#AP Physics 1: Fluids & Conservation Laws 🌊

Hey! Let's make sure you're totally prepped for the exam. We're diving into fluids and how conservation laws govern their behavior. Think of this as your final pit stop before the big race. Let's get started!

#Flow of Incompressible Fluids

#Pressure Differences in Fluids

• Fluid flow happens when there's a pressure difference between two spots. It's like water flowing downhill—it goes from high to low pressure.

- The **cross-sectional area** and **fluid speed** control how much fluid flows. Bigger areas and faster speeds mean more flow. - The **continuity equation** describes how mass flow rate is conserved in incompressible fluids. It's all about keeping the flow steady.

$$A_{1} v_{1}=A_{2} v_{2}$$

• $A_1$ and $A_2$ are the cross-sectional areas at two different points.

• $v_1$ and $v_2$ are the fluid velocities at those points.

• Incompressible fluids keep their density constant, so the volume flow rate stays the same throughout the system.

#Continuity Equation for Fluids

• The continuity equation is all about mass conservation in fluid dynamics. It's a fundamental principle.
• Mass flow rate ($dot{m}$) is density ($ho$) times area ($A$) times velocity ($v$): $$dot{m} = \rho A v$$
• For incompressible fluids, density is constant, so mass flow rate depends only on area and velocity.
• This applies to steady-state flow, where fluid properties don't change over time.

- If the pipe narrows, the fluid speeds up. If it widens, the fluid slows down. It's all about keeping the flow rate constant. - This helps analyze systems like water pipes and blood vessels by predicting how velocity changes with area.

Practice Question

Multiple Choice Questions

1. A fluid flows through a pipe of varying cross-sectional area. At point 1, the area is $A_1$ and the velocity is $v_1$. At point 2, the area is $A_2 = 2A_1$. What is the velocity $v_2$ at point 2?

   (A) $v_1/4$ (B) $v_1/2$ (C) $v_1$ (D) $2v_1$

2. An incompressible fluid flows through a pipe. If the radius of the pipe is halved, what happens to the fluid velocity to maintain the same flow rate?

   (A) It is reduced to one-fourth. (B) It is halved. (C) It doubles. (D) It quadruples.

Free Response Question

A horizontal pipe narrows from a radius of 10 cm to a radius of 5 cm. Water flows through the wider section at a speed of 2 m/s. Assume the water is an ideal fluid.

(a) Calculate the speed of the water in the narrower section. (3 points) (b) If the pressure in the wider section is 150 kPa, what is the pressure in the narrower section? (4 points) (c) What is the volume flow rate in the pipe? (2 points)

Scoring Rubric

(a) (3 points)

• 1 point: Using the continuity equation $A_1v_1=A_2v_2$
• 1 point: Correctly calculating the areas $A_1=\pi(0.1)^2$ and $A_2=\pi(0.05)^2$
• 1 point: Correctly solving for $v_2 = 8 m/s$

(b) (4 points)

• 1 point: Using Bernoulli's equation $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$ (since $y_1=y_2$)
• 1 point: Correctly substituting values
• 1 point: Correctly using the density of water $\rho = 1000 kg/m^3$
• 1 point: Correctly solving for $P_2 = 118 kPa$

(c) (2 points)

• 1 point: Using the volume flow rate equation $Q = A_1v_1$
• 1 point: Correctly calculating $Q = 0.0628 m^3/s$

#Energy Differences in Fluid Flow

#Gravitational Potential Energy in Fluids

• Differences in gravitational potential energy lead to changes in kinetic energy and pressure. It's like a rollercoaster—potential energy turns into motion.
• Fluid at a higher elevation has more gravitational potential energy than fluid at a lower elevation. 🎢
• As fluid flows down, potential energy becomes kinetic energy, increasing the fluid's velocity.
• Pressure also changes with height due to the weight of the fluid above (hydrostatic pressure).

#Bernoulli's Equation

• Bernoulli's equation expresses conservation of mechanical energy in fluid flow. It's a cornerstone concept.

$$P_{1}+\rho g y_{1}+\frac{1}{2} \rho v_{1}^{2}=P_{2}+\rho g y_{2}+\frac{1}{2} \rho v_{2}^{2}$$

• $P$ is pressure.

• $\rho$ is fluid density.

• $g$ is gravitational acceleration.

• $y$ is height.

• $v$ is velocity.

• Each term represents energy per unit volume: pressure energy, gravitational potential energy, and kinetic energy.

• This equation assumes steady, incompressible, and inviscid flow along a streamline. It's an ideal scenario.

• It applies to various situations like aircraft wings, venturi meters, and pitot tubes.

- Pressure differences can create lift (airplanes) or measure fluid velocity (flow meters).

#Torricelli's Theorem

• Torricelli's theorem links fluid exit velocity to the height difference between the exit and the fluid's surface. It's a practical application of energy conservation.
• It's derived from energy conservation, assuming no viscous losses and negligible surface velocity.

$$v=\sqrt{2 g \Delta y}$$

• $v$ is the exit velocity.

• $g$ is gravitational acceleration.

• $\Delta y$ is the height difference.

• It applies to fluid discharge from tanks, reservoirs, and other containers. 🚰

• It predicts how fast fluid escapes from a hole based on its depth below the surface.

• Deeper holes mean higher exit velocities because of more gravitational potential energy conversion.

• This helps design systems like water towers, dams, and fountains by estimating discharge rates and velocities.

Practice Question

Multiple Choice Questions

1. Water flows from a large tank through a small hole at the bottom. If the height of the water is doubled, what happens to the exit velocity?

   (A) It remains the same. (B) It is halved. (C) It doubles. (D) It increases by a factor of $\sqrt{2}$.

2. According to Bernoulli's principle, what happens to the pressure of a fluid as its velocity increases?

   (A) It increases. (B) It decreases. (C) It remains constant. (D) It fluctuates randomly.

Free Response Question

A large water tank is open to the atmosphere and has a small hole near the bottom. The water level is 5 meters above the hole. Assume the water is an ideal fluid.

(a) Calculate the speed at which the water exits the hole. (3 points) (b) If the hole has a radius of 1 cm, what is the volume flow rate of the water exiting the hole? (3 points) (c) If a horizontal tube is attached to the hole and the radius of the tube is doubled, what is the new exit velocity? (2 points)

Scoring Rubric

(a) (3 points)

• 1 point: Using Torricelli's theorem $v = \sqrt{2g\Delta y}$
• 1 point: Correctly substituting values $v = \sqrt{2 * 9.8 * 5}$
• 1 point: Correctly solving for $v = 9.9 m/s$

(b) (3 points)

• 1 point: Calculating the area of the hole $A = \pi r^2 = \pi (0.01)^2$
• 1 point: Using the volume flow rate equation $Q = Av$
• 1 point: Correctly calculating $Q = 0.0031 m^3/s$

(c) (2 points)

• 1 point: Understanding that the exit velocity is independent of the tube's radius
• 1 point: Stating that the new exit velocity is the same as in part (a), $9.9 m/s$

#

• High-Priority Topics: Focus on the continuity equation, Bernoulli's equation, and Torricelli's theorem. These are often tested in both MCQs and FRQs.
• Common Question Types: Expect questions that involve calculating fluid velocities, pressures, and flow rates. Be ready to apply these equations in different scenarios.
• Time Management: Quickly identify the relevant equations and apply them. Practice solving problems under timed conditions.