DESIGN OF WELL TYPE FALL AT 6110M OF LEFT CANAL

OF 3 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 3m
2 General ground level 287.250 m
3 Full supply depth 0.5 m
4 Upstream CBL 287.711
5 Discharge 14.23 cusec
Q 0.40 cumec
Bed width upstream
6 1.2 m
and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.40 = 2.22*(0.5)^(3/2)*(l+0.4*n*0.5)
0.51 = l+0.2*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.33 m
Q' = 0.20 cumec

0.20 = 2.22*(0.33)^(3/2)*(l+0.4*n*0.33)
0.48 = l+0.132*n ……… ii)

Subtracting (ii) from (i), we get

0.03 = 0.068 n
n = 0.51

2 tan (α/2) = 0.51

α/2 = 14.3

Substituting this value of n in ii), we get

L = 0.37
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.4 m
bottom width and each side inclined to an angle of 14.3° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.63 m Say 0.7

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.60 m/s

Let us now assume

the diameter of the 1.1
pipe used = m

Velocity V3 through the pipe

= 0.4/(3.14/4*1.1^2)
= 0.42 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.4/(3.14/4*0.4^2)
= 3.18 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.4/(3.14/4*0.4^2)^2)/(2*9.81)] + [(0.4/
HL1 = (3.14/4*0.4^2)-0.42)^2/ (2*9.81)] +
[(0.012*10*(0.42)^2)/(2*9.81*1)] + [(0.42)^2/
(2*9.81)]
= 0.66

R.L. of water surface in the inlet well

= d/s FSL + 0.66
= 285.871
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 287.711+0.5/3
= 287.88

Height Y of CP above water level in the inlet well

= 2.01 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 1.02 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.53 m

Keep the dia of the d/s outlet well

= 1.13 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining

Scour Depth
q= 0.33 m
f= 1
Scour Depth R = 1.35*(q2/f)1/3 0.65 m
Depth Below bed 0.15 m
DESIGN OF WELL TYPE FALL AT 6110M OF LEFT CANAL
OF 3 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 3m
2 General ground level 283.815 m
3 Full supply depth 0.5 m
4 Upstream CBL 284.535
5 Discharge 14.23 cusec
Q 0.40 cumec
Bed width upstream
6 ssss m
and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.40 = 2.22*(0.5)^(3/2)*(l+0.4*n*0.5)
0.51 = l+0.2*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.33 m
Q' = 0.20 cumec

0.20 = 2.22*(0.33)^(3/2)*(l+0.4*n*0.33)
0.48 = l+0.132*n ……… ii)

Subtracting (ii) from (i), we get

0.03 = 0.068 n
n = 0.51

2 tan (α/2) = 0.51

α/2 = 14.3

Substituting this value of n in ii), we get

L = 0.37
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.4 m
bottom width and each side inclined to an angle of 14.3° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.63 m Say 0.7

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.60 m/s

Let us now assume

the diameter of the 1.1
pipe used = m

Velocity V3 through the pipe

= 0.4/(3.14/4*1.1^2)
= 0.42 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.4/(3.14/4*0.4^2)
= 3.18 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.4/(3.14/4*0.4^2)^2)/(2*9.81)] + [(0.4/
HL1 = (3.14/4*0.4^2)-0.42)^2/ (2*9.81)] +
[(0.012*10*(0.42)^2)/(2*9.81*1)] + [(0.42)^2/
(2*9.81)]
= 0.66

R.L. of water surface in the inlet well

= d/s FSL + 0.66
= 282.695
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 284.535+0.5/3
= 284.70

Height Y of CP above water level in the inlet well

= 2.01 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 1.02 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.53 m

Keep the dia of the d/s outlet well

= 1.13 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining

Scour Depth
q= #VALUE! m
f= 1
Scour Depth R = 1.35*(q2/f)1/3 #VALUE! m
Depth Below bed #VALUE! m
DESIGN OF WELL TYPE FALL AT 6925 M OF LEFT CANAL
OF 3 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 3m
2 General ground level 281.810 m
3 Full supply depth 0.5 m
4 Upstream CBL 281.507
5 Discharge 14.23 cusec
Q 0.40 cumec
Bed width upstream
6 1.2 m
and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.40 = 2.22*(0.5)^(3/2)*(l+0.4*n*0.5)
0.51 = l+0.2*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.33 m
Q' = 0.20 cumec

0.20 = 2.22*(0.33)^(3/2)*(l+0.4*n*0.33)
0.48 = l+0.132*n ……… ii)

