This is a lecture on the hydraulics of gradually varied flow in open channels. It shows the profiles common in the open channels and some numerical examples using numerical integration.

1. Gradually Varied Flow
Computations (GVF)
Amro Elfeki, PhD

2. GVF examples in Open
Channel

3. Water Surface Profiles

4. Let’s evaluate H, total energy, as a function of x.
 
 
2
2
/2
/2
H z y v g
dH d
z y v g
dx dx


  
    
Where H = total energy head
z = elevation head,
v2/2g = velocity head
Take derivative,
Derivation of Equation of GVF
Bernoulli Equation

5. Replace terms for various values of S and
So. - solve for dy/dx, the slope of the water
surface
 
 
2
2/3
2
2
2
2/3
2
2/3
, 1 (SI units) 1.49( .)
.
/2
/2
.
.
m
m
m
o
m
dH nQ
S C or Eng
dx C A R
dH d
z y v g
dx dx
nQ dz dy d
v g
C A R dx dx dx
nQ dy v dv
S
C A R dx g dx



 
    
 
    
 
    
 
   
       
  

6. 2
2
2/3 2
2 2
2/3 3
( )
0
( )
.
.
o
m
o
m
Q AV
dQ d
AV
dx dx
dV dA
A V
dx dx
dA Tdy
dV VT dy QT dy
dx A dx A dx
nQ dy v QT dy
S
C A R dx g A dx
nQ dy Q T dy
S
C A R dx gA dx




 

   
 
     
 
     
         
    

7. 2 2
1 1
2
2/3
2
3
2
3
2
2/3
2
3
2
2/3
.
1
1
.
1
.
o
m
o
m
x y
x y
o
m
nQ
S
C A Rdy
Q Tdx
gA
Q T
gA
dx dy
nQ
S
C A R
Q T
gA
L dx dy
nQ
S
C A R



 
 
 



 
 
 

 
 
 
 
 

8. How to solve this integral?
2 2
1 1
2
3
2
2/3
1
.
x y
x y
o
m
Q T
gA
L dx dy
nQ
S
C A R

 
 
 
 
 

9. Solution with Numerical
Integration
2
1
2
3
2
2/3
12
1
( )
.
( ) Area under the curve
o
m
y
y
Q T
gA
f y
nQ
S
C A R
L f y dy


 
 
 
 
f(y)
y
y
1
y
2
y
3
h h
f0 f1 f3 f(y)

10. f(y)
y
y
1
y
2
y
3
h h
f0 f1 f3 f(y)
(I) Divide integration region into strips of width h
(II) Approximate curve over each strip by a straight line
(III) Sum areas of resulting trapezia - n strips means (n+1) function
evaluations
   1 0 1 1
1 1,...,
2 2n n nA f f h A f f h   
   
0
0 1 1
1
1
2 ... 2
2
ny n
i n n
iy
f y dy A h f f f f

     
Trapezoidal Rule

11. f
y
yi-1 yi yi+1
h h
fi-1 fi fi+1 f(y)
A B C
(I) Divide integral region into an even number of strips, each of width h
(II) Approximate curve over two adjacent strips by a quadratic-
interpolation through A, B, C
       
2
1 1 1 12
1 1
2
2 2
i i i i i i i if y f f f y y f f f y y
h h
          
(III) Find area underneath interpolating curve
     
3
1 1 1 12
1 2
2 0 2 4
2 3 3
i
i
y h
i i i i i i i
y h
h h
f y dy f h f f f f f f
h

   

       
Simpson’s Rule

12. (IV) Sum:
 ...yyyyyy
h
A  432210 44
3
 nnn yyy...yyyy
h
  123210 42424
3
Simpson’s Rule (Cont’d)

13. Analytical Solution in Special Case
(Horizontal Channel of Great Width)
2
1
2
3
2 2
10
2 3
for 0 and unit width:
1, , ,
1
o
y
y
m
S
T Q q A y R y
q
gy
L dy
n q
C y

   

 
 
 
 
 
 


