Reservoir routing through spillway, 6-hr inflow

Use the storage-indication (modified Puls) method. The reservoir is assumed to start full to the spillway crest (elevation 140 m), so $O = 0$ and $S = 0$ at $t = 0$.

Method

Continuity over one interval $\Delta t$: $\dfrac{I_1 + I_2}{2} - \dfrac{O_1 + O_2}{2} = \dfrac{S_2 - S_1}{\Delta t}$, which rearranges to

$$\left(\frac{S_2}{\Delta t} + \frac{O_2}{2}\right) = \frac{I_1 + I_2}{2} + \left(\frac{S_1}{\Delta t} - \frac{O_1}{2}\right)$$

$\Delta t = 6 \times 3600 = 21600$ s.

Step 1: Storage-indication curve

Step 2: Routing

At each step, $\psi_2$ is computed from the right-hand side, then $O_2$ and $S_2$ are found by linear interpolation in the table above ($S_1/\Delta t - O_1/2 = \psi_1 - O_1$).

Results

• Peak inflow = 2130 m³/s at 36 h.
• Peak outflow occurs at t = 54 h (where the outflow curve meets the inflow curve, $O \approx I$).
• Storage at that time $\approx$ 101.8 × 10⁶ m³, which lies between 144 m (95) and 145 m (140).

$$\text{Max. level} = 144 + \frac{101.8 - 95}{140 - 95} = 144.15\ \text{m}$$

Answer:

• i) Maximum reservoir level $\approx$ 144.15 m
• ii) Maximum outflow rate $\approx$ 1442 m³/s
• iii) Reduction in peak $= 2130 - 1442 =$ 688 m³/s (32.3 % of the peak inflow)