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Question ANSWER

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									Question 1: [ANSWER]
CHAOS 2007 (A): Mathematical problems:

Part I

Topography is the recording of terrain and relief, which is the contours of the land, thus giving an
overall representation of the shape and configuration of the land. This allows for easy observation of
the three-dimensional aspect and the elevation of the surface, thus aiding also in identification of
specific landforms. While it can be used as a general term to describe such contour features of the
Earth’s surface, it is more commonly associated with land which is assigned positive elevation values
with reference to the sea level. Bathymetry is the equivalent of topography in that it is the study of
underwater depth relative to the sea leve, and involves the recording of contour features of the ocean
floor beneath the sea surface, which is then used to generate a three-dimensional image of the sea
floor.

Contour maps are maps characterized by large-scale detail and quantitative representation of the relief
of the land, using contour lines in mapping, connecting areas of similar elevation. It is used to
represent the general relief, i.e. slope, elevation, orientation, of the land. A geographical information
system is a database for recording and storing geographical and associated data spatially referenced to
the earth, which are then analyzed and integrated to solve planning and management problems.

Taking all these into consideration, the planimetric data of the sea bed in the Marina Bay area will be
needed. Located at the lower course of the river, the proposed Marina Bay lagoon will comprise part
of the Singapore River and associated streams within the Kallang Basin, which are expected to have
the geographical characteristic of typical lower-course rivers with the vertical cross-section in a wide
vertical cross-section with a horizontal river bed and gently sloping channel walls due to the constant
scouring of the river bed. Bathymetric data of the river bed, perhaps in hypsometric tinting form and
contour data, are needed to determine the exact dimensions and relief conformation of the riverbed,
from where the volume can be calculated at certain heights by sectioning parts of the river bed into
known polygonal shapes from where the volume can be derived from the appropriate formulas. The
cross-sectional area is likely to be a trapezoidal shape since the river sides have been concreted.
Another obvious polygonal shape present is the waterfront area in front of Raffles Place Interchange,
which is in the shape of a trapezium, narrowing towards the southwest. The volume there can be thus
calculated by utilising the geometric formula of a truncated rectangular pyramid where the dimensions
of the base and the rate of narrowing has to be obtained to determine the height of the actual pyramid
whereby the entire volume of the pyramid as well as the truncated portion can be calculated and hence
the volume of that catchment zone.

Alternatively, the various depths and volumes and different areas can be obtained from the
bathymetric and contour data of the river bed, from where they can be entered into a typical
Geographical Information System (GIS) computer database and inserted into various function tables to
calculate the volume as a function of height using gradient data and depth data. One example is to plot
the volume versus height graph and using computational programs, determine the relation y = f(h)
from the graph curvature where h refers to height from the lowest depth in the Marina Basin.

Likewise, the region covered by the new Marina Catchment area requires geographical information
such as the topography of the area, including contour lines and the three-dimensional orientation and
gradient of the entire basin. Via topographical data, the horizontal cross-sectional area can be
calculated throughout the basin. This cross-sectional area, taking the fluvial characteristics of a lower-



                                                    1
course river, will vary linearly with depth, especially since the likely vertical cross-sectional shape is
that of a trapezoid, hence a linear equation

SA = mh + c                                                          (2)

will be obtained, and hence the surface area of the water body can be calculated given a specified
height.

Furthermore, if a function for the volume of the reservoir as a function of the height, v = f(h) is
obtained, then the total surface area of the water body can be calculated via the following formula:

Volume =    A(h)dh , and therefore, A(h) = dV/dh                            (3)

This area function includes the wetted perimeter of the river channel as well as the surface area of the
river, and, as the surface area of the water body can be obtained through equation (2), the wetted
perimeter can thus be calculated by subtracting the above value from the differential equation (3),
deriving

R = dV/dh – mh – c                                                           (4)

Where R is the region, or wetted perimeter, V is the volume, m is the constant of proportionality and c
is an arbitrary constant. The geographical location of this water body can be obtained by comparing
the height of the water body with a contour map and joining areas having the same height as that of
the water body.

The geographical information can be obtained through satellite mapping or radar altimetry of the
entire basin to measure depths, such as through the service provided by CRISP, or the Centre for
Remote Imaging, Sensing and Processing, or international services such as WorldSat International Inc.
for cloud-free satellite imaging. Triangulation of the entire basin can also be made to determine the
topography of the surrounding land and this can be carried out through contractors.

Part II

Dynamic model refers to representational framework that expresses and describes the behaviour of a
system, including the different states the system can have, its transitions between these states and its
different pathways the system can follow under specified parameters. Catchment area is the area
drained by a water body such as lakes and rivers, bounded by the watersheds which are the highest
points of elevation forming the boundary of the catchment area, within which all water runoff will
drain into the river, forming the “catchment volume” of water.

Having a basin area of 240 ha and serving the Kallang Basin, a catchment area of 10 000 ha, the water
volume flowing into the Marina Reservoir will be very huge. However this would be affected by
several factors including the rainfall amount and rain rate and intensity, the dry flow of the river, the
amount of surface runoff, land use, soil type, drainage area and slope, surface evaporation and
groundwater inflow, all of which affect the flow directions and characteristics of the input flow into
the Marina reservoir through the various channels.

As rainfall occurs, some of the water is caught by leaf interception, and the rest falls to the ground as
raindrops. Normally some of the water infiltrates into the ground and goes into the throughflow and



                                                    2
groundwater above the aquifer, which travels according to ground elevation from higher to lower
areas, finally draining into water bodies, with that in the Kallang Basin being the various tributaries of
the Kallang and Singapore River and the Marina Basin. The rest of the water travels aboveground as
runoff, draining into drains and storm canals and finally emptying into the Marina Basin. However due
to the built-up urban area that Marina Catchment Area consists of, the largely impermeable concrete
surfaces will result in large amounts of river runoff and hence a huge peak discharge in rivers as
compared to a staggered, lower peak discharge typical of a hydrograph of rivers in wooded areas.

With major rivers having a return period (the period of occurrence of the worst storms that
infrastructure are designed to withstand) of 50-100 years, hence our calculations of the peak discharge
and the resulting enginnering requirements imposed on the dams revolve around a worst-case scenario
of 1-day rainfall amount, which, in the past 10 years, fell on 27/12/2001 with a rainfall of 211.1 mm,
hence for ease of calculations this value shall be approximated to 240.0mm, while air temperature is
30oC. We base our calcuations on all runoff entering the start of Kallang River with the entire length
of Kallang River being 10km, the typical hydraulic radius being 150/56 =2.6786, and the elevation
being 20m (figures provided by Longman Atlas). The computation of peak discharge rate follows the
Rational Formula assuming 100% runoff drains into the drainage canals and finally into the Marina
Basin.

