Physics · Fluid dynamics

Continuity equation for steady flow

Q: How can I check a calculation involving the continuity equation?

For steady water flow at equal density, $A_1=4.0\,\mathrm{cm^2}$ and $\bar v_1=2.0\,\mathrm{m\,s^{-1}}$ narrowing to $A_2=1.0\,\mathrm{cm^2}$ gives $\bar v_2=\frac{A_1\bar v_1}{A_2}=8.0\,\mathrm{m\,s^{-1}}$. Check by converting both areas to square metres: 4.0 × 10⁻⁴ and 1.0 × 10⁻⁴ m². Both products $A\bar v$ equal 8.0 × 10⁻⁴ m³/s, or 0.80 L/s. At $\rho=1000\,\mathrm{kg\,m^{-3}}$, mass flow is 0.80 kg/s. These inputs are areas rather than diameters. Halving circular pipe diameter quarters area and quadruples speed under the same conditions. This confirms mass balance, but does not calculate required pump pressure.

Q: How do I check units and scale for the continuity equation?

Volume flow $Q=A\bar v$ has units $\mathrm{m^2}\frac{\mathrm m}{\mathrm s}=\frac{\mathrm{m^3}}{\mathrm s}$. Multiplication by density in kg/m³ produces mass flow in kg/s. One square centimetre is 10⁻⁴ m² because the centimetre factor is squared; one litre is 10⁻³ m³. For 18 L/min, divide by 60 to obtain $Q=0.00030\,\mathrm{m^3\,s^{-1}}$. Through 3.0 cm², this gives $\bar v=1.0\,\mathrm{m\,s^{-1}}$. Check the units of $\frac QA$ and $\rho Q$ separately. In a narrowing pipe, speed follows area ratio alone only at equal density. A drawn pipe height does not fully specify area without a cross-sectional shape.

Steady mass balance connects density, cross-sectional area and mean normal speed at one inlet and one outlet.

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Formula

Calculate

ρ_{1} A_{1} v_{1} = ρ_{2} A_{2} v_{2}

$\rho A\bar v_n$ in kg/s; A in m²; v in m/s; ρ in kg/m³.

Variables & units

Variables & units
Symbol	Meaning	Unit
$ρ₁,ρ₂$	local fluid density uniform over each cross-section	kg/m³
$A₁,A₂$	cross-sectional area normal to flow direction	m²
$v₁,v₂$	average normal velocity component at each cross-section	m/s

LaTeX

\rho_1 A_1 v_1 = \rho_2 A_2 v_2

QUANTA STUDYquanta-study.de

Circular pipe with halved diameter: left drawn height80 and right40 units. Mean speed on the right is four times larger, with four-times-longer arrows. Diameters and constant volume flows are labelled. — Equal density, steady flow and circular sections: D₁ = 2D₂ gives A₁ = 4A₂ and v̄₂ = 4v̄₁. Velocity arrows are drawn in ratio 1:4; Q₁ = Q₂.

Continuity equation calculator

Find the volume flow rate, flow speed or cross-section in a pipe for steady flow without branches, for gases also with different densities.

Calculation model

Find

Inlet density ρ₁

Density at the inlet, e.g. 1000 kg/m³ for water; for liquids usually the same as at the outlet.

Inlet area A₁

Cross-sectional area at the inlet, e.g. 4 cm².

Inlet speed v₁

Mean flow speed at the inlet, e.g. 2 m/s.

Outlet density ρ₂

Density at the outlet; 1000 kg/m³ again for water, while for gases it can change.

Outlet area A₂

Cross-sectional area at the narrow point or outlet, e.g. 1 cm².

Filled with example values. Type your own and the result updates straight away.

Result: Outlet speed v₂

v₂ = 8 m/s

With ρ₁ = 1000 kg/m³, A₁ = 4 cm², v₁ = 2 m/s, ρ₂ = 1000 kg/m³ and A₂ = 1 cm², v₂ = 8 m/s.

