# Hydrosystems Hydrulics open channel flow iit nptel _34_

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```					Hydraulics                                                                                                                      Prof. B.S. Thandaveswara

12.2 Problems
There are three types of problems in critical flows as shown in table.

Type                        m                Q                       yc                  b or d
I                                          ?
II                                                                ?
III                                                                                       ?

Types I and II are easy to solve.

Type III problem requires a different approach.

Type I problem
myc
is known.
b
Qm3/ 2
∴ From the graph yc Vs Z =
b 2 gb
Q can be determined.
Type II problem
Here the solution is for obtaining critical depth. There are different methods. Graphically

Qm3/ 2                                            myc
Z=        2
can be computed and value of          can be obtained from which
b       gb                                         b
yc can be computed.

Type III problem
This problem can be solved using simultaneous solution of two algebraic equations

which is illustrated below.

Defining

myc                                                       yc
Y1 =            ( for trapezoidal channel )         or                 ( for circular channel ) .
b                                                        d0
Qm3/ 2                                            Q
and X1 =           2
or       2
b       gb                                   d0    gd o

Then

myc
b =
Y1

Hydraulics                                                                                                           Prof. B.S. Thandaveswara

Q m3/ 2                  Q m3/ 2 Y15 / 2
X1 =                          =
2
m 2 yc     g ( myc )            2
m 2 yc m1/ 2 y1/ 2 g
c
Y12             1/
Y1 2
Q Y15 / 2
=        2
= M1Y15 / 2
m   yc     gyc

X1 = M1Y15 / 2
In which

Q
M1 =            2
my   c       gyc
Given Q, yc , m,
Qm3 / 2                         myc
X1 =           ,              Y1 =
b 2 gb                           b

y'c (1 + y'c )
3           3
2    3
Qm
=
gb5                 1 + 2y'c
Q
M1 =         2
and is known.
myc gyc
X1 = M1Y15 / 2
Substitutingin the above equation
5/ 2
Qm3 / 2      ⎛ myc ⎞
= M1 ⎜     ⎟
2
b gb         ⎝ b ⎠
Qm3 / 2 ( myc ) gyc
2

= b 2.5 y1 / 2
5

Q g
Solve the equation and obtain the solution for bed width b for trapezoidal channel.
Similarly solve for diameter for the pipe line.

Hydraulics                                                                                                                            Prof. B.S. Thandaveswara

Problem

Compute the critical depth in a trapezoidal channel for flow of 30 m3 s-1. The channel

bottom width is 10.0 m, side slope m =2. The bottom slope is negligible and α = 1

T
1                                                 1
m                                        m y

b

.
Trapezoidal
Solution

Given

Bottom width b = 10m
Sideslope      m =2
Flow             Q = 30 m3s -1
α =1
Critical Depth y c = ?
For finding the critical depth,
Cross sectional area of the channel A= ( b + 2 y C ) * y C
= ( 10 + 2 y C ) * y C
Section factor                             Z=A D
in which D = A / T
for trapezoidal channel the top width T = ( b + 2m y C )
D = ( 10 + 2y C ) * yC /(10 + 2 * 2 * yC )
then the section factor                                       Z=A      ( 10 + 2 yC ) * yC / (10 + 2 * 2 * yC )
30
by u sin g the equation                                      = 9.578
A D =Q / g, =
9.81
Substituting all the parameters A, P, T, D, and Q in the above equation and solving for y c one gets
3
⎡(10 + 2 y c ) y c ⎤
⎣                  ⎦
2
1
= 9.578
(10 + 4y c )        2

3                               1
⎡(10 + 2 y c ) y c ⎤
⎣                  ⎦
2   − 9.578 (10 + 4 y c )       2   =0

⎡(10 + 2 y c ) y c ⎤ − 91.743 (10 + 4 y c ) = 0
3
⎣                  ⎦
by trial and error, y c = 0.91 m

Hydraulics                                                                                           Prof. B.S. Thandaveswara

Problems

1. A trapezoidal channel with side slopes of 2 horizontal to 1 vertical is to carry a flow of

16.7 m3/s. For a bottom width of 3.65 m, calculate (a) the critical depth and, (b ) the

critical velocity.

2. A rectangular channel carries 5.60 m3/s. Find the critical depth yc and critical velocity

Vc for

(a) a width of 3.65m and, (b) a width of 2.75m ,

(c) What slope will produce the critical velocity in (a) if n = 0.020 ?

3. Find the diacharge over a broad crested weir of 5.0m length and head 1.0m above

the crest. Assume coefficient of discharge to be 0.9.

___

V1
H
y1
yc

P

B

Hydraulics                                                                                       Prof. B.S. Thandaveswara

P is the height of weir, B is the breadth of the weir. Assume the approach velocity V1 to

be very small.

⎛             3 yc            2                      ⎞
⎜ Answer: H =             ⎛Q⎞ 1
, yc = ⎜ ⎟   ,   Q = 0.544L gH3/2 ⎟
⎜               2         ⎝L⎠ g                      ⎟
⎝                                                    ⎠