What is the design discharge for full regulation of the annual hydrograph?

Mean available flow, 0.9 x 75 = 67.5 m3/s, so the plant runs constantly

Which gives the installed capacity and annual energy?

9.81 x 0.855 x 150 x 67.5 = 84.9 MW and E = P x 8760

2071 Chaitra · 3+2+3 marks

Full regulation: reservoir, capacity, annual energy

Short answer

Reservoir capacity ≈ 437.6 Mm³, installed capacity ≈ 84.9 MW, annual energy ≈ 744 GWh.

A hydropower project is planned to develop in a Nepalese river having net head of 150 m, turbine efficiency of 90% and generator efficiency of 95% with the monthly hydrograph as shown below:

Months	Oct	Nov	Dec	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep
Q (m³/sec)	100	80	60	50	40	30	40	50	70	110	150	120

As an environmental flow, a minimum flow of 10% of each month is mandatory. If the storage project is designed with full regulation of the annual hydrograph, find out the capacity of the reservoir, installed capacity of the power plant, and annual energy generation.

Data

$H=150$ m, $\eta=\eta_t\eta_g = 0.90\times0.95=0.855$ . Environmental flow = 10% of each month's flow, so the flow available for power is $0.9\,Q$ . One month $=2.628\times10^6$ s.

Sum of flows $=900$ m³/s, mean $=75$ m³/s. Mean available flow $=0.9\times75 = 67.5$ m³/s. With full regulation the design discharge is this mean flow, so the plant runs at a constant $Q_d=67.5$ m³/s.

Reservoir capacity (mass-curve method)

The reservoir is full at the start of December, when the inflow drops below the draft. Cumulative of (0.9Q − 67.5):

Month	Q (m³/s)	0.9Q	0.9Q − mean	Cumulative (m³/s-month)
Dec	60	54.0	-13.5	-13.5
Jan	50	45.0	-22.5	-36.0
Feb	40	36.0	-31.5	-67.5
Mar	30	27.0	-40.5	-108.0
Apr	40	36.0	-31.5	-139.5
May	50	45.0	-22.5	-162.0
Jun	70	63.0	-4.5	-166.5
Jul	110	99.0	+31.5	-135.0
Aug	150	135.0	+67.5	-67.5
Sep	120	108.0	+40.5	-27.0
Oct	100	90.0	+22.5	-4.5
Nov	80	72.0	+4.5	+0.0

The cumulative reaches its lowest value of $-166.5$ m³/s-months in June; from July the surplus refills the reservoir.

V = 166.5\times2.628\times10^6 = 437.6\times10^6\ \text{m}^3 \approx 438\ \text{Mm}^3

Installed capacity

P = 9.81\,\eta\,H\,Q_d = 9.81\times0.855\times150\times67.5 = 84,924\ \text{kW} = 84.92\ \text{MW}

Annual energy

E = P\times8760 = 743.9\ \text{GWh}

(The head is taken as constant at 150 m; in practice the head varies with the reservoir level.)

Answer: reservoir capacity ≈ 437.6 Mm³; installed capacity ≈ 84.9 MW; annual energy ≈ 744 GWh.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Full regulation: reservoir, capacity, annual energy

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