How is a power duration curve obtained from the flow duration data?

By converting each flow to power with P = 9.81 Q H eta

By ranking power in ascending order

Which flow gives primary energy?

100 m^3/s available 100% of time

How is total energy found?

Area under the curve capped at 380 m^3/s, giving mean used flow 323 m^3/s

2070 Magh · 10 marks

Power duration curve and primary/secondary energy

Short answer

Installed capacity at 45% exceedance flow (380 m³/s) is 331.8 MW. Primary energy from minimum flow (100 m³/s) is 764.8 GWh/yr; secondary energy 1705.6 GWh/yr from area under the capped power duration curve.

From the following data of flow at a given site for an average year compute and draw a power duration curve. Assume an average net available head of 100 m and a combined turbine generator efficiency of 89%. Determine the primary and secondary energy available during a year if the plant capacity is fixed at power corresponding to the flow available for 45% of time.

Flow (m³/s)	900	600	500	450	400	360	340	300	280	200	140	100
% time	1	10	20	30	40	50	60	70	80	90	93	100

A power duration curve is the flow duration curve with the flow axis converted to power, $P = 9.81\,Q H \eta$ . The area under it gives the energy.

Power for each flow

P = 9.81 \times Q \times 100 \times 0.89 = 873.09\,Q\ \text{kW}

% time	Q (m³/s)	Power (MW)
1	900	785.78
10	600	523.85
20	500	436.55
30	450	392.89
40	400	349.24
50	360	314.31
60	340	296.85
70	300	261.93
80	280	244.47
90	200	174.62
93	140	122.23
100	100	87.31

P (MW)
 786 |*
     | \
 524 |  *
 437 |   *
 393 |    *
 349 |     *
 332 |======*=====================  <- installed capacity 331.8 MW (Q45)
 314 |       *
 297 |        *
 262 |          *      secondary
 244 |            *
 175 |               *
 122 |                  *
  87 |=====================*=====  <- firm power 87.3 MW
     |        primary (firm)
     +------------------------------>
     0      45             93  100   % time

Plant capacity (flow at 45% time)

\begin{aligned} Q_{45} &= 400 - \frac{45-40}{50-40}(400-360) = 380\ \text{m}^3/\text{s} \\ P_{inst} &= 873.09 \times 380 = 331{,}774\ \text{kW} \approx 331.8\ \text{MW} \end{aligned}

Primary energy

Flow available 100% of the time = 100 m³/s.

\begin{aligned} P_f &= 873.09 \times 100 = 87{,}309\ \text{kW} \\ E_p &= 87{,}309 \times 8760 = 764.83\ \text{GWh} \end{aligned}

Total energy (area under capped curve, trapezoidal rule)

Flow used is capped at 380 m³/s from 0% to 45%.

Interval (%)	Mean Q used	Δt (%)	Area
0–45	380	45	17,100
45–50	370	5	1,850
50–60	350	10	3,500
60–70	320	10	3,200
70–80	290	10	2,900
80–90	240	10	2,400
90–93	170	3	510
93–100	120	7	840
Total			32,300

Mean flow used $= 32{,}300/100 = 323$ m³/s.

\begin{aligned} E_{total} &= 873.09 \times 323 \times 8760 = 2470.39\ \text{GWh} \\ E_s &= 2470.39 - 764.83 = 1705.56\ \text{GWh} \end{aligned}

Plant factor $= 323/380 = 85\%$ .

Answer: Installed capacity ≈ 331.8 MW; primary energy ≈ 764.8 GWh/yr; secondary energy ≈ 1705.6 GWh/yr.

✓ Answer checked by Claude Opus 5.5 (high effort)

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