Handbook of Civil Engineering Calculations

2. Apply the given values and solve for D
   Thus, He = 4.5 fHb/lb (1.37 J/N); Qu = 500/20 = 25 ft^3 /(s-ft) [2323 L/(s-m)]. Rearrange
   Eq. 26 to obtain

&(He -D)= -^- (26«)

Or, D
2
(4.5 -D) = 25
2
/64.4 = 9.705. This cubic equation has two positive roots, D = 1.95
ft (59.436 cm) and D = 3.84 ft (117.043 cm). There are therefore two stages of flow that
accommodate the required discharge with the given energy. [The third root of the equa-
tion is D = -1.29 ft (-39.319 cm), an impossible condition.]

3. Compute the slope associated with the computed depths
   Using Eq. 22, at the lower stage we have D = 1.95 ft (59.436 cm); A = 20(1.95) = 39.0 ft
   2

(36,231.0 cm^2 ); WP = 20 + 2(1.95) = 23.9 ft (728.47 cm); R = 39.0/23.9 = 1.63 ft (49.682
cm); V= 25/1.95 = 12.8 ft/s (390.14 cm/s); s = [«F/(1.486tf2/3)]^2 = (0.013 x 12.8/1.486 x
1.63^273 )^2 = 0.00654.
At the upper stage D = 3.84 ft (117.043 cm); A = 20(3.84) = 76.8 ft
2
(71,347.2 cm
2
);
WP = 20 + 2(3.84) = 27.68 ft (843.686 cm); R = 76.8/27.68 = 2.77 ft (84.430 cm); V =
25/3.84 = 6.51 ft/s (198.4 cm/s); s = [0.013 x 6.517(1.486 x 2.77
2/3
)]
2
= 0.000834. This
constitutes the solution to part a.

4. Plot the D-QU curve
   For part b, consider He as remaining constant at 4.5 ft-lb/lb (1.37 J/N) while Qu varies.
   Plot the D-QU curve as shown in Fig. 1 Ia. The depth that provides the maximum potential
   discharge is called the critical depth with respect to the given specific energy.
   5. Differentiate Eq. 26 to find the critical depth; then
   evaluate QUetnax
   Differentiating Eq. 26 and setting dQJdD = O yield

Critical depth Dc =

2

AHe (27)

Qu= Unit discharge, L/(s»m) H 6 = Specific energy, J/N

(a) Diagram for H 6 = 4.5ft (1.4m) (b) Diagram for Qu = 25 ft^3 /(s»ft)

[2323 L/(s-m)]

FIGURE 11

H 6 = Specific energy, ft • Ib/lb

Critical depth

Qu = Unit discharge, ft

3

/(s«ft)

Critical depth

D =

Depth,

ft

D =

Depth,

m

D =

Depth,

ft