The resulting power equation is:

v/0.738 = 1.37 (y/yo)0.24

Using the above information calculate the shear stress on the bed using the several methods given

in Section 2.4.5.

(a) Average Shear Stress on the Bed

Calculate the average shear stress on the bed using Equation 2.97

τ0 = γ R Sf = 9800 x 2.38 x 0.000206 = 4.805 N/m2

Using R = yo (for a wide channel)

(b) Local Shear Stress on the Bed Using Single Point Velocity near the Bed

Using the above information and the logarithmic velocity distribution equation derived from Equation

2.75 (v/V* = 2.564 ln (y/0.009)) the shear stress on the bed using a single point velocity near the

bed is:

ρv2

1000 (0.518)2

= 5.11N / m 2

τ0 =

=

2

2

y

0.152

2.564 ln

2.564 ln

0.009

0.009

(c) Local Shear Stress on the Bed Using Two Point Velocities near the Bed

For this determination use Equation 2.99

(

)

(

)

2

2

ρ v1 - v 2

1000 0.633 - 0.518

= 4.37 N / m 2

τ0 =

=

2

2

y1

0.305

5.75 log

5.75 log

0.152

y2

(d) Shear Stress on the Bed Using Average Vertical Velocity

Using Equation 2.100 and the average depth 2.38 m, average velocity 0.738 m/sec, and a value

for ks determine the average shear stress on the bed.

Taking ks as (30.2) y' = 0.27 m

()

(

)

2

2

ρV

1000 0.738

= 3.98 N / m 2

τ0 =

=

2

2

12.27 y 0

12.27 x 2.38

5.75 log

5.75 log

0.27

ks

2.73

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