hydr-design of settling basin new

8/3/2019 Hydr-Design of Settling Basin new

http://slidepdf.com/reader/full/hydr-design-of-settling-basin-new 1/36

1 of 36

Hydraulic Design of Settling Basin

(1) Design conditions

Design intake discharge per 1 (one) settling basin Qo = 3.004 m3 /s

No. of sedimentation channels per 1 settling basin Ns = 2 channels

Design discharge per channel Q = 1.502 m3 /s = Qo / 2

Flushing discharge per channel Qs = 1.802 m3 /s = Q x 120 %

Content of suspended load in the river water Rs = 0.01 %Minimum particle size to be settled dc = 0.3 mm

Content of suspended load with particle size of 0.3 mm or more

Rc = 50 %

Flushing interval Fi = 14 days (= twice a month)

Fall velocity of sediment particle Vg = 0.04 m/s (for dc = 0.3 mm)

Bed slope of channel i = 1 / 100 ( > 1/100)

Average velocity in cedimentation channel U = 0.20 m/s

Thickness of sediment (= 0.2 to 0.25 m) D = 0.25 m

(at a point where the minimum particle is settled completely)

Thickness of partition wall tp = 0.30 m

Wall angle of inlet transition against flow direction q = 12.0 deg. ( < 12.5 deg.)

Width at beginning section of inlet transition Bo = 1.500 m

Water depth at beginning section of inlet transition ho = 2.069 m

Sill elevation at beginning section of inlet transition ELo = 29.250 m

(2) Width of sedimentation channelb = Q / (h U) = 3.00 m

and, b = {h2

+ a Q2

/(k h2)}

0.5 - h = 2.34 m say, 3.00 m

where, b : width of channel (m)

h : water depth above sediment h = 2.50 m


Q : discharge in channel Q = 1.502 m3 /s

U : average velocity in cedimentation channel U = 0.20 m/s

a : velocity coefficient (= 1.0 to 1.2) a = 1.2

k : coefficient k = tc / (r i) = 0.0252 = tc / (r i)

tc : critical tractive force tc = 0.000252 t/m2

= 8.41 r dc11/32

/ 104

r : unit weight of water r = 1.0 t/m3

dc : particle size to be settled dc = 0.03 cm

i : bed slope of channel i = 1 / 100

(3) Width of settling basin

B = Ns b + (Ns-1) t = 6.30 m

where, B : width of settling basin (m)

b : width of channel b = 3.00 m

Ns : no. of sedimentation channels Ns = 2 channels

tp : thickness of partition wall tp = 0.30 m

h=2.50m

D=0.25m

b=3.00m tp=0.30m b=3.00m

B=6.30m

(4) Design of inlet transition

Width at beginning section of inlet transition Bo = 1.50 m

Water depth at beginning section of inlet transition ho = 2.069 m

Sill elevation at beginning section of inlet transition ELo = 29.25 m

Width at end section of inlet transition Bi = 6.30 m = B

Length of transition Li = 11.00 m = (Bi - Bo)/(2 tanq)

Wall angle of inlet transition against flow direction q = 12.0 deg.

Water level difference between beginning and end sections

Dh = 0.006 m (see hydaulic calculation)

dimension



2 of 36

Hydraulic elements of inlet transition

Distance from beginning point (m) 0.000 0.500 1.500 3.000 5.500 8.000 11.000

Width (m) 1.500 1.718 2.155 2.809 3.900 4.991 6.300

Water depth (m) 2.069 1.806 1.440 1.105 0.796 0.622 0.493

Flow area (m) 3.103 3.103 3.103 3.103 3.103 3.103 3.103

Sill elevation (m) 29.250 29.512 29.878 30.213 30.520 30.693 30.820

(5) Length of sedimentation channel

(a) Length from sedimentation theory

La = K h U / Vg = K Q / (b Vg) = 25.03 m

where, La : length of channel (m)

K : safety factor (= 1.5 to 2.0) K = 2.0




U : average velocity in cedimentation channel U = 0.20 m/s

Vg : Fall velocity of sediment particle Vg = 0.04 m/s

Q : discharge in channel Q = 1.502 m3 /s

(b) Length from a detaching length of flow on gapped bed

Lb = L1 + L2 + L3 = 35.00 m

where, La : length of channel (m)

L1 : length of sediment terrace L1 = 12.80 mL2 : distance between downstream of terrace L2 = 19.40 m = 10 W'

and a point where the minimum particle is settled completely

L3 : excess length of channel (m) L3 = 2.80 m

(about gap height at end of sedimentation channel)

W : drop height at beginning section of chann W = 1.94 m = h+D-h1-(L1+L2)i

W' : downstream height of sediment terrace W' = 1.94 m = W OK

D : thickness of sediment D = 0.25 m




h1 : water depth above crest h1 = 0.49 m

(at beginning section of sedimentation channel) (see hydraulic elements of inlet transition)

i : bed slope of channel i = 1 / 100

We : step height at end section of channel We = 2.33 m

(c) Design length of sedimentation channel

La = 25.03 m < Lb 35.00 m say, L = Lb = 35.00 m

(6) Capacity of settling basin

Qr = 86400 Q Rs Rc Fi = 90.8 m3

Qv = b {W' L1 + D (L - L1) = 91.1 m3

> Qr OK

where, Qr : required capacity of settling basin (m3)

Qv : design capacity of settling basin (m3)

Q : Design discharge per channel Q = 1.502 m3 /s

Rs : Ratio of suspended material Rs = 0.01 %

Fi : flushing interval Fi = 14 days


L : length of channel (m) L = 35.00 m

W : drop height at beginning section of chann W = 1.94 mWe : step height at end section of channel We = 2.33 m

h1=0.49m

W'=1.94m h=2.50m h=2.50m

W=1.94m i=1/100 We=2.33m

D=0.25m D=0.25m

L1=12.80m L2=19.40m L3=2.80m b=3.00m tp=0.30m b=3.00m

L=35.00 m

B=6.30m

dimension



3 of 36

Hydraulic Design of Flushing Conduit/Flume


Design discharge for flushing Qso = 3.605 m3 /s = Qo x 120 %

Number of sedimentation channels Ns = 2 nos.

Flushing discharge per channel Qs = 1.803 m3 /s = Qso / Ns

Width of sedimentation channel B = 3.000 m/channel

Bed slope of sedimentation channel I = 1/100

Acceleration of gravity g = 9.8 m/s2

(2) Hydraulic calculation in the sedimentation channel at flushing time

Critical water depth in the channel hc = 0.333 m = { Qs2

/ ( g B2

) }1/3

Flushing velocity V1 = 4.149 m/s = f { 2 g ( Z + 1.5 hc ) }1/2

where, Z : height of drop Z = 1.940 m

f : coefficient ( 0.6 to 0.7 ) f = 0.6

Water depth in the channel h1 = 0.145 m = Qs / ( B V1 ) (supercritical flow)

Froude number Fr1 = 3.481 = V1 / ( g h1 )1/2

(3) Transition and flushing conduit

Length of transition l = 3.00 m

Assumed width of flushing conduit b = 1.49 m (rectangular)

Assumed b/B ratio b/B = 0.498

where, B: width of sedimentation channel B = 3.00 m

b/B ratio from equation below b/B = 0.498 = assumed b/B OK

b/B = 2{1+(2l /B)2} / [2+(2l /B)

