An aqueduct has a weir that is 5 feet narrower than the distance across the aqueduct.
Assuming a constant weir overflow rate of 28.75 gpm/ft, an average depth of 12 feet, a
distance across the bottom of 8 feet, a length of 22 miles, and a daily flow of 0.85 MG,
what is the capacity of the aqueduct in AF?

To find the capacity of the aqueduct in acre-feet (AF), we need to use the given data and follow a series of steps. The goal is to calculate the cross-sectional area and the volume of the aqueduct in acre-feet.

### Given Data:
- **Weir overflow rate**: 28.75 gallons per minute per foot (gpm/ft) (not needed for this problem).
- **Average depth of aqueduct**: 12 feet.
- **Distance across the bottom**: 8 feet.
- **Length of aqueduct**: 22 miles.
- **Daily flow**: 0.85 million gallons (MG).
- **The weir is 5 feet narrower than the top of the aqueduct**.

### Step 1: Determine the shape of the cross-section of the aqueduct
Since we know that the top of the aqueduct is 5 feet wider than the weir, let's assume the top width of the aqueduct is \( x \) feet, and the width of the weir is \( x - 5 \) feet. For simplicity, let's assume the aqueduct cross-section is trapezoidal.

The area \( A \) of a trapezoid is calculated using the formula:
\[
A = \frac{1}{2} \times (b_1 + b_2) \times h
\]
Where:
- \( b_1 = 8 \, \text{feet} \) (bottom width),
- \( b_2 = x \, \text{feet} \) (top width),
- \( h = 12 \, \text{feet} \) (depth).

Thus, the area becomes:
\[
A = \frac{1}{2} \times (8 + x) \times 12
\]

### Step 2: Convert the length of the aqueduct to feet
The length of the aqueduct is given as 22 miles. We need to convert this into feet:
\[
22 \, \text{miles} = 22 \times 5280 \, \text{feet} = 116,160 \, \text{feet}
\]

### Step 3: Calculate the volume of the aqueduct
The volume of the aqueduct is the cross-sectional area multiplied by the length. So the volume in cubic feet \( V \) is:
\[
V = A \times \text{length} = \frac{1}{2} \times (8 + x) \times 12 \times 116,160
\]
This gives us the volume in cubic feet.

### Step 4: Convert cubic feet to acre-feet
To convert cubic feet to acre-feet, we use the conversion factor:
\[
1 \, \text{acre-foot} = 43,560 \, \text{cubic feet}
\]

Thus, the capacity of the aqueduct in acre-feet (AF) is:
\[
\text{Capacity (AF)} = \frac{V}{43,560}
\]

Let me now compute the exact value of the capacity in AF.

An aqueduct has a weir that is 5 feet narrower than the distance across the aqueduct. Assuming a constant weir overflow rate of 28.75 gpm/ft, an average depth of 12 feet, a distance across the bottom of 8 feet, a length of 22 miles, and a daily flow of 0.85 MG, what is the capacity of the aqueduct in AF?

Learn how to calculate the capacity of an aqueduct with a trapezoidal cross-section. This problem involves key geometry concepts, volume calculation, and unit conversion. With given dimensions and flow data, you can compute the aqueduct's capacity in acre-feet (AF). Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation