Conversion Factor for Flow Between Various Gases

[redsox9887]

Hi Everyone,

I am trying to understand how a certain flow conversion factor was reached. The factor is used to convert Oxygen to Air and vice versa. The source of this conversion factor states:

"The theoretical ratio of Air flow to Oxygen flow is as follows:

(Flow (Air))/(Flow (O_2 ) )=1.0512

1.0512= √(32/28.96)

Where... Molecular Weight of O_2 = 32.00 g/mol
Molecular Weight of Air = 28.96 g/mol"


Based off of the Ideal Gas Law it makes sense that the ratio of flows would be based off of the ratios of molecular weights. I am not sure where the square root term comes in though?

Some further information: I am measuring the flow rate through an orifice using air and now I am trying to find what the equivalent flow rate in oxygen would be. My temperature is the same for both gases, specified at 70°F. My upstream pressure is 69.7 psia and my downstream pressure is 14.7 psia. I am trying to find flow rate in LPM with a specified Temp and Pressure.

 

[boneh3ad]

It comes, most likely, from the fact that the speed of the sound in an ideal gas depends on, among other things, the square root of the specific gas constant. The speed of sound in an ideal gas is
$$a = \sqrt{\gamma R T}$$
where $R = \bar{R}/m$ is the specific gas constant and $m$ is the molecular weight of the gas. So, naturally speaking, if you have two gases flowing at a constant Mach number, $M=u/a$, then the velocity is $u = Ma$ and the ratio of the two velocities are
$$\dfrac{u_1}{u_2} = \dfrac{a_1}{a_2} = \dfrac{\sqrt{\gamma R_1 T} }{\sqrt{\gamma R_2 T}} = \dfrac{\sqrt{\gamma \bar{R} T} }{\sqrt{\gamma \bar{R} T}}\dfrac{m_2}{m_1} = \dfrac{m_2}{m_1}.$$

Be careful here, though, because this all assumes that the gases have the save value for $\gamma = c_p/c_v$, which is approximately true for air and oxygen since both are diatomic (or at least, with air, mostly diatomic), but is not true in general. For the case of, say, air and carbon dioxide, the value of $\gamma$ will change and so not only will the numerator and denominator above not cancel out so nicely, but the Mach number wouldn't be the same in the first place since the isentropic relations governing the flow involve various incarnations of that quantity.