Exploring Effects of Compressