Chapter 12: Problem 44
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
Expert verified
(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
01
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.
02
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/sĀ²) and \( h \) is the height of the water above the hole (14 m). Substitute these values into the equation: \( v = \sqrt{2 \times 9.81 \times 14} \).
03
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.
04
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: \( A = \pi (0.003)^2 \).
05
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.
06
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 \( Q = 0.00002827 \times 16.57 \).
07
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 mĀ³ = 1000 L), to get approximately 0.468 liters per second.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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/sĀ²), 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.
\[ v = \sqrt{2gh} \]
Here, \( v \) is the efflux speed, \( g \) is the acceleration due to gravity (approximately 9.81 m/sĀ²), 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.
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.
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:
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: \( A = \pi (0.003)^2 = 0.00002827 \text{ m}^2 \).
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:
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.