---
title: "RoR and PRoR from hydrograph equation (Q40)"
subject: "Hydropower Engineering"
chapter: 3
key: "semester_7/hydropower_engineering:ror-pror-hydrograph-equation-q40"
source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-pror-hydrograph-equation-q40
site: BCT Engineering
---

# RoR and PRoR from hydrograph equation (Q40)

Hydropower Engineering · Chapter 3: Power and Energy Potential study

- Asked in: 2074 Ashwin, 3+3+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/ror-pror-hydrograph-equation-q40
- Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-3

## Question

The hydrograph of a typical river of Nepal follows the equation $Q_t = 5.589t^2 - 51.275t + 139.94$, where $Q_t$ is the mean monthly discharge in m³/s and $t$ is time in months counted with October as the 1st month and so on. A hydropower plant has to be developed in this river with net head of 150 m and overall efficiency of 85%; the environmental flow is not considered.
a) Calculate the installed capacity and firm energy for the RoR project that will be developed for design discharge $Q_{40}$.
b) If the project has to be designed as a Peaking Run of the River (PRoR) project for 6 hrs daily peaking (3 hrs in morning and 3 hrs in evening) and with design discharge $Q_{40}$, what is the installed capacity of the PRoR project? Assume that the project is designed in such a way that 50% of the available flow is used during the off-peak hours and the remaining 50% of the available flow is stored for peak hour generation. Neglect all the losses.

## Short answer

(a) RoR: installed capacity = 116.2 MW, firm energy = 255.2 GWh. (b) PRoR: installed capacity = 232.4 MW.

## Answer

#### Monthly flows from $Q_t = 5.589t^2 - 51.275t + 139.94$
| t | Month | Q (m³/s) |
|---|---|---|
| 1 | Oct | 94.25 |
| 2 | Nov | 59.75 |
| 3 | Dec | 36.42 |
| 4 | Jan | 24.26 |
| 5 | Feb | 23.29 |
| 6 | Mar | 33.49 |
| 7 | Apr | 54.88 |
| 8 | May | 87.44 |
| 9 | Jun | 131.17 |
| 10 | Jul | 186.09 |
| 11 | Aug | 252.18 |
| 12 | Sep | 329.46 |

$H=150$ m, $\eta=0.85$: $9.81\,\eta\,H = 1250.77$ kW per m³/s.

#### Design discharge $Q_{40}$
Descending flows: 329.46, 252.18, 186.09, 131.17, 94.25, 87.44, 59.75, 54.88, 36.42, 33.49, 24.26, 23.29. Weibull position for 40%: $m=0.4\times13=5.2$ (between the 5th flow 94.25 and the 6th flow 87.44):

$$
Q_{40} = 94.25 - 0.2\,(94.25-87.44) = 92.89\ \text{m}^3/\text{s}
$$

#### a) RoR project
$$
P = 1250.77\times92.89 = 116,185\ \text{kW} = 116.18\ \text{MW}
$$

The minimum monthly flow is 23.29 m³/s (February), the flow available in all months, so

$$
E_{firm} = 1250.77\times23.29\times8760 = 255,183,616\ \text{kWh} = 255.2\ \text{GWh}
$$

(Total energy with flows limited to $Q_{40}$ is 715.8 GWh; the part above firm energy is secondary energy.)

#### b) Peaking run-of-river (6 h daily peaking)
Daily water volume at $Q_{40}$: $Q_{40}\times24\times3600$. Half of it (50%) is used during the 18 off-peak hours and the other half is stored (pond volume $=0.5\times92.89\times86400=4,012,865$ m³) and used in the 6 peak hours:

$$
Q_{peak} = \frac{0.5\,Q_{40}\times24}{6} = 2\,Q_{40} = 185.78\ \text{m}^3/\text{s}
$$

$$
P_{PRoR} = 1250.77\times185.78 = 232,370\ \text{kW} = 232.37\ \text{MW}
$$

Off-peak the plant uses $0.5\,Q_{40}\times24/18 = 61.93$ m³/s. The installed capacity is therefore twice that of the RoR plant.

**Answer: (a) RoR: installed capacity = 116.2 MW, firm energy = 255.2 GWh; (b) PRoR: installed capacity = 232.4 MW.**

## Check yourself

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

1. How are the monthly flows obtained?
   A. Substitute t = 1 to 12 (October = 1) in Q = 5.589t^2 - 51.275t + 139.94
   B. Take an average of the coefficients
   C. Integrate the equation
   D. Use t = 0 for October
   Answer: A. t counts months from October, so t = 1 gives 94.25 m3/s and so on. (Key point: Monthly flows from the equation)

2. Q40 is found by
   A. Ranking flows and interpolating at m = 0.4 x 13 = 5.2 between 94.25 and 87.44, giving 92.89 m3/s
   B. Taking the 4th flow
   C. Taking the mean flow
   D. Taking the minimum flow
   Answer: A. Weibull position 5.2 lies between the 5th and 6th flows. (Key point: Q40)

3. The firm energy of the RoR plant is based on
   A. Q40
   B. The minimum monthly flow 23.29 m3/s, available in all months
   C. The maximum flow
   D. The average flow
   Answer: B. E_firm = 1250.77 x 23.29 x 8760 = 255.2 GWh. (Key point: Firm energy)

4. Half of the Q40 volume is used off-peak and half stored for 6 peak hours. What is the peak discharge?
   A. Q40
   B. 2 x Q40 = 185.78 m3/s
   C. 4 x Q40
   D. 0.5 x Q40
   Answer: B. 0.5 Q40 x 24 h / 6 h = 2 Q40. (Key point: PRoR peak discharge)

5. How does PRoR installed capacity compare with RoR?
   A. Equal
   B. Half
   C. Twice: 232.4 MW against 116.2 MW
   D. Ten times
   Answer: C. Peak discharge is twice Q40 with the same head and efficiency, so capacity doubles. (Key point: PRoR capacity vs RoR)

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