Why does a pneumatic motor run faster on helium than air?

[darrelu]

I have observed that a pneumatic motor runs much faster on helium than air at the same tmeperature. Why is this? Is the power output the same for both gasses or is power just proportional to speed?

 

[Daiquiri]

It can be explained in simple terms, imho, by considering a general equation: m = rho * V
A time-derivative of this equation yields:

dm/dt = rho * dV/dT​

or (since dV/dT = f * V0)

dm/dt = f * rho * V0​

where:

dm/dt = mass flow of the gas
V0 = motor displacement
f = frequency of rotation
rho = gas density (0.176 kg/m3 helium; 1.225 kg/m3 air)​

by re-arranging, you get:

f = dm/dt / ( rho * V0 )

So, everything else being equal (mass flow, motor displacement) the frequency of rotation is inversely proportional to the density of the gas. Hence the higher rotation frequency for helium, which is much less dense than air.

 

[darrelu]

Thank you very much

 

[darrelu]

I have been thinking more about your reply. It seems to me that the issue is that dm/dt is not equal when we run a pneumativ motor at a constant pressure. In fact, if we consider the gases to be ideal doesn't dm/dt and 1/rho exactly compensate for each other?

 

[Daiquiri]

Not really. :)
Let's carry on from the second equation:

dm/dt = f * rho * V0​

where:

dm/dt = mass flow of the gas
V0 = motor displacement
f = frequency of rotation
rho = gas density (0.176 kg/m3 helium; 1.225 kg/m3 air)​

If you want to see the pressure in this relationship, the easiest way is to consider the ideal gas equation:

P = rho * R * T​

where:

R is the Specific (or Individual) Gas Constant, in J/(kg K)
T is the temperature (K)
p is the pressure (Pa)​

You can find values of R for some common gasses here: http://www.engineeringtoolbox.com/individual-universal-gas-constant-d_588.html .
As you can see, for air R = 287 J/(kg K), whilst for helium it is R = 2077 J/(kg K), seven times higher.
Now, by substituting "rho" from the latter equation, you have:

dm/dt = ( f * p * V0 ) / ( R * T )​

or

f = ( R * T * dm/dt ) / ( p * V0 )​

If you suppose that P, T and V0 are constant, you obtain that frequency of rotation is proportional to the mass flow dm/dt and to the R:

f = const. * R * dm/dt​

And since R value is, as seen before, several times higher for helium than for air, you again obtain higher rotation frequency for helium - and it would be higher even if helium mass flow was up to 6-7 times smaller then the mass flow of air.

Cheers!

 

[darrelu]

Maybe I am being dense here, but isn't this the very same issue I asked about earlier? The product of R and dm/dt is approximately a constant in all cases - lower weight is compensated for by a higher value for the specific gas constant. That is why the R values you quote are different by about a factor of seven - 29/4 which reflects the weight of air to helium. Using the numbers from your reference the product is 8323 verssus 8308 - a small difference on the scale of my interest. I think the product would exactly be a constant if the gases were ideal. If R*dm/dt a constant then f does not seem to be a function of the gas. Where am I wrong?

 

[Daiquiri]

Well, no... dm/dt and R are independent values.

The mass flow of a gas (dm/dt) is decided by you. For example, you can control it with a ball valve, which can be set more or less open.

R is a physical property of the gas used, and you have no control over it (if not by choosing another gas).

 

[darrelu]

I think you are missing my point. If we assume the resistance of the syststem is the same for air and He, and we fix the pressure for both, then dm/dt for He will always be about seven times less than for air. When you multiply each by the specific gas constant for that gas, it compensates for this and the product is about the same - hence this predicts little difference in motor speed. However, when we do the experiment we observe a difference. Motor speed is about 20% larger for He than air if I remember correctly.

 

[Daiquiri]

Hi, here I am again. :)
I was overwhelmed with work in the last few days, so couldn't reply.

Yes, now I do understand what you intend. And your analysis is correct. If we put the pressure loss due to friction in the equations then - everything else being equal (piping, pressure difference, temperature) - the math says that RPM for two gasses should be equal.

The difference is probably to be sought in the assumptions used to derive the said formulae. One thing that comes to my mind is that a difference in the rate of heat transfer between the two gasses and piping could void the assumption of constant temperature. There will surely be some other real-gas issues too. It would require a more in-depth analysis which would probably soon branch into loads of thermodynamic variables and unknowns, and I really don't have time to do it right now (see the initial note about my current work load).

So... Good luck in your search for the Truth. Let me know when you have found it. ;)

Cheers!

 

[darrelu]

Thank you. I appreciate your effort very much. It helps a lot to just have had you work through this with me. Like you, I have also been wondering how big a role heat transfer plays in this problem. There is also a difference between air and nitrogen and I am pretty sure that is do to oxidation of the polymer rotor vanes because of the oxygen. The observation is that the motors run on air wear out much more quickly than those run on nitrogen. Thanks again for your help.