weir design

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MODEL CALCULATION FOR DESIGN OF H.C. WEIR

The Discharge per Meter length of weir

Q= 1.933 Q= discharge per meterCd= 2.264 proposed by poondy research station

L= 1 Length of weir in meterh= 0.9 Head of Flow in meterH= 1.4 Height of Body wall

SCOUR DEPTH CALCULATIONS

Nominal Scour Depth= R=R= 1.663 Q= discharge per meter

f= 2 Laceys silt factor

Down Stream cut off is taken up to 2R depth =2xR= 3.325 m below MFL

Depth Below apron level=2R-d2= 1.961

Provide: 2.20 m which is safe

STABILITY CALCULATIONS: CASE I: Where there is water up to crest level on up stream side and no water on down

FTLL5= 0.22 W7 W5

W4c= 2.25 Specific gravity of concrete

W6L4= 1.18

W1 W2 W3

A 0.45 0.45 0.7L1 L2 L3

Figure I

S.No. Force Details Magnitude L.A. Moment1 W1 1/2xL1xL3x2.25 0.5974 0.3000 0.17922 W2 L2xL3x2.25 1.1948 0.6750 0.80653 W3 1/2xL3x(L3+L4)x2.25 1.1025 1.1333 1.24954 W4 1/2xL2xL5x2.25 0.1114 0.7500 0.08355 W5 1/2xL2xL5x1 0.0495 0.6000 0.0297

Q= Cd x L x h3/2

1.35(Q2/F)1/3 metres

Top width of Body wall is taken as 0.45m for all heads of flow and height of body wall

stream side FIG I

TAKING MOMENTS ABOUT (A)


6 W6 1/2xL1xL4x1 0.2655 0.1500 0.03987 W7 L1xL5x1 0.0990 0.2250 0.0223

8 0.4573TOTAL 3.4200 2.8678

Position of Resultant= 0.838548

The resultant falls within middle third of 0.533333 and 1.066667 Hence safe.

MWL

h= 0.9 W8 W9

FTLL1= 0.22 W7 W5 TWL

W4

W6L2= 1.18

W1 W2 W3

L3 0.45 L5

Figure II

WH3/6 = 1x1.43/6 =

E M/ EW=


MODEL CALCULATION FOR DESIGN OF H.C. WEIR


DESIGN OF F.O.F WEIR











FTL

c= 2.25 Specific gravity of concrete

L3= 2.5W1 W2

A 0.45 1.55L1 L2

2

Figure I

S.No. Force Details Magnitude L.A. Moment

Name of Work:

Q= Cd x L x h3/2



stream side FIG I


1 W1 L1xL3x2.25 2.5313 #VALUE! #VALUE!2 W2 1/2xL2xL3x2.25 4.3594 #VALUE! #VALUE!3 W3 1/2xL3x(L4+L5)x2.25 4.3594 #VALUE! #VALUE!4 W4 1/2xL2xL5x2.25 0.0000 #VALUE! #VALUE!5 W5 1/2xL2xL5x1 0.0000 #VALUE! #VALUE!6 W6 1/2xL1xL4x1 #VALUE! #VALUE!7 W7 L1xL5x1 #VALUE! #VALUE! #VALUE!

8 2.6042TOTAL #VALUE! #VALUE!

Position of Resultant= #VALUE!

Middle third is between #VALUE! and #VALUE!#VALUE!

MWL

h= 0.6 W5 W4

FTL

W3

L3= 2.5W1 W2

A 0.45 1.55L1 L2

H=L1+L2 Figure II

S.No. Force Details Magnitude L.A. Moment1 W1 L4xL1x1.25 1.4063 0.2250 0.31642 W2 1/2xL1xL5x1.25 2.4219 0.2250 0.54493 W3 1/2xL1xL5x1 1.9375 0.9667 1.87294 W4 hxL5x1 0.9300 0.3000 0.27905 W5 1/2xL1xL4x1 0.0000 0.1500 0.00006 W6 1/2xL2xL3x1 0.0000 0.0000 0.00007 W7 L1xL3x1 0.0000 0.0000 0.00008 W8 hxL3x1 0.0000 0.0000 0.00009 W9 hxL4x1 0.2700 0.2250 0.0608

10 W10 1/2xL5x(L1+L2)x1 1.9375 1.4833 2.8740

2.81258.9031 8.7605

WH3/6 =

E M/ EW=


MD=(H3+3xdxH2-D3)/6

Position of the Resultant( EM/EW)= 0.9840

The resultant falls within middle third of 0.666667 and 1.333333333333

The resultant falls between middle Third, Hence safe

FIXING OF LENGTH OF SOLID APRON BASED ON HYDRAULIC JUMP THEORY:

H

Lb=Ld+Lj

0.005

2.509

0.137

0.969

The Length of the Apron after the Jump,Lj = 5x(d2-d1) 4.159

The Total Length of the Apron= Lb = Ld+Lj = 6.668 m say 7.000 m

Rough Stone Apron

Scour Depth, D= 0.80 mLength to be covered on a slope of 2:1= Sqrt(5)xD 1.789

Thickness of Settled stones after Scour=1.24xT 0.567(Assuming side slope thickness as T 1.5 feets i.e 0.4572 metres)

Quantity/m Run= 1.014Length of Launching Apron=1.5D= 1.200 metres

d2

d1

Ld Lj

The Drop Number=Dn= q2/(gH3)=

The Length of the Apron before Jump,Ld=Hx4.3x(Dn)0.27

The Jump depths are d1=Hx0.54xDn0.425

d2=Hx1.66x(Dn)0.27

Thickness= 0.845 m or say 0.900 m

DESIGN OF F.O.F WEIR


proposed by poondy research stationLength of weir in meter

STABILITY CALCULATIONS:

