Design of WTP

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 = = (Considering (1)

2.6 x 1.05 x 24 / 4100 cum/d = 170.83 cum/hr = 2.847 cum/minute = 5 % wastage in back washing and desludging and 16 hrs. pumping, Re

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 inv

(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 (4)(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 a 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: Diameter R.L. Lip of aeration fountain

1.30 m

180.00 m

1st Cascade 1.72 m tray

179.75 m

2nd Cascade 2.15 m tray

179.50 m

3rd Cascade 2.57 m tray

179.25 m

4th Cascade 3.00 m tray

179.00 m

Collection channel (top 4.20 m water level)

(3)

178.55 m

FLOW MEASURING FLUM AND INLET CHANNEL

4.10x1.5 = 6.15 mld = 0.0712 m3/s

(a)

Flow measurin : g capacity

(Consider ing 50 % overloadi ng as per NIT condition 6.2.1, page 73)

5.0 m

(b)

Length of : channel

(as per prevailing practice)

(c )

Width of the : channel

0.60 m

(c )

Width of the : channel

0.60 m

(d)

Liquid depth at : d/s of the weir

0.10 m

(0.0712) (0.60x0.1 0) = 1.19 m/s

(e)

Velocity of flow @ : ultimate flow

(f)

Free board

(g)

Width of the Rectangul : ar Weir (Notch) Plate

:

(which is less than maximum permissib le velocity of 2.0 m/s , hence O. K.)

0.3 m

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)

(j)

Free fall :

Total Depth of : the channel

(k)

R. L. of the channel top

(l)

R. L. of the channel bottom

(m)

R.L. of the notch : lip

(4)

FLASH MIXER

0.10 m 0.10+0.12 +0.10+0.3 0 = 0.62 m Say 0.65 m

:

179.10 m

:

179.100.65 = 178.45 m

178.45+0. 10+0.10 = 178.65 m

30 to 60 seconds (a)

Detention time

:

(as per NIT condition 6.3.1, page 73)

(b)

Diameter

:

1.20 m

(c )

Minimum depth to diameter ratio

:

(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

1:01

(x(2)2x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

Flow = 0.0474 cum/s Velocity of flow = 1.2 m/s

(k)

Diameter of outlet pipe to : clarifloccul ator

(0.9 m/s to 1.8 m/s is permissible) Diameter = ((4)x(0.04741.2))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 0.0474(( 4)x(0.6)2) = 0.17 m/s

(iii) Flow velocity

:

(iii) Flow velocity

:

(iv) Ope nin g port s in the inle t

which is less than the max. permissible velocity of 0.6 m/s, hence O.K.

Velocity of flow through the ports = 0.2 m/s Cross sectional area of ports = 0.04740.2

shaft

: = 0.237 sqm Provide 4 Nos. ports each of 225 mm width and 300 mm height

(b)

Central Flocculato r: 30 minutes (i) Detention time

:

(i) Detention time

:

(as per NIT condition 6.4.3, page 74) 3.50 m

(ii) Liquid : depth

(as per CPHEEO Manual, 7.5.5, page 226)

(iii) Volume of : flocculator

2.847x30 = 85.41 cum

(iv) Surface : area of flocculator

85.413.50 = 24.40 sqm

(v) Diameter of : flocculator

(4x24.4)0 .5 = 5.57 say 5.60 m

(vi) Cross sectional area of

water in the flocculator

:

5.60x3.50 = 19.60 sqm

(vii) Providing 15 % of this area for paddles

(as per CPHEEO Manual, 7.4.3.2(2), page 181), then (viii) Area : of paddles

19.60x15100 = 2.94 sqm

Providing two paddle arms with six paddles each, then

(c)

(ix) Area of each : paddle

2.94(2x6) = 0.245 sqm

(x) Height of : each paddle

3.0 m

(xi) Width of each : paddle

0.2453.0 = 0.082 m say 85 mm

Clarifier: 2.5 hours (i) Detention time

:

(as per NIT condition 6.4.4, page 74) 3.50 m

(ii) Liquid depth

:

(as per CPHEEO Manual 7.5.5, page 227)

(iii) Volume of : clarifier

170.83x2.5 = 427.07 cum

(iv) Surface area of clarifier

427.073.50 = 122.02 sqm

:

Dia met er ‘D’ :

((4)x(D)2 ((4)x(5.60)2) = 122.02 D = 13.66 Say 13.70 m

clarifloccul ator

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

clarifloccul : ator at side

3.50+0.30 = 3.80 m

(vii) Tota l dept h of :

3.80+0.50 = 4.30 m

clarifloccul ator at centre

(viii) Surface loading

:

4100122.02 = 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)

Out let wei r (lau : nde

4100(x13.7) = 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. Vel ocit y thro ugh bottom opening of :

