Sheet1

HYDRAULIC DESIGN OF THE PROPOSED 2.6 MLD (16 HR) WATER TREATMENT PLANT AT PUNJAPURA, DISTRICT: DEWAS (MP)

Proposed Water Treatment Plant Units:

(1) Inlet Chamber(2) Cascade Aerator(3) Inlet Channel(4) Flow Measuring Flume(5) Flash Mixer(6) Clariflocculator(7) Coagulant Solution Tank, Feeding Equipment etc.(8) Rapid Sand Gravity Filters(9) Wash Water Tank(10) Chlorination Arrangement(11) Clear Water Storage Tank / Sump(12) Sludge Collection Tank(13) Sludge Drying Beds

Design Flow:

Design flow = 2.6x1.05x24/16= 4.1mld = 4100cum/d = 170.83 cum/hr = 2.847cum/minute = 0.0475cum/s (Considering 5% wastage in back washing and desludging and 16 hrs. pumping, Ref. : NIT condition & CPHEEO Manual Type Design)

(1) INLET CHAMBER

(a)Diameter of Chamber:1.00 m(as per prevailing practice)

(b )R. L. of Ground Level:175.00 m (As specified on site plan)

(c)R. L. of Top of the Chamber:180.00 m

(d)R. L. of Bottom of the Chamber:It shall be 0.2 m lower than the invert R. L. of inlet pipe (Tentatively 174.00 m)

(2) CASCADE AERATIOR

The Surface Area Required @ 0.030 m2/m3/hr = 170.83x0.03 = 5.12 m2(As per CPHEEO manual, 7.2.3.3, page 192)

Diameter of the Bottom-most tray of the cascade is given by

(p4)(D2-1.302) = 5.12, On Solving D = 2.86 Say 3.00 m

(This takes into account the outside diameter of the inlet chamber also. The outside diameter of the inlet chamber is assumed as 1.30 m.)

Number of trays = 4 (As per CPHEEO manual, 7.2.3.3, page 192)

Rise of each step = 300 mm

Hence, providing 4 cascade trays with dimensions and levels as below:

DiameterR.L.

Lip of aeration fountain1.30 m180.00 m

1st Cascade tray1.72 m179.75 m

2nd Cascade tray2.15 m179.50 m

3rd Cascade tray2.57 m179.25 m

4th Cascade tray3.00 m179.00 m

Collection channel (top water level)4.20 m178.55 m

(3) FLOW MEASURING FLUM AND INLET CHANNEL (a)Flow measuring capacity:4.10x1.5 = 6.15 mld = 0.0712 m3/s(Considering 50 % overloading as per NIT condition 6.2.1, page 73)

(b)Length of channel:5.0 m(as per prevailing practice) (c )Width of the channel:0.60 m

(d)Liquid depth at d/s of the weir:0.10 m

(e)Velocity of flow @ ultimate flow:(0.0712)(0.60x0.10)= 1.19 m/s (which is less than maximum permissible velocity of 2.0 m/s , hence O. K.)

(f)Free board:0.3m

(g)Width of the Rectangular Weir (Notch) Plate:0.60 m

(h)Maximum head over the notch=(0.0712/(0.67x0.68((2x9.81)0.5)0.604))0.67 = 0.12 m

(i)Free fall:0.10 m

(j)Total Depth of the channel:0.10+0.12+0.10+0.30 = 0.62 mSay 0.65 m

(k)R. L. of the channel top:179.10 m

(l)R. L. of the channel bottom :179.10-0.65 = 178.45 m

(m) R.L. of the notch lip:178.45+0.10+0.10 = 178.65 m

(4) FLASH MIXER

(a)Detention time:30 to 60 seconds(as per NIT condition 6.3.1, page 73)

(b)Diameter:1.20 m

(c )Minimum depth to diameter ratio:1:01(as per CPHEEO manual, 7.4.2.2, page 206)

(d) Liquid depth:1.50 m

(e)Free board:0.60 m

(f)Total depth:2.10 m

(g) Detention time actually provided:(px(1.2)2x1.50)/(4x0.0474)= 36 seconds (more than 30 seconds, hence OK)

Hence provide flash mixer of 1.2 m diameter and 1.5 m liquid depth.