Working · rearrange for v₂, substitute, evaluate

1. Rearrange for v₂
$v_2=\frac{\rho_1A_1v_1}{\rho_2A_2}$
2. Substitute values (in SI units)
- A₁ = 4 cm² = 0.0004 m²
- A₂ = 1 cm² = 0.0001 m²
$\frac{1000\cdot 0.0004\cdot 2}{1000\cdot 0.0001}=8$
3. Result
v₂ = 8 m/s

Quick tips

With equal density A₁ · v₁ = A₂ · v₂: if the cross-section is four times smaller, the water flows four times faster.
Half the diameter means a quarter of the area, so four times the speed. Always turn a diameter into an area first with A = π · d² / 4.

Assumptions and limits

Steady flow, one inlet and one outlet, no accumulation or side stream; specified positive uniform density at each section and mean normal speed magnitudes. Positive areas; no equation of state or energy balance. Different densities are retained.

Derivation & background

For a fixed pipe section, change in stored mass equals inflow minus outflow. Steady flow without additional connections gives $\rho_1A_1\bar v_1=\rho_2A_2\bar v_2$ . For density uniform over each cross-section, v̄ denotes the average normal velocity component. Cancel density only when $\rho_1=\rho_2$ . This mass balance does not require frictionless motion.

Exam blueprint

Validity range

Steady flow with one inlet and one outlet, no storage change or additional connections. $\rho A\bar v_n$ assumes uniform cross-sectional density; constant $A\bar v_n$ additionally requires equal densities. Use $\int_A\rho\vec v\cdot\hat{\vec n}\,\mathrm dA$ for varying profiles and the full balance for storage changes.

Derivation steps

Mass conservation for a fixed pipe section makes entering and leaving mass-flow magnitudes equal.

1Choose a fixed section with one inlet and one outlet.
2Balance stored mass: $\frac{\mathrm dM}{\mathrm dt}=\dot m_{\mathrm{in}}-\dot m_{\mathrm{out}}$ .
3Steady flow gives $\frac{\mathrm dM}{\mathrm dt}=0$ ; both positive connection flows match.
4Uniform cross-sectional density gives $\dot m=\rho A\bar v_n$ using the mean normal component.
5Substitute: $\rho_1A_1\bar v_1=\rho_2A_2\bar v_2$ ; do not cancel different local gas densities.
6Only $\rho_1=\rho_2>0$ gives $A_1\bar v_1=A_2\bar v_2=Q$ .

Rearrangements

Speed in the narrower pipe

v_2=\frac{\rho_1 A_1 v_1}{\rho_2 A_2}

Only when $\rho_1=\rho_2$ does this reduce to $\bar v_2=\frac{A_1\bar v_1}{A_2}$ . Both areas and densities must be positive.

Task variant

An incompressible fluid flows at 1.5 m/s through 6 cm² and then 2 cm². Find v₂.

$A_1v_1=A_2v_2$ .
Thus $v_2=\frac{A_1}{A_2}v_1=\frac62\cdot1.5\,\mathrm{m\,s^{-1}}=4.5\,\mathrm{m\,s^{-1}}$ .

Common mistakes

Applying the constant-density shortcut to a compressible gas.

Retain both state densities when they differ: $\rho_1A_1\bar v_1=\rho_2A_2\bar v_2$ .

Exam context

Pipe contractions, nozzles and combination with Bernoulli for pressure and speed questions.

Formula cluster

Fluid dynamics

Mass conservation is the basis of steady pipe-flow balance.

p + ½ρv² + ρgh = const.

