An emergency spillway is being considered for the
East Demerera Water Conservancy, in Guyana, to safeguard the integrity of the dam under conditions of climate change (similar to Fig. 920).
Assume that the existing relief sluices would be inoperable during a major flood due to high tailwater.
Determine the length of the freeoverflow spillway
required to pass the Probable Maximum Flood (PMF). The following data is applicable:
 PMF 1day duration: 428 mm
 Hydrologic abstraction: 18 mm
 Contributing drainage area: 582 km^{2}
 Time base of the flood hydrograph: 3 days
 Embankment crest elevation: 18.288 m
 Spillway crest elevation: 17.526 m
 Freeboard: 0.3 m
 Weir discharge coefficient: 1.45
For simplicity, assume a triangularshaped flood hydrograph. Use all the freeboard to contain the PMF.
Fig. 920 The 8000ft weir,
Boerasirie Conservancy, Guyana.


The volume of runoff is:
V_{r} = (428  18) mm × 582 km^{2} × (1 m / 1000 mm) × (1000 m / 1 km)^{2}
V_{r} = 238,620,000 m^{3}
Time base T_{b} = 3 d × 86,400 s/d = 259,200 s
Assuming a triangularshaped flood hydrograph:
Q_{p}
= 2 V_{r} / T_{b} = (2) × (238,620,000) /
(259,200)
Q_{p} = 1841.2 m^{3}
When the entire freeboard is used, the head H is:
H = 18.288  17.526 = 0.762 m
L = Q / (C H^{ 3/2}) = 1841.2 / (1.45 × 0.762^{3/2}) = 1909 m. ANSWER.
Design an overflowspillway section having a vertical
upstream face and a crest length L = 150 ft.
The design discharge is Q = 50,000 ft^{3}/s.
The upstream water surface at design discharge is at Elev. 750 ft,
and the average channel floor is at Elev. 650 ft (see Fig. 96 for graphical example).
Q = 50,000 cfs; L = 150 ft.
h + H_{d} = 750  650 = 100 ft.
Assume h / H_{d} > 1.33: C_{d} = 4.03
Q = C_{d} L H_{e}^{1.5}
H_{e} = [50000/(4.03 × 150)]^{2/3} = 18.98 ft.
V_{a} = Q / [L (h + H_{d})] = 50000/(150 × 100) = 3.33 fps.
H_{a} = V_{a}^{2}/(2g) = 3.33^{2}/(2 × 32.17) = 0.17 ft.
H_{d} = H_{e}  H_{a} = 18.98  0.17 = 18.81 ft.
h = 100  H_{d} = 100  18.81 = 81.19 ft.
h/ H_{d} = 81.19 / 18.81 = 4.31 > 1.33 OK!
Crest elevation: 750  18.81 = 731.19 ft.
Crest shape: X^{ 1.85} = 2 H_{d}^{ 0.85} Y
X^{1.85} = 2 (18.81)^{ 0.85} Y
Y = 0.04128 X^{ 1.85}
Y' = 1.0/0.6 = 1.85 × 0.04128 X^{ 0.85}
1.66667 = 0.0764 X^{ 0.85}
21.82 = X^{ 0.85}
X = (21.82)^{1.176} = 37.54 ft. ANSWER.
Y = 0.04128 X^{ 1.85} = 33.77 ft. ANSWER.
0.282 H_{d} = 5.30 ft. ANSWER.
0.175 H_{d} = 3.29 ft. ANSWER.
0.5 H_{d} = 9.40 ft. ANSWER.
0.2 H_{d} = 3.76 ft. ANSWER.
Use ONLINE OGEE RATING to
determine the rating for an ogee spillway with length L = 15 m,
design head H_{d} = 2 m, spillway crest elevation = 1045 m,
river bed elevation = 1000 m, and freeboard F_{b} = 1 m.
Neglect the approach velocity.
What should be the spillway length to pass the Probable Maximum Flood
Q_{PMF} = 250 m^{3}/s while taking all the freeboard?
Express spillway length to nearest 0.1 m by excess.
Run ONLINE OGEE RATING
with spillway length L = 15 m, design head H_{d} = 2 m, dam height
P = 45 m, freeboard F_{b} = 1 m, and spillway crest
elevation
E = 1045 m, to give: (1) design discharge at Elev. 1047 is Q = 92.647 m^{3}/s;
(2) discharge to use all freeboard, at Elev. 1048, is Q_{Fb} = 180.246 m^{3}/s. ANSWER.
