Slope area method - nptel - Indian Institute of Technology Madras

Hydraulics Prof. B.S. Thandaveswara Rao
Indian Institute of Technology Madras
Slope area method - Uniform flow
Area-Slope Method
In the event of infeasibility of area velocity method due to either rapid rise and fall
of stage or lack of equipment, the slope area method is adopted for rough
estimation of the discharge.
The requirements of the site are mostly similar to those for area velocity method.
The cross-sectional area is measured adopting the procedure as in case of area
velocity method. The velocity formula used is that of Manning's, the slope
entering the formula being the energy slope which allows for slight non-uniformity
of flow. The roughness coefficient value to be used is related to bed material size
and condition of the channel. These recommendations are given in Indian
Standards Institutions IS: 2912 - 1964 ( 51 ).
Using the slope area method compute the flood discharge through a river reach
of 150m apart, having a fall in the water surface of 150mm. Water areas,
conveyances and energy coefficients of upstream and downstream end sections
are given below:
A u=1107sq.m K u=86857 αu
= 1.134
A =1099sq.m K =88832 α = 1.177
d d d
Au
Solution: Slope -area method
Ad

Hydraulics Prof. B.S. Thandaveswara Rao
Indian Institute of Technology Madras
Kd − Ku
1975
(i) * 100 = * 100 = 2.2270
Kd 88832
less than 30% hence, slope area method can be applied.
(ii) Average value of conveyance K = Ku. K d
Assuming initially zero velocity head,
= 86857 * 88832 = 87839
(1) fall in water surface in the reach (F) 0.15 1
energy slope
Sf
= = = = 0.001
length of the reach (L) 150 1000
( 1 ) (1)
corresponding discharge Q = K. S f = 87839
1
3 -1
= 2777.713 m s
1000
3 -1
For second iteration , taking discharge =2777. 713 m s ,
Velocity heads at the upstream and downstream section respectively
Q
V u =
Au
-1
= 2.5089 ms
Q
V d =
Ad
V u < Vd
∴ K =1.0
-1
= 2.5270 ms
if V u > V d K=0.5
2
(1)
2 ⎛2777.713 ⎞
( Vu
)
⎜ ⎟
α =
⎝ 1107
(i)
⎠
u.
1.134 *
= 0. 364 m
2g2 * 9.81
2
(1)
2 ⎛2777.713 ⎞
( Vd
)
⎜ ⎟
α =
⎝ 1099
(ii)
⎠
d.
1.177 * = 0.383 m
2g2 * 9.81
The difference in the velocity head is equal to
0.364 -0.383 = -0.019 m
(1 )
f
∴ h = 0.15 - 0.019 = 0.131m
Revised Energy slope = S
( 2 )
f
1
Q = 87839 x = 2596
m s
1145
( 2 ) 3 -1
0.131
1
= =
150 1145

Hydraulics Prof. B.S. Thandaveswara Rao
Indian Institute of Technology Madras
For third approximation, taking discharge = 2596 m s
velocity heads:
(2)
( Vu
)
() i α
u
(2)
( Vd
)
( ii)
α
D
2 ⎛2596 ⎞
⎜ ⎟
=
⎝1107 1.134 *
⎠
= 0.318 m
2g2 * 9.81
2 ⎛2598 ⎞
⎜ ⎟
=
⎝1099 . 1.177 *
⎠
= 0.335 m
2g2 * 9.81
0.318 -0.335 = -0.017 m
( 2 )
f
∴ h = 0.15 - 0.017 = 0.133 m
Energy slope = S
( 3 )
f
2
2
0.133 1
= =
150 1128
1
Q = 87839 x = 2615 m s
1128
( 3 ) 3 -1
S.No.
⎛ (2)
2
⎞
⎜Vu⎟ α
⎝ ⎠
u
.
2g
⎛ (2)
2
⎞
⎜V⎟ α
⎝ d ⎠
d
.
2g
⎛V 2 ⎞
∆⎜ ⎟
⎜ 2g ⎟
⎝ ⎠
h
f
s
f
Q
1 - - - 1 / 1000 2777.713 2596
2 0.364 0.383 - 0.019 0.131 1/1145 2596
3 0.318 0.335 - 0.017 0.133 1/1128 2615
4 0.3225 0.34 - 0.0175 0.1325 1/1132 2611
∴
3 -1
Flood discharge through the given river reach Q =2611 m s
Example:
Estimate the conveyances of two sections and the average conveyance given
the following data:
A u = 143 sqm Pu = 76m n u = 0.035
A = 120 sqm P = 94m n = 0.035
d d d
u
1
( nu 23)
u
1 ⎡
⎢
0.035 ⎢⎣ 23
⎛143 ⎞ ⎤
⎜ ⎟ ⎥
⎝ 76 ⎠ ⎥⎦
d
1
( nd 23)
d
1 ⎡
⎢
0.035
23
⎛120 ⎞ ⎤
⎜ ⎟ ⎥
⎝ 94 ⎠
K = AR = 143* = 6226.79
K = AR = 120* = 4648.85
⎢⎣ ⎥⎦
K = K *K = 28947382 = 5380.28
u d
3 -1
,