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Drainage of a fluid. Torricelli’s Theorem + 6 Solved Exercises

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La última actualización de esta entrada fue hecha el 9 enero, 2024 por Hernán R. Gómez

Introduction

In this Fluid Mechanics article, we will see drainage of a fluid and how to calculate the exit velocity of a fluid through a container. The interesting thing about this topic is that we can, applying the same logic and reasoning, handle cases where we have a liquid container with an opening at the bottom (such as a tank with a hole in its lower part, as shown in Figure 1) or a container that has a hole in its walls (as shown in Figure 2).

Drainage of a fluid. Torricelli's Theorem + 6 Solved Exercises
Figure 1: Drainage of a fluid.
Drainage of a fluid. Torricelli's Theorem + 6 Solved Exercises
Figure 2: Exit velocity in a pierced container.

Separemos nuestro estudio en ambos casos.

What is fluid mechanics?

Fluid mechanics is a branch of physics that deals with the study of the behavior of fluids (liquids and gases) at rest or in motion. It aims to describe how fluids move and interact with their surroundings.

Fluid mechanics is applied in a wide variety of fields, from civil and mechanical engineering to medicine and biology. Some of the topics studied in fluid mechanics include fluid dynamics (movement of fluids), fluid kinematics (velocity and acceleration of the fluid), fluid thermodynamics (heat transfer in fluids), and computational fluid mechanics (use of computational tools to simulate the behavior of fluids).

Exit velocity in a container with a hole (in side walls or bottom)

Figure 1 shows a container with a cross-sectional area A, filled with a fluid up to a height that we will call h. The space above the tank is at atmospheric pressure $latex p_{1}$, and the fluid exits through a tube with area $latex A_{2}$.

Let’s try to obtain an expression for the exit velocity. We can locate point 1 on the surface of the liquid. There, the pressure will be called $latex p_{1}$, and the fluid velocity will be $latex v_{1}$. Point 2 will be located at the tank outlet.

There, we will call $latex v_{2}$ to the exit velocity, and its pressure will be denoted as $latex p_{2}$. We can set the level of point 2 as $latex h=0$, so the height of point 1 will be equal to h. Since area 1 is much larger than area 2, we can say that the fluid level in the tank will drop very slowly.

We could say that the velocity at point 1 is zero (practically, $latex v_{1}=0$). Another important consideration is the following: The exit velocity will depend on several factors. But… which ones? To answer this, let’s apply the Bernoulli’s Principle.

Bernoulli’s Principle

$latex p_{1}+\delta \cdot g\cdot h_{1}+\frac{1}{2}\cdot \delta \cdot v_{1}^{2}=p_{2}+\delta \cdot g\cdot h_{2}+\frac{1}{2}\cdot \delta \cdot v_{2}^{2} $

Puedes encontrar más información sobre Teorema de Bernoulli en nuestro post de Ensamble de Ideas.

We see that, of course, the velocity $latex v_{2}$ will depend on the gravity and density of the fluid. What is not so evident and we must always keep in mind is that said velocity will depend on the pressure difference $latex \Delta p $ between the upper and lower points, that is, it depends on $latex \Delta p = p_{2}-p_{1}$, and also on the height h of the liquid.

At this point, we have a lot of information that will make it easy for us to work with the Bernoulli Equation. Let’s go step by step.

First, let’s start from the original equation:

$latex p_{1}+\delta \cdot g\cdot h_{1}+\frac{1}{2}\cdot \delta \cdot v_{1}^{2}=p_{2}+\delta \cdot g\cdot h_{2}+\frac{1}{2}\cdot \delta \cdot v_{2}^{2} $

We solve for the square of the velocity at the bottom of the container:

$latex \frac{p_{1}+\delta \cdot g\cdot h_{1}+\frac{1}{2}\cdot \delta \cdot v_{1}^{2}-p_{2}-\delta \cdot g\cdot h_{2}}{\frac{1}{2}\cdot \delta }=v_{2}^{2}$

We know that $latex h_{2}=0$ and that $latex v_{1}=0$, so:

