Problem 64 - Fundamentals of Physics

Question

Water flows through a horizontal pipe and then out into the atmosphere at a speed \(v_{1}=15\) \(\mathrm{m} / \mathrm{s}\). The diameters of the left and right sections of the pipe are \(5.0 \mathrm{~cm}\) and \(3.0\) \(\mathrm{cm}\). (a) What volume of water flows into the atmosphere during a 10 min period? In the left section of the pipe, what are (b) the speed \(v_{2}\) and (c) the gauge pressure?

Step-by-Step Solution

Answer

(a) 6.36 m³, (b) 5.41 m/s, (c) 97850 Pa.

Step 1: Calculate Area of the Pipe Sections

The cross-sectional area of a pipe can be calculated using the formula for the area of a circle, \( A = \pi r^2 \). For the diameter given, the radius is half of the diameter. Hence, for the left section of the pipe (\( d_2 = 5.0 \) cm), the area \( A_2 \) is: \[ A_2 = \pi \left( \frac{5.0}{2} \times 10^{-2} \right)^2 \approx 19.63 \times 10^{-4} \, \text{m}^2 \]Similarly, for the right section of the pipe (\( d_1 = 3.0 \) cm), the area \( A_1 \) is: \[ A_1 = \pi \left( \frac{3.0}{2} \times 10^{-2} \right)^2 \approx 7.07 \times 10^{-4} \, \text{m}^2 \]

Step 2: Determine Volume Flow Rate Using Continuity Equation

The continuity equation for incompressible fluids states that the flow rate must be constant across any cross-section of the pipe: \[ A_1 v_1 = A_2 v_2 \] We know \( A_1 \), \( A_2 \), and \( v_1 = 15 \text{ m/s} \). Plug in the values:\[ 7.07 \times 10^{-4} \times 15 = 19.63 \times 10^{-4} \times v_2 \]\[ v_2 = \frac{7.07 \times 15}{19.63} \approx 5.41 \text{ m/s} \]

Step 3: Calculate the Volume of Water Flowed into Atmosphere

The volume flow rate \( Q \) is given by \( Q = A_1 v_1 \). Multiply by the time duration to find the total volume:\[ Q = A_1 v_1 = 7.07 \times 10^{-4} \times 15 = 0.0106 \text{ m}^3/\text{s} \]For a 10-minute period (600 seconds):\[ V = Q \times 600 = 0.0106 \times 600 \approx 6.36 \text{ m}^3 \]

Step 4: Apply Bernoulli's Principle to Find Gauge Pressure

Bernoulli's equation relates pressure and flow speed for incompressible fluids:\[ P_2 + \frac{1}{2} \rho v_2^2 = P_1 + \frac{1}{2} \rho v_1^2 \]Assuming the atmospheric pressure \( P_1 \) is the reference, the gauge pressure \( P_g = P_2 - P_1 \). Simplifying for \( P_g \):\[ P_g = \frac{1}{2} \rho (v_1^2 - v_2^2) \]Using \( \rho = 1000 \text{ kg/m}^3 \):\[ P_g = \frac{1}{2} \times 1000 \times (15^2 - 5.41^2) \]\[ P_g \approx \frac{1}{2} \times 1000 \times (225 - 29.3) \]\[ P_g \approx \frac{1}{2} \times 1000 \times 195.7 \]\[ P_g \approx 97850 \text{ Pa} \]

Key Concepts

Continuity Equation

The continuity equation is a fundamental concept in fluid dynamics, particularly for incompressible fluids like water. It ensures that, within a closed system, the amount of fluid remains constant even if the cross-sectional area of the flow changes. This is mathematically expressed as:\[ A_1 v_1 = A_2 v_2 \]where:

• \( A_1 \) and \( A_2 \) are the cross-sectional areas of two different sections of the pipe.
• \( v_1 \) and \( v_2 \) are the respective flow velocities at these sections.

This principle allows us to calculate the unknown velocity or area, given the known values in a pipe. If a pipe narrows, the velocity must increase to ensure the flow rate remains the same, just like traffic speed adjusts in a narrower lane. This ensures that what flows into the system must flow out, maintaining a constant volume flow rate.

Bernoulli's Principle

Bernoulli's Principle is a cornerstone of fluid dynamics that explains the behavior of a fluid in motion, taking into account velocity, pressure, and height. Bernoulli's equation for a horizontal flow simplifies to:\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]Here:

• \( P_1 \) and \( P_2 \) are the pressures at two points along the flow.
• \( \rho \) is the fluid density, which is constant for incompressible fluids like water.
• \( v_1 \) and \( v_2 \) are the fluid velocities at these points.

This principle illustrates that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure. Think of it like when air rushes over a car, reducing pressure above it and helping the car to stay on the ground at high speeds. This is crucial for understanding how fluids behave under different conditions.

Volume Flow Rate

The concept of volume flow rate quantifies the amount of fluid that passes through a section of pipe per unit time. It is given by the formula:\[ Q = A v \]where:

• \( Q \) is the volume flow rate.
• \( A \) is the cross-sectional area of the pipe.
• \( v \) is the fluid velocity.

The flow rate tells us how much fluid moves within a certain period, crucial for determining efficiency in pipes and systems. For engineers and hobbyists alike, understanding this tells you how quickly a tank may empty or fill through a pipe. It's like measuring how fast a bath fills based on the tap's width and how much you've turned it on.

Gauge Pressure

Gauge pressure measures the pressure of a fluid relative to the atmospheric pressure. Unlike absolute pressure, it does not include atmospheric pressure in its value, focusing solely on the pressure added by the fluid itself:\[ P_g = \frac{1}{2} \rho (v_1^2 - v_2^2) \]This equation reflects the change in dynamic pressure due to velocity changes along the flow. It tells us:

• How much pressure is exerted by a fluid inside pipes beyond just gravity and atmospheric pressure.
• Reliable for comparing pressures in different scenarios by eliminating local atmospheric influences.

Understanding gauge pressure is essential when designing systems that have to withstand varying pressures, such as plumbing systems or hydraulic machinery. It lets you know how much pressure the fluid is exerting, crucial for safety and performance assessments.