((0.0474)(( x5.60x0.4)(8x0.2x0.4))

= 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, Flo w in half port : ion

0.0474/2 = 0.0237 cum/s

channel

Are a (we tted cros

:

0.02370.6 = 0.0395 sqm

sectional area) of channel

Wid th of coll : ecti

0.45 m (assumed)

channel Dep th of wat er in : channel at d/s end

0.03950.45 = 0.09 m

(v) Clear fall in the : channel

(vi) Total depth of channel

0.15 m (as per prevailing practice) 0.09+0.15 = 0.24 m

:

Say 0.30 m at downstream end

S= (0.013xV)2 (R)1.33

where, S = slope of the channel bed

V= velocity of flow in channel R= hydraulic radius (vii) Slope of the : channel bed

= (0.45x0.09) (0.45+(2x 0.09)) = 0.064 m

 S= (0.0 13x 0.6) 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 (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: R.L . of top of clarifloccul ator walls

:

178.30 m

(ii) R.L . of full wat er : leve l in clarifloccul ator

178.30-0.30 = 178.00 m

(iii) R.L . of clar iflo ccul : ator bottom at sides

178.00-3.50 = 174.50 m

(iv) R.L . of clar iflo ccul : ator

174.50-0.50 = 174.00 m

bottom at the centre (v) R.L . of bott om of : slud ge collection pit

174.00-0.30 = 173.70 m

R. L. of top of the ope : nin

177.70 m

in the central inlet shaft (vii) R.L . of bott om of : cent ral inlet

173.70 0.45 = 173.25 m

shaft (viii ) R.L . of inve rt of slud : ge

173.70 0.15 = 173.55 m

drainage pipe (ix) R.L . of wat er leve : l in collecting channel

178.00 0.15 = 177.85 m

(x) R.L . of coll ecti ng : cha nnel

178.00.30 = 177.70 m

bed at D/S end (6) ALUM SOLUTION PREPARATION TANK 100 mg/l (a)

Maximum : alum dose

(as per normal practice) 10%

(b)

Solution strength

:

(as per NIT condition 6.14.3, page 82) 8 hours

(c)

Capacity of : tanks

(as per NIT condition 6.14.3, page 82) Two

(d)

(e)

Nos. of tanks

:

Liquid volume of : each tank

(as per NIT condition 6.14.3, page 82)

(100x4.10x 8/24)x(100/ (10x1000))

(e)

(f)

Liquid volume of : each tank

Liquid depth of each tank

:

Providing square tank, then (g)

: length and width of tank

= 1.37 cum

0.90 m

(1.37/0.90)0 .5 = 1.23 Say 1.25 m

0.3 m (h)

Free board :

(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: 80 lpm/sqm of filter area (i) Filtration rate

:

(ii) Number of : filter beds

(as per NIT condition 6.5.1, page 77)

2 nos.

(iii) Total area of : filters (iv) Area of each : filter

(4100x1000)(80x60x24) = 35.59 sqm

35.592 = 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

(17.81.25) 0.5 = 3.77 Say 3.80 m

(vi) Length of : each filter bed

17.8/3.80 = 4.68 Say 4.70 m 0.75 m

(vii) Depth of sand bed

(as per CPHEEO Manual 7.6.3.7, page 245)

0.5 m

(viii) Depth of : gravel layer

(ix) Depth of : water over sand

(as per CPHEEO Manual 7.6.3.10, page 249)

1.3 m

(as per CPHEEO Manual 7.6.3.7, page 245)

0.5 m (x) Free : board

(b)

(x) Free : board

(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

Under Drain System: The under drain system shall comprise of perforated pipe laterals and the centrally located manifol (i) Are a of orifi ces in eac h

0.3 % of filte r bed area

filter :

(as per CPHEEO Manual 7.6.3.9, page 247) = (0.3100)x 3.80x4.70 = 0.054 sqm

2 times area of orifices

(ii) Area of : laterals

(as per CPHEEO Manual 7.6.3.9,

(ii) Area of : laterals

page 247 for 12 mm dia. Orifices) = 2x0.054 = 0.11 sqm 2 times area of laterals

(iii) Area of manifold : channel

(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

Adopting lateral of 63 mm OD, HDPE Area of cross section of one lateral = (4)x(0.0 5)2 = 0.002 sqm

 number of laterals

= 0.110.002 = 55

Number of laterals on each side of manifold = 552 = 28

(v) Design : of laterals

 spacing of laterals = 470028 = 168 mm c/c Ratio of length to diameter of lateral = 3700(2x5 0) = 37

Which is less than the maximum permissible value of 60 (as per CPHEEO Manual 7.6.3.9, page 247), hence O.K.

