2
$\begingroup$

Is 1 kg/s of air flow rate equivalent to 1 m3/s?

I am calculating supply air flow rate into a zone for an air conditioning system. The simulation software gives the result in m3/s but the mathematical formula that I used takes in kg/s.

Do I need to convert the flow rate, and if so, how do I convert between kg/s and m3/s?

Improve this question
$\endgroup$
1
  • $\begingroup$ 1 kg/s of air flow rate is about equivalent to 1 m^3/s at around 1km altitude... $\endgroup$
    – SF.
    Jul 3, 2019 at 10:41

4 Answers 4

Reset to default
8
$\begingroup$

You need to consider the density of the air, which varies with temperature and air pressure. At 15 degrees Celsius, at sea level, the density of air is 1.225 ${kg}/{m^3}$. The table here gives air densities at 5 degree intervals.

Now density is mass divided by volume,

${\rho} = m/v$

Hence, to get the volume flow rate (in ${m}^3/s$), for a know mass flow rate, divide the mass flow rate (in $kg/s$) by the density (in $kg/{m}^3$).

Thus, a flow rate of $1 kg/s$ is,

$1 / 1.225 = 0.8163 {m}^3/s$

To get the mass flow rate for a know volume flow rate multiple the volume flow rate by the air density.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Are four significant digits warranted? $\endgroup$
    – Peter Mortensen
    Apr 21, 2015 at 10:29
  • 3
    $\begingroup$ @PeterMortensen Take a look at the density values in the table Fred linked. A 5 °C change in temperature corresponds to roughly a 0.02 $kg/m^3$ change in density. Since the values given go to 4 sig figs, I'd say staying at that level of accuracy is appropriate. $\endgroup$
    – Trevor Archibald
    Apr 21, 2015 at 13:56
5
$\begingroup$

HVAC-Systems usually give a Volume-Flow-Rate, hence $m^3/s$. If you need Mass-Flow-Rate ($kg/s$) you simply need to multiply with the density ($\rho$) of the fluid. The density can be calculated using the ideal gas law (see 1):

$$ \rho_s = \frac{p_s}{R T_s}$$

Please observe that you need static values for the pressure ($p_s$, see 2) and temperature ($T_s$, see 3). The specific gas constant ($R$, see 4) for air is depending on the humidity of the air but $287.058 J/kg/K$ is a good starting point. For low velocities static and absolute values are so close to each other that you should be able to use the absolute/ambient pressure and Temperature. However, the density is changing with pressure and temperature, neglecting either will decrease the accuracy of the measurement.

From my experience the HVAC system has a measurement-error ($5\%$) which is larger than the error which is made by assuming static=total.

All values in SI units!

Improve this answer
$\endgroup$
1
$\begingroup$

Constant Mass Flow:

$$ \dot{m} = \rho A V $$

(Hence mass flow is equal to density times area times velocity.

So assuming you have an area of $1$ m^2 with a velocity of $1$ m/s, air with a density of $1.225$ kg/m^3 will equate to a mass flow of $1.225$ kg/s. For incompressible flows, the mass flow rate is constant.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Caution! Incompressibility (Velocity<<Ma=0.3) only shows constant density as far as the flow field is concerned. When pressure, temperature or humidity change due to weather or other process influences (flight altitude etc.) density changes as well. $\endgroup$
    – rul30
    Apr 21, 2015 at 22:16
0
$\begingroup$

You need to understand that whilst the flow rates may remain static, the mass will not be so because warmer air is lighter and less dense whilst chilled aircon air is substantially heavier. The difference between them determines the power required to transfer the air by the machine to the zone your looking into.

The difference in calculations is to answer two different questions: power consumption or airflow {m^3 \cdot s}

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.