# Continuity Equation (Fluids)

**Continuity** uses the conservation of matter to describe the relationship between the velocities of a fluid in different sections of a system. The simple observation that the volume flow rate, $Av$, must be the same throughout a system provides a relationship between the velocity of the fluid through a pipe and the cross-sectional area.

$A_1 v_1 = A_2 v_2$

Continuity works in tandem with Bernoulli's principle in the design and construction of systems of irrigation, plumbing, etc.

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## Flow Rate

The first rate of change of interest for fluid in motion is the **mass flow rate**: the amount of mass that passes through a checkpoint in one unit time. Or, mathematically,

$\text{mass flow rate} = \frac{\Delta m}{\Delta t}.$

However, the mass of a fluid is strange to calculate, since there is not necessarily a feesibly measurable amount of, say, water flowing through a pipe. As such, relations to density, $\rho = \frac{m}{V}$ are preferred. Combining this equation with the mass flow rate equation above gives

$\frac{\Delta m}{\Delta t} = \frac{\rho \Delta V}{\Delta t}.$

Provided the pipe is prismatic, its volume can be expressed $V = AL$ where $A$ is the cross-sectional area and $L$ is a length perpendicular to that area.

$\frac{\rho \Delta V}{\Delta t} = \frac{\rho A \Delta L}{\Delta t} = \rho A \frac{\Delta L}{\Delta t} = \rho A v.$

Therefore,

$\text{mass flow rate} =\dot{m} = \frac{\Delta m}{\Delta t} = \rho A v.$

Also of interest is **volume flow rate**.

$\text{volume flow rate} = \frac{\Delta V}{\Delta t} = \frac{A \Delta L}{\Delta t} = A \frac{\Delta L}{\Delta t} = Av.$

$\text{volume flow rate} =\dot{V} = \frac{\Delta V}{\Delta t} = Av$

The average volume flow of blood through an aortic valve is $9.0 \times 10^{-5} \text{m}^3/\text{s}$. If the cross-sectional area of a healthy aortic valve is $3.0 \times 10^{-4} \text{m}^2$, what is the average speed of the blood?

The volume flow rate equation is $\text{volume flow rate} = \frac{\Delta V}{\Delta t} = Av$. Since the problem provides the rate and $A$ and asks for $v$, use the second definition.

$\text{volume flow rate} = Av$

$v = \frac{\text{volume flow rate} }{A} = \frac{9.0 \times 10^{-5} \text{m}^3/\text{s}}{3.0 \times 10^{-4} \text{m}^2} = 3.0 \times 10^{-1} \text{m/s}$.

An engine's piston moves at an average speed of 10 $\text{m/s}$ while pulling the air-fuel mixture through a 3 $\text{cm}$ by 2 $\text{cm}$ rectangular intake valve. Find the average volume flow rate for the air-fuel mixture entering the piston in $\frac{\text{m}^3}{s}$. (Assume the piston has the same cross-sectional dimensions as the intake valve.)

## Continuity Equation

Imagine two pipes of different diameters connected so that all the matter that passes through the first section must pass through the second. This means the mass flow rate of each section must be *equal*, otherwise some mass would be disappearing between the two sections. Mathematically, this can be expressed as

$\frac{\Delta m_1}{\Delta t}_{out} = \frac{\Delta m_2}{\Delta t}_{in}.$

According to the definition of density, $m=\rho V$, so

$\frac{\rho \Delta V_1}{\Delta t}_{out} = \frac{\rho \Delta V_2}{\Delta t}_{in}.$

As long as the fluid is incompressible, the density will not change from one section to the next, so $\rho$ cancels out of both sides. Additionally, if the pipes are prismatic (as most conventional pipes are), then the volume can be expressed in terms of the cross-sectional area and length: $V=AL.$ The equation is now

$\frac{A\Delta L_1}{\Delta t}_{out} = \frac{A\Delta L_2}{\Delta t}_{in}$

$A_1\frac{\Delta L_1}{\Delta t}_{out} = A_2\frac{\Delta L_2}{\Delta t}_{in}.$

But since $\frac{\Delta L}{\Delta t} = v$,

$A_1 v_1 = A_2 v_2.$

Water flows from a cylindrical pipe of radius 4 $\text{cm}$ into another cylindrical pipe of radius 2 $\text{cm}$. If the water exhibits a velocity of 20 $\text{m/s}$ in the second pipe, what was the velocity in the first?

Solution: Fluid velocity in joined pipes requires the continuity equation:$A_1 v_1=A_2 v_2.$

Since the pipes are cylindrical, each cross-section is a circle with area $A=\pi r^2$. Additionally, $v_2 = 20 \text{ m/s}$ is given, so

$\pi r_1^2 v_1 = \pi r_2^2 v_2$

$(4 \text{cm})^2 v_1 = (2 \text{cm})^2 (20 \text{m/s})$

$v_1 = 5 \text{m/s}.$

**Cite as:**Continuity Equation (Fluids).

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/continuity-equation-fluids/