Adopting 12 mm dia. Orifices Area of one orifice = (4)x(0.0 12)2 = 0.000113 sqm

Total number of orifices for each filter bed = 0.0540.00 0113 = 478 nos.

Number of orifices on each lateral (vi) Design of orifices

:

= 47856 = 9 say 10 nos. Spacing of orifices = 190010 = 190 mm c/c

which 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. 5 minutes (i) Duration of : air wash

(as per NIT condition 6.5.9, page 78)

45 m/hr @ 0.35 kg/sqcm pressure (ii) Air flow rate

:

(as per NIT condition 6.5.9, page 78)

Tota l qua ntit y of :

45x3.80x4. 70

= 804 cum/hr @ 0.35 kg/sqcm

required for one bed/min.

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

30 m/hr (iv) Wash water : supply rate (v) Ma xim um Dur atio : n of wat er was hCap acit y of was h wat tank

:

(as per NIT condition 6.5.9, page 78)

10 minutes

(as per NIT condition 6.5.9, page 78)

30x10x3.80x4.70/60 = 89.3 m3

Say 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) Des ign of was h wat er

troughs

Adopting three number of troughs for each filter running parallel to the width of the filter bed.

Quantity of water required per bed per second = (30x3.80x4. 70)(60x60 ) = 0.15 cum/s Wash water flow per trough

= 0.153 = 0.05 cum/s Bottom slope of trough = 1 in 50 = 0.02 Width of each trough = 250 mm

If 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 mm Free Board = 100 mm

Hence, provide total depth of 230 mm at upstream end and that of 300 mm at downstrea m 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/s Velocity of flow = 0.90 m/s Diameter = ((4)x(0.0 240.90))0.5 = 0.18 m Say 200 mm

(ii) Filtered : water outlet pipe

same as settled water inlet = 200 mm

Out let pipe fro m was tank

Discharge = (30x3.80x4. 70)(60x60 )

water

= 0.15 cum/s

Flow velocity = 3.6 m/s :

(as per prevailing practice) Diameter = ((4)x(0.1 53.6))0.5 = 0.23 m Say 250 mm

Spe nt was h wat er outl (drain)

Discharge = 0.15 cum/s

pipe Flow velocity = 2.4 m/s (as per prevailing practice)

:

Diameter :

= ((4)x(0.1 52.4))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 Calculatio ns for Checking Wash Water Tank Height

Hei ght of was h wat er tank bott : om abo

6.00 m (minimum)

:

6.00 m (minimum)

the wash water trough lip Hei ght of was h wat er trou gh : lip abo ve filter under drainage Hei ght of was h wat er tank bott : om abo

0.5+0.75+0 .75

= 2.0 m

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 

(vi) Flow through : wash water inlet pipe Fric tion al hea d loss in the was h wat : er pipe per metre length

0.15 cum/s

0.04 m/m length

(from CPHEEO Manual appendix 6.3, page 601)

Tota l frict iona l hea d loss : in the

40x0.04 = 1.60 m

water inlet pipe Net hea d avai labl e at filte r und : er drai nag

8.001.60 = 6.40 m

:

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 wate r : level in

177.40 m

bed (ii) R.L. of top of filter : bed R.L. of top of unex : pand sand bed

177.40+0.5 = 177.90 m

177.401.3 = 176.10 m

(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 : troughs and gullet

176.10+0.75 = 176.85 m

R.L. of botto m of wash : wate r tank

(8)

174.85+8.00 = 182.85 m (minimum)

CHLORINATION ARRANGEMENT 5 mg/l

(a)

Chlorine dose : (maximu m)

(as per NIT condition 6.15, page 82)

5x4.1

(b)

Amount of chlorine : required per day

= 20.5 kg/day

(20.5x2) 24 = 1.7 kg/hr

(c)

Capacity of chlorinato r : including 100 % stand-by

hence, provide chlorinato r of

1.7 kg/hr capacity.

(9) UNDERGROUND CLEAR WATER SUMP 435 kl (a)

Capacity :

(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(2/1 00) = 82 kl

(as per prevailing practice)

(b)

R.L. of full water : level

175.00 m

(11) SLUDGE DRYING BEDS

Total surface area @2% sludge volume, 5 % suspended solids, 0.3 : m depth of sludge & 7 days cycle

(4100x(2/1 00)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 m2

(a)

Hence, 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

LANT AT

16

= 4.1 mld um/minute = 0.0475 cum/s nd 16 hrs. pumping, Ref. : NIT condition & CPHEEO Manual Type Design)

as per prevailing practice)

As specified on site plan)

.2 m lower than the invert R. L. of inlet pipe (Tentatively 174.00 m)

r of the inlet chamber is assumed as 1.30 m.)

m 1.2 m wide foot bridge shall be provided. The tip velocity of the scraper shall be 0.3 m/min.

e centrally located manifold channel.