---
title: "Linear kinematic wave, 200 ft channel"
subject: "Computational Techniques in Civil Engineering"
chapter: 5
key: "semester_8/computational_techniques_in_civil_engineering:kinematic-wave-linear-200ft"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/kinematic-wave-linear-200ft
site: BCT Engineering
---

# Linear kinematic wave, 200 ft channel

Computational Techniques in Civil Engineering · Chapter 5: Finite difference method

- Asked in: 2078 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/kinematic-wave-linear-200ft
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

The following are data pertaining to a rectangular channel: width of channel = 200 ft, length of channel = 15000 ft, bed slope $S_0 = 1\%$, Manning's $n = 0.035$. At $t = 0$, there is a uniform flow of 2000 cfs along the channel. The discharge value at the upstream boundary from the inflow hydrograph at time $t = 3$ min is obtained as 2250 cfs. Determine the discharge at a distance of 3000 ft downstream along the channel. Use the linear kinematic wave model. Take $\Delta x = 3000$ ft and $\Delta t = 3$ min. There is no lateral inflow ($q = 0$).

## Short answer

Apply linear kinematic wave model with backward difference. Manning’s α=3.514, β=0.6; Δt/Δx=0.06. Upstream Q=2250 cfs, downstream initial 2000 cfs, average 2125 cfs, coefficient=0.09841. Solving implicit formula yields Q_{i+1}^{n+1}=2094.7 cfs ≈ 2095 cfs.

## Answer

The linear kinematic wave model writes the kinematic wave equation as $\dfrac{\partial Q}{\partial x}+\alpha\beta Q^{\beta-1}\dfrac{\partial Q}{\partial t}=q$, where $A=\alpha Q^{\beta}$ comes from Manning's equation. It is solved with backward differences at the unknown point $(i+1,\ n+1)$, and the coefficient $Q^{\beta-1}$ is evaluated at the average $\bar Q=(Q_{i+1}^n+Q_i^{n+1})/2$, so the equation is linear in the unknown.

#### Step 1: Initial uniform flow (US units, Manning constant 1.49)
$$
Q=\frac{1.49}{n}AR^{2/3}S_0^{1/2},\qquad A=200y,\ P=200+2y,\ R=A/P
$$
Trial for $Q=2000$ cfs gives $y=1.680$ ft, so $A=336.09$ ft$^2$, $P=203.36$ ft, $R=1.653$ ft and $Q=2000$ cfs (check).

#### Step 2: Kinematic coefficients
$$
\alpha=\left(\frac{nP^{2/3}}{1.49\sqrt{S_0}}\right)^{0.6}=\left(\frac{0.035\times203.36^{2/3}}{1.49\times\sqrt{0.01}}\right)^{0.6}=3.514,\qquad \beta=0.6
$$
(Taking $P\approx B=200$ ft gives $\alpha=3.49$ and the same answer to the nearest cfs.)

#### Step 3: Finite difference equation
$$
\frac{Q_{i+1}^{n+1}-Q_i^{n+1}}{\Delta x}+\alpha\beta\,\bar Q^{\beta-1}\,\frac{Q_{i+1}^{n+1}-Q_{i+1}^n}{\Delta t}=\bar q
$$
Solving for the unknown:
$$
Q_{i+1}^{n+1}=\frac{\dfrac{\Delta t}{\Delta x}Q_i^{n+1}+\alpha\beta\,Q_{i+1}^n\,\bar Q^{\beta-1}+\Delta t\,\bar q}{\dfrac{\Delta t}{\Delta x}+\alpha\beta\,\bar Q^{\beta-1}}
$$
Here $\bar q=0$.

