---
title: "Muskingum routing, k=10 h, x=0.2 (table missing)"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:muskingum-k10-x0-2-10cumec"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-k10-x0-2-10cumec
site: BCT Engineering
---

# Muskingum routing, k=10 h, x=0.2 (table missing)

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2072 Chaitra, 8 marks
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-k10-x0-2-10cumec
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

Route the following flood hydrograph through a river reach for which Muskingum coefficient k = 10 h and x = 0.2. At the start of inflow flood, the outflow discharge is 10 m$^3$/sec. [Inflow table, time 0 to 54 h at 6 h interval, is cut off in the scan.]

## Short answer

Muskingum routing with C0=0.0909, C1=0.4545, C2=0.4545 and initial outflow 10 m³/s yields peak outflow 51.33 m³/s at 30 h, peak attenuation 8.67 m³/s, and lag 12 h.

## Answer

**Assumption:** the inflow table is cut off in the question. The inflow ordinates are assumed to be the standard set for this problem (10, 20, 50, 60, 55, 45, 35, 27, 20, 15 m³/s at 6 h intervals from 0 to 54 h). If your table differs, use the same coefficients and the same steps with your values.

#### 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 | 10.00 | - | - | - | 10.00 |
| 6 | 20.00 | 1.82 | 4.55 | 4.55 | 10.91 |
| 12 | 50.00 | 4.55 | 9.09 | 4.96 | 18.60 |
| 18 | 60.00 | 5.45 | 22.73 | 8.45 | 36.63 |
| 24 | 55.00 | 5.00 | 27.27 | 16.65 | 48.92 |
| 30 | 45.00 | 4.09 | 25.00 | 22.24 | 51.33 |
| 36 | 35.00 | 3.18 | 20.45 | 23.33 | 46.97 |
| 42 | 27.00 | 2.45 | 15.91 | 21.35 | 39.71 |
| 48 | 20.00 | 1.82 | 12.27 | 18.05 | 32.14 |
| 54 | 15.00 | 1.36 | 9.09 | 14.61 | 25.06 |

#### Peak attenuation and lag

- Peak inflow = 60 m³/s at t = 18 h
- Peak outflow = 51.33 m³/s at t = 30 h
- Attenuation of peak = 60 − 51.33 = **8.67 m³/s** (14.5 % 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 = 10 h, x = 0.2, dt = 6 h, C0, C1, C2 are:
   A. 0.0909, 0.4545, 0.4545
   B. 0.0476, 0.4286, 0.5238
   C. 0.5, 0.25, 0.25
   D. 0.2, 0.2, 0.6
   Answer: A. Denominator is 11. (Key point: Coefficients)

2. The peak outflow is:
   A. 60 m^3/s at 18 h
   B. 48.92 at 24 h
   C. 46.97 at 36 h
   D. 51.33 m^3/s at 30 h
   Answer: D. Routing gives 51.33 at 30 h. (Key point: Peak outflow)

3. Attenuation and lag of peak are:
   A. 10.42 m^3/s and 12 h
   B. 8.67 m^3/s and 6 h
   C. 8.67 m^3/s and 12 h
   D. 60 m^3/s and 30 h
   Answer: C. 60 - 51.33 = 8.67; 30 - 18 = 12. (Key point: Attenuation and lag)

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