Spillway Discharge

8/11/2019 Spillway Discharge

1/8

x y

0 0

2149 131.09

15729 152.4

x y

0 0

2149 131.09

15723 152.4

20886 157.01

x y z

0 0 0

2149 131.09 131.09

15723 152.4 152.4

20886 157.01 157.01

0

20

40

60

80

100

120

140

160

180

0 5000

Elevationinm

0

20

40

60

80

100

120

140

160

180

0 5000 10000

Elevationinm

capacity in h

0

20

40

60

80

100

120

140

160

180

0 5000 10000 15000 2

c

a

p

a

c

i

t

y

Elevation

chandan


2/8


3/8

19.5 5.95 576.1361 2680 482.4 93.73609 22692.47 158.3 "

20 5.9 568.8892 2580 464.4 104.4892 22587.98 158.28 "

20.5 5.88 565.9989 2420 435.6 130.3989 22457.59 158.2 "

21 5.8 554.4873 2340 421.2 133.2873 22324.3 158.1 "

21.5 5.7 540.2091 2160 388.8 151.4091 22172.89 157.9 "

22 5.5 512.028 2120 381.6 130.428 22042.46 157.8 "

22.5 5.4 498.1272 1960 352.8 145.3272 21897.13 157.75 "

23 5.35 491.2248 1840 331.2 160.0248 21737.11 157.7 "23.5 5.3 484.3546 1700 306 178.3546 21558.75 157.6 "

24 5.2 470.7113 1640 295.2 175.5113 21383.24 157.4 "

24.5 5 443.8177 1560 280.8 163.0177 21220.23 157.3 "

25 4.9 430.57 1440 259.2 171.37 21048.86 157.2 "

25.5 4.8 417.4568 1420 255.6 161.8568 20887 157.1 "

26 4.7 404.4794 1400 252 152.4794 20734.52 157 "

And so on.

Now the reservoir level is on decreasing order.

The maximum storage level attained by the reservoir is 158.85 m

Above caslculations have been done in following ways:-1. The time interval is 0.5 hour and opening interval of height is 0.5 m

2. In orifice flow condition,outflow has been calculated by the orifice formula=Q=

=Cd xa x L1 X no. of gates x(2gH)

Where Cd=0.65, L1=effective length of one gate as given by consultant,

no. of gates =11 , H= effective head causing flow

3. In free low condition, out flow has been calculated by the formula,

Q= C X LX H^1.5=1.48 X149.01 XH^1.5

Where C=1.48, L=effective length of water ways as given by consultant= 149.01 m

H= Head causing the flow

For example, taking the time of 1 hour and head causing flow= 1 m

Q= 0.65 X1 mx13.52 m x(157.0-152.40-1.0) x11 x1800/10000=146.2366 ha-m

Next example, taking time of 4.5 Hour and head causing flow =3.6 m

Q=1.48 X149.01 X((156.0-152.40)^1.5)*1800/10000=271.1461 hect.-m

the clarification of the formula used and Cd value have been given on another place.

XXXXXX


4/8

ASSESSMENT OF Cd VALUE FOR THE DISCHARGE CALCULATION

1Discharge Co-efficient of various weirs are function of its geometry,viscosity and surface

tension. However ,effect of viscocity and surface tension are secondary for large heads.

Table 5.1 of Garde gives the discharge co-efficient of sharp edged orifice of rectangular

shape as 0.622 to 0.626 depending upon u/s slope as below:-

Cd for rectangular shape (2:1)= 0.624

Cd for rectangular shape (3:1)= 0.635 In case of Chandan ,it is 2.25:1

Cd for rectangular shape (4:1)= 0.622

Cd for rectangular shape (1.5:1)= 0.626

2.Garde again has given equation (5.17a) to find the Cd value for sharp crested weir as :-

Cd = 0.611+0.075H/W, where H= head over crest

W= Crest heightIn case of Chandan, Cd=0.611+0.075(157.01-152.40)/(152.4-149.25)

=0.611+0.075 x 4.61/3.15=0.611+0.109=0.72

Ii is remarkable that all the above values of Cd are for Sharp crested weir and free flow

conditions. , Here the weir is behaving like broad crested weir and old spillway

is obstructing the flow creating case of submergence. So its value should be around 0.62.

3. Third clarification has been given by Garde in fig 5.20 (Page-122), which shows the

variation of Cd with head causing flow and opening height of gate .It shows that maxi.

value of Cd is 0.60 for a sharp crested weir (fi. 5.20 attached)

Although here we are taking the value of Cd as 0.65 which is in higher side .

4. Further expression for Cd value is in terms of C, and the value of C , taking all other

constants , has been given by RUNKAN KASAUTI, Water Resources Department, Gov.

of Bihar,as follows :-

Weir MKS FPS

For Broad crested weir Q=1.7L H^1.5 Q=3.1L H 1.5

For sharp crested weir Q=1.84L H^1.5 Q=3.3 L H 1.5

This expression is for free flow condition. But here it is the case of submergence ,

for which a correction is needed.