Bernoulli Equation

ρ = m/V

Density

F_A = ρ·V·g

Buoyant Force (Archimedes)

Worked example

Water flows steadily at $v_1=2.0\,\mathrm{m/s}$ through $A_1=4.0\,\mathrm{cm^2}$ , narrowing to $A_2=1.0\,\mathrm{cm^2}$ . Equal density gives $v_2=\frac{A_1v_1}{A_2}=8.0\,\mathrm{m/s}$ . Check: $A_1v_1=A_2v_2=8.0\cdot10^{-4}\,\mathrm{m^3/s}$ .

Applications

Pipe contractions, nozzles, volume and mass flow, and as a prerequisite for Bernoulli problems.

How do you calculate with Continuity equation for flow?

Here is how to work through a typical Continuity equation for flow (ρ₁A₁v₁ = ρ₂A₂v₂) task step by step:

1
Task
An incompressible fluid flows at 1.5 m/s through 6 cm² and then 2 cm². Find v₂.
Solution path
1. $A_1v_1=A_2v_2$ .
2. Thus $v_2=\frac{A_1}{A_2}v_1=\frac62\cdot1.5\,\mathrm{m\,s^{-1}}=4.5\,\mathrm{m\,s^{-1}}$ .

Understand and apply the equation

If the same mass of water flows through a pipe each second, it must cross a narrower section faster. This idea leads to the continuity equation. First distinguish mass per time from volume per time; then explore pipe diameters, gas with changing density and a filling tank.

Which mass crosses the boundary

Imagine collecting discharged water in a container for ten seconds. Collected volume divided by time gives volume flow rate. Density tells you how much mass this represents. This distinction prevents incorrectly assuming equal volume flows when dealing with gases.[1]

Choose a fixed pipe section with one inlet and one outlet. Density ρ is mass per local volume. Mass flow ṁ is the mass crossing a cross-section per unit time, in kg/s. Volume flow Q instead counts local volume per time, in m³/s. A mass flow of 1 kg/s is not also 1 L/s without a density specification.[1]

\begin{aligned}&\dot m=\rho A\overline v_n\\&Q=A\overline v_n\end{aligned}

For uniform cross-sectional density, v̄n is the average normal velocity component.[1]

Choose the velocity component normal to the area. Motion tangent to a measuring surface carries no mass through it. Straight-pipe cross-sections are normally perpendicular to mean flow. The v in the short form then denotes cross-sectional mean velocity, not necessarily velocity at the pipe center.[1]

How does mass conservation give the steady continuity equation?

For the fixed section, change in contained mass equals inflow minus outflow. Sources, additional connections and leaks would add balance terms. In steady flow, stored mass is constant. With exactly one inlet and one outlet, positive incoming and outgoing mass-flow magnitudes must therefore match.[2]

\begin{aligned}\frac{\mathrm dM_{\mathrm{CV}}}{\mathrm dt}&=\dot m_{\mathrm{in}}-\dot m_{\mathrm{out}}=0\\\rho_1A_1\overline v_1&=\rho_2A_2\overline v_2\end{aligned}

The steady two-section form requires no mass accumulation between the areas.[1]

This derivation uses mass conservation and an appropriate flow model. It does not require frictionless motion. The same steady mass balance can hold in a real pipe with pressure loss. Friction alters energy and momentum balances but creates no additional fluid mass in an ordinary pipe model.[1]

When is volume flow conserved and when must density be retained?

\rho_1=\rho_2>0\quad\Longrightarrow\quad A_1\overline v_1=A_2\overline v_2=Q

Equal density also makes volume flow equal.[1]

The simplification often fits water under moderate state changes, but is not an automatic guarantee for every liquid. Gas density can change substantially with temperature and pressure. Mass flow then remains constant while $Q=\frac{\dot m}{\rho}$ can vary by location. With changing density, area alone does not determine the velocity ratio.[1]

\frac{\overline v_2}{\overline v_1}=\frac{\rho_1}{\rho_2}\frac{A_1}{A_2}

Compression can change the velocity effect of a narrowing cross-section.[1]

