Theoretical Background (Corrected)

The Governing Equation for the kinematic wave model is the continuity equation:

\frac{\partial Q}{\partial x} + \frac{\partial A}{\partial t} = 0

The $A = \alpha Q^\beta$ Relationship

Using Manning's equation for a wide channel ($R \approx y = A/P$), we can relate Area and Discharge as:

Q = \frac{1}{n} A \left(\frac{A}{P}\right)^{2/3} \sqrt{S_0} = \frac{\sqrt{S_0}}{n P^{2/3}} A^{5/3}

Rearranging this to solve for $A$ gives the specific formulation used in your course:

A = \left( \frac{n P^{2/3}}{\sqrt{S_0}} \right)^{0.6} Q^{0.6}

This is in the form $A = \alpha Q^\beta$, where $\beta = 0.6$ and $\alpha = \left( \frac{n P^{2/3}}{\sqrt{S_0}} \right)^{0.6}$.

Discretizing the Equation

Using the chain rule, $\frac{\partial A}{\partial t} = \frac{dA}{dQ} \frac{\partial Q}{\partial t} = \alpha \beta Q^{\beta-1} \frac{\partial Q}{\partial t}$. The PDE becomes:

\frac{\partial Q}{\partial x} + \alpha \beta Q^{\beta-1} \frac{\partial Q}{\partial t} = 0

We approximate the derivatives and evaluate the coefficient $\alpha \beta Q^{\beta-1}$ at the average known flow $\overline{Q} = \frac{Q_i^{n+1} + Q_{i+1}^n}{2}$:

\frac{Q_{i+1}^{n+1} - Q_i^{n+1}}{\Delta x} + \alpha \beta \overline{Q}^{\beta-1} \frac{Q_{i+1}^{n+1} - Q_{i+1}^n}{\Delta t} = 0

5.4.2 Non-Linear Scheme (Precise Geometry)

The Kinematic Wave Equation is inherently non-linear because Area and Discharge are related by a power law ($A = \alpha Q^\beta$, where $\beta = 0.6$).

In simpler linear problems (like Questions 2-5), we assume a "wide rectangular channel" where the wetted perimeter $P$ is roughly equal to the width $B$. This allows us to treat $P$ as a constant, thus making $\alpha$ a constant.

In the formal Non-Linear Scheme, we do not make this simplifying assumption. Instead, we compute the true geometry of the channel iteratively:

Find Normal Depth: Use Manning's equation iteratively to calculate the exact normal depth ($y$) for the initial base discharge ($Q_{initial}$).
Exact Wetted Perimeter: Compute the exact Wetted Perimeter ($P$) using this depth (e.g., $P = B + 2y\sqrt{1+z^2}$ for a trapezoid).
Precise $\alpha$: Calculate a highly precise $\alpha$ coefficient using this exact $P$.
Linearized Application: Apply the same finite difference fractional formula to advance the solution in time, utilizing the non-linear relationship via the localized $C$ coefficient.

The Exact Course Formula

Rearranging the discretized equation to solve for the unknown future downstream flow $Q_{i+1}^{n+1}$ yields the exact formula found in your CH-V.pdf notes:

Q_{i+1}^{n+1} = \frac{\frac{\Delta t}{\Delta x}Q_{i}^{n+1} + \alpha \beta Q_{i+1}^n \left(\frac{Q_{i+1}^n + Q_i^{n+1}}{2}\right)^{\beta-1}}{\frac{\Delta t}{\Delta x} + \alpha \beta \left(\frac{Q_{i+1}^n + Q_i^{n+1}}{2}\right)^{\beta-1}}

The Simplified 5-Step Process

To make calculations easier and less prone to calculator errors, we can break that massive formula down into clean, intermediate variables:

1 Calculate $\alpha$ and $\beta$

$$ \alpha = \left( \frac{n \cdot P^{2/3}}{\sqrt{S_0}} \right)^{0.6} \quad \text{and} \quad \beta = 0.6 $$

