: LP = LW- 10 log A  Area of sphere =

4 π r 2

Q= Area of sphere / Actual area of noise propagation[A] [4 marks] ===Model answer==  Actual area of noise propagation = half a sphere = A=

2 π r 2

LP = LW- 10 log A LP = LW- 10 log

2 π r 2

90 = LW - 10 log [2 π (20) 2 ] --- Eqn 1 LP = LW - 10 log [2 π (50) 2 ] --- Eqn 2 Eqn 1 - Eqn 2,

90 - LP

=10 log [2 π (50) 2 ] - 10 log [2 π (20) 2 ]

= 10 log (2.5)2 = 7.9 dB LP =82.1dB QUESTION TWO [23 marks] a. A laboratory test has established that a wastewater sample has the following parameters: ● Total Dissolved Solids [TDS] = 100 mg/l ● Total Volatile Solids [TVS] =170 mg/l ● Total Solids [TS] = 350 mg/l ● Volatile Suspended Solids [VSS] = 120 mg/l Calculate the following parameters: ● ● ● ● ●

Total Fixed Solids [TFS] Total Suspended Solids [TSS] Fixed Suspended Solids [FSS] Volatile Dissolved Solids [VDS] Fixed Dissolved Solids [FDS]

[5 marks] ===MODEL ANSWER===

Dissolved Solids

Suspended Solids

Total Solids

Volatile Solids

VDS = 50 mg/l

VSS = 120 mg/l

TVS =170 mg/l

Fixed Solids

FDS = 50 mg/l

FSS =130 mg/l

TFS = 180 mg/l

Total Solids

TDS = 100 mg/l

TSS = 250 mg/l

TS = 350 mg/l

b) Compare and contrast slow sand filters[SSF] with rapid sand filters[RSF] [6 marks]

===MODEL ANSWER [7 marks for 4 points]==== Parameter

SSF

RSF

Space requirements

Occupies large area

Occupies little space

Effective particle size

0.2-0.3 mm

0.4-0.7 mm

Required pretreatment

Plain sedimentation tank

Chemical coagulation and flocculation

Filtration rate

Slow

More than 100 faster than RSF

Maintenance

Scrap off cake layers

Backwashing

Head loss before maintenance

Half of RSF

High

Removal of turbidity

Good

Good

Removal of pathogens

Good [Upto 98%]

Excellent [Upto 99.99%]

Operation

Does not require skilled labour

Requires skilled labour

Initial costs

High

Low

Running costs

Low

High

c) Describe the purpose of each of the following water treatment unit operations: ●   Pre-chlorination ●   Re-chlorination ●

Post chlorination

[6 marks] ==Model answer===

Prechlorination is the addition of chlorine to raw water The purpose of prechlorination is: ●

Disinfection against microbes that can cause disease to those working at the water treatment plant

●

Prevention of algal bloom within the water treatment plant

●

Odour control

●

Helps in decolourization of raw water

Post chlorination is the addition of chlorine to water that has already undergone the primary unit operations of softening, coagulation, flocculation, sedimentation and filtration.

The main purpose of post chlorination is:

●

To disinfect the water from disease causing microbes [pathogens]

●

To create a residual effect so that this disinfection process will continue to occur in the chlorinated for 24- 48 hours.

Re-chlorination is the post treatment addition of chlorine to water. The main purpose of re-chlorination is: ●

To disinfect the water that has overaged in the water distribution network [eg stagnated water in dead-end networks]

●

To boost the chlorination residual effect

●

To offer some disinfection to the water against point-of-use pollution

d) Explain using a sketch diagram the phenomenon called Breakpoint Chlorination [6 marks]

====MODEL ANSWER=== Explanation [4 marks]  Adding sufficient free available chlorine to completely oxidize all organic matter and ammonia [ nitrogen compounds in raw water]. All chlorine added after that point is free available

chlorine. Breakpoint chlorination is the minimum chlorination concentration that will cause lasting residual effect in the treated water. Diagram [3 marks]

QUESTION THREE [23 marks] Using a sketch diagram of the Ngethu Water Treatment plant answer the following questions[ 5 marks]: a. b. c. d. e. f.

