Expanding gases and temperatures
Discussion
I remember small amounts my GCSE and A-Level physics education, and know roughly that gases cool as they expand. So if compressed air from, say, a 200 bar cylinder vents to atmosphere it's going to get colder. How can I work out how much colder it will get? How about when filling a tyre?
Do I need to use P1.V1/T1 = P2.V2/T2? (Which is about all I can remember from my school days.)
Do I need to use P1.V1/T1 = P2.V2/T2? (Which is about all I can remember from my school days.)
All I can remember from school is the ideal gas law: pV=nRT.
Never used it since but just like the first line to a Kippling poem it is stuck in my head.
ETA
Thinking about it a bit further - I can see where your equation comes from. My gut feeling is that the hardest part in terms of inflating a tyre would be determining the change in volume.
Leading me to a quick Google and T2 = T1(V2/V1) - which is just a rearrangement of the equation you posted. That's the answer I think - how you measure V2 is the tricky bit.
However this assumes that Pressure remains constant which of course it doesn't!
ETA-2
There's a wiki on cold inflation pressure - but it is looking at pressure changes with temp change, not temp changes with pressure change. Still quite interesting though.
Never used it since but just like the first line to a Kippling poem it is stuck in my head.
ETA
Thinking about it a bit further - I can see where your equation comes from. My gut feeling is that the hardest part in terms of inflating a tyre would be determining the change in volume.
Leading me to a quick Google and T2 = T1(V2/V1) - which is just a rearrangement of the equation you posted. That's the answer I think - how you measure V2 is the tricky bit.
However this assumes that Pressure remains constant which of course it doesn't!
ETA-2
There's a wiki on cold inflation pressure - but it is looking at pressure changes with temp change, not temp changes with pressure change. Still quite interesting though.
Edited by TheExcession on Monday 11th April 17:53
Edited by TheExcession on Monday 11th April 17:59
When you vent a pressurised gas, the drop in temperature will depend on the rate - a slow leak will result in a smaller drop in temperature than if you release it quickly.
As for a tyre, you're going the other way - starting at atmospheric temperature and increasing which will result in an increase in temperature inside the tyre (using a compressor rather than a compressed gas cylinder). If you inflate a tyre from a cylinder then it'll get really complicated as the filling rate will change as the pressure inside the tyre increases, so the expansion of gas will not be linear.
As for a tyre, you're going the other way - starting at atmospheric temperature and increasing which will result in an increase in temperature inside the tyre (using a compressor rather than a compressed gas cylinder). If you inflate a tyre from a cylinder then it'll get really complicated as the filling rate will change as the pressure inside the tyre increases, so the expansion of gas will not be linear.
It isn't ideal gas laws as the gas is not bound by a container in its initial and final states.
Here: http://www.google.co.uk/url?sa=t&rct=j&q=&...
Essentially, you would have to determine whether the decompression occurred with constant enthalpy, constant entropy or constant temperature and then do the correct energy balance to work out the final state (this would assume no mixing).
For the bike tyre it depends on the isentropic efficiency of the pump. In practice, both operations would be difficult to calculate accurately as the processes will not take place at constant anything but a mixture of all.
Are you all still awake?
Hello?
Mike…
Here: http://www.google.co.uk/url?sa=t&rct=j&q=&...
Essentially, you would have to determine whether the decompression occurred with constant enthalpy, constant entropy or constant temperature and then do the correct energy balance to work out the final state (this would assume no mixing).
For the bike tyre it depends on the isentropic efficiency of the pump. In practice, both operations would be difficult to calculate accurately as the processes will not take place at constant anything but a mixture of all.
Are you all still awake?
Hello?
Mike…
xRIEx said:
It's sounding a bit complicated for me now! I think I'll categorise this as "rocket science without the combustion" 
It's not as bad as it sounds; if you know the initial temperature and pressure of the gas you can find its initial entropy or enthalpy from tables. You then know its final pressure and its final entropy or enthalpy (the same as at the start, depending on which was constant) from which you can look up its corresponding temperature from tables again.(But, as I said earlier, difficult to calculate accurately as it won't be truly constant. In practice, you just want to work out the worst case, whichever that may be.)
Mike...
xRIEx said:
It's sounding a bit complicated for me now! I think I'll categorise this as "rocket science without the combustion"
Not a bad analogy - the rocket accelerates not just due to the combustion but also its mass reduces (ergo under F = ma the acceleration increases to compensate) as the gas is expelled.Unfortunately with your tyres resisting expansion we get a whole new set of calculations that ultimately effect the temperature.
I think about it like this.
You have a big balloon with a certain amount of heat in it.
You compress the balloon. It still has the same amount of heat, except it is contained within a smaller area. So, the temperature mist be higher.
If you allow the balloon to expand, then the same amount of heat will be spread across a greater area, so the temperature will be lower.
You have a big balloon with a certain amount of heat in it.
You compress the balloon. It still has the same amount of heat, except it is contained within a smaller area. So, the temperature mist be higher.
If you allow the balloon to expand, then the same amount of heat will be spread across a greater area, so the temperature will be lower.
don4l said:
I think about it like this.
You have a big balloon with a certain amount of heat in it.
You compress the balloon. It still has the same amount of heat, except it is contained within a smaller area. So, the temperature mist be higher.
If you allow the balloon to expand, then the same amount of heat will be spread across a greater area, so the temperature will be lower.
Far too close to the phlogiston theory don4l. Need to talk more about kinetic energy rather than some imaginary "heat" property.You have a big balloon with a certain amount of heat in it.
You compress the balloon. It still has the same amount of heat, except it is contained within a smaller area. So, the temperature mist be higher.
If you allow the balloon to expand, then the same amount of heat will be spread across a greater area, so the temperature will be lower.
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