How does the valve temperature change after the bicycle tire is deflated

You must not have learned gas dynamics, after deflation, the air flow accelerates in the valve to reach supersonic speed, and in the acceleration process, the friction factor is also taken into account, and there is the following relationship:

t2/t1=(1+

T2 is the outlet temperature, T1 is the temperature in the tire, M1, M2 are related to the inlet and outlet speed, the greater the speed, the greater the m. m is the Mach number. Obviously, m2 is much larger than m1, so the temperature drop (in k, can reach a change of a few tenths of a degree).

In contrast, the frictional effect of the gas (a few degrees) is completely negligible.

However, the outlet will generally form a shock wave (that is, the violent deflation sound heard), and the air flow passes through the shock wave, and the kinetic energy is violently lost, which is converted into heat energy, that is, the air temperature rises again after a certain distance from the air nozzle.

You can find a book related to gas dynamics, go directly to the conclusion of "Friction Pipe Flow with Equal Section", and check it.

Experiments are the most correct.

The temperature drops. The gas expands in volume, does work externally, and absorbs heat.

The temperature will decrease. I used to do this experiment to verify it. First, the tire is fully inflated (the pressure increases), and then the valve core is quickly pulled out, and the "poof" gas is exhausted (the pressure decreases), and the valve core will freeze.

Pay attention to the valve core and stay a little away from the valve, don't believe it and try it yourself.

The high-speed air passing through the valve will quickly take away the heat of the valve, causing the temperature to drop, and observing that it will frost, and the air will intensify the water vapor that knots on the surface of the valve after the temperature is lowered. (My family does tire repair, and when the tire is released, it will definitely bring down the temperature of the valve, and I do this job every day, and sometimes the temperature will freeze.)

It is raised first (the air rubs against the valve) and then descends (the heat transfer between the valve and the air).

When the cylinder is pumped, the piston of the cylinder does work on the air in the cylinder, the internal energy of the air in the cylinder increases, the temperature of the air in the cylinder increases, the internal energy of the cylinder wall is changed through heat transfer, and the temperature of the cylinder wall rises At the same time, due to the friction between the piston movement of the cylinder and the cylinder wall, heat is also generated by the way of work, so that the cylinder wall is heated;

When the tire is exposed to sunlight, the air temperature in the tire increases, the pressure increases, and the gas expands, so it is easy to burst

The wall of the cylinder will heat up because the piston of the pump compresses the air to do work, the internal energy of the air increases, and the internal energy of the gas is transmitted.

Give the cylinder wall to increase the temperature of the cylinder wall.

At the same time, the pumping process is also a process of overcoming the friction between the piston and the cylinder wall, which can also make the cylinder.

The internal energy of the wall increases, and the temperature increases.

For the same object, the greater the internal energy, the higher the temperature The process of the pump doing work on the air in the cylinder will lead to an increase in the internal energy of the air and an increase in temperature

The heat transfer phenomenon refers to the phenomenon of heat transfer from a high-temperature object to a low-temperature object, so the temperature of the cylinder wall will also rise

So the answer is: elevated, enlarged

For the same object, the greater the internal energy, the higher the temperature The process of the pump doing work on the air in the cylinder will lead to an increase in the internal energy of the air and an increase in temperature

The heat transfer phenomenon refers to the phenomenon of heat transfer from a high-temperature object to a low-temperature object, so the temperature of the cylinder wall will also rise

So the answer is: elevate, do work

For the same object, the greater the internal energy, the higher the temperature The process of the pump doing work on the air in the cylinder will lead to an increase in the internal energy of the air and an increase in temperature

The heat transfer phenomenon refers to the phenomenon of heat transfer from a high-temperature object to a low-temperature object, so the temperature of the cylinder wall will also rise

So the answer is: elevate, do work