---
title: "Muskingum routing, K=12 h, X=0.20"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:muskingum-k12-x0-2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-k12-x0-2
site: BCT Engineering
---

# Muskingum routing, K=12 h, X=0.20

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2075 Asoj, 10 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-k12-x0-2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

Route the following hydrograph through a river reach for which K = 12 h and X = 0.20. At the start of the inflow flood, the outflow discharge is 10 m$^3$/s. Also find lag of peak and attenuation.

| Time (h) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 |
|---|---|---|---|---|---|---|---|---|---|---|
| Inflow (m$^3$/s) | 10 | 20 | 50 | 60 | 55 | 45 | 35 | 27 | 20 | 15 |

## Short answer

Using Muskingum routing with coefficients C0=0.0476, C1=0.4286, C2=0.5238 and initial outflow 10 m³/s, peak outflow is 49.58 m³/s at 30 h. Peak inflow 60 m³/s at 18 h gives attenuation 10.42 m³/s and lag 12 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 $K = 12$ h, $x = 0.2$, $\Delta t = 6$ h, so $K - Kx + 0.5\Delta t = 12.6$ h.

$$
\begin{aligned} C_0 &= \frac{-(12)(0.2) + 3}{12.6} = 0.0476\\ C_1 &= \frac{(12)(0.2) + 3}{12.6} = 0.4286\\ C_2 &= \frac{9.6 - 3}{12.6} = 0.5238\end{aligned}
$$

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

#### Routing table

| t (h) | I | C0·I2 | C1·I1 | C2·O1 | O (m³/s) |
|---|---|---|---|---|---|
| 0 | 10.00 | - | - | - | 10.00 |
| 6 | 20.00 | 0.95 | 4.29 | 5.24 | 10.48 |
| 12 | 50.00 | 2.38 | 8.57 | 5.49 | 16.44 |
| 18 | 60.00 | 2.86 | 21.43 | 8.61 | 32.90 |
| 24 | 55.00 | 2.62 | 25.71 | 17.23 | 45.57 |
| 30 | 45.00 | 2.14 | 23.57 | 23.87 | 49.58 |
| 36 | 35.00 | 1.67 | 19.29 | 25.97 | 46.92 |
| 42 | 27.00 | 1.29 | 15.00 | 24.58 | 40.86 |
| 48 | 20.00 | 0.95 | 11.57 | 21.41 | 33.93 |
| 54 | 15.00 | 0.71 | 8.57 | 17.77 | 27.06 |

#### Peak attenuation and lag

- Peak inflow = 60 m³/s at t = 18 h
- Peak outflow = 49.58 m³/s at t = 30 h
- Attenuation of peak = 60 − 49.58 = **10.42 m³/s** (17.4 % of the inflow peak)
- Lag of peak = 30 − 18 = **12 h** (to the nearest time step)

## Check yourself

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

1. For K = 12 h, x = 0.2, dt = 6 h, what are C0, C1, C2?
   A. 0.0909, 0.4545, 0.4545
   B. 0.0476, 0.4286, 0.5238
   C. 0.4286, 0.0476, 0.5238
   D. 0.1, 0.4, 0.5
   Answer: B. Denominator 12.6: C0 = 0.6/12.6, C1 = 5.4/12.6, C2 = 6.6/12.6. (Key point: Coefficients)

2. Which equation does Muskingum routing use?
   A. O2 = C0 I2 + C1 I1 + C2 O1
   B. O2 = C0 (I1+I2) + C2 O1
   C. O2 = I2 - O1
   D. O2 = K(xI + (1-x)O)
   Answer: A. It is the three-term recursive equation in I2, I1, O1. (Key point: Routing equation)

3. What is the attenuation of peak?
   A. 60 - 10 = 50 m^3/s
   B. 49.58 - 10 = 39.58 m^3/s
   C. 0
   D. 60 - 49.58 = 10.42 m^3/s
   Answer: D. Peak inflow is 60 and peak outflow 49.58. (Key point: Attenuation)

4. What is the lag of peak?
   A. 6 h
   B. 18 h
   C. 12 h (18 h to 30 h)
   D. 30 h
   Answer: C. Outflow peak at 30 h minus inflow peak at 18 h = 12 h. (Key point: Lag)

## Close variants set in other exams

- [Muskingum routing, K=10, x=0.20](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Amuskingum-k10-x0-2-5to40) — Engineering Hydrology, chapter 7, asked 2×

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