---
title: "Annual firm and secondary energy (900 m³/s, Nov 650)"
subject: "Hydro Power"
chapter: 3
key: "semester_6/hydro_power:firm-secondary-energy-2079"
source: https://www.bctengineer.com/q/semester_6/hydro_power/firm-secondary-energy-2079
site: BCT Engineering
---

# Annual firm and secondary energy (900 m³/s, Nov 650)

Hydro Power · Chapter 3: Planning of Hydropower Projects

- Asked in: 2079 Chaitra, 8 marks
- Video: [Calculating Power, Primary & Secondary Energy | Power and Energy potential || Hydropower Engineering](https://www.youtube.com/watch?v=3v4OCvSD9fU) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/firm-secondary-energy-2079
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3

## Question

Following table shows the mean monthly flow of a typical river in Nepal. Calculate annual firm and secondary energy if design discharge selected is 900 m³/s. Take overall efficiency = 90% and head 100 m.

| Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Discharge (m³/s) | 120 | 100 | 90 | 150 | 350 | 1800 | 2000 | 2500 | 2000 | 900 | 650 | 500 |

## Short answer

Annual firm energy 696.1 GWh from minimum flow 90 m³/s. Secondary energy 3467.5 GWh, total 4163.6 GWh. Installed capacity 794.6 MW at design discharge 900 m³/s. Power per unit discharge 882.9 kW per m³/s. Monthly energy summed using flows capped at 900 m³/s over 730 hours per month.

## Answer

**Firm energy** comes from the minimum flow, which is available throughout the year. **Secondary energy** is the extra energy from flows above the minimum, up to the design discharge (900 m³/s).

#### Power per unit discharge

$$
P = 9.81 \times Q \times 100 \times 0.90 = 882.9\,Q\ \text{kW}
$$

Installed capacity $= 882.9 \times 900 = 794{,}610$ kW ≈ 794.6 MW.

#### Firm energy
Minimum flow = 90 m³/s (March).

$$
\begin{aligned}
P_f &= 882.9 \times 90 = 79{,}461\ \text{kW} \\
E_f &= 79{,}461 \times 8760 = 696.08\ \text{GWh}
\end{aligned}
$$

#### Total energy (flow used = smaller of Q and 900; 730 h per month)

| Month | Q (m³/s) | Q used | Power (MW) |
|---|---|---|---|
| Jan | 120 | 120 | 105.95 |
| Feb | 100 | 100 | 88.29 |
| Mar | 90 | 90 | 79.46 |
| Apr | 150 | 150 | 132.44 |
| May | 350 | 350 | 309.02 |
| Jun | 1800 | 900 | 794.61 |
| Jul | 2000 | 900 | 794.61 |
| Aug | 2500 | 900 | 794.61 |
| Sep | 2000 | 900 | 794.61 |
| Oct | 900 | 900 | 794.61 |
| Nov | 650 | 650 | 573.89 |
| Dec | 500 | 500 | 441.45 |
| **Sum** | | **6460** | |

$$
\begin{aligned}
E_{total} &= 882.9 \times 6460 \times 730 = 4163.58\ \text{GWh} \\
E_s &= 4163.58 - 696.08 = 3467.50\ \text{GWh}
\end{aligned}
$$

**Answer: Annual firm energy ≈ 696.1 GWh; annual secondary energy ≈ 3467.5 GWh (total ≈ 4163.6 GWh).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which flow fixes the firm energy?
   A. Design discharge 900 m^3/s
   B. Minimum flow 90 m^3/s (March)
   C. Mean flow
   D. Flow of October
   Answer: B. Firm energy comes from the minimum flow available all year. (Key point: Firm energy flow)

2. How is annual firm energy computed?
   A. 882.9 x 900 x 730
   B. 882.9 x 90 x 730
   C. 882.9 x 90 x 8760 kWh
   D. 882.9 x 900 x 8760
   Answer: C. Firm power 882.9 x 90 = 79,461 kW over 8760 h gives 696.08 GWh. (Key point: Firm energy calculation)

3. Why is the Q used in June to September taken as 900 m^3/s?
   A. Efficiency changes in the monsoon
   B. Flow above design discharge cannot pass through the turbines and is spilled
   C. It is the average flow
   D. The head drops in those months
   Answer: B. Used flow is the smaller of the river flow and the design discharge. (Key point: Design discharge cap)

## Close variants set in other exams

- [Annual firm and secondary energy (900 m³/s, Nov 100)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afirm-secondary-energy-2078) — Hydro Power, chapter 3, asked 2×

## Related in Hydro Power

- [Firm and secondary energy from average monthly flows](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afirm-secondary-energy-2075) — Hydro Power, chapter 3
- [Primary and secondary energy with 10% environmental flow](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aenergy-with-environmental-flow-950) — Hydro Power, chapter 3
- [Primary and secondary energy at 45% exceedance](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aprimary-secondary-energy-45pct) — Hydro Power, chapter 3
- [Primary and secondary energy (Nepali calendar months)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aenergy-nepali-months-155m) — Hydro Power, chapter 3
- [Primary and secondary energy (102 m head)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aprimary-secondary-energy-102m) — Hydro Power, chapter 3

## Related in other subjects

- [Flow duration curve; firm and secondary energy](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Afdc-firm-secondary-energy-69) — Hydropower Engineering, chapter 3