encroachment of backwater, or some other factor. The deceleration of liquid requires an
increase in hydrostatic pressure, but only part of the drop in velocity head is accounted
for by a gain in pressure head. The excess head is dissipated in the formation of a
turbulent standing wave. Thus, the phenomenon of hydraulic jump resembles the be-
havior of liquid in a pipe at a sudden enlargement, as analyzed in an earlier calculation
procedure.
Let D 1 and D 2 denote the depth of flow immediately upstream and downstream
of the jump, respectively. Then D 1 < Dc < D 2. Refer to Fig. lib. Since the hydraulic
jump is accompanied by a considerable drop in energy, the point on the D-He dia-
gram that represents D 2 lies both above and to the left of that representing D 1. There-
fore, the upstream depth is less than the depth that would exist in the absence of any
loss.
Using literal values, apply Eq. 11 to find the difference in hydrostatic forces per unit
width of channel that is required to decelerate the liquid. Solve the resulting equation for
D 1 :
D 2 12VlD 2 D\ \o.s
"'-T+
I-T^T) <29
>- Substitute numerical values in Eq. 29
Thus, Qu = 7500/100 = 75 ft^3 /(s-ft) [6969 L/(s-m)]; V 2 = 75/9 = 8.33 ft/s (2.538 m/s);
D 1 = -Y 2 + (2 x 8.33^2 x 9/32.2 + 92 /4)°-^5 = 3.18 ft (0.969 m).
RATE OF CHANGE OF DEPTH
IN NONUNIFORM FLOW
The unit discharge in a rectangular channel is 28 ft^3 /(s-ft) [2602 L/(s-m)]. The energy
gradient is 0.0004, and the grade of the channel bed is 0.0010. Determine the rate at
which the depth of flow is changing in the downstream direction (i.e., the grade of the
liquid surface with respect to the channel bed) at a section where the depth is 3.2 ft
(0.97 m).Calculation Procedure:- Express H as a function of D
Let H = total specific energy at a given section as evaluated by selecting a fixed horizon-
tal reference plane; L = distance measured in downstream direction; z = elevation of given
section with respect to datum plane; sb - grade of channel bed = -dz/dL; se = energy gra-
dient = -dHldL.
Express H as a function of D by annexing the potential-energy term to Eq. 26.
Thus,
H=
^+D+Z (30)