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Variation of Which Is Given by The Moody Diagram As:: V RS N

1. Open channel flow involves boundary shear stress, uniform flow resistance formulas like Manning's and Chezy equations, and the Darcy-Weisbach equation. 2. Relationships exist between the Manning's n, Chezy C, and Darcy-Weisbach friction factor f. Uniform flow can be computed using discharge, geometric properties, and slope. 3. For maximum velocity and discharge in a circular channel, the hydraulic radius is 0.29 times the depth and the depth is 0.95 times the diameter.
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100% found this document useful (1 vote)
515 views3 pages

Variation of Which Is Given by The Moody Diagram As:: V RS N

1. Open channel flow involves boundary shear stress, uniform flow resistance formulas like Manning's and Chezy equations, and the Darcy-Weisbach equation. 2. Relationships exist between the Manning's n, Chezy C, and Darcy-Weisbach friction factor f. Uniform flow can be computed using discharge, geometric properties, and slope. 3. For maximum velocity and discharge in a circular channel, the hydraulic radius is 0.29 times the depth and the depth is 0.95 times the diameter.
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Open channel flow

1. Boundary shear: τ 0 = γ RS0 , 𝛄=unit wt. of water, R=hydraulic radius


2. Uniform flow resistance formula:
1 2/ 3 1/ 2
a) Manning’s eqn: V = R S b) Chezy formula: V = C RS0
N

C) Darcy-Weisbach resistance eqn. V = (8g / f ) RS0 where f = Darcy-Weisbach friction factor the

 4RV   4R  
variation of which is given by the Moody Diagram=
as: f fn
=  Re  , 
 ν   ε s  

1 1/ 6 R 1/ 6 f R 1/ 6
3. Relationship between n, c and
= f: C =
8g / f R= ; n = ;n
n 8g C

1
4. Uniform flow Computation: Q = AR 2 / 3S01 / 2
n

Problem Given Required Method of solution


type
1 y0, n, S0, geometric elements(GE) Q and V Direct(explicit)
2 Q, y0 , n, GE S0 Direct
3 Q, y0 , S0, GE n Direct
4 Q, n, S0 , GE y0 Trial and error
5 Q, y0 ,n, S0 GE Trial and error

5. Maximum velocity and discharge (for circular channel, for max. discharge, R = 0.29D, y = 0.95D)
6. Hydraulically efficient channel section: For a given n, S0 and c/s A, a minimum perimeter channel section
represents the hydraulically efficient section that will convey maximum discharge. For a hydraulically efficient
section A = Constant, and dP/dy = 0.

 Of all the various possible open channel sections, the semi-circular has the least amount of perimeter for a
given area.
ye
 For a rectangular section : Be = 2ye and Re =
2

 For a triangular section the vertex angle 2θe = 900


 For a trapezoidal section of side slope:- m horz: 1vert

B
= e 2ye  1 + m 2 − m  Ae =  2 1 + m 2 − m  ye2 Re = ye / 2 , If the side slope can also be varied then the
   

1
optimum values of m = gives the most efficient section.
3

7. Computation of specific energy & critical depth:

V2
i. Specific energy: E= y + α
2g

3
Q 2 Ac
ii. Critical depth: = , at critical depth Froude no. F = 1. At any other depth Froude no. is given by
g Tc

V V
=F =
gy (
g A/T )
1/ 3 1/ 5
 q2   2Q 2 
(a) Rectangular channel: yc =   (b) Triangular channel yc =  
 g   gm 2 
   

8. Rapidly varied flow:

− Q2 − Q2
i) Hydraulic jump: General expression to determine y1 and y2: A1 y1 + = A2 y2 + E=
L E1 − E 2
gA1 gA2

( )
3
y 1 V y2 − y1
a) 2 =  −1 + 1 + 8F12 =, F1 = Froude no. of approaching flow , E L = E1 − E2 = for F1 > 4.5 , length
y1 2   gy1 4y1y2

of the jump is L f = 6.1y2

b) The power dissipated in the jump is P = γ QE L

dy S0 − S f dE
9. Gradually varied flow: = , in terms of specific energy = S0 − S f ,
dx QT2 dx
1−
gA3
3.33
y 
1− 0 
dy y 
For a wide rectangular channel: =  c
dx  yc 
3

1− 
 y 

10. Categories of channel flow

11. Classification of GVF profiles

dE
12. Computation of GVF profiles: Direct step method = S0 − S f is written in finite difference form as:
dx

∆E − ∆E − Sf 1 +Sf 2
S0 − S f or, ∆x =
= −
, where S f = = average friction slope for the reach
∆x S −S 2
0 f

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