Download Solution of manning equation by Newton Raphson Method

Hydraulics
Prof. B.S. Thandaveswara
Solution of manning equation by Newton Raphson Method
There is no general analytical solution to manning equation for determining the flow
depth given the flow rate because the area A and hydraulic radius R may be
complicated functions of the depth. Newton Raphson method can be applied iteratively
to obtain a numerical solution. Suppose that at iteration k the depth yk is selected and
the flow rate Qn, is computed using manning formula using the area and hydraulic
radius corresponding to yk. This Qk is compared with actual flow Qn; then the objective is
to chose y such that the error.
f (yk) = Qk - Qn is within the tolerance limit. The gradient of f with respect to y is
df(yk ) dQk
=
dyk
dyk
because Qn is constant. Hence, assuming manning roughness is constant,
2 ⎞
⎛ df ⎞
⎛1 1
3
⎜
⎟ = ⎜ So 2 A k R k ⎟
dy
n
⎝
⎠
⎝
⎠k
⎛
⎞
-1
2
⎜
1 1 2A R 3 dR
dA ⎟⎟
= So 2 ⎜
+R 3
n
3
dy
dy ⎟
⎜
⎜
⎟
⎝
⎠k
2 ⎛ 2 dR
1 1
1 dA ⎞
= So 2 A k R k 3 ⎜
+
⎟
n
⎝ 3R dy A dy ⎠k
⎛ 2 dR
⎛ df ⎞
1 dA ⎞
+
⎜ ⎟ = Qk ⎜
A dy ⎟⎠k
⎝ dy ⎠k
⎝ 3R dy
in which the subscript k out side the bracket indicates that the
quantities in the bracket computed for y = yk.
In Newton's method,
given a choice of yk , yk+1 is chosen to satisfy
0- f (y)k
⎛ df ⎞
=
⎜
⎟
⎝ dy ⎠k yk + yk+1
This yk+1 is the value of yk ,
f (yk )
yk+1 = yk ( df / dy ) k
Indian Institute of Technology Madras
Hydraulics
Prof. B.S. Thandaveswara
Which is the fundamental equation of the Newton's method. Iterations are continued
until there is no significant change in yn; this will happen when the error is nearly zero or
an acceptable prescribed tolerance.
Thus for manning equation it may be written as
y k+1 = y k -
1 - Q / Qk
⎛ 2 dR 1 dA ⎞
+
⎜
⎟
⎝ 3R dy A dy ⎠ k
For rectangular channel A = bo y and R = bo y / ( bo + 2y ) where bo is the
channel width; The quantity in denominator can be for rectangular channel
( )
d
d
(R ) = A P
dy
dy
1 dA A dP
=
−
P dy P 2 dy
⎡ T R dP ⎤
=⎢ −
⎥
⎣ P P dy ⎦
consider
2 dR 1 dA
+
3R dy A dy
2 P ⎡ T R dP ⎤ T
−
+
3 A ⎢⎣ P P dy ⎥⎦ A
dP ⎤ T
21⎡
T −R ⎥+
⎢
dy ⎦ A
3 A⎣
2 T 2 R dP T
−
+
3 A 3 A dy A
⎡ 5 T 2 1 dP ⎤
⎢ 3 A − 3 P dy ⎥
⎣
⎦
For rectangular channel
5 bo 2
1
−
2
3 bo y 3 ( bo + 2 y )
51 4
1
−
3 y 3 ( bo + 2 y )
5 ( bo + 2 y ) − 12 y
3 y ( bo + 2 y )
=
=
5bo + 10 y − 4 y
3 y ( bo + 2 y )
5bo + 6 y
3 y ( bo + 2 y )
y k+1 = y k -
1 - Q /Qk
5 bo + 6 y k
⎛
⎞
⎜
⎟
⎜ 3y b + 2 y ⎟
k ⎠
⎝ k o
Similarly the channel shape function ⎡⎣( 2/3R )( dR/dy ) + ( 1/A ) (dA/dy)⎤⎦
(
for other cross sections can be derived.
Indian Institute of Technology Madras
)
Hydraulics
Prof. B.S. Thandaveswara
(1 + m2 ) + 4my 2 (1 + m2 )
3y ( bo + my ) ⎛⎜ bo + 2 y (1 + m 2 ) ⎞⎟
⎝
⎠
( bo + 2my ) + 6 y
Trapezoided Channel
8
3y
Triangular Channel
Circular Conduit
4 ( 2sinθ + 3θ − 5θ cos θ )
⎛θ ⎞
3do (θ )(θ − sin θ ) sin ⎜ ⎟
⎝2⎠
in which
⎛
θ = 2 cos −1 ⎜1 −
⎝
2y ⎞
⎟
do ⎠
Example:
Compute the flow depth in a 0.6 m wide rectangular channel having n= 0.015, S0 =
0.025, and Q = 0.25 m3s-1.
y
B
Solution:
Let wide bo = 0.6m
Manning coefficient n = 0.015
bed slope Sο = 0.025
discharge Q = 0.25m3 s −1
normal depth y = ?
Hyraulic mean radius R =
2
1
1
Q = AR 3 Sο 2
n
2
⎞3
1
⎛ byk
1
Q = bo y ⎜
⎟ Sο 2
+
2
n
b
y
k ⎠
⎝
Indian Institute of Technology Madras
bo
A
=
p bo + 2 y
Hydraulics
Prof. B.S. Thandaveswara
5 ⎤
⎡
1
by
(
)
1⎢
k 3 ⎥
Q= ⎢
S 2
2⎥ ο
n
⎢⎣ ( b + 2 yk ) 3 ⎥⎦
5 ⎤
⎡
1
1
⎢ ( 0.6 × yk ) 3 ⎥
×⎢
×
0.025
Qk =
2
(
)
2⎥
0.015
⎢⎣ ( 0.6 + 2 yk ) 3 ⎥⎦
5
53
53
0.6 3 y
4.4993 y
k
k
=
Qk = 10.5409*
23
23
0.6 + 2 yk
0.6 + 2 yk
(
)
Shape function =
=
(
(1)
)
5bo + 6 yk
3 yk ( b + 2 yk )
5 ( 0.6 ) + 6 yk
3 yk ( 0.6 + 2 yk )
=
3 + 6 yk
1 + 2 yk
=
3 yk ( 0.6 + 2 yk ) yk ( 0.6 + 2 yk )
⎛ 0.25 ⎞
⎜1 −
⎟ yk ( 0.6 + 2 yk )
Qk ⎠
⎝
y k+1 = yk −
(1 + 2 yk )
(2)
Iteration (k)
yk ( m )
1
0.100
2
0.1815
3
0.1727
Q(m3s -1 )
0.1125
0.2684
0.2488
Froude number F =
F=
0.2488 / ( 0.6* 0.1727 )
( 9.81*0.1727 )
∴ super critical flow
Indian Institute of Technology Madras
Q/ A
V
=
gy
gy
= 1.844