#### Step 4: Substitution
$Q_{i+1}^n=2000$ cfs (uniform flow at $t=0$), $Q_i^{n+1}=2250$ cfs (upstream, $t=3$ min); the point at 3000 ft is $i+1$. $\Delta t/\Delta x=180/3000=0.06$.
$$
\bar Q=\frac{2000+2250}{2}=2125\ \text{cfs},\qquad \alpha\beta\,\bar Q^{-0.4}=3.514(0.6)(2125)^{-0.4}=0.09841
$$
$$
Q_{i+1}^{n+1}=\frac{0.06(2250)+0.09841(2000)}{0.06+0.09841}=\frac{135.00+196.82}{0.15841}=2094.7\ \text{cfs}
$$

**Answer:** the discharge at 3000 ft downstream after 3 min is $Q\approx2095$ cfs. It lies between the initial 2000 cfs and the upstream 2250 cfs, because the flood wave is starting to arrive.

**Note:** if the second-order four-point linear scheme with a constant celerity $c=\tfrac53V=9.918$ ft/s ($V=5.951$ ft/s, $C=c\Delta t/\Delta x=0.5951$) is used instead, $Q_{i+1}^{n+1}=Q_i^n+\dfrac{1-C}{1+C}\left(Q_{i+1}^n-Q_i^{n+1}\right)=1936.5$ cfs. The dip below 2000 cfs is a numerical oscillation (dispersion) of that scheme, not a physical effect.

## Check yourself

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

1. How is the kinematic wave celerity found from Manning's equation?
   A. c = V
   B. c = (3/5) V
   C. c = sqrt(g y)
   D. c = (5/3) V
   Answer: D. With Q = alpha A^(5/3), c = dQ/dA = (5/3)V, giving 9.918 ft/s for V = 5.951 ft/s. (Key point: Kinematic celerity)

2. Which formula gives the unknown Q_{i+1}^{n+1} in the linear four-point scheme?
   A. Q_i^n + (1 - C)/(1 + C) (Q_{i+1}^n - Q_i^{n+1})
   B. Q_i^n - C (Q_{i+1}^n - Q_i^n)
   C. Q_{i+1}^n + C (Q_i^{n+1} - Q_i^n)
   D. Q_i^{n+1} + (1 + C)/(1 - C)(Q_i^n - Q_{i+1}^n)
   Answer: A. Solving the box scheme for the one unknown gives this expression with Courant number C. (Key point: Four-point solution)

3. What is the Courant number for c = 9.918 ft/s, dt = 180 s, dx = 3000 ft?
   A. 1.65
   B. 0.595
   C. 5.95
   D. 0.168
   Answer: B. C = 9.918 x 180 / 3000 = 0.595. (Key point: Courant number value)

## Related in Computational Techniques in Civil Engineering

- [Non-linear kinematic wave discharge at 1 km](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Akinematic-wave-implicit-numerical) — Computational Techniques in Civil Engineering, chapter 5
- [Linear kinematic wave, 30 m channel](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Akinematic-wave-linear-30m) — Computational Techniques in Civil Engineering, chapter 5
- [Non-linear kinematic wave, trapezoidal river](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Akinematic-wave-trapezoidal-river) — Computational Techniques in Civil Engineering, chapter 5
- [Maximum time step, kinematic wave routing](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Akinematic-wave-max-time-step) — Computational Techniques in Civil Engineering, chapter 5
- [Second-order scheme, linear kinematic wave](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asecond-order-linear-kinematic-scheme) — Computational Techniques in Civil Engineering, chapter 5
- [Second-order non-linear kinematic wave; FDM principle](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asecond-order-nonlinear-kinematic-principle) — Computational Techniques in Civil Engineering, chapter 5

## Related in other subjects

- [Uniform flow in rectangular channel (Chezy)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-channel-chezy-flow) — Hydro Power, chapter 1
- [Economical rectangular condition and trapezoidal bed slope](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atrapezoidal-channel-bed-slope) — Hydro Power, chapter 1
- [Slope break: jump depth and jump length by step](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aslope-break-jump-downstream-depth-and-length) — Hydraulics, chapter 9
- [Flood discharge in rectangular channel](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aflood-discharge-slope-area) — Hydro Power, chapter 2