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Conservation of Energy in Fluid Flow

Elijah Ramirez

Elijah Ramirez

8 min read

Next Topic - Conservation of Mass Flow Rate in Fluids

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Study Guide Overview

This study guide covers fluid dynamics with a focus on Bernoulli's equation and its applications. It explains the equation's components (pressure, velocity, height), assumptions (incompressible fluid, streamline flow, negligible viscosity), and provides problem-solving strategies. The guide also connects Bernoulli's equation to the continuity equation and the concept of energy conservation, and offers practice questions and exam tips.

#Fluid Dynamics: Energy Conservation and Bernoulli's Principle 🌊

Hey there, future AP Physics 2 master! Let's dive into the world of fluid dynamics, focusing on how energy is conserved when fluids are in motion. This is a crucial topic, so let's make sure you've got it down pat!

#Bernoulli's Equation: The Heart of Fluid Flow

#What is it?

Bernoulli's equation is essentially a statement of energy conservation for fluids. It relates pressure, velocity, and height at different points in a flowing fluid. Think of it as the fluid version of KE + PE = constant.

Key Concept

It's a powerful tool, but it comes with some assumptions:

  • Incompressible Fluid: The fluid's density remains constant.
  • Streamline Flow: The flow is smooth, without turbulence.
  • Negligible Viscosity: Internal friction within the fluid is minimal.

Don't sweat these assumptions too much; the AP exam usually provides scenarios where these conditions are met.

#The Equation

Here's the star of the show:

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2P1​+21​ρv12​+ρgh1​=P2​+21​ρv22​+ρgh2​

Where:

  • PPP is pressure (in Pascals, Pa)
  • ρ\rhoρ is fluid density (in kg/m³)
  • vvv is fluid velocity (in m/s)
  • ggg is the acceleration due to gravity (9.8 m/s²)
  • hhh is the height above a reference point (in meters)
Memory Aid

Remember this as "Pressure + Kinetic Energy + Potential Energy = Constant"

Bernoulli's Equation

#Key Concepts

  • Pressure: Force per unit area. Higher pressure can push the fluid faster.
  • Kinetic Energy: Energy of motion, related to fluid velocity. Faster fluid = more KE.
  • Potential Energy: Energy of position, related to height. Higher fluid = more PE.
Quick Fact

Bernoulli's principle is a direct consequence of the conservation of energy. It states that an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy.

#Applying Bernoulli's Equation: Problem-Solving Strategies

#Step-by-Step

  1. Identify the points: Choose two points in the fluid flow you want to compare.
  2. List knowns and unknowns: Note the pressure, velocity, and height at each point.
  3. Simplify: Are there any terms that can be ignored (e.g., if the height doesn't change, ρgh\rho g hρgh terms cancel out)?
  4. Solve: Plug in the values and solve for the unknown.

#Example: Pipe with Changing Diameter

Let's revisit the example from the notes. Water flows through a pipe, and the diameter changes. Here's how we apply Bernoulli's equation:

Problem: Water flows at 2 m³/s through a 1m diameter pipe with a pressure of 80 kPa. What's the pressure after the pipe narrows to 0.5m?

Solution:

  1. Bernoulli's Equation: P1+12ρv12=P2+12ρv22P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2P1​+21​ρv12​=P2​+21​ρv22​ (since height is constant, we can ignore the potential energy terms)
  2. Continuity Equation: We need to find the velocities first. The continuity equation is V=vAV = vAV=vA, where VVV is the volume flow rate and AAA is the cross-sectional area. Rearranging, we get v=V/Av = V/Av=V/A.
  3. Area Calculations: Calculate the areas: A1=π4m2A_1 = \frac{\pi}{4} m^2A1​=4π​m2 and A2=π16m2A_2 = \frac{\pi}{16} m^2A2​=16π​m2.
  4. Velocity Calculations: v1=2π4=8πv_1 = \frac{2}{\frac{\pi}{4}} = \frac{8}{\pi}v1​=4π​2​=π8​ m/s and v2=2π16=32πv_2 = \frac{2}{\frac{\pi}{16}} = \frac{32}{\pi}v2​=16π​2​=π32​ m/s.
  5. Solve for P2P_2P2​: P2=P1+12ρ(v12−v22)=80000+12∗1000∗(8π2−32π2)≈76.2P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) = 80000 + \frac{1}{2} * 1000 * (\frac{8}{\pi}^2 - \frac{32}{\pi}^2) \approx 76.2P2​=P1​+21​ρ(v12​−v22​)=80000+21​∗1000∗(π8​2−π32​2)≈76.2 kPa
Exam Tip

Always double-check your units! Make sure everything is in SI units (Pascals, m/s, kg/m³, etc.) before plugging into the equation.

#Special Case: Leaking Tank

Another classic application is a leaking tank. Here, we can make some simplifications:

  • The velocity at the top of the tank is negligible (v1≈0v_1 \approx 0v1​≈0).
  • Both the top and the leak are at atmospheric pressure (P1=P2P_1 = P_2P1​=P2​).

Leaking Tank

This simplifies Bernoulli's equation to:

ρgh=12ρv2\rho g h = \frac{1}{2} \rho v^2ρgh=21​ρv2

Solving for the velocity of the leak, vvv, we get:

v=2ghv = \sqrt{2gh}v=2gh​

Memory Aid

Notice the similarity to the velocity of an object in free fall! This is a great example of how physics concepts are interconnected. 💡

#Connecting to Other Concepts

#Continuity Equation

Remember that V=vAV=vAV=vA where VVV is the volume flow rate, vvv is the velocity and AAA is the cross-sectional area. This is crucial for understanding how fluid velocity changes when the pipe diameter changes.