(h)R. L. of full water level in flash mixer:178.30 m

(i)R. L. of flash mixer top:178.90 m

(j)R. L. of flash mixer bottom:178.30-1.50 = 176.80 m

(k)Diameter of outlet pipe to clariflocculator:Flow = 0.0474 cum/s Velocity of flow = 1.2 m/s(0.9 m/s to 1.8 m/s is permissible)\Diameter = ((4p)x(0.04741.2))0.5 = 0.22 m Say 250 mm

(5) CLARIFLOCCULATOR

A circular radial flow clariflocculator with the central flocculator and the outer annular clarifier shall be provided.

(a)Central Inlet Shaft:

(i) Diameter:0.60 m

(ii) Discharge:0.0474 cum/s

(iii) Flow velocity:0.0474((p4)x(0.6)2) = 0.17 m/swhich is less than the max. permissible velocity of 0.6 m/s, hence O.K.

(iv) Opening ports in the inlet:Velocity of flow through the ports = 0.2 m/s shaftCross sectional area of ports = 0.04740.2 = 0.237 sqm Provide 4 Nos. ports each of 225 mm width and 300 mm height (b)Central Flocculator:

(i) Detention time:30 minutes(as per NIT condition 6.4.3, page 74)

(ii) Liquid depth:3.50 m(as per CPHEEO Manual, 7.5.5, page 226)

(iii) Volume of flocculator :2.847x30 = 85.41 cum

(iv) Surface area of flocculator:85.413.50 = 24.40 sqm

(v) Diameter of flocculator:(4x24.4p)0.5 = 5.57 say 5.60 m

(vi) Cross sectional area of :5.60x3.50 = 19.60 sqm water in the flocculator

(vii) Providing 15 % of this area for paddles (as per CPHEEO Manual, 7.4.3.2(2), page 181), then

(viii) Area of paddles:19.60x15100 = 2.94 sqm Providing two paddle arms with six paddles each, then

(ix) Area of each paddle:2.94(2x6) = 0.245 sqm

(x) Height of each paddle:3.0 m

(xi) Width of each paddle:0.2453.0 = 0.082 m say 85 mm

(c)Clarifier:

(i) Detention time:2.5 hours(as per NIT condition 6.4.4, page 74)

(ii) Liquid depth:3.50 m(as per CPHEEO Manual 7.5.5, page 227)

(iii) Volume of clarifier:170.83x2.5 = 427.07 cum

(iv) Surface area of clarifier:427.073.50 = 122.02 sqm

(v) Diameter D of :((p4)x(D)2-((p4)x(5.60)2) = 122.02 clariflocculator\D = 13.66 Say 13.70 m Providing 1 in 12 bottom slope (as per CPHEEO manual, 7.5.9 page 230) and 0.3 m free board, then (vi) Total depth of clariflocculator at side:3.50+0.30 = 3.80 m

(vii) Total depth of:3.80+0.50 = 4.30 m clariflocculator at centre

(viii) Surface loading:4100122.02 = 33.6 cum/sqm/day

which is within the typical range of 30 to 40 cum/sqm/day (as per NIT condition 6.4.4, page 74)

(ix) Outlet weir (launder):4100(px13.7) = 95 cum/m/day loading

which is less than the normally permissible value of 300 cum/m/day (as per CPHEEO Manual, 7.5.8, page 230). Note: Orifices of 50 mm dia. @ 150 mm c/c shall be provided all along the periphery of the outlet weir.(x) Velocity through the:((0.0474)((p x5.60x0.4)-(8x0.2x0.4)) bottom opening of = 0.008 m/s flocculator partition wall

which is less than the permissible velocity of 0.3 m/s, hence OK

(d)Peripheral Collecting Channel:

Let velocity of flow in the channel be 0.6 m/s (as per prevailing practice). Half portion of the channel for collecting settled water shall collect half the quantity. Therefore,

