civil engineering design

8/8/2019 Civil Engineering Design

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Civil Engineering Design

YearPopulation in thousand

1951 951961 1701971 2201981 2501991 2802001 350

Per capita water supply rate = 115 lpcd.

Design period:

o Pumping machinery = 15yr.o Treatment unit = 20yr.o Conveyance pipe = 30yr.

Parameter

o BOD(mg/l) = 225o PH = 7.2o TSS(mg/l) =225o COD(mg/l)=330

Ig = (0.78940.29410.13630.120.25) 51

Ig= 0.248

P15=P01(1+Ig) 5.1


2/21

=350K(1+0.248) 5.1

=487.96K

P20=P01(1+Ig) 2

=350K(1+0.248) 2=545.12K

P30=P01(1+Ig) 3

=350K(1+0.248) 3

=680.31K

Q15 (avg)=4900001150.810^-3=45080m 3 /day

Q20 (avg)=541001150.810^-3=50149.2 m 3 /day

Q30 (avg)=6803001150.810^-3=62587 m 3 /day

Consider for monsoon

Q15 (avg)=450801.1 = 49588 m 3 /day=0.573 m 3 /sec

Q20 (avg)=50149.21.1=55164.12 m 3 /day

=0.63 m3

/secQ30 (avg)=62587.61.1=68846.36 m 3 /day

=0.79 m 3 /sec

Q15 (peak)=(Q 15avg) (P.F)+O.1(sewage)


3/21

=45080(2.25)/(243600)+0.1(45080)/(243600)

=1.22 m 3 /sec

Q20 (peak)=(Q 20avg) (P.F)+O.1(sewage)=50149.2(2.25)/

(243600)+0.1(50149.2)/(243600)=1.36 m 3 /sec

Q30 (peak)=(Q 30avg) (P.F)+O.1(sewage)=62587(2.25)/

(243600)+0.1(62587)/(243600)

=1.69 m3

/sec

Detention Time =20min

V=Qt=1.361200=1632m 3

Assume liquid Depth inwet well as 6m.

Area= 61632

=272m 2

Assume diameter of dry well

A= 4 (d2 -7 2 )


4/21

d =20.05Design of rising mainV =0.8 to 2.5 m/sAssume V= 1 m/sQ30 (peak) =1.69 m 3 /secA= V

Q = 169.1 =1.69m 3 /sec

Assume of 4 n.o.s. of pipe. Diameter of Pipe (d) =4/4(d 2)=1.6

d=0.73 mHead loss

Now, Hf=10.7852.1

C Q

C Q

89.4 D L

Where, C=Chazzysconstant(135)

L=15m

=10.7852.1

413569.1

( ) 89.473.015

=0.017mH=Hf+Hs

=0.017+15=15.017 mHp of Pumps,

=n

H QSw75

15 Where, n=0.7.

= 7.075017.1522.11000

=348.96 HpProvide 4 pump of 90 Hp

Provide 4 pump for stand by Provide 8 Pumps

1.Design Of Screen Chamber


5/21

Assume Velocity of flow throughscreen=0.8m/sec.

Q peak 30 =1.69 m 3 /secOpen area of screen = V

Q = 8.069.1 =2.112m 2

Rectangular Bars of 1 cm width are providedat 5cm clear distance.

Total Area =2.112 56 =2.534m 2

Provide 2 Screen with one as stand by,

Assume screen bar are placed at 60 withhorizontal.

Total Area=60sin

534.2 =2.92 m 2

Assume Width of screen chamber =2.5mHeight of screen chamber=1.17m

Velocity of flow above screenV=0.8 6

5 =0.67 m/sHead loss through screen

=0.0729(V 2 -v 2 )=0.0729(0.8 2 -0.6 2 )

=0.014 mHead loss when screen is half clog=0.0729(1.6 2 -0.67 2 )=0.15

2.Design of grit chamber


6/21

Minimum size of grit particle to be removed is0.2mm with Specific gravity of 2.65

Vs=( )

18

12 s sd g

;=

( ))1001.1(18

165.2)102.0(81.96

23

;

=0.036 m/sec.

Re=

d V s ;

=01.1

)102.0(036.0 3

=7.128>1.Transcent flow condition,Vs= (0.707(Ss-1) d 6.06.1 ) 714.0 Where, d=0.2

310 =0.025 m/sec

Vc= (8K (Ss-1) f gd ) 21 ; Where, K=0.04;

=0.186 m/sec f =0.03;Keep Vh=0.15 m/sec.As Q peak 20 =1.36 m 3 /sec.

