# Compressed air

1. Apr 30, 2005

### tonyzzztony

If I compress air in 1/2 of the original volume the ideal gas equation predicts either
1-pressure remains the same temperature doubles
2-temperature remains the same pressure doubles
My questions is..what would I really observe during the compression..a increase in temp, increase in pressure or both???
(To make things easier consider air as ideal gas and ignore heat transfer through chamber/piston walls)
Thank you

2. Apr 30, 2005

### chroot

Staff Emeritus
The question really can't be answered as given. Is the system thermally isolated from the environment? Is the compression done quickly (adiabatically), so that there is no heat transfer to/from the environment?

- Warren

3. Apr 30, 2005

### Staff: Mentor

If one compresses a gas, one is doing work on it, so one is changing the internal energy - temperature will increase. Also, since one is decreasing the volume, the particle (molecular) density is increasing, so pressure has to increase.

If the system is adiabatic, i.e. no heat transfer to the out of the gas, the temperature will increase.

In a normal system however, heat energy is dissipated, and so the compressed gas will assume the temperature of the environment, and if the gas is in 1/2 the volume as before, the pressure will be double as before.

4. Apr 30, 2005

### Q_Goest

Tony,
Ignoring heat transfer, the compression is not just adiabatic, it's isentropic, meaning the change in entropy is zero.

Compression can be looked at as a polytropic process, which simply means it follows the relationship where PV^n = constant.

For an isentropic process, the exponent n is equal to the ratio of specific heats, k.

So for the case you described, if you knew the initial pressure and volume, and you knew the type of gas (ie: it's ratio of specific heats) and if you knew either the final pressure or final volume, you could solve for the final state using the polytropic formula using k for the exponent.

5. Apr 30, 2005

### tonyzzztony

The compression will happen in a fraction of the second so even though the system is not totally isolated from the environment the heat transfer can be ignored and the process can be considered adiabatic. Q_Goest can you explain the isentropic proces please ?
Actually what I'm trying to do is figure out how much power the AC compressor uses to compress the gas before it send it to the condenser. I already found a website which has the properties of the gas
http://tranhaz-mat.sctc.edu/Air_Conditioning/r-134a.htm [Broken]

Thank you,
Tony

Last edited by a moderator: May 2, 2017
6. Apr 30, 2005

### Q_Goest

Isentropic simply means the compression follows a line of constant entropy. For this to be true, the system must not transfer heat, which is one of the assumptions you've provided.

For power, you can apply the first law which will reduce to:
Hin + Win = Hout

where Hin = enthalpy of the fluid entering the compressor
Hout = enthalpy of the fluid leaving the compressor
Win = Work put into fluid

To find the outlet enthalpy, you'll need the fluid conditions, and a database that provides enthalpy. You already know the inlet conditions, so you should be able to get enthalpy directly from your database. You can find the outlet conditions if you know the outlet pressure (usually you'll know that). Then, assuming the compression is an isentropic process, calculate temperature from the ideal gas law and polytropic process.

With the outlet conditions, your database should give you enthalpy, which you can put into the first law equation above to determine work or energy input.

Note also, most compressors are only about 85% efficient, though some are close to 95% and some much lower so you may want to consider this. Contact the manufacturer of the compressor for efficiency. Efficiency is simply the theoretical work as determined above from the first law divided by the actual work. For that matter, you should be able to contact the manufacturer for power requirements.

7. May 1, 2005

### Peter.E

Would it not be feasable to use the equation:

(Pressure1 x Volume1)/Temperature1 (Kelvin) = (Pressure2 x Volume2) /Temperature2

But I guess in reality you would probably see variations in all :)

8. May 1, 2005

### Andrew Mason

If you use that relationship, you have to find the final temperature. For an adiabatic compression (where the work done on the gas does not leak to the surroundings as heat) the adiabatic condition applies:

$$P_1V_1^\gamma = P_2V_2^\gamma$$ where $\gamma = C_p/C_v$

Since P = nRT/V:

$$T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}$$

AM

9. May 2, 2005

### tonyzzztony

Thanks for your input. Actually what I was trying to do is find out the efficiency of my AC by measuring the current, voltage and calculating the amount of work done by the compressor. With so many variables involved I dont think I will come up with an accurate number. Andrew..if the work done on the gas was immediately transfered to the surroundings as the gas is compressed (by cooling the chamber) how would you re-write your euqations??
Thank you

10. May 2, 2005

### Andrew Mason

In that case, temperature would be constant (ie remains at T1). It is just:

$$P_1V_1 = nRT_1 = P_2V_2$$

AM