2[{1+8Fr1

2 /(1+(2l /B)

2)}

1/2 -1]]

Design width of flushing conduit b = 1.50 m (rectangular)

Critical water depth in the conduit hc2 = 0.528 m = { Q2

/ ( g b2

) }1/3

Water depth of shock wave h2 = 0.436 m

Check h2 /hc2 h2 /hc2 = 0.826 < 1 OK

h2 /hc2 = [[1+8Fr12 /{1+(2l /B)

2}]

1/2 -1] h1 / [2{Q

2 /(g b

2)}

1/3]

Bed slope of flushing conduitBackwater effect from the river 1 ( 1 : not to be considered )

( 2 : to be considered )

Bed slope of flushing conduit Ic = 1/100 = I

In case, considering backwater effect from the river

Velocity in the conduit V2 = 2.757 m/s = Qs / (b h2 )

Froude number in the conduit Fr2 = 1.334 = V2 / ( g h2 )1/2

h3 /h2 ratio h3 / h2 = 1.634

Water depth in the conduit h3 = 0.712 m

Required height of flushing conduit H = 0.60 m > hc2 OK

(4) Design dimensions of flushing flume

Internal width Bo = 1.50 m

Internal height Ho = 1.00 m (> H + Fbo) OK

No. of barrels No = 1 bareels

Freeboard Fbo = 0.25 m (for Qo = 1.803 m3/s < 5.0 m3/s)

(5) Design dimensions of flushing gate and conduit

Clear width bg = 1.50 m

Clear height hg = 1.00 m (> H + Fbg) OK

Freeboard Fbg = 0.25 m (for Qo = 1.803 m3/s < 5.0 m3/s)

No. of gate/conduit Ng = 2 sets/barrels

dimension



4 of 36

(6) Flushing conduit/flume (in case, considering backwater effect from the river)

Flushing discharge Qo = 1.803 m3 /s = Qs

Average velocity Vo = 1.69 m/s (about 1.5 m/s for coarse sand)

Width of flushing conduit bo = 1.50 m = b

Water depth ho = 0.712 m = h3

Bed slope of flushing conduit/flume

Vo = Ko x Ro2/3 x Io1/2

\ Io = {Vo / (Ko x Ro2/3

)} = 0.00892 (= 1/112.1)

where,

Roughness coefficient Ko = 35

Flow area Ao = 1.068 m2

= bo ho

Wetted perimeter Po = 2.924 m = bo + 2 ho

Hydraulic radius Ro = 0.365 m = Ao / Po

dimension



5 of 36

Hydraulic Design of Broad-crested Weir


Type of water measuring device Broad-crested weir with rounded entrance

and downstream slope of 1:1

Shape of control Rectangular control

Design discharge Q = 1.502 m3 /s = Qo / 2

Crest elevation EL0 = 30.860 mCanal bed elevation upstream of weir EL1 = 28.530 m (end section of sedimentation channel)

Canal bed elevation downstream of weir EL2 = 29.960 m (Bed EL at BP of main camal)

Water level upstream of weir WL1 = 31.321 m (end section of sedimentation channel)

Water level downstream of weir WL2 = 31.140 m (WL at BP of main camal + 0.05m)

Width of crest b0 = 3.000 m = b1

Width of upstream canal b1 = 3.000 m = b (sedimentation channel)

Width of downstream canal b2 = 3.000 m = b1

Upstream water depth above crest h1 = 0.461 m = WL1 - EL0

Downstream water depth above crest h2 = 0.280 m = WL2 - EL0

Upstream height of weir p1 = 2.330 m = EL0 - EL1

Downstream height of weir p2 = 0.900 m = EL0 - EL2


(2) Hydraulic elements

Critical water depth above crest h0 = 0.295 m = {Q2

/ (g b02

)}1/3

Flow area above crest A* = 0.885 m2

= b0 h0

Flow area upstream of weir A1 = 8.373 m = b1 (h1 + p1)

Flow area downstream of weir A2 = 3.540 m = b2 (h2 + p2)

Velocity of upstream canal v1 = 0.179 m/s = Q / A1

Velocity of downstream canal v2 = 0.424 m/s = Q / A2

Velocity head of upstream canal hv1 = 0.002 m = v12

/ (2 g)

Velocity head of downstream canal hv2 = 0.009 m = v22

/ (2 g)

Upstream energy head above crest H1 = 0.463 m = h1 + hv1

Downstream energy head above crest H2 = 0.289 m = h2 + hv2

(3) Check to complete overflow

in case, (2/3) H1 > H2 : Complete overflowin case, (2/3) H1 < H2 : Incomplete/submerged overflow

(2/3) H1 = 0.308 m > H2 = 0.289 m Complete overflow

OK

(4) Length and radius of crest

L > 1.75 H1 + 0.2 H1 = 1.95 H1 = 0.902 m say, 1.50 m

R = 0.2 H1 = 0.093 m say, 0.15 m

where, L : length of crest (m)

R : radius at upstream face of crest (m)

H1 : Upstream energy head above crest H1 = 0.463 m

(5) Discharge above crest

Q0 = Cd Cv (2/3) {(2/3) g}0.5 bc h11.5 = 1.546 m3

/s

> Q = 1.502 m3 /s OK

where, Q0 : discharge (m /s)

Cd : discharge coefficient

Cd = 0.93 + 0.10 (H1 /L) (for 0.1 < H1 /L <1.0)

H1 / L = 0.31

Cd = 0.96

Cv : approach velocity coefficient

Cd A* / A1 = 0.10

Cv = 1.005 (from Figure 2.3 of KP-04)

dimension



6 of 36

Check of Hydraulic Design of Broad-crested WeirIntake Water Level 31.550 m (Crest EL. )

Intake Discharge 3.605 m3 /s (120% of required intake discharge)


Type of water measuring device Broad-crested weir with rounded entrance

and downstream slope of 1:1

Shape of control Rectangular control

Design discharge Q = 1.802 m3

/s = Qs (per sedimentation channel)Crest elevation EL0 = 30.860 m

Canal bed elevation upstream of weir EL1 = 28.530 m (end section of sedimentation channel)

Canal bed elevation downstream of weir EL2 = 29.960 m (Bed EL at BP of main camal)

Water level upstream of weir WL1 = 31.385 m (end section of sedimentation channel)

Water level downstream of weir WL2 = 31.140 m (WL at BP of main camal + 0.05m)

Width of crest b0 = 3.000 m = b1

Width of upstream canal b1 = 3.000 m = b (sedimentation channel)

Width of downstream canal b2 = 3.000 m = b1

Upstream water depth above crest h1 = 0.525 m = WL1 - EL0

Downstream water depth above crest h2 = 0.280 m = WL2 - EL0

Upstream height of weir p1 = 2.330 m = EL0 - EL1

Downstream height of weir p2 = 0.900 m = EL0 - EL2


(2) Hydraulic elements

Critical water depth above crest h0 = 0.333 m = {Q2

/ (g b02

)}1/3

Flow area above crest A* = 0.999 m2

= b0 h0

Flow area upstream of weir A1 = 8.565 m2

= b1 (h1 + p1)

Flow area downstream of weir A2 = 3.540 m = b2 (h2 + p2)