        1
Qr        CIA                                                                      (1)
       360

where Qr       = peak runoff (m3 s-1)
      C        = runoff coefficient (chosen value is 0.80 – describing residential/industrial areas
               densely built up)
       I       = average rainfall intensity (mm hr-1) – chosen value is 240  10
                                                                           24
       A       = catchment area (ha) – which is 10 000 ha

However this discharge will be tempered by the rate of evaporation from the open water surfaces in
the canals and rivers draining into the Marina Basin, and also from Marina Basin itself. The rate of
evaporation over the whole catchment area will hence reduce the amount of water collected. This rate
of evaporation is described using the Penman equation:


                                                                                    (2)

where:
m      = Slope of the saturation vapor pressure curve (Pa K-1)
Rn     = Net irradiance (W m-2) – in this case assuming 0
sunshine during rain
ρa     = density of air (kg m-3) – 1.164 kg m-3
cp     = heat capacity of air (J kg-1) – 1006 J kg-1
Cat    = atmospheric conductance (m s-1) – 5
Δe     = vapor pressure deficit (Pa)
λv     = latent heat of vaporization (J kg-1) – 2 260 000 J kg-1
γ      = psychrometric constant (Pa K-1)
E      = kg of water evaporated every second for each m2 of
area.




                                                    3
And γ given by:

      Cp Po
                                                       (3)
       L

where Cp       = Specific heat of air (J Kg-1 oC -1) – 1006 J Kg-1 oC -1
      Po       = Atmospheric pressure (hPa) – 1013 hPa
      L        = Latent heat of vaporization (J kg-1) – 2 260 000 J kg-1
              = Ratio of molecular weight of water vapour over dry air – 0.622

Therefore total discharge Q of rivers =

                                   0.80(10)(10000)                 1.164(1006)(5)(0.21229)
1.0 103 Qr  1.0 108 E  (1.0 103 )              (1.0 108 )
                                          360                    2.26 106 (0.02207  0.72495)
                  = 1.4860 105 kg s-1 (5 sf)

In comparison, this rate of evaporation is much smaller than the runoff rate, mainly due to the high
humidity levels during rainfall that minimizes the evaporation rate. However, this amount calculated is
still likely to have a significant effect on the final value of the force of onrushing water since it is to
the tune of about 700 kg.

                                                                 dm
Theoretically, by conserving the rate of flow of mass, i.e.           Av where ρ is density of water, A is
                                                                 dt
cross-sectional area and v is the velocity of water, the velocity can be calculated, but this is not
feasible in practical situation since it is affected to a great extent by the roughness of the channel walls
and also the different rate of flows that water at different parts of the river cross-section will have.
Tying these factors together, the velocity is then calculated by the Manning equation:

                                                                                    Estimated cross section of
                                                         (4)                             Kallang River

where V        = cross-sectional average velocity (m s-1)
      n        = Manning coefficient of roughness 0.0150 for concrete 3 m
      Rh       = hydraulic radius (m) – (50x3)/(50+3+3) = 2.6786
      S        = slope of pipe/channel – 20/10 000 = 0.002
                                                                                              50 m
                                          -1
Therefore the calculated v = 5.7503 m s
Force exerted by this water = Q v = 1.4860 x 105 x 5.7503 = 8.5449 x 105 N (5 sf)
As the dimensions of the dam are 350m x 9.6m,
                            8.5449 105
Pressure exerted on dam =                254.31 Pa
                              350  9.6

In fluvial geography, the maximum flow rate, and hence maximum force, occurs near the surface,
away from the sides and bottom which is slower due to friction and also due to viscosity, so we take a
case study on a patch of width 1.0 m in the middle of the dam for simplicity:                5.1 m

Total F = 254.31 (1.0)2 + (1000)(10)(4.8)(1.0) = 4.8254 x 104 N (5 sf)                        Dam wall

Thus by Principle of Moments                                            F = 4.8254 x 104N

                                                                                                                 4.8 m
                                                     4
 4.8254 10   4.8  W (2.55)
            4


W  9.0831104 N

Having a length of 350m,
The dam has to be at least 350 x 90831 = 3.1791 x 107 kg.                      Dam wall

Now, considering the area near the base of the dam,
Assuming       force     of     incoming     water     is     the    same      as     in    above      and
force is equal on both sides of dam wall as is possible during high tides:
                                                        F = 1.9251 x 105N                      F = 1.9251 x 105N
Total compressive forces =
2   254.311.02  10 103  9.6  1.9251105 N

Therefore the dam has to have a high compressive strength of at least 2.7392 x 106 Pa. Concrete or
reinforced concrete can provide this since they can be made to have a high compressive strength of
20.7 MPa.

The Ultrasonic Doppler Flowmeter can be used to measure fluid velocities by making use of the
principle of Doppler shifts – that the motion of a source will effect the frequency of reflected or
emitted sound, or electromagnetic waves. If a fluid is moving towards a transducer, the frequency of
the returning signal will increase. As fluid moves away from a transducer, the frequency of the
returning signal decrease. Hence by aiming ultrasonic waves at it and comparing the returned
frequency with the transmitted one, the flow velocity of a liquid can be ascertained:
      f  VT 
VT                   Kf                                                        (5)
      f T  sin  T 
                    

where VT = sonic velocity of transmitter material,
      T = angle of transmitted beam,
      K = calibration factor
      Vf = flow velocity
      f = Doppler frequency shift
      Vs = sonic velocity of fluid                                                   Since A1 > A2,
      fT = transmitter frequency                                                           v1 < v2
       = angle of fT entry into liquid

Ways of measuring flow rates make use of Bernoulli’s equation which states that an increase in
velocity is accompanied by a reduction in pressure. Hence the flow rate is calculated by measuring
velocity and/or pressure changes by fluids through orifices that narrow the cross-sectional area of flow
of the fluid.This is appliled using an orifice plate or venturi meter that narrows the cross-sectional flow
area by introducing a plate with a small hole, or orifice, that forces the fluid to converge and flow
through the hole. Bernoulli’s equation is then applied:
     1               1
 P1  v1  P2  v 2
           2              2
                                                        (6)
     2               2
Hence, since discharge (Q) = volume (V) multiplied by cross-sectional area (A), and so V = Q/A and
the discharge can be calculated by manipulating the two equations:




                                                     5
         2P1  P2  / 
QA                                                                                 (7)
         1   A2  A1 
                        2


Instead of using venture meters or orifice plates, a weir can also be constructed across the river, which
is a small overflow-type dam over which water flows, and since the dimensions of the top of the weir
are known, the cross-sectional area can be calculated by measuring depth of water flow over the weir,
since it has to maintain an equal discharge as upstream although it encounters an obstruction. This is
then multiplied with the velocity at that point to give the flow rate of the river                   (m3 s-1)
                                        Water flowing over           Depth h

                                               Weir                                 Where x and y are
                                                                       x            known lengths
                                    Question 1: [ANNEX]
                                          y




