AP Physics 1- 8.4 Fluids and Conservation Laws- Study Notes- New Syllabus
AP Physics 1-8.4 Fluids and Conservation Laws – Study Notes
AP Physics 1-8.4 Fluids and Conservation Laws – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Flow of an Incompressible Fluid through a Cross-Sectional Area — Mass Conservation
- Flow of a Fluid as a Result of Energy Difference
- Bernoulli’s Equation
- Torricelli’s Theorem
Flow of an Incompressible Fluid through a Cross-Sectional Area — Mass Conservation
Mass conservation (continuity) states that mass cannot be created or destroyed. For a fluid flowing through a pipe or channel, the mass of fluid entering per unit time must equal the mass leaving per unit time.
Important Concepts:
- A difference in pressure between two locations causes a fluid to flow.
- The rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which matter exits the tube (steady flow condition).
- The flow rate (mass or volume per unit time) is proportional to the cross-sectional area of the tube and the fluid speed through it.
Mass Flow Rate (General):
\( \dot{m} = \rho A v \)
- \( \dot{m} \) = mass flow rate (kg/s)
- \( \rho \) = fluid density (kg/m\(^3\))
- \( A \) = cross-sectional area (m\(^2\))
- \( v \) = average flow speed (m/s)
Continuity Equation for Incompressible Fluids:
\( A_1 v_1 = A_2 v_2 \)
This shows that if the area decreases, the speed must increase to maintain mass conservation (and vice versa).
Alternate Form — Volume Flow Rate:
\( Q = A v \), where \( Q \) is the volumetric flow rate (m\(^3\)/s).
For steady, incompressible flow: \( Q_1 = Q_2 \).
Example :
Water (incompressible, \( \rho = 1000 \, \text{kg/m}^3 \)) flows in a pipe that narrows from diameter 0.10 m to 0.05 m. If the speed in the wider section is \( v_1 = 2.0 \, \text{m/s} \), find the speed in the narrow section \( v_2 \).
▶️Answer/Explanation
Area: \( A = \dfrac{\pi D^2}{4} \).
\( A_1 = \dfrac{\pi (0.10)^2}{4} = 7.854 \times 10^{-3} \, \text{m}^2 \)
\( A_2 = \dfrac{\pi (0.05)^2}{4} = 1.963 \times 10^{-3} \, \text{m}^2 \)
Continuity: \( A_1 v_1 = A_2 v_2 \Rightarrow v_2 = \dfrac{A_1}{A_2} v_1 \)
\( v_2 = \dfrac{7.854 \times 10^{-3}}{1.963 \times 10^{-3}} \times 2.0 \approx 8.0 \, \text{m/s} \)
Answer: \( v_2 \approx 8.0 \, \text{m/s} \).
Example :
A pump delivers water at a volumetric flow rate \( Q = 0.01 \, \text{m}^3/\text{s} \) through a hose of radius 0.04 m. Find (i) the average speed in the hose and (ii) the mass flow rate.
▶️Answer/Explanation
(i) Speed: \( v = \dfrac{Q}{A} \) with \( A = \pi r^2 = \pi (0.04)^2 = 5.027 \times 10^{-3} \, \text{m}^2 \).
\( v = \dfrac{0.01}{5.027 \times 10^{-3}} \approx 1.99 \, \text{m/s} \)
(ii) Mass flow rate: \( \dot{m} = \rho Q \). For water \( \rho = 1000 \, \text{kg/m}^3 \):
\( \dot{m} = 1000 \times 0.01 = 10 \, \text{kg/s} \)
Answer: \( v \approx 1.99 \, \text{m/s} \), \( \dot{m} = 10 \, \text{kg/s} \).
Flow of a Fluid as a Result of Energy Difference
Fluid flows between two locations due to differences in mechanical energy (pressure energy, kinetic energy, or gravitational potential energy) within the fluid–Earth system. Fluid naturally moves from regions of higher energy to lower energy.
Forms of Energy per Unit Volume in a Fluid:
- Pressure Energy: \( P \) (Pa)
- Kinetic Energy: \( \dfrac{1}{2} \rho v^2 \)
- Gravitational Potential Energy: \( \rho g h \)
Bernoulli’s Principle (Definition):
Bernoulli’s principle states that for an ideal, incompressible, non-viscous fluid flowing along a streamline, the sum of pressure energy, kinetic energy, and gravitational potential energy per unit volume is constant:
\( P + \dfrac{1}{2} \rho v^2 + \rho g h = \text{constant} \)
- If pressure decreases, velocity or height increases, and vice versa.
- Explains phenomena such as lift on airplane wings, faster flow through narrow pipes, and siphoning of fluids.
Torricelli’s Theorem (Definition):
Torricelli’s theorem is a special case of Bernoulli’s principle that describes the speed of fluid flowing out of an opening under the influence of gravity:
\( v = \sqrt{2 g h} \)
- \( h \) is the vertical height of the fluid column above the outlet.
- Predicts that the exit velocity of a fluid equals the speed of a freely falling object from the same height.
Example :
Water flows through a horizontal pipe that narrows from 0.05 m² to 0.01 m². If the speed in the wide section is 2 m/s and pressure is 1.5 × 10⁵ Pa, find the pressure and velocity in the narrow section.
▶️Answer/Explanation
Step 1: Continuity equation: \( A_1 v_1 = A_2 v_2 \)
\( 0.05 \times 2 = 0.01 \times v_2 \implies v_2 = 10 \, \text{m/s} \)
Step 2: Bernoulli’s equation:
\( P_1 + \dfrac{1}{2} \rho v_1^2 = P_2 + \dfrac{1}{2} \rho v_2^2 \)
\( 1.5 \times 10^5 + 0.5 \times 1000 \times 2^2 = P_2 + 0.5 \times 1000 \times 10^2 \)
\( P_2 = 1.5 \times 10^5 + 2000 – 50,000 = 1.02 \times 10^5 \, \text{Pa} \)
Answer: Velocity = 10 m/s, Pressure = 1.02 × 10⁵ Pa
Example :
Water flows out of a hole at the bottom of a tank 5 m above the outlet. Find the exit speed.
▶️Answer/Explanation
Using Torricelli’s theorem: \( v = \sqrt{2 g h} \)
\( v = \sqrt{2 \times 9.8 \times 5} \approx 9.9 \, \text{m/s} \)
Answer: Exit speed ≈ 9.9 m/s