---
title: "Annual firm and secondary energy (900 m³/s, Nov 100)"
subject: "Hydro Power"
chapter: 3
key: "semester_6/hydro_power:firm-secondary-energy-2078"
source: https://www.bctengineer.com/q/semester_6/hydro_power/firm-secondary-energy-2078
site: BCT Engineering
---

# Annual firm and secondary energy (900 m³/s, Nov 100)

Hydro Power · Chapter 3: Planning of Hydropower Projects

- Asked in: 2078 Chaitra, 7 marks
- Video: [Unit:2 | Power Regulation | Calculation of firm energy & secondary energy | Prashant YT | PU2023Qn2a](https://www.youtube.com/watch?v=UlAQB6VU0Lk) — Prashant YT
- 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-2078
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3

## Question

Following are the mean monthly flow of the river. Where head (H) = 100 m, overall efficiency = 90% and design discharge selected is 900 m³/s. Calculate the annual firm and secondary energy.

| Month | Jan | Feb | Mar | Apr | May | Jun | July | Aug | Sep | Oct | Nov | Dec |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Flow (m³/s) | 100 | 120 | 140 | 300 | 320 | 1800 | 2000 | 2510 | 2000 | 900 | 100 | 300 |

## Short answer

Annual firm energy 773.4 GWh from minimum flow 100 m³/s. Secondary energy 3016.3 GWh, total 3789.8 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** is the energy from the minimum flow (available all year). **Secondary energy** is the extra energy from flows above the minimum, limited by the design discharge of 900 m³/s.

#### Power per unit discharge

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

Installed capacity $= 882.9 \times 900 = 794.61$ MW.

#### Firm energy
Minimum flow = 100 m³/s (January and November).

$$
\begin{aligned}
P_f &= 882.9 \times 100 = 88{,}290\ \text{kW} \\
E_f &= 88{,}290 \times 8760 = 773.42\ \text{GWh}
\end{aligned}
$$

#### Total energy (730 h per month)

| Month | Q | Q used | Power (MW) |
|---|---|---|---|
| Jan | 100 | 100 | 88.29 |
| Feb | 120 | 120 | 105.95 |
| Mar | 140 | 140 | 123.61 |
| Apr | 300 | 300 | 264.87 |
| May | 320 | 320 | 282.53 |
| Jun | 1800 | 900 | 794.61 |
| Jul | 2000 | 900 | 794.61 |
| Aug | 2510 | 900 | 794.61 |
| Sep | 2000 | 900 | 794.61 |
| Oct | 900 | 900 | 794.61 |
| Nov | 100 | 100 | 88.29 |
| Dec | 300 | 300 | 264.87 |
| **Sum** | | **5880** | |

$$
\begin{aligned}
E_{total} &= 882.9 \times 5880 \times 730 = 3789.76\ \text{GWh} \\
E_s &= 3789.76 - 773.42 = 3016.34\ \text{GWh}
\end{aligned}
$$

**Answer: Annual firm energy ≈ 773.4 GWh; annual secondary energy ≈ 3016.3 GWh (total ≈ 3789.8 GWh).**

## Check yourself

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

1. What is the minimum flow and the firm power here?
   A. 100 m^3/s; 88,290 kW
   B. 900 m^3/s; 794,610 kW
   C. 120 m^3/s; 105,948 kW
   D. 2510 m^3/s; 2,216,000 kW
   Answer: A. The minimum is 100 m^3/s (Jan and Nov), so Pf = 882.9 x 100 = 88,290 kW. (Key point: Minimum flow)

2. How is total energy found from the table?
   A. Sum all flows without capping, times 882.9 x 730
   B. Use only the monsoon months
   C. 882.9 x sum of used flows (capped at 900) x 730 h
   D. Multiply firm power by 12
   Answer: C. Capped flows sum to 5880, so E = 882.9 x 5880 x 730 = 3789.76 GWh. (Key point: Total energy method)

3. How is secondary energy obtained?
   A. Total energy x 0.9
   B. Total energy minus firm energy
   C. Firm energy minus total energy
   D. Energy of the flood months only
   Answer: B. Es = 3789.76 - 773.42 = 3016.34 GWh. (Key point: Secondary energy)

## Close variants set in other exams

- [Annual firm and secondary energy (900 m³/s, Nov 650)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Afirm-secondary-energy-2079) — 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 (102 m head)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aprimary-secondary-energy-102m) — 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/secondary energy and pondage volume](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aenergy-and-pondage-volume) — Hydro Power, chapter 3

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