$latex \frac{p_{1}+\delta \cdot g\cdot h_{1}-p_{2}}{\frac{1}{2}\cdot \delta }=v_{2}^{2} $

If we consider that both the top and bottom of the container are at atmospheric pressure (that is, $latex p_{1}=p_{2}=p_{atm}$, so $latex \Delta p =0$), we can cancel out the pressures:

$latex \frac{\delta \cdot g\cdot h_{1}}{\frac{1}{2}\cdot \delta }=v_{2}^{2}$

Rearranging (remembering that $latex \frac{1}{\frac{1}{2}}=2$) and eliminating $latex \delta$ because it appears in both the numerator and denominator:

$latex 2\cdot g\cdot h_{1}=v_{2}^{2}$

We obtain the square root to solve for the exit velocity:

$latex v_{2}=\sqrt{2\cdot g\cdot h_{1}}$

And that’s it! We have obtained the expression for the exit velocity of a fluid! Do you notice something similar? In free fall, the speed with which an object hits the ground presents the same expression, with h being the height from which said object is dropped. This result is known as Torricelli’s theorem and is valid not only for a tank that has an opening at the bottom but also for the sides of the container.

Flow rate of volume in a container with drain

Calculating the volume flow rate in a container that has a drain, whether it is at the bottom or on the side, is very simple, by applying Torricelli’s theorem and the Principle of Continuity.

We know that the flow rate must remain constant over time, meaning that the flow rate at the top point must be equal to the flow rate at the bottom point. Therefore:

$latex C_{1}=C_{2}$

As $latex C=A\cdot v$, we can establish the relationship:

$latex C= A_{2} \cdot v_{2}$

Given that $latex v_{2}=\sqrt{2\cdot g\cdot h_{1}}$ (as we deduced before), then:

$latex C= A_{2} \cdot \sqrt{2\cdot g\cdot h_{1}}$

Activities about drainage of a fluid

  1. A tank has a capacity of 1000 liters and is filled with water. The tank drain has a diameter of 5 cm and a length of 50 cm. If the flow of water coming out of the drain is 2 liters per minute, how long will it take for the tank to empty completely?
  2. A conical tank has a height of 5 meters and a base diameter of 4 meters. The drain of the tank is at the bottom and has a diameter of 5 cm. If the flow of water coming out of the drain is 1 liter per minute, what is the volume flow rate of water leaving the tank?
  3. A tank has a capacity of 500 liters and is filled with a saline solution. The tank drain has a diameter of 2 cm and a length of 20 cm. If the concentration of salt in the solution is 0.1 g/ml and the flow of water coming out of the drain is 1 liter per minute, what is the flow rate of salt leaving the tank?
  4. A cylindrical tank with a height of 2 meters and a diameter of 1 meter is filled with water up to a height of 1.5 meters. If there is a circular hole with a diameter of 3 centimeters at the bottom of the tank, at what rate will the water flow out of the hole? Assume that the velocity of the water at the hole is the same as that of the fluid at the surface of the tank.
  5. A conical tank with a height of 5 meters and a base radius of 3 meters is filled with oil up to a height of 4 meters, and there is a circular hole with a diameter of 2 centimeters at the bottom of the tank. At what rate will the oil flow out of the hole? Assume that the velocity of the oil at the hole is the same as that of the fluid at the surface of the tank.
  6. A rectangular tank has a height of 1 meter, a length of 2 meters, and a width of 0.5 meters. The tank is filled up to a height of 80 centimeters with water, and there is a rectangular hole with a width of 5 centimeters and a height of 10 centimeters in one of the side walls of the tank. If the velocity of the water at the hole is the same as that at the surface of the water in the tank, at what rate will the water flow out of the hole?

Answer key:

  1. The volume flow rate of water exiting the drain is 2 liters per minute. If the tank has a capacity of 1000 liters, it will take 500 minutes (or 8 hours and 20 minutes) to completely empty. This can be calculated by dividing the tank capacity by the volume flow rate: 1000 L / 2 L/min = 500 min.

