With a design discharge of 45 m3/s, what flow is used for power in a month when the river carries 72 m3/s?

45 m3/s, because the turbines cannot pass more than the design discharge

58.5 m3/s, the average of the two

On what flow is firm energy based?

The minimum flow of 16 m3/s, available all the time

The flow exceeded 50% of the time

How is secondary energy obtained?

Total energy (with Q used) minus firm energy

Energy from flows above 45 m3/s

Which calculation gives the installed capacity?

9.81 x 0.82 x 60 x 45, about 21.7 MW, using the design discharge

9.81 x 0.82 x 60 x 35 = 16.9 MW

9.81 x 60 x 45 = 26.5 MW, ignoring efficiency

2069 Chaitra · 8 marks

Flow duration curve; firm and secondary energy

Short answer

With head 60 m, design discharge 45 m³/s, and 82% efficiency, firm energy is 67.6 GWh, secondary energy is 80.3 GWh, and total annual energy is about 148.0 GWh.

The stream flow record for a hydropower development site is given below. Draw a flow duration curve and determine firm and secondary energy if the available head is 60 m, design discharge capacity is 45 m³/s and overall efficiency is 82%.

Month	Jan	Feb	Mar	Apr	May	Jun	Jul	Aug	Sep	Oct	Nov	Dec
Q (m³/s)	30	38	28	22	16	32	56	72	54	46	38	36

Flow duration curve

Rank the monthly flows in descending order and take $p=\dfrac{m}{N+1}\times100$ with $N=12$ .

Rank	Q (m³/s)	Exceedance (%)	Q used = min(Q, 45)
1	72	7.7	45
2	56	15.4	45
3	54	23.1	45
4	46	30.8	45
5	38	38.5	38
6	38	46.2	38
7	36	53.8	36
8	32	61.5	32
9	30	69.2	30
10	28	76.9	28
11	22	84.6	22
12	16	92.3	16

 Q (m³/s)
  72 |*
  56 | *
  45 |-+--*-*-----  design flow
  38 |        * *
  30 |           * *
  22 |               *
  16 |                  *
     +----------------------
      0  20  40  60  80 100 %

The FDC is flat at 45 m³/s for about 31% of the time (months with flow above the design flow).

Energy

Constants: $H=60$ m, $\eta=0.82$ , $Q_d=45$ m³/s, 730 h per month.

Installed capacity $P=9.81\times0.82\times60\times45 = 21,719$ kW $=21.72$ MW
Firm energy is based on the flow available all the time (minimum flow 16 m³/s):

P_{firm} = 9.81\times0.82\times60\times16 = 7,722\ \text{kW},\quad E_{firm} = 7,722\times8760 = 67.65\ \text{GWh}

Total energy with $Q_{used}=\min(Q,45)$ : sum of used flows = 420 m³/s-months, so mean used flow = 35 m³/s

E_{total} = 9.81\times0.82\times60\times35\times8760 = 147.98\ \text{GWh}

Secondary energy $=E_{total}-E_{firm} = 147.98-67.65 = 80.33$ GWh

Answer: firm energy ≈ 67.6 GWh; secondary energy ≈ 80.3 GWh; total ≈ 148.0 GWh per year.

✓ Answer checked by Claude Opus 5.5 (high effort)

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