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.

Short answer

(a) The speed of efflux is approximately 16.57 m/s. (b) The volume discharged per second is about 0.468 L/s.

Step-by-step solution

Understanding the Problem

We have a large water tank with a hole located 14 meters below the water level. The hole has a diameter of 6 mm, which we will need to convert to meters later. We need to calculate (a) the speed at which water flows out of the hole (efflux speed) and (b) the volume of water discharged per second.

Apply Torricelli's Law

To find the speed of efflux of the water, we use Torricelli's Law: \( v = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity (9.81 m/s²) and \( h \) is the height of the water above the hole (14 m). Substitute these values into the equation: \( v = \sqrt{2 \times 9.81 \times 14} \).

Calculate the Speed of Efflux

Compute the value from the equation: \( v = \sqrt{2 \times 9.81 \times 14} = \sqrt{274.68} \). Calculate this to obtain \( v \approx 16.57 \text{ m/s} \). This is the speed of efflux.

Convert Hole Diameter to Meters and Find Area

Convert the diameter of the hole from millimeters to meters: 6 mm = 0.006 m. Calculate the cross-sectional area \( A \) of the hole using the formula for the area of a circle: \( A = \pi \left(\frac{d}{2}\right)^2 \). Substitute \( d = 0.006 \) m into the formula: \( A = \pi (0.003)^2 \).

Calculate the Area of the Hole

Compute the area: \( A = \pi (0.003)^2 = \pi \times 0.000009 = 0.00002827 \text{ m}^2 \). This is the cross-sectional area of the hole.

Calculate the Volume Discharged per Second

The volume discharged per second \( Q \) is given by \( Q = Av \), where \( v \) is the speed of efflux (16.57 m/s) and \( A \) is the area of the hole. Substituting these values, we get \( Q = 0.00002827 \times 16.57 \).

Find the Numerical Value of the Volume Discharge

Compute the volume: \( Q = 0.000468 \text{ m}^3/\text{s} \). Convert this to a more common unit if necessary, such as liters per second (1 m³ = 1000 L), to get approximately 0.468 liters per second.

Key concepts

Torricelli's Law

When trying to determine how fast water exits a hole in a tank, Torricelli's Law is instrumental. It describes the speed of fluid flowing out of an opening under the influence of gravity. Torricelli's Law is given by:
\[ v = \sqrt{2gh} \]
Here, \( v \) is the efflux speed, \( g \) is the acceleration due to gravity (approximately 9.81 m/s²), and \( h \) is the height of the fluid above the hole. This formula emerges from principles of energy conservation, where gravitational potential energy converts to kinetic energy as fluid moves downward.

• Utilize the formula to predict efflux speed without needing complex calculations of pressure differences.
• In our task, \( h \) is 14 meters, which allows us to substitute directly into the equation to find \( v \).
• This simplifies the problem-solving process greatly and is fundamental in fluid dynamics.

Efflux Speed

Efflux speed refers to how fast a fluid exits through an opening due to pressure differences. It is crucial for various applications, from irrigation to hydraulic machines.

• In our exercise, the efflux speed is calculated using the formula derived from Torricelli's Law.
• The calculated speed is \( v \approx 16.57 \text{ m/s} \), which indicates how rapidly water shoots out from the hole.
• This speed is dependent on both the height of the water above the hole and the gravitational pull acting on it.

Understanding efflux speed is essential for designing systems where precise fluid measurements are critical. Designing efficient water systems requires knowing how fast fluids travel, and this is where understanding efflux speed becomes incredibly significant.

Volume Discharge

Volume discharge quantifies how much fluid exits through an opening per unit of time. This is particularly useful in scenarios like managing water supply or drainage systems.
The formula for volume discharge \( Q \) is:
\[ Q = Av \]
Here, \( A \) is the cross-sectional area of the hole, and \( v \) is the efflux speed. Thus, volume discharge combines geometric and kinetic factors.

• After applying values from our step-by-step solution, the volume discharge is \( Q = 0.000468 \text{ m}^3/ ext{s} \) or 0.468 liters per second.
• This means approximately 0.468 liters of water exits the tank every second.
• Knowing the volume discharge is crucial for water management, ensuring adequate flow rates in pipes and outlets.

Cross-sectional Area

The cross-sectional area is the area of the slice through which fluid flows. This physical parameter significantly impacts the fluid flow rate.
To determine the cross-sectional area of the tank hole:

• Convert the diameter of 6 mm to meters, yielding 0.006 m.
• Use the formula for the area of a circle: \( A = \pi \left(\frac{d}{2}\right)^2 \).
• Substitute to calculate: \( A = \pi (0.003)^2 = 0.00002827 \text{ m}^2 \).

This calculation ensures accuracy in determining the speed and volume of water exiting. Cross-sectional area plays a key role in determining how much fluid can pass through an opening, affecting everything from sprinkler systems to medical equipment.

Water Pressure

Water pressure in a tank plays a pivotal role in determining how water flows through openings. Pressure is directly influenced by the height of the fluid above the point in question.
In fluid dynamics, the pressure difference propels water out of the tank hole. For our example:

• The higher the water above the hole, the greater the pressure exerted on the water at the hole, resulting in faster efflux.
• Understanding this allows engineers to design systems that harness gravitational potential to achieve desired flow rates.
• In open-top tanks like ours, atmospheric pressure often balances out, leaving gravitational potential as the primary force.

This concept is vital for comprehending how variances in water pressure can alter fluid dynamics behavior.