Problem 44 - College Physics

Question

Water flowing through a garden hose of diameter \(2.74 \mathrm{~cm}\) fills a \(25.0\) - L bucket in \(1.50 \mathrm{~min}\). (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

Step-by-Step Solution

Answer

The speed of the water leaving the hose without the nozzle is approximately 0.503 m/s and after the nozzle is attached is approximately 4.52 m/s.

Step 1: Determining the Speed of the Water Without Nozzle

First, calculate the volume of the water in cubic meters which is \(25.0 \, L = 25.0 \times 10^{-3} \, m^3 \). Then, calculate the time in seconds which is \(1.50 \, min = 90.0 \, s \). The volume flow rate (Q) is volume/time, so \(Q = \frac{25.0 \times 10^{-3}}{90} \, m^3/s\). The speed of the fluid (v) is given by \(v = \frac{Q}{A} \) where A is the cross-sectional area of the hose. Substituting for A using the formula for the area of a circle (\(A= π (d/2)^2 \)), where d is the diameter of the hose, yields the speed.

Step 2: Determining the Speed of the Water With Nozzle

Now, calculate the diameter of the nozzle which is \(\frac{d}{3}\). Then calculate the area of the nozzle cross-section using the same formula for the area of a circle as before. Apply the equation of fluid continuity, \(A_1v_1=A_2v_2\), where \(A_1\) is the area of the hose, \(v_1\) is the speed at the hose, \(A_2\) is the area of the nozzle, and \(v_2\) is the speed at the nozzle, to find the speed of the water leaving the nozzle.

Key Concepts

Volume Flow Rate

Volume flow rate is a fundamental concept in fluid dynamics. It describes the quantity of fluid moving through a given area per unit time. Imagine water rushing through a garden hose. Volume flow rate helps us to determine how much water passes a certain point in the hose in a specific time period.

To calculate the volume flow rate (\(Q\)), use the formula:

• \(Q = \frac{\text{Volume}}{\text{Time}}\)

In the example of water filling a bucket, the volume of water is \(25.0\) liters, which is converted to cubic meters for calculation. And the time taken is \(1.50\) minutes, converted into seconds for uniformity in units.

By understanding volume flow rate, we can determine how quickly fluids are being transported, which is crucial in designing systems that rely on efficient fluid movement.

Continuity Equation

The continuity equation is a principle used in fluid dynamics to understand how fluid flows through varying cross-sections. It's based on the conservation of mass, meaning that the mass of fluid entering a system equals the mass leaving it. This is especially useful when a fluid passes through pipes of different sizes.

The equation itself is expressed as:

• \(A_1v_1 = A_2v_2\)

Where \(A_1\) and \(v_1\) are the cross-sectional area and fluid velocity at one point, and \(A_2\) and \(v_2\) are those at another point.

In our garden hose example, when a nozzle is attached, the nozzle's reduced diameter changes the cross-sectional area. The continuity equation allows us to calculate the new speed of water exiting the nozzle based on the known speed and area of water before the nozzle was attached.

Cross-Sectional Area

Cross-sectional area is crucial for understanding how fluid flows through a conduit like a hose or a pipe. It refers to the surface area of the opening through which the fluid moves. In a cylindrical object, such as a hose or pipe, this area is shaped like a circle.

To find the cross-sectional area (\(A\)), for objects with circular openings, we use:

• \(A = \pi \left( \frac{d}{2} \right)^2\)

Where \(d\) is the diameter of the circle. Knowing the cross-sectional area helps find out the speed of the fluid by applying it in the formula for speed: \(v = \frac{Q}{A}\).

Once the nozzle with reduced diameter is attached, the equation must be used again to determine how the reduction affects the speed of the water exiting the hose.