Problem 51 - Fundamentals of Physics

Question

A garden hose with an internal diameter of \(1.9 \mathrm{~cm}\) is connected to a (stationary) lawn sprinkler that consists merely of a container with 24 holes, each \(0.13 \mathrm{~cm}\) in diameter. If the water in the hose has a speed of \(0.91 \mathrm{~m} / \mathrm{s}\), at what speed does it leave the sprinkler holes?

Step-by-Step Solution

Answer

The water leaves the sprinkler holes at approximately 8.089 m/s.

Step 1: Find the Cross-Sectional Area of the Hose

The formula for the cross-sectional area of a circle is \( A = \pi r^2 \). The diameter of the hose is given as \( 1.9 \) cm, so the radius \( r \) is \( 0.95 \) cm or \( 0.0095 \) m. Thus, the area of the hose is:\[ A_h = \pi (0.0095)^2 \approx 2.835 \times 10^{-4} \text{ m}^2 \]

Step 2: Find the Cross-Sectional Area of Each Sprinkler Hole

Similarly, for each sprinkler hole, the diameter is \( 0.13 \) cm, so the radius \( r \) is \( 0.065 \) cm or \( 0.00065 \) m. The area of one hole is:\[ A_s = \pi (0.00065)^2 \approx 1.327 \times 10^{-6} \text{ m}^2 \]

Step 3: Calculate the Total Area of All Sprinkler Holes

There are 24 holes in the sprinkler, hence the total area for water to flow out is:\[ A_t = 24 \times 1.327 \times 10^{-6} \approx 3.1848 \times 10^{-5} \text{ m}^2 \]

Step 4: Apply the Continuity Equation

The continuity equation states that the volumetric flow rate must be constant, so \( A_h v_h = A_t v_t \) where \( v_h \) is the speed in the hose and \( v_t \) is the speed at the holes. We have:\[ (2.835 \times 10^{-4}) \times 0.91 = (3.1848 \times 10^{-5}) \times v_t \]

Step 5: Solve for the Speed at the Sprinkler Holes

Rearrange the equation to solve for \( v_t \):\[ v_t = \frac{(2.835 \times 10^{-4}) \times 0.91}{3.1848 \times 10^{-5}} \approx 8.089 \text{ m/s} \]

Key Concepts

Continuity Equation

The continuity equation is a fundamental principle in fluid dynamics and is essential in understanding how fluid flows through different sections of a system. It's based on the law of conservation of mass. It asserts that for an incompressible fluid, the mass flow rate must remain constant from one cross-section of a flow to another. This principle can be expressed as follows: \[ A_1v_1 = A_2v_2 \] Where: - \( A_1 \) is the cross-sectional area of the first section. - \( v_1 \) is the flow speed at the first section. - \( A_2 \) is the cross-sectional area of the second section. - \( v_2 \) is the flow speed at the second section. In simple terms, what goes in must come out, and the rate of fluid flow must be the same on either side of a barrier. In the context of a garden hose and a sprinkler, this means that the water entering the hose must exit out of the sprinkler holes at a consistent flow rate. Understanding the continuity equation is crucial to solving fluid flow problems in various scenarios, as it dictates that the product of the cross-sectional area and velocity remains constant.

Cross-Sectional Area Calculation

Cross-sectional area calculation is an important step in analyzing fluid dynamics problems. The cross-sectional area is usually the area of a circle when dealing with pipes or hoses, as these are typically cylindrical. To find the cross-sectional area of a circular pipe, you use the formula:\[ A = \pi r^2 \]Where \( r \) is the radius of the circle. To calculate the radius from a given diameter, remember: - Radius \( r = \frac{ ext{diameter}}{2} \) In the case of the garden hose exercise: - The diameter of the hose is \( 1.9 \text{ cm} \), so the radius is \( 0.95 \text{ cm} \) or \( 0.0095 \text{ m} \). - For the sprinkler holes, the diameter is \( 0.13 \text{ cm} \), giving a radius of \( 0.065 \text{ cm} \) or \( 0.00065 \text{ m} \). Once you have the radius, you can calculate the area for both cases, as was shown in the exercise solution. Correct cross-sectional area calculations are essential for applying the continuity equation and solving for the velocity of fluid through different sections.

Volumetric Flow Rate

Understanding the volumetric flow rate helps us grasp how much fluid passes through a conduit per unit time. It's crucial in applications like piping systems, irrigation, or any scenario involving moving fluids. The volumetric flow rate \( Q \) is given by the product of the cross-sectional area \( A \) and the velocity \( v \):\[ Q = A \times v \]In essence, it measures how quickly volume is passing through a point in the system. Here's why it's important: - **Conservation Principle**: In the continuity equation, the volumetric flow rate must be maintained across sections of a flow that are connected. - **Design and Analysis**: Engineers use it to design piping systems and analyze the fluid movement. Understanding the relationship between area and velocity helps in predicting system behavior. In the garden hose scenario, using the given speed and cross-sectional area of the hose, the volumetric flow rate is calculated so that it can then be set equal to the flow rate at the sprinkler holes, ensuring consistent water output. This allows us to adjust our systems to meet specific requirements, ultimately giving us control over fluid distribution.