Areas, diameters and units

\begin{aligned}&A=\frac\pi4D^2\\&\frac{A_1}{A_2}=\left(\frac{D_1}{D_2}\right)^2\end{aligned}

For circular pipes, area ratio is the square of diameter ratio.[1]

Half the diameter means one quarter of the area. At equal density and mass flow, mean velocity then quadruples. For a planar slit of unchanged width, halving height instead halves area. A height shown in a sketch therefore does not fully specify area without the cross-sectional shape.[1]

$1\,\mathrm{cm^2}=10^{-4}\,\mathrm{m^2}$ and $1\,\mathrm L=10^{-3}\,\mathrm{m^3}$ . For L/min, convert both liters and minutes. Units of $\rho A\bar v_n$ are $\left(\frac{\mathrm{kg}}{\mathrm{m^3}}\right)\mathrm{m^2}\left(\frac{\mathrm m}{\mathrm s}\right)=\frac{\mathrm{kg}}{\mathrm s}$ . Solving for v₂ requires positive ρ₂ and A₂; a closed zero area cannot be handled by the same finite-flow calculation. Density connects local volume flow to mass flow.[1]

Example: water through a narrowing

Check volume flow and velocity together

Given: Steady water with $\rho=1000\,\mathrm{kg\,m^{-3}}$ ; $D_1=30\,\mathrm{mm}$ , $D_2=15\,\mathrm{mm}$ and $\bar v_1=0.80\,\mathrm{m\,s^{-1}}$ . No other connections.
Find: Find v̄₂, Q and ṁ.

Approach: For steady flow at equal density use $Q=A_1\bar v_1=A_2\bar v_2$ , with $A_i=\frac{\pi D_i^2}{4}$ and $\dot m=\rho Q$ . Calculate circular-area ratio from diameters.
$\begin{aligned}\frac{A_1}{A_2}&=\left(\frac{30}{15}\right)^2=4\\{} \overline v_2&=4\cdot0.80=3.20\,\mathrm{m/s}\end{aligned}$
Evaluate the larger area in SI units and multiply by its mean velocity.
$Q=\frac\pi4(0.030)^2\cdot0.80=0.00018\pi\,\mathrm{m^3/s}\approx5.65487\cdot10^{-4}\,\mathrm{m^3/s}$
Use density to obtain mass flow; convert both factors for L/min.
$\begin{aligned}\dot m&=\rho Q\approx0.565487\,\mathrm{kg/s}\\{} Q&\approx33.9292\,\mathrm{L/min}\end{aligned}$

Result: $\bar v_2=3.20\,\mathrm{m\,s^{-1}}$ ; Q ≈ 0.5655 L/s; ṁ ≈ 0.5655 kg/s.

Check: The smaller area $A_2=\frac{\pi(0.015\,\mathrm m)^2}{4}$ times $3.20\,\mathrm{m\,s^{-1}}$ gives the same Q. In 10 s approximately 5.655 L passes both sections. The required pressure difference is still undetermined.

A second direct area check: 4.0 cm² at 2.0 m/s with a 1.0 cm² outlet gives $\bar v_2=8.0\,\mathrm{m\,s^{-1}}$ and $Q=8.0\cdot10^{-4}\,\mathrm{m^3\,s^{-1}}$ . Both calculations require equal density and unchanged volume flow. They do not determine pump power. Bernoulli’s equation adds a mechanical energy balance under its own assumptions, allowing pressure changes to be examined.[1][3]

Half diameter, four times the speed

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Water example with unchanged volume flow and equal density. Diameter increases along the axis: wider pipes need smaller mean speeds for the same volume flow.[1]

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Example: changing density in a gas model

Same mass flow, different volume flow

Given: Steady model: $\rho_1=1.20\,\mathrm{kg\,m^{-3}}$ , $A_1=25\,\mathrm{cm^2}$ , $\bar v_1=6.0\,\mathrm{m\,s^{-1}}$ ; $\rho_2=1.80\,\mathrm{kg\,m^{-3}}$ and $A_2=10\,\mathrm{cm^2}$ . Densities are specified state values.
Find: Calculate ṁ, v̄₂ and both volume flows.