(i) Flow in half portion of :0.0474/2 = 0.0237 cum/s channel

(ii) Area (wetted cross:0.02370.6 = 0.0395 sqm sectional area) of channel

(iii) Width of collecting :0.45 m (assumed) channel

(iv) Depth of water in the:0.03950.45 = 0.09 m channel at d/s end

(v) Clear fall in the channel:0.15 m(as per prevailing practice)

(vi) Total depth of channel:0.09+0.15 = 0.24 m Say 0.30 m at downstream end

(vii) Slope of the channel bed:S = (0.013xV)2 (R)1.33

where, S = slope of the channel bed V = velocity of flow in channel R = hydraulic radius = (0.45x0.09)(0.45+(2x0.09)) = 0.064 m

\ S = (0.013x0.6)2(0.064)1.33 = 0.00234 Say 1 in 400

(e) Mechanical Scraper:

A mechanical sludge scraper driven by electric motor and suspended from 1.2 m wide foot bridge shall be provided. The tip velocity of the scraper shall be 0.3 m/min. (as per CPHEEO Manual, 7.5.9, page 230).

(f) Sludge Drainage: Sludge shall be finally collected in the centrally located pit of 0.3 m depth and 2.0 m diameter (as per prevailing practice). It shall be carried by 150 mm dia.cast iron pipe under hydrostatic pressure to sludge collection tank. (as per CPHEEO manual, 7.5.9, page 230).

(g) Reduced Levels:

(i) R.L. of top of the :178.30 m clariflocculator walls

(ii) R.L. of full water level in:178.30-0.30 = 178.00 m clariflocculator

(iii) R.L. of clariflocculator:178.00-3.50 = 174.50 m bottom at sides

(iv) R.L. of clariflocculator:174.50-0.50 = 174.00 m bottom at the centre

(v) R.L. of bottom of sludge:174.00-0.30 = 173.70 m collection pit

(vi) R. L. of top of the opening :177.70 m in the central inlet shaft

(vii) R.L. of bottom of central:173.70-0.45 = 173.25 m inlet shaft

(viii) R.L. of invert of sludge:173.70-0.15 = 173.55 m drainage pipe

(ix) R.L. of water level in:178.00-0.15 = 177.85 m collecting channel

(x) R.L. of collecting channel:178.0-0.30 = 177.70 m bed at D/S end(6) ALUM SOLUTION PREPARATION TANK

(a)Maximum alum dose:100 mg/l(as per normal practice)

(b)Solution strength:10%(as per NIT condition 6.14.3, page 82)

(c)Capacity of tanks:8 hours(as per NIT condition 6.14.3, page 82)

(d)Nos. of tanks:Two(as per NIT condition 6.14.3, page 82)

(e) Liquid volume of each tank:(100x4.10x8/24)x(100/(10x1000))= 1.37 cum

(f)Liquid depth of each tank:0.90 m

(g)Providing square tank, then:(1.37/0.90)0.5 = 1.23 Say 1.25 m length and width of tank(h)Free board:0.3 m(as per NIT condition 6.14.3, page 82)

(i)Total depth of each tank:0.90+0.30 = 1.20 m

(7) RAPID SAND FILTER

(a)Filter Beds:

(i) Filtration rate:80 lpm/sqm of filter area(as per NIT condition 6.5.1, page 77)

(ii) Number of filter beds:2 nos.

(iii) Total area of filters:(4100x1000)(80x60x24) = 35.59 sqm

(iv) Area of each filter:35.592 = 17.80 sqm

Taking length to width ratio of filter bed as 1.25 (as per CPHEEO Manual 7.6.3.5, page 211), then

(v) Width of each filter bed:(17.81.25)0.5 = 3.77 Say 3.80 m

(vi) Length of each filter bed:17.8/3.80 = 4.68 Say 4.70 m

(vii) Depth of sand bed0.75 m(as per CPHEEO Manual 7.6.3.7, page 245)

(viii) Depth of gravel layer:0.5 m (as per CPHEEO Manual 7.6.3.10, page 249)

(ix) Depth of water over sand:1.3 m(as per CPHEEO Manual 7.6.3.7, page 245)

(x) Free board:0.5 m(as per CPHEEO Manual 7.6.3.7, page 245)

(xi) Total depth of each filter:0.75+0.5+1.3+0.5 = 3.05 m

(b) Under Drain System:The under drain system shall comprise of perforated pipe laterals and the centrally located manifold channel.

