---
title: "Reservoir routing through spillway, 6-hr inflow"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:reservoir-routing-spillway-6hr"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/reservoir-routing-spillway-6hr
site: BCT Engineering
---

# Reservoir routing through spillway, 6-hr inflow

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2073 Shrawan, 8 marks
- Video: [Modified Puls Method](https://www.youtube.com/watch?v=jvOgwyTWDBs) — Jagadish Torlapati
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/reservoir-routing-spillway-6hr
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

The ordinates of the inflow hydrograph at 6 hr interval are as follows:

| Time (hrs) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 | 72 | 78 | 84 | 90 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Discharge (m$^3$/s) | 0 | 50 | 280 | 610 | 1290 | 1900 | 2130 | 1900 | 1600 | 1440 | 1060 | 780 | 500 | 370 | 220 | 130 |

The discharge over the spillway crest and the surcharge storage above the crest for different water surface elevations are as follows:

| Water surface elevation (m) | 140 | 141 | 142 | 143 | 144 | 145 | 146 |
|---|---|---|---|---|---|---|---|
| Outflow discharge (m$^3$/s) | 0 | 170 | 482 | 883 | 1360 | 1905 | 2500 |
| Storage ($\times 10^6$ m$^3$) | 0.00 | 15.0 | 35.0 | 60.0 | 95.0 | 140.0 | 240.0 |

Determine: i) Maximum reservoir level, ii) Maximum outflow rate, iii) Reduction in the peak.

## Short answer

Using the modified Puls method with Δ t=6 h, the maximum reservoir level is ≈144.15 m, maximum outflow rate ≈1442 m³/s, and reduction in peak is 688 m³/s (32.3% of peak inflow).

## Answer

Use the **storage-indication (modified Puls) method**. The reservoir is assumed to start full to the spillway crest (elevation 140 m), so $O = 0$ and $S = 0$ at $t = 0$.

#### Method

Continuity over one interval $\Delta t$: $\dfrac{I_1 + I_2}{2} - \dfrac{O_1 + O_2}{2} = \dfrac{S_2 - S_1}{\Delta t}$, which rearranges to

$$
\left(\frac{S_2}{\Delta t} + \frac{O_2}{2}\right) = \frac{I_1 + I_2}{2} + \left(\frac{S_1}{\Delta t} - \frac{O_1}{2}\right)
$$

$\Delta t = 6 \times 3600 = 21600$ s.

#### Step 1: Storage-indication curve

| Elevation (m) | O (m³/s) | S (10⁶ m³) | ψ = S/Δt + O/2 (m³/s) |
|---|---|---|---|
| 140 | 0 | 0 | 0.0 |
| 141 | 170 | 15 | 779.4 |
| 142 | 482 | 35 | 1861.4 |
| 143 | 883 | 60 | 3219.3 |
| 144 | 1360 | 95 | 5078.1 |
| 145 | 1905 | 140 | 7434.0 |
| 146 | 2500 | 240 | 12361.1 |

#### Step 2: Routing

At each step, $\psi_2$ is computed from the right-hand side, then $O_2$ and $S_2$ are found by linear interpolation in the table above ($S_1/\Delta t - O_1/2 = \psi_1 - O_1$).

| t (h) | I1 | I2 | (I1+I2)/2 | S1/Δt − O1/2 | ψ2 = S2/Δt + O2/2 | O2 (m³/s) | S2 (10⁶ m³) |
|---|---|---|---|---|---|---|---|
| 6 | 0 | 50 | 25 | 0.0 | 25.0 | 5.5 | 0.48 |
| 12 | 50 | 280 | 165 | 19.5 | 184.5 | 40.3 | 3.55 |
| 18 | 280 | 610 | 445 | 144.3 | 589.3 | 128.5 | 11.34 |
| 24 | 610 | 1290 | 950 | 460.8 | 1410.8 | 352.1 | 26.67 |
| 30 | 1290 | 1900 | 1595 | 1058.7 | 2653.7 | 716.0 | 49.59 |
| 36 | 1900 | 2130 | 2015 | 1937.7 | 3952.7 | 1071.2 | 73.81 |
| 42 | 2130 | 1900 | 2015 | 2881.5 | 4896.5 | 1313.4 | 91.58 |
| 48 | 1900 | 1600 | 1750 | 3583.1 | 5333.1 | 1419.0 | 99.87 |
| 54 | 1600 | 1440 | 1520 | 3914.1 | 5434.1 | 1442.4 | 101.80 |
| 60 | 1440 | 1060 | 1250 | 3991.8 | 5241.8 | 1397.9 | 98.13 |
| 66 | 1060 | 780 | 920 | 3843.9 | 4763.9 | 1279.4 | 89.08 |
| 72 | 780 | 500 | 640 | 3484.6 | 4124.6 | 1115.3 | 77.05 |
| 78 | 500 | 370 | 435 | 3009.3 | 3444.3 | 940.7 | 64.24 |
| 84 | 370 | 220 | 295 | 2503.5 | 2798.5 | 758.7 | 52.25 |
| 90 | 220 | 130 | 175 | 2039.8 | 2214.8 | 586.4 | 41.51 |

#### Results

- Peak inflow = 2130 m³/s at 36 h.
- Peak outflow occurs at t = 54 h (where the outflow curve meets the inflow curve, $O \approx I$).
- Storage at that time $\approx$ 101.8 × 10⁶ m³, which lies between 144 m (95) and 145 m (140).

$$
\text{Max. level} = 144 + \frac{101.8 - 95}{140 - 95} = 144.15\ \text{m}
$$

**Answer:**

- i) Maximum reservoir level $\approx$ **144.15 m**
- ii) Maximum outflow rate $\approx$ **1442 m³/s**
- iii) Reduction in peak $= 2130 - 1442 = $ **688 m³/s** (32.3 % of the peak inflow)

## Check yourself

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

1. Which method routes the flood through the reservoir?
   A. Muskingum with K and x
   B. Storage-indication (modified Puls) using S/dt + O/2
   C. Rational method
   D. Clark time-area
   Answer: B. Elevation-storage-outflow data suit modified Puls. (Key point: Method)

2. What is the initial condition?
   A. Reservoir full to crest at 140 m, O = 0, S = 0
   B. Reservoir empty
   C. Outflow equal to peak inflow
   D. Level at 146 m
   Answer: A. The reservoir is assumed to start at spillway crest. (Key point: Initial condition)

3. When does peak outflow occur?
   A. At 36 h, the inflow peak
   B. At 90 h
   C. At 6 h
   D. At 54 h, where outflow meets inflow
   Answer: D. Peak outflow is where O = I. (Key point: Time of peak outflow)

4. What are max level, max outflow and reduction in peak?
   A. 145 m, 2130 m^3/s, 0
   B. 146 m, 1905 m^3/s, 225 m^3/s
   C. 144.15 m, 1442 m^3/s, 688 m^3/s
   D. 144.15 m, 2130 m^3/s, 688 m^3/s
   Answer: C. Storage 101.8 x 10^6 m^3 gives 144.15 m; reduction is 2130 - 1442 = 688. (Key point: Results)

## Related in Engineering Hydrology

- [Muskingum routing, K=36 hr, x=0.15](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Amuskingum-k36-x0-15) — Engineering Hydrology, chapter 7
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