*(If specified as "Non-Linear Scheme", find $P$ using normal depth first. Otherwise, $P \approx B$)*

2 Calculate Average Flow ($\overline{Q}$)

$$ \overline{Q} = \frac{Q_i^{n+1} + Q_{i+1}^n}{2} $$

3 Calculate Celerity Coefficient ($C$)

This represents the entire $\alpha \beta \overline{Q}^{\beta-1}$ term from the formula.

$$ C = \alpha \cdot \beta \cdot (\overline{Q})^{\beta-1} = \alpha \cdot 0.6 \cdot (\overline{Q})^{-0.4} $$

4 Calculate Grid Ratio ($R$)

$$ R = \frac{\Delta t}{\Delta x} $$

5 Compute Final Discharge

$$ Q_{i+1}^{n+1} = \frac{R \cdot Q_i^{n+1} + C \cdot Q_{i+1}^n}{R + C} $$

Step-by-Step Solutions

Question 2 (2077 Chaitra, Q2, Group A)

Question Text:

For a $30 \text{ m}$ wide and $0.015$ bed slope rectangular channel following flow rates are given: $Q_i^n = 20 \text{ m}^3\text{/s}$, $Q_i^{n+1} = 28 \text{ m}^3\text{/s}$ and $Q_{i+1}^n = 18 \text{ m}^3\text{/s}$. Taking Manning's $n = 0.025$, $\Delta x = 1200 \text{ m}$ and $\Delta t = 10 \text{ min}$, determine $Q_{i+1}^{n+1}$ using finite difference scheme for linear kinematic wave model. Assume lateral inflow to be zero. Take wetted perimeter is approximately equal to the width of the channel. [6]

Given Data:

$B \approx P = 30 \text{ m}$
$S_0 = 0.015$
$n = 0.025$
$Q_i^{n+1} = 28 \text{ m}^3\text{/s}$
$Q_{i+1}^n = 18 \text{ m}^3\text{/s}$
$\Delta x = 1200 \text{ m}$
$\Delta t = 10 \text{ min} = 600 \text{ s}$

Step 1: Calculate $\alpha$

\alpha = \left( \frac{0.025 \times 30^{2/3}}{\sqrt{0.015}} \right)^{0.6} = \left( \frac{0.025 \times 9.655}{0.12247} \right)^{0.6} = (1.9708)^{0.6} = 1.5076

Step 2: Average Flow ($\overline{Q}$)

\overline{Q} = \frac{28 + 18}{2} = 23 \text{ m}^3\text{/s}

Step 3: Celerity Coefficient ($C$)

C = 1.5076 \times 0.6 \times (23)^{-0.4} = 0.90456 \times 0.28475 = 0.2576

Step 4: Grid Ratio ($R$)

R = \frac{600}{1200} = 0.5

Step 5: Final Discharge

$$ Q_{i+1}^{n+1} = \frac{0.5(28) + 0.2576(18)}{0.5 + 0.2576} = \frac{14 + 4.6368}{0.7576} $$

$$ Q_{i+1}^{n+1} = 24.60 \text{ m}^3\text{/s} $$

Question 3 (2073 Bhadra, Q2, Group A)

Question Text:

A river which is generalized as trapezoidal channel is $300 \text{ m}$ wide with side slope $5:1$, has bed slope $1\%$ and Manning's $n = 0.04$. Initially discharge through the river is $100 \text{ m}^3\text{/s}$. Due to a flood the discharge observed at the upstream section of the river is $300\text{m}^3\text{/s}$, compute discharge at $2650\text{m}$ downstream from upstream section. Take $\Delta x = 2650\text{m}$ and $\Delta t = 1\text{hr}$. Use linear kinematic wave solution. [6]

Note on Geometry: Because the channel is exceptionally wide ($300\text{m}$) relative to typical depths, the friction from the side slopes is negligible. It is standard practice to use the wide channel approximation ($P \approx B = 300\text{m}$) here to avoid solving complex implicit equations for normal depth without a computer.

