adiabatic gas extension 2

Your equations are correct, but your example sucks. Read your fist line

grakson said:

Is termodynamic equation for calculating gas temperature, [red]when at constant mass [/red]the pressure or volume has change correct in an adiabatic process?

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Do you really think that you are not removing mass when you empty a tire? The appropriate equation is Joule-Thomson cooling which would reduce the temperature of the air in the tire by 5-6C.

David

Maybe your surrounding air temperature is just too cool to notice or your 'skin thermometer' isn't calibrated too well.

Not that I carried a thermometer around in my pocket to check it, but knowing that the adiabatic expansion would produce cooling when I deflated a tire, I have often placed my hand in the air stream as a tire deflated to enjoy a little free momentary air conditioning.

But then again, I live in a part of my country where the ambient around me was probably between 30-40C when I was seeking such cooling.

rmw

Hi grakson,
The equations you posted are for an ideal gas undergoing an adiabatic expansion or compression where work is being done by the gas or on the gas. That is, the process is adiabatic as you've noted, but the equation you are quoting should only be used when gas is doing work (if the pressure is decreasing) or conversely for gas which has work being done on it (if the pressure is increasing).

In the case of gas expanding across a valve such as a tire valve in your example, the air is not doing any work as it expands. Saying a process is adiabatic doesn't account for work. In the case of expansion across the valve, there is no work being done by the air. This is an "isenthalpic" process, which means the air undergoes an expansion but no change in enthalpy. For an ideal gas, an isenthalpic expansion results in no change in temperature. Real gasses are generally not ideal however, and air will undergo a slight cooling as it expands.

See also the writeup on a similar question by Q_Goest (post #6) here:

Zdas04 is dead on. However if you carry a thermometer, don't just measure the temperature of the air coming out. If you've been rolling, probably the tire's air is warmer than the surroundings.
iainuts, the gas expansion against the ambient pressure, would it not be considered work thus cooling the air? I do remember when I was 9 or 10 I broke the gas line going into my parent's place. Just a crack in a joint, but it had ice buildup all around it after a while, so it did cool quite a bit expanding vs the ambient pressure (probably cooled at the same rate my behind warmed up after my dad found out what I had done)

iainuts, the gas expansion against the ambient pressure, would it not be considered work thus cooling the air?

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Imagine a control volume around the tire valve where air is entering the CV at tire pressure and exiting the CV at ambient.

We can follow the typical first law equation:
dU = dQ + dW
In this case, dQ=0 and we're left with Uin and pVin with Uout and pVout. See IIIc (just less than half way down this page from an MIT web site)

The result is an isenthalpic process.

Just a crack in a joint, but it had ice buildup all around it after a while, so it did cool quite a bit expanding vs the ambient pressure

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While natural gas will cool a bit during an isenthalpic process (maybe 5 to 10 F under the conditions you mention) I think the real reason you saw ice around the crack has to do with what we often neglect when we apply the first law.

The first law as written above doesn’t take into account the change in kinetic energy (or potential energy) as the fluid changes velocity. In the case of a localized high velocity flow stream (near sonic conditions at the crack) the pressure energy is temporarily being converted to kinetic energy by Bernoulli’s law which will be converted back again as the gas slows. At the location where velocity is highest, you can have a temperature far below the temperature predicted by an isenthalpic process.

In fact, if you were to put gas through a nozzle, temperature would approach that of the isentropic case as indicated in the OP. You won’t get this kind of efficiency through a tire valve or crack in a pipe due to turbulence, but the cooling at the vena contracta could be very significant.

iainuts, thank you for removing a little of the ignorance cloud from me. Took me a while to grasp the full meaning but now is somewhat better.

iainuts (Mechanical)Isn't the velocity entering the control volume different than the jet velocity exiting the valve? If so, the process is not isenthalpic. If there is negligible difference in KE, then the process is with Q=0, approximately isenthalpic.

Also what does the CV look like upstream of the valve? Doesn't the mass of the reservoir change? If so, how is this taken into account?

Isn't the velocity entering the control volume different than the jet velocity exiting the valve?