Subtracting (ii) from (i), we get

0.03 = 0.068 n
n = 0.51

2 tan (α/2) = 0.51

α/2 = 14.3

Substituting this value of n in ii), we get

L = 0.37
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.4 m
bottom width and each side inclined to an angle of 14.3° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.63 m Say 0.7

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.60 m/s

Let us now assume

the diameter of the 1.1
pipe used = m

Velocity V3 through the pipe

= 0.4/(3.14/4*1.1^2)
= 0.42 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.4/(3.14/4*0.4^2)
= 3.18 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.4/(3.14/4*0.4^2)^2)/(2*9.81)] + [(0.4/
HL1 = (3.14/4*0.4^2)-0.42)^2/ (2*9.81)] +
[(0.012*10*(0.42)^2)/(2*9.81*1)] + [(0.42)^2/
(2*9.81)]
= 0.66

R.L. of water surface in the inlet well

= d/s FSL + 0.66
= 279.667
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 281.507+0.5/3
= 281.67

Height Y of CP above water level in the inlet well

= 2.01 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 1.02 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.53 m

Keep the dia of the d/s outlet well

= 1.13 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining
DESIGN OF WELL TYPE FALL AT 6925 M OF LEFT CANAL
OF 3 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 3m
2 General ground level 278.730 m
3 Full supply depth 0.5 m
4 Upstream CBL 278.483
5 Discharge 14.23 cusec
Q 0.40 cumec
Bed width upstream
6 1.2 m
and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.40 = 2.22*(0.5)^(3/2)*(l+0.4*n*0.5)
0.51 = l+0.2*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.33 m
Q' = 0.20 cumec

0.20 = 2.22*(0.33)^(3/2)*(l+0.4*n*0.33)
0.48 = l+0.132*n ……… ii)

Subtracting (ii) from (i), we get

0.03 = 0.068 n
n = 0.51

2 tan (α/2) = 0.51

α/2 = 14.3

Substituting this value of n in ii), we get

L = 0.37
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.4 m
bottom width and each side inclined to an angle of 14.3° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.63 m Say 0.7

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.60 m/s

Let us now assume

the diameter of the 1.1
pipe used = m
54
Velocity V3 through the pipe
= 0.4/(3.14/4*1.1^2)
= 0.42 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.4/(3.14/4*0.4^2)
= 3.18 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.4/(3.14/4*0.4^2)^2)/(2*9.81)] + [(0.4/
HL1 = (3.14/4*0.4^2)-0.42)^2/ (2*9.81)] +
[(0.012*10*(0.42)^2)/(2*9.81*1)] + [(0.42)^2/
(2*9.81)]
= 0.66

R.L. of water surface in the inlet well

= d/s FSL + 0.66
= 276.643
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 278.483+0.5/3
= 278.65

Height Y of CP above water level in the inlet well

= 2.01 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 1.02 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.53 m

Keep the dia of the d/s outlet well

= 1.13 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining
DESIGN OF WELL TYPE FALL AT 6632 M OF RIGHT CANAL
OF 2 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 2m
2 General ground level 288.500 m
3 Full supply depth 0.49 m
4 Upstream CBL 288.534
5 Discharge 14.82 cusec
Q 0.42 cumec
Bed width upstream
6 1.25 m
and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.42 = 2.22*(0.49)^(3/2)*(l+0.4*n*0.49)
0.55 = l+0.196*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.32 m
Q' = 0.21 cumec

0.21 = 2.22*(0.32)^(3/2)*(l+0.4*n*0.32)
0.52 = l+0.128*n ……… ii)

Subtracting (ii) from (i), we get

0.03 = 0.068 n
n = 0.43

2 tan (α/2) = 0.43

α/2 = 12.13

Substituting this value of n in ii), we get

L = 0.44 SAY 0.5
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.5 m
bottom width and each side inclined to an angle of 12.13° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.65 m Say 0.7

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.58 m/s

Let us now assume

the diameter of the 1.1
pipe used = m
54
Velocity V3 through the pipe
= 0.42/(3.14/4*1.1^2)
= 0.44 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.42/(3.14/4*0.4^2)
= 3.34 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.42/(3.14/4*0.4^2)^2)/(2*9.81)] +
HL1 = [(0.42/(3.14/4*0.4^2)-0.44)^2/ (2*9.81)] +
[(0.012*10*(0.44)^2)/(2*9.81*1)] + [(0.44)^2/
(2*9.81)]
= 0.72

R.L. of water surface in the inlet well

= d/s FSL + 0.72
= 287.744
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 288.534+0.49/3
= 288.70

Height Y of CP above water level in the inlet well

= 0.95 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 0.70 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.05 m

