---
title: "Muskingum routing, K=10, x=0.20"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:muskingum-k10-x0-2-5to40"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-k10-x0-2-5to40
site: BCT Engineering
---

# Muskingum routing, K=10, x=0.20

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2078 Bhadra, 6+2 marks
- Video: [Flood Routing | Muskingum Method | Problem 01](https://www.youtube.com/watch?v=9PB7dEdav6c) — mrtaosifexplains
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-k10-x0-2-5to40
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

Route the following hydrograph through a river reach with routing parameters K and x as 10 and 0.20, respectively. Take outflow discharge at the start of inflow flood as 5 m$^3$/s. Also estimate attenuation and lag of peak.

| Time (hr) | 0 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 |
|---|---|---|---|---|---|---|---|---|---|
| Inflow (m$^3$/s) | 5 | 10 | 30 | 45 | 40 | 30 | 18 | 12 | 8 |

## Short answer

Muskingum coefficients: C0=0.0476, C1=0.4286, C2=0.5238 for K=10h, x=0.2, Δt=5h. Routing yields peak inflow 45 m³/s at 15h, peak outflow 35.00 m³/s at 25h. Attenuation = 10.00 m³/s, lag = 10 h.

## Answer

#### Method

Muskingum routing: $O_2 = C_0 I_2 + C_1 I_1 + C_2 O_1$, where

$$
C_0 = \frac{-Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_1 = \frac{Kx + 0.5\Delta t}{K - Kx + 0.5\Delta t},\quad C_2 = \frac{K - Kx - 0.5\Delta t}{K - Kx + 0.5\Delta t}
$$

Here $K = 10$ h, $x = 0.2$, $\Delta t = 5$ h, so $K - Kx + 0.5\Delta t = 10.5$ h.

$$
\begin{aligned} C_0 &= \frac{-(10)(0.2) + 2.5}{10.5} = 0.0476\\ C_1 &= \frac{(10)(0.2) + 2.5}{10.5} = 0.4286\\ C_2 &= \frac{8 - 2.5}{10.5} = 0.5238\end{aligned}
$$

Check: $C_0 + C_1 + C_2 = 1.0000$.

#### Routing table

| t (h) | I | C0·I2 | C1·I1 | C2·O1 | O (m³/s) |
|---|---|---|---|---|---|
| 0 | 5.00 | - | - | - | 5.00 |
| 5 | 10.00 | 0.48 | 2.14 | 2.62 | 5.24 |
| 10 | 30.00 | 1.43 | 4.29 | 2.74 | 8.46 |
| 15 | 45.00 | 2.14 | 12.86 | 4.43 | 19.43 |
| 20 | 40.00 | 1.90 | 19.29 | 10.18 | 31.37 |
| 25 | 30.00 | 1.43 | 17.14 | 16.43 | 35.00 |
| 30 | 18.00 | 0.86 | 12.86 | 18.33 | 32.05 |
| 35 | 12.00 | 0.57 | 7.71 | 16.79 | 25.07 |
| 40 | 8.00 | 0.38 | 5.14 | 13.13 | 18.66 |

#### Peak attenuation and lag

- Peak inflow = 45 m³/s at t = 15 h
- Peak outflow = 35.00 m³/s at t = 25 h
- Attenuation of peak = 45 − 35.00 = **10.00 m³/s** (22.2 % of the inflow peak)
- Lag of peak = 25 − 15 = **10 h** (to the nearest time step)

## Check yourself

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

1. With K = 10 h, x = 0.2, dt = 5 h, which coefficients result?
   A. C0 = 0.0909, C1 = 0.4545, C2 = 0.4545
   B. C0 = 0.0476, C1 = 0.4286, C2 = 0.5238
   C. C0 = 0.2, C1 = 0.3, C2 = 0.5
   D. C0 = 0.5238, C1 = 0.4286, C2 = 0.0476
   Answer: B. Denominator K - Kx + 0.5dt = 10.5 gives the listed values, and they sum to 1. (Key point: Coefficients)

2. What is the attenuation of the peak?
   A. 45 - 35 = 10 m^3/s
   B. 45 - 31.37 = 13.63 m^3/s
   C. 35 - 5 = 30 m^3/s
   D. 45 m^3/s
   Answer: A. Peak inflow 45 and peak outflow 35.00 give 10 m^3/s. (Key point: Peak attenuation)

3. What is the lag of peak?
   A. 5 h
   B. 15 h
   C. 0 h
   D. 10 h (inflow peak at 15 h, outflow peak at 25 h)
   Answer: D. Outflow peak at 25 h minus inflow peak at 15 h = 10 h. (Key point: Lag)

## Close variants set in other exams

- [Muskingum routing, K=12 h, X=0.20](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Amuskingum-k12-x0-2) — Engineering Hydrology, chapter 7, asked 2×

## 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
- [Muskingum routing, k=10 h, x=0.2 (table missing)](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Amuskingum-k10-x0-2-10cumec) — Engineering Hydrology, chapter 7
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- [Muskingum routing at Chovar, Bagmati River](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Amuskingum-chovar-k10-x0-2) — Engineering Hydrology, chapter 7
- [Muskingum routing, K=8 h, X=0.25](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Amuskingum-k8-x0-25) — Engineering Hydrology, chapter 7
- [Attenuation and lag of peak in routing](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aattenuation-lag-peak) — Engineering Hydrology, chapter 7, asked 2×

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