How are the points of the flow duration curve obtained from the monthly flows?

Flows are sorted in descending order and the exceedance p = m/(N+1) x 100 is assigned to each rank m

Flows are kept in calendar order and plotted against month

Flows are sorted in ascending order and plotted against m/N

Flows are averaged in groups of three

On the FDC, the point (150 m3/s, 7.7%) means that

150 m3/s is equalled or exceeded about 7.7% of the time

150 m3/s occurs exactly 7.7% of the time

150 m3/s is the average flow in 7.7% of months

What is the mean flow of the river?

72.5 m3/s, the sum of 870 divided by 12 months

Which expression gives the power at the mean flow with H = 100 m and efficiency 85%?

P = 9.81 x 72.5 x 100 x 0.85, about 60.45 MW

P = 9.81 x 72.5 x 100 = 71.1 MW, efficiency ignored

2079 Bhadra · 4+4 marks

Flow duration curve and power at mean flow

Short answer

The mean monthly flow is 72.5 m³/s. Power at mean flow with 100 m head and 85% overall efficiency is 60.45 MW.

In a Nepali river the mean monthly flow in the year 2021 is given below.

Month	Discharge (m³/s)	Month	Discharge (m³/s)
Jan	50	July	125
Feb	40	Aug	150
March	30	Sept	120
April	25	Oct	100
May	10	Nov	75
June	75	Dec	70

a) Draw the flow duration curve. b) Find the power available at the mean flow of water if the available head is 100 m at the site and the overall efficiency of the plant is 85%.

a) Flow duration curve

Arrange the 12 monthly flows in descending order and assign the exceedance probability by the Weibull formula $p = \dfrac{m}{N+1}\times100$ with $N = 12$ .

Rank	Q (m³/s)	Exceedance (%)
1	150	7.7
2	125	15.4
3	120	23.1
4	100	30.8
5	75	38.5
6	75	46.2
7	70	53.8
8	50	61.5
9	40	69.2
10	30	76.9
11	25	84.6
12	10	92.3

Plot Q (vertical) against percentage of time (horizontal) and join the points with a smooth curve:

 Q (m³/s)
 150 |*
 125 | *
 100 |   *
  75 |      * *
  50 |           *
  25 |               * *
  10 |                    *
     +--------------------------
      0  20  40  60  80  100 %
         time flow is exceeded

b) Power at the mean flow

Mean flow $\bar Q = \dfrac{\sum Q}{12} = \dfrac{870}{12} = 72.5$ m³/s

P = \gamma Q H \eta = 9.81\times72.5\times100\times0.85 = 60,454.12\ \text{kW} = 60.45\ \text{MW}

The mean flow of 72.5 m³/s is exceeded about 46% of the time (six of the twelve months have flows above it).

Answer: mean flow = 72.5 m³/s; power at mean flow ≈ 60.45 MW (the FDC is tabulated above).

✓ Answer checked by Claude Opus 5.5 (high effort)

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