Velocity of upstream canal v1 = 0.210 m/s = Q / A1

Velocity of downstream canal v2 = 0.509 m/s = Q / A2

Velocity head of upstream canal hv1 = 0.002 m = v12

/ (2 g)

Velocity head of downstream canal hv2 = 0.013 m = v22

/ (2 g)

Upstream energy head above crest H1 = 0.527 m = h1 + hv1

Downstream energy head above crest H2 = 0.293 m = h2 + hv2

(3) Check to complete overflow

in case, (2/3) H1 > H2 : Complete overflow

in case, (2/3) H1 < H2 : Incomplete/submerged overflow

(2/3) H1 = 0.352 m > H2 = 0.293 m Complete overflow

OK

(4) Length and radius of crest

L > 1.75 H1 + 0.2 H1 = 1.95 H1 = 1.028 m say, 1.50 m

R = 0.2 H1 = 0.105 m say, 0.15 m

where, L : length of crest (m)

R : radius at upstream face of crest (m)

H1 : Upstream energy head above crest H1 = 0.527 m

(5) Discharge above crest

Q0 = Cd Cv (2/3) {(2/3) g}0.5

bc h11.5

= 1.889 m3 /s

> Q = 1.802 m3 /s OK

where, Q0 : discharge (m /s)

Cd : discharge coefficient

Cd = 0.93 + 0.10 (H1 /L) (for 0.1 < H1 /L <1.0)

H1 / L = 0.35

Cd = 0.97

Cv : approach velocity coefficient

Cd A* / A1 = 0.11

Cv = 1.006 (from Figure 2.3 of KP-04)

dimension



Hydrauric Design Statement of Lanang Intake (between Inlet and End Section of Settling Basin)

Intake Water Level 31.450 m (Crest EL. 31.550 m - 0.10 m)


Energy Energy Velocity Water Water Bed

Structure Station Distance Discharge Loss Line EL. Velocity Head Level Depth Elevation Rem

L (m) Q (m3 /s) hl (m) EH (m) V (m/s) Vh (m) WL (m) d (m) BL (m)River

0 0.00 3.004 31.450 0.000 0.000 31.450 6.095 25.355

Inlet 0.00 0.011 hl0 = hi

1 0.00 3.004 31.439 0.934 0.045 31.394 2.144 29.250

Inlet Flume 3.00 0.033 hl1 = hf 22 3.00 3.004 31.406 0.949 0.046 31.360 2.110 29.250

Inlet of Conduit 0.00 0.000 hl2 = hsc

3 3.00 3.004 31.406 0.949 0.046 31.360 2.110 29.250

Conduit (Free flow) 13.60 0.006 hl4 = hf 44 16.60 3.004 31.400 0.952 0.046 31.354 2.104 29.250

Outlet of Conduit 0.00 0.000 hl5 = hse

5 16.60 3.004 31.400 0.952 0.046 31.354 2.104 29.250

Transition 0.00 0.000 hl5' = hg

5' 16.60 3.004 31.400 0.952 0.046 31.354 2.104 29.250

Outlet Flume 72.60 0.033 hl6 = hf 66 B.P. of Settling Basin 89.20 3.004 31.367 0.968 0.048 31.319 2.069 29.250

Inlet Transition 11.00 0.006 hl7 = hge

7 100.20 3.004 31.361 0.967 0.048 31.313 0.493 30.820

Inlet Crest 2.00 0.003 hl8 = hsc

8 102.20 3.004 31.358 1.039 0.055 31.302 0.482 30.820

Drop at Beginning Sect. 0.00 0.035 hl9 = hse

9 102.20 3.004 31.323 0.205 0.002 31.320 2.440 28.880

Sedimentation Channel 35.00 0.000 hl10 = hf

10 137.20 3.004 31.323 0.179 0.002 31.321 2.791 28.530

(Broad crested weir) 30.860 (Crest el

Total 137.20 m 0.127 m SDWL = 0.129 m

Water Level at Upstream of Broad-crested Weir

31.450 -0.129 = 31.321 m



Hydrauric Design Statement of Lanang Intake (between Inlet and End Section of Settling Basin)

Intake Water Level 31.550 m (Crest EL. )


Energy Energy Velocity Water Water BedStructure Station Distance Discharge Loss Line EL. Velocity Head Level Depth Elevation Rem

L (m) Q (m3 /s) hl (m) EH (m) V (m/s) Vh (m) WL (m) d (m) BL (m)River

0 0.00 3.605 31.550 0.000 0.000 31.550 6.195 25.355

Inlet 0.00 0.015 hl0 = hi

1 0.00 3.605 31.535 1.080 0.060 31.475 2.225 29.250

Inlet Flume 3.00 0.043 hl1 = hf 22 3.00 3.605 31.492 1.102 0.062 31.430 2.180 29.250

Inlet of Conduit 0.00 0.000 hl2 = hsc

3 3.00 3.605 31.492 1.102 0.062 31.430 2.180 29.250

Conduit (Free flow) 13.60 0.008 hl4 = hf 44 16.60 3.605 31.484 1.107 0.063 31.421 2.171 29.250

Outlet of Conduit 0.00 0.000 hl5 = hse

5 16.60 3.605 31.484 1.107 0.063 31.421 2.171 29.250

Transition 0.00 0.000 hl5' = hg

5' 16.60 3.605 31.484 1.107 0.063 31.421 2.171 29.250

Outlet Flume 72.60 0.045 hl6 = hf 66 B.P. of Settling Basin 89.20 3.605 31.439 1.132 0.065 31.373 2.123 29.250

Inlet Transition 11.00 0.009 hl7 = hge

7 100.20 3.605 31.430 1.031 0.054 31.375 0.555 30.820

Inlet Crest 2.00 0.003 hl8 = hsc

8 102.20 3.605 31.427 1.106 0.062 31.363 0.543 30.820

Drop at Beginning Sect. 0.00 0.038 hl9 = hse

9 102.20 3.605 31.389 0.240 0.003 31.384 2.504 28.880

Sedimentation Channel 35.00 0.000 hl10 = hf

10 137.20 3.605 31.389 0.210 0.002 31.385 2.855 28.530

(Broad crested weir) 30.860 (Crest el

Total 137.20 m 0.161 m SDWL = 0.165 m

Water Level at Upstream of Broad-crested Weir

31.550 -0.165 = 31.385 m



9 of 36

Hydraulic Calculation of Lanang IntakeIntake Water Level 31.450 m (Crest EL. 31.550 m - 0.10 m)


1. Design Conditions and Input Data

(1) Design discharge and no. of channels per one (1) settling basin

- Design intake discharge Q = 3.004 m3 /s

- Design discharge per inlet flume channel or conduit barrel Q1 = 3.004 m3 /s/channel

- Design discharge per outlet flume channel Q2 = 3.004 m3 /s/barrel

- Design discharge per sedimentation channel Q3 = 1.502 m3 /s/channel

- Design discharge per approach flume channel Q4 = 3.004 m3 /s/channel

- No. of inlet flume channel or conduit barrels N1 = 1 nos.

- No. of outlet flume channels N2 = 1 nos.

- No. of sedimentation channels N3 = 2 nos.

- No. of approach flume channels N4 = 1 nos.

