exercise (4): open channel flow - gradually varied flo (4): open channel flow - gradually varied...

CAIRO UNIVERSITY HYDRAULICS

Faculty of Engineering 3rd

year Civil Eng.

Irrigation & Hydraulics Department 2011 – 2012

1

Exercise (4): Open Channel Flow - Gradually Varied Flow

1) A wide channel consists of three long reaches and has two gates located midway

of the first and last reaches. The bed slopes for the three reaches are S1 = 0.008,

S2= 0.00309, and S3 = 0.0005. The water discharges to the channel from a lake

where the water surface is higher than the normal depth at the inlet. If the

discharge is 0.675 m3/sec/m, and n = 0.015, sketch the possible water surface

profiles in the channel for the following cases:

a) The last reach terminates in a sudden fall.

b) The last reach discharges to a lake of water level higher than the normal depth

at the exit.

Also trace the variation in water depth through both gates on the S.E.D.

2) Water flows with constant discharge q=1.0 m3/s/m into a wide rectangular channel

that consists of two very long reaches where the bed slope changes from

So1=6.87x10-3 to So2=3.24x10-4, Calculate:

a) The depth of both uniform and critical flow in both reaches (take n=0.018)

b) Draw a neat sketch of the water profile at the transition zone and calculate the

water depths and the head losses wherever appropriate.

3) A barrage is constructed across a wide river whose discharge is 6 m3/sec/m’, and bed

slope is 10 cm/km. If the afflux produced at the site of the barrage is 3 m, find the

length of the water surface profile produced from the site of the barrage till a point

where the water depth is 6.25m. Use the direct step method considering 3 points,

Chezy coefficient C = 50.

4) A trapezoidal channel of bed width of 7 m, side slopes 3:2, Manning coefficient

n of 0.02 is laid on a slope of 0.001 and carries a discharge of 30m3/sec. The

channel terminates to a free fall. It is required to compute and plot the water

surface profile from downstream to a water depth of 0.9 the normal depth. Use the

direct step method (use 3 steps).

5) A steep long channel takes its water from a lake. Prove that the discharge per unit

width in the channel is given by:

2

3

27

8Hgq

Two lakes are joined by a wide concrete channel A-B as shown in the figure below, n

= 0.02. The water levels in the lakes are constant.

a) Find the flow rate into the channel for each of the following bed slopes:



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2.0m

A

B n = 0.02, Bed Slope = 0.01

1.5m

i. So = 10 cm/km

ii. So = 0.003924

iii. So = 0.01

Compare your results and explain how the flow depth at A affects the flowing

discharge.

b) Sketch the Water surface profile for the above cases and calculate the water

depths wherever appropriate.

c) Does a hydraulic jump occur? If so, How far upstream point B does it occur?

Use the direct step method, considering 5 points.

d) If a gate is located at point A (Cc=0.65 and Cd=0.6) to control the discharge

into the channel as in the figure below. Find the flow rate into the channel for

the following gate openings (So=0.01):

i. d = 0.5m

ii. d = 0.8m

iii. d = 1.2m

2.0m

A

B n = 0.02, Bed Slope = So

1.5m



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6) A 10 m wide, rectangular concrete-lined canal (n = 0.015) has a bottom slope of 0.01

and a constant level lake at the upstream end. The water level in the lake is 6m above

the bottom of the canal at the entrance. If the entrance losses are negligible,

determine:

a) The flow depth 600 m downstream of the channel entrance, use the improved

Euler method.

b) The distance from the lake where the flow depth is 3.0 m. Use the direct step

method.

7) The figure below shows a longitudinal section in a channel of wide cross section.

Manning’s n = 0.02 and the bed slope is 0.001.

A gate is located at point A. The discharge equation for the gate in case of free

outflow is given by:

)(2 dCHgdCq cd

and for a submerged outflow is:

)(2 yHgdCq d

Where q is the discharge per unit width (m2/sec), Cd is the discharge coefficient and

is equal to 0.6, d is the height of the gate opening (m), H is the water depth just

upstream of the gate (m), y is the water depth downstream the gate (m), and g is the

gravitational acceleration (m/sec2). The coefficient of contraction of the gate Cc =

0.63.

A B

So= 1/1000, n = 0.02

q



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a) If q = 3 m3/sec/m’, and H = 3m, find the gate opening d, and check if a free

hydraulic jump will form downstream of the gate. Assume the channel

downstream A is long enough to allow for uniform flow to develop. If a jump will

form, how far downstream of A will it be? (Use an appropriate method to

calculate the distance)

b) Point B is located 600m upstream of the gate. Find the water depth at B. Use the

improved Euler method considering two steps. If the bed level at A = 20.75m,

find the water level at B.

c) The gate at A is used to maintain a constant water level at B for different values

of discharge. What is the required gate opening to keep the water level at B

unchanged, for a low flow of 1 m3/sec/m’? Use the improved Euler method with

two steps.

8) A trapezoidal channel having bottom slope 0.001 is carrying a flow of 30m3/s. The

bottom width is 10.0m and side slope 2H to 1V. A control structure is built at the

downstream end which raises depth at the downstream end to 5.0m. Compute and

draw the water surface profile. Manning n is 0.013.



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Problems to be solved by computer:

Water flows in a rectangular channel of 10m bed width, n = 0.018, at a flow rate of 15m3/sec.