Approach: At different densities conserve $\dot m=\rho_1A_1\bar v_1=\rho_2A_2\bar v_2$ ; $Q_i=\frac{\dot m}{\rho_i}$ need not be equal. Convert inlet area and find mass flow.
$\dot m=1.20\cdot25\cdot10^{-4}\cdot6.0=0.0180\,\mathrm{kg/s}$
Solve outlet mass flow for velocity.
$\overline v_2=\frac{0.0180}{1.80\cdot10\cdot10^{-4}}=10.0\,\mathrm{m/s}$
Calculate both local volume flows separately.
$\begin{aligned}Q_1&=0.0150\,\mathrm{m^3/s}\\{} Q_2&=0.0100\,\mathrm{m^3/s}\end{aligned}$

Result: $\bar v_2=10.0\,\mathrm{m\,s^{-1}}$ ; $\dot m=0.0180\,\mathrm{kg\,s^{-1}}$ remains equal while Q decreases from 15 to 10 L/s.

Check: Check: $\rho_1Q_1=\rho_2Q_2=0.0180\,\mathrm{kg\,s^{-1}}$ . Incorrectly cancelling density gives outlet speed 15 m/s and the inconsistent mass flow $0.0270\,\mathrm{kg\,s^{-1}}$ , violating the balance.

This calculation checks mass balance for specified states. It does not prove that those states jointly occur in a passive nozzle. Energy, momentum, an equation of state and possibly technical work are additional requirements.[1]

Advanced: storage, branches and profiles

\frac{\mathrm dM_{\mathrm{CV}}}{\mathrm dt}=\sum\dot m_{\mathrm{in}}-\sum\dot m_{\mathrm{out}}

Count multiple connections by direction; retain the left-hand term during filling.[1]

A tank with 5 kg/s inflow and 3 kg/s outflow gains 2 kg each second. Mass is conserved although inlet and outlet rates differ: stored mass changes. At a steady branch, outgoing mass flows instead sum to the incoming rate. A leak is an additional outlet, not an exception to mass conservation.[1]

Explore further: Mass flow using a surface integral

\dot m=\int_A\rho\,\vec v\cdot\vec n\,\mathrm dA

n⃗ is the chosen unit normal of the cross-section. The integral gives signed mass flow and accounts for local density and normal-velocity variation. The inlet/outlet balance then uses the corresponding positive magnitudes.[1]

If density varies across the section, multiplying separately averaged density and velocity does not automatically equal the average of their product. Constant ρ can be taken outside the integral. Turbulent time averages likewise require appropriate treatment of fluctuations. Continuity replaces neither a pressure balance nor checking actual operating flow conditions.[1]

Five exercises with checked solutions

Calculate · Foundation

In steady equal-density flow through a circular pipe without branches, diameter halves. Inlet mean normal speed is $\bar v_1=1.5\,\mathrm{m\,s^{-1}}$ . Find $\bar v_2$ .

Show hint

First calculate area ratio.

Learning goal: Translate diameter change quadratically into velocity ratio.

Show solution and calculation

Approach: Steady flow without branches at equal density requires $A_1\bar v_1=A_2\bar v_2$ .
Circular pipes use the squared diameter ratio.
Area decreases to one quarter.
Thus $\bar v_2=4\cdot1.5\,\mathrm{m\,s^{-1}}=6.0\,\mathrm{m\,s^{-1}}$ .

Check: $\frac{A_1}{4}\cdot6.0\,\mathrm{m\,s^{-1}}=A_1\cdot1.5\,\mathrm{m\,s^{-1}}$ .

