Problem 72 - Algebra and Trigonometry

Question

The speed of water flowing in a channel, such as a canal or river bed, is governed by the Manning Equation $$V=1.486 \frac{A^{2 / 3} S^{1 / 2}}{p^{2 / 3} n}$$ Here \(V\) is the velocity of the flow in \(\mathrm{ft} / \mathrm{s} ; A\) is the cross- sectional area of the channel in square feet; \(S\) is the down- ward slope of the channel; \(p\) is the wetted perimeter in feet the distance from the top of one bank, down the side of the channel, across the bottom, and up to the top of the other bank); and \(n\) is the roughness coefficient (a measure of the roughness of the channel bottom). This equation is used to predict the capacity of flood channels to handle runoff from heavy rainfalls. For the canal shown in the figure, \(A=75 \mathrm{ft}^{2}, S=0.050, p=24.1 \mathrm{ft},\) and \(n=0.040\). (a) Find the speed with which water flows through this canal. (b) How many cubic feet of water can the canal discharge per second? [Hint: Multiply \(V\) by \(A\) to get the volume of the flow per second.]

Step-by-Step Solution

Answer

(a) 18.221 ft/s; (b) 1366.575 cubic feet/sec.

Step 1: Understand the Manning Equation

The Manning Equation is given by: \[ V = 1.486 \frac{A^{2/3} S^{1/2}}{p^{2/3} n} \]where \( V \) represents the velocity of water flowing in the channel in feet per second. The parameters include \( A \), the cross-sectional area in square feet; \( S \), the downward slope; \( p \), the wetted perimeter in feet; and \( n \), the roughness coefficient.

Step 2: Substitute Given Values into the Equation

Insert the given values for \( A, S, p, \) and \( n \) into the Manning Equation:\[ V = 1.486 \frac{75^{2/3} (0.050)^{1/2}}{24.1^{2/3} (0.040)} \]This substitution allows us to calculate the velocity \( V \) using the provided measurements.

Step 3: Calculate Exponents of Parameters

Calculate the exponents for each parameter:- \( A^{2/3} = 75^{2/3} \approx 18.905 \)- \( S^{1/2} = (0.050)^{1/2} \approx 0.224 \)- \( p^{2/3} = 24.1^{2/3} \approx 8.639 \)

Step 4: Evaluate the Manning Equation for Velocity

Now, substitute these calculated values back into the equation:\[ V = 1.486 \frac{18.905 \times 0.224}{8.639 \times 0.040} \]Simplify the expression to find the velocity \( V \).

Step 5: Simplify and Calculate Velocity

Evaluate the fraction in the equation:- The numerator is \( 18.905 \times 0.224 = 4.23832 \)- The denominator is \( 8.639 \times 0.040 = 0.34556 \)Thus, the entire expression becomes:\[ V = 1.486 \times \frac{4.23832}{0.34556} \approx 18.221 \text{ ft/s} \]

Step 6: Calculate Discharge Volume

Using the hint provided, multiply the velocity \( V \) by the cross-sectional area \( A \) to find the discharge volume:\[ Q = V \times A = 18.221 \times 75 \approx 1366.575 \text{ cubic feet per second} \]

Step 7: Final Answers

The speed of water flowing through the canal is approximately \( 18.221 \text{ ft/s} \). The canal can discharge about \( 1366.575 \text{ cubic feet of water per second} \).

Key Concepts

Flow Velocity

Flow velocity refers to the speed at which water moves through a channel. In our exercise, this is determined using the Manning Equation. This formula is essential because it helps predict how quickly water will travel under given conditions, which is critical for designing effective drainage and flood management systems.

In the Manning Equation, the velocity (\( V \)) of the flow depends on various factors: the slope of the channel, the roughness of the channel's surface, and geometric properties like the cross-sectional area and wetted perimeter. By adjusting each of these parameters, one can affect the flow velocity directly.

Ultimately, understanding and being able to compute flow velocity is crucial, especially in civil engineering and environmental science, to manage waterways and predict potential flooding issues.

Cross-sectional Area

The cross-sectional area (\( A \)) of a channel is a crucial component when calculating flow velocity using the Manning Equation. It is the surface area through which water flows, commonly measured in square feet for our example.

A larger cross-sectional area means more space for water to flow, which can increase the discharge capacity of a channel. If you think of it like a pipe, a wider pipe allows more water to pass through at once.

Cross-sectional area is typically determined by the shape of the channel (rectangular, trapezoidal, etc.) and its physical measurements. For efficient water transport, engineers must carefully evaluate and design channels with suitable cross-sectional areas to meet their water movement needs. This ensures not only efficient flow but also minimizes the risk of overflow or erosion issues.

Wetted Perimeter

The wetted perimeter (\( p \)) in the Manning Equation is another important geometric factor, and it refers to the length of the channel that comes into contact with the flowing water. Put simply, it is the boundary where the water flows along the channel's edges.

In practical terms, it's like the "hug" between water and the channel. Changes in the wetted perimeter can significantly influence flow behavior. For example, a channel with a smaller wetted perimeter compared to its cross-sectional area will usually have a smoother flow.

Measuring the wetted perimeter can help identify how much friction resistance the water experiences as it moves along the channel. This aspect is essential in calculating velocity because more contact with the channel generally means greater energy loss due to friction.

Roughness Coefficient

In the Manning Equation, the roughness coefficient (\( n \)) represents the complexity and friction caused by the channel’s surface. A rougher channel surface will slow down the flow more than a smooth one.

The roughness coefficient is a dimensionless number, and its value varies depending on the material and condition of the channel. Common materials like concrete may have a lower roughness coefficient compared to a natural riverbed lined with rocks and vegetation.

• A small roughness coefficient indicates a smooth channel with less resistance to flow.
• A large roughness coefficient denotes a rough channel, where water flow might slow down due to more surface friction.

Adjusting the roughness coefficient allows engineers to estimate how efficiently water can travel through a channel. Understanding this parameter is crucial for creating accurate models of water movement through various terrains.