Handbook of Civil Engineering Calculations

Or, Dc =

2

/ 3 (4.5) = 3.0 ft (91.44 cm); gMmax = [64.4(4.5 x 3.0

2

• 3.O
  2
  )]
  0
  
  
  • 5
    = 29.5 ft
    3
    /(s-ft)
    (2741 L/(s-m)]; gmax = 29.5(20) = 590 ftVs (16,704.2 L/s). This constitutes the solution to
    part b.
  
  
  
  

6. Plot the D-H 0 curve
   For part c, consider Qu as remaining constant at 25 ft^3 /(s-ft) [2323 L/(s-m)] while H 0
   varies. Plot the D-H 6 curve as shown in Fig. UZ?. (This curve is asymptotic with the
   straight lines D = H 6 and D = 0.) The depth at which the specific energy is minimum is
   called the critical depth with respect to the given unit discharge.


7. Differentiate Eq. 26 to find the critical depth; then
   evaluate f/e,m/n
   Differentiating gives

/ Ql \

1/3

MT)

<28)

Then Dc = (25^2 /32.2)1/3 = 2.69 ft (81.991 cm). Then Hemin = 252 /[64.4(2.69)^2 ] + 2.69 =
4.03 ft-lb/lb(l.229 J/N).
The values of D as computed in part a coincide with those obtained by referring to the
two graphs in Fig. 11. The equations derived in this procedure are valid solely for rectan-
gular channels, but analogous equations pertaining to other channel profiles may be de-
rived in a similar manner.

DETERMINATION OF HYDRAULIC JUMP

Water flows over a 100-ft (30.5-m) long dam at 7500 ft^3 /s (212,400 L/s). The depth of

tailwater on the level apron is 9 ft (2.7 m). Determine the depth of flow immediately up-

stream of the hydraulic jump.

Calculation Procedure:

1. Find the difference in hydrostatic forces per unit width
   of channel required to decelerate the liquid
   Refer to Fig. 12. Hydraulic jump designates an abrupt transition from lower-stage to
   upper-stage flow caused by a sharp decrease in slope, sudden increase in roughness,

FIGURE 12. Hydraulic jump on apron of dam.