References:
http://library.thinkquest.org/28022/case/spore.html

http://supit.net/main.php?q=aXRlbV9pZD02NA==

http://www.michigan.gov/cgi/

http://www.fao.org/docrep/R4082E/r4082e04.htm

http://www.linz.govt.nz/core/topography/topographicmaps/index.html

http://www.pub.gov.sg/downloads/DR7.aspx

http://www.sensorsmag.com/articles/1097/flow1097/main.shtml

http://en.wikipedia.org/wiki/Weir

http://www.faqs.org/faqs/geography/infosystems-faq/

http://erg.usgs.gov/isb/pubs/gis_poster/

http://sealevel.jpl.nasa.gov/science/bathymetry.html

http://www1.uts.com/Physics/flowmetering/flowmeter.htm




                                                      6
  Acknowledgment:


Question 2: [ANSWER]
   (B) Biology Problems
   [A]
   Biodiversity refers in this context to the variety of species of organisms which exist in a
particular area. The biodiversity in the Singapore River 1, which is considered as a natural open
country flowing water body, would hence be gauged by the array of different organisms that live
in the river, the completeness of the food chain as well as the degree to which these thrive.
Organisms in the Singapore River are covered under a few main classes. Under plants, we have
the green algae (Pediastrum boryanum) and small clusters of duckweed (Lemna minor). We have
waterfowl such as the Rallidae Family (Rallus striatus), (Ardea purpurea) and (Porzana fusca)
which have periodical breeding grounds on the banks of the Singapore River.
Short Term
       The initial process of dam construction would be damaging enough in itself. The scouring
and armouring of the dam would destroy the sedimental layers on the lower surface of the
riverbed, which would release dust and relevant pollutants into the water thus leading to an
increase in the concentration of particulate matter in the water. In addition to this, the dam walls,
when constructed across the river basins, would cause the accumulation of floodwaters and silt in
the dam, thus increasing the rate and magnitude of overall sedimentation. Over prolonged
periods of time, the particulate matter, if failing to settle, would result in lower light intensity and
turgid waters in the river, the effect of which would be to reduce the sunlight received by the
plankton and aquatic plants, ending in eutrophication and the unsuitability of the river to further
sustain marine life. Hence, biodiversity would be reduced as the river can only support a small
segment of life forms with lower oxygen levels.
       Tarbela Dam, located in the Indus River Basin is an example of such a case study. The
predicted rate of sediment inflow was 0.294 bcm per year meaning that the dam would silt up to
90% capacity in 50 years. This has caused the waters in the area to have sedimented tributaries
that block water flow, such as the tributaries near Islamabad and Peshawar where drought due to
poor water distribution is soon becoming a problem.
       The Singapore River is one that opens directly into the South China Sea. The dam will
create a blockage, thus causing erosion to take place along coastal regions by the sea beyond
the mouth of the river. Hence, the river will also tend to become narrower and deeper
downstream, which will also reduce the diversity of animal and plant life that it can support. This
was witnessed in the case of the Akosombo Dam in Ghana. The presence of the Lake Volta has
actually disrupted the tributaries which convey water to the jungles in the area, causing them to
transport less water to these areas, which is a detrimental change that could adversely impact
populations reliant on small streams like the common guppy, (Poecilia reticulate).
       Dams also release pollutants into the water and affect multiple aspects of water quality
downstream by their very existence. River temperature, nutrient load, turbidity, dissolved gases,
concentration of heavy metals and minerals are among the factors that can be altered. These
could be affected by the production and deposition of waste effluents by factories and inhabited
areas along the river without conclusive treatment. The storage of river water in the reservoirs
further up the dam would cause a rise in the water level over the forests and former inhabited

  1 The specific conditions of the Singapore River are further elaborated on under the annexe segments.


                                                       7
areas in the river. Thus, the net decomposition of these items would culminate in the growth of
many aquatic plants and would also take up oxygen as the bacteria go about the process, which
would deplete oxygen levels further. The lower levels of oxygen in the water would bring about
eutrophication, a potentially deadly scenario in which the river is no longer able to support the
aquatic life in it due to insufficient oxygen levels in the water, thus rendering the river ‘dead’. The
maintenance of stagnant and not moving water in the dams would also cause a rise in the
amount of toxins and deadly substances released by industries into the water, as there is no
outlet in the still waters for the removal of these wastes. These wastes could come in the form of
biotoxins, which when accumulated would pose serious threats to the biodiversity higher up the
food chain. Commercial fish which fishermen gather from the fringes of the river, such as the
common snakehead (Channa striata) and the tilapia (Oreochromis mossambicus) which are local
culinary favourites are highly susceptible to this as they are tertiary predators which absorb the
multiplied effect of toxins lower down the food chain. Biodiversity would be limited once again as
not only are the top predators eliminated; their low quantity also causes the species below them
in the food chain to thrive unnaturally without predation, which will exhaust the population of the
subsequent species below them and so forth. This vicious cycle will result in the dominance of
certain species, which will reduce biodiversity dramatically.
        Water contamination could result in the form of industrial pollutants such as mercury.
Studies in Canada have shown that methylmercury accumulated in river salmon increased
substantially after the construction of artificial dams, which create reservoirs in the middle of the
river’s course. Hence, the change to the reservoir’s water flow pattern would only result in few
species being able to acclimatize, thus narrowing the scope of biodiversity.
Long Term
        The macro scale impact of the damming of the river will affect the sequence and timeframe
in which floods take place. This has detrimental effects twofold by delaying the rate at which
nutrients from the land are transferred into the river, in addition to preventing the creation of
backwater areas in which the young of marine organisms, such as fish and crab larvae, can
mature without being consumed by predators.
    The generations of organisms would also face the problem of continuity and species
reproduction because the larvae will have high mortality rates when introduced together with
predators. With reproduction cycles that do not coincide with favourable river conditions,
temperature and nutrients, larvae will be brought into existence at a point in time in which they
cannot possibly survive. In extreme conditions, the harmful effects of the dams would be
exemplified fully as the organisms are deprived of proper bypass systems to enter breeding
grounds and will thus encounter localized species extinction. For example, one immediate impact
of settlement on Lake Ontario fish populations in the early to mid-19th century, was elimination of
dozens of salmon runs in streams flowing into Lake Ontario, as these streams were blocked by
milldams. When river ecosystems are fragmented and the migration process is ceased, it can
have severe impacts on the wildlife in the area. The combination of these factors would limit
biodiversity as there is reduced interaction between the predator, the prey and living habitat,
which causes an imbalance in the proportions in the ecosystem. This hence produces the
dominance of some species over others, upsetting the fragile ecological balance and reducing
biodiversity.
           If the water in the dammed up area remains stagnant for excessively long periods of
time, assuming that the dam lacks any filtration or water aeration mechanisms, it will become
stagnant and filled with algae as the slow moving waters are breeding grounds for organisms.
Mosquitoes, snails, and flies, the vectors that carry malaria, schistosomiasis, and river blindness,
are examples of organisms that thrive in slow moving, stagnant water. The negative impact would
be that the dam’s water would not be fresh and would promote the spread of water borne
diseases to individuals who consume the dam’s water.


