Control-volume mass balance
The integral continuity equation balances stored mass and mass flow across an open-system boundary.
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Formula
ρ in kg/m3; V in m3; ṁ in kg/s
Variables & units
| Symbol | Meaning | Unit |
|---|---|---|
| Local fluid density | kg/m3 | |
| Fixed control volume | defined region | |
| Mass flow rate across a section | kg/s |
LaTeX
\frac{d}{dt}\int_{CV}\rho\,dV+\sum_{out}\dot m-\sum_{in}\dot m=0
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Derivation & background
The control volume is fixed in space, while mass may cross its surface. In steady operation accumulation vanishes. With one inlet and one outlet, ṁin=ṁout; ṁ=ρAv applies directly only for a uniform profile, otherwise integrate the flux over the section. Density changes must be included for compressible media.
Exam blueprint
Validity range
The control volume is fixed in space, while mass may cross its surface. In steady operation accumulation vanishes. With one inlet and one outlet, ṁin=ṁout; ṁ=ρAv applies directly only for a uniform profile, otherwise integrate the flux over the section. Density changes must be included for compressible media.
Derivation steps
The integral continuity equation balances stored mass and mass flow across an open-system boundary.
- 1State the model assumptions and list quantities with units.
- 2Substitute into Akkumulation + Abfluss − Zufluss = 0; convert quantities to SI first.
- 3Water with ρ=1000 kg/m3 enters at A1=0.010 m2 and v1=2 m/s: ṁ1=20 kg/s. With A2=0.005 m2 and the same density, v2=4 m/s. Check: ρA2v2=20 kg/s.
Rearrangements
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Task variant
Work through the numerical example for “Control-volume mass balance” and check its unit and magnitude.
Water with ρ=1000 kg/m3 enters at A1=0.010 m2 and v1=2 m/s: ṁ1=20 kg/s.
With A2=0.005 m2 and the same density, v2=4 m/s.
Check: ρA2v2=20 kg/s.
Common mistakes
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: ρ in kg/m3; V in m3; ṁ in kg/s.
Exam context
- Analyze nozzles, pipes, tanks and steady systems with inlets and outlets.
Formula cluster
Fluid mechanics
The integral continuity equation balances stored mass and mass flow across an open-system boundary.
Worked example
Water with ρ=1000 kg/m3 enters at A1=0.010 m2 and v1=2 m/s: ṁ1=20 kg/s. With A2=0.005 m2 and the same density, v2=4 m/s. Check: ρA2v2=20 kg/s.
Applications
Analyze nozzles, pipes, tanks and steady systems with inlets and outlets.
How do you calculate with Control-volume mass balance?
Here is how to work through a typical Control-volume mass balance (Akkumulation + Abfluss − Zufluss = 0) task step by step:
- 1
Task
Work through the numerical example for “Control-volume mass balance” and check its unit and magnitude.
Solution path
Water with ρ=1000 kg/m3 enters at A1=0.010 m2 and v1=2 m/s: ṁ1=20 kg/s.
With A2=0.005 m2 and the same density, v2=4 m/s.
Check: ρA2v2=20 kg/s.
Frequently asked questions about Control-volume mass balance
When may I use ṁ=ρAv?+
The shortcut applies when density and normal velocity are uniform over the section or correctly averaged. In general, ṁ=∫ρ(v·n)dA. In a compressible gas, density and speed may vary along a duct while steady mass flow remains constant. Check the section rather than comparing volume flow alone. The balance uses a defined control surface; at steady state, mass accumulation vanishes. Application: Analyze nozzles, pipes, tanks and steady systems with inlets and outlets. The numerical example Water with ρ=1000 kg/m3 enters at A1=0.010 m2 and v1=2 m/s: ṁ1=20 kg/s. With A2=0.005 m2 and the same density, v2=4 m/s. Check: ρA2v2=20 kg/s anchors the idea in a concrete measurement case.
What does negative accumulation mean?+
A negative d/dt∫ρdV means the mass stored in the control volume is decreasing. Outflow can then temporarily exceed inflow. In steady state the storage term is zero, but that simplification does not apply to a filling or draining tank. The balance uses a defined control surface; at steady state, mass accumulation vanishes. The corresponding units are ρ in kg/m3; V in m3; ṁ in kg/s. Water with ρ=1000 kg/m3 enters at A1=0.010 m2 and v1=2 m/s: ṁ1=20 kg/s. With A2=0.005 m2 and the same density, v2=4 m/s. Check: ρA2v2=20 kg/s. This checks that the reference quantities are compatible.
Why does equal volume flow not always mean equal mass flow?+
Mass flow is density times volume flow. Two streams with equal Q=Av carry different mass per time when their densities differ. This matters especially for gases undergoing pressure or temperature changes. Continuity enforces mass conservation, not generally constant volume flow. The balance uses a defined control surface; at steady state, mass accumulation vanishes. Scope: The control volume is fixed in space, while mass may cross its surface. In steady operation accumulation vanishes. With one inlet and one outlet, ṁin=ṁout; ṁ=ρAv applies directly only for a uniform profile, otherwise integrate the flux over the section. Density changes must be included for compressible media. A practical application is Analyze nozzles, pipes, tanks and steady systems with inlets and outlets.
How do I check the result and units for Control-volume mass balance?+
Write the quantity and unit from the variable list next to every value: ρ in kg/m3; V in m3; ṁ in kg/s. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example Water with ρ=1000 kg/m3 enters at A1=0.010 m2 and v1=2 m/s: ṁ1=20 kg/s. With A2=0.005 m2 and the same density, v2=4 m/s. Check: ρA2v2=20 kg/s. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. The balance uses a defined control surface; at steady state, mass accumulation vanishes.
Which model limit matters for Control-volume mass balance?+
The background states this starting condition: The control volume is fixed in space, while mass may cross its surface. In steady operation accumulation vanishes.. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. The balance uses a defined control surface; at steady state, mass accumulation vanishes.
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