Load duration curve; diesel plant; ROR flow duration

Each month is taken as 1/12 of the year (8.33% of the time). The months are already in descending order of power, so the table is the load duration curve (LDC).

a) Load duration curve and diesel plant

 MW
 550 |*
 500 | *
 450 |  *
 380 |   *
 330 |    *
 280 |     *
 250 |      *
 220 |       *
 200 |--------*---- Magh level (target)
 150 |         *
 145 |          *
 100 |           *
     +------------------------
      0   25   50   75   100 %

Target: at least the Magh power, 200 MW, in every month. Months below 200 MW and their deficits:

The import in these three months is exactly equal to the deficits, so on the LDC the supply is at or above 200 MW for 100% of the time and no further deficit is left to be covered by the diesel plant. The minimum diesel capacity needed is therefore 0 MW for the stated imports. If the imports were not available, or fell short by $x$ MW in a month, the minimum diesel plant would be the largest remaining deficit: with no import at all it would be the largest gap, 100 MW (in Baisakh), and it would run only for about 25% of the time (energy ≈ (50+55+100) MW × 730 h = 149,650 MWh/year).

b) Flow duration curve for a RoR hydro plant

A constant (future) demand equal to the table has to be met by the hydropower plant, so the required flow follows from

$$Q = \frac{P}{9.81\,\eta\,H}$$

Head and efficiency are not given, so assume $H=100$ m and $\eta=0.85$: $9.81\times0.85\times100 = 833.85$ kW per m³/s, i.e. $Q=P/833.85$ with $P$ in kW. The FDC has the same shape as the LDC.

So the flow required is 659.6 m³/s for 8.3% of the time, 263.8 m³/s for about 67% of the time and at least 119.9 m³/s all the year (flow scales as $1/(\eta H)$ for other values). The plant must be able to deliver 550 MW; for the minimum months the river must provide at least the flow of the Baisakh row.

Answer: (a) diesel capacity = 0 MW with the stated imports (deficits 50, 55, 100 MW are fully covered); (b) FDC as tabulated, with Q = P/(9.81ηH).