---
title: "Muskingum routing, x=0.28, k=38.4 hr"
subject: "Engineering Hydrology"
chapter: 7
key: "semester_5/engineering_hydrology:muskingum-x0-28-k38-4"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-x0-28-k38-4
site: BCT Engineering
---

# Muskingum routing, x=0.28, k=38.4 hr

Engineering Hydrology · Chapter 7: Flow Routing

- Asked in: 2082 Baisakh, 8 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/muskingum-x0-28-k38-4
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-7

## Question

The storage in the reach of a stream has been studied. The values of x and k in Muskingum equation have been identified as 0.28 and 38.4 hr. If the inflow hydrograph to reach is as given below, compute the outflow hydrograph. Assume the outflow from the reach at t = 0 is 35 m$^3$/s.

| Time (hr) | 0 | 6 | 12 | 18 | 24 | 30 |
|---|---|---|---|---|---|---|
| Inflow (m$^3$/s) | 35 | 55 | 92 | 130 | 160 | 140 |

## Short answer

Muskingum routing with K=38.4 h, x=0.28, Δt=6 h gives C₀=−0.2529, C₁=0.4487, C₂=0.8042 using O₂=C₀I₂+C₁I₁+C₂O₁. Starting from O=35 m³/s, outflow at 0,6,12,18,24,30 h is 35.00, 29.94, 25.49, 28.90, 41.10, 69.44 m³/s; peak not yet reached within given data.

## 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 $K = 38.4$ h, $x = 0.28$, $\Delta t = 6$ h, so $K - Kx + 0.5\Delta t = 30.648$ h.

$$
\begin{aligned} C_0 &= \frac{-(38.4)(0.28) + 3}{30.648} = -0.2529\\ C_1 &= \frac{(38.4)(0.28) + 3}{30.648} = 0.4487\\ C_2 &= \frac{27.648 - 3}{30.648} = 0.8042\end{aligned}
$$

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

$C_0$ is negative because $\Delta t < 2Kx$ ($6 < 21.5$ h). This is acceptable in the computation, but it makes the outflow dip at the start of the rise, as the table shows.

#### Routing table

| t (h) | I | C0·I2 | C1·I1 | C2·O1 | O (m³/s) |
|---|---|---|---|---|---|
| 0 | 35.00 | - | - | - | 35.00 |
| 6 | 55.00 | -13.91 | 15.70 | 28.15 | 29.94 |
| 12 | 92.00 | -23.27 | 24.68 | 24.08 | 25.49 |
| 18 | 130.00 | -32.88 | 41.28 | 20.50 | 28.90 |
| 24 | 160.00 | -40.47 | 58.33 | 23.24 | 41.10 |
| 30 | 140.00 | -35.41 | 71.79 | 33.06 | 69.44 |

**Answer:** the outflow hydrograph is 35.00, 29.94, 25.49, 28.90, 41.10 and 69.44 m³/s at t = 0, 6, 12, 18, 24 and 30 h. The outflow is still rising at 30 h (the given inflow record ends there), so the outflow peak is not reached within the data.

## Check yourself

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

1. With K = 38.4 h, x = 0.28, dt = 6 h, what is K - Kx + 0.5dt?
   A. 30.65 h
   B. 41.15 h
   C. 35.15 h
   D. 27.65 h
   Answer: A. 38.4 - 10.752 + 3 = 30.648 h. (Key point: Denominator)

2. Why is C0 negative here?
   A. Because outflow is less than inflow
   B. Because dt < 2Kx (6 < 21.5)
   C. Because the inflow is increasing
   D. Because the coefficients do not sum to 1
   Answer: B. C0 = (-Kx + 0.5dt)/(...) is negative when 0.5dt < Kx, i.e. dt < 2Kx. (Key point: Negative C0)

3. Which computation gives O at t = 6 h?
   A. C0(55) + C1(35) + C2(35) = 29.94
   B. C0(35) + C1(55) + C2(35) = 40
   C. C0(55) + C1(35) + C2(55) = 50
   D. C1(55) + C2(35) only
   Answer: A. O2 = C0 I2 + C1 I1 + C2 O1 with I2 = 55, I1 = 35, O1 = 35, giving about 29.94. (Key point: First routed value)

4. What is true of the computed outflow at 30 h?
   A. It is the peak
   B. It is still rising, so the peak is not reached within the data
   C. It equals the inflow
   D. It is zero
   Answer: B. Outflow rises from 41.10 to 69.44 and the record ends, so the peak is not yet reached. (Key point: Interpretation of result)

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