critical flow depth computations

Hydraulics Prof. B.S. Thandaveswara

Indian Institute of Technology Madras

12.1 Critical flow depth computations

One of the important aspects in Hydraulic Engineering is to compute the critical depth if

discharge is given.

Following methods are used for determining the critical depth.

(i) Algebraic method.

(ii) Graphical method.

(iii) Design chart.

(iv) Numerical method. Bi section method/ Newton Raphson method.

(v) Semi empirical approach - a method has introduced by Strarb.

12.1.1 Algebraic method

In this method the algebraic equation is formulated and then solved by trial and error.

The following example illustrates the method.

1. Consider a trapezoidal channel:

2.

( )( )( )

( ) ( )( )

( ) ( )

c c

c c

c

c 1

1/ 2c c

1 c cc

32 31 c c c

6 5 4 3c c c c c

A b my y

b my yD

b 2myQZ constant C knowng

b my yC b my y (1)

b 2my

C b 2my b my yleads to

y py qy ry sy t 0in which the cons tan ts p,q, r,s and t are known.

= +

+=

+

= = = =

⎧ ⎫+⎪ ⎪= + ⎨ ⎬+⎪ ⎪⎩ ⎭

+ = +

+ + + + + =

Solve this by polynomial or by trial and error method.It would be easier to solve the equation (1) by trial and error procedure.After obtaining the answer check for the Froude number which should be eq

ual to 1.



Example:

Consider a Rectangular channel and obtain the critical depth for a given discharge.

Solution:

1/ 2

3/ 2c

2 / 3 2 2

c

A b yArea b y D yT b

QZ b y yg

Qyb g

Q q qygb g g

3

= = = =

∴ = =

=

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

12.1.2 Trial and error method

For a given trapezoidal channel obtain the critical depth by trial and error method.

Solution:

( )( )

( )( )

( )

3/ 2c c

1/ 2c

3 23

c3

c3 3

c c

c

For trapezoidal channel

b my yA D

b 2my

b my y QSquaring y constantb 2my g

For a given b, m,Q, select a value of y

Assume b 6 m, m 2m, Q 12 m / s Solve for y

6 2y y 16 4y

⎡ ⎤+⎣ ⎦=+

⎛ ⎞+= =⎜ ⎟⎜ ⎟+⎝ ⎠

= = =

+=

+

( )3 3c c

c

44 14.6799.81

3 y y 36 3.66973 2y 9.81

=

+= =

+

Assume a value of yc and compute A D and compare with the value obtained by Qg

.

yc A D A D Remarks 1.2 23.708 too high 0.5 1.339 low 0.8 6.170 high

0.65 3.10 0.70 3.94



Remarks column indicate that the values are high or low when compared to the given

value. The improvement is done till it converges.

In the above table yc lies between 0.65 and 0.70.

This could be improved further by selecting the values in between these two.

12.1.3 Graphical method

For natural channels and complicated channels, the graphical method is adopted. A

curve is generated assuming different values of yc and Z. The value of Qg

is computed

and yc is obtained from the chart. A one meter diameter culvert carries a discharge of

0.7 m3/s. Determine the critical depth.

yd0θ

T

( )

o

1.50.500.5

1 sinD dg sin

2

2 sinZ d

32 sin2

θ θθ

θ θ

θ

⎡ ⎤⎢ ⎥−

= ⎢ ⎥⎢ ⎥⎣ ⎦

−=

⎡ ⎤⎢ ⎥⎣ ⎦

Knowing the value of d0 for different values of depth A and D could be obtained from the

table.

Example:

A one meter diameter pipe carries a discharge of 0.7 m3/s. Determine the critical depth.

c

c c

c

Q 0.7Z 0.22353.132g

Construct a graph of y Vs and obtain the value of yFrom the graph y 0.4756

Z =

= = =

From the design chart determine the critical depth for a circular channel of 0.9 m diameter. Discharge 0.71 m3/s.



Solution:

( )2.50

cc

0

0.71Z 0.226699.81

Z 0.29499dy 0.56, y 0.49527d

from table

m

= =

=

= =

( )

( )

22

0.27

c 0.75 1.25

1.0 17Q 29.5g 9.81

29.5 6y 0.81 0.8630 22 6

m

αψ = = =

⎡ ⎤= − =⎢ ⎥⎣ ⎦



12.1.4 Graphical Procedure

Straub proposed several semi empirical equations to obtain the critical depth. The

advantage of this is a quick estimation of the critical depth. However, the equations are

non homogenous.

yc

Graph showing variation of section factorwith critical depth for a given pipe of diameter do

Z=A D

yc

d0

__ yc

b__or

DA_____d

0

2.5

DA_____b2.5

or



Reference:

Straub W.O, Civil Engineering, ASCE, 1978 Dec, pp 70 - 71 and Straub 1982.

