Then
1 ro V 2
∆Z = ri
dr
(2.158)
g
r
To solve Equation 2.158, the transverse velocity distribution along the radius of the bend must be
known or assumed. The results obtained assuming various velocity distributions follow.
Woodward (1920) assumed V equal to the average velocity Q/A and r equal to the radius to the
center of the stream rc, and obtained:
1 ro V 2
∆Z = ri
dr
(2.159)
g
rc
or
V2
∆Z = z o - z i =
(ro - ri )
(2.160)
g rc
in which zi and ri are the water surface elevation and the radius at the inside of the bend, and zo
and ro are the water surface elevation and the radius at the outside of the bend.
By assuming the velocity distribution to approximate that of a free vortex (Vθ = C1/r), Shukry
(1950) obtained:
1
1
2
2
1 ro C1
C1
∆Z = ri 3 dr =
2 - 2
(2.161)
g
2g
ri
ro
r
in which C1 = rV is the free vortex constant. By assuming flow depth of flow upstream of the
bend equal to the average depth in the bend, Ippen and Drinker (1962) reduced Equation 2.161
to:
V 2W
2
1
∆Z =
(2.162)
2
2g rc W
1 - 2r
c
For situations where high velocities occur near the outer bank of the channel, a forced vortex
may approximate the flow pattern. With this assumption and assuming a constant average
specific head, Ippen and Drinker (1962) obtained:
2.48
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