(2) Hydraulic conditions at beginning section (river)

- Water level in the river NWL WL0 = 31.450 m = (Crest EL. 31.550 m - 0.10 m)

- Bed elevation in the river EL0 = 25.355 m

- Water depth in the river h0 = 6.095 m

- Velocity V0 = 0.000 m/s

- Roughness coefficient n = 1 / K = 0.015 (Concrete)

K = 65

- Acceleration of gravity g = 9.8 m/s2

(3) Dimensions in each section

Structure Bed EL. Width Length Nos. Thickness of No. of Bed slope

(m) (m) (m) partition wall (mpartition wall

Inlet flume

1 Beginning section 29.250 1.500 0.000 1 0.000 02 End section 29.250 1.500 3.000 1 0.000 0 LevelConduit (free flow) (Clear height 2.770 )

3 Beginning section 29.250 1.500 0.000 1 0.000 0

4 End section 29.250 1.500 13.600 1 0.000 0 Level

Outlet flume

5 Beginning section 29.250 1.500 0.000 1 0.000 05' Transition 29.250 1.500 0.000 1 0.000 0 Level


Settling basin

7 Inlet transition (E.P.) 30.820 6.300 11.000 1 0.000 0 Reverse

7' Inlet crest (B.P.) 30.820 3.000 0.000 2 0.300 1

8 Inlet crest (E.P.) 30.820 3.000 2.000 2 0.300 1 Level

9 Drop at beginning secti 28.880 3.000 0.000 2 0.300 1(Gap height 1.940 )

10 End section of 28.530 3.000 35.000 2 0.300 1 0.010000sedimentation channel (Gap height 2.330 )

(Broad crested weir) 30.860 3.000 0.000 2 0.300 1

hydro-calc Q100%



10 of 36

(4) Head loss factor

(a) Head loss coefficient at inlet

Shape of inlet Coefficient fi Shape of inlet : edge cut

square edge 0.5 fi = 0.25

edge cut 0.25

circular round 0.1

squared round 0.2

bell-mouth 0.01 - 0.05

(b) Head loss factor at trashrack

ItemHeight of trash a = min. (0.1h or 0.25m) h: upstream water depth

Unit weight of wetted trash ga = 200 kg/m3

Thickness of bar t = 0.01 m

Clear space between bars b = 0.150 m

Inclined angle of trash rack q = 90 deg.

(c) Head loss coefficient at open transition

Changing condition of Gradual contraction Gradual enlargement

open transition formation coefficient fgc coefficient fge

rectangular to rectangular 0.10 0.20

trapezoidal to trapezoidal 0.10 0.20rectangular to circular with fillet 0.20 0.30

trapezoidal to rectangular with twisted wall 0.20 0.30

trapezoidal to circular with twisted wall 0.30 0.40

trapezoidal to rectangular with bended wall 0.30 0.50

trapezoidal to circular with bended wall 0.40 0.70

Transition at B.P. of outlet flume Formation of open transition : Rectangular to rectangular

fgc = 0.10

Inlet transition of settling basin Formation of open transition : Rectangular to rectangularfge = 0.20

Outlet transition of settling basin Formation of open transition : Rectangular to rectangular

fgc = 0.10

hydro-calc Q100%



11 of 36

2. Inlet Flume

(1) Design discharge and no. of channels

- Design discharge Q = 3.004 m3 /s

- Design discharge per inlet channel Q1 = 3.004 m3 /s/channel

- No. of inlet channels N1 = 1 nos.

(2) Hydraulic conditions at beginning section (river)- Water level WL0 = 31.450 m

- Base elevation EL0 = 25.355 m

- Water depth h0 = 6.095 m


- Velocity head V02 / 2g = 0.000 m

(3) Change of water level due to inflow

Dhi = hi + ( V12

/ 2g - V02

/ 2g )

hi = fi ( V12

/ 2g - V02

/ 2g )

where,

hi : Head loss due to inflowfi : Head loss coefficient at inlet fi = 0.25 (for edge cut)

(a) Assumed Dhi Dhi = 0.056 m

(b) Beginning section of inlet flume (after inflow)


- Width b1 = 1.500 m

- Water depth h1 = 2.144 m = WL0 - EL1 - Dhi

- Flow area A1 = 3.216 m = N1 (b1 h1)

- Wetted perimeter P1 = 5.788 m = N1 (b1 + 2h1)

- Hydraulic radius R1 = 0.556 m = A1 / P1

- Velocity V1 = 0.934 m/s = Q / A1


(c) Calculation of hi and Dhi hi = 0.011 m

\ Dhi = 0.056 m = Assumed Dhi OK

(4) Water level at beginning section of inlet flume (after inflow)

WL1 = 31.394 m = WL0 - Dhi

(5) Length and bed slope of inlet flume

- Length L2 = 3.000 m

- Bed slope I2 = Level

(6) Change of water level due to friction

Dhf 2 = hf 2 + ( V22

/ 2g - V12

/ 2g )

hf 2 = ( n Vm2 / Rm22/3

)2

L2

where,

hf 2 : Head loss due to friction

Vm2 : Mean velocity Vm2 = ( V1 + V2 ) / 2

Rm2 : Mean hydraulic radius Rm2 = ( R1 + R2 ) / 2

hydro-calc Q100%



12 of 36

(a) Assumed Dhf 2 Dhf 2 = 0.001 m

(b) upstream section of trashrack


- Width b2 = 1.500 m

- Water depth h2 = 2.143 m = WL1 - EL2 - Dhf 2- Flow area A2 = 3.215 m = N1 (b2 h2)

- Wetted perimeter P2 = 5.786 m = N1 (b2 + 2h2)- Hydraulic radius R2 = 0.556 m = A2 / P2

- Velocity V2 = 0.934 m/s = Q / A1


(c) Calculation of Dhf 2 Vm2 = 0.934 m/s

Rm2 = 0.556 m

hf 2 = 0.001 m

\ Dhf 2 = 0.001 m = Assumed Dhf 2 OK

(7) Water level at upstream section of trashrack

WL2 = 31.393 m = WL1 - Dhf 2

(8) Change of water level due to trashrack

Dht = ht + ( V2'2

/ 2g - V22

/ 2g )

where,

ht : Head loss due to trashrack ht = 6.69 sinq (t/b)4/3

exp(0.074 ga a/h2) V22 / 2g

(a) Head loss due to trashrack

- Water depth (u/s of trash rack) h2 = 2.143 m

- Velocity V2 = 0.935 m/s = Q / {N1 (h2 b2)}


- Height of trash a = 0.200 m = min. (0.1h2 or 0.25m)

- Unit weight of wetted trash ga = 200 kg/m

3

- Thickness of bar t = 0.01 m

- Clear space between bars b = 0.150 m

- Inclined angle of trash rack q = 90 deg.

Head loss due to trashrack ht = 0.032 m

(b) Assumed Dht Dht = 0.033 m

(c) Downstream section of trashrack

- Base elevation EL2' = 29.250 m = EL2

- Width b2' = 1.500 m = b2

- Water depth h2' = 2.110 m = WL2 - EL2' - Dht

- Flow area A2' = 3.165 m = N1 (b2' h2')

- Wetted perimeter P2' = 5.720 m = N1 (b2' + 2h2')

- Hydraulic radius R2' = 0.553 m = A2' / P2'

- Velocity V2' = 0.949 m/s = Q / A2'

- Velocity head V2'2 / 2g = 0.046 m

(d) Calculation of Dht Dht = 0.033 m = Assumed Dht OK

(9) Water level at downstream section of trashrack

WL2' = 31.360 m = WL2 - Dht

hydro-calc Q100%



13 of 36

3. Conduit (Free flow)

(1) Design discharge and no. of barrels


- Design discharge per conduit barrels Q1 = 3.004 m3 /s/barrel

- No. of conduit barrels N1 = 1 nos.