                                                  8
        The construction of a dam would also hold back the amounts of sediment in the flow,
preventing the water downstream from gaining sufficient nutrients and particulate matter in the
form of sediments. Upstream habitats would be damaged as rising water levels would destroy
breeding colonies and marshes across the shoreline and banks, which are crucial to waterfowl
and amphibians that inhabit the area. Furthermore, the increased sedimentation of silt layers at
the bottom of the dam would bury the riverbed, thus destroying the habitats of fish such as gobies
and water plants that root at the base of the river. The downstream water would thus erode its
channels and banks, thus lowering the riverbed. The presence of the dam also negates the
beneficial impacts of flooding through its spreading of nutrients onto land near the river. The
impact on biodiversity in this case, for rivers in general, is that the wildlife further down the river
will be in great decline as the waters that emerge from the dam, devoid of sediment, will then
recapture sediments by eroding downstream banks and river beds. Eventually, when most of the
riverbed sediments have been eroded, the only surface is the rocky riverbed, which is a poor
habitat for bottom dwelling organisms such as crayfish, molluscs and shellfish. Species such as
the walking catfish, (Clarias batrachus) and the 6 banded tiger barb (Puntius johorensis) which
make their homes amongst river plants or on the river bed would be severely affected. The
habitat destruction both upstream and downstream that damming brings about deprives animals
of their breeding grounds. The river’s biodiversity is thus reduced by the lack of a variety of
habitats, replaced by the aquatic habitat of the reservoir which can only support the limited range
of wildlife seen in the deeper regions of the water.
    To conclude, the overall impact of water flow in the upper half of the river has great impact on
what goes on the lower half. Therefore, a disruption in the upper half of the river such as a dam
will definitely produce devastating results on the river in the lower half.
[B]
   The condition with excessive algal growth in the water is known as eutrophication. The root
cause of the problem of eutrophication would be the sudden increase in the addition of large
quantities of nutrients into the water from sewage effluent, which would cause excessive plant
growth and commensurate reductions in water quality. Also, in the case of dams in which the
existing features of the landscape and inhabited areas are covered by the rising level of the river
waters, the decomposition of these areas and the trees would also deprive the water of its oxygen
quality, producing a similar impact of a vicious cycle involving the decomposition of plant matter
thus depleting the oxygen levels in the water. The prevention of algal growth and the occurrence
of a stagnant pool could be handled by 2 different approaches, one which involves the continuous
circulation and aeration of water so as to prevent the particles in the water from accumulating
excessively, thus causing pollution, and secondly, involving the stringent control of an input of
contents into the water so as to prevent any pollutants from infiltrating the water initially.
        First, the aeration of the water in the reservoir would aid the process of preventing
eutrophication. However, this would have to be carried out in combination with the other
measures implemented for the removal of algal growth, as the aeration of water only serves to
increase the oxygen content of the water. At the base of the dam, air pipes and filters could be
introduced so as to drive in oxygen from the base of the dam and increase the oxygen content of
the water in the reservoir as a whole. Surface aeration, in the form of electrical water pumps or
diffusion aerators which function underwater, can be used in conjunction with mechanical and
biological filtration systems. Water pumps, targeted at removing algae, fish wastes and other
sediments, can also be installed to filter off sediments.
        Second, before allowing the construction of the dam to take place and the creation of a
reservoir, existing biodegradable materials and human inhabited areas must be removed before
the water submerges. This prevents them from decomposing in the water when the reservoir is
formed, thus conserving the precious oxygen content that we require in the water. The pollutants
that enter the water from dumping and industrial outputs also need to be carefully monitored and


                                                  9
checked. Outputs from industries could be checked by incentives to contain the waste efficiently
underneath the ground, as well as through the usage of taxation on polluters who grossly exceed
the limits. The major problem, which could originate from non-point sources and can be found
along the Singapore River, does not occur as the Singapore River is situated along industrial and
developed areas, where the main sources of pollution come in the form of point sources.
        The solutions to the problem are not restricted to merely preventive measures. The
problem can be solved via remediation. These can be through the usage of cleanup process,
such as using the Finnish method of putting in phosphorus / nitrogenous reactive compounds
which can remove these substances. Measures to intercept non-point as well as point sources
could be established, through the implementation of buffers between the source and the river
could be enforced. This is known as the Riparian buffer system, an interface between land and
water which filters off pollutants. Restrictions on animal waste and nitrogenous fertilizers, such as
those in Japan, should also be imposed to prevent excessive nutrients from leaching into the
water.
   Third, a preliminary biological barrier which controls the hard objects and relatively large solid
pollutants coming into the reservoir could be established. At the entry point of the Singapore
River before the tributaries converge into the reservoir, a barrier made from metal bars with fine
wire mesh in between will be erected. This would allow large decomposing objects and dead
aquatic organisms to be screened out and removed from polluting the waters in of the reservoir.
Fish runs, similar to the salmon runs at dams, will be established beside the barrier to facilitate
the movement of fish that seek to enter the reservoir to breed. Through the prevention of large
objects from leaving the reservoir or entering it, the barrier keeps large commercial fish in the
reservoir while preventing large decomposing organisms from polluting the water.
   Fourth, the maintenance of the large scale water purification in the lagoon and the reservoir
cannot be maintained by purely automated mechanisms alone. The maintenance of the clean and
clear water using natural mechanisms is crucial if it is to be spread over the entire lagoon. Aquatic
plants can be introduced into the river, which can remove pollutants from the water and release
healthy amounts of oxygen when photosynthesis takes place. Plants such as the hydrilla,
cabomba and the Indian lotus, which are able to remove smaller pollutants through their leaves
by absorbing smaller sediments, will be introduced onto the river bed. The byproducts of stagnant
water, inclusive of wrigglers and algae, can be combated by the introduction of their natural
predators. Species such as the mosquito fish (Gambusia holbrookii) which fulfill this function are
useful inclusions into the ecosystem. These also maintain the gaseous composition of the water
and do not cause unnecessary releases of plant matter as they are rooted to the ground. Further
to that, we will introduce the bighead carp (Hypopthalmichthys nobilis)into the lagoon in the
confines of floating net cages. These do not interfere with the overall process of water purification
as they are kept in cages, but serve to control the microscopic floating plankton population in the
water.
   Fifth, periodical releases of water and sediment, so as to facilitate the inflow and exchange of
water, can be instituted via vents at the base of the reservoir so as to prevent the reservoir from
being infested with algae and bacteria. Continual flow of water is requires in order to relay the
nitrogenous wastes out to the sea and other areas where they will not amass and promote algal
growth. Fishes such as the Harlequin Rasbora (Rasbora heteromorpha) are an excellent gauge
of the water's potability quality, and can be introduced into the lagoon for monitoring purposes.
   The maintenance of the freshwater ecosystem can be improved if we include a wide variety of
organisms that perform different roles in the food chain. Detritivores, top-of-the-chain predators,
decomposers and plants are examples of organisms which fulfill important roles in the
ecosystem. Species which do not compete with local freshwater species (such as loaches,
halfbeaks, mussels and some crabs) can be introduced. Loaches are particularly useful as they
aid the process of decomposition by turning solid and liquid biodegradable waste into smaller