(i) Area of orifices in each:0.3 % of filter bed area filter(as per CPHEEO Manual 7.6.3.9,page 247)= (0.3100)x3.80x4.70 = 0.054 sqm

(ii) Area of laterals:2 times area of orifices(as per CPHEEO Manual 7.6.3.9,page 247 for 12 mm dia. Orifices)= 2x0.054 = 0.11 sqm(iii) Area of manifold channel: 2 times area of laterals(as per CPHEEO Manual 7.6.3.9, page 247) = 2x0.11 = 0.22 sqm

(iv) Size of manifold channel:0.60 m width x 0.40 m depth

(v) Design of laterals :Adopting lateral of 63 mm OD, HDPEArea of cross section of one lateral= (p4)x(0.05)2 = 0.002 sqm\ number of laterals = 0.110.002 = 55Number of laterals on each side of manifold = 552 = 28\ spacing of laterals= 470028 = 168 mm c/cRatio of length to diameter of lateral= 3700(2x50) = 37Which is less than the maximum permissible value of 60 (as per CPHEEO Manual 7.6.3.9, page 247), hence O.K. (vi) Design of orifices:Adopting 12 mm dia. OrificesArea of one orifice= (p4)x(0.012)2 = 0.000113 sqmTotal number of orifices for each filter bed = 0.0540.000113 = 478 nos.Number of orifices on each lateral= 47856 = 9 say 10 nos.\Spacing of orifices= 190010 = 190 mm c/cwhich is less than maximum permissible value of 200 mm for 12 mm dia. orifices(as per CPHEEO Manual 7.6.3.9, page 247) , hence O.K.

(c) Filter Back-wash System:The filter back-wash system shall comprise of air-water wash system.

(i) Duration of air wash:5 minutes(as per NIT condition 6.5.9, page 78)

(ii) Air flow rate:45 m/hr @ 0.35 kg/sqcm pressure(as per NIT condition 6.5.9, page 78)

(iii) Total quantity of air:45x3.80x4.70 required for one bed/min.= 804 cum/hr @ 0.35 kg/sqcm

Hence, provide an air blower of 804 cum/hr capacity @ 0.35 kg/sqcm pressure.

(iv) Wash water supply rate:30 m/hr (as per NIT condition 6.5.9, page 78)

(v) Maximum Duration of water wash:10 minutes(as per NIT condition 6.5.9, page 78)

(vi) Capacity of wash water:30x10x3.80x4.70/60 = 89.3 m3 tankSay 90000 litre(as per NIT condition 6.6, page 78)

The bottom of wash water tank shall be minimum 6.0 m above the lip of wash water troughs.

(vii) Design of wash water:Adopting three number of troughs for each filter running parallel to the width of the filter bed. troughs Quantity of water required per bed per second = (30x3.80x4.70)(60x60)= 0.15 cum/sWash water flow per trough= 0.153 = 0.05 cum/sBottom slope of trough = 1 in 50 = 0.02Width of each trough = 250 mmIf y (in metres) is depth of water in trough, then(B)1.67((B+2y)0.67) = (0.013xq)((y)1.67x(s)0.5)where, B = Width of trough in metres = 0.25 m q = Discharge through trough in cum/s = 0.05 cum/s\(0.25)1.67((0.25+2y)0.67) = (0.013x0.05)((y)1.67x(0.02)0.5)On solving, y = 0.13 m = 130 mmFree Board = 100 mmHence, provide total depth of 230 mm at upstream end and that of 300 mm at downstream end.Height of the lip of the trough above the unexpanded sand bed= (Depth of sand bed x 0.5) + Depth of trough + Thickness of trough= (0.75x0.5) + 0.30 + 0.08= 0.75 m

(viii) A 600 mm wide gullet @ 1 in 100 bottom slope shall be provided along length of the bed.

