A trapezoidal canal lined with cement concrete ($n = 0.01$) is designed to carry a discharge of $20 \text{ } m^3/s$ at a bed slope 1 in 400. If the bed width is twice of the depth of flow and side slope of the canal section is 2 (1 vertical:2 horizontal) then the corresponding depth of flow will be ________ (in m, rounded off to two decimal places).
Trapezoidal Canal Depth Calculation Using Manning's Formula
This problem requires calculating the depth of flow ($y$) in a trapezoidal canal using Manning's formula, given the discharge ($Q$), roughness coefficient ($n$), bed slope ($S$), bed width ($b$) in terms of depth, and side slope ($z$).
Given Parameters and Formulas
- Discharge, $Q = 20 \text{ } m^3/s$
- Manning's coefficient, $n = 0.01$
- Bed slope, $S = 1/400 = 0.0025$
- Bed width, $b = 2y$
- Side slope, $z = 2$ (meaning 2 horizontal for 1 vertical)
Key formulas for a trapezoidal channel:
- Area, $A = (b + zy)y$
- Wetted Perimeter, $P = b + 2y\sqrt{1+z^2}$
- Hydraulic Radius, $R = A/P$
- Manning's Formula, $Q = \frac{1}{n} A R^{2/3} S^{1/2}$
Calculating Flow Area and Hydraulic Radius
First, express the Area ($A$) and Wetted Perimeter ($P$) in terms of the depth ($y$):
- Top width, $T = b + 2zy = 2y + 2(2)y = 2y + 4y = 6y$
- Area, $A = \frac{b+T}{2} \times y = \frac{2y+6y}{2} \times y = 4y^2$
- Wetted Perimeter, $P = b + 2y\sqrt{1+z^2} = 2y + 2y\sqrt{1+2^2} = 2y + 2y\sqrt{5} = 2y(1+\sqrt{5})$
- Hydraulic Radius, $R = \frac{A}{P} = \frac{4y^2}{2y(1+\sqrt{5})} = \frac{2y}{1+\sqrt{5}}$
Applying Manning's Equation
Substitute the expressions for $A$ and $R$ into Manning's formula:
$Q = \frac{1}{n} A R^{2/3} S^{1/2}$
Plugging in the values:
$20 = \frac{1}{0.01} (4y^2) \left( \frac{2y}{1+\sqrt{5}} \right)^{2/3} (0.0025)^{1/2}$
Simplify the equation:
$20 = 100 \times 4y^2 \times \left( \frac{2y}{1+\sqrt{5}} \right)^{2/3} \times 0.05$
$20 = 20 \times y^2 \times y^{2/3} \times \left( \frac{2}{1+\sqrt{5}} \right)^{2/3}$
$1 = y^{8/3} \times \left( \frac{2}{1+\sqrt{5}} \right)^{2/3}$
Determining the Depth of Flow
Rearrange to solve for $y^{8/3}$:
$y^{8/3} = \frac{(1+\sqrt{5})^{2/3}}{2^{2/3}} = \left( \frac{1+\sqrt{5}}{2} \right)^{2/3}$
Calculate the right side:
$\left( \frac{1+\sqrt{5}}{2} \right)^{2/3} \approx (1.618034)^{0.666667} \approx 1.3922$
So, $y^{8/3} \approx 1.3922$
Now, solve for $y$:
$y = (1.3922)^{3/8} = (1.3922)^{0.375}$
$y \approx 1.1267 \text{ m}$
Rounding the depth of flow to two decimal places gives $1.13$ m.
Important Questions from Open Channel Flow
- A trapezoidal channel carries $5\text{ m}^3\text{.s}^{-1}$ of water under uniform flow condition. The channel has a bottom width of $2\text{ m}$, side slope of $2:1$ (horizontal: vertical), and a bed slope of $1\%$. If the Manning roughness coefficient is $0.03$, the conveyance of the channel, in $\text{m}^3\text{.s}^{-1}$, is
- Two rectangular channels, A and B, join to form a large rectangular channel, C. Each rectangular channel is lined with the same material and has the same bottom slope. The bottom widths of Channels A and B are $2\text{ m}$ and $1\text{ m}$, respectively. If the flow depth in each channel is $2\text{ m}$, the bottom width of Channel C (in m) is ________. (Rounded off to two decimal places)
- Water is flowing at the rate of 15 m$^3$/s through a rectangular channel of 5 m width.
If the acceleration due to gravity ($g$) is 9.81 m/s$^2$, the critical velocity for the flow is ______ m/s (rounded off to three decimal places). A Hydraulic Jump is said to be Oscillating Jump if its Froude number is ________.
The Francis formula for the discharge over cippoletti weir is