\overline v_2=\left(\frac{D_1}{D_2}\right)^2\overline v_1=6.0\,\mathrm{m/s}

Calculate · Foundation

2. A flow of 18 L/min passes 3.0 cm². What is mean velocity?

Show hint

Convert litres to cubic metres, minutes to seconds, and square centimetres to square metres.

Learning goal: Convert volume-flow and area units to find mean velocity.

Show solution and calculation

Approach: Volume flow equals area times mean normal velocity, so $\bar v=\frac QA$ .
Convert both inputs to SI units first.
$Q=\frac{18\cdot10^{-3}}{60}\,\mathrm{m^3\,s^{-1}}=3.0\cdot10^{-4}\,\mathrm{m^3\,s^{-1}}$ .
$A=3.0\cdot10^{-4}\,\mathrm{m^2}$ .
Thus $\bar v=\frac QA=1.0\,\mathrm{m\,s^{-1}}$ ; reversing $A\bar v$ gives 18 L/min again.

\overline v=\frac QA=1.0\,\mathrm{m/s}

Transfer · Application

3. A steady branch receives 0.90 kg/s. One outlet carries 0.35 kg/s. The other has $\rho=1000\,\mathrm{kg\,m^{-3}}$ and $A=2.0\,\mathrm{cm^2}$ . Find its velocity.

Show hint

First balance the missing mass flow, then divide by ρA.

Learning goal: Solve a branch balance using mass flows before calculating velocity.

Show solution and calculation

Approach: At a steady branch, outgoing mass flows sum to incoming mass flow.
For the second outlet then use $\bar v_2=\frac{\dot m_2}{\rho A}$ .
The second outlet carries $\dot m_2=0.55\,\mathrm{kg\,s^{-1}}$ .
Thus Check: $(0.35+1000\cdot2.0\cdot10^{-4}\cdot2.75)\,\mathrm{kg\,s^{-1}}=0.90\,\mathrm{kg\,s^{-1}}$ .

\begin{aligned}\dot m_2&=(0.90-0.35)\,\mathrm{kg\,s^{-1}}=0.55\,\mathrm{kg\,s^{-1}}\\\bar v_2&=\frac{\dot m_2}{\rho A}\\&=\frac{0.55}{1000\cdot2.0\cdot10^{-4}}\,\mathrm{m\,s^{-1}}\\&=2.75\,\mathrm{m\,s^{-1}}\end{aligned}

Interpret · Application

Gas enters a steady pipe section without branches at $\rho_1=2.0\,\mathrm{kg\,m^{-3}}$ and $Q_1=0.030\,\mathrm{m^3\,s^{-1}}$ . Outlet density is $\rho_2=1.5\,\mathrm{kg\,m^{-3}}$ . Find Q₂. A diagram draws outlet volume flow greater than inlet volume flow: does this violate mass conservation? The stated conditions are prescribed for this balance.

Show hint

Balance mass flow. The same mass flow requires more volume per second at lower density.

Learning goal: Recognize density change as the reason for unequal volume flows at constant mass flow.

Show solution and calculation

Approach: calculate incoming mass flow and divide by outlet density.
Mass flow stays 0.060 kg/s while volume flow increases to 0.040 m³/s.
The diagram does not violate mass conservation.

Check: outlet density times outlet volume flow again gives 0.060 kg/s. This balance alone does not establish that the states occur without additional operating constraints.

\begin{aligned}\dot m&=\rho_1Q_1=(2.0)(0.030)\,\mathrm{kg/s}=0.060\,\mathrm{kg/s}\\Q_2&=\frac{\dot m}{\rho_2}=\frac{0.060\,\mathrm{kg/s}}{1.5\,\mathrm{kg/m^3}}=0.040\,\mathrm{m^3/s}\\\rho_2Q_2&=\rho_1Q_1\end{aligned}

[1][2]

Error analysis · Application

A tank initially holds 5 kg of liquid. For 20 s, constant flow is 0.30 kg/s inward and 0.10 kg/s outward, with no other connections. A solution calls the unequal mass flows impossible and sets them equal. Find final mass and identify the missing assumption in that solution.