(2) Change of water level due to sudden contraction A3 / A2' fsc0.0 0.50

Dhsc = hsc + ( V32

/ 2g - V2'2

/ 2g ) 0.1 0.48

hsc = fsc V32 / 2g 0.2 0.45

0.3 0.41

where, 0.4 0.36

hsc : Head loss due to sudden contraction 0.5 0.29

fsc : Head loss coefficient of sudden contraction 0.6 0.21

(Refer to the right table) 0.7 0.13

0.8 0.07(a) Assumed Dhsc Dhsc = 0.000 m 0.9 0.01

1.0 0.00

(b) Before contraction

- Width per channel b2' = 1.500 m

- Water depth h2' = 2.110 m

- Flow area A2' = 3.165 m

- Wetted perimeter P2' = 5.720 m

- Hydraulic radius R2' = 0.553 m

- Velocity V2' = 0.949 m/s


(c) After contraction


- Width of channel b3 = 1.500 m Clear height hi= 2.770 m

- Water depth h3 = 2.110 m = WL2' - EL2' - Dhsc

- Flow area A3 = 3.165 m = N1 (b3 h3)- Wetted perimeter P3 = 5.720 m = N1 (b3 + 2h3)


- Velocity V3 = 0.949 m/s = Q / A3


(d) Calculation of Dhsc A3 / A2' = 1.00

fsc = 0.00 (Refer to the table above)

hsc = 0.000 m\ Dhsc = 0.000 m = Assumed Dhsc OK

(3) Water level at beginning section of conduit

WL3 = 31.360 m = WL2' - Dhsc

(4) Length and bed slope of conduit

- Length L4 = 13.600 m


hydro-calc Q100%



14 of 36


Dhf 4 = hf 4 + ( V42

/ 2g - V32

/ 2g )

hf 4 = ( n Vm4 / Rm42/3

)2

L4

where,





(b) Beginning section of conduit


- Width of channel b3 = 1.500 m


- Flow area A3 = 3.165 m

- Wetted perimeter P3 = 5.720 m

- Hydraulic radius R3 = 0.553 m


- Velocity head V32

/ 2g = 0.046 m

(c) End sections of conduit


- Width b4 = 1.500 m




- Velocity V4 = 0.952 m/s = Q / A4


(d) Calculation of Dhf 4 Vm4 = 0.951 m/sRm4 = 0.553 m

hf 4 = 0.006 m


(6) Water level at end section of conduit

WL4 = 31.354 m = WL3 - Dhf 4

hydro-calc Q100%



15 of 36

4. Beginning Section of Outlet Flume



- Design discharge per conduit barrel Q1 = 3.004 m3 /s/barrel

- Design discharge per flume channel Q2 = 3.004 m3 /s/channel


- No. of flume channels N2 = 1 nos.

(2) Change of water level due to sudden enlargement A4 / A5 fse

0.0 1.00

Dhse = hse + ( V52

/ 2g - V42

/ 2g ) 0.1 0.81

hse = fse V4 2 / 2g 0.2 0.64

0.3 0.49

where, 0.4 0.36hse : Head loss due to sudden enlargement 0.5 0.25

fse : Head loss coefficient of sudden enlargement 0.6 0.16


0.8 0.04

(a) Assumed Dhse Dhse = 0.000 m 0.9 0.01

1.0 0.00(b) Before enlargement

- Width per channel b4 = 1.500 m







(c) After enlargement



- Water depth h5 = 2.104 m = WL4 - EL5 - Dhse

- Flow area A5 = 3.156 m = N2 (b5 h5)



- Velocity V5 = 0.952 m/s = Q / A5


(d) Calculation of Dhse A4 / A5 = 1.00

fse = 0.00 (Refer to the table above)

hse = 0.000 m

\ Dhse = 0.000 m = Assumed Dhse OK

(3) Water level at beginning section of outlet flume

WL5 = 31.354 m = WL4 - Dhse

hydro-calc Q100%



16 of 36

5. Transition at Beginning Section of Outlet Flume





(2) Length and bed slope

- Length L5' = 0.000 m

- Bed slope I5' = Level

(3) Change of water level due to gradual contraction

Dhgc = hgc + hf 5' + ( V5'2

/ 2g - V52

/ 2g )

hgc = fgc ( V5'2

/ 2g - V52

/ 2g )

hf 5' = ( n Vm5' / Rm5'2/3

)2

L5'

where,

hgc : Head loss due to gradual contraction

hf 5' : Head loss due to frictionVm5' : Mean velocity Vm5' = ( V5 + V5' ) / 2

Rm5' : Mean hydraulic radius Rm5' = ( R5 + R5' ) / 2

fgc : Head loss coefficient of gradual contraction

fgc = 0.10 (for Rectangular to rectangular)

(a) Assumed Dhgc Dhgc = 0.000 m

(b) End section of inlet transition

- Bed elevation EL5' = 29.250 m

- Bed width b5' = 1.500 m

- Water depth h5' = 2.104 m = WL5 - EL5' - Dhgc

- Flow area A5' = 3.156 m = N2 (b5' h5')

- Wetted perimeter P5' = 5.708 m = N2 (b5' + 2h5')- Hydraulic radius R5' = 0.553 m = A5' / P5'

- Velocity V5' = 0.952 m/s = Q / A5'


(c) Calculation of Dhgc Vm5' = 0.952 m/s

Rm5' = 0.553 m

hf 5' = 0.000 m

hgc = 0.000 m

\ Dhgc = 0.000 m = Assumed Dhgc OK

(4) Water level at end section of inlet transition

WL5' = 31.354 m = WL5 - Dhgc

hydro-calc Q100%



17 of 36

6. End Section of Outlet Flume






- Length L6 = 72.600 m



Dhf 6 = hf 6 + ( V62

/ 2g - V5'2

/ 2g )

hf 6 = ( n Vm6 / Rm62/3

)2

L6

where,


Vm6 : Mean velocity Vm6 = ( V5' + V6 ) / 2

Rm6 : Mean hydraulic radius Rm6 = ( R5' + R6 ) / 2


(b) End section of outlet flume

- Bed elevation EL6 = 29.250 m

- Bed width b6 = 1.500 m

- Water depth h6 = 2.069 m = WL5' - EL6 - Dhf 6- Flow area A6 = 3.103 m = N2 (b6 h6)



- Velocity V6 = 0.968 m/s = Q / A6



Rm6 = 0.552 m

hf 6 = 0.033 m


(4) Water level at end section of outlet flume

(BP of settling basin)