Keep the dia of the d/s outlet well

= 0.65 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining
DESIGN OF WELL TYPE FALL AT 6870 M OF RIGHT CANAL
OF 3 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 3m
2 General ground level 286.500 m
3 Full supply depth 0.49 m
4 Upstream CBL 286.466
5 Discharge 14.82 cusec
Q 0.42 cumec
Bed width upstream
6 1.25 m
and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.42 = 2.22*(0.49)^(3/2)*(l+0.4*n*0.49)
0.55 = l+0.196*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.32 m
Q' = 0.21 cumec

0.21 = 2.22*(0.32)^(3/2)*(l+0.4*n*0.32)
0.52 = l+0.128*n ……… ii)

Subtracting (ii) from (i), we get

0.03 = 0.068 n
n = 0.43

2 tan (α/2) = 0.43

α/2 = 12.13

Substituting this value of n in ii), we get

L = 0.44 SAY 0.5
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.5 m
bottom width and each side inclined to an angle of 12.13° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.65 m Say 0.7

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.58 m/s

Let us now assume

the diameter of the 1.1
pipe used = m
54
Velocity V3 through the pipe
= 0.42/(3.14/4*1.1^2)
= 0.44 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.42/(3.14/4*0.4^2)
= 3.34 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.42/(3.14/4*0.4^2)^2)/(2*9.81)] +
HL1 = [(0.42/(3.14/4*0.4^2)-0.44)^2/ (2*9.81)] +
[(0.012*10*(0.44)^2)/(2*9.81*1)] + [(0.44)^2/
(2*9.81)]
= 0.72

R.L. of water surface in the inlet well

= d/s FSL + 0.72
= 284.676
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 286.466+0.49/3
= 286.63

Height Y of CP above water level in the inlet well

= 1.95 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 1.00 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.50 m

Keep the dia of the d/s outlet well

= 1.10 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining
DESIGN OF WELL TYPE FALL AT 8415 M OF RIGHT CANAL
OF 3 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 3m
2 General ground level 281.010 m
3 Full supply depth 0.49 m
4 Upstream CBL 283.023
5 Discharge 14.82 cusec
Q 0.42 cumec
Bed width upstream
6 1.25 m
and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.42 = 2.22*(0.49)^(3/2)*(l+0.4*n*0.49)
0.55 = l+0.196*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.32 m
Q' = 0.21 cumec

0.21 = 2.22*(0.32)^(3/2)*(l+0.4*n*0.32)
0.52 = l+0.128*n ……… ii)

Subtracting (ii) from (i), we get

0.03 = 0.068 n
n = 0.43

2 tan (α/2) = 0.43

α/2 = 12.13

Substituting this value of n in ii), we get

L = 0.44 SAY 0.5
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.5 m
bottom width and each side inclined to an angle of 12.13° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.65 m Say 0.7

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.58 m/s

Let us now assume

the diameter of the 1.1
pipe used = m
54
Velocity V3 through the pipe
= 0.42/(3.14/4*1.1^2)
= 0.44 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.42/(3.14/4*0.4^2)
= 3.34 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.42/(3.14/4*0.4^2)^2)/(2*9.81)] +
HL1 = [(0.42/(3.14/4*0.4^2)-0.44)^2/ (2*9.81)] +
[(0.012*10*(0.44)^2)/(2*9.81*1)] + [(0.44)^2/
(2*9.81)]
= 0.72

R.L. of water surface in the inlet well

= d/s FSL + 0.72
= 281.233
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 283.023+0.49/3
= 283.19

Height Y of CP above water level in the inlet well

= 1.95 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 1.00 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.50 m

Keep the dia of the d/s outlet well

= 1.10 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining
DESIGN OF WELL TYPE FALL AT 9610 M OF RIGHT CANAL
OF 3 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 3m
2 General ground level 280.030 m
3 Full supply depth 0.49 m
4 Upstream CBL 279.683
5 Discharge 14.82 cusec
Q 0.42 cumec
Bed width upstream
6 1.25 m
and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.42 = 2.22*(0.49)^(3/2)*(l+0.4*n*0.49)
0.55 = l+0.196*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.32 m
Q' = 0.21 cumec

0.21 = 2.22*(0.32)^(3/2)*(l+0.4*n*0.32)
0.52 = l+0.128*n ……… ii)

Subtracting (ii) from (i), we get

0.03 = 0.068 n
n = 0.43

2 tan (α/2) = 0.43

α/2 = 12.13

Substituting this value of n in ii), we get

L = 0.44 SAY 0.5
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.5 m
bottom width and each side inclined to an angle of 12.13° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.65 m Say 0.7