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To be perfectly rigerous about it, we have to equate the enthalpy of the air inside the tire where there is no movement (ie: kinetic energy = 0) to the enthalpy of the air outside the tire where there is no movement. In that situation, and assuming adiabatic conditions, the process is isenthalpic. For velocities much less than sonic velocity, as found in most piping systems or where gas velocity is low such as it would be an inch away from the tire valve, we can neglect kinetic energy and consider the case to be isenthalpic.

Also what does the CV look like upstream of the valve? Doesn't the mass of the reservoir change? If so, how is this taken into account?

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As the reservoir mass drops (air leaves the tire) the first law has to be applied to what remains inside the tire. Drawing a CV around the inside of the tire, the first law reduces to:
dU = Qin – Qout – Hout

If there is no heat flow (adiabatic) then the equation reduces to dU = Hout. In other words, the change in internal energy of the gas inside the CV changes (drops) according to the enthalpy leaving. Since enthalpy changes as pressure drops (because the gas remaining in the tire cools) one needs to account for this change in enthalpy. I’ve done that in two ways, first by applying the first law and doing the calculation iteratively, and then by noting that this is the same as an isentropic expansion. Any gas remaining inside the tire as it expands is doing work by pushing air out of the tire, so the air inside the tire also follows a line of constant entropy if the process is considered adiabatic. The result is that as the tire pressure decays, the temperature of the air inside the tire follows a line of constant entropy.

Of course, even small changes in temperature result in some heat flux, so the reality is that the air inside any tank or tire won’t expand isentropically. It is more accurate to model some heat flux into the air and not assume adiabatic conditions. At any rate, the air will cool to some degree as pressure decays which is dependant on heat flux.

To summarize, as the air pressure decays, the air inside is doing work which causes it to cool, but as it goes through the valve it expands isenthalpically which causes some very small amount of cooling due to the fact it is not an ideal gas, but nothing like the cooling the air undergoes inside the tire.

iainuts (Mechanical)
The enthalpy of the air outside the tire, (the outside universe) for all practical purposes is not affected during the process. The temperature in question is that of the jet from the tire. That jet has a kinetic energy.
If that KE is negligible, them for a quasi static (slow) process from the reservoir to the jet, constant enthalpy may be approximated.

"Drawing a CV around the inside of the tire, the first law reduces to:......
Your application of the first law does not include mass flow across the CV.
Applying the first law with no Q
dU=ho*dM Where U is the internal energy within the CV
ho is specific enthalpy within the CV
M is mass within the CV
Resulting process is isentropic
For perfect gas pv^=constant.

Addition to my previous post.
"Applying the first law with no Q"
and assuming the inner walls of the tire as rigid.

the resulting process within the CV is isentropic.

The enthalpy of the air outside the tire, (the outside universe) for all practical purposes is not affected during the process. The temperature in question is that of the jet from the tire. That jet has a kinetic energy.
If that KE is negligible, them for a quasi static (slow) process from the reservoir to the jet, constant enthalpy may be approximated.

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I agree. Note that the air temperature must change as it accelerates and KE increases, then decelerates and KE decreases. The net result however is still an isenthalpic process.

Note that for an ideal gas, specific internal energy u is independant of pressure, so the change in u for an ideal gas should not change as the gas expands across the valve. Similarly, pV should be a constant for an ideal gas since pV = mRT. So the isenthalpic expansion across a valve should not result in any change in temperature for an ideal gas since:
H = U + pV = m Cv T + m R T

Your application of the first law does not include mass flow across the CV.

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Note that H = mh which is the mass flow times specific enthalpy. See also:

"Drawing a CV around the inside of the tire, the first law reduces to:......
Your application of the first law does not include mass flow across the CV.
Applying the first law with no Q
dU=ho*dM Where U is the internal energy within the CV
ho is specific enthalpy within the CV
M is mass within the CV
Resulting process is isentropic
For perfect gas pv^=constant.

Click to expand...

Right. If the air inside the tire is expanding adiabatically, it is undergoing an isentropic process. The air expanding across a valve is undergoing an isenthalpic process, but there are points within that isenthalpic process which more closely approach an isentropic process before that kinetic energy in the flow stream is lost such that the overall expansion across the valve is isenthalpic.

The steady process across the valve with( Q=0 and const elevation) approaches isenthalpic if and only if the change in KE across the valve is negligible.