For the present case , IS Code-6966 -1-1966 has given the exact solution for Cd value in

terms of Drowning Ratio ( D.R.) considering the case of submergence.

D.R.=(D/S WATER LEVEL-CREST LEVEL)/( U/S WATER LEVEL-CREST LEVEL)

The code gives the maximum value of Cd as 1.8 and minimum value as 0.9. Taking the

value of D.R. , we can calculate the value of C.Taking head loss = ha=0.5 m , D.R.=(157.01-152.4)/(157.01-0.5-152.4)=4.11/4.61=0.90

For D.R. Value=0.9 or 90 % submergence, C=1.48

The consultant has taken the C value for Ogee type and sharp crested

weir . That is not the case here. Given type weir is barrage type and will be guided by

the IS-CODE- 6966 -1-1996 and C value will be 1.48 for the present case.


5/8

Period/

time inhrs

Head

causingflow

Outflow-

Hect-m in0.5 hr.

Cumulativ

e.inflowcumec

Inflow in

hec-m in0.5 hour

Net

outflow in

hec-

mCol.3-col.5

Residual

Storage inhec-m

ResrvoirLevel in m Remarks

1 2 3 4 5 6 7 8 9

0 0 0 0 0 0 20886.55 157.01

Orifice

flow

condi.

0.5 0.5 78.70387 60 10.8 67.90387 20818.65 157 "

1 1 146.2366 120 21.6 124.6366 20694.01 156.9 "

1.5 1.5 198.7616 200 36 162.7616 20531.25 156.8 "

2 2 235.2707 280 50.4 184.8707 20346.38 156.5 "

2.5 2.5 238.3195 380 68.4 169.9195 20176.46 156.4 "

3 3 224.3795 500 90 134.3795 20042.08 156.2 "3.5 3.5 142.1785 580 104.4 37.77853 20004.3 156.15

4 3.75 288.2681 640 115.2 173.0681 19831.23 156

Free flow

condition

4.5 3.6 271.1461 800 144 127.1461 19704.09 155.9 '

5 3.5 259.9272 980 176.4 83.52718 19620.56 155.85 '

5.5 3.45 254.3773 1080 194.4 59.97726 19560.58 155.8 '

6 3.4 248.8674 1200 216 32.8674 19527.71 155.75 '

6.5 3.35 243.3979 1440 259.2 -15.8021 19543.52 155.8 '

7 3.4 248.8674 1660 298.8 -49.9326 19593.45 155.85 '

7.5 3.45 254.3773 1900 342 -87.6227 19681.07 155.9 '

8 3.5 259.9272 2180 392.4 -132.473 19813.54 156 '

8.5 3.6 271.1461 2560 460.8 -189.654 20003.2 156.2 '

9 3.8 294.0526 2700 486 -191.947 20195.15 156.4 '

9.5 4 317.5701 3000 540 -222.43 20417.58 156.6 '

10 4.2 341.6832 3340 601.2 -259.517 20677.09 156.80 '

10.5 4.4 366.3774 4000 720 -353.623 21030.71 157 '

11 4.6 391.6394 4400 792 -400.361 21431.08 157.4 '

11.5 5 443.8177 4500 810 -366.182 21797.26 157.8 '

12 5.4 498.1272 4531 815.58 -317.453 22114.71 158 '

12.5 5.6 526.0556 4440 799.2 -273.144 22387.85 158.2 '

13 5.8 554.4873 4300 774 -219.513 22607.37 158.3 '13.5 5.9 568.8892 4100 738 -169.111 22776.48 158.4 '

14 6 583.4135 4000 720 -136.586 22913.06 158.6 '

14.5 6.2 612.826 3880 698.4 -85.574 22998.64 158.7 '

15 6.3 627.712 3780 680.4 -52.688 23051.33 158.75 '

15.5 6.35 635.1996 3660 658.8 -23.6004 23074.93 158.8 '

16 6.4 642.7167 3600 648 -5.28328 23080.21 158.85 Max Elev.

16.5 6.45 650.2632 3400 612 38.26325 23041.95 158.8 Free flow

17 6.4 642.7167 3380 608.4 34.31672 23007.63 158.6 "

ROUTING CHART FOR HALF HOURLY INFLOW OF PMF-160000 CUSEC

AND FOR HALF HOURLY OUTFLOW FOR GATE OPENING OF HALF METRE


6/8


7/8

10000 15000 20000

capacity in hect.m

Series1

15000 20000 25000 ect.m

Series1

2 per. Mov. Avg. (Series1)

Log. (Series1)

000 25000

Series1

Series2


8/8

Spillway Discharge

Documents

Transcript of Spillway Discharge