---
title: "Muskingum routing, K=8 h, X=0.25"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:muskingum-k8-x0-25"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-k8-x0-25
site: BCT Engineering
---

# Muskingum routing, K=8 h, X=0.25

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2081 Bhadra, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-k8-x0-25
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

Route the flood hydrograph through a river reach for which Muskingum coefficient K = 8 h and X = 0.25. Initial outflow discharge from the reach is 8.0 m$^3$/s.

| Time (hour) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 |
|---|---|---|---|---|---|---|---|---|
| Inflow (m$^3$/s) | 8 | 16 | 30 | 30 | 25 | 20 | 15 | 10 |

## Short answer

Muskingum routing with K=8 h, x=0.25, Δt=6 h gives C₀=0.1111, C₁=0.5556, C₂=0.3333. Starting outflow 8 m³/s, outflows at 0,6,12,18,24,30,36,42 h: 8, 8.89, 15.19, 25.06, 27.80, 25.38, 21.24, 16.52 m³/s. Peak outflow 27.80 m³/s at 24 h; attenuation 2.20 m³/s, 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 $K = 8$ h, $x = 0.25$, $\Delta t = 6$ h, so $K - Kx + 0.5\Delta t = 9$ h.

$$
\begin{aligned} C_0 &= \frac{-(8)(0.25) + 3}{9} = 0.1111\\ C_1 &= \frac{(8)(0.25) + 3}{9} = 0.5556\\ C_2 &= \frac{6 - 3}{9} = 0.3333\end{aligned}
$$

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

#### Routing table

| t (h) | I | C0·I2 | C1·I1 | C2·O1 | O (m³/s) |
|---|---|---|---|---|---|
| 0 | 8.00 | - | - | - | 8.00 |
| 6 | 16.00 | 1.78 | 4.44 | 2.67 | 8.89 |
| 12 | 30.00 | 3.33 | 8.89 | 2.96 | 15.19 |
| 18 | 30.00 | 3.33 | 16.67 | 5.06 | 25.06 |
| 24 | 25.00 | 2.78 | 16.67 | 8.35 | 27.80 |
| 30 | 20.00 | 2.22 | 13.89 | 9.27 | 25.38 |
| 36 | 15.00 | 1.67 | 11.11 | 8.46 | 21.24 |
| 42 | 10.00 | 1.11 | 8.33 | 7.08 | 16.52 |

#### Peak attenuation and lag

- Peak inflow = 30 m³/s at t = 12 h
- Peak outflow = 27.80 m³/s at t = 24 h
- Attenuation of peak = 30 − 27.80 = **2.20 m³/s** (7.3 % of the inflow peak)
- Lag of peak = 24 − 12 = **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 = 8 h, x = 0.25, dt = 6 h, what are C0, C1, C2?
   A. 0.1111, 0.5556, 0.3333
   B. 0.3333, 0.5556, 0.1111
   C. 0.2, 0.5, 0.3
   D. 0.1111, 0.3333, 0.5556
   Answer: A. The denominator is 9, so C0 = 1/9, C1 = 5/9, C2 = 3/9. (Key point: Coefficients)

2. Which equation advances the outflow?
   A. O2 = C0 I1 + C1 I2 + C2 O1
   B. O2 = C0 I2 + C1 I1 + C2 O1
   C. O2 = C0 O2 + C1 I1 + C2 I2
   D. O2 = C0 + C1 + C2
   Answer: B. C0 multiplies the new inflow I2, C1 the old inflow I1, C2 the previous outflow O1. (Key point: Routing equation)

3. What is the peak attenuation?
   A. 2.20 m3/s
   B. 30 m3/s
   C. 27.80 m3/s
   D. 12 m3/s
   Answer: A. Inflow peak 30 minus outflow peak 27.80 = 2.20 m3/s. (Key point: Peak attenuation)

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