Given Data:

$B \approx P = 300 \text{ m}$
$S_0 = 1\% = 0.01$
$n = 0.04$
$Q_i^{n+1} = 300 \text{ m}^3\text{/s}$
$Q_{i+1}^n = 100 \text{ m}^3\text{/s}$
$\Delta x = 2650 \text{ m}$
$\Delta t = 1 \text{ hr} = 3600 \text{ s}$

Step 1: Calculate $\alpha$

\alpha = \left( \frac{0.04 \times 300^{2/3}}{\sqrt{0.01}} \right)^{0.6} = \left( \frac{0.04 \times 44.814}{0.1} \right)^{0.6} = (17.9256)^{0.6} = 5.645

Step 2: Average Flow ($\overline{Q}$)

\overline{Q} = \frac{300 + 100}{2} = 200 \text{ m}^3\text{/s}

Step 3: Celerity Coefficient ($C$)

C = 5.645 \times 0.6 \times (200)^{-0.4} = 3.387 \times 0.1201 = 0.4068

Step 4: Grid Ratio ($R$)

R = \frac{3600}{2650} = 1.3585

Step 5: Final Discharge

$$ Q_{i+1}^{n+1} = \frac{1.3585(300) + 0.4068(100)}{1.3585 + 0.4068} = \frac{407.55 + 40.68}{1.7653} $$

$$ Q_{i+1}^{n+1} = 253.9 \text{ m}^3\text{/s} $$

Question 4 (2072 Magh, Q1b, Group A)

Question Text:

Water flows through a rectangular channel $25 \text{ m}$ wide, having bed slope $0.015$ and Manning's $n = 0.035$. The following flow rates are given: $Q_i^{n+1} = 30 \text{ m}^3\text{/s}$, $Q_i^n = 22 \text{ m}^3\text{/s}$ and $Q_{i+1}^n = 20 \text{ m}^3\text{/s}$, where $\text{i} =$ index for space and $\text{n} =$ index for time. Taking $\Delta x = 1500\text{m}$ and $\Delta t = 10 \text{ min}$, determine $Q_{i+1}^{n+1}$ using first order accurate linear kinematic wave model. Assume lateral flow to be zero and wetted perimeter is equal to width of the channel. [6]

Given Data:

$B \approx P = 25 \text{ m}$
$S_0 = 0.015$
$n = 0.035$
$Q_i^{n+1} = 30 \text{ m}^3\text{/s}$
$Q_{i+1}^n = 20 \text{ m}^3\text{/s}$
$\Delta x = 1500 \text{ m}$
$\Delta t = 10 \text{ min} = 600 \text{ s}$

Step 1: Calculate $\alpha$

\alpha = \left( \frac{0.035 \times 25^{2/3}}{\sqrt{0.015}} \right)^{0.6} = \left( \frac{0.035 \times 8.5499}{0.12247} \right)^{0.6} = (2.4435)^{0.6} = 1.7145

Step 2: Average Flow ($\overline{Q}$)

\overline{Q} = \frac{30 + 20}{2} = 25 \text{ m}^3\text{/s}

Step 3: Celerity Coefficient ($C$)

C = 1.7145 \times 0.6 \times (25)^{-0.4} = 1.0287 \times 0.2759 = 0.2838

Step 4: Grid Ratio ($R$)

R = \frac{600}{1500} = 0.4

Step 5: Final Discharge

$$ Q_{i+1}^{n+1} = \frac{0.4(30) + 0.2838(20)}{0.4 + 0.2838} = \frac{12 + 5.676}{0.6838} $$

$$ Q_{i+1}^{n+1} = 25.85 \text{ m}^3\text{/s} $$

Question 5 (2071 Bhadra, Q1b, Group A)

Question Text:

