1. What is Hydraulic Radius Calculator?

Definition: This calculator computes the hydraulic radius (\( R \)), area (\( A \)), and wetted perimeter (\( P \)) for fluid flow in channels. The hydraulic radius is defined as the cross-sectional area of the flow (\( A \)) divided by the wetted perimeter (\( P \)). It supports five configurations: a full pipe, a partially filled pipe, a rectangular channel, a trapezoidal channel, and a triangular channel, with customizable output units.

Purpose: The hydraulic radius, area, and wetted perimeter are used in fluid dynamics to assess channel efficiency, calculate flow velocity (e.g., via Manning’s equation), and determine head loss (e.g., via Darcy-Weisbach equation). These parameters are critical for designing pipelines, drainage systems, and water supply networks.

2. How Does the Calculator Work?

The calculator computes three values:

• Area (\( A \)): The cross-sectional area of the flow (displayed in selected unit).
• Wetted Perimeter (\( P \)): The perimeter of the channel in contact with the fluid (displayed in selected unit).
• Hydraulic Radius (\( R \)): The ratio \( R = \frac{A}{P} \) (displayed in selected unit).

Shape-Specific Formulas:

• Full Pipe:
  • Area: \( A = \pi r^2 \)
  • Wetted Perimeter: \( P = 2 \pi r \)
  • Hydraulic Radius: \( R = \frac{r}{2} \), where \( r \) is the radius of the pipe.
• Partially Filled Pipe:
  • Central angle: \( \theta = 2 \times \arccos\left(\frac{r - h}{r}\right) \)
  • Area: \( A = \frac{r^2 \times (\theta - \sin(\theta))}{2} \)
  • Wetted Perimeter: \( P = r \times \theta \)
  • Hydraulic Radius: \( R = \frac{\frac{r^2 \times (\theta - \sin(\theta))}{2}}{r \times \theta} \)
• Rectangular Channel:
  • Area: \( A = b \times y \)
  • Wetted Perimeter: \( P = b + 2y \)
  • Hydraulic Radius: \( R = \frac{b \times y}{b + 2y} \), where \( b \) is the base width and \( y \) is the depth.
• Trapezoidal Channel:
  • Slope: \( z = \frac{B - b}{2y} \)
  • Area: \( A = b y + y^2 z \)
  • Wetted Perimeter: \( P = b + 2 y \sqrt{1 + z^2} \)
  • Hydraulic Radius: \( R = \frac{b y + y^2 z}{b + 2 y \sqrt{1 + z^2}} \), where \( b \) is the bottom width, \( B \) is the top width, and \( y \) is the height.
• Triangular Channel:
  • Slope: \( z = \frac{B}{2y} \)
  • Area: \( A = y^2 z \)
  • Wetted Perimeter: \( P = 2 y \sqrt{1 + z^2} \)
  • Hydraulic Radius: \( R = \frac{y z}{2 \sqrt{1 + z^2}} \), where \( B \) is the top width and \( y \) is the height.

Steps:

• Select the shape configuration (full pipe, partially filled pipe, rectangular channel, trapezoidal channel, or triangular channel).
• Enter the required dimensions (\( r \) for full pipe; \( r \) and \( h \) for partially filled pipe; \( b \) and \( y \) for rectangular channel; \( b \), \( B \), and \( y \) for trapezoidal channel; \( B \) and \( y \) for triangular channel) with their units.
• Convert all inputs to meters for calculation.
• Validate inputs (e.g., positive values, \( h \leq 2r \) for partial pipes, \( B \geq b \) for trapezoidal channels).
• Calculate \( A \), \( P \), and \( R \) in base units (m² for area, m for lengths).
• Convert results to the selected output units independently for \( A \), \( P \), and \( R \).
• Display the results, formatted in scientific notation if less than 0.001, otherwise with 5 decimal places.

3. Importance of Area, Wetted Perimeter, and Hydraulic Radius

These parameters are essential for:

• Flow Rate Calculation: The area (\( A \)) is used to compute the flow rate (\( Q = A \times v \)), where \( v \) is the velocity.
• Frictional Resistance: The wetted perimeter (\( P \)) determines the surface area in contact with the fluid, affecting frictional losses.
• Flow Efficiency: The hydraulic radius (\( R \)) indicates channel efficiency; a larger \( R \) means less frictional resistance.
• System Design: These values help optimize the design of pipelines and drainage systems to minimize energy losses.

4. Using the Calculator

Example 1 (Full Pipe, Output in ft², ft, ft): Calculate \( A \), \( P \), and \( R \) for a full pipe:

• Radius: \( r = 0.5 \, \text{m} \);
• Area: \( A = \pi \times 0.5^2 \approx 0.785 \, \text{m}^2 \), in ft²: \( 0.785 \times 10.76391 \approx 8.451 \, \text{ft}^2 \);
• Wetted Perimeter: \( P = 2 \pi \times 0.5 \approx 3.142 \, \text{m} \), in ft: \( 3.142 \times 3.28084 \approx 10.308 \, \text{ft} \);
• Hydraulic Radius: \( R = \frac{0.5}{2} = 0.25 \, \text{m} \), in ft: \( 0.25 \times 3.28084 \approx 0.820 \, \text{ft} \);
• Result: \( A = 8.45133 \, \text{ft}^2 \), \( P = 10.30840 \, \text{ft} \), \( R = 0.82021 \, \text{ft} \).