Show hint

Mass conservation permits storage in the tank. Equality of inflow and outflow follows only when stored mass is unchanged.

Learning goal: Use the full control-volume balance when storage changes with time.

Show solution and calculation

Approach: use the difference in mass flows as the rate of mass increase and integrate over the stated time.
The tank gains 4 kg and finally holds 9 kg.
The incorrect solution assumes steady storage without justification.

Check: total inflow is 6 kg and outflow 2 kg; 5+6−2 gives the same final mass.

\begin{aligned}\frac{\mathrm dM}{\mathrm dt}&=\dot m_{\mathrm{in}}-\dot m_{\mathrm{out}}=0.20\,\mathrm{kg/s}\\M_f&=M_i+(\dot m_{\mathrm{in}}-\dot m_{\mathrm{out}})\Delta t\\&=5\,\mathrm{kg}+(0.20\,\mathrm{kg/s})(20\,\mathrm s)=9\,\mathrm{kg}\end{aligned}

[2]

Key points

Choose the system boundary and count every connection. Without mass accumulation, inflow equals outflow. Use mean normal speeds and each local density; cancel density only when the values match. Circular diameter enters quadratically through area.

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 2 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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Change history

October 2, 2026Highlighted steady mass balance and the density condition in question headings; added paragraph connections to mass density and the conditional Bernoulli energy balance.
October 1, 2026Clarified the entry and learning transitions; checked and refined existing exercise solutions, typeset mathematical prose and added an original quantitative article figure.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.

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Quanta. Continuity equation for flow. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/kontinuitaetsgleichung-stroemung

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Frequently asked questions about Continuity equation for flow

What does the continuity equation describe physically?+

The continuity equation is a mass balance for a chosen pipe section. In steady flow with one inlet and one outlet, stored mass is constant, so incoming and outgoing mass-flow magnitudes match. For density uniform over each cross-section, the short form is $\rho_1A_1\bar v_1=\rho_2A_2\bar v_2$ . Velocity denotes the average normal component, not necessarily the centerline value. Equal density can be cancelled, giving constant volume flow $Q=A\bar v$ . With changing gas density, volume flow can instead vary while mass flow remains constant. The equation alone determines neither pressure nor temperature; further state relations and energy balances are needed.

How can I check a calculation involving the continuity equation?+

For steady water flow at equal density, $A_1=4.0\,\mathrm{cm^2}$ and $\bar v_1=2.0\,\mathrm{m\,s^{-1}}$ narrowing to $A_2=1.0\,\mathrm{cm^2}$ gives $\bar v_2=\frac{A_1\bar v_1}{A_2}=8.0\,\mathrm{m\,s^{-1}}$ . Check by converting both areas to square metres: 4.0 × 10⁻⁴ and 1.0 × 10⁻⁴ m². Both products $A\bar v$ equal 8.0 × 10⁻⁴ m³/s, or 0.80 L/s. At $\rho=1000\,\mathrm{kg\,m^{-3}}$ , mass flow is 0.80 kg/s. These inputs are areas rather than diameters. Halving circular pipe diameter quarters area and quadruples speed under the same conditions. This confirms mass balance, but does not calculate required pump pressure.

What assumptions and limits apply to the continuity equation?+

The two-section form assumes a steady section without additional connections or accumulating mass. The general balance also applies to a filling tank: 5 kg/s entering and 3 kg/s leaving means stored mass increases by 2 kg/s. At branches, add all appropriate connection flows. Friction does not invalidate mass conservation; it changes energy and pressure balances. The simplified form $\rho A\bar v_n$ requires density uniform over each cross-section. If density and velocity vary together, integrate their product over the area; separate averages generally do not suffice. A model with prescribed gas densities also does not prove that a passive nozzle can actually produce those states.