Q avg 20 =0.58 m 3 /sec.Assume Q min20 =0.41 m 3 /sec.Cross sectional area of grit chamber

A peak =h

peak

V

Q

chamber gritof No.

Provide 4 Grit chamber.A peak = 15.04

36.1 = 2.26 m

2

A Avg = 15.0458.0

=0.967 m2


7/21

A min = 15.044.0

=0.67 m2

A parabola = 32

W Z0

Assume width W= 2 m.2.26= 3

2 2 Z 0 ;Z 0 =1.695 m.

Now, Z 0 =CW 2 ;1.695=C 2 2 ;

C = 0.42Z 0 =0.4 W 2

A parabola = 32 W Z 0 ;

= 32 W 0.4 W 2

A parabola =0.28 W 2 ,

Flow Area Z 0 WPeak =1.36 2.26 1.695 2Avg. =0.58 0.967 0.948 1.53Min. =0.41 0.67 0.396 1.27

sh V D

V L =

max D

LV

V

s

h =

L=1.695 025.015.0

=10.17;

Efficiency =1-n

s AV n L

1

+

;

Where = measure of settling basin


8/21

Take = 81

for good performance,n =75%;

0.75=1-8

...025.0

81

+ ROS L ;

S.O.R. = 0.01653 m/sec.Q peak 20 =1.36m 3 /sec.

Actual S.O.R. =

217.10436.1 =0.020 m/sec.

3.Design Of Partial Flume:

Provide 4 Grit chamber so that maximumflow rate in each is 436.1 =0.34.

Throat width h =0.305,C=0.69,

n =1.52,St=0.62,

Flow in partial flumeQ=Cha n

0.4=0.69Ha 52.1Ha=0.71 m.

Assume width of 2.5m,Vh=0.25 m/s.

At peak flow depth of flow in each is,d= VhChamber Gritof Width

Chamber GriteachinQ

;

= 0.252.50.41

; =0.66+0.3;

=0.99m Provide 1md = H +a

0.656 = 0.7+ =-0.044 m


9/21

Q 441.0

min = =0.1m 3 /sec.Q=C ha H

0.1=0.69Ha 52.1

Ha=0.28 m d += a H min

=0.28-0.044=0.236 m

Vh at minimum flow,

Vh=min

min

dChamber Gritof WidthChamber GriteachinQ

;

= 0.2362.50.1

=1.69 m/s.Surface area of grit chamber;= S.O.R.

Q ;= 0.016531.36 ;

=82.27 m 2Gross area of each,

= 482.27 ; =20.57.

DIMENSION OF GRIT CHAMBER

So, Width =2.5 mLength= 8.5 m

4.Design of primary settling tank:

Q design = Q avg = 0.580For primary settling tank S.O.R. = 35 to 50

m 3 /m 2 /dAssume S.O.R. = 40 m 3 /m 2 /d.

S.O.R. = AQ ;

A = 4058.0 ; =1253.73 m 2

Provide 2 clarifierEach A= 2

73.1253 ; =626.86m 2 .


10/21

A= 24 d

;d=28.25 m 30 m.

Assume side water depth (SWD) =3.5 mVolume of each =2194.0 m 3

Detention time = QV =

58.0

0.2194 =3782.77 sec=21.050=2.1 hr.

Weir load ratry (WLR) = DQ

WLR= ( )25.2828640058.0

; ( As 2clarifiers)

= 282.464 m 3 /m 2 /d


11/21

=0.27m 0.30 m

Assume 0.1 m head loss +0.5 m freeboard.

Total depth = 0.6 m

Assume suspended solid concentrationin sewage = 225 mg/lit.

Further assume that 70% of removal of excess occurs in primary clarifier and outof total suspended solids 70% are volatile and 30% arein organic

Or fixed.Q avg 20 =0.58

Mass of sludge settled in eachprimmaryclarifier=2257.0

258.0

=45.67gm/sec.

=164.43kg/hr.

Assume specific gravity of inorganic solids =1.05Specific gravity of inorganic solids = 2.5.

Overall Specific gravity of suspended solids=

+=

05.17.0

5.23.01

ST S

ST S =1.27.

+= 27.1 05.0195.01SC S ; SC S =1.01.

Volume of primary sludge,

s P M

V SLw

SC

.=

;


12/21

05.001.1100043.164

= ;=3.25 m 3 /hr.