Example 2 (Partially Filled Pipe, Output in cm², cm, cm): Calculate \( A \), \( P \), and \( R \) for a partially filled pipe:

• Radius: \( r = 0.5 \, \text{m} \);
• Fill Height: \( h = 0.3 \, \text{m} \);
• Central Angle: \( \theta = 2 \times \arccos\left(\frac{0.5 - 0.3}{0.5}\right) = 2 \times \arccos(0.4) \approx 2.498 \, \text{radians} \);
• Area: \( A = \frac{0.5^2 \times (2.498 - \sin(2.498))}{2} \approx 0.217 \, \text{m}^2 \), in cm²: \( 0.217 \times 10000 \approx 2169.2 \, \text{cm}^2 \);
• Wetted Perimeter: \( P = 0.5 \times 2.498 \approx 1.249 \, \text{m} \), in cm: \( 1.249 \times 100 \approx 124.9 \, \text{cm} \);
• Hydraulic Radius: \( R = \frac{0.217}{1.249} \approx 0.174 \, \text{m} \), in cm: \( 0.174 \times 100 \approx 17.37 \, \text{cm} \);
• Result: \( A = 2169.16199 \, \text{cm}^2 \), \( P = 124.90485 \, \text{cm} \), \( R = 17.37433 \, \text{cm} \).

Example 3 (Rectangular Channel, Output in m², m, m): Calculate \( A \), \( P \), and \( R \) for a rectangular channel:

• Base Width: \( b = 2 \, \text{m} \);
• Depth: \( y = 1 \, \text{m} \);
• Area: \( A = 2 \times 1 = 2 \, \text{m}^2 \);
• Wetted Perimeter: \( P = 2 + 2 \times 1 = 4 \, \text{m} \);
• Hydraulic Radius: \( R = \frac{2 \times 1}{2 + 2 \times 1} = \frac{2}{4} = 0.5 \, \text{m} \);
• Result: \( A = 2.00000 \, \text{m}^2 \), \( P = 4.00000 \, \text{m} \), \( R = 0.50000 \, \text{m} \).

Example 4 (Trapezoidal Channel, Output in m², m, m): Calculate \( A \), \( P \), and \( R \) for a trapezoidal channel:

• Bottom Width: \( b = 2 \, \text{m} \);
• Top Width: \( B = 4 \, \text{m} \);
• Height: \( y = 1 \, \text{m} \);
• Slope: \( z = \frac{4 - 2}{2 \times 1} = 1 \);
• Area: \( A = 2 \times 1 + 1^2 \times 1 = 3 \, \text{m}^2 \);
• Wetted Perimeter: \( P = 2 + 2 \times 1 \times \sqrt{1 + 1^2} = 2 + 2 \times \sqrt{2} \approx 4.828 \, \text{m} \);
• Hydraulic Radius: \( R = \frac{3}{2 + 2 \times \sqrt{2}} \approx 0.621 \, \text{m} \);
• Result: \( A = 3.00000 \, \text{m}^2 \), \( P = 4.82843 \, \text{m} \), \( R = 0.62132 \, \text{m} \).

Example 5 (Triangular Channel, Output in m², m, m): Calculate \( A \), \( P \), and \( R \) for a triangular channel:

• Top Width: \( B = 2 \, \text{m} \);
• Height: \( y = 1 \, \text{m} \);
• Slope: \( z = \frac{2}{2 \times 1} = 1 \);
• Area: \( A = 1^2 \times 1 = 1 \, \text{m}^2 \);
• Wetted Perimeter: \( P = 2 \times 1 \times \sqrt{1 + 1^2} = 2 \times \sqrt{2} \approx 2.828 \, \text{m} \);
• Hydraulic Radius: \( R = \frac{1 \times 1}{2 \times \sqrt{1 + 1^2}} = \frac{1}{2 \times \sqrt{2}} \approx 0.354 \, \text{m} \);
• Result: \( A = 1.00000 \, \text{m}^2 \), \( P = 2.82843 \, \text{m} \), \( R = 0.35355 \, \text{m} \).

5. Frequently Asked Questions (FAQ)

Q: Why is the area important in pipe flow?
A: The area (\( A \)) determines the cross-sectional area available for flow, directly affecting the flow rate (\( Q = A \times v \)).

Q: What does the wetted perimeter represent?
A: The wetted perimeter (\( P \)) is the perimeter of the channel in contact with the fluid, influencing frictional losses in the flow.

Q: How do area and wetted perimeter relate to hydraulic radius?
A: The hydraulic radius (\( R \)) is the ratio of the area to the wetted perimeter (\( R = \frac{A}{P} \)), indicating the efficiency of the channel’s flow.

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