How do I check units and scale for the continuity equation?+

Volume flow $Q=A\bar v$ has units $\mathrm{m^2}\frac{\mathrm m}{\mathrm s}=\frac{\mathrm{m^3}}{\mathrm s}$ . Multiplication by density in kg/m³ produces mass flow in kg/s. One square centimetre is 10⁻⁴ m² because the centimetre factor is squared; one litre is 10⁻³ m³. For 18 L/min, divide by 60 to obtain $Q=0.00030\,\mathrm{m^3\,s^{-1}}$ . Through 3.0 cm², this gives $\bar v=1.0\,\mathrm{m\,s^{-1}}$ . Check the units of $\frac QA$ and $\rho Q$ separately. In a narrowing pipe, speed follows area ratio alone only at equal density. A drawn pipe height does not fully specify area without a cross-sectional shape.

When should I combine the continuity equation with another relation?+

Continuity supplies mass balance, while Bernoulli supplies a mechanical energy balance along a streamline under its own assumptions. Together they can connect speed and pressure in an idealised narrowing pipe. Real systems may require friction losses, pump work or compressible state changes. For a gas, an equation of state links local density to pressure and temperature; those quantities do not follow from $\rho A\bar v_n$ alone. A tank with a branch also requires a time-dependent storage balance instead of simply equating two flows. First choose the system boundary, count connections and define reference directions. Then check the assumptions of each additional equation.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 1, §14.5: Fluid Dynamics

More formulas: Mechanics

\sum \vec F_{\mathrm{ext}} = \vec F_{\mathrm{res}} = m \vec a

Newton's Second Law

F_G = G \cdot \frac{m_1 \cdot m_2}{r^2}

Newton's Law of Universal Gravitation

E_{kin} = \frac{1}{2} m v^2

Kinetic Energy

E_{pot} = m \cdot g \cdot h

Potential Energy (Gravitational)

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How do you calculate with Continuity equation for flow?

Which mass crosses the boundary

How does mass conservation give the steady continuity equation?

When is volume flow conserved and when must density be retained?

Areas, diameters and units

Example: water through a narrowing

Check volume flow and velocity together

Half diameter, four times the speed

Example: changing density in a gas model

Same mass flow, different volume flow

Advanced: storage, branches and profiles

Five exercises with checked solutions

In steady equal-density flow through a circular pipe without branches, diameter halves. Inlet mean normal speed is vˉ1=1.5 m s−1\bar v_1=1.5\,\mathrm{m\,s^{-1}}vˉ1​=1.5ms−1. Find vˉ2\bar v_2vˉ2​.

2. A flow of 18 L/min passes 3.0 cm2. What is mean velocity?

3. A steady branch receives 0.90 kg/s. One outlet carries 0.35 kg/s. The other has ρ=1000 kg m−3\rho=1000\,\mathrm{kg\,m^{-3}}ρ=1000kgm−3 and A=2.0 cm2A=2.0\,\mathrm{cm^2}A=2.0cm2. Find its velocity.

A tank initially holds 5 kg of liquid. For 20 s, constant flow is 0.30 kg/s inward and 0.10 kg/s outward, with no other connections. A solution calls the unequal mass flows impossible and sets them equal. Find final mass and identify the missing assumption in that solution.

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More formulas: Mechanics

Retain Continuity equation for flow for exams

In steady equal-density flow through a circular pipe without branches, diameter halves. Inlet mean normal speed is $\bar v_1=1.5\,\mathrm{m\,s^{-1}}$ . Find $\bar v_2$ .

2. A flow of 18 L/min passes 3.0 cm². What is mean velocity?

3. A steady branch receives 0.90 kg/s. One outlet carries 0.35 kg/s. The other has $\rho=1000\,\mathrm{kg\,m^{-3}}$ and $A=2.0\,\mathrm{cm^2}$ . Find its velocity.