Problem 69 - Fundamentals of Physics

Question

A liquid of density \(900 \mathrm{~kg} / \mathrm{m}^{3}\) flows through a horizontal pipe that has a cross-sectional area of \(1.90 \times 10^{-2} \mathrm{~m}^{2}\) in region \(A\) and a cross-sectional area of \(9.50 \times 10^{-2} \mathrm{~m}^{2}\) in region \(B\). The pressure difference between the two regions is \(7.20 \times 10^{3} \mathrm{~Pa}\). What are (a) the volume flow rate and (b) the mass flow rate?

Step-by-Step Solution

Answer

The volume flow rate is \(6.40 \times 10^{-2} \text{ m}^3/\text{s}\) and the mass flow rate is \(57.6 \text{ kg/s}\).

Step 1: Understanding Volume Flow Rate

To find the volume flow rate, we apply the principle of continuity which states that the volume flow rate must remain constant throughout the pipe. This can be expressed with the equation: \( A_A v_A = A_B v_B \), where \( A \) is the cross-sectional area and \( v \) is the fluid velocity. We are first tasked with finding the velocities \( v_A \) and \( v_B \).

Step 2: Applying Bernoulli's Equation

We use Bernoulli's equation to relate the pressure difference to fluid velocities. Bernoulli’s equation is \( P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2 \). We substitute \( P_A - P_B = 7.20 \times 10^3 \text{ Pa} \), and rearrange the equation to find \( v_A \) and \( v_B \).

Step 3: Solving for Velocity in Region A

Since we know the pressure difference, rearrange Bernoulli's equation to solve for \( v_A \). You can express the difference in velocities \( v_A^2 - v_B^2 = \frac{2(P_A - P_B)}{\rho} \) and combine with the continuity equation to solve: \( v_A = \sqrt{\frac{2(P_A - P_B)}{\rho \frac{A_A^2}{A_B^2}-\rho}} \). Substituting known values gives \( v_A \approx 3.37 \text{ m/s} \).

Step 4: Calculating Volume Flow Rate

With \( v_A \approx 3.37 \text{ m/s} \), we use the continuity equation \( Q = A_A v_A = A_B v_B \) to find the volume flow rate. Substituting \( A_A \) and \( v_A \), we calculate: \( Q = 1.90 \times 10^{-2} \text{ m}^2 \times 3.37 \text{ m/s} = 6.40 \times 10^{-2} \text{ m}^3/\text{s} \).

Step 5: Understanding Mass Flow Rate

Mass flow rate can be found by multiplying the volume flow rate by the fluid's density, \( \dot{m} = \rho \cdot Q \). Here, \( \rho = 900 \text{ kg/m}^3 \) and we already calculated the volume flow rate \( Q = 6.40 \times 10^{-2} \text{ m}^3/\text{s} \).

Step 6: Calculating Mass Flow Rate

Using the formula for mass flow rate, we have \( \dot{m} = 900 \text{ kg/m}^3 \times 6.40 \times 10^{-2} \text{ m}^3/\text{s} = 57.6 \text{ kg/s} \). This provides our final answer for the mass flow rate.

Key Concepts

Continuity Equation

The continuity equation is fundamental in fluid dynamics. It states that within a closed system, the volume flow rate must be constant. This is particularly relevant when dealing with incompressible fluids, such as liquids.
The equation can be expressed as: \[ A_A v_A = A_B v_B \] where \( A \) denotes the cross-sectional area and \( v \) represents the fluid velocity.

• This equation assumes there is no leakage or deposit in the system.
• It ensures that the product of the area and velocity remains stable across different points in the pipeline.

When solving problems, we use this principle to find unknown variables, such as velocities at different points in the pipe.

Bernoulli's Equation

Bernoulli's equation is another key concept in fluid dynamics, linking pressure and velocity in a flowing fluid. It is crucial when there's a need to calculate how these quantities affect each other in various parts of the same fluidic system.
Bernoulli’s equation can be written as: \[ P_A + \frac{1}{2} \rho v_A^2 = P_B + \frac{1}{2} \rho v_B^2 \] Here, \( P \) represents pressure, \( \rho \) is fluid density, and \( v \) is velocity.

• It helps in understanding the energy conservation within a fluid stream.
• The equation indicates how a pressure differential can drive changes in velocity across different points.

In practical applications, it allows us to determine fluid speeds when other variables like pressure and density are known. This concept is instrumental in solving for the velocity in regions A and B in our exercise.

Volume Flow Rate

The volume flow rate represents the volume of fluid passing through a given point per unit time. It is denoted as \( Q \), and can be calculated when the cross-sectional area of the pipe and the velocity of the fluid are known.
Mathematically, it is given by: \[ Q = A v \] where \( Q \) is the flow rate, \( A \) is the area, and \( v \) is the velocity.

• It is a measure of the total movement of fluid over time.
• Consistent volume flow rate means the same amount of fluid material is present at different parts of the system notwithstanding any change in area or velocity.

Our solution involves using this formula to compute the flow rate given velocity data obtained from Bernoulli's equation.

Mass Flow Rate

Mass flow rate defines how much mass flows through a section of the system per time unit. It is denoted by \( \dot{m} \) and is calculated by multiplying the volume flow rate \( Q \) by the fluid density \( \rho \).
The equation is expressed as:\[ \dot{m} = \rho \, Q \] where \( \rho \) is the fluid density.

• It provides insight into how much actual material (by mass) is moving through the pipeline.
• Understanding mass flow is important for applications like chemical processing and hydraulic systems.

In our exercise, the mass flow rate helps ascertain how effective a pipeline system is in transporting a specific amount of liquid in a certain timeframe.