Lecture 11,12

Date: 08,09-03-2020

FLOW PROFILES IN SERIAL ARRANGEMENT OF CHANNELS

Flow profiles in a mild slope channel followed by a steep slope channel

Flow profile in a steep slope channel followed by a mild slope channel

Lecture 13

Date: 16-03-2020

Example 6.1

A trapezoidal channel with b = 6m, n = 0.025, s

= 2 and S0 = 0.001 carries a discharge of 28 m3/s.

At a certain section A of the channel the depth of

flow is 1.30 m. (i) Determine the type of channel

slope. (ii) Determine the type of flow profile. (iii)

If at another section B, the depth of flow is 1.50 m,

state whether section B is located upstream or

downstream of section A.

Example 6.1

A trapezoidal channel with b = 6m, n = 0.025, s = 2 and Sn = 0.001 carries a discharge of 28

m3/s. At a certain section A of the channel the depth of flow is 1.30 m. (i) Determine the type of

channel slope. (ii) Determine the type of flow profile. (iii) If at another section B, the depth of

flow is 1.50 m, state whether section B is located upstream or downstream of section A.

b= 6

s= 2

S0= 0.001

Q= 28

α= 1

n= 0.025

22

h A P R AR2/3

hn 1.91 1.91 18.76 14.54 1.29 22

9

h A B D Z = AD

hc 1.14 1.14 9.44 10.56 0.89 9

068.11001.0

14025.0

0

3/2

S

nQAR 068.11

001.0

14025.0

0

3/2

S

nQAR

/g

QZ c /g

QZ c

i) hn>hc Mild slope channel

ii) M1 flow profile.

iii) As, hA=1.30m, hB=1.50m

Lecture 14

Example 6.2: A rectangular channel b=10 m

wide and having = 1.10 and n = 0.025 has

three reaches arranged serially. The bottom

slopes of these reaches are S0=0.0040,

0.0065 and 0.0090, respectively. For a

discharge of Q=35 m3/s in this channel,

sketch the resulting flow profiles.

PROBLEM

Solution The critical depth for the given conditions is obtained as

mgb

QhC 11.1

1081.9

3510.13

2

2

32

2

Since critical slope is the slope for which flow in the channel is both uniform and critical, hence

hn = hc = 1.11 m

Therefore, A = 10 1.11 = 11.12 m2, P = 10 + 2 1.11 = 12.22 m and R = A/P = 0.9 m, and

0070.091.012.11

35025.022

3/2

AR

nQSc

Thus, the bottom slopes of the three reaches are mild, steeper mild and steep, respectively. The resulting

flow profiles are M2, M2 and S2, as shown in the following figure. Mild (S0 < Sc)

Steep (S0 > Sc)

0.0040Mild

0.0065 Steeper mild

0.0090>0.0070->Steep

COMPUTATION OF GRADUALLY VARIED FLOW PROFILES

)18.6.......()/(1

)/(10 M

c

N

n

hh

hhS

dx

dh

Dynamic equation of gradually varied flow

COMPUTATION OF GRADUALLY VARIED FLOW PROFILES

The computation of the gradually varied flow basically involves the integration of

the dynamic equation of gradually varied flow, presented earlier.

This equation is a non linear ordinary differential equation of the first order and its

solution requires one boundary condition for depth, i.e. the depth at the section

where the computation begins must be given.

This equation can be easily integrated (i) for a wide channel, and (ii) for a

horizontal channel. For other channels, the integration of the gradually varied flow

equation has to be performed either graphically or numerically.

The computation of gradually varied flow profile must begin with the known depth

of flow at a control and proceed in the direction in which the control operates.

the computation of flow profiles is usually terminated at a section where the depth

of flow is about 5% greater or less than the normal depth.

Information are generally required:

1. The discharge Q for which the flow profile is desired.

2. The depth of flow or stage at the control section where the computation

begins.

3. The channel shape at various channel sections.

4. The bottom slope S0 of the channel.

5. The energy coefficient .

6. The Manning’s n or Chezy’s C.

There are many methods for computing gradually varied flow profiles.