Using the Finite Difference equation developed in question (a) (First order accurate implicit equation for kinematic wave model), compute the discharge at $1 \text{ km d/s}$ of location X at time $14:00 \text{ hrs}$, for the following data: [6]

Rectangular channel, width $= 20 \text{ m}$, Bed slope $= 0.001$, Manning's $n = 0.03$
Discharge at location X at time $14:00 \text{ hrs} = 14 \text{ m}^3\text{/s}$
Discharge at location X at time $13:45 \text{ hrs} = 12 \text{ m}^3\text{/s}$
Discharge at $1 \text{ km d/s}$ of location X at time $13:45 \text{ hrs} = 11 \text{ m}^3\text{/s}$
No lateral flow, wetted perimeter approximately equal to width of channel.

Given Data:

$B \approx P = 20 \text{ m}$
$S_0 = 0.001$
$n = 0.03$
$Q_i^{n+1} \text{ (14:00)} = 14 \text{ m}^3\text{/s}$
$Q_{i+1}^n \text{ (13:45)} = 11 \text{ m}^3\text{/s}$
$\Delta x = 1 \text{ km} = 1000 \text{ m}$
$\Delta t = 15 \text{ min} = 900 \text{ s}$

Step 1: Calculate $\alpha$

\alpha = \left( \frac{0.03 \times 20^{2/3}}{\sqrt{0.001}} \right)^{0.6} = \left( \frac{0.03 \times 7.368}{0.03162} \right)^{0.6} = (6.99)^{0.6} = 3.22

Step 2: Average Flow ($\overline{Q}$)

\overline{Q} = \frac{14 + 11}{2} = 12.5 \text{ m}^3\text{/s}

Step 3: Celerity Coefficient ($C$)

C = 3.22 \times 0.6 \times (12.5)^{-0.4} = 1.932 \times 0.364 = 0.703

Step 4: Grid Ratio ($R$)

R = \frac{900}{1000} = 0.9

Step 5: Final Discharge

$$ Q_{i+1}^{n+1} = \frac{0.9(14) + 0.703(11)}{0.9 + 0.703} = \frac{12.6 + 7.733}{1.603} $$

$$ Q_{i+1}^{n+1} = 12.68 \text{ m}^3\text{/s} $$

Question 6 (2070 Magh, Q1b, Group B)

Question Text:

A flood of $150 \text{ m}^3\text{/s}$ peak discharges passed a gaging station at 12:00 noon on a river. There is a community adjacent to the river $7.2 \text{ km}$ downstream. What will be the value of peak discharge at that community at 12:00 noon of the velocity of flow is $1.2\text{m/s}$ and peak discharge at that community at 9:00 A.M is $100\text{m}^3\text{/s}$. Assume width of river as inside and use first order accurate numerical scheme of kinematic wave equation, Take $\Delta x = 7.2\text{km}$ and $\Delta t = 1 \text{ hrs}$. [6]

The "Trick" in this Question: We are given the Velocity ($V = 1.2 \text{ m/s}$) directly. Because the celerity coefficient $C = \alpha \beta \overline{Q}^{\beta-1}$ is mathematically exactly equal to $\frac{1}{\text{celerity } (c)}$, and we know $c = \frac{5}{3}V$, we can completely bypass calculating $\alpha$ and area!

Given Data:

$Q_i^{n+1} = 150 \text{ m}^3\text{/s}$
$Q_{i+1}^n = 100 \text{ m}^3\text{/s}$
$V = 1.2 \text{ m/s}$
$\Delta x = 7.2 \text{ km} = 7200 \text{ m}$
$\Delta t = 1 \text{ hr} = 3600 \text{ s}$

Steps 1-2: Skipped

Because Velocity ($V$) is given directly, we don't need channel geometry.