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.58 m/s

Let us now assume

the diameter of the 1.1
pipe used = m
54
Velocity V3 through the pipe
= 0.42/(3.14/4*1.1^2)
= 0.44 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.42/(3.14/4*0.4^2)
= 3.34 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.42/(3.14/4*0.4^2)^2)/(2*9.81)] +
HL1 = [(0.42/(3.14/4*0.4^2)-0.44)^2/ (2*9.81)] +
[(0.012*10*(0.44)^2)/(2*9.81*1)] + [(0.44)^2/
(2*9.81)]
= 0.72

R.L. of water surface in the inlet well

= d/s FSL + 0.72
= 277.893
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 279.683+0.49/3
= 279.85

Height Y of CP above water level in the inlet well

= 1.95 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 1.00 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.50 m

Keep the dia of the d/s outlet well

= 1.10 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining
 DESIGN OF WELL TYPE FALL AT 4350M OF BUDHUA CANAL
OF 4 M HEIGHT

(Reference : Irrigation Engineering & Hydraulic Structures by SK Garg, Page 650-652)

DATA
1 Height of fall 4.5 m
2 General ground level 278.625 m
3 Full supply depth 0.6 m
4 Upstream CBL 279.093
5 Discharge 20 cusec
Q 0.57 cumec

6 Bed width upstream 2m

and downstream

DESIGN

For a trapezoidal notch, we have the discharge eqn. (12.4) as

Q=2.22*H^(3/2) * (l + 0.4nH)

At full supply discharge, we have

0.57 = 2.22*(0.6)^(3/2)*(l+0.4*n*0.6)
0.55 = l+0.24*n ………….. i)

At 50% full discharge, we have

Q'=2.22*H'^(3/2) * (l + 0.4nH')
where
H' = 0.66*FSD m
= 0.40 m
Q' = 0.29 cumec

0.29 = 2.22*(0.4)^(3/2)*(l+0.4*n*0.4)
0.51 = l+0.16*n ……… ii)

Subtracting (ii) from (i), we get

0.04 = 0.08 n
n = 0.56

2 tan (α/2) = 0.56

tan (α/2) 0.28 0.28
α/2 = 7.5

Substituting this value of n in ii), we get

L = 0.39
 Hence, provide a trapezoidal notch in the steining of the inlet well, with 0.4 m
bottom width and each side inclined to an angle of 7.5° with the vertical.

Now, the width of water (at FSL) flowing over notch

= 0.73 m Say 0.5

Velocity (V1) over the notch

= FSQ
Area of flow over notch

= 1.69 m/s

Let us now assume

the diameter of the
pipe used = 1.1 m

Velocity V3 through the pipe

= 0.57/(3.14/4*1.1^2)
= 0.60 m/s

Let us now assume that

the diameter of the opening at the inlet of pipe
= 0.4 m

The velocity of entry into the pipe (V2)

= 0.57/(3.14/4*0.4^2)
= 4.54 m/s

Now
Loss of head between the inlet well and the d/s FSL is
= [0.5*(V2^2)/(2g)] + [(V2-V3)^2/(2g)] + [(f'L(V3)^2)/(2gd)] + [(V3)^2/ (2g)]

Let us assume
Length of pipe = 10 m
f' (Darcey's coeff of
friction) = 0.012 m

[0.5*(0.57/(3.14/4*0.4^2)^2)/(2*9.81)] +
HL1 = [(0.57/(3.14/4*0.4^2)-0.6)^2/ (2*9.81)] +
[(0.012*10*(0.6)^2)/(2*9.81*1)] + [(0.6)^2/
(2*9.81)]
= 1.34

R.L. of water surface in the inlet well

= d/s FSL + 1.34
= 276.533
Approximate RL of centre of pressure (CP) of the trapezoidal waterway through
notch
= u/s canal bed level + FSD/3
= 279.093+0.6/3
= 279.29

Height Y of CP above water level in the inlet well

= 2.76 m

Now using Eqn., we have

V1 = √(g*X^2/(2*Y))
X = √(V1^2*2*Y/g)
= 1.26 m

Now the dia of the inlet well may be kept at about 1.5X
i.e., X = 1.90 m

Keep the dia of the d/s outlet well

= 1.50 m

Also provide a water cushion at the bottom of the inlet well.

Bed and sides of channel for suitable lengths on the u/s as well as d/s side are
protected by CC Lining

Scour Depth
q= 0.29 m
f= 0.9
Scour Depth (R) = 1.35*(q2/f)1/3 0.61 m
Depth Below bed 1.5R-d 0.32 m