CASE I: Where there is water up to crest level on up stream side and no water on down

FTLL5= 0.22 W7 W5


W6L4= 2.28

W1 W2 W3

A 0.65 0.9 1.25L1 L2 L3

2.8

Figure I

S.No. Force Details Magnitude L.A. Moment1 W1 1/2xL1xL3x2.25 1.6672 0.4333 0.72252 W2 L2xL3x2.25 4.6170 1.1000 5.07873 W3 1/2xL3x(L4+L5)x2.25 3.5156 1.9667 6.91414 W4 1/2xL2xL5x2.25 0.2228 1.2500 0.27845 W5 1/2xL2xL5x1 0.0990 0.9500 0.09406 W6 1/2xL1xL4x1 0.7410 0.2167 0.16057 W7 L1xL5x1 0.1430 0.3250 0.0465

8 2.6042TOTAL 11.0056 15.8989


Middle third is between 0.933333 and 1.866667The resultant falls between middle Third, Hence safe

stream side FIG I


WH3/6 =

E M/ EW=

MWL

h= 1.5 W8 W9


W4 W10

W6L2= 2.28 1.6

W1 W2 W3

A 0.65 0.9 1.25L3 L4 L5

H=L1+L2 Figure II

S.No. Force Details Magnitude L.A. Moment1 W1 1/2xL2xL3x1.25 0.9262 0.4333 0.40142 W2 L2xL4x1.25 2.5650 1.1000 2.82153 W3 1/2xL2xL5x1.25 1.9531 1.9667 3.84114 W4 1/2xL1xL4x1.25 0.1238 1.2500 0.15475 W5 1/2xL1xL4x1 0.0990 0.9500 0.09406 W6 1/2xL2xL3x1 0.7410 0.2167 0.16057 W7 L1xL3x1 0.1430 0.3250 0.04658 W8 hxL3x1 0.9750 0.3250 0.31699 W9 hxL4x1 1.3500 1.1000 1.4850

10 W10 1/2xL5x(L1+L2)x1 1.5625 2.3833 3.7240

5.671510.4386 18.7171





MD=(H3+3xdxH2-D3)/6

DESIGN OF H.C. WEIR











FTLL5= 0.22 W7 W5


W6L4= 2.48

W1 W2 W3

A 0.85 0.9 1.35L1 L2 L3

3.1

Figure I

Name of Work: Constructing checkdam across Bhyrampalli vagu, Bhyrampally(v), Maganoor(M), Mahaboobnagar District

Q= Cd x L x h3/2



stream side FIG I

S.No. Force Details Magnitude L.A. Moment1 W1 1/2xL1xL3x2.25 2.3715 0.5667 1.34392 W2 L2xL3x2.25 5.0220 1.3000 6.52863 W3 1/2xL3x(L4+L5)x2.25 4.1006 2.2000 9.02144 W4 1/2xL2xL5x2.25 0.2228 1.4500 0.32305 W5 1/2xL2xL5x1 0.0990 1.1500 0.11386 W6 1/2xL1xL4x1 1.0540 0.2833 0.29867 W7 L1xL5x1 0.1870 0.4250 0.0795

8 3.2805TOTAL 13.0569 20.9893


Middle third is between 1.0333333 and 2.0666667The resultant falls between middle Third, Hence safe

MWL

h= 1.2 W8 W9


W4 W10

W6L2= 2.48 1.6

W1 W2 W3

A 0.85 0.9 1.35L3 L4 L5

H=L1+L2 Figure II

S.No. Force Details Magnitude L.A. Moment1 W1 1/2xL2xL3x1.25 1.3175 0.5667 0.74662 W2 L2xL4x1.25 2.7900 1.3000 3.62703 W3 1/2xL2xL5x1.25 2.2781 2.2000 5.01194 W4 1/2xL1xL4x1.25 0.1238 1.4500 0.17945 W5 1/2xL1xL4x1 0.0990 1.1500 0.11386 W6 1/2xL2xL3x1 1.0540 0.2833 0.29867 W7 L1xL3x1 0.1870 0.4250 0.07958 W8 hxL3x1 1.0200 0.4250 0.43359 W9 hxL4x1 1.0800 1.3000 1.4040

10 W10 1/2xL5x(L1+L2)x1 1.8225 2.6500 4.8296

6.971811.7719 23.6958




WH3/6 =

E M/ EW=


MD=(H3+3xdxH2-D3)/6

FIXING OF LENGTH OF SOLID APRON BASED ON HYDRAULIC JUMP THEORY:

H

Lb=Ld+Lj

0.046

5.052

0.393

1.950

The Length of the Apron after the Jump,Lj = 5x(d2-d1) 7.784

The Total Length of the Apron= Lb = Ld+Lj = 12.836 m say 13.000 m

Rough Stone Apron

Scour Depth, D= 3.30 mLength to be covered on a slope of 2:1= Sqrt(5)xD 7.379

Thickness of Settled stones after Scour=1.24xT 0.567(Assuming side slope thickness as T 1.5 feets i.e 0.4572 metres)

Quantity/m Run= 4.183Length of Launching Apron=1.5D= 4.950 metres

Thickness= 0.845 m or say 0.900 m

d2

d1

Ld Lj

The Drop Number=Dn= q2/(gH3)=

The Length of the Apron before Jump,Ld=Hx4.3x(Dn)0.27

The Jump depths are d1=Hx0.54xDn0.425

d2=Hx1.66x(Dn)0.27

weir design

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