What is the purpose of the water softening unit operation ? [3 marks] What is the purpose of the coagulation-flocculation unit operation? [3 marks] What is the purpose of the sedimentation unit operation? [3 marks] What is the purpose of the filtration unit operation? [3 marks] What is the purpose of disinfection unit operation? [3 marks] What is the purpose of water neutralization [pH correction] unit operation? [3 marks]

===Model answer===

[5 marks] a. Purpose of softening during water treatment [3 marks] ●

The purpose of water softening is to remove compounds of calcium and magnesium from water [eg sulphates, bicarbonates and chlorides]

●

These compounds prevent the action of soap [lathering] and also precipitate in water pipes leading to pipe clogging

b. Purpose of coagulation during water treatment [1.5 marks] ●

Coagulation mainly targets organic and soil particles less than 0.002 mm diameter which remain in solution due to surface negative charge ● The purpose of coagulation is to introduce particles with strong positive surface charge to neutralize the negative particles so as to make them coalesce Purpose of flocculation during water treatment [1.5 marks] ● ●

The purpose of flocculation is to continue the work began by coagulation of coalescing small particles into bigger particles so as to remove water turbidity and colour. Flocculation creates favourable conditions for small particles to coalesce into bigger

particles , called flocs. c. Purpose of sedimentation during water treatment [3 marks] ●

The purpose of sedimentation is to create the settlement of large flocs[macro-flocs] by gravity onto the bottom of the tank where they form sludge that is continuously removed and disposed. d. Purpose of filtration during water treatment [3 marks] ●

The purpose of filtration is to mechanically and biologically remove micro-flocs that do not settle in the sedimentation tank. Biochemical reactions in the filter media also help remove some microbes.

e. Purpose of disinfection during water treatment [3 marks] ●

The purpose of disinfection is to kill-off disease-causing microbes that may still be present in the water. f. The purpose of pH Correction in water treatment [3 marks] ●

The purpose of pH correction is to ensure the water transported to water distribution networks is not corrosive to the pipes.

●

pH must also be strictly controlled in order to prevent the production of Disinfection By Products [DPBs], some which can cause cancer. DBPs formation increases as pH is lowered, and especially at around pH=3.

QUESTION FOUR [23 marks] a) Using a graphical sketch differentiate between the terms BOD(remaining) and BOD exerted (consumed) [ 3 marks] ==Model answer====

b) Using a sketch diagram, briefly give the explanations for the following phenomena as observed during the oxygen sag curve: ● Increasing oxygen deficit phase ● The critical point phase ● The stream recovery phase [6 marks] ===Model answer====

●

●

●

Initially the oxygen deficit increases because the oxygen demand by BOD consuming microbes exceeds the natural stream aeration ability, giving a net negative oxygen concentration change in the stream water. At the critical point the stream natural aeration ability becomes equal to the rate of oxygen demand by microbes consuming BOD. The net oxygen concentration in the stream becomes zero. The DO concentration in the stream is lowest during this phase After the critical point, the oxygen concentration of the stream water continually increases towards saturation because the rate of oxygen removal by microbes becomes less than the rate of oxygen addition by natural stream aeration process.

c) The BOD(remaining) at 3 days was found to be 350 mg/l. The BOD(remaining) at 5 days was found to be 100 mg/l. Calculate the following parameters: ● UBOD ● BOD decay constant ● BOD( remaining) at 8 days. ● BOD (consumed at 8 days) [8 marks] ===Model answer==

BODt = UBOD e -Kt 350 = UBOD e -3K .... Equation 1 100 = UBOD e -5K ..... Equation 2 Eqn 1 divided by Eqn 2, 3.5 = e

2K

Ln 3.5 = 2K = 1.25 k= 0.625 Using Eqn 1, 350 = UBOD e -3K .... Equation 1 350 = UBOD e -[3x o.625] = UBOD e - 1.875 =UBOD/6.5 UBOD = 350x 6.5 = 2282 mg/l The first order kinetics model, BODt = 2282 e -0.625t t= 8 BOD8 = 2282 e

-0.625x8

= 15.4 mg/l

BOD8 [Consumed] = UBOD- BOD 8[Remaining] = 2282 - 15.4 = 2266.6 mg/l d) The Gaussian Plume Model [given below] is widely used to estimate air pollution due to industrial flue gas emissions:

Wind power law,

Explain how you would obtain the following parameters: ● ● ● ●

The standard deviation parameters The wind speed Effective stack height Ground level pollutant gas concentrations

[6 marks] ==Model answer === (i) Standard deviation parameters are obtained from locally developed graphs [or equations] for various atmospheric stability classes and various gases, eg

(ii) Wind speed is calculated from wind power law,

(iii) The effective height can be found using Plume rise formula eg Briggs’’s or Holland”s,

[iv) Ground level center line pollutant gas concentrations occur when,

Z= 0 Y =0 Substituting this in the formula we obtain,