Mass Flux Density Converter
Convert mass flux density between kg/m²·s, g/cm²·s, kg/m²·h, lb/ft²·s, and lb/ft²·h. Enter a value, then choose the units to convert from and to.
Convert Mass Flow Rate Between Units
Mass Flux Density Formula
Where:
- \(v_{\mathrm{from}}\) — the entered mass flux density in the selected From unit.
- \(v_{\mathrm{to}}\) — the converted mass flux density in the selected To unit.
- \(f_{\mathrm{from}}\) and \(f_{\mathrm{to}}\) — factors that convert each selected unit to kg/m²·s.
The converter first expresses the entered value in kg/m²·s, then divides by the factor for the destination unit. This changes the units without changing the physical mass flux density.
Mass Flux Density – Calculation Example
Example: convert the default value of 5 kg/m²·s to g/cm²·s. One g/cm²·s equals 10 kg/m²·s, so the destination unit has a factor of 10.
\[v_{\mathrm{to}}=5\times\frac{1}{10}=0.5\ \mathrm{g/(cm^{2}\cdot s)}\]
Result: 5 kg/m²·s = 0.5 g/cm²·s.
How to use the calculator
- Enter a nonnegative number in Mass flux density value. Enter the number only; select its unit separately.
- Choose the unit of that number in From unit.
- Choose the unit you need in To unit.
- Read the Converted Mass Flux Density result. Its unit is shown alongside the number.
The default selections convert from kg/m²·s to g/cm²·s. You can also reverse the direction or choose the same unit in both menus to check an entered value.
What mass flux density means
Mass flux density describes how much mass passes through an area in a given time. For a flow passing uniformly and perpendicular to a surface, it is the mass flow rate divided by the surface area: \(j_m=\dot m/A\). For example, if 2 kg passes through each square meter every second, the mass flux density is 2 kg/m²·s. The unit includes both area and time, so kg/m²·h describes an hourly rate rather than a rate per second.
This quantity helps compare flows through surfaces of different sizes. It is useful for membrane transport, filtration, evaporation, fluid flow, and other mass-transfer problems. If a uniform mass flux density and the relevant area are known, their product gives the total mass flow rate. Dividing by area makes clear whether a larger total flow reflects more intense transport or simply a larger surface.
Interpret the converted result as the same rate per unit area, written in another unit. A smaller numerical value does not mean less material is moving: 0.5 g/cm²·s and 5 kg/m²·s describe identical mass flux densities. In thermal analysis, the similarly named heat flux density instead measures energy transfer per unit area per unit time and cannot be converted directly into mass flux density.
Conversion factors and practical limits
The table shows the factors used by this calculator. Multiply a value by its factor to express it in kg/m²·s; divide a value in kg/m²·s by the factor to express it in the listed unit.
| Unit | Factor to kg/m²·s |
|---|---|
| kg/m²·s | 1 |
| g/cm²·s | 10 |
| kg/m²·h | 1/3600 |
| lb/ft²·s | 4.88243 |
| lb/ft²·h | 4.88243/3600 |
The pound-to-kilogram and square-foot-to-square-meter conversion is represented here by the rounded factor 4.88243. The displayed result is rounded to no more than eight decimal places, so very small converted values may appear as zero and repeated conversions may not recover every digit. Use additional precision if a calculation is sensitive to small differences.
The input accepts zero or positive values. A signed flux can be useful when direction matters, but this converter does not accept a negative input or specify a surface-normal direction. It also does not offer mg/cm²·min; choose one of the five units shown in its menus. For nonuniform flow, a single mass flux density represents only the point or area average to which it applies. To find total flow across a surface with varying flux, account for how the flux changes across that surface rather than multiplying one local value by the entire area.
Frequently Asked Questions
How do I convert lb/ft²·h to kg/m²·s by hand?
Multiply by 4.88243 to convert pounds per square foot to kilograms per square meter, then divide by 3,600 to change the hourly rate to a per-second rate. For example, 2 lb/ft²·h is approximately \(2\times4.88243/3600=0.00271246\) kg/m²·s. The pound-and-foot factor is rounded, so the result is approximate.
What is a normal or good mass flux density?
There is no universal good value. A useful target depends on the material, process, operating conditions, and area used to report the flux. For example, membrane performance may be judged against a specified throughput and separation requirement, while an evaporation rate depends strongly on temperature and airflow. Compare values only when their units, measurement conditions, and area definitions match.
How do I convert mg/cm²·min if it is not in the unit list?
Convert it to kg/m²·s first, then use kg/m²·s as the input unit. One milligram is \(10^{-6}\) kilograms, one square centimeter is \(10^{-4}\) square meters, and one minute is 60 seconds. Therefore, multiply a value in mg/cm²·min by \(1/6000\) to obtain kg/m²·s.
Why does converting a per-second flux to a per-hour flux multiply the number by 3,600?
A per-hour value counts the mass crossing the same area over 3,600 seconds, so its numerical value is 3,600 times the per-second value. For instance, 0.01 kg/m²·s equals 36 kg/m²·h. The reverse conversion divides by 3,600. Changing the time unit does not change the physical flow.
How is mass flux density related to flow velocity?
For a fluid of uniform density flowing perpendicular to a surface, mass flux density equals density times the velocity component normal to that surface: \(j_m=\rho v_n\). Density must be in kg/m³ and velocity in m/s to give kg/m²·s. If the flow crosses at an angle, using its full speed instead of the normal component overstates the flux.
Can I convert mass flux density to molar flux?
Yes, if you know the molar mass of the substance whose mass flux is being measured. Divide its mass flux in kg/m²·s by its molar mass in kg/mol to get mol/m²·s. For a mixture, use the mass flux of the particular component; a total mixture mass flux alone does not determine each component’s molar flux.
Which area should I use when calculating flux through a porous membrane?
Use the area specified by the measurement or performance standard you are comparing against. Flux is often reported per unit of the membrane’s exposed, projected area, not per unit of the much smaller open-pore area. Dividing the same mass flow rate by different area definitions produces different flux values, even though the total flow is unchanged.
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