Design of Sarda Fall
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Design of Sarda Fall: Step 1: Total fall ( H w )=U / S ( Full supply level) –D / S( Full supply level ) Step 2: Crestlength of fall( L) =Width of channel Crest widthof fall( B) =0.55 √ H+ W ( For Rectangular fall ) Crest widthof fall ( B) =0.45 √ H+ W ( For Trapezoidal fall ) Where, H=Headabovethecrest <Total fall ( H w ) D=Normal waterdepth ( Full supply) Step 3: Discharge ( Q )=0.415 √ 2 gL×H 3 2 × ¿ Step 4: Velocity of reach ( V )= Q A Step 5: Velocity Head( h a )= V 2 2 g Effective Head( H e )=H + V 2 2 g Step 6: Total energy level( U/ S)=( U / S) Waterlevel + V 2 2 g Step 7: Crestlevel =Totalenergylevel ( U / S) –H Step 8:
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Transcript of Design of Sarda Fall
Design of Sarda Fall:
Step 1:
Total fall (Hw )=U / S(Full supply level) – D / S(Full supply level)
Step 2:Crest lengthof fall ( L )=Widthof channel
Crest width of fall (B )=0.55√H+W (For Rectangular fall )Crest width of fall (B )=0.45√H+W (For Trapezoidal fall )
Where,
H=Head above the crest<Total fall (H w )D=Normal water depth(Full supply)
Step 3:
Discharge (Q)=0.415√2 g L× H32 × ¿
Step 4:
Velocity of reach(V )=QA
Step 5:
Velocity Head (ha)=V 2
2g
Effective Head (H e)=H+ V 2
2g
Step 6:
Total energy level (U / S)=(U /S )Water level+V 2
2g
Step 7:Crest level=Total energy level (U / S)– H
Step 8:
Height of crest above (U / S)bed level (hc)=Crest level – (U /S)Bed level
Step 9:Hydraulic Head (H h)=Crest level – (D /S )Bed level
Step 10:
Total Length (LT )=Hydraulic Head (H h)×Bligh ’ scoefficient (c )
Step 11:Length of cistern(L)=5×√Hw ×H e
Step 12:Depthof cistern(x)=0.25׿
Step 13:Cistern level=(D /S)Bed level – Depth of cistern( x)
Step 14:(U /S)Pakka floor length=Total Length(LT )– Creep Length(Lc )
Step 15:(D / S)Stone pitching length=10 H e+2H w
Step 16:
The (U /S )wingwalls are provided∈acircular shapewithRadius
R=6×H e
Step 17:Length of (D /S )Wingwall=8×√H ×Hw
Example:
Design a Sarda type fall for the following set of data
Full supply discharge=14m3/ sec
Bedwidth=18m
Full supply depth(FSD )=1.5m
Full supply level (U /S)=101.00 m
Full supply level (D /S)=100.00m
(U /S)Bed level=99.5m
(D / S)Bed level=98.5m
Natural surface level=99.5m(D /S )
Bligh’ sCoefficient (c)=8
Solution:
Step 1:
Total fall ( Hw )=U /S ( Full supply level ) – D /S ( Full supply level )¿101– 100=1m
Step 2:
Crest lengthof fall ( L )=Widthof channel
¿m18
Crest width of fall (B)=0.55√H+D(For Rectangular fall )
Where,
H=Head abovethe crest<Total fall (Hw )D=Normal water depth(Full supply)
Assumingsuitable value for (H)e .g .0.7m<(H w)1m
(U /S)Depth(D)=Full supply level (U /S) – (U / S)Bed LevelD=101 – 99.5=1.5m
Crest width of fall (B)=0.55√0.7+1.5=0.815m≈1m
Step 3:
Discharge (Q)=0.415√2 g×L×H32 ׿14=0.415√2×9.81×18×H
32 ׿14=33.09 H
106 H=0.6m
Step 4:
Velocity of reach(V )=QA
A=(18+18+1.5+1.5)
2×1.5=29.25m
V= 1429.25
=0.48 m/ sec
Step 5:
Velocity Head (ha)=V 2
2g= 0.482
2×9.81=0.012m
Effective Head (H e)=H+ V 2
2g=0.6+0.012=0.612m
Step 6:
Total energy level (U / S)=(U /S )Water level+V 2
2g
¿101+0.012=101.012m
Step 7:Crest level=Total energy level (U / S)– H
¿101.012– 0.6=100.41m
Step 8:Height of crest above (U / S)bed level (hc)=Crest level – (U /S)Bed level
¿100.41– 99.5=0.91m
Step 9:Hydraulic Head (H h)=Crest level – (D /S )Bed level
¿100.41– 98.5=1.91m
Step 10:
Total Length(LT )=Hydraulic Head (H h)×Bligh ’ sconstant (c)
¿1.91×8=15.28m
Step 11:Length of cistern(L)=5×√Hw ×H e
¿5×√1×0.612=3.91m
Step 12:Depthof cistern(x)=0.25׿
¿0.25׿
Step 13:Cistern level=(D /S)Bed level – Depth of cistern( x)
¿98.5 – 0.18=98.32m
Step 14:(U /S)Pakka floor length=Total Length(LT )– Creep Length(Lc )
¿15.28 – (1.2+0.3+1+0.3+4+0.2)=8.28m≈8.5m
At end theoretically the impervious floor should be finished butfor safety we extend thisdistance by x .
Step 15:(D / S)Stone pitching length=10 H e+2H w
¿(10×0.612)+(2×1)=8.12≈8.5m
Step 16:
R=6×H e=6×0.612=3.7m
Step 17:
Length of (D /S )wingwalls are kept straight up¿adistance of ,
Length of (D /S )Wingwall=8×√H ×Hw
¿8×√0.6×1=6.2m
