---
title: "Muskingum routing at Chovar, Bagmati River"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:muskingum-chovar-k10-x0-2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-chovar-k10-x0-2
site: BCT Engineering
---

# Muskingum routing at Chovar, Bagmati River

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2080 Baisakh, 7 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-chovar-k10-x0-2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

Route the following flood hydrograph at Chovar of Bagmati River using Muskingum method having coefficient k = 10 hr and x = 0.2. Following is the inflow hydrograph with outflow of Chovar at the start is 60 m$^3$/s.

| t (hr) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Inflow (m$^3$/s) | 60 | 120 | 480 | 660 | 460 | 260 | 160 | 100 | 90 | 60 | 40 |

## Short answer

Using Muskingum routing with K=10 h, x=0.2, Δt=6 h yields coefficients C0=0.0909, C1=C2=0.4545. The outflow peak is 494.70 m³/s at 24 h, an attenuation of 165.30 m³/s and a 6-h lag from the inflow peak of 660 m³/s at 18 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, $x = 0.2$, $\Delta t = 6$ h, so $K - Kx + 0.5\Delta t = 11$ h.

$$
\begin{aligned} C_0 &= \frac{-(10)(0.2) + 3}{11} = 0.0909\\ C_1 &= \frac{(10)(0.2) + 3}{11} = 0.4545\\ C_2 &= \frac{8 - 3}{11} = 0.4545\end{aligned}
$$

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

#### Routing table

| t (h) | I | C0·I2 | C1·I1 | C2·O1 | O (m³/s) |
|---|---|---|---|---|---|
| 0 | 60.00 | - | - | - | 60.00 |
| 6 | 120.00 | 10.91 | 27.27 | 27.27 | 65.45 |
| 12 | 480.00 | 43.64 | 54.55 | 29.75 | 127.93 |
| 18 | 660.00 | 60.00 | 218.18 | 58.15 | 336.33 |
| 24 | 460.00 | 41.82 | 300.00 | 152.88 | 494.70 |
| 30 | 260.00 | 23.64 | 209.09 | 224.86 | 457.59 |
| 36 | 160.00 | 14.55 | 118.18 | 208.00 | 340.72 |
| 42 | 100.00 | 9.09 | 72.73 | 154.87 | 236.69 |
| 48 | 90.00 | 8.18 | 45.45 | 107.59 | 161.22 |
| 54 | 60.00 | 5.45 | 40.91 | 73.28 | 119.65 |
| 60 | 40.00 | 3.64 | 27.27 | 54.39 | 85.29 |

#### Peak attenuation and lag

- Peak inflow = 660 m³/s at t = 18 h
- Peak outflow = 494.70 m³/s at t = 24 h
- Attenuation of peak = 660 − 494.70 = **165.30 m³/s** (25.0 % of the inflow peak)
- Lag of peak = 24 − 18 = **6 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 = 10 h, x = 0.2, dt = 6 h, what are C0, C1, C2?
   A. 0.0909, 0.4545, 0.4545
   B. 0.4545, 0.0909, 0.4545
   C. 0.1, 0.5, 0.4
   D. 0.0909, 0.5, 0.4091
   Answer: A. Denominator 11: C0 = 1/11, C1 = 5/11, C2 = 5/11. (Key point: Coefficients)

2. What does K represent in the Muskingum method?
   A. Weighting factor of inflow
   B. Storage time constant, approximately the travel time through the reach
   C. Reach length
   D. Peak attenuation
   Answer: B. K is the storage time constant, roughly the flood wave travel time through the reach. (Key point: Storage terms)

3. What are the peak outflow and its time?
   A. 494.70 m3/s at 24 h
   B. 660 m3/s at 18 h
   C. 457.59 m3/s at 30 h
   D. 336.33 m3/s at 18 h
   Answer: A. The routed hydrograph peaks at 494.70 m3/s at t = 24 h. (Key point: Peak outflow)

4. What is the peak attenuation and lag?
   A. 165.30 m3/s and 6 h
   B. 494.70 m3/s and 6 h
   C. 165.30 m3/s and 12 h
   D. 60 m3/s and 6 h
   Answer: A. 660 - 494.70 = 165.30 m3/s; peak shifts from 18 h to 24 h, a 6 h lag. (Key point: Attenuation)

## 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=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×
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