                                                 10
particles for the detritivores to break down. This is the crucial expedient link that most ecosystems
lack as the process for detritivores to break down solid waste in huge chunks is extremely slow,
as the surface area of the waste is extremely small relative to the mass in which the waste is
deposited. Other niche roles such as that of the top predator – keeping the populations of the
herbivores lower down the food chain in check – prevent the decomposition of these herbivores
which will take up additional oxygen.
   The water in the Singapore River is fresh and slow moving, and due to its intended potable
purposes will have relatively few major pollutants such as heavy metals. The suitability of aquatic
life in adapting to these conditions can hence be gauged. Fishes that are suited to freshwater
near estuaries could be considered, such as the Common Molly (Poecilia sphenops) and the
Common Carp (Cyprinus carpio). Also, the maintenance of indigenous, endemic species which
deserve to be conserved could be introduced, the chief of which would be the Forest Halfbeak
(Hemirhamphodon pogonognathus).
   Finally, a careful monitoring and implementation of conservation related measures needs to be
instituted so as to maintain the state of the environment and promote the abundance of wildlife.
The maintenance of this ecosystem has to be supported by careful monitoring of the changes in
its temperature, qualities and surroundings. Based on surveys of the water, such as excessive
algal growth or insufficient sunlight in the lower reaches, we can then implement the relevant
changes required to keep this almost self sustaining ecosystem in balance. Furthermore, in order
to encourage greater biodiversity in the area, the area beside the reservoir should be cultivated
with shrubs and low-lying trees so as to provide the materials with which waterfowl require to
construct their nests. The availability of nesting sites would definitely contribute towards a better
rounded environment.
          Our proposed mechanism to make the ecosystem in the Singapore River one that is
   teeming with marine life and has a flourishing ecosystem one that involves establishing a
   resource base at the bottom of the entire ecosystem so that the organisms further up on the
   ecosystem will be able to tap on it and take in nutrients from the base of the entire food
   chain. By fortifying the foundation the food pyramid, we actually enable the organisms further
   up the food pyramid to thrive as they have an increased number of primary producers to feed
   on. The introduction of plant phytoplankton into the reservoir at suitable quantities as a
   keystone species, and removing algae, will allow a more rapid conversion of nutrients and
   energy further up the food chain as they are more immediate mechanisms as compared to
   the rest, due to the wide array of organisms that can consume phytoplankton, and also the
   fact that phytoplankton can reproduce more efficiently using less resources than algae do,
   and do not absorb that much oxygen upon decomposition. Question 2: [ANNEX]

  Background Information
         The Singapore River has tributaries from both open country and secondary forest
  streams, which means its water has reasonable shade, generally higher temperatures and
  the riverbed is almost always muddy. The waters in the Singapore river are relatively
  challenging compared to conventional streams, as it has waters that are constantly flowing,
  with rife and unpredictable pollution sources, resulting in extreme temperature and pH
  fluctuations.
         In the area in which the proposed dam is to be built, the water quality changes
  dramatically. The water becomes slightly brackish with increased proximity to the sea, and
  thus has more rocky substrates instead of a muddy riverbed.




                                                 11
Planned Species for Introduction: Indigenous Fish Forest Halfbeak




Diagram elucidating on the impacts of eutrophication

References:

Tan, K.S. (2006).The Invasive Carribbean Bivalve MYTILOPSIS SALLEI (DREISSENIDAE)
INTRODUCED TO SINGAPORE, AND JOHOR BAHRU, MALAYSIA . The Raffles Bulletin of
Zoology. 6.

Alfred, E.R. 1961. Singapore Freshwater Fishes. Malayan Nature Journal, Volume 15, pages
1-19.

Axelrod, H. R., W. E. Burgess, N. Pronek & J.G. Walls, 1989. Dr. Axelrod's Atlas of
Freshwater Fishes, Third Edition. T.F.H. Publications, 797 pages.

Johnson, D.S., 1973. Brackish Waters. In: Animal Life and Nature in Singapore. Edited by
S.H. Chuang, Singapore University Press, pages 103-127.

Smith, H.M. 1945. The Freshwater Fishes of Siam, or Thailand. Smithsonian Institution,
United States of America National Museum Bulltein 188, 622 pages.

Mohsin, A. K. M. & A. Ambak, 1983, Freshwater Fishes of Peninsular Malaysia. Penerbit
Universiti Pertanian Malaysia, 284 pages.



                                            12
Acknowledgment:

      http://www.dams.org/report/wcd_overview.htm

      http://www.eawag.ch/media/20060323/index_EN

      http://www.environment.fi/default.asp?node=6024&lan=en

http://www.dams.org/kbase/studies/pk/pk_exec.htm Question              3: [ANSWER]
Waste management
Regarding waste management for the water entering the lagoon, gantries with raked screens
and skimming devices will be erected at the various confluences of the tributaries. They will
trap and filter off solid visible objects (litter and dead leaves) and floating oils, leaving the rest
of the river water into the lagoon. The trapped litter can be collected and treated like
household refuse. The organic waste and litter makes up most of the unsightly pollutants in
the river, so after its removal the water can be deemed clean enough to be released into the
lagoon. The purpose of such mechanical filtration is to reduce and/or control the natural
organic matter and particulate load entering treatment facility. Further purification of water is
carried out in the water treatment facility which will be discussed in detail in part (ii).
Membrane filtration vs biofilm treatment
Membrane filtration technology is preferred because of its many advantages over biofilm
treatment, it has been developed for a relatively longer period of time, therefore the
technology is mature, widely-understood and practised, also, it is versatile because it can
filter off all kinds of impurities of all sizes (from 1mm to 1 μm), and works without the addition
of chemicals, with a relatively low energy use. There are also doubts about the effectiveness
of biofilm treatment; and biodegradation of substances in the water may also produce
harmful products, which may require additional effort to remove.
(i)
Classification of pollutants
The pollutants in the river water can be classified into the following categories: Biological,
chemical and microbial wastes. Biological wastes include sewage, dead plant and animal
matter. Chemical wastes include heavy metal ions (lead, mercury .etc), plastics, waste oils
and solvents. Microbial wastes include waterborne bacteria, viruses and protozoa. Potable
water and industrial water should not contain any of the three categories of pollutants listed
above.
Samples of water should be taken at different areas of Kallang river and Marina Bay before
and after dam construction and analysis of their contents compared. During construction of
Marina Barrage, water samples should also be taken to make sure water quality is
maintained and there is not pollution form the construction site entering the river.
Chemical waste should be focused on because it encompasses a wider range of harmful
substances which may cause water poisoning. It differs from biological waste in terms of
biodegradability, most biological wastes are biodegradable, but chemical wastes may include
inorganic chemicals (metals) and organic chemicals (plastics) which are non-biodegradable.
It differs from microbial waste in terms of source of pollution. Microbes, including pathogens,