#Energy Conservation

Bernoulli's equation is a direct application of energy conservation. It's all about how kinetic and potential energies transform within a fluid system.

#Example Problem: Pumping Water

Let's break down the example problem step-by-step:

Problem: A pump moves water from a 2m depth to a 10m height, with a power output of 5 kW. How much work is done?

Solution:

  1. Potential Energy Change: ΔPE=mgh=VρgΔh=AhρgΔh\Delta PE = mgh = V\rho g \Delta h = Ah\rho g \Delta hΔPE=mgh=VρgΔh=AhρgΔh
  2. Work-Energy Theorem: Work done by the pump equals the change in potential energy of the water.
  3. Power and Time: W=P∗tW = P*tW=P∗t, so t=W/P=ΔPE/Pt = W/P = \Delta PE/Pt=W/P=ΔPE/P
  4. Calculations: t=A∗2∗ρ∗9.81∗(10−2)5000≈7.84t = \frac{A * 2 * \rho * 9.81 * (10 - 2)}{5000} \approx 7.84t=5000A∗2∗ρ∗9.81∗(10−2)​≈7.84 seconds
  5. Work Done: W=5000∗7.84=39200W = 5000 * 7.84 = 39200W=5000∗7.84=39200 J or 39.2 kJ
Common Mistake

Don't confuse power and work! Work is energy transferred, while power is the rate of energy transfer.

#Final Exam Focus

#High-Priority Topics

  • Bernoulli's Equation: Understand its assumptions and be able to apply it to various scenarios.
  • Continuity Equation: Know how it relates to fluid velocity and pipe diameter.
  • Energy Conservation: Recognize how energy transforms in fluid systems.

#Common Question Types

  • Conceptual questions about the relationship between pressure, velocity, and height.
  • Quantitative problems involving calculations using Bernoulli's equation and the continuity equation.
  • Applications to real-world scenarios like pipes, tanks, and wings.

#Last-Minute Tips

  • Review your notes and practice problems.
  • Focus on understanding the concepts rather than memorizing formulas.
  • Manage your time wisely during the exam; don't spend too much time on a single question.
  • Stay calm and confident; you've got this!
Practice Question

#Practice Questions

#Multiple Choice Questions

  1. Water flows through a horizontal pipe of varying cross-sectional area. At one point, the pressure is P1P_1P1​ and the velocity is v1v_1v1​. At a second point where the pipe is narrower, the pressure is P2P_2P2​ and the velocity is v2v_2v2​. Which of the following is true? (A) P1<P2P_1 < P_2P1​<P2​ and v1<v2v_1 < v_2v1​<v2​ (B) P1>P2P_1 > P_2P1​>P2​ and v1<v2v_1 < v_2v1​<v2​ (C) P1<P2P_1 < P_2P1​<P2​ and v1>v2v_1 > v_2v1​>v2​ (D) P1>P2P_1 > P_2P1​>P2​ and v1>v2v_1 > v_2v1​>v2​

  2. A fluid flows through a pipe that is elevated from the ground. If the fluid's velocity increases as it moves to a higher elevation, what must be true about the pressure of the fluid? (A) The pressure increases. (B) The pressure decreases. (C) The pressure remains the same. (D) The pressure cannot be determined without more information.

  3. A large water tank has a small hole near the bottom. If the height of the water in the tank is doubled, the speed at which the water exits the hole will: (A) Double. (B) Increase by a factor of 2\sqrt{2}2​. (C) Remain the same. (D) Decrease by a factor of 2\sqrt{2}2​.

#Free Response Question

A large cylindrical tank of height HHH is filled with water. A small hole of area AAA is opened at the bottom of the tank. The area of the tank is ATA_TAT​, and AT>>AA_T >> AAT​>>A. The density of water is ρ\rhoρ.

(a) Derive an expression for the speed vvv at which the water exits the hole, in terms of ggg and HHH.

(b) Derive an expression for the volume flow rate QQQ at which water exits the hole, in terms of ggg, HHH, and AAA.

(c) If the tank is initially filled to a height of H0H_0H0​, derive an expression for the time ttt it takes for the tank to empty completely, in terms of H0H_0H0​, ATA_TAT​, AAA, and ggg.

#Scoring Rubric:

(a) (3 points)

  • 1 point: Correctly applying Bernoulli's equation to the top of the tank and the hole.
  • 1 point: Recognizing that the velocity at the top of the tank is negligible (v1≈0v_1 \approx 0v1​≈0).
  • 1 point: Correctly deriving v=2gHv = \sqrt{2gH}v=2gH​.

(b) (2 points)

  • 1 point: Using the volume flow rate equation Q=vAQ = vAQ=vA.
  • 1 point: Correctly expressing Q=A2gHQ = A\sqrt{2gH}Q=A2gH​.

(c) (4 points)

  • 1 point: Recognizing that the volume of water leaving the tank is equal to the change in volume of the tank.
  • 1 point: Setting up the differential equation ATdHdt=−A2gHA_T \frac{dH}{dt} = -A\sqrt{2gH}AT​dtdH​=−A2gH​.
  • 1 point: Integrating the differential equation.
  • 1 point: Correctly deriving t=ATA2H0gt = \frac{A_T}{A} \sqrt{\frac{2H_0}{g}}t=AAT​​g2H0​​​.

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Question 1 of 11

🚀 When a fluid flows through a pipe and its velocity increases, what happens to the pressure, assuming height remains constant?

The pressure increases

The pressure decreases

The pressure remains the same

The pressure fluctuates randomly