Question 14.2: A reservoir has a constant surface area AWS of 0.15 km². The......
A reservoir has a constant surface area A_{WS} of 0.15 km². The inflow hydrograph to the reservoir is shown in the first two columns of Table 14.2. The outflow from the reservoir is over a broad crested weir that has a coefficient C = 1.6 and a breadth b = 10 m. Assuming that initially Q_{I} = Q_{O}, using a time interval Δt = 30 min, calculate the routed outflow from the reservoir and determine the attenuation and time lag between the two hydrograph peaks.
Equation (14.4) can be rearranged as:
A_{ WS } \Delta H=\left(Q_{ MI }-Q_{ MO }\right) \Delta t (14.4)
\Delta H=\left(Q_{M I}-Q_{M O}\right) \Delta t / A_{W S} (14.6)
so \Delta H=\left(Q_{M I}-Q_{M O}\right) \times 30 \times 60 / 0.15 \times 10^6
\Delta H=0.012\left(Q_{M 1}-Q_{M O}\right) (1)
Equation (14.5) is Q_O=C b H^{3 / 2} which becomes Q_O=1.6 \times 10 \times H^{3 / 2}
so Q_O=16 H^{3 / 2} (2)
and H =\left(Q_O / 16\right)^{2 / 3} (3)
The calculations are conducted using a table, as in Table 14.2. Computer spreadsheets are ideal for obtaining quick and accurate trial and error solutions. Although great precision is not justified, below 3 or 4 decimals are used to avoid what would otherwise look like obvious numerical errors or discrepancies in the calculations. Because a spreadsheet was used, the guessed values look unbelievably precise. No incorrect guesses are recorded.
Time = 0
Initially at t = 0 the instantaneous discharge is 1.42 m³/s (column 2) and under steady flow conditions this equals the instantaneous outflow recorded in column 9. From equation (3) the head over the outflow weir H = (1.42/16)^{{2}/{3}} = 0.1990 m, which is recorded in column 8.
Time period Δt = 0 to 0.5 h
During Δt, the mean inflow Q_{MI} in column 4 = (1.42 + 2.03)/2 = 1.725 m³/s.
Guess that in column 5 the corresponding mean outflow Q_{MO} = 1.439 m³/s.
The net mean inflow (Q_{MI} − Q_{MO}) in column 6 = (1.725 − 1.439) = 0.286 m³/s.
From equation (1) this increases the water level by ΔH = 0.012 × 0.286 = 0.0034 m (column 7).
This is added to the previous H value in column 8, so now H = 0.0034 + 0.1990 = 0.2024 m.
From equation (2), the instantaneous routed outflow at the end of Δt is Q_O=16 H^{3 / 2}= 16 \times 0.2024^{3 / 2}=1.457 m ^3 / s (column 9).
In column 10 the calculated Q_{MO} is the average of the values in column 9 at the start and end of the time period, so Q_{MO} = (1.420 + 1.457)/2 = 1.439 m³/s. This is the same as the guessed value in column 5, so the solution is valid (otherwise, guess again and repeat the calculations).
This procedure is repeated for the other time periods until the outflow hydrograph is complete.
The calculations up to t = 6 h are in Table 14.2 and the instantaneous hydrographs are plotted in Fig. 14.8. The maximum attenuation is 11.11 − 4.95 = 6.16 m³/s and the time lag between the hydrograph peaks is (3.0 − 2.0) = 1.0 h.
As an illustration of the validity of equation (14.4), if, for example, the precise values from the spreadsheet solution are adopted for the time period 1.5 – 2.0 h, then:
A_{W S} \Delta H=\left(Q_{M I}-Q_{M O}\right) \Delta t
0.15 \times 10^6 \times 0.0815 \approx 6.7927 \times 30 \times 60
12 225 m³ ≈ 12 227 m³
