Design of WTP

1. design flows residential area= population density= population (P) = water consumption (WC) = service area = water consumption in service area = Qav = P * WC + service= Qmm = 1.5 * Qav = Qmd = 1.8 * Qav = Qmh = 2.5 * Qav = Qav = 0.5 Qav (2nd stage) = Qmm = 1.5 * Qav = Qmd = 1.8 * Qav = Qmh = 2.5 * Qav = 2. design of intake Source canal wdith = HWL = LWL = Bedl L = at 2nd stage: Qmm = design flow rate (Qd) = 1.1 * Qmm = a. design of conduits: assume velocity (V) = cross sectional area of conduit (A) = Qd/V = assume No of conduit = diamter of conduit = take diamter = actual velocity = at 1st stage: Qd = No of conduits under opertaion = check velocity = b. design of screens: No of screens = width of screen (W) = 0.2m + diam + 0.2m = water depth = LWL - Bed L - 0.5m = width of screen (W) = n * S + (n+1) * b n = No of spacings S = width of spacing = 1.5 --> 5.0 cm = b = bar size or width = 1.5 --> 3.0 cm then get n = check head loss through screen hl = 1.4 *(V2 - v2 / 2g) V = velocity through screen = (Qd/No) / (d * n *S) = v = velocity approacg = (Qd/No) / (d * W ) = hl =

877.5 75 65812.5 300 202.5 50 10125 0.12 0.18 0.21 0.29 0.06 0.09 0.11 0.15

50 52 50 45

acre person/acre cap l/c/day acre m3/d/acre m3/d m3/s m3/s m3/s m3/s m3/s m3/s m3/s m3/s

m m m m

0.18 m3/s 15187.829 m3/d 0.19 m3/s 1.2 0.161 2 320 350 1.01

m/s m2 mm mm m/s

0.10 m3/s 1.00 1.0 m/s

2 0.85 m 1.5 m

2 cm 1.5 cm 24

0.12 m/s 0.07 m/s 0.0007 m

c. design of Sump of LLP: assume No of Pumps = assume spacing between pumps = total length of sump (L) = assume width of sump (B) = assume RT (in case of design flow) = assume RT (in case of min flow) = volume of sump = RT * Qd = or = RT * Qmin = take it = water depth of sump = V / (B * L) = min water depth = LWL - Bed L - hl = take it = d. design of main fold Qd = assume velocity (V) = get Area (A) = Qd/V = assume No of pipes = diamter = take it = actual velocity = calculation of frictional head loss (Hf) V = 0.355 * C * D0.63 * S0.54 assume C = get S = Hf = S * pipe length = S * 21000m =

4 1.5 6 2 2 5 21.1 28.1 28.1 2.34 5.00 5.0

0.19 1.5 0.13 1.00 405 450 1.2

m m m min min m3 m3 m3 m m m

m3/s m/s m2 mm mm m/s

120 0.0035 73.6 m

e. design of pupms: No of pumps = No of working pumps = flow rate per pump = calculation of power of pumps TDH = Hst + Hf + Hm Hst = LWL in sump - WL at WTP LWL in sump = assume WL at WTP = Hst = assume Hm = TDH = Hst + Hf + Hm =

50.0 85 35.0 1 109.6

g=

1000 kg / m3

h = efficiency = 0.8 * 0.8 =

0.64

Power per pump =

221 HP

Power per pump= g

* H * Qd / 75 * h

3. design of water treatment units 3.1 design of concentration tanks: Qd = alum dose = alum quantity = Qd * dose = assume alum solution concentration = volume of tanks = Qd * dose / g * conc * 1000 =

4 2 0.10 m3/s

159840.00 50 7992 10 79.9

m m m m m

m3/d ppm kg/d % m3

1.85

assume No of working tanks = assume No of stand-py tanks = assume water depth = get wedth = length = 3.2 design of Distrbution chamber: assume RT = Volume = Qd * RT = assume water depth = surface area = Vol / depth = diameter = 3.3 design of Clariflocculators: Qd = assume RT (for flocculation) = assume RT (for sedimentation) = total RT = total volume = Qd * RT =

3 0 2 3.65 m

60 111.0 3 37.00 6.9

6660.0 0.5 2.5 3 19980.0

sec m3 m m2 m

m2/hr hr hr hr m3

assume outer depth (dout) = get Surface area = Qd / d =

3.5 m 5708.6 m2

assume outer diameter Dout = get No of tanks = take it = volume of flocculation zone =

45 m 3.59 4 3330.0 m3

assume inner depth (din) = inner surface area = calculate inner diameter Din = check Surface loading rate = Qd/area = check weir loading rate = Qd/weir length =

3 1110.0 19 30.5 282.8

3.4 design of flash mixer: No of tanks = assume RT = volume of tank = assume water depth = get width = length =

3.5 design of Rapid sand filtersr: design flow = 1.07* Qmm = assume rate of filtration (ROF) = surface area = Qd / ROF = assume length of filter = assume filter width = get No of filters = take it =

4 30 13.88 2 2.63

3.000 120 2160.00 10.0 8.0 27.00 26

m m2 m m3/m2/d m3/m/d

sec m3 m m

m3/s m3/m2/d m2 m m

ok ok

take total No =

3.6 design of ground reservoirs: design flow = Qmm = C1 = Qmd - Qmm = C2 = 8 hr * Qmm = C3 = 0.5 hr * Qmm = take Cmax = C4 = 0.8 * fire demand = volume = Cmax + C4 = assume water depth = surface area = assume width = assume width = get No of tanks = take it =

30

0.18 3037.57 5062.61 316.4 5062.6 631.80 5694.4 4 1423.6024 14 28 3.63 4

m3/s m3 m3 m3 m3 m3 m3 m m2 m m

0.855

1.2

159840