Create an Excel Spread Sheet to plot to a suitable horizontal and vertical scale the bed slope,

water surface profile, and the total energy line (TEL) between points A and B for the following

cases (water flows at point A with uniform flow depth).

In order to perform the required plots, calculate the bed elevation, water surface elevation, and

the TEL elevation along the profile. You may assume bed elevation at the start of your profile

with any appropriate value.

Comment on the relation between the bed slope So and the slope of the TEL Se for the three

cases then calculate and compare the length between A and B.

A

So= 15 cm/km

q

B

H = 3m

A

So= 15 cm/km

q

B

Free Fall

A

So= 0.02

q

B

H = 3m



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Model Answer

Question (1)

q = 0.675 m3/sec/m’, n = 0.015, So1 = 0.008, So2 = 0.00309, So3 = 0.0005.

√

√

(

√ )

(

√ )

(

√ )

(

√ )

(

√ )

(

√ )



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Question (2)

q = 1.0 m3/sec/m’ , So1 = 6.87 x 10-3, So2 = 3.24 x 10-4, n =0.018

(a) √

√

(

√ )

(

√ )

(

√ )

(

√ )

(b)

( √

)

( √

)

then the hydraulic jump will occur on the steep slope

( √

)

( √

)



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Question (3)

q = 6 m3/sec/m’, So = 10 cm/km, afflux = 3 m, C = 50 .

(

√ )

(

√ )

√

√



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calculate the length of M1 curve using direct step method from y = 8.24 m

to y = 6.25 m (3 points).

(

)

(

)

y (m) yavg (m) Fr^2 Se (1-Fr^2)/(So-

Se) dy dx

6.25

7.25 6.750 0.012 0.000047 18580.431 1.00 18580.431

8.24 7.745 0.008 0.000031 14377.332 0.99 14233.559

∑dx= 32813.990



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Question (4)

Trapezoidal channel

b = 7 m, t = 1.5, n = 0.02, So = 0.001, Q = 30 m3/sec.

√

√

√

then yo > yc (Mild channel)



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calculate the length of M2 curve using direct step method from y =

0.9x1.623 = 1.46 m to y = yc = 1.13 m (4 points).

(

)

(

)

y (m) yavg (m)

A P R B Fr^2 Se (1-Fr^2)/(So-Se) dy dx

1.46

1.35 1.405 12.796 12.066 1.061 11.215 0.491 0.002033 -492.695 -0.11 54.196

1.24 1.295 11.581 11.669 0.992 10.885 0.643 0.002712 -208.546 -0.11 22.940

1.13 1.185 10.401 11.273 0.923 10.555 0.861 0.003704 -51.575 -0.11 5.673

∑dx= 82.810



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Question (5)

neglect the head loss at the inlet (relatively short distance )

But √

then

√

√



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a – i )

n = 0.02 , So = 10 cm / km

assume steep channel

√ √

(

√ )

(

√ )

yo > yc (wrong assumption)



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assume mild channel

√

√

from 1, 2 then



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a – ii )

n = 0.02 , So = 0.003924


√ √

(

√ )

(

√ )

yo = yc (Critical channel)

As the water depth at the inlet remains yc , then

is valid

and the discharge (q) will remain the same



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a – iii )

n = 0.02 , So = 0.01


√ √

(

√ )

(

√ )

yo < yc (Steep channel)



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case i ii iii

So 0.0001 0.003924 0.01 type Mild Critical Steep

q (m3/sec/m’) 0.959 3.132 3.132

"The discharge flowing into the channel from the lake depends on the

conditions at the inlet (water depth at point A). For a constant specific

energy available (water head H in the lake), the maximum discharge that

can be diverted to the channel occurs when water depth at the inlet is at

Ycritical, otherwise, the flowing discharge decreases. In this problem,

changing the channel slope from critical to steep slope did not affect the

water depth at A and thus the flowing discharge remained te same. On the

other hand, when the channel slope changed to be Mild, water depth at A

increased to Normal depth and thus the flowing discharge decreased. In

addition to the above, when the water depth at A in Ycritical, the flowing

discharge in to the channel depends only on the available head in the lake,

while in case of mild slope, other channel parameters that affect normal

depth (n, So) also affect the flowing discharge in addition to H."



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c )

For case i , and ii no hydraulic jump will occur

For case iii a hydraulics jump will occur

( √

)

( √

)



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calculate the length of S1 curve using direct step method from y = 1.293 m

to y = 2.0 m (5 points).

(

)

(

)

y (m) yavg (m) Fr^2 Se (1-Fr^2)/(So-Se) dy dx

1.293

1.45 1.372 0.388 0.001369 70.953 0.157 11.140

1.6 1.525 0.282 0.000961 79.441 0.150 11.916

1.8 1.700 0.204 0.000669 85.359 0.200 17.072

2 1.9 0.146 0.000462 89.558 0.200 17.912

∑dx= 58.039 m

∑



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d )

Cd = 0.6 , Cc = 0.65

(i) d = 0.5 m

√ √

(

√ )

(

√ )

√

√



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(ii) d = 0.8 m

√ √

(

√ )

(

√ )

√

√

(iii) d = 1.2 m

q = qmax = 3.132 m3/sec/m’

(

√ )

(

√ )



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CAIRO UNIVERSITY HYDRAULICS Faculty of Engineering 3rd year Civil Eng. Irrigation & Hydraulics Department 2011 – 2012

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exercise (4): open channel flow - gradually varied flo (4): open channel flow - gradually varied...

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