Pondage surface area and weekly output

Data

Inflow $Q=40$ m³/s; daily inflow volume $=40\times86400 = 3,456,000$ m³. The plant works 6 days (Sunday-Friday) and on Saturday the whole flow is stored. $H_{full}=25$ m, maximum fluctuation $=1$ m, $\eta_t=0.80$, $\eta_g=0.90$.

(i) Surface area of the pond

1. Weekly pondage (Saturday flow): all of Saturday's inflow is stored for use on the other days.

$$V_1 = 40\times86400 = 3,456,000\ \text{m}^3$$

2. Daily pondage: two requirements are given.

• To take care of the daily flow fluctuation: $0.30\times3,456,000 = 1,036,800$ m³
• For the 50% daily load factor: mean flow to the turbines on working days $=40\times\frac{7}{6}=46.67$ m³/s, so the pondage $=0.2\times46.67\times86400 = 806,400$ m³

The pond must satisfy both, so the larger value is taken: $V_2=1,036,800$ m³.

3. Total live storage:

$$V = V_1+V_2 = 3,456,000+1,036,800 = 4,492,800\ \text{m}^3$$

The allowed draw-down is 1 m, so

$$A = \frac{V}{\Delta h} = \frac{4,492,800}{1} = 4,492,800\ \text{m}^2 \approx 4.49\ \text{km}^2$$

(ii) Weekly output at the switchboard

The whole weekly inflow is used: $40\times7\times24 = 6,720$ m³/s·h. The pond level goes down by 1 m, so the average head is $25-0.5 = 24.5$ m.

$$E = 9.81\,\eta_t\eta_g\,H\,\sum Q\,t = 9.81\times0.80\times0.90\times24.5\times6,720 = 1,162,885\ \text{kWh}$$

(With the full-pond head of 25 m, the figure would be 1,186,618 kWh.)

Answer: pond surface area ≈ 4.49 km² (4,492,800 m²); weekly output ≈ 1.16 GWh (1,162,885 kWh).