Hydraulics Prof. B.S. Murty
Indian Institute of Technology Madras
26.1 Standard Step Method
In the standard step method, flow depth at a specified location,
yd is determined, given
the flow depth, Yu at another specified location. Consider the
channel shown in Figure
26.1. In this channel, say Yu occurs at a distance Xu from the
reference point.
Discharge, Q, Channel bottom slope, S0, the roughness
coefficient, n and cross-
sectional shape parameters (which relate A, P and R to y) are
also known. The problem
now is to determine the flow depth, Yd at the specified location
Xd (figure 26.1).
Fig. 26.1: Definition sketch for standard step method
yu (known)yd (unknown)
Water surfaceFlow
Datum
Channel Bed
Xd (known)
Zu Zd
ud
u
d
Equation (25.3) ( ) ( ) ( )2 2d u fd d u u d u 0 d u
V Vy y S x x S x x 25.32g 2g
+ + + =
can be rewritten as
( )( ) ( ) ( )d u
2 2f d u fd u
d u d u 0 d u2 2d u
S x x SQ Qy y x x S x x2.0 22gA 2gA
+ + = + + 26.1
In Equation 26.1, the flow rate (Q), the roughness coefficient
(n), distances Xd and Xu,
the channel slope (S0), the flow conditions at section u ( u u
uy , and A ) are known.
Therefore the right hand side of Eq. (26.1) can be determined.
On the left hand side, the
area, Ad and the friction slope, dfS are functions of the flow
depth Yd. Thus we have
one equation (Eq. 26.1) in one unknown Yd. Therefore, Yd can be
determined by solving
Hydraulics Prof. B.S. Murty
Indian Institute of Technology Madras
Equation (26.1). Equation (26.1) is a non-linear equation.
Either trial and error or
numerical techniques such as bisection, Newton -Raphson
techniques etc. can be used
for solving Eq. (26.1).
For example, for a wide rectangular channel (assuming u d 1.0= =
), Eq. (26.1)
becomes
( ) ( ) ( )2 22 2 2 2
d ud u d u 0 d u2 10 / 3 2 10 / 3
d d u u
n q x xq q n qy y x x S x x2gy 2y 2gy 2y
+ + = + + 26.2
In Eq. (26.2), u d u 0q, n, y , x , x , S and g are known, and
so Yd can be determined by
solving this equation. Note that Eq. (26.2) is non-linear.