WL6 = 31.319 m = WL5' - Dhf 6

hydro-calc Q100%



18 of 36

7. Inlet Transition of Settling Basin






- Length L7 = 11.000 m

- Bed slope I7 = Reverse

(3) Change of water level due to gradual enlargement

Dhge = hge + hf 7 + ( V72

/ 2g - V62

/ 2g )

hge = fge ( V62

/ 2g - V72

/ 2g )

hf 7 = ( n Vm7 / Rm72/3

)2

L7

where,

hge : Head loss due to gradual enlargement

hf 7 : Head loss due to frictionVm7 : Mean velocity Vm7 = ( V6 + V7 ) / 2


fge : Head loss coefficient of gradual enlargement

fge = 0.20 (for Rectangular to rectangular)

(a) Assumed Dhge Dhge = 0.006 m




- Water depth h7 = 0.493 m = WL6 - EL7 - Dhge

- Flow area A7 = 3.106 m = N2 (b7 h7)


- Velocity V7 = 0.967 m/s = Q / A7


(c) Calculation of Dhge Vm7 = 0.968 m/s

Rm7 = 0.488 m

hf 7 = 0.006 m

hge = 0.000 m

\ Dhge = 0.006 m = Assumed Dhge OK


WL7 = 31.313 m = WL6 - Dhge

hydro-calc Q100%



19 of 36

8. Inlet Crest




- Design discharge per crest channel Q3 = 1.502 m3 /s/channel


- No. of inlet crest channels N3 = 2 nos.

(2) Change of water level due to sudden contraction A7' / A7 fsc

0.0 0.50

Dhsc = hsc + ( V7'2

/ 2g - V7 2

/ 2g ) 0.1 0.48

hsc = fsc V7'2 / 2g 0.2 0.45

0.3 0.41

where, 0.4 0.36hsc : Head loss due to sudden contraction 0.5 0.29



0.8 0.07

(a) Assumed Dhsc Dhsc = 0.008 m 0.9 0.01

1.0 0.00(b) Before contraction









- Base elevation EL7' = 30.820 m

- Width of channel b7' = 3.000 m

- Water depth h7' = 0.485 m = WL7 - EL7' - Dhsc

- Flow area A7' = 2.910 m = N3 (b7' h7')



- Velocity V7' = 1.032 m/s = Q / A7'


(d) Calculation of Dhsc A7' / A7 = 0.94


hsc = 0.001 m

\ Dhsc = 0.008 m = Assumed Dhsc OK

(3) Water level at beginning section of inlet crest

WL7' = 31.305 m = WL7 - Dhsc


- Length L8 = 2.000 m


hydro-calc Q100%



20 of 36


Dhf 8 = hf 8 + ( V82

/ 2g - V7'2

/ 2g )

hf 8 = ( n Vm8 / Rm82/3

)2

L8

where,





(b) Beginning section of inlet crest








- Velocity head V7'2

/ 2g = 0.054 m

(c) End sections of inlet crest


- Width b8 = 3.000 m




- Velocity V8 = 1.039 m/s = Q / A8



hf 8 = 0.002 m


(6) Water level at end section of inlet crest

WL8 = 31.302 m = WL7' - Dhf 8

hydro-calc Q100%



21 of 36

9. Drop at Beginning Section of Sedimentation Channel



- Design discharge per channel Q3 = 1.502 m3 /s/channel


(2) Height and length of drop

- Height of drop W = 1.940 m

- Length of drop L9 = 0.000 m


0.0 1.00

Dhse = hse + ( V92

/ 2g - V82

/ 2g ) 0.1 0.81

hse = fse V8 2 / 2g 0.2 0.64

0.3 0.49

where, 0.4 0.36

hse : Head loss due to sudden enlargement 0.5 0.25



0.8 0.04(a) Assumed Dhse Dhse = -0.018 m (rising) 0.9 0.01

1.0 0.00

(b) Before enlargement









- Base elevation EL9 = 28.880 m = EL8 - W



- Flow area A9 = 14.640 m = N3 (b9 h9)



- Velocity V9 = 0.205 m/s = Q / A9




hse = 0.035 m

\ Dhse = -0.018 m = Assumed Dhse OK

(4) Water level at beginning section of sedimentation channel

WL9 = 31.320 m = WL8 - Dhse

hydro-calc Q100%



22 of 36

10. Sedimentation Channel






- Length L10 = 35.000 m

- Bed slope I10 = 0.010000 = 1/100


Dhf 10 = hf 10 + ( V102

/ 2g - V92

/ 2g )

hf 10 = ( n Vm10 / Rm102/3

)2

L10

where,




(a) Assumed Dhf 10 Dhf 10 = -0.001 m (rising)

(b) End section of sedimentation channel

- Bed elevation EL10 = 28.530 m = EL9 - I10 L10


- Water depth h10 = 2.791 m = WL9 - EL10 - Dhf 10

- Flow area A10 = 16.746 m = N3 (b10 h10)



- Velocity V10 = 0.179 m/s = Q / A10



Rm10 = 0.953 m

hf 10 = 0.000 m

\ Dhf 10 = -0.001 m = Assumed Dhf 10 OK

(4) Water level at end section of sedimentation channel

WL10 = 31.321 m = WL9 - Dhf 10

hydro-calc Q100%



23 of 36

Hydraulic Calculation of Lanang IntakeIntake Water Level 31.550 m (Crest EL. )


1. Design Conditions and Input Data

(1) Design discharge and no. of channels per one (1) settling basin

- Design intake discharge Qs = 3.605 m3 /s

- Design discharge per inlet flume channel or conduit barrel Q1 = 3.605 m3 /s/channel

- Design discharge per outlet flume channel Q2 = 3.605 m3 /s/barrel

- Design discharge per sedimentation channel Q3 = 1.802 m3 /s/channel

- Design discharge per approach flume channel Q4 = 3.605 m3 /s/channel

- No. of inlet flume channel or conduit barrels N1 = 1 nos.

- No. of outlet flume channels N2 = 1 nos.


- No. of approach flume channels N4 = 1 nos.

(2) Hydraulic conditions at beginning section (river)

- Water level in the river NWL WL0 = 31.550 m = (Crest EL.)

- Bed elevation in the river EL0 = 25.355 m

- Water depth in the river h0 = 6.195 m


- Roughness coefficient n = 1 / K = 0.015 (Concrete)

K = 65

- Acceleration of gravity g = 9.8 m/s2

(3) Dimensions in each section

Structure Bed EL. Width Length Nos. Thickness of No. of Bed slope

(m) (m) (m) partition wall (mpartition wall

Inlet flume

1 Beginning section 29.250 1.500 0.000 1 0.000 02 End section 29.250 1.500 3.000 1 0.000 0 LevelConduit (free flow) (Clear height 2.770 )

3 Beginning section 29.250 1.500 0.000 1 0.000 0


Outlet flume

5 Beginning section 29.250 1.500 0.000 1 0.000 05' Transition 29.250 1.500 0.000 1 0.000 0 Level


Settling basin

7 Inlet transition (E.P.) 30.820 6.300 11.000 1 0.000 0 Reverse

7' Inlet crest (B.P.) 30.820 3.000 0.000 2 0.300 1

8 Inlet crest (E.P.) 30.820 3.000 2.000 2 0.300 1 Level

9 Drop at beginning secti 28.880 3.000 0.000 2 0.300 1(Gap height 1.940 )

10 End section of 28.530 3.000 35.000 2 0.300 1 0.010000sedimentation channel (Gap height 2.330 )

(Broad crested weir) 30.860 3.000 0.000 2 0.300 1

hydro-calc Q120%



24 of 36

(4) Head loss factor

(a) Head loss coefficient at inlet

Shape of inlet Coefficient fi Shape of inlet : edge cut

square edge 0.5 fi = 0.25

edge cut 0.25

circular round 0.1

squared round 0.2

bell-mouth 0.01 - 0.05

(b) Head loss factor at trashrack

ItemHeight of trash a = min. (0.1h or 0.25m) h: upstream water depth

Unit weight of wetted trash ga = 200 kg/m3

Thickness of bar t = 0.01 m

Clear space between bars b = 0.150 m

Inclined angle of trash rack q = 90 deg.

