A 4-m wide rectangular channel is carrying 10 m³/s at a depth of 2.5 m. There is a step rise of 0.2 m in the channel bottom. Assuming there are no losses at the transition, determine the flow depth downstream of the bottom step. Does the water surface rise or fall at the step?

Given:

Q = 10 m³/s
B = 4 m
$y_{1}$ = 2.5 m
$Δ_{z}$ = 0.2 m
No head losses at the transition.

Determine:

$y_{2}$ = ?
Change in water-surface level = ?

Question

A 4-m wide rectangular channel is carrying 10 m³/s at a depth of 2.5 m. There is a step rise of 0.2 m in the channel bottom. Assuming there are no losses at the transition, determine the flow depth downstream of the bottom step. Does the water surface rise or fall at the step?

Given:

Q = 10 m³/s
B = 4 m
$y_{1}$ = 2.5 m
$Δ_{z}$ = 0.2 m
No head losses at the transition.

Determine:

$y_{2}$ = ?
Change in water-surface level = ?

Given:

Q = 10 m³/s
B = 4 m
[latex]y_{1}[/latex] = 2.5 m
[latex]Δ_{z}[/latex] = 0.2 m
No head losses at the transition.

Determine:

[latex]y_{2}[/latex] = ?
Change in water-surface level = ?

Accepted Answer

The flow velocity and the specific energy at section 1 are

[latex]V_{1} =\frac{Q}{A_{1} }[/latex]

[latex]=\frac{10}{4 imes 2.5}[/latex]

= 1 m/s

[latex]E_{1} =y_{1} +\frac{V^{2}_{1} }{2g}[/latex]

[latex]=2.5 +\frac{1}{2 imes 9.81}[/latex]

= 2.55 m

Since there are no losses, referring to Fig. 2-10

[latex]E_{2} =E_{1} -0.2[/latex]

= 2.55-0.2

= 2.35 m

Now

[latex]E_{2} =y_{2} +\frac{Q^{2} }{2gA^{2}_{2} }[/latex]

By substituting values of [latex]E_{2}[/latex], Q, and [latex]A_{2}[/latex] = 4[latex]y_{2}[/latex] into this equation and simplifying, we obtain

[latex]y^{3}_{2} -2.35y^{2}_{2} +0.32=0[/latex]

Solution of this equation by trial and error yields three roots: 2.29, 0.405, and −0.345 m. The third root is physically impossible because of the negative depth. In addition, only the first root is possible, since the upstream flow is subcritical, i.e., [latex]F_{r}[/latex] < 1; the second root requires that the flow has to pass through the critical depth at the step. Hence, [latex]y_{2}[/latex] = 2.29 m is the only possible downstream depth. Let us use the channel bottom upstream of the transition as the datum. Then

Water level downstream of the transition = 0.2 + 2.29

= 2.49 m

Thus, the water-surface level drops by 2.5 - 2.49 = 0.01 m.

Question 2.1: A 4-m wide rectangular channel is carrying 10 m³/s at a dept...

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