2. TITLE OF PRESENTATION
Hydraulic Exponent for Critical-flow Computation
& Computation of Critical-flow
By
Engr. Zeeshan Ahmed Soomro
2k19-ID-32
Masters Scholar at
Faculty of Agricultural Engineering
Sindh Agriculture University Tandojam Pakistan
3. WHAT IS HYDRAULIC EXPONENT FOR
CRITICAL FLOW
• Hydraulic exponents are the parameters M and N in the mathematical relations for
open-channel flows as expressed in Z²=CyM and K²=CyN, in which Z = the section
factor for critical-flow computation; K = the conveyance of the channel flow cross
section; y = the flow depth; and C = a coefficient.
4. • This clarifies the mathematical derivation of the hydraulic exponents and their
implication in application to the computation of flow profiles of gradually varied
flow in a prismatic channel. The relationships between these hydraulic exponents,
which are assumed to depend only on the average flow depth, and those, which are
assumed to vary with the flow depth of a given range, are also explained.
5. HYDRAULIC COMPONENT FOR CRITICAL
FLOW COMPUTATION
• Since the section factor for critical flow computation (Zc) is a function of depth of
flow; it may be assume that
Where additionally C is coefficient, M is hydraulic exponent for critical flow computation
Eq…..1
6. Taking log of both sides of equation 1, we get
2 log Z═ Log C + M log y
By differentiating Eq 2
2
𝑍
𝑑𝑦
𝑑𝑧
= 0 +
𝑀
𝑦
𝑀 =
2𝑦
𝑍
𝑑𝑍
𝑑𝑦
Eq…..2
Eq…..3
12. PROBLEM
• A trapezoidal channel having bed width of 20 ft, side slope 2:1 and depth of flow 5
ft. caries a discharge od 400 cusecs. Find the hydraulic exponent for critical flow
computation using algebraic method and graph / chart?
Solution:
DATA;
Bed width = 20ft
Side slope 2:1
Depth of flow = 5ft
Discharge = 400 cusecs
14. GRAPHICAL METHOD
• y/b = 5/20
= 0.25
From figure at y/b = 0.25,
the value of M be read
against z= 2 thus
M= 3.5
Curve for M values
15. COMPUTATION OF CRITICAL FLOW
• Computation of critical flow involves the determination of critical depth and
velocity when the discharge of channel section are known.
• On the other hand, if the critical depth and channel section are known, the
critical discharge can be determined by method described in example given
below.
16. • 1. Algebraic method
• For a simple gematric channel section, the critical flow can be determined by an
algebraic computation using the basic equations.
• 2. method of design chart/Graph
• The design chart for determining the critical depth can be used with great expediency.
17. PROBLEM
• Compute the critical depth and velocity of the trapezoidal channel having bed
width 20 ft. and side slope 2:1 and carrying a discharge of 400 cusecs using
algebraic method and design chart/graph?
DATA:
Bed width = 20 ft
Side slope = 2:1
Discharge = 400 cusecs
18. BY ALGEBRAIC METHOD
• The hydraulic depth and water area of
the trapezoidal section in terms of the
depth (y) as:
• 𝐷 =
𝐴
𝑇
• 𝐴 = 𝑦 𝑏 + 𝑧𝑦
• 𝑇 = 𝑏 + 2𝑧𝑦
• 𝐷 =
𝑦 𝑏+𝑧𝑦
𝑏+2𝑧𝑦
The velocity is
𝑉 ≡
𝑄
𝐴
≡
400
𝑦 20 + 2𝑦
Substitute the above expression for D and V
in Eq:
𝑉2
2𝑔
=
𝐷
2
4002
2𝑔 𝑦 20 + 2𝑦 2
=
𝑦 10 + 𝑦
10 + 2𝑦
19. • Simplification of above expression yields;
• 2484(5+ y) = 𝑦 10 + 𝑦 3
• Solving this equation for y by a trial and error procedure the value of required
parameters is;
• Yc =2.15 ft
• Ac = 52.2 ft
• 𝑉𝑐 =
400
52.2
= 7.66
𝑓𝑡
𝑠𝑒𝑐
20. BY DESIGN CHART/GRAPH
• 𝑍 =
𝑄
𝑔
• 𝑍 =
400
32.2
•
𝑍
𝑏2.5 = 0.0394
For this value of
𝑍
𝑏2.5 , and z = 2 from
graph the value of y/b = 0.108
Then
𝑌𝑐 =
3 𝑄2
𝑏2 𝑔
𝑌𝐶 =
3 4002
(0.1402× 32.2
Yc =2.16 ft