                                                 13
sometimes are naturally occurring, but most chemical wastes, excluding naturally metal ions,
are released into water by industries and households. Thus chemical wastes can be
considered as a major source of human pollution. Of the chemical wastes, agriculture
wastewater should not be a concern because Singapore is not an agricultural country and
Kallang river flows through urban areas. So pollution from that re4spect is not expected to be
a major problem. Instead, the focus should be on waste oils and solvents because it is often
dumped from boats travelling in the river. This problem has been persistent in the history of
Singapore river, and therefore should be of a major concern for the Marina reservoir.
Gas chromatography mass spectrometry (GC/MS) is useful for analyzing organic chemicals
present in the river water and electrothermal atomic absorption spectrometry (EAAS) to
determine amounts of inorganic chemicals and metal ions. Both technologies are chosen due
to their ease of usage, versatility, sensitivity, range of chemicals tested and above all,
accuracy of their results.
GC/MS identifies and quantifies trace organic compounds. A gas is used as the mobile
phase, and the stationary phase is a sample of river water coated either on an inert granular
solid or on the walls of a capillary column. When the sample is injected into the column, the
organic compounds are vaporized and moved through the column by the carrier gas at
different rates. The gas exiting the column is passed to a mass spectrometer, where it is
ionized, fragmented and separated by a magnetic field to according to their mass. The
separation ability for GC/MS is very good, so mixtures of organic substances with similar
structure can be separated, identified and determined quantitatively in a single operation.
Therefore, GC/MS is very suitable for carrying out an analysis of waste oils in the river.
EAAS uses an electrically heated atomizer or graphite furnace to atomize metals and
measures it absorption spectrum, thereby determining the identity of the metal. An
advantage of EAAS is that it has a higher sensitivity and lower detection limit than other
systems, so minute traces of harmful metals can be detected by EAAS.
(ii)
Physical processes include sedimentation and filtration. Sedimentation is a water treatment
process whereby sewage water is passed into large tanks for a long enough period such
that, under gravitational influence, suspended solids (or flocs)settle down and deposits at the
bottom while oil and grease float tot eh top. This leaves a layer of liquid generally free of
solids in the middle which can be isolated is treated separately. Total time for sedimentation
may range from 1-4 hours. Filtration is a process whereby a filter is used to physically
separate a mixture of solids and liquids. In the case of water purification, the filter media is a
bed of granular material, such as sand or carbon, which attracts and retains the solid
particles in the water as it passes through the filter. Reverse osmosis, on one hand, is also a
filtration process. In reverse osmosis, a solvent is hydraulically forced through a
semipermeable membrane, which acts as a filter here. Solutes are trapped on the filter and
pure solvent passes through to be collected.
Chemical processes include coagulation chlorination, water conditioning (pH adjustment).
Coagulation and flocculation are a series of water treatment processes to separate
suspended solids portion from sewage water, step prior to sedimentation. Many of the
suspended water particles are negatively charged, which keeps them suspended because
the charges repel one another. Coagulants (such as aluminum sulphate or iron (III) chloride)
are added to the water to neutralize charges in the suspended particles. Once the charge is
neutralized, the small suspended particles then can stick together to form larger particles
(though still not visible to the naked eye) called microflocs. Rapid high energy mixing is often



                                               14
required to thoroughly disperse coagulants as well as allow microflocs to collide and to
increase in size and deposit on the bottom of the flocculation basin (or sedimentation tank),
where sedimentation takes place. The total time for coagulation and flocculation may vary
from 20-60 minutes.Chlorination is a chemical process of adding sodium hypochlorite into
water for water purification. It releases free chlorine when dissolved in water. Chlorination is
used to disinfect the water and prevent the spread of waterborne diseases by killing
pathogens in the water. Water conditioning involves adding chemicals to balance the pH of
water and/or reducing its hardness. pH of potable water is 6.8 to 8.1, if the water is acidic,
liming material such as lime or soda ash is added to raise the pH. If the water is hard i.e.
contains a lot of calcium or magnesium carbonate, it needs to be treated with soda ash to
precipitate out the excess salts.
Some of the above processes, except reverse osmosis and chlorination, can be applied to
the water management programme before and after the building of the dam. Reverse
osmosis cannot be applied prior to dam construction because the feed water entering the
processing plant must be free of large particles (sand and grit) which may clog the
membranes, so sedimentation and microfiltration must take place before going through
reverse osmosis. Chlorination may cause problems chlorine ions reacts with substances
dissolved in the water to form carcinogenic disinfection byproducts (DBPs). Furthermore,
since the water in Singapore is not hard, water conditioning is not of a major necessity,
periodic monitoring of the hardness of the water will suffice.
The actual membrane filtration is a three-stage water purification process. Firstly, treated
water goes through a ceramic membrane filter pipe, and then the water is pumped through a
reverse osmosis filter with a thin film composite membrane in a polymer matrix at 17 bar. The
first membrane filter removes heavy metals, cyanides, organic chemicals, while the reverse
osmosis filter removes anything with molecule size larger than 200 daltons. Through this, we
can expect total removal of bacteria, viruses and protozoa. The last stage of water treatment
is disinfection, application of an adequate level of disinfection is an essential element for
most treatment systems to achieve the necessary level of microbial risk reduction (even
though there will be minimal/zero amount of microbes present after reverse osmosis,
disinfection process is there as a safety precaution. Ultraviolet Germicidal Irradiation is used
to sterilize the treated water of pathogens and moulds, unlike chlorination, it does not involve
addition of chemicals and leaves no residue in the water.
Reverse osmosis is necessary because water treatment for drinking and industry use is a
matter of public health and social concern, therefore there should not be any margin of error.
Treated water for industrial usage must be extremely pure (deionised water is required)
because the precision and high level of technology of electronics industry in Singapore. Only
water treated with reverse osmosis can achieve such purity and quality.




Logistics




                                              15
Logistical considerations include
power        consumption,       which
contributes to the cost of production.
The average power needed to treat
1 m3 of water using reverse osmosis
is about 1kWh. So for a large scale
water treatment plant (about 1million
m3 of water per day) the power
consumption is around 42MW/h.
The power consumption of such a
water treatment plant is very high,
the cost of which ultimately goes
into the elevated prices of the water.
Another logistical concern is the
land on which the water treatment
plant will be built on. The land prices
in CBD is extremely high, so it is not
cost effective to build a water
treatment facility which will occupy a                               Location of
                                                                     treatment plant
large land area. However, if the
water treatment plant is built far
away from the Marina Barrage,
there will be a problem installing the
pipes conducting water from the
lagoon. So the better choice is to
build it on newly reclaimed land near
Marina Bay, in this case, the land
prices is not as high as those in the
Raffles Place area and at the same
time the distance to the Marina
Barrage is short for easy transport of
water. A part of the visitor centre of Marina Barrage project can also include the water
treatment plant.
Sludge disposal is also a logistical concern. Since Singapore has developed a very thorough
system of waste incineration. The sludge can be filtered, thickened and finally dried into
sludge cakes and sent to incineration plants to be burnt and generate electricity. This not
only solves a lot of problems regarding the possible pollution of sludge from treatment plants,
but also provide an alternative fuel for power generation.




