- Use the flow mass curve to determine monthly deficits.
The total annual volume is 156,000 ac-ft. Therefore, the average monthly release (draft) is:
156,000 / 12 = 13,000 ac-ft.
Find the cumulative draft for each month, and subtract the cumulative draft minus the cumulative inflow to determine the
monthly deficit.
| Month | October | November | December | January | February | March | April | May | June | July | August | September
|
| Inflow volume (ac-ft × 10-3) | 5 | 10 | 12 | 20 | 32 | 28 | 15 | 12 | 8 | 5 | 5 | 4
|
| Cumulative inflow volume (ac-ft × 10-3) | 5 | 15 | 27 | 47 | 79 | 107 | 122 | 134 | 142 | 147 | 152 | 156
|
| Cumulative draft (ac-ft × 10-3) | 13 | 26 | 39 | 52 | 65 | 78 | 91 | 104 | 117 | 130 | 143 | 156
|
| Deficit (ac-ft × 10-3) | +8 | +11 | +12 | +5 | -14 | -29 | -31 | -30 | -25 | -17 | -9 | 0
|
The required reservoir storage volume is the sum of the maximum positive deficit (+12) and maximum negative deficits (-31).
The required storage volume is: V = 43,000 ac-ft.
- Fetch: F = 37 mi × 5280 ft/mi = 195,360 ft.
Wind speed: U = 40 mph × (5280 ft/mi( / (3600 s/hr) = 58.7 fps.
U2/(gD) = (58.7)2/(32.2 × 7) = 15.29
Setup: S = 0.000002025 F [U2/(gd)] = 0.000002025 × 195,360 × 15.29 = 6.05 ft. ≅ 6 ft.
From Fig. 6-31, the significant wave height for F = 37 mi and U = 40 mph: Hs = 9.0 ft.
From Table 6-4, for 2% wave exceedance, the ratio H/Hs = 1.4
The specific wave height is: Ho = 1.4 Hs = 1.4 × 9.0 = 12.6 ft.
The wave period is: T = 0.429 U0.44 F0.28 / g0.72
The wave period is: T = 0.429 (58.7)0.44 (195,360)0.28 / (32.2)0.72 = 6.4 s.
The wave length is: Lo = 0.159 g T2 = 0.159 × 32.2 × (6.4)2 = 209.7 ft.
Ho / Lo = 12.6 / 209.7 = 0.06
From Fig. 6-33, for embankment slope 0.33 and Ho / Lo = = 0.06: R/Ho = 0.6
Runup: R = 0.6 Ho = 0.6 × 12.6 = 7.6 ft = 8 ft
Freeboard: F = S + R + settlement + contingencies = 6 + 8 + 2 + 1 = 17 ft.
- The Dendy and Bolton formula, for Q ≥ 2 in is:
S = 1965 e-0.055 Q (1.43 - 0.26 log A)
with S in ton/mi2/yr, A in mi2, and Q in inches.
S = 1965 e (-0.055 × 3.5) [ 1.43 - 0.26 log (188) ]
S = 1965 × 0.825 × ( 1.43 - 0.26 × 2.274 )
S = 1359.7 ton/mi2/yr.
S = 1359.7 × 188 = 255,624 ton/yr.
- Use ONLINE RESERVOIR LIFE with given data.
The number of years to fill 80% of the volume (20% reserved for dead storage) of the reservoir is 679 yr.
Rerun ONLINE RESERVOIR LIFE with S = 1.3 × 1350 = 1755.
With S = 1755, the useful life of the reservoir is 522 yr.