Rework example 5.5.3 with a flow rate of 450 cfs at a normal depth of 3.2 ft. All other da...

Rework example 5.5.3 with a flow rate of 450 cfs at a normal depth of 3.2 ft. All other data remain the same.

EXAMPLE 5.5.3

A rectangular channel is 10.0 ft wide and carries a flow of 400 cfs at a normal depth of 3.00. Manning’s n = 0.017. An obstruction causes the depth just upstream of the obstruction to be 8.00 ft deep. Will a jump form upstream from the obstruction? If so, how far upstream? What type of curve will be present?

SOLUTION

First a determination must be made whether a jump will form by comparing the normal depth and critical depth. Using equation (5.2.14), we find

and yc > yn = 3.0 ft, therefore the channel is steep. Because yn < yc < 8 ft, a subcritical flow exists on a steep channel, a hydraulic jump forms upstream of the obstruction. If the depth before the jump is considered to be normal depth, yn = y1 = 3, then the conjugate depth y2 can be computed using equation (5.5.4), where the Froude number is

so

Next the distance Δx from the depth of 4.45 ft to the depth of 8 ft is determined using equation (5.3.1):

which can be rearranged to yield

so

To solve for Δx, first compute E2 and E3:

Now compute Sf from Manning’s equation using equation (5.1.24) with Vave and Rave, and rearrange to yield

Compute S0 using Manning’s equation with the normal depth:

Thus, S0 = 0.0100. Now, using , the distance from the conjugate depth of the jump y2 = 4.45 ft downstream to the depth y3 of 8 ft (location of the obstruction) is 322 ft. In other words, the hydraulic jump occurs approximately 322 ft upstream of the obstruction. The type of water surface profile after the jump is an S1 profile.