14. Special Case (cont’d)
 
2
1
10
3
13 13
4 43 3
3 3
2
3
2 2
2
2 2
2 1 2 1
1
3 3
13 4
1 for metric units,
1.49 for English units,
y
y
m
m m
m
m
q
gy
L dy
n q
C y
C C
L y y y y
nq g n
C
C

 
 
  
 
    
        
    




16. Exercise
After contracting a flow below a sluice gate, water
flows onto a wide horizontal floor with a velocity
of 15 m/s and a depth of 0.7 m. Find:
1. The equation for the surface water profile,
n=0.015 [x=f(y)] up to 0.9 of the critical depth.
2. Calculate the length of the curve from the gate
till 0.9 of the critical depth analytically.
3. Draw the y-f(y) curve.
4. Calculate the length of the curve using numerical
integration (Trapezoidal and Simpson rules)
(take number of strips 5 and 10). Check with
analytical method.
5. Draw the water surface profile from the gate to
0.9 of the critical depth and plot the critical and
the normal depth.

17. Exercise (Cont’d)

18. Solution
2
3
2
2
3
2
14 * 0.75
10.5
Q
yc
gb
q
yc
g
q vy
q
q m s






19. Solution Cont.
2
3
(10.5)
2.24
9.81
0.9 2.02
yc m
yc m
 


20. Analytical Solution
 
13 13
4 43 3
3 3
2 2
2 1 2 1
2 213 13 4 4
3 3 3 3
3 3
13 4
3 1 3 1
2.02 0.75 2.02 0.75
13 0.015*10.5 4*9.81 0.015
442.36
m mC C
L y y y y
nq g n
L
L m
    
        
    
      
          
      


21. Solution by Numerical
Integration
2
1
10
3
2
3
2 2
2
( )
1
( )
y
y
m
L f y dy
q
gy
f y
n q
C y


 
 
  
 


22. y vs f(y) with 6 steps
150
200
250
300
350
400
450
0.5 0.75 1 1.25 1.5 1.75 2 2.25
f(y)
y
y-f(y) Curve
Index # y f(y) Length (m)
Cumulative
Length (m)
1 0.75 396.5937802 0 0
2 0.96 412.2380141 85.37436193 85.37436193
3 1.17 409.7187839 86.75974123 172.1341032
4 1.38 386.4198588 84.0345658 256.168669
5 1.59 338.9271114 76.56231518 332.7309841
6 1.81 263.3125597 63.56801008 396.2989942
7 2.02 155.2754959 44.18309025 440.4820845

23. WS Profile
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
0 50 100 150 200 250 300 350 400 450
Depth(m)
Distance/Length of the Flow (m)
Water Surface Profile from gate upto 0.9Yc
Water Surface
Critical Depth

24. Final Answer
TRAPEZOIDAL Rule SIMPSON’S Rule
440.48 443.11
Analytical Solution
443.12

25. y vs f(y) with 10 steps
150
200
250
300
350
400
450
0.5 0.75 1 1.25 1.5 1.75 2 2.25
f(y)
y
y-f(y) Curve
Index # y f(y) Length (m)
Cumulative Length
(m)
1 0.75 396.5937802 0 0
2 0.88 408.0514075 50.95947263 50.95947263
3 1.00 413.2558976 52.01471131 102.9741839
4 1.13 411.7872209 52.25130638 155.2254903
5 1.26 403.0701328 51.60622552 206.8317158
6 1.38 386.4198588 49.99966972 256.8313856
7 1.51 361.0693869 47.33969499 304.1710805
8 1.64 326.1868273 43.52504033 347.6961209
9 1.76 280.8870382 38.44696334 386.1430842
10 1.89 224.239756 31.99049151 418.1335757
11 2.02 155.2754959 24.03531071 442.1688864

26. WS Profile
0
0.25
0.5
0.75
1
1.25
1.5
1.75
2
2.25
2.5
0 50 100 150 200 250 300 350 400 450
Depth(m)
Distance/Length of the Flow (m)
Water Surface Profile from gate upto 0.9Yc
Water Surface

27. Final Answer
TRAPEZOIDAL Rule SIMPSON’S Rule
442.17 443.12
Analytical Solution
443.12