Problem 42 - University Physics with Modern Physics

Question

A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second.

Step-by-Step Solution

Answer

Speed of efflux is 16.6 m/s, and the discharge rate is 0.000469 m³/s.

Step 1: Understanding the Situation

We have a water tank with a hole 14.0 m below the water surface. The water will flow out of the hole due to gravity. We need to determine the speed at which water exits (efflux) and the discharge rate (volume per second).

Step 2: Applying Bernoulli's Principle for Speed

According to Bernoulli's principle, the speed of efflux for a hole in the side of an open tank can be given by Torricelli’s theorem, which is essentially derived from energy conservation. The speed \( v \) of water leaving the hole is given by \( v = \sqrt{2gh} \), where \( g \approx 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h = 14.0 \, \text{m} \) is the height of the water column above the hole.

Step 3: Calculating the Speed of Efflux

Using the formula \( v = \sqrt{2gh} \), we substitute the given values: \[ v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 14.0 \, \text{m}} \approx 16.6 \, \text{m/s}. \] Thus, the speed of efflux is approximately 16.6 m/s.

Step 4: Determining the Area of the Hole

The volume of water discharged depends on the area of the hole. The diameter of the hole is 6.00 mm, which is equal to 0.006 m. First, find the radius: \( r = \frac{0.006}{2} = 0.003 \, \text{m} \). The area \( A \) is then given by \( A = \pi r^2 \).

Step 5: Calculating the Area of the Hole

Using the formula for the area of a circle, substitute in the radius: \[ A = \pi (0.003 \, \text{m})^2 = \pi \times 9 \times 10^{-6} \, \text{m}^2 \approx 2.827 \times 10^{-5} \, \text{m}^2. \]

Step 6: Finding the Volume Discharged per Second

The volume flow rate \( Q \) can be calculated using \( Q = Av \), where \( A \) is the area and \( v \) is the speed. Substitute in the values: \[ Q = 2.827 \times 10^{-5} \, \text{m}^2 \times 16.6 \, \text{m/s} \approx 4.69 \times 10^{-4} \, \text{m}^3/ ext{s}. \] Therefore, the volume discharged per second is approximately 0.000469 cubic meters per second.

Key Concepts

Bernoulli's principle

Bernoulli's principle is a cornerstone of fluid dynamics. It describes how the speed of a fluid relates to its pressure and potential energy. Essentially, it states that in a streamline flow, the sum of the pressure energy, kinetic energy, and potential energy per unit volume is constant. In a simplified form, \[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}, \]where:

• \( P \) is the pressure exerted by the fluid,
• \( v \) is the velocity of the fluid,
• \( \rho \) is the density of the fluid,
• \( g \) is the acceleration due to gravity, and
• \( h \) is the height above a reference point.

This principle allows us to understand how fluid speeds up in constrictions or when changing heights. It is instrumental in calculating fluid flow in various applications, like airplane wings, Venturi meters, and in our case, the speed of water leaving a tank.

Torricelli’s theorem

Torricelli’s theorem simplifies Bernoulli’s principle for a specific situation—a fluid discharging from an orifice. It states that the speed \( v \) at which a fluid exits a hole in a container only depends on the height of the fluid above the hole, not on the fluid amount or container shape. The formula derived from this theorem is:\[ v = \sqrt{2gh}, \]where:

• \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and
• \( h \) is the height of the fluid column above the outlet.

This implies that fluid exits faster when the tank is fuller. In the exercise, this theorem was used to calculate the speed of water efflux when height \( h \) was 14 meters, resulting in a speed of approximately 16.6 m/s. This concept illustrates that the potential energy of the fluid is transformed into kinetic energy of motion.

Volume flow rate

The volume flow rate is a measure of the volume of fluid that passes a given surface per unit time. It's essential in applications ranging from irrigation systems to cardiovascular physiology. The volume flow rate, typically denoted by \( Q \), is calculated as:\[ Q = Av, \]where:

• \( A \) is the cross-sectional area through which the fluid flows, and
• \( v \) is the velocity of the fluid.

In the context of our exercise, after calculating the speed of water exiting the tank and the area of the hole, we can determine how much water escapes per second. This was computed to be approximately 0.000469 cubic meters per second. Understanding volume flow rate is crucial for designing systems that manage fluid flow efficiently.

Conservation of energy in fluids

The conservation of energy in fluids is the underlying principle for understanding how fluids behave in motion. Fluid motion involves the transformation of potential energy into kinetic energy. Bernoulli's equation is a direct application of this principle. For a non-viscous, incompressible fluid in steady flow, the energy is conserved along a streamline. Therefore, fluid rising in a pipe loses kinetic energy to gain potential energy, and vice versa when falling. In our exercise, the energy from the 14-meter high column of water is converted into the kinetic energy of water flowing out of the tank. This principle provides a foundation for fields ranging from aerospace to engineering applications and explains phenomena like fluid flow rates in natural and man-made systems.