---
title: "Primary and secondary energy (102 m head)"
subject: "Hydro Power"
chapter: 3
key: "semester_6/hydro_power:primary-secondary-energy-102m"
source: https://www.bctengineer.com/q/semester_6/hydro_power/primary-secondary-energy-102m
site: BCT Engineering
---

# Primary and secondary energy (102 m head)

Hydro Power · Chapter 3: Planning of Hydropower Projects

- Asked in: 2070 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/primary-secondary-energy-102m
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-3

## Question

Estimate primary and secondary energy from the data given below. Net head available = 102 m, Discharge capacity = 55 m³/s, Overall efficiency = 76%.

| Month | Jan | Feb | Mar | Apr | May | Jun | Jul | Aug | Sep | Oct | Nov | Dec |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Flow (m³/s) | 30 | 28 | 24 | 29 | 18 | 37 | 65 | 90 | 74 | 56 | 42 | 39 |

## Short answer

Primary energy 119.9 GWh/yr from minimum flow 18 m³/s. Secondary energy 139.3 GWh/yr, total 259.3 GWh/yr. Installed capacity 41.8 MW at design discharge 55 m³/s. Calculated using power per unit discharge (760.47 kW per m³/s), monthly flows capped at 55 m³/s, 730 hours per month.

## Answer

**Primary energy** is generated by the flow available 100% of the time (minimum flow). **Secondary energy** is the additional energy from flows above the minimum, limited by the plant (design) discharge of 55 m³/s.

#### Power per unit discharge

$$
P = 9.81\,Q\,H\,\eta = 9.81 \times Q \times 102 \times 0.76 = 760.47\,Q\ \text{kW}
$$

Installed capacity $= 760.47 \times 55 = 41{,}826$ kW ≈ 41.8 MW.

#### Primary energy
Minimum flow = 18 m³/s (May).

$$
\begin{aligned}
P_f &= 760.47 \times 18 = 13{,}688\ \text{kW} \\
E_p &= 13{,}688 \times 8760 = 119.91 \times 10^6\ \text{kWh} = 119.91\ \text{GWh}
\end{aligned}
$$

#### Monthly energy (flow used ≤ 55 m³/s, 730 h per month)

| Month | Q (m³/s) | Q used | Power (kW) | Energy (GWh) |
|---|---|---|---|---|
| Jan | 30 | 30 | 22,814 | 16.65 |
| Feb | 28 | 28 | 21,293 | 15.54 |
| Mar | 24 | 24 | 18,251 | 13.32 |
| Apr | 29 | 29 | 22,054 | 16.10 |
| May | 18 | 18 | 13,688 | 9.99 |
| Jun | 37 | 37 | 28,137 | 20.54 |
| Jul | 65 | 55 | 41,826 | 30.53 |
| Aug | 90 | 55 | 41,826 | 30.53 |
| Sep | 74 | 55 | 41,826 | 30.53 |
| Oct | 56 | 55 | 41,826 | 30.53 |
| Nov | 42 | 42 | 31,940 | 23.32 |
| Dec | 39 | 39 | 29,658 | 21.65 |
| **Total** | | **467** | | **259.25** |

$$
\begin{aligned}
E_{total} &= 760.47 \times 467 \times 730 = 259.25\ \text{GWh} \\
E_s &= 259.25 - 119.91 = 139.34\ \text{GWh}
\end{aligned}
$$

**Answer: Primary energy ≈ 119.9 GWh/yr; secondary energy ≈ 139.3 GWh/yr; total ≈ 259.3 GWh/yr.**

## Check yourself

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

1. Which flow is used for primary energy here?
   A. The design discharge 55 m^3/s
   B. The mean of the monthly flows
   C. The minimum monthly flow, 18 m^3/s (May)
   D. The maximum monthly flow, 90 m^3/s
   Answer: C. Primary energy comes from the flow available 100% of the time, the minimum, 18 m^3/s. (Key point: Primary energy flow)

2. Which expression gives power in kW for this plant?
   A. 9.81 x Q x 102 x 0.76 = 760.47 Q
   B. 9.81 x Q x 102 / 0.76
   C. 1000 x Q x 102 x 0.76
   D. 9.81 x Q x 55 x 0.76
   Answer: A. P = 9.81 Q H eta with H = 102 m and eta = 0.76 gives 760.47 Q kW. (Key point: Power per unit discharge)

3. For July (65 m^3/s) which flow is used in the total energy sum?
   A. 65 m^3/s
   B. 18 m^3/s
   C. 10 m^3/s
   D. 55 m^3/s
   Answer: D. The plant cannot pass more than its 55 m^3/s capacity; the excess is spilled. (Key point: Limiting by design discharge)

4. How is secondary energy obtained?
   A. Total energy (limited to 55 m^3/s) minus primary energy
   B. Total energy plus primary energy
   C. Primary energy x load factor
   D. Energy at 55 m^3/s for 8760 h
   Answer: A. Es = Etotal - Ep = 259.25 - 119.91 = 139.34 GWh. (Key point: Secondary energy)

## Related in Hydro Power

- [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 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/secondary energy and pondage volume](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aenergy-and-pondage-volume) — Hydro Power, chapter 3
- [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×
- [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×
- [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

## 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
- [Installed capacity, energy and daily peaking reservoir (2079)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-capacity-plant-factor-18cumec-peaking-79) — Hydropower Engineering, chapter 3, asked 2×
- [Installed capacity, energy and daily peaking reservoir (2072)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-capacity-plant-factor-18cumec-peaking-72) — Hydropower Engineering, chapter 3, asked 2×
- [RoR best installed capacity (NRs, 45 years)](https://www.bctengineer.com/question.md?key=semester_7%2Fhydropower_engineering%3Aror-best-capacity-nrs-10-percent) — Hydropower Engineering, chapter 3