6.2 Equation of Continuity (Copy)

1. Derivation of the Equation (SLO 6.2.1)

The Equation of Continuity is essentially a statement of the Law of Conservation of Mass applied to fluid dynamics. It assumes that the fluid is ideal: incompressible (constant density, $rho$), non-viscous, and undergoing steady flow.

The Step-by-Step Logic

Consider a fluid flowing through a pipe of varying cross-section. Let $A_1$ and $v_1$ be the cross-sectional area and fluid velocity at the lower end, and $A_2$ and $v_2$ be the area and velocity at the upper end.

Distance Covered: In a time interval $Delta t$, the fluid at the lower end moves a distance $Delta x_1 = v_1 Delta t$.

Volume Flowing In: The volume of fluid entering the pipe in time $Delta t$ is $V_1 = A_1 Delta x_1 = A_1 v_1 Delta t$.

Mass Flowing In: The mass of this fluid is $m_1 = rho_1 V_1 = rho_1 A_1 v_1 Delta t$.

Mass Flowing Out: Similarly, the mass of fluid leaving the upper end in time $Delta t$ is $m_2 = rho_2 A_2 v_2 Delta t$.

Since no fluid can be created or destroyed within the pipe, the mass entering must equal the mass leaving:

$$m_1 = m_2$$

$$rho_1 A_1 v_1 Delta t = rho_2 A_2 v_2 Delta t$$

Because the fluid is incompressible, its density is constant ($rho_1 = rho_2 = rho$). Canceling the density ($rho$) and time ($Delta t$) from both sides yields the Equation of Continuity:

This shows that the product $Av$ is a constant. This constant represents the Volume Flow Rate (Volume/Time), measured in $m^3/s$.

2. Application: Rocket Motion (SLO 6.2.2)

The equation of continuity is beautifully demonstrated in the design of rocket engines. In a rocket, solid or liquid fuel burns in a large combustion chamber, creating a massive volume of high-pressure gas.

The combustion chamber has a relatively large cross-sectional area ($A_1$). The gas is then forced to exit through a specially designed, narrow exhaust nozzle ($A_2$). Because $A_2$ is significantly smaller than $A_1$, the continuity equation ($A_1 v_1 = A_2 v_2$) mandates that the exit velocity ($v_2$) of the gas must be incredibly high. This high-velocity exhaust generates the massive momentum required to produce the upward thrust for liftoff.

3. Solving Word Problems (SLO 6.2.3)

Example Problem: Water is flowing through a garden hose with an internal cross-sectional area of $3.0 times 10^{-4} text{ m}^2$ at a velocity of $1.5 text{ m/s}$. A nozzle is attached to the hose, which reduces the exit area to $0.5 times 10^{-4} text{ m}^2$. Calculate the velocity at which the water exits the nozzle.

Solution Strategy:

Identify Variables:

$A_1 = 3.0 times 10^{-4} text{ m}^2$

$v_1 = 1.5 text{ m/s}$

$A_2 = 0.5 times 10^{-4} text{ m}^2$

$v_2 = ?$

Apply the Equation:

$$A_1 v_1 = A_2 v_2$$

$$(3.0 times 10^{-4})(1.5) = (0.5 times 10^{-4}) v_2$$

Solve for $v_2$:

$$v_2 = frac{4.5 times 10^{-4}}{0.5 times 10^{-4}}$$

$$v_2 = 9.0 text{ m/s}$$

The water exits the nozzle at $9.0 text{ m/s}$, explaining why placing your thumb over a hose makes the water spray much further!

Interactive Fluid Continuity Simulator

To truly understand how area affects velocity, test the principle yourself. Use the slider in the widget below to change the diameter of the nozzle at the right end of the pipe and observe the effect on particle speed.

📝 Concept Check