Question

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070 m\(^2\), and the magnitude of the fluid velocity is 3.50 m/s. (a) What is the fluid speed at points in the pipe where the cross- sectional area is (a) 0.105 m\(^2\) and (b) 0.047 m\(^2\)? (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

Short answer

(a) 2.33 m/s, (b) 5.21 m/s, (c) 882 m³

Step-by-step solution

Understand the Continuity Equation

The principle of conservation of mass in fluids is illustrated by the Continuity Equation. It states that the mass flow rate must remain constant from one cross-section to another in a closed system. The equation can be written as:\[ A_1V_1 = A_2V_2 \]where \( A \) is the area of the cross-section and \( V \) is the fluid velocity. This equation can help us find the velocity when the area changes.

Calculate Fluid Speed at Point (a) 0.105 m²

For point (a), we use the Continuity Equation:\[ A_1V_1 = A_2V_2 \]Given: \( A_1 = 0.070 \, \text{m}^2 \), \( V_1 = 3.50 \, \text{m/s} \), and \( A_2 = 0.105 \, \text{m}^2 \).Substitute the given values to find \( V_2 \):\[ (0.070)(3.50) = (0.105)(V_2) \]Calculate \( V_2 \):\[ V_2 = \frac{0.070 \times 3.50}{0.105} \approx 2.33 \, \text{m/s} \]

Calculate Fluid Speed at Point (b) 0.047 m²

For point (b), apply the Continuity Equation again:\[ A_1V_1 = A_3V_3 \]Given: \( A_3 = 0.047 \, \text{m}^2 \).Substitute the known values:\[ (0.070)(3.50) = (0.047)(V_3) \]Calculate \( V_3 \):\[ V_3 = \frac{0.070 \times 3.50}{0.047} \approx 5.21 \, \text{m/s} \]

Calculate Volume of Water Discharged in 1 Hour

To find the volume of water discharged, use the formula:\[ \text{Volume} = \text{Area} \times \text{Velocity} \times \text{Time} \]Choose the correct A and V for the outlet pipe. Assume the pipe ends at the original cross-section \( A_1 \):Use \( A_1 = 0.070 \, \text{m}^2 \), \( V = 3.50 \, \text{m/s} \), and \( \text{Time} = 1 \, \text{hour} = 3600 \, \text{seconds} \).\[ \text{Volume} = 0.070 \times 3.50 \times 3600 \approx 882 \, \text{m}^3 \]

Key concepts

Conservation of Mass in Fluids

The Conservation of Mass in Fluids, beautifully illustrated by the Continuity Equation, is a fundamental concept in fluid dynamics. Imagine you're squeezing toothpaste from a tube, and despite the varying thickness of the tube, each segment contains the same amount of paste. Similarly, in any closed fluid system, the amount of fluid mass entering must equal the amount of fluid mass exiting. This is because mass cannot magically appear or disappear; it just gets redistributed along the flow path. When it comes to fluids traversing through pipes or channels, this principle is represented by the Continuity Equation:

• \( A_1V_1 = A_2V_2 \)

Here, \( A \) stands for the cross-sectional area, and \( V \) for velocity. It means that if water flows through a narrower section of a pipe, it has to move faster to maintain the mass flow, like traffic speeding up to avoid jams. Understanding this equation allows us to predict changes in speed as pipe diameters fluctuate, making it a powerful tool in engineering and fluid dynamics education.

Fluid Speed Calculation

When tasked with Fluid Speed Calculation, the Continuity Equation becomes our trusty guide. Let's break this down with a practical example. Suppose we have a pipe with an initial cross-sectional area \( A_1 = 0.070 \, \text{m}^2 \) and an initial fluid velocity \( V_1 = 3.50 \, \text{m/s} \). If the section expands, say to \( 0.105 \, \text{m}^2 \), the water must slow down, and we compute the new speed like this:
Plugging into the equation:

• \( 0.070 \times 3.50 = 0.105 \times V_2 \)
• Solve for \( V_2 \), yielding \( V_2 \approx 2.33 \, \text{m/s} \)

Likewise, if the pipe narrows to \( 0.047 \, \text{m}^2 \), the water speeds up:

• \( 0.070 \times 3.50 = 0.047 \times V_3 \)
• From this, \( V_3 \approx 5.21 \, \text{m/s} \)

It’s like adjusting the nozzle on a garden hose; you decrease the opening size and the water jets out faster. Calculating fluid speed with changing areas informs us how different pipe sections affect flow speeds.

Volume Discharge Calculation

Volume Discharge Calculation is about determining how much fluid exits a system over a set period. For instance, if we wish to calculate how much water is discharged from the end of a pipe in an hour, we utilize the relationship:

• Volume = Area \( \times \) Velocity \( \times \) Time

Assuming the water flows through a section with area \( 0.070 \, \text{m}^2 \) at a velocity of \( 3.50 \, \text{m/s} \), in an hour (3600 seconds), the volume can be computed as:
\[ \text{Volume} = 0.070 \times 3.50 \times 3600 \approx 882 \, \text{m}^3 \]
Think of it like filling a container; the area corresponds to the opening size, the velocity to the flow speed, and the time to how long you let it flow. This calculation is indispensable in determining capacities and can affect everything from irrigation system design to municipal water supply processes.