Problem 59 - College Physics

Question

Water flows through a horizontal tube similar to that in Fig. 9.20. However in this case, the constricted part of the tube is half the diameter of the larger part. If the water speed is \(1.5 \mathrm{~m} / \mathrm{s}\) in the larger parts of the tube, by how much does the pressure drop in the constricted part? Express the final answer in atmospheres.

Step-by-Step Solution

Answer

The pressure drops by approximately 0.166 atmospheres.

Step 1: Identify the Given Data and Required Result

We know that the diameter of the constricted part is half of the larger part. This implies the radius of the smaller section, \( r_2 \), is half the radius of the larger section, \( r_1 \). The speed of water in the larger section, \( v_1 \), is given as 1.5 m/s. We are tasked with finding the pressure drop in the constricted part relative to the larger part, expressed in atmospheres.

Step 2: Relate Diameters and Velocities

From the problem, \( d_2 = \frac{1}{2}d_1 \). Since \( v_1 = 1.5 \, \mathrm{m/s} \), we use the equation of continuity, \( A_1v_1 = A_2v_2 \), where \( A = \pi r^2 \). Since \( r_2 = \frac{1}{2}r_1 \), we find:\[v_2 = \frac{A_1}{A_2}v_1 = \left(\frac{rac{\pi (r_1)^2}{\pi ((\frac{1}{2}r_1)^2)}}\right) v_1 = \left( \frac{1}{0.25} \right) v_1 = 4v_1 = 4 \times 1.5 \, \mathrm{m/s}\]This gives \( v_2 = 6 \, \mathrm{m/s} \).

Step 3: Apply Bernoulli's Equation

Supply Bernoulli's principle: \( P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \), to find the pressure difference:\[ P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) \]Water's density, \( \rho = 1000 \, \mathrm{kg/m^3} \). Substitute \( v_1 = 1.5 \, \mathrm{m/s} \) and \( v_2 = 6 \, \mathrm{m/s} \):\[P_1 - P_2 = \frac{1}{2} \times 1000 \, \mathrm{kg/m^3} ((6 \, \mathrm{m/s})^2 - (1.5 \, \mathrm{m/s})^2)\]Calculate:\[P_1 - P_2 = 500 \times (36 - 2.25) = 16875 \, \mathrm{N/m^2}\]

Step 4: Convert Pressure Drop to Atmospheres

Convert the pressure difference from pascals to atmospheres. We know that 1 atmosphere is equivalent to about 101325 N/m\(^2\):\[\text{Pressure Drop in atm} = \frac{16875}{101325}\]Calculate:\[\text{Pressure Drop in atm} \approx 0.166 ext{ atm}\]

Key Concepts

Understanding Fluid Dynamics

Fluid dynamics is a branch of physics focused on the movement of liquids and gases. It provides models that predict how fluids behave under different conditions. In this exercise, we examine water flowing through a tube with varying diameter. The speed at which the fluid moves and changes in tube size are predictable with fluid dynamics principles.

Key factors influencing fluid movement include:

• Velocity: The speed of the fluid. It's a central variable in fluid dynamics equations.
• Density: Mass per unit volume of the fluid, often assumed to be constant in calculations unless otherwise noted.
• Pressure: The force exerted per unit area in the fluid, affected by elevation and velocity changes in fluid paths.

Fluid dynamics helps solve problems involving these factors to determine how variations in a tube's structure and size impact the flow of the fluid.

Calculating Pressure Drop Using Bernoulli's Principle

One important aspect of fluid dynamics is predicting pressure changes in a flowing fluid. Bernoulli's Principle is key to this, stating that in a streamline flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume is constant. This principle can predict how a fluid's speed and pressure change with varying diameters in the tube.

By applying Bernoulli's equation:\[ P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2 \]We find the pressure drop \( P_1 - P_2 \) by subtracting kinetic energies:\[ P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2) \]Given that \( \rho = 1000 \, \mathrm{kg/m^3} \), \( v_1 = 1.5 \, \mathrm{m/s} \), and \( v_2 = 6 \, \mathrm{m/s} \), the pressure difference calculates to \( 16875 \, \mathrm{N/m^2} \) or \( 0.166 \) atm. This decrease occurs because the fluid speeding up in the constricted section loses pressure energy, evidenced by higher speed and lower pressure in the faster flowing section.

Exploring the Continuity Equation

The continuity equation is vital in understanding how fluids move consistently through varying sections of a tube or pipe. It demonstrates that the mass flow rate of a fluid is constant, so velocity and cross-sectional area are inversely related, given by:\[ A_1v_1 = A_2v_2 \]where \( A \) represents the cross-sectional area and \( v \) is the velocity.

In this exercise:

• The larger tube has half the speed of the smaller, constricted section due to its larger diameter.
• Since the constricted part's diameter is half, its area is one-fourth, leading the speed in the smaller section to quadruple, as calculated: \( v_2 = 4 \times v_1 \).

This equation is essential in pipe systems, ensuring efficient fluid transport. Whenever pipes narrow, fluid speeds up, necessitating careful consideration of material constraints and fluid behavior to predict performance accurately.