    The volume of the conical tank can be calculated using the formula V = (1/3)πr²h, where r is the radius of the base and h is the height of the cone. In this case, the radius is 2 meters (half the diameter of the base) and the height is 5 meters. Therefore, the volume of the tank is approximately 20.94 cubic meters. The volume flow rate of water exiting the tank can be calculated by multiplying the area of the drain by the velocity of the water: A = (π/4)d² = (π/4)(0.05²m²) ≈ 0.00196 m². Therefore, the volume flow rate is 0.00196 m²/minute * 1 liter/1000 cm³ * 60 seconds/minute ≈ 0.0702 liters per second.

    The volume of salt in the solution can be calculated by multiplying the salt concentration by the volume of the solution: V_salt = 0.1 g/ml * 500 liters = 50 grams. The flow rate of salt exiting the tank can be calculated by multiplying the volume flow rate of water by the salt concentration: 1 liter/minute * 0.1 g/ml = 0.1 grams/minute. Therefore, the flow rate of salt is 0.1 grams per minute, or approximately 0.00167 grams per second.

    The exit velocity of the water can be calculated using Torricelli’s equation, which relates the exit velocity to the height of the liquid in the tank: v = √(2gh), where v is the exit velocity, g is the acceleration due to gravity (9.81 m/s²), and h is the height of the liquid in the tank measured from the center of the hole. In this case, h = 1.5 meters. Therefore, the exit velocity is v = √(2 * 9.81 m/s² * 1.5 m) ≈ 3.83 m/s. Converted to units of liters per second (L/s), the exit velocity is approximately 38.37 L/s.

    As in the previous exercise, the exit velocity of the oil can be calculated with Torricelli’s equation. In this case, h = 4 meters, so the exit velocity is v = √(2 * 9.81 m/s² * 4 m) ≈ 8.85 m/s. Converted to units of liters per second (L/s), the exit velocity is approximately 185.37 L/s.

    The speed at which water exits can be calculated using the Bernoulli equation, which relates the water’s velocity at the hole with the liquid pressure on the surface of the tank: v = √(2gh), where v is the exit velocity, g is the acceleration due to gravity (9.81 m/s²), h is the height of the liquid above the hole, and the pressure on the surface of the water is P = ρgh, where ρ is the density of water (1000 kg/m³). In this case, h = 0.8 meters (the height difference between the water and the hole), so the pressure on the surface of the water is P = 1000 kg/m³ * 9.81 m/s² * 0.8 m ≈ 7848 Pa. The exit velocity is then v = √(2 * 7848 Pa / 1000 kg/m³) ≈ 3.53 m/s. Converted to liters per second (L/s), the exit velocity is approximately 17.67 L/s.

 
 

Teaching resources

Here are five teaching resources on Torricelli’s theorem along with their links:

  1. Torricelli’s Law – Conceptual Physics by Paul G. Hewitt: This video explains the principles behind Torricelli’s theorem in a simple and intuitive way, making it suitable for high school students. https://www.youtube.com/watch?v=sJ7VjHsGVgU

  2. The Torricelli Theorem: This website provides an interactive demonstration of Torricelli’s theorem, which allows students to visualize the theorem in action. http://www.mathematik.com/Torricelli.html

  3. Torricelli’s Law Experiment: This website provides a detailed experiment that teachers can use to demonstrate Torricelli’s theorem in a laboratory setting. https://www.teachengineering.org/activities/view/cub_fluid_lesson02_activity1

  4. Torricelli’s Theorem – A Simple Proof: This PDF provides a clear and concise explanation of Torricelli’s theorem, including a proof of the theorem that can be used in a classroom setting. https://mathshelper.co.uk/uploads/1/5/5/0/15505214/torricelli.pdf

  5. Torricelli’s Theorem: This website provides a brief overview of Torricelli’s theorem along with examples of its applications, making it useful for students who are interested in learning more about the practical applications of this theorem. https://www.pleacher.com/mp/mlessons/calculus/torricel.html

 

 

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