Table: Semi empirical equations for the estimation of yc (Straub, 1982) MKS units

Channel type

Equation for yc in terms of

2Q / gψ α=

bRectanglar

1/ 3

2bψ⎛ ⎞

⎜ ⎟⎝ ⎠

b

m1

Trapezoidal

0.27

0.75 1.25b0.81

30mm bΨ⎛ ⎞ −⎜ ⎟

⎝ ⎠

2.5

2.5

Range of applicabilityQ0.1 4.0

bQFor 0.1

buse equation for rectangular channel

< <

<

lm

TRIANGULAR

0.20

22mΨ⎛ ⎞

⎜ ⎟⎝ ⎠

y

y = cx2

xParabolic

( )0.250.84cΨ

2y cx=



d0

Circular

[ ][ ]

0.250.260

0.52

c 0.30

c3 1

0

1.01d

Qy 0.053d

y m

Q m s d m , −

⎛ ⎞Ψ⎜ ⎟⎜ ⎟

⎝ ⎠

=

=

= =

c

o

Range of applicabilityy0.02 0.85d

≤ ≤

y

x

a

b

Elliptical

0.250.22

20.84baψ⎛ ⎞

⎜ ⎟⎝ ⎠

c

Range of applicabilityy0.05 0.852b

a = major axis b = minor axis

≤ ≤

y

y = cx

1m-1____

xExponential

( )1/ 2m 13 2m 2m c4

ψ+−⎛ ⎞

⎜ ⎟⎜ ⎟⎝ ⎠

( )1/ m 1y cx −=

Example:

3cb 6.0 m, m 2, Q 17m / s determine y = = =

Solution:

From table

0.27

0.75 1.25 2.5

2

b Q0.81 for 0.1 4.030mm b b

Qwhereg

17The value of 0.19,

c

2.5 2.5

y

Q b 6

ψ

αψ

⎛ ⎞= − < <⎜ ⎟⎝ ⎠

=

= =

It is in the range of the equation. Substituting the appropriate values,

( )

( )

2

0.27

c 0.75 1.25

1 1729.5

9.8

29.5 6y 0.81 0.8630 22 6

m

ψ = =

⎛ ⎞= − =⎜ ⎟⎝ ⎠



Problem:

Non rectangular channel involves trial and error solution.

Obtain the critical depth for the trapezoidal channel of bottom width 6 m with a side

slope of 2.5: 1, which carries a discharge of 20 m3/s.

6 m

m1

m1 yc

Solution:

Trial and error procedure

( ) ( )

( )

[ ]( ) ( )

c c

c

c c

c

c

0.52 2 2c c

c c2cc

c

c

y 6 2.5 y yT 6 5y

6 2.5y yADT 6 5yQZ A Dg

6 2.5y yV Q / A 20*20 6 2.5y y2g 2g 6 5y6 2.5 y 19.62y

gyc

A= b+my b+2my=

v

?

?

= +

= +

+= =

+

= =

⎧ ⎫+= = = + ⎨ ⎬++ ⎩ ⎭

=

= =

Solution of Algebraic or Transcendental Equations by the Bisection Method

In the algebraic expression F(x) =0, when a range of values of x is known that contains

only one root, the bisection method is a practical way to obtain it. It is best shown by an

example.

The critical depth in a trapezoidal channel is to be determined for given flow Q and

channel dimensions.

2

3Q T1 0gA

− =

The formula must be satisfied by some positive depth yc greater than 0 (a lower bound)

and less than, an arbitrarily selected upper bound say, 10 m.



T is the free surface width cb 2my+ . The interval is bisected and this value of yc tried. If

the value is positive, then the root is less than the midpoint and the upper limit is moved

to the midpoint and the remaining half bisected, etc.

This method gives the solution very quickly.

F(x)

0100

Bisection

T

b

m1

m1

Trapezoidal

y

Newton Raphson Method is discussed elsewhere.

critical flow depth computations

Documents