If I squeeze air, how much hotter does it get?
for example in a bicycle pump
Answers:
You actually have two effects in a bicycle pump. The first one we would look at the ideal gas law:
PV = nRT
In a bicycle pump the volume is decreased and the pressure increases proportionally -- in an ideal system the temperature would not change. However, air is not ideal and there is a very slight increase in temperature (but not enough to be significant).
The second effect is the venturi effect. As the air is pushed out of the bicycle pump, the pressure causes the air to move faster. In essence we are removing energy from the air to give it velocity and the air cools down slightly (once again, not enough to be significant in a bicycle pump).
There is yet a third effect -- friction. As you continue to use the pump it heats up from friction (rubbing parts together). The volume of air in the pump will be heated by the pump itself. The exact increase in temperature is dependant on the temperature of the pump.
It is all relevant to how much pressure is being built up.
Tell us a pressure and initial volume, and we will give a temperature
Not sure about through a bike pump, but when I squeeze air between my buttocks, it can get pretty warm, especially after a vindaloo!
It's not the squeeze, it's the friction of the air passing through a small aperture at high speed. I'm not sure how hot it would get but I know what you mean.
It is the squeeze! that is how a Diesel engine works, hence the alternative name, Compression/ignition engine.
If you keep compressing it (ie the container that you have can hold that much pressure), its temperature would drop (how you get liquid oxygen, nitrogen and so forth
As roy said its a compression ignition engine, usuallly they have a compression ratio of 20 to 1 so its going to squeeze it quite considerably to the point of when fuel is injected it combusts so its very hot.
and it doesnt explode as people say this would cause pinking and can wreck engines very quickly it expands at a fantastic rate hence the pressure build up which pushes the piston down..
Using si units Charles Law
n (constant for the gas)
------------------------------. X Temperature = Volume
. Pressure
It is directly proportional double the pressure double the temp, is you use Standard Units
Boyles & Charles
PV=MRC
OK you'd be looking at using the ideal gas laws. This is for a single gas eg Oxygen as opposed to air. To actually use air you would need van der Waals equation and my A level memory is fresh enough to remember the molarity of all the gases.. OK I can't be bothered.
For an ideal gas
(p1 * v1) / t1 = (p2 *v2) / t2
so to rearrange
t2 = p2 x v2 x t1 / v1 x p2
The pressure is remaining constant so these can be removed
t2 = (v2 / v1) x t1
so in your example t2 would be a tenth of t1, but this would be in Kelvins not degrees Centigrade.
PV = nrT
It's pretty much plug 'n play from there.
Doug
It all depends upon how well the 'squeezed' air is insulated during the squeezing process. If it is perfectly insulated so that no heat escapes then compression is said to be adiabatic. On the other hand, if there is no insulation so that all of the heat of compression is lost and there is no rise of temperature, squeezing is said to be isothermal. In between these extremes you have 'polytropic' compression.
To calculate the rise of temperature in each case it is necessary to launch into mathematics (for which purpose you should examine any book on thermodynamics)
if you squeeze it between your cheeks it gets hot enough to burn your pants
Let's assume its a bike pump that's plugged where the outlet should be.
If the compression is done very slowly; or quickly + waiting a minute, then the gas temperature you measure will be the same as the starting temperature (almost).
However, in acoustics or diesel engines or this bike pump problem (compressed quickly) the compression is called adiabatic. Adiabatic means thermally insulated. It is adiabatic by virtue of the speed of compression and the fact that air doesn't conduct heat very fast.
The problem with using the formula PV = nRT is that you don't know the final pressure or the final temperature. I.e. you have two unknowns & only one equation. So we need another eq.!
For adiabatic compressions a formula relating pressure & volume changes is P(V^gamma) = Constant. Gamma is the ratio of isobaric to isochoric specific heats for the gas.
gamma = cp/cv = 1.4 approx. for air.
So P2/P1 = (V1/V2)^gamma = 25.1 (if V1/V2 = 10).
Now the gas law can be applied.
T2/T1 = (P2*V2/P1*V1) = 2.51.
If T1 = 10 deg. C = 283 K, then T2 = 711 K = 438 deg. C.
438 deg. C is pretty hot, but not quite hot enough to ignite diesel fuel. Also, a 10-to-1 compression is the upper limit of the capability of a bike pump.
Well done Tom H, I haven't done that calc in sometime but you have certainly nailed it.
My bicycle pump gets hot as I pump up a tyre. I suppose it has something to do with friction.