(c) Head loss coefficient at open transition

Changing condition of Gradual contraction Gradual enlargement

open transition formation coefficient fgc coefficient fge

rectangular to rectangular 0.10 0.20

trapezoidal to trapezoidal 0.10 0.20rectangular to circular with fillet 0.20 0.30

trapezoidal to rectangular with twisted wall 0.20 0.30

trapezoidal to circular with twisted wall 0.30 0.40

trapezoidal to rectangular with bended wall 0.30 0.50

trapezoidal to circular with bended wall 0.40 0.70

Transition at B.P. of outlet flume Formation of open transition : Rectangular to rectangular

fgc = 0.10

Inlet transition of settling basin Formation of open transition : Rectangular to rectangularfge = 0.20

Outlet transition of settling basin Formation of open transition : Rectangular to rectangular

fgc = 0.10

hydro-calc Q120%



25 of 36

2. Inlet Flume



- Design discharge per inlet channel Q1 = 3.605 m3 /s/channel

- No. of inlet channels N1 = 1 nos.

(2) Hydraulic conditions at beginning section (river)- Water level WL0 = 31.550 m





(3) Change of water level due to inflow

Dhi = hi + ( V12

/ 2g - V02

/ 2g )

hi = fi ( V12

/ 2g - V02

/ 2g )

where,

hi : Head loss due to inflowfi : Head loss coefficient at inlet fi = 0.25 (for edge cut)

(a) Assumed Dhi Dhi = 0.075 m

(b) Beginning section of inlet flume (after inflow)


- Width b1 = 1.500 m

- Water depth h1 = 2.225 m = WL0 - EL1 - Dhi

- Flow area A1 = 3.338 m = N1 (b1 h1)



- Velocity V1 = 1.080 m/s = Q / A1


(c) Calculation of hi and Dhi hi = 0.015 m

\ Dhi = 0.075 m = Assumed Dhi OK

(4) Water level at beginning section of inlet flume (after inflow)

WL1 = 31.475 m = WL0 - Dhi

(5) Length and bed slope of inlet flume

- Length L2 = 3.000 m



Dhf 2 = hf 2 + ( V22

/ 2g - V12

/ 2g )

hf 2 = ( n Vm2 / Rm22/3

)2

L2

where,




hydro-calc Q120%



26 of 36


(b) upstream section of trashrack


- Width b2 = 1.500 m



- Velocity V2 = 1.081 m/s = Q / A1



Rm2 = 0.561 m

hf 2 = 0.002 m


(7) Water level at upstream section of trashrack

WL2 = 31.473 m = WL1 - Dhf 2

(8) Change of water level due to trashrack

Dht = ht + ( V2'2

/ 2g - V22

/ 2g )

where,

ht : Head loss due to trashrack ht = 6.69 sinq (t/b)4/3

exp(0.074 ga a/h2) V22 / 2g

(a) Head loss due to trashrack

- Water depth (u/s of trash rack) h2 = 2.223 m

- Velocity V2 = 1.081 m/s = Q / {N1 (h2 b2)}


- Height of trash a = 0.200 m = min. (0.1h2 or 0.25m)

- Unit weight of wetted trash ga

= 200 kg/m3

- Thickness of bar t = 0.01 m

- Clear space between bars b = 0.150 m

- Inclined angle of trash rack q = 90 deg.

Head loss due to trashrack ht = 0.041 m

(b) Assumed Dht Dht = 0.043 m

(c) Downstream section of trashrack

- Base elevation EL2' = 29.250 m = EL2

- Width b2' = 1.500 m = b2

- Water depth h2' = 2.180 m = WL2 - EL2' - Dht

- Flow area A2' = 3.270 m = N1 (b2' h2')



- Velocity V2' = 1.102 m/s = Q / A2'


(d) Calculation of Dht Dht = 0.043 m = Assumed Dht OK

(9) Water level at downstream section of trashrack

WL2' = 31.430 m = WL2 - Dht

hydro-calc Q120%



27 of 36

3. Conduit (Free flow)

(1) Design discharge and no. of barrels


- Design discharge per conduit barrels Q1 = 3.605 m3 /s/barrel


(2) Change of water level due to sudden contraction A3 / A2' fsc0.0 0.50

Dhsc = hsc + ( V32

/ 2g - V2'2

/ 2g ) 0.1 0.48

hsc = fsc V32 / 2g 0.2 0.45

0.3 0.41

where, 0.4 0.36

hsc : Head loss due to sudden contraction 0.5 0.29



0.8 0.07(a) Assumed Dhsc Dhsc = 0.000 m 0.9 0.01

1.0 0.00

(b) Before contraction

- Width per channel b2' = 1.500 m









- Width of channel b3 = 1.500 m Clear height hi= 2.770 m

- Water depth h3 = 2.180 m = WL2' - EL2' - Dhsc

- Flow area A3 = 3.270 m = N1 (b3 h3)- Wetted perimeter P3 = 5.860 m = N1 (b3 + 2h3)


- Velocity V3 = 1.102 m/s = Q / A3


(d) Calculation of Dhsc A3 / A2' = 1.00


hsc = 0.000 m\ Dhsc = 0.000 m = Assumed Dhsc OK

(3) Water level at beginning section of conduit

WL3 = 31.430 m = WL2' - Dhsc

(4) Length and bed slope of conduit

- Length L4 = 13.600 m


hydro-calc Q120%



28 of 36


Dhf 4 = hf 4 + ( V42

/ 2g - V32

/ 2g )

hf 4 = ( n Vm4 / Rm42/3

)2

L4

where,





(b) Beginning section of conduit








- Velocity head V32

/ 2g = 0.062 m

(c) End sections of conduit


- Width b4 = 1.500 m




- Velocity V4 = 1.107 m/s = Q / A4



hf 4 = 0.008 m


(6) Water level at end section of conduit

WL4 = 31.421 m = WL3 - Dhf 4

hydro-calc Q120%



29 of 36

4. Beginning Section of Outlet Flume



- Design discharge per conduit barrel Q1 = 3.605 m3 /s/barrel





0.0 1.00

Dhse = hse + ( V52

/ 2g - V42

/ 2g ) 0.1 0.81

hse = fse V4 2 / 2g 0.2 0.64

0.3 0.49

where, 0.4 0.36hse : Head loss due to sudden enlargement 0.5 0.25



0.8 0.04

(a) Assumed Dhse Dhse = 0.000 m 0.9 0.01

1.0 0.00(b) Before enlargement












- Flow area A5 = 3.257 m = N2 (b5 h5)