Let the sludge be removed after every 4 hr,

Volume of sludgestorage=(3.254)=13.02 m 3

V= ( )( )

++ 21

3r r

hr ; Assume, h=1.5.

13.02 = ( )( )

++ 21

35.1

r r r ;4.144= r+0.5 ( )( )21 + r r

r = 2.74.

Calculation of bottom slope clarifier,Bottom slope of D.C.= 8

1to 12

1 Assume 10

1

Accurate sludge unit:No. of aeration tank minimum 2. Side

water depth i.e. liquid depth in (diffused)aeration tank =4.5 to 7.5 m. 1 to 6 m forsurface aeration . Usually 3 to 4.5. Freeboard . 0.3 to 0.6 m. For diffused aerationand 1 to 2.5 for surface aeration.Ratio of width to length of aeration tankB:L= 1:3 to1:4.

B.O.D. 5 Of settle sewage =225 mg/lOutlet B.O.D. of treated sewage= 30 mg/l.Concentration of bio mass in treatedsewage =25 mg/lWith 65% bio-mass as degradable.

SO=225 mg/lB.O.D. total =B.O.D. Sol +B.O.D. SS


13/21

Bio degradable bio mass in treatedeffluent= 0.6525

=16.25 mg/lB.O.D. ult of bio mass= 1.4216.25,In treated effluent = 23.1 mg/l.

703.0...... 5 =ult DO B

DO B ; B.O.D. due to biomass=23.10.7

=16.17 mg/l30=B.O.D. Sol 5 +16.17;B.O.D. Sol 5 =13.83 mg/l;

Efficiency of treatment1. Based on Total B.O.D.= %95.801005.157

305.157 =

2. Based on Sol B.O.D.= %23.911005.157

83.135.157 =

Volume of aeration tank required:

V=( )

( )cd esol cavg

K

S S Q

+

1

0

;Where: = 0.5 mg/l c =10days

d K =0.06 day 1

Q d avg /544328640063.0 ==

V=( )

( )( );

1006.01300083.135.157105.054332

+

=8146 m 3Provide 2 aeration tank with 4m side water

depth(SWD)S.A. of each A.T. = 42

8146 =1018.3 m

2

B4B=1018.3 m 2


14/21

B =15.95m say 16mm L = 64 m

Free board =1.0 mProvide 2 aeration tanks of 15.3m width and 64mlength & 5m total depth.

=

544324649.15

2...

T R H

=0.149 day= 3.58 hr.

Calculate sludge wasting rate;

+

= ew

c QQVx

Q ;

Assume concentration of MLSS insludge recycle line as 10,000 mg/lit. Assumeconcentration of biomass in treated effluentas 80% of S.S.

8.0=MLSS

MLVSS ;X w =0.810000= 8000mg/l

Q we QQ =X e =0.82520 mg/l.V=15.96442

=8140.8 m 3

+=

eewwc QQ

V Q

;

( )

+=

2054432800030008.8140

10ww QQ

;

Q w =169.62 m 3/d. =1696.2 kg /day

Calculate O 2 required and Hp of surfaceaeration;

B.O.D. 5 Removed= 157.5-13.83=143.67 mg/l


15/21

B.O.D. L = 27.21168.067.143 = mg/l

Kg of B.O.D. removed;=54432211.27 1000

1 ;

=11500 Kg/dBio mass waste per day;=1696.20.8;=1356.96 Kg/d.Net O 2 required = (11500-1.42)

1356.96=9573.1168 Kg/d=365.56 Kg/hr.

Computation of power requirement for surfaceaeration,

Mt= ( ) ;02.1 2010

T

s

l sw

C C BC

R

Mt= ( ) ;95.002.108.9224.81

5.1 2025

=1.08r P

g

H H Ok 2 ;

Hp aeration required =08.1

87.398 =369.33h p;

Hp required in Aeration tank =66.186

233.369 = h p

So provide 8 aeration of 25 HpCalculate the re circulation ratio

R= 3000)100008.0(3000

== X X X

Q

Q

w

R ;

QQ R 6.0=Calculate M

F Ratio;

M F = X

OS .. = 151.030005.157

=0.34 day1 (O.K.)

5. Design Of secondary clarifier :


16/21

Total flow of secondary clarifier = Q + RQ=1.6Q= 1.654432=87091.2 m 3 /d

Take S.O.R. = 20 m 3 / m 2 /day.S.A. = 20

2.87091 =4354.46Provide 4 Secondary clarifier.

