---
title: "Muskingum routing, K=36 hr, x=0.15"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:muskingum-k36-x0-15"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-k36-x0-15
site: BCT Engineering
---

# Muskingum routing, K=36 hr, x=0.15

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2080 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-k36-x0-15
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

The inflow hydrograph readings for a stream reach are given below for which the Muskingum coefficients of K = 36 hr and x = 0.15. Route the flood through the reach and determine the outflow hydrograph. Also, determine the reduction in peak and the time of peak of outflow. Outflow at the beginning of the flood may be taken as same as inflow.

| Time (hr) | 0 | 12 | 24 | 36 | 48 | 60 | 72 | 84 | 96 | 108 | 120 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| Inflow (cumec) | 42 | 45 | 88 | 272 | 342 | 288 | 240 | 198 | 162 | 133 | 110 |

## Short answer

Muskingum routing with K=36 h, x=0.15, Δt=12 h yields C₀=0.0164, C₁=0.3115, C₂=0.6721. Initial O=42 m³/s. Outflows at times 0–120 h (12‑h steps): 42, 42.05, 43.72, 61.26, 131.50, 199.63, 227.82, 231.12, 219.67, 200.29, 177.85 m³/s. Peak outflow 231.12 m³/s at 84 h; reduction 110.88 m³/s, lag 36 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 $K = 36$ h, $x = 0.15$, $\Delta t = 12$ h, so $K - Kx + 0.5\Delta t = 36.6$ h.

$$
\begin{aligned} C_0 &= \frac{-(36)(0.15) + 6}{36.6} = 0.0164\\ C_1 &= \frac{(36)(0.15) + 6}{36.6} = 0.3115\\ C_2 &= \frac{30.6 - 6}{36.6} = 0.6721\end{aligned}
$$

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

#### Routing table

| t (h) | I | C0·I2 | C1·I1 | C2·O1 | O (m³/s) |
|---|---|---|---|---|---|
| 0 | 42.00 | - | - | - | 42.00 |
| 12 | 45.00 | 0.74 | 13.08 | 28.23 | 42.05 |
| 24 | 88.00 | 1.44 | 14.02 | 28.26 | 43.72 |
| 36 | 272.00 | 4.46 | 27.41 | 29.39 | 61.26 |
| 48 | 342.00 | 5.61 | 84.72 | 41.17 | 131.50 |
| 60 | 288.00 | 4.72 | 106.52 | 88.38 | 199.63 |
| 72 | 240.00 | 3.93 | 89.70 | 134.18 | 227.82 |
| 84 | 198.00 | 3.25 | 74.75 | 153.12 | 231.12 |
| 96 | 162.00 | 2.66 | 61.67 | 155.35 | 219.67 |
| 108 | 133.00 | 2.18 | 50.46 | 147.65 | 200.29 |
| 120 | 110.00 | 1.80 | 41.43 | 134.62 | 177.85 |

#### Peak attenuation and lag

- Peak inflow = 342 m³/s at t = 48 h
- Peak outflow = 231.12 m³/s at t = 84 h
- Attenuation of peak = 342 − 231.12 = **110.88 m³/s** (32.4 % of the inflow peak)
- Lag of peak = 84 − 48 = **36 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 = 36 h, x = 0.15, dt = 12 h, which is correct?
   A. C0 = 0.0164, C1 = 0.3115, C2 = 0.6721
   B. C0 = 0.3115, C1 = 0.0164, C2 = 0.6721
   C. C0 = 0.1, C1 = 0.3, C2 = 0.6
   D. C0 = 0.0164, C1 = 0.6721, C2 = 0.3115
   Answer: A. Denominator 36.6: C0 = 0.6/36.6, C1 = 11.4/36.6, C2 = 24.6/36.6. (Key point: Coefficients)

2. What starting outflow is used?
   A. Zero
   B. Equal to the initial inflow, 42 m3/s
   C. The peak inflow
   D. Average inflow
   Answer: B. The problem states initial outflow equals inflow, 42 m3/s. (Key point: Initial condition)

3. What are the peak outflow and its time?
   A. 231.12 m3/s at 84 h
   B. 342 m3/s at 48 h
   C. 227.82 m3/s at 72 h
   D. 219.67 m3/s at 96 h
   Answer: A. The outflow column peaks at 231.12 at t = 84 h, lagging the inflow peak by 36 h. (Key point: Peak outflow and lag)

4. What is the reduction in peak?
   A. 110.88 m3/s
   B. 231.12 m3/s
   C. 36 m3/s
   D. 342 m3/s
   Answer: A. 342 - 231.12 = 110.88 m3/s (about 32.4 percent). (Key point: Attenuation)

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