Solutions manual Open-Channel Flow, 3rd Edition, by M. Hanif Chaudhry (COMPLETE)

Chapter 1
BASIC CONCEPTS



1-1 The flow is:
i- Unsteady
ii- Unsteady Y
iii- Steady
iv- Unsteady Bo
v- Unsteady

1-2 i- False
ii – True

1-3 i- Nonuniform
ii- Nonuniform
iii – Nonuniform
iv- Uniform
v- Nonuniform

1-4
(i) Rectangular section
A = B0 Y
P = 2Y+B0
B = B0
R = A/P = B0Y/(2Y+B0)
D = A/B = B0Y/ B0 = Y

(ii) Trapezoidal section
A = B0Y+ 2Y(SY/2)
= Y(B0 + SY)
1
Y
P = B0 +2Y S(S2+1)
S

Bo

1

, R = A/P = Y(B0 + SY)/ [B0 +2Y S(S2+1)]


B = B0 +2SY
D = A/B = Y(B0 + SY)/(B0 + 2SY)

(iii) Triangular section
We may use the same equation as that in the case of trapezoidal section with B0 = 0.
Thus, D = Y/2



(iv) Partially full circular section

A = r2θ/2 +2 (rCos α)/2 (Y-D0/2)
= D02 / 8 + (Y − D)(rCos )

Y = D0/2 + (D0/2) Sin α
A = Do2 θ /8 + Do Sin α Cos α




D o2 θ D o2
A = + sinα cosα
8 4
B
D o2
= (θ + sin 2α )
8
But θ = 2α + π
D0and sin θ = sin (2α +  ) = − sin 2α
θθ DY2
A = o (θ − sin θ ) 0  θ  2π
8
Doθ sin 
P = rθ = ( 1 − )
4 
  
B = 2 r cos  = 2 r cos ( − ) = D o sin
2 2 2
A Do  - sin 
D= = ( 1 − )
B 8 Sin 
2


2

,(v) Standard horseshoe section:
Length KB
_____ _____ _____
OM = MC − OM
_____
OM = d 0 2 2
 1
2 2 2
_____
 _____   _____   d 
MC =  CG  −  GM  = d 0 −  0  = d 0  1 − 
2

    2 2  8
_____
7
MC = d0
8
____
 1 
OC = d 0  7 8 −  = 0.58186d 0 − − − − − − − (1)
 2 2
2
 
_____ _____ _____ ____
KC = FC 2 − KF 2 = d 0 −  d 0 − KB  − − − − − −(2)
2

 
2
 ____ ____ 
____ _____ _____ ____
KC = FC − OK = OC −  OB − KB 
2 2 2 2

 
2
d 
____
KC 2 = (0.58186d 0 ) −  0 − KB  − − − − − − − −(3)
2

 2 


2 2
 _____
  d ____ 
d 0 −  d 0 − KB  = 0.33856 d 0 −  0 − KB 
2 2
(2) and (3)
   2 
2
 ____ 
____ ____ ____ 2
d
d 0 − d 0 + 2d 0 KB − KB = 0.38856 d 0 − 0 + d 0 KB −  KB 
2 2 2 2

4  
____ ____
d
KB = 0.088562 d 0 OK = 0 − 0.088562 d 0 = 0.411438 d 0
2
_____
KC 2 = d 0 − (d 0 − 0.088562d 0 ) = 0.169281 d 0
2 2 2


KC = 04114377 d 0 CC = 0.822875d 0
L L
Sin = 0.4114377 = 24.2950  L = 48.590
2 2



The standard horse shoe section is divided into three sections, i.e., upper section, middle
section and lower section.


(a) Upper section
3

,π ≤ θu ≤ 2 π

D02 D02
Flow area, A = ( u − Sin u ) −
8 8
D
Wetted perimeter, P = 0
2
D  Sin 
Hydraulic radius, R = A/P = 0 1 − 
4   
Top water surface, B = D0 Sin (θ/2)
D   − Sin 
Hydraulic depth, D = A/B = 0  
8  Sin( / 2) 


(b) Lower section

0 ≤ θL ≤ 48.590

D02 d2
Flow area, A = ( L − Sin L ) = 0 ( − Sin )
8 2
D
Wetted perimeter, P = 0 = d 0
2
d  Sin 
Hydraulic radius, R = A/P = 0 1 − 
2   
Top water surface, B = 2d0 Sin (θ/2)
d   − Sin 
Hydraulic depth, D = A/B = 0  
4  Sin( / 2) 

(c) Middle section

Assume trapezoidal section, S = 0.215
Area, A = Y(B0 + SY)
A = Y(0.8229 d0 + 0.215Y)

P = B0 + 2Y S 2 −1




4

, P = 0.8229 d 0 + 2Y 0.2152 + 1 = 0.8229 d 0 + 2.05Y
Y (0.8229d0 + 0.2154)
R=A =
P 0.8229 d 0 + 2.05Y
B = B0 − 254 = 0.8229d0 + 2  0.215Y = 0.8229d0 + 0.43Y
Y(0.8229d0 + 0.215Y)
D=A =
B 0.8229d0 + 0.43Y



1-5
Q = K A R2/3
A = (θ – Sinθ) D2/8
R = A/P
P = Dθ/2
10

( − Sin )3 ( − Sin )3
5 5
3
KD
Q= 5 2
=C 2

83 (D / 2 ) 3  3
2
 3


10
3
KD
Where C =
(D / 2 ) 3
5 2
3
8

dQ
= 0 will give the angle  corresponding to Qmax .
d
 2 −5 5 −2 
= C −  3 ( − Sin ) 3 +  3 ( − Sin ) 3 (1 − Cos ) = 0
dQ 5 2

d  3 3 
−2
 3
dQ
d
=C
3
2

( − Sin ) 3 − 2 −1 ( − Sin ) + 5(1 − Cos ) = 0 
 2( − Sin ) = 5 (1 − Cos )
θ – Sinθ = 5/2θ – 5/2θ Cosθ
5/2 Cos θ – Sin θ / θ – 1.5 = 0

Solving by trial and error or numerically θ = 302.41
D D 
From the figure Y = + Sin and  = −  / 2
2 2 2
D D  
Y = + Sin −  / 2 
2 2 2 
D 
Y = 1 − Cos 
2 2

Substituting θ value for the Qmax


5