Minimum reservoir capacity and wasted water

Water needed by the turbine

Take 1 HP = 0.7457 kW. Output $=1000\times0.7457 = 745.7$ kW, net head 30 m, overall efficiency 0.80:

$$Q = \frac{P}{9.81\,\eta\,H} = \frac{745.7}{9.81\times0.80\times30} = 3.167\ \text{m}^3/\text{s}$$

Monthly demand volume $D = 3.167\times2.628\times10^6 = 8.32\times10^6$ m³ $=8.32$ Mm³ (constant draft).

With the data as given (1000 HP)

The demand (8.32 Mm³/month) is far below the smallest monthly inflow (70 Mm³), so the inflow exceeds the demand in every month and the reservoir is never short of water.

• (i) Minimum capacity required $=0$ Mm³ (no storage is needed for this load).
• (ii) Water wasted $=\sum(\text{inflow}) - 12D = 1067 - 99.88 = 967.1$ Mm³ per year.

If the turbine is 10,000 HP (probable intended load)

The inflow data (mean 88.9 Mm³/month) match a turbine about ten times larger, so the working for 10,000 HP is also shown. $P=7,457$ kW, $Q=31.67$ m³/s, monthly demand $D=83.24$ Mm³.

The reservoir (full at the start of the year) falls to empty in May, so the minimum capacity is the deficit accumulated from February to May: 29.9 Mm³. The inflow after June refills it, and the water that cannot be stored (spill) adds up to 76.4 Mm³ per year.

(The same table applies if the load is 1000 HP but the inflows are really in units of 10⁵ m³; all volumes are then divided by 10, giving a capacity of about 2.99 Mm³ and wasted water of about 7.64 Mm³.)

Answer: for the 1000 HP load stated, minimum capacity = 0 and wasted water = 967 Mm³. For a 10,000 HP turbine, minimum capacity ≈ 29.9 Mm³ and wasted water ≈ 76.4 Mm³.