---
title: "Primary and secondary energy with 10% environmental flow"
subject: "Hydro Power"
chapter: 3
key: "semester_6/hydro_power:energy-with-environmental-flow-950"
source: https://www.bctengineer.com/q/semester_6/hydro_power/energy-with-environmental-flow-950
site: BCT Engineering
---

# Primary and secondary energy with 10% environmental flow

Hydro Power · Chapter 3: Planning of Hydropower Projects

- Asked in: 2074 Bhadra, 10 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/energy-with-environmental-flow-950
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3

## Question

Mean monthly flow for a Nepalese river is given below, the net head of river 150 m, overall efficiency of 85%. Assume 10% of water of minimum flow is required to be left for environment and downstream user in the river. Determine the primary and secondary energy produced by the plant, if design discharge is set to 950 m³/s. Assume any other data suitably if required.

| Month | Jan | Feb | Mar | Apr | May | Jun | July | Aug | Sept | Oct | Nov | Dec |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Q (m³/s) | 100 | 120 | 140 | 300 | 350 | 1800 | 2000 | 2520 | 2000 | 900 | 650 | 500 |

## Short answer

Primary energy 986.1 GWh/yr, secondary 5204.5 GWh/yr. Installed capacity 1188 MW. Minimum usable flow after 10 m³/s env. release is 90 m³/s, firm power 112.6 MW; energy above that from flows capped at 950 m³/s.

## Answer

Primary energy comes from the minimum usable flow; secondary energy from the usable flow above it, up to the design discharge. An environmental flow must first be subtracted from the river flow.

#### Environmental flow
Minimum river flow = 100 m³/s (January).

$$
Q_{env} = 0.10 \times 100 = 10\ \text{m}^3/\text{s}\ \text{(released every month)}
$$

Usable flow = Q − 10, limited to the design discharge 950 m³/s.

#### Power per unit discharge

$$
P = 9.81 \times Q \times 150 \times 0.85 = 1250.78\,Q\ \text{kW}
$$

Installed capacity $= 1250.78 \times 950 = 1{,}188{,}236$ kW ≈ 1188.2 MW.

#### Primary energy
Minimum usable flow = 100 − 10 = 90 m³/s.

$$
\begin{aligned}
P_f &= 1250.78 \times 90 = 112{,}570\ \text{kW} \\
E_p &= 112{,}570 \times 8760 = 986.11\ \text{GWh}
\end{aligned}
$$

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

| Month | Q | Q − 10 | Q used | Power (MW) |
|---|---|---|---|---|
| Jan | 100 | 90 | 90 | 112.57 |
| Feb | 120 | 110 | 110 | 137.59 |
| Mar | 140 | 130 | 130 | 162.60 |
| Apr | 300 | 290 | 290 | 362.72 |
| May | 350 | 340 | 340 | 425.26 |
| Jun | 1800 | 1790 | 950 | 1188.24 |
| Jul | 2000 | 1990 | 950 | 1188.24 |
| Aug | 2520 | 2510 | 950 | 1188.24 |
| Sep | 2000 | 1990 | 950 | 1188.24 |
| Oct | 900 | 890 | 890 | 1113.19 |
| Nov | 650 | 640 | 640 | 800.50 |
| Dec | 500 | 490 | 490 | 612.88 |
| **Sum** | | | **6780** | |

$$
\begin{aligned}
E_{total} &= 1250.78 \times 6780 \times 730 = 6190.59\ \text{GWh} \\
E_s &= 6190.59 - 986.11 = 5204.47\ \text{GWh}
\end{aligned}
$$

**Answer: Primary energy ≈ 986.1 GWh/yr; secondary energy ≈ 5204.5 GWh/yr (installed capacity ≈ 1188 MW).**

## Check yourself

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

1. How is the environmental release determined?
   A. 10% of each month's flow
   B. 10% of the minimum flow, 10 m^3/s, every month
   C. 10% of design discharge
   D. 10% of mean flow
   Answer: B. 10% of the minimum flow (100) is 10 m^3/s, released every month. (Key point: Environmental flow)

2. What flow is used for primary energy?
   A. 100 m^3/s
   B. 950 m^3/s
   C. 90 m^3/s (minimum minus environmental flow)
   D. 10 m^3/s
   Answer: C. Usable minimum flow is 100 - 10 = 90 m^3/s. (Key point: Primary energy flow)

3. For June (1800 m^3/s) what flow is used?
   A. 1790 m^3/s
   B. 950 m^3/s, since usable flow 1790 exceeds the design discharge
   C. 1800 m^3/s
   D. 10 m^3/s
   Answer: B. Usable flow Q-10 is capped at the design discharge 950 m^3/s. (Key point: Order of subtraction and capping)

4. What is the power per unit discharge?
   A. 9.81 x 150 x 0.85 = 1250.78 kW per m^3/s
   B. 9.81 x 950 x 0.85
   C. 9.81 x 100 x 0.85
   D. 9.81 x 150 / 0.85
   Answer: A. P = 9.81 Q H eta with H = 150 m and eta = 0.85. (Key point: Power coefficient)

## Related in Hydro Power

- [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
- [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
- [Power duration curve and primary/secondary energy](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apower-duration-curve-energy) — Hydro Power, chapter 3
- [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

## 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
- [River flow required for 1500 population](https://www.bctengineer.com/question.md?key=semester_6%2Fsanitary_engineering%3Ariver-flow-for-1500-population) — Sanitary Engineering, chapter 6
- [Full regulation: reservoir, capacity, annual energy](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Astorage-full-regulation-monthly-hydrograph) — Hydropower Engineering, chapter 3
- [Micro hydro net head, power and energy](https://www.bctengineer.com/question.md?key=semester_6%2Fenergy_resources%3Amicro-hydro-power-output-numerical) — Energy Resources, chapter 5