(d)Conduit Diameters: (i) Settled water inlet pipe:Flow = 0.0474/2 = 0.024 m3/sVelocity of flow = 0.90 m/s\Diameter = ((4p)x(0.0240.90))0.5 = 0.18 m Say 200 mm

(ii) Filtered water outlet pipe:same as settled water inlet = 200 mm

(iii) Outlet pipe from wash:Discharge = (30x3.80x4.70)(60x60) water tank = 0.15 cum/s

Flow velocity = 3.6 m/s(as per prevailing practice)

\Diameter = ((4p)x(0.153.6))0.5 = 0.23 m Say 250 mm

(iv) Spent wash water outlet :Discharge = 0.15 cum/s (drain) pipeFlow velocity = 2.4 m/s(as per prevailing practice)

Diameter = ((4p)x(0.152.4))0.5= 0.28 m Say 400 mm

This spent wash water outlet (drain) pipe shall be of RCC NP2 class pipe.

(e)Head Loss Calculations for Checking Wash Water Tank Height

(i) Height of wash water tank bottom above:6.00 m (minimum) the wash water trough lip

(ii) Height of wash water trough lip above the:0.5+0.75+0.75 filter under drainage= 2.0 m

(iii) Height of wash water tank bottom above:6.00+2.0 the filter under drainage = 8.0 m

(iv) Equivalent length of wash water inlet pipe:40 m (assumed)

(v) Diameter of wash water inlet pipe:250 mm f

(vi) Flow through wash water inlet pipe:0.15 cum/s

(vii) Frictional head loss in the wash water inlet:0.04 m/m length pipe per metre length(from CPHEEO Manualappendix 6.3, page 601)

(viii) Total frictional head loss in the wash:40x0.04 = 1.60 m water inlet pipe

(ix) Net head available at filter under drainage:8.00-1.60 = 6.40 m during back washing

which is more than the minimum permissible head of 5.0 m. (as per CPHEEO Manual 7.6.3.12, page 250)

(e)Reduced Levels:

(i) R.L. of full water level in filter:177.40 m bed

(ii) R.L. of top of filter bed:177.40+0.5 = 177.90 m

(iii) R.L. of top of unexpanded:177.40-1.3 = 176.10 m sand bed

(iv) R.L. of bottom of sand bed:176.10-0.75 = 175.35 m

(v) R.L. of bottom of filter bed:175.35-0.5 = 174.85 m

(vi) R.L. of top of wash water:176.10+0.75 = 176.85 m troughs and gullet

(vii) R.L. of bottom of wash water tank :174.85+8.00 = 182.85 m (minimum)(8) CHLORINATION ARRANGEMENT

(a)Chlorine dose (maximum):5 mg/l(as per NIT condition 6.15, page 82)

(b)Amount of chlorine required per day:5x4.1 = 20.5 kg/day (c)Capacity of chlorinator including 100 % stand-by:(20.5x2)24 = 1.7 kg/hr hence, provide chlorinator of 1.7 kg/hr capacity.

(9) UNDERGROUND CLEAR WATER SUMP

(a)Capacity:435 kl(as per NIT condition 1.31, page 71)

(b)R.L. of full water level :175.50 m

(10) SLUDGE COLLECTION TANK

(a)Capacity @ 2 % sludge volume & 1 day detention time:4100x(2/100) = 82 kl(as per prevailing practice)

(b)R.L. of full water level :175.00 m

(11) SLUDGE DRYING BEDS

(a)Total surface area @ 2 % sludge volume, 5 % suspended solids, 0.3 m depth of sludge & 7 days cycle:(4100x(2/100)x(5/100)x7)/0.3(as per prevailing practice) = 96 m2

(b)Number of sludge drying beds:Four

(c) Surface area of each sludge drying bed=96/4 = 24 m2Hence, four nos. sludge drying beds, each of 6.0 m x 4.0 m shall be provided.

(d)R.L. of full sludge level :176.00 m

Sheet2

Sheet3