8.4 Fluids and Conservation Laws

Learning objectives

8.4A—Describe the flow of an incompressible fluid through a cross-sectional area by using mass conservationDescribe the flow of an incompressible fluid through a cross-sectional area by using mass conservation.• A difference in pressure between two locations causes a fluid to flow.- i. The rate at which matter enters a fluid-filled tube open at both ends must equal the rate at which matter exits the tube.- ii. The rate at which matter flows into a location is proportional to the crosssectional area of the flow and the speed at which the fluid flows. Derived equation: V =Avt• The continuity equation for fluid flow describes conservation of mass flow rate in incompressible fluids. Relevant equation: Av11 =Av22 TOPIC 8.4 Fluids and Conservation Laws 8.4B—Describe the flow of a fluid as a result of a difference in energy between two locations within the fluid– Earth…Describe the flow of a fluid as a result of a difference in energy between two locations within the fluid– Earth system.• A difference in gravitational potential energies between two locations in a fluid will result in a difference in kinetic energy and pressure between those two locations that is described by conservation laws.• Bernoulli’s equation describes the conservation of mechanical energy in fluid flow. Relevant equation:• Torricelli’s theorem relates the speed of a fluid exiting an opening to the difference in height between the opening and the top surface of the fluid and can be derived from conservation of energy principles. Derived equation: BOUNDARY STATEMENT All fluids will be assumed to be ideal, and all pipes are assumed to be completely filled by the fluid, unless otherwise stated. AP Physics 1: Algebra-Based Course and Exam Description Laboratory Investigations AP PHYSICS 1

Connect pressure difference to flow

A pressure difference between two locations can drive fluid flow. In a completely filled tube carrying steady incompressible flow, matter cannot accumulate, so the rate entering equals the rate leaving.

Calculate volume flow rate

$\frac{V}{t}=Av$

Quantity	Meaning	SI unit
$V/t$	Volume flow rate through a cross section	$\text{m}^3\,\text{s}^{-1}$
$A$	Cross-sectional area perpendicular to the flow	$\text{m}^2$
$v$	Average fluid speed through that cross section	$\text{m s}^{-1}$

Apply mass conservation

$A_1v_1=A_2v_2$

For an incompressible fluid, density is constant. Equal mass flow rates therefore mean equal volume flow rates: $A_1v_1=A_2v_2$ . A smaller cross-sectional area requires a greater speed to carry the same volume each second.

Relate area and speed

Example: water moves at $2.0\,\text{m s}^{-1}$ through area $A_1=6.0\,\text{cm}^2$ and enters a section with $A_2=3.0\,\text{cm}^2$ .

$v_2=\dfrac{A_1v_1}{A_2}=\dfrac{(6.0)(2.0)}{3.0}=4.0\,\text{m s}^{-1}$ .

The area halves, so the speed doubles; the volume flow rate is unchanged.

Use the incompressible-flow condition

Continuity does not say pressure stays constant. It conserves mass flow. The simplified form $A_1v_1=A_2v_2$ requires incompressible flow; if density changes, conserve $\rho Av$ instead.

Conserve mechanical energy

In ideal fluid flow, pressure, kinetic, and gravitational energy can change between two locations while their total mechanical energy is conserved.

$P_1+\rho gy_1+\frac12\rho v_1^2=P_2+\rho gy_2+\frac12\rho v_2^2$

Track three energy stores

Bernoulli term	Energy represented per unit volume
$P$	Pressure energy
$\rho gy$	Gravitational potential energy
$\tfrac12\rho v^2$	Kinetic energy

Compare the same three terms at locations 1 and 2. At equal height, a larger kinetic term must be balanced by a smaller pressure term. When height changes, keep all three terms.

Derive the exit-speed relationship

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

Torricelli example: a small opening is $1.25\,\text{m}$ below the liquid surface. The surface and opening are both exposed to the same pressure, and the wide surface moves negligibly. Bernoulli’s equation reduces to

$\rho g\Delta y=\tfrac12\rho v^2$ , so $v=\sqrt{2g\Delta y}$ .

$v=\sqrt{2(9.8)(1.25)}=4.95\,\text{m s}^{-1}\approx5.0\,\text{m s}^{-1}$ .

Check the model conditions

Do not use ‘faster means lower pressure’ without checking height and Bernoulli’s assumptions. Unless stated otherwise here, treat the fluid as ideal and the pipe as completely filled. In Torricelli’s theorem, $\Delta y$ is the vertical height difference, not the path length.

Back to course directory

—AP Physics 1Read