Step 3: Celerity Coefficient ($C$)

c = \frac{5}{3} V = \frac{5}{3}(1.2) = 2.0 \text{ m/s} $$ $$ C = \frac{1}{c} = \frac{1}{2.0} = 0.5 \text{ s/m}

Step 4: Grid Ratio ($R$)

R = \frac{\Delta t}{\Delta x} = \frac{3600}{7200} = 0.5 \text{ s/m}

Step 5: Final Discharge

$$ Q_{i+1}^{n+1} = \frac{0.5(150) + 0.5(100)}{0.5 + 0.5} = \frac{75 + 50}{1.0} $$

$$ Q_{i+1}^{n+1} = 125 \text{ m}^3\text{/s} $$

Question 7 (2079 Shrawan, Q1b, Group A) - Non-Linear Solution

Question Text:

A river which can be generalized as a trapezoidal channel is $350 \text{ m}$ wide with side slope $6:1$, has a bed slope $1.5\%$, and Manning’s $n=0.0359$. The initial discharge through the river is $420 \text{ cumecs}$. Due to a flood observed at upstream, the value of discharge rises to $580 \text{ cumecs}$. Calculate the discharge that will occur at $4.65 \text{ km}$ downstream. Take $\Delta x=4650 \text{ m}$ and $\Delta t=1.5 \text{ hours}$. Use non-linear kinematic wave solution. [6]

Given Data:

$B = 350 \text{ m}$
$z = 6 \text{ (Side Slope 6H:1V)}$
$S_0 = 1.5\% = 0.015$
$n = 0.0359$
$Q_i^{n+1} = 580 \text{ m}^3\text{/s}$
$Q_{i+1}^n = 420 \text{ m}^3\text{/s}$
$\Delta x = 4650 \text{ m}$
$\Delta t = 1.5 \text{ hr} = 5400 \text{ s}$

Step 0: Exact Geometry (Non-Linear Scheme Preparation)

Unlike previous questions, the instruction "Use non-linear kinematic wave solution" requires us to compute the exact normal depth ($y$) and exact wetted perimeter ($P$) for the initial flow ($Q = 420 \text{ m}^3\text{/s}$), rather than assuming $P \approx B$.

1. Setting up Manning's Equation for $y$:

Q = \frac{\sqrt{S_0}}{n} \frac{(By + zy^2)^{5/3}}{(B + 2y\sqrt{1+z^2})^{2/3}} $$ $$ 420 = \frac{\sqrt{0.015}}{0.0359} \frac{(350y + 6y^2)^{5/3}}{(350 + 2y\sqrt{37})^{2/3}}

2. Solving iteratively yields the exact normal depth:

y \approx 0.533 \text{ m}

3. Calculate Exact Wetted Perimeter ($P$):

P = B + 2y\sqrt{1+z^2} = 350 + 2(0.533)\sqrt{37} = 356.48 \text{ m}

Step 1: Calculate precise $\alpha$

\alpha = \left( \frac{n \cdot P^{2/3}}{\sqrt{S_0}} \right)^{0.6} = \left( \frac{0.0359 \times 356.48^{2/3}}{\sqrt{0.015}} \right)^{0.6} = (14.736)^{0.6} = 5.024

Step 2: Average Flow ($\overline{Q}$)

\overline{Q} = \frac{580 + 420}{2} = 500 \text{ m}^3\text{/s}

Step 3: Celerity Coefficient ($C$)

C = \alpha \beta (\overline{Q})^{-0.4} = 5.024 \times 0.6 \times (500)^{-0.4} = 3.0144 \times 0.08325 = 0.2510

Step 4: Grid Ratio ($R$)

R = \frac{\Delta t}{\Delta x} = \frac{5400}{4650} = 1.1613

Step 5: Final Discharge

$$ Q_{i+1}^{n+1} = \frac{1.1613(580) + 0.2510(420)}{1.1613 + 0.2510} = \frac{673.55 + 105.42}{1.4123} $$

$$ Q_{i+1}^{n+1} = 551.56 \text{ m}^3\text{/s} $$