Rework example 5.2.3 for a 30-cm high hump and a side wall constriction that reduces the c...

Rework example 5.2.3 for a 30-cm high hump and a side wall constriction that reduces the channel width to 1.6 m.

EXAMPLE 5.2.3

A rectangular channel 2 m wide has a flow of 2.4 m3/s at a depth of 1.0 m. Determine whether critical depth occurs at (a) a section where a hump of Δz = 20 cm high is installed across the channel bed, (b) a side wall constriction (with no humps) reducing the channel width to 1.7 m, and (c) both the hump and side wall constrictions combined. Neglect headlosses of the hump and constriction caused by friction, expansion, and contraction.

SOLUTION

(a) The computation is focused on determining the critical elevation change in the channel bottom (hump) Δzcrit that causes a critical depth at the hump. The energy equation is E = Emin + Δzcrit or Δzcrit = E − Emin, where E is the specific energy of the channel flow and Emin is the minimum specific energy, which is at critical depth by definition. If Δzcrit ≤ Δz then critical depth will occur. Using equation (5.2.2) yields

which can be solved for q:

Differentiating this equation with respect to y because maximum q and minimum E are equivalent (see Figure 5.2.4) yields

To compute specific energy, use

Next compute Emin using Emin = 3/2yc, where yc = (q2/g)1/3 (equation (5.2.14)):

So Emin = 3/2(0.528 m) = 0.792 m. Then Δzcrit = E − Emin = 1.073 − 0.792 = 0.281 m. In this case Δz = 20 cm = 20/100 m = 0.2 m < Δzcrit = 0.281 m. Therefore, yc does not occur at the hump.

(b) The critical depth at the side wall constriction is

Thus Emin = (3/2)yc = (3/2)(0.588) = 0.882 m. E is computed above as E = 1.073 m. Because Emin = 0.882 m < E = 1.073 m, critical depth does not occur at the constriction. Remember that energy losses are negligible so that the specific energy in the constriction and upstream of the constriction must be equal. For critical flow to occur, the constriction width can be computed as follows: Emin = E = 1.073 m = (3/2)yc, so that yc = 0.715 m. Then using equation (5.2.14), 0.715 = [(2.4/Bc)2/9.81]1/3 and Bc = 1.267 m.

(c) With both the hump and the side wall constriction, yc is 0.588 m, Emin = 0.882 m. Then Δzcrit = E − Emin = 1.073 − 0.882 = 0.191 m.

Because Δz = 20/100 m = 0.20 m > Δzcrit = 0.191 m, critical depth will occur at the hump with a constriction.

Figure 5.2.4 Specific energy curve and y versus q for constant E.