By trial and error, for spillway length L = 20.9 m, the discharge to use all freeboard, at Elev. 1048,
is Q_{Fb} = 251.143 m^{3}/s. ANSWER.
Prove Eq. 914.
ΔE = E_{1}  E_{2} =
y_{1} + [V_{1}^{ 2}/(2g)]  y_{2}  [V_{2}^{ 2}/(2g)]
 (1) 
ΔE =
y_{1} + y_{1} [F_{1}^{ 2}/2]  y_{2}  y_{1} [F_{1}^{ 2}/2] (V_{2} / V_{1})^{2}
 (2) 
ΔE =
y_{1} + y_{1} [F_{1}^{ 2}/2]  y_{2}  y_{1} [F_{1}^{ 2}/2] (y_{1} / y_{2})^{2}
 (3) 
The hydraulic jump equation, Eq. 913:
y_{2} 1 ^{____} = ^{____} [
(1 + 8 F_{1}^{ 2} )^{1/2}  1 ]
y_{1} 2
 (4) 
Combining Eqs. 3 and 4, after some algebraic manipulation:
(y_{2}  y_{1})^{3}
ΔE = ^{ _____________ }
4 y_{1} y_{2}
 (5) 
Prove Eq. 919.
h_{j} / E_{1} =
(y_{2}  y_{1}) / E_{1}
h_{j} / E_{1} = [(y_{2} / y_{1})  1 ] / (E_{1}/y_{1})
h_{j} / E_{1} = [(2 y_{2} / y_{1})  2 ] / [2 (E_{1}/y_{1})]
h_{j} / E_{1} = [(1 + 8F_{1}^{2})^{1/2}  3 ] / [2 (E_{1}/y_{1})]
(E_{1}/y_{1}) = 1 + (1/2)F_{1}^{2}
(2 E_{1}/y_{1}) = 2 + F_{1}^{2}
h_{j} / E_{1} = [(1 + 8F_{1}^{2})^{1/2}  3 ] /
[ 2 + F_{1}^{2} ]
Calculate the energy loss in a
hydraulic jump, given the sequent depths
y_{1} = 0.58 m and y_{2} = 2.688 m.
Using Eq. 914: ΔE = 1.5 m.
ANSWER.
Calculate the relative
height of the hydraulic jump h_{j} / E_{1} for F_{1} = 3.
Using Eq. 919: h_{j} / E_{1} = 0.504
ANSWER.
Use ONLINE CHANNEL 11
to calculate the sequent depth y_{2} through a hydraulic jump when the
discharge is q = 5 m^{2}/s and the initial depth y_{1} = 0.58 m.
Run ONLINE CHANNEL 11 with v_{1} = 5.0 / 0.58 = 8.621 m^{2}/s and
y_{1} = 0.58 m to give y_{2} = 2.688 m. ANSWER.
Use ONLINE CHANNEL 12
to determine the sequent depth y_{2} and energy loss ΔE
through a hydraulic jump when the
discharge is q = 10 ft^{2}/s and the upstream flow depth is y_{1} = 0.5 ft.
v_{1} = q / y_{1} = 10 / 0.5 = 20 ft/s
Run ONLINE CHANNEL 12 with y_{1} = 0.5 ft and v_{1} = 20 ft/s to give: y_{2} = 3.285 ft and ΔE = 3.287 ft. ANSWER.
Use ONLINE CHANNEL 16
to calculate the sequent depths through a hydraulic jump when the
discharge is q = 10 ft^{2}/s and the energy loss is ΔE = 3.287 ft.
Run ONLINE CHANNEL 16 with q = 10 ft^{2}/s and
ΔE = 3.287 ft to give: y_{1} = 0.5 ft and y_{2} = 3.284 ft. ANSWER.
Use ONLINE CHANNEL 16
to calculate the sequent depths through a hydraulic jump when the
discharge is q = 5 m^{2}/s and the energy loss is ΔE = 1.5 m.
Run ONLINE CHANNEL 16 with q = 5 m^{2}/s and
ΔE = 1.5 m to give: y_{1} = 0.58 m and y_{2} = 2.687 m. ANSWER.
Use ONLINE CHANNEL 18 to
calculate the efficiency of the hydraulic jump E_{2} / E_{2} for
q = 10 ft^{2}/s and
y_{1} = 0.5 ft.
v_{1} = q / y_{1} = 10 / 0.5 = 20 ft/s
Run ONLINE CHANNEL 18 with y_{1} = 0.5 ft and v_{1} = 20 ft/s to give: E_{2} / E_{2} =
0.51. ANSWER.