Question 3: [ANNEX]




                                              16
Above picture- A ceramic membrane filter pipe




Above picture- A commercial reverse osmosis system

References:

Dr J. Bartram, Mr P. Callan, (2006). Guidelines for Drinking-water Quality (First Addendum
To Third Edition ed., Vol. 1). 20 Avenue Appia, 1211 Geneva 27, Switzerland: World Health
Organization Press.

Acknowledgment:

www.mrwa.com
www.agapewater.com/ROSystems.htm
http://www.eca-water.gov.uk/page_seq.asp?contentType=criteria&pagecode=00030013
http://www.lenntech.com/ceramic-membranes-features.htm
Map taken from http://www.marina-bay.sg/images/map_01.jpg




                                                     17
Question 4: [ANSWER]
Part 1
Width of the dam
There is no specific width of the dam that has to be satisfied in order to have a single-lane road and a
pedestrian walkway on it. However, there is a minimum width that the dam must fulfill according to
industry regulations and we shall attempt to find out the value of that minimum width.
Width of a single-lane road
According to industry guidelines in the standard details of road elements posted on the Land Transport
Authority of Singapore website, the width of a standard Singapore single-lane road, taking into
account road shoulder, is 3.6m.
Width of a pedestrian walkway
According to industry regulations taken from the Road Data Hub Data Collection Specification - Part
2      posted     on      the     Land      Transport       Authority      of    Singapore       website
(http://www.lta.gov.sg/dbc/doc/rdh/Requirements/RDHDataCollectionSpecificationV1.4Part2.pdf),
the standard width of the pedestrian walkway would be about 1.5m which falls into Width category 2 (


                                                  18
1.2m < width < 2m). In this case, the width of the pedestrian walkway is assumed to have taken into
account the width needed for extra rail guards as a safety barrier, which is negligible.
Minimum width of dam
Therefore, the minimum width of the dam is:

Width of road + width of pedestrian walkway = 3.6m + 1.5m
                                                      = 5.1m
Therefore, the dam has to be at least 5.1m wide. This is just a lower limit estimate that shows the
practical minimum width of the dam required for a single lane road and pedestrian walkway.
Height of the dam
In order to determine the height of the dam, we have to find the distance between the seabed and the
sea level, taking into consideration sea level changes due to high and low tides and monsoon season,
and the extra height between water level and ground level for safety reasons such that the dam is safe
enough for human and vehicle traffic.
Assuming that the dam will not be used for berthing of boat and vessels, which means that the
difference in level between the dam and the deck of boat at high tide and low tide need not be taken
into consideration, the height of the dam, H will be:
From the side exposed to the sea,

H = depth of sea water at marina channel + highest high tide level + maximum wave height + safety
buffer

Sea level changes due to tidal changes is the main consideration here as we all know that the sea level
fluctuates 4 times over a day. There are 2 high tides and 2 low tides every day due to the differential
gravitational pull of the moon and the sun on the earth's seas. It is important that we ensure that we
build a dam high enough to keep the seawater out even during high tide. We certainly do not want the
dam to be flooded by seawater during high tide, thus the need to build the dam higher than the highest
sea level due to high tides.
Depth of sea water at marina channel = 8.1 to 9.6m.
(Source: http://www.ura.gov.sg/waterfront/q_a.htm) We assume that this depth has already taken into
consideration the changes in sea level due to tides. Thus, 8.1m corresponds to the depth at low tide
while 9.6m corresponds to the depth at high tide. Thus, there is no need to add in the highest high tide
level now.
Maximum wave height = 0.3m (we made the assumption that typical wave height approximately equal
to maximum wave height, source: http://www.windfinder.com/forecast/singapore_changi)
Safety buffer = 0.5m (own approximation)

Therefore, H = 9.6 m + 0.3m + 0.5m
            = 10.4m (1 d.p.)

(Please refer to fig. 4.1 for schematic drawing of height of dam)

From the reservoir side,
H = highest water level that reservoir is expected to store up to + maximum wave height + buffer

The main purpose of the marina barrage is to maintain a constant water level in the reservoir and
prevent tidal changes in the sea to affect water level. A brief explanation of how this works is as
follows:
     Current situation without the barrage is that low-lying areas in the city such as Boat Quay,
       Geylang, Chinatown and Jalan Besar would be flooded during high tide.


                                                   19
       When the dam is built, under normal conditions, the crest gates would remain closed to keep
        high tides out. As such, we can safely conclude that the normal water level in the reservoir
        would be lower than that of sea water level during high tide.
     At low tide, when there is heavy rain inland causing water level in the reservoir to rise rapidly,
        as the water level in the reservoir is higher than that of sea level outside, the dam will open its
        crest gates to release excess rainwater.
     At high tide, when there is heavy rainfall such as during monsoon season, the crest gates have
        to remain upright as opening the gates would let seawater in instead and increase the water
        level further. Pumps will be used to pump excess water out to the sea.
Therefore, in order to prevent flooding upstream during monsoon seasons or periods of high rainfall,
the height of the dam has to be low enough so that water level is kept at a desired low level but, at the
same time, the height of the dam has to be high enough to keep sea water out during high tide.
Therefore, the equation stated above can be simplified to the one below as ultimately, the dam has to
be high enough to keep out sea water during high tide.

H = highest seawater level + maximum wave height + buffer
  = depth at high tide + maximum wave height + buffer
  = 10.4m

Which is essentially the same as that derived from the seawater side.
Thus, the height of the dam has to be 10.4m.

Part 2
For the following section, we shall assume that the dam is of uniform rectangular cross section in
order to simplify calculations, g (acceleration due to gravity) is taken to be 9.8m s-2 .
Physical properties for the materials used to build the dam are as follows:

Erosion resistant
Fast moving water carrying rocks, pebbles and silt emptying into the reservoir has the potential to
erode the dam. Rate of erosion increases greatly when the water is moving faster and in much greater
volume. As shown in question 1 part 2, the fast flow rate of the water as well as the great volume of
water flowing into the reservoir from its various sources means that the dam would be subjected to a
lot of erosion by water and the load that it carries. As the dam will be constantly exposed to water, it
would be wise to use hard materials that are erosion resistant to prevent the erosion of material over
long periods of time.

Chemically inert
(We understand that this is a chemical property and not a physical property but we still believe that
this is an important property for the material used to build the dam to have.)
Seawater contains high amount of salts, with compounds such as chloride, sodium, sulphate, calcium,
magnesium, potassium and other minor salts present. Thus, it is important that the materials used to
build the dam are not chemically reactive so that they will not react with the chemicals found in the
sea so as to prevent weakening of the structure.

Low coefficient of thermal expansion
The material would be exposed to extreme temperature changes as it right under the sun with no
shelter. Thus, there is a need to have a low coefficient of thermal expansion so that the volume
changes due to cycles of expansion and contraction would be minimal so as to reduce internal stress
which prevents fatigue fractures from developing.



                                                    20
High compressive strength
The dam will be subjected to huge amount of compressive forces due to water pressure exerted on
both sides of the dam. Please see Fig 4.2 in annex for detailed illustration of the mechanism.

Calculation of compressive forces on the dam
First, let us assume that the body of water is still, thus compressive forces on the dam would be due to
water pressure or the weight of the water itself. As water pressure increases proportionately as depth
increases, we would expect the greatest compression force on the dam due to water pressure to be at
the point of greatest depth or the bottom of the reservoir/sea.

Formula to calculate water pressure is

P = ρgh
Where p = pressure, ρ = density of fluid, g = acceleration due to gravity, h = distance from surface of
water

At the point of greatest compressive stress for reservoir side,
p = 1000 kg m-3(density of freshwater) x 9.8 m s-2 x 9.6m
  = 94 000 Pa (2 s.f.)
At the point of greatest compressive stress for seawater side,
p = 1025 kg m-3 (average density of seawater) x 9.8 m s-2 x 9.6m
  = 96 000 Pa (2.s.f)

Total compressive stress due to pressure on both sides = 94 000 + 96 000 = 190 000 Pa

However, in reality, the body of water is not still. There is compressive stress due to water colliding
with dam walls when emptying into reservoir. From the discharge of the marina channel and the
velocity of the water emptying into the bay, the pressure due to the water impacting the wall can be
calculated to be 254.31 Pa. Note that we used the middle of the water body for the calculation as flow
rate decreases for water at the bottom and sides due to friction and viscosity. Please refer to Question 1
part 2 for detailed calculations.

Maximum possible compressive stress on dam = 2 x 254.31 Pa + 190 000Pa
                                          = 190 508.62
                                          =190 508 Pa (0 d.p.)

Therefore, the material must be able to withstand at least 190 508 Pa of compressive stress. Concrete
would be a good candidate as it is able to withstand up to 20.7 MPa.

High tensile strength
The dam must also be able to withstand quite a high amount of tensile stress due to the torque
experienced by the dam when the water level is different on both sides. Please see fig. 4.3.

Calculation of tensile stress experienced by dam:
Derivation of formula for calculating resultant moment about pivot of dam:
We consider a section of wall of length 1.0m to simplify calculations.
m  F (Y1  y)

Where m represents moment at a point, F represents force at that point, y represents distance from
surface of water and Y1 represents the height of the dam immersed in water.


                                                   21
m  gyy (Y1  y )                                          Neutral axis
     Y1

M   gy(Y1  y )dy
      0
     Y1                                                 y
 g  yY1  y 2 dy                                                       m          Y1
     0                                                       F
      y 2Y1 y 3 Y1
 g (      )0
        2     3
        3    3
      Y    Y
 g ( 1  1 ) Y1
                0
       2    3
                                                                                    Pivot
  gY13

    6

                                                 1025  9.8  9.6 3
Clockwise moment due to water pressure on the left =                =1 481 196 Nm
                                                         6
                                                               1000  9.8  8.13
Anticlockwise moment due to water pressure on the right =                        = 868 020 Nm
                                                                       6
Net moment = 1 481 196 – 868 020 = 613 176 Nm clockwise

Calculation of bending stress
For a symmetrical beam (the dam can be thought of as a beam) with uniform rectangular cross-
sectional area, the formula for determining the bending stress is: (source:
http://en.wikipedia.org/wiki/Bending)
             12M
        
              bh 3

Where: σ is the bending stress, M is the moment at the neutral axis, b is the width of the section being
analyzed , h is the depth of the section being analyzed. The external moment induces a countering
internal moment in the dam. The greatest internal moment is at the bottom of the dam which is equal
to the resultant external moment in this case. Thus, the bending stress

     12  613176
               = 55 469.7 = 55 000 N m-2 (2 s.f.)
      1.0  5.13



The dam must have a tensile strength capable of withstanding 55 000 N m-2 of tensile stress due to
bending. As illustrated above, it is important for the material used to build the dam to be able to
withstand both compressive stress and tensile stress. Concrete is good at withstanding compressive
stress but not when tensile stress is applied due to its brittle nature. It has low tensile strength of
around 2.2 MPa. Steel, on the other hand, can withstand tensile stress (ultimate tensile strength of 440
MPa) and steel rods can bend without breaking. Thus, a combination of steel rods and concrete in
reinforced concrete would be an ideal material to build the dam.

Ability to withstand high shear stress
The maximum shear stress at a point on a 1m by 1m section of the dam wall is the point where the
water pressure is the greatest. Therefore,



                                                  22
Maximum shear stress = ( gh + force due to flow of water) / Area
             = (1025 x 9.8 x 9.6)/ 1m2 + 254.31 Pa (taken from Question 1 part 2)
             = 96 686 Pa (0 d.p.)
The material must be able to withstand a shear stress of up to 96 686 Pa. Such a material must be rigid
so that it does not slide laterally when a shear stress is applied. Such materials must have a high shear
modulus which is defined as the shear stress over shear strain. Steel has a high shear modulus of 79.3
GPa which means it takes a lot of shear stress before it yields or break.
Question 4: [ANNEX]
                  Sea                           Reservoir           Fig 4.1: Schematic
                                           Road and                 drawing of the height of
                                           pedestrian               the dam.
                         0.5m buffer                  walkway
                                              Dam
                                                           Safety buffer

H                        Wave height 0.3m                          Wave height 0.3m


                                                       Water level in
                       Low tide 8.1m                   reservoir


    High tide                          Sea bed
    9.6m

                                                                     Dam
Water pressure exerted perpendicularly                                                               Fig 4.2: Diagram
on the dam walls results in huge                                                                     showing compressive
compressive forces on the dam.                                                                       forces on dam



    Seawater                                                            Freshwater




       Water pressure on the left side is much greater than water pressure on the right side. With
       the bottom of the dam as the pivot, a net clockwise moment is produced. The dam will be          Fig 4.3: Diagram showing
       forced to bend towards the right.                                                                torque on dam due to
                         Clockwise moment                                                               different in water level on
                                                                                                        both sides of dam.
                                                                     Freshwater level is
                                                         Dam         lower than seawater
            Seawater at                                              level.
9.6m        high tide
                                                                                                       8.1m
                     Tensile stress                                Compressive stress




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References:

List of materials properties. Retrieved April 7, 2007, from Wikipedia Web site:
http://en.wikipedia.org/wiki/List_of_Materials_Properties

Marina Barrage-creating a reservoir in the city. Retrieved April 7, 2007, from Wikipedia Web site:
http://www.pub.gov.sg/Marina/Index.htm



Acknowledgment:

Table       showing       the       relative      proportion            of        dissolved   salts
http://www.physicalgeography.net/fundamentals/8p.html)

Road Data Hub Data Collection Specification – Part 2
http://www.lta.gov.sg/dbc/doc/rdh/Requirements/RDHDataCollectionSpecificationV1.4Part2.pdf

Depth of Marina Bay: http://www.ura.gov.sg/waterfront/q_a.htm

Singapore Tide tables. http://app.nea.gov.sg/cms/htdocs/article.asp?pid=2293)

Wind and wave heights forecasts: http://www.windfinder.com/forecast/singapore_changi




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