- Velocity V5 = 1.107 m/s = Q / A5




hse = 0.000 m

\ Dhse = 0.000 m = Assumed Dhse OK

(3) Water level at beginning section of outlet flume

WL5 = 31.421 m = WL4 - Dhse

5. Transition at Beginning Section of Outlet Flume



hydro-calc Q120%



30 of 36




- Length L5' = 0.000 m

- Bed slope I5' = Level

(3) Change of water level due to gradual contraction

Dhgc = hgc + hf 5' + ( V5'2

/ 2g - V52

/ 2g )

hgc = fgc ( V5'2

/ 2g - V52

/ 2g )

hf 5' = ( n Vm5' / Rm5'2/3

)2

L5'

where,

hgc : Head loss due to gradual contraction

hf 5' : Head loss due to friction

Vm5' : Mean velocity Vm5' = ( V5 + V5' ) / 2

Rm5' : Mean hydraulic radius Rm5' = ( R5 + R5' ) / 2

fgc : Head loss coefficient of gradual contraction

fgc = 0.10 (for Rectangular to rectangular)

(a) Assumed Dhgc Dhgc = 0.000 m


- Bed elevation EL5' = 29.250 m

- Bed width b5' = 1.500 m

- Water depth h5' = 2.171 m = WL5 - EL5' - Dhgc

- Flow area A5' = 3.257 m = N2 (b5' h5')



- Velocity V5' = 1.107 m/s = Q / A5'


(c) Calculation of Dhgc Vm5' = 1.107 m/s

Rm5' = 0.558 m

hf 5' = 0.000 m

hgc = 0.000 m

\ Dhgc = 0.000 m = Assumed Dhgc OK


WL5' = 31.421 m = WL5 - Dhgc

hydro-calc Q120%



31 of 36

6. End Section of Outlet Flume






- Length L6 = 72.600 m



Dhf 6 = hf 6 + ( V62

/ 2g - V5'2

/ 2g )

hf 6 = ( n Vm6 / Rm62/3

)2

L6

where,





(b) End section of outlet flume






- Velocity V6 = 1.132 m/s = Q / A6



Rm6 = 0.556 m

hf 6 = 0.045 m


(4) Water level at end section of outlet flume

(BP of settling basin)

WL6 = 31.373 m = WL5' - Dhf 6

hydro-calc Q120%



32 of 36

7. Inlet Transition of Settling Basin






- Length L7 = 11.000 m

- Bed slope I7 = Reverse

(3) Change of water level due to gradual enlargement

Dhge = hge + hf 7 + ( V72

/ 2g - V62

/ 2g )

hge = fge ( V62

/ 2g - V72

/ 2g )

hf 7 = ( n Vm7 / Rm72/3

)2

L7

where,

hge : Head loss due to gradual enlargement

hf 7 : Head loss due to frictionVm7 : Mean velocity Vm7 = ( V6 + V7 ) / 2


fge : Head loss coefficient of gradual enlargement

fge = 0.20 (for Rectangular to rectangular)

(a) Assumed Dhge Dhge = -0.002 m (rising)




- Water depth h7 = 0.555 m = WL6 - EL7 - Dhge

- Flow area A7 = 3.497 m = N2 (b7 h7)


- Velocity V7 = 1.031 m/s = Q / A7


(c) Calculation of Dhge Vm7 = 1.082 m/s

Rm7 = 0.513 m

hf 7 = 0.007 m

hge = 0.002 m

\ Dhge = -0.002 m = Assumed Dhge OK


WL7 = 31.375 m = WL6 - Dhge

hydro-calc Q120%



33 of 36

8. Inlet Crest




- Design discharge per crest channel Q3 = 1.802 m3 /s/channel


- No. of inlet crest channels N3 = 2 nos.

(2) Change of water level due to sudden contraction A7' / A7 fsc

0.0 0.50

Dhsc = hsc + ( V7'2

/ 2g - V7 2

/ 2g ) 0.1 0.48

hsc = fsc V7'2 / 2g 0.2 0.45

0.3 0.41

where, 0.4 0.36hsc : Head loss due to sudden contraction 0.5 0.29



0.8 0.07

(a) Assumed Dhsc Dhsc = 0.009 m 0.9 0.01

1.0 0.00(b) Before contraction











- Water depth h7' = 0.546 m = WL7 - EL7' - Dhsc

- Flow area A7' = 3.276 m = N3 (b7' h7')



- Velocity V7' = 1.100 m/s = Q / A7'


(d) Calculation of Dhsc A7' / A7 = 0.94


hsc = 0.001 m

\ Dhsc = 0.009 m = Assumed Dhsc OK

(3) Water level at beginning section of inlet crest

WL7' = 31.366 m = WL7 - Dhsc


- Length L8 = 2.000 m


hydro-calc Q120%



34 of 36


Dhf 8 = hf 8 + ( V82

/ 2g - V7'2

/ 2g )

hf 8 = ( n Vm8 / Rm82/3

)2

L8

where,





(b) Beginning section of inlet crest








- Velocity head V7'2

/ 2g = 0.062 m

(c) End sections of inlet crest


- Width b8 = 3.000 m




- Velocity V8 = 1.106 m/s = Q / A8



hf 8 = 0.002 m


(6) Water level at end section of inlet crest

WL8 = 31.363 m = WL7' - Dhf 8

hydro-calc Q120%



35 of 36

9. Drop at Beginning Section of Sedimentation Channel





(2) Height and length of drop

- Height of drop W = 1.940 m

- Length of drop L9 = 0.000 m


0.0 1.00

Dhse = hse + ( V92

/ 2g - V82

/ 2g ) 0.1 0.81

hse = fse V8 2 / 2g 0.2 0.64

0.3 0.49

where, 0.4 0.36

hse : Head loss due to sudden enlargement 0.5 0.25



0.8 0.04(a) Assumed Dhse Dhse = -0.021 m (rising) 0.9 0.01

1.0 0.00

(b) Before enlargement









- Base elevation EL9 = 28.880 m = EL8 - W



- Flow area A9 = 15.024 m = N3 (b9 h9)



- Velocity V9 = 0.240 m/s = Q / A9




hse = 0.038 m

\ Dhse = -0.021 m = Assumed Dhse OK

(4) Water level at beginning section of sedimentation channel

WL9 = 31.384 m = WL8 - Dhse

hydro-calc Q120%



36 of 36

10. Sedimentation Channel






- Length L10 = 35.000 m

- Bed slope I10 = 0.010000 = 1/100


Dhf 10 = hf 10 + ( V102

/ 2g - V92

/ 2g )

hf 10 = ( n Vm10 / Rm102/3

)2

L10

where,




(a) Assumed Dhf 10 Dhf 10 = -0.001 m (rising)

(b) End section of sedimentation channel

- Bed elevation EL10 = 28.530 m = EL9 - I10 L10


- Water depth h10 = 2.855 m = WL9 - EL10 - Dhf 10

- Flow area A10 = 17.130 m = N3 (b10 h10)



- Velocity V10 = 0.210 m/s = Q / A10



Rm10 = 0.961 m

hf 10 = 0.000 m

\ Dhf 10 = -0.001 m = Assumed Dhf 10 OK

(4) Water level at end section of sedimentation channel

WL10 = 31.385 m = WL9 - Dhf 10

hydr-design of settling basin new

Documents