No Bugaboo, it's mainly the compression.
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Answers:
You actually have two effects in a bicycle pump. The first one we would look at the ideal gas law:
PV = nRT
In a bicycle pump the volume is decreased and the pressure increases proportionally -- in an ideal system the temperature would not change. However, air is not ideal and there is a very slight increase in temperature (but not enough to be significant).
The second effect is the venturi effect. As the air is pushed out of the bicycle pump, the pressure causes the air to move faster. In essence we are removing energy from the air to give it velocity and the air cools down slightly (once again, not enough to be significant in a bicycle pump).
There is yet a third effect -- friction. As you continue to use the pump it heats up from friction (rubbing parts together). The volume of air in the pump will be heated by the pump itself. The exact increase in temperature is dependant on the temperature of the pump.
It is all relevant to how much pressure is being built up.
Tell us a pressure and initial volume, and we will give a temperature
Not sure about through a bike pump, but when I squeeze air between my buttocks, it can get pretty warm, especially after a vindaloo!
It's not the squeeze, it's the friction of the air passing through a small aperture at high speed. I'm not sure how hot it would get but I know what you mean.
It is the squeeze! that is how a Diesel engine works, hence the alternative name, Compression/ignition engine.
If you keep compressing it (ie the container that you have can hold that much pressure), its temperature would drop (how you get liquid oxygen, nitrogen and so forth
As roy said its a compression ignition engine, usuallly they have a compression ratio of 20 to 1 so its going to squeeze it quite considerably to the point of when fuel is injected it combusts so its very hot.
and it doesnt explode as people say this would cause pinking and can wreck engines very quickly it expands at a fantastic rate hence the pressure build up which pushes the piston down..
Using si units Charles Law
n (constant for the gas)
------------------------------. X Temperature = Volume
. Pressure
It is directly proportional double the pressure double the temp, is you use Standard Units
Boyles & Charles
PV=MRC
OK you'd be looking at using the ideal gas laws. This is for a single gas eg Oxygen as opposed to air. To actually use air you would need van der Waals equation and my A level memory is fresh enough to remember the molarity of all the gases.. OK I can't be bothered.
For an ideal gas
(p1 * v1) / t1 = (p2 *v2) / t2
so to rearrange
t2 = p2 x v2 x t1 / v1 x p2
The pressure is remaining constant so these can be removed
t2 = (v2 / v1) x t1
so in your example t2 would be a tenth of t1, but this would be in Kelvins not degrees Centigrade.
PV = nrT
It's pretty much plug 'n play from there.
Doug
It all depends upon how well the 'squeezed' air is insulated during the squeezing process. If it is perfectly insulated so that no heat escapes then compression is said to be adiabatic. On the other hand, if there is no insulation so that all of the heat of compression is lost and there is no rise of temperature, squeezing is said to be isothermal. In between these extremes you have 'polytropic' compression.
To calculate the rise of temperature in each case it is necessary to launch into mathematics (for which purpose you should examine any book on thermodynamics)
if you squeeze it between your cheeks it gets hot enough to burn your pants
Let's assume its a bike pump that's plugged where the outlet should be.
If the compression is done very slowly; or quickly + waiting a minute, then the gas temperature you measure will be the same as the starting temperature (almost).
However, in acoustics or diesel engines or this bike pump problem (compressed quickly) the compression is called adiabatic. Adiabatic means thermally insulated. It is adiabatic by virtue of the speed of compression and the fact that air doesn't conduct heat very fast.
The problem with using the formula PV = nRT is that you don't know the final pressure or the final temperature. I.e. you have two unknowns & only one equation. So we need another eq.!
For adiabatic compressions a formula relating pressure & volume changes is P(V^gamma) = Constant. Gamma is the ratio of isobaric to isochoric specific heats for the gas.
gamma = cp/cv = 1.4 approx. for air.
So P2/P1 = (V1/V2)^gamma = 25.1 (if V1/V2 = 10).
Now the gas law can be applied.
T2/T1 = (P2*V2/P1*V1) = 2.51.
If T1 = 10 deg. C = 283 K, then T2 = 711 K = 438 deg. C.
438 deg. C is pretty hot, but not quite hot enough to ignite diesel fuel. Also, a 10-to-1 compression is the upper limit of the capability of a bike pump.
Well done Tom H, I haven't done that calc in sometime but you have certainly nailed it.
My bicycle pump gets hot as I pump up a tyre. I suppose it has something to do with friction.
No Bugaboo, it's mainly the compression.
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