However, these methods can be broadly classified into the following two categories:

i) Methods used for computing flow profiles in prismatic channels.

ii) Methods used for computing flow profiles in non-prismatic channels.

The methods used for computing flow profiles in prismatic or regular or uniform

channels compute a longitudinal distance x for a given h explicitly without involving

any trial. The direct step method and the direct integration method fall in this

category.

On the other hand, the methods used for computing flow profiles in non-prismatic or

irregular or non-uniform channels compute h from a given x. In this case, a trial-

and-error procedure is necessary. The standard step method falls in this second

category.

Methods for computing GVF

Computation of Flow Profiles in Prismatic Channels

(a) Direct Integration Method

Direct integration of the gradually varied flow equation for computing flow profiles in a

wide channel and in a horizontal channel is simple and considered here.

(i) Flow Profile a Wide Channel: Breese Method

Equation (6.18) can be integrated exactly for a wide rectangular channel with the

conveyance expressed in terms of the Chezy equation. For this case, M = N = 3 and

3

3

0)/(1

)/(1

hh

hhS

dx

dh

c

n

Putting u = h/hn, so that du = dh/hn in above Eq. and rearranging yields

duuh

h

S

hdx

n

cn

33

3

0 1

111

which on integration gives 13

3

0

1 Ch

hu

S

hx

n

cn

)18.6.......()/(1

)/(10 M

c

N

n

hh

hhS

dx

dh

where is the Breese function given by

u

uu

uu

u

du

0

1

2

2

3 12

3tan

3

1

1

1ln

6

1

1

and C1 is a constant of integration.

This integration was first performed by J.A. Ch. Breese in 1860.

A determination of the flow profile by this solution is widely known as the Breese

method.

For a wide channel the critical depth hc is given by 3

2

g

qhc

and, using the Chezy formula, the normal depth hn is given by2

32

0

n

qh

C S

where C is Chezy’s C and q is the discharge per unit width, and hence

g

SC

h

h

n

c 0

23

The length of the flow profile between two consecutive sections of depth h1 and h2 is

123

3

12

0

12 1 n

cn

h

huu

S

hxxL

13

3

0

1 Ch

hu

S

hx

n

cn

3

1 1 1 13

0

1n c

n

h hx u C

S h

3

2 2 2 13

0

1n c

n

h hx u C

S h

where 1 and 2 are the values of corresponding to u1 = h1/hn and u2 = h2/hn, respectively

211 1

1 2

11

212 2

2 2

21

11 1 3ln tan

6 2 131

11 1 3ln tan

6 2 131

u u

uu

u u

uu

h2h1

L

Example 6.3

A wide rectangular channel with Chezy’s C = 47 m1/2/s and S0 = 0.0001

carries a discharge of 2 m2/s. A dam raises the water level by 0.50 m above

the normal depth at the dam site. Compute the length of the resulting flow

profile between the dam site and the location where the depth is 2.90 m.

0.50 h2=2.90

h1=hn+0.50

L

mSC

qh

mg

qh

n

c

626.20001.00.47

2

742.081.9

2

32

2

3

0

2

2

3

2

3

2

Since hn > hc, the channel slope is mild.

Now, h1 = 2.626 + 0.50 = 3.126 m, h2 = 2.90 m, Since h1 or h2 > hn > hc, the

profile is M1.

So, u1 = h1/hn = 1.190

u2 = h2/hn = 1.104.

6687.02858.09545.0

12

3tan

3

1

1

1ln

6

1

4933.02734.07667.012

3tan

3

1

1

1ln

6

1

2

1

2

1

2

2

22

1

1

2

1

1

2

11

uu

uu

uu

uu

Hence, the length of the profile is obtained using Eq. (6.33) as

3

2 1 2 13

0

3

3

1

0.7422.6261.104 1.190 1 0.6687 0.4933 6760.06

0.0001 2.626

n c

n

h hL u u

S h

m

Flow Profile in a Horizontal Channel

*By yourself