Surface area of each clarifier= 1088.64 m 2

1088.64= 24 d

;

d= 37.23 m 37.5 m.S.W.D.= 3.5 mWeir loading rate = d

Q 4 ;

= 5.37450112

;=115.5

Q Peak =1.36 m 3 /secS.O.R. = A

Q ;

= 25.374

4

8640036.1

;

=26.59W.L.R. = D

Q

; =

5.3748640036.1

=249.35m


17/21

2)Excess sludge generated for activated sludgeprocess

=1696.2kg/day

3) Volume of primary sludge =4.46 m 3 /hr=107.04 m 3 /day.

4)Volume of excess sludge generated fromactivated sludge

process Q w =169.62 m 3 /day.5)Total volume of raw mixed sludge

=169.62+107.04

=276.66 km 3 /d6)Total quantity of raw mixed

sludge=3946.32+1696.2=5642.

5 kg/d7)Suspended solid cone of raw mixed sludge =

66.27652.5642

=20.39 kg/m 3

8)Approximate % of volatile matters in mixedsludge=70%

0.7 ..sludgeActivated8.0

S A P P

s

s

++

( ) ( )

2.169632.3946

1696.20.832.39467.0

++

0.73 73%9)Quantity of volatile matters in raw mixed sludge

is

=0.73 55642.52

=4035.46 kg/day


18/21

10)Quantity of non volatile matter = 100-73%=27%

0.275642.52=1523.48 kg/day.

7.Design of Low Rate sludge:.1)Approximate % of destruction of V m =50 %( Designvalue).2)For achieving 50% of volatile matter destruction

under mesophilic condition the hydraulic retentiontime required is 40day (manual).3)Quantity of Vm in digestive sludge=0.54119.0396

=2059.51 kg/day.4)Total quantity of non volatile or inorganicindigestive sludge is 1523.48 kg/day.5)Total quantity of solid in digestive sludge

=2059.51+1523.48=3582.99 Kg/day.6)% of volatile matter in digestive sludge = 99.3582

51.2059

=57.48%7)% of inorganic matter = 99.3582

48.1523 =42.51%8)Depending upon frequency of sludge withdrawn

consistency of

digestive sludge from low rate digestion is excepctedto be inrange of 4-6%% consistency = 10000 kg/day

=10kg/m 3 .


19/21

9)For an average consistency of 5%(50kg/day)volume of digestive

sludge is 365.715099.3582

m= /day,

10)Volume of digester:V= ( ) 13

2T V V V d f f ;

check = 40 ( ) 65.7166.27632

66.276 ;=5427.06 m 3 .

8.Check for volatile solids:

Loading rate kg VSS/daym 3 .

= 06.5427 039.4119 =0.74 kg VSS/daym 3 .9.Design for gas generation:

1)Gas generation /kg of volatile matter destroyed =0.9 3 /kg

2)Total gas generation 0.92059.51.73=1853.559 m3 .

3)To avoid foaming minimum surface area requiredto meet condition 9 m 3 of gas generated/ day.Per m 3 surface area will be = 9

559.1853 =205.951 m 2 .4)For operational flexibility and construction reasonsit is suggested to install 2 digester of followingdimensions = 2

06.5427 ; =2713.53 m3

5)Minimum surface area = 2951.205 ; =102.97 m 2 .


20/21

Choosing digester shape as low for vertical cylinderand for Q of 30m the surface area of each digestion= ( )2304

; =706.8 m 2 .

Effective digester depth = m AV

8.36.141306.5427

== o f depth.Addition volume of sludge storage during monsoonperiod when sludge drying bed option is not use forsludge dewatering =VdT 2 For a period of 12 day=35.25512;

=423.06 m 3 ;Equivalent to 8.706

06.423 =0.59 m;

Addition allowance for grit and slump accumulation is0.6m free board 0.6 total addition =3.8+0.59+0.6+0.6=5.59 5.6m

10. Design of sludge drying bed:

1. Volume of digested sludge= 71.65 daym3

;

2. Dewatering, drying and sludge removalcycle=10 days.

3. Depth applical of sludge solution= 0.3m

4. Total plan area of sludge drying=3.0

1065.71

= 2388.33= 23.88 m 2


21/21

Provide 50m long and 10m width of sludge drying bed No. of bed required =

50033.2388 ;

=4.77 say 5bedsBed provide with single with singledischarge 5% slope for collection of super native sludge.

civil engineering design

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