Hydraulic Conduit Laminar Volumetric Flow Rate (Civil & Environmental Engineering) Solution

A civil and environmental water resources engineer at the UNLV Howard R. Hughes College of Engineering is designing a pressurized subsurface drainage culvert to convey stormwater and treated municipal effluent under Tropicana Avenue in Las Vegas.

The culvert is constructed as a circular reinforced concrete conduit of internal radius $R = 0.5\text{ meters}$. Due to viscous fluid friction and boundary layer adhesion along the conduit wall, the stormwater exhibits a fully developed axisymmetric parabolic velocity profile, modeled in Cartesian coordinates as:

Step-by-Step Solution:

1. Convert Domain Boundaries to Polar Coordinates:
   State the Circle Boundary Equation in Polar Form:

   $$x^2 + y^2 \le R^2 \implies r^2 \le (0.5)^2 \implies 0 \le r \le 0.5\text{ m}$$
   Because the conduit cross-section is a complete circle enclosing the origin, the angular sweep encompasses the full circumference:
   $$0 \le \theta \le 2\pi\text{ rad}$$
   Thus, the domain is the polar rectangle $D = \{(r, \theta) \mid 0 \le r \le 0.5, \, 0 \le \theta \le 2\pi\}$.

2. Transform Velocity Profile and Area Element:
   Apply the Polar Coordinate Substitution Rule:

   $$x^2 + y^2 = r^2$$
   Substitute into the Cartesian velocity distribution:
   $$v(r) = 4\left(0.25 - r^2\right) = 1 - 4r^2\text{ m/s}$$
   State the Polar Area Element Formula with Jacobian:
   $$dA = r \, dr \, d\theta$$
   Formulate the double integral for total volumetric discharge $Q$:
   $$Q = \iint_D v \, dA = \int_{0}^{2\pi} \int_{0}^{0.5} (1 - 4r^2) \, r \, dr \, d\theta = \int_{0}^{2\pi} \int_{0}^{0.5} (r - 4r^3) \, dr \, d\theta$$

3. Evaluate the Double Integral:
   Apply Fubini's Separability Theorem in Polar Coordinates (defined in Separability in Polar Integrals):

   $$Q = \left( \int_{0}^{2\pi} d\theta \right) \left( \int_{0}^{0.5} (r - 4r^3) \, dr \right)$$

   Evaluate the radial integral:

   $$\int_{0}^{0.5} (r - 4r^3) \, dr = \left[ \frac{r^2}{2} - r^4 \right]_{0}^{0.5} = \left( \frac{(0.5)^2}{2} - (0.5)^4 \right) - 0$$
   $$= \frac{0.25}{2} - 0.0625 = 0.125 - 0.0625 = 0.0625 = \frac{1}{16}\text{ m}^3/\text{s}$$

   Evaluate the angular integral:

   $$\int_{0}^{2\pi} d\theta = [\theta]_{0}^{2\pi} = 2\pi\text{ rad}$$

   Multiply the evaluated components:

   $$Q = 2\pi \times \frac{1}{16} = \frac{\pi}{8}\text{ m}^3/\text{s} \approx 0.3927\text{ m}^3/\text{s}$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The total volumetric flow rate through the circular conduit is exactly $\frac{\pi}{8}\text{ m}^3/\text{s} \approx 0.393\text{ m}^3/\text{s}$ (approximately $393\text{ liters per second}$).
• Engineering Application: In civil flood-control engineering, evaluating volumetric discharge across circular pipe cross-sections is fundamental to Hagen-Poiseuille pipe flow analysis and stormwater master planning. By converting Cartesian coordinate profiles into polar form, engineers eliminate cumbersome radical boundaries ($y = \pm\sqrt{R^2 - x^2}$) and verify whether gravity drainage or lift pump stations can handle peak convective desert storm runoffs without exceeding culvert hydraulic capacity.