Population Year1980199020002010

Population3000570064009600

YearPopulationAverage people growth rate per decade (I)% of Average people growth per decade (i)Average incremental growth rate per decade (m)Average decreases of increasing per decade (k)

19803000----

19905700270090--

2000640070012.28-20002000

201096003200502500+2500

Total 6600152.28500-500

Average 220050.76166.67-166.67

The WTP is designed for the usage for 20 years, thus, we need to estimate the population at the year of 2030. By using the method of Incremental increase:Pn = Pi + n ( I + m ) Pi = P2010= 9600n=(2030-2010)/10 =2P2030 = 9600+2(2200 + 166.67)=14334P2030 =14334 people

Water Demand

The daily water demand for the whole development is 15,566,182 L/day and is computed based on guideline provided by JBA Perlis:

Housing and Public DevelopmentPeople/Unit of DevelopmentWater Demand per person (L/capita/day)Total Water Demand (L/day)

1.Terrace house600515004,500,000.00

2.Semi-D 500520005,000,000.00

3.Bungalow200620002,400,000.00

4.Apartment400515003,000,000.00

5.Mosque1 unit60% per population508,600.40

6.Daily school1 unit20% per population50143,340.00

7.Boarding school 1 unit 40% per population2505,733.60

8.Kindergarten2 unit20% per population30172,008.00

Total15,229,682

Trade Development

Unit of developmentWater Demand per unit (L/day)Total Water Demand (L/day)

1.Shop lots (single-storey)1502000300,000

2.Petrol station25000010,000

3.Wet market12500025,000

4. Light industry workshop115001,500

Total336,500

Water Demand FluctuationsAnnual average demand = 5,681,656,430 L/ year

Maximum daily demand = 180% of annual average demand= 180/100 x 15,566,182 L/day= 28,019,127.60 L/day

Maximum weekly demand = 148% of annual average demand= 148/100 x 15,566,182 L/day= 23,037,949.36 L/day

Maximum monthly demand = 128% of annual average demand= 128/100 x 15,566,182 L/day= 19,924,712.96 L/day

Maximum hourly demand = 150% of maximum daily demand= 150/100 x 28,019,127.6 L/day= 42,028,691.40 L/day

Water intake ( about 5km from pauh ) The intake is located near Simpang Empat town just upstream of the second tidal gate at the Arau irrigation canal. It is understood that Arau canal would have a capacity to accommodate uo ti 540mgd of flow. The water mainly comes from the existing Muda. Pedu and Ahning reservoirs, a number of rivers joining upstream and rainfall. The sophisticated irrigation scheme under MADA would allow for efficient redirection and control of water flow to the canal to meet the requirement at all times of the season.

Structure---- tower..

Coagulation Design flow, Qpeak =28,019,127.60 L/day = 28,019.13 m3/day =19.45 m3/min Detention time, t = 25 min (range between 15 40 minutes) Velocity gradient, G = 950 s-1 Number of proposed unit = 2 Geometry information: Assume the basins with a ratio of length : width = 1 : 1 D/W = 1.5 L x W x D = W x W x 1.5W = 1.5 W3 Design calculationVolume of tank = (19.45/2) x 25= 244 m31.5 W3 = 244W = (244/1.5)1/3 = 5.5m L = 5.5 m D = 8 m

For each tank Actual volume = 5.5 x 5.5 x 8 = 242 m3 Actual detention time provided, tact = V/Q=242/9.7=24.9 min=25 min

Equipment design*Water temperature=5 OC , =1.518 x 10-3 N-s/m2Mixed power,P=G2V=(950s-1)2x(1.518 x 10-3 N-s/m2) x 242 m3=331,538.79 N-m/s (W)=331.54 kW

Water temperature,T=5CEfficiency of mixer=90%Efficiency Power = 331.54 kW/ 0.9= 368.40 kW Impeller size and rotational speed (rpm)Design considerationImpeller type:Discturbine 4 flat bladesw/d=0.16Np=2.6

diameter of the impeller is of diameter of basin

Check the Reynolds number

Therefore turbulent condition exist

F. Flocculation Type of flocculator; Mechanical or hydraulics. (Why??) Purposes; Design Parameter: Design flow, Qpeak =28,019,127.60 L/day = 28,019.13 m3/day =19.45 m3/min

Detention time, t = 45min (range between 20 to 60 minutes) Number of proposed unit; 2 Geometry information: Assume the basins with a ratio of length : width: depth = 1 : 1: 1 Volume of square = L x L x L = L3 Design calculations Volume of tank = (19.45/2) x 45= 436.95 mL3 = 436.95L = (436.95)1/3 = 7.58 m = 7.6 m W = 7.6 m D = 7.6 m

For each tank , Actual volume = 7.6 x 7.6 x 7.6 = 438 mDetention of time, tact = V/Q= 438 / 9.7= 45 min

For first tank, Water temperature, T = 5C Agitation requirements: 1st tank, G = 90 s-1, 2nd tank, G = 60 s-1

Mixed power,P=G2V=(90s-1)2x(1.518 x 10-3 N-s/m2) x 438 m3=5,385.56 N-m/s (W)=5.39 kWFor second tank, Mixed power,P=G2V=(60s-1)2x(1.518 x 10-3 N-s/m2) x 438 m3=2,393.58N-m/s (W)=2.39 kW

Sedimentation Design Parameters: Design flow, Qpeak = 28,019,127.60 L/day = 28,019.13 m3/day =19.45 m3/min Detention time, t = 4 hr Number of proposed unit; 4 Surface loading rate: 30 m3/m2.day Weir loading rate: 250 m3/m.day Geometry information: The length to width ratio is 4 : 1 The depth is in range of 3 to 5 meters.

Design calculations L = 4W Area= L x W = (28019.13/4)/304W x W = 233.494W2 = 233.49W = (233.49/4)1/2 =7.6 mL = 30.4 m (28019.13/4)/24 = (7.6 x 30.4 x D )/ 4D = 5m

Actual volume = 30.4 x 7.6 x 5 = 1155.2 m3 Actual detention time, tact = 1155.2/ (28019.13/4/24)= 3.9 hour= 4.0 hour Horizontal velocity = Q/ A= Q/ (W x D)= (28019.13/4/24)/ (7.6 x 5)= 7.68 m2/ hr Length of overflow weir, L = 0.2 Q/ (D x u)= (0.2 x 28019.13/4)/ (5 x 30)= 9.33 m= 9.4 m

Filtration Design parameter: Design flow, Qpeak = 28,019,127.60 L/day = 28,019.13m3/day =1,167.46m3/hr Filtration rate = 5 m3/m2.h Number of proposed unit = 2 Assume a one filter unit is out of service for backwash at the design capacity. Provide dimension of filter units; Assume: square size filter units

Design calculation:For two filter units functioned well,Area = Q/V = (1,167.46/2)/5= 116.75 m2 L2 = 116.75L = (116.75)1/2 = 10.8 m W = 10.8 mD = 10.8 mFor one filter unit is out of service for backwash,Area = Q/V = (1,167.46)/5= 233.50 m2 L2 = 233.50L = (233.50)1/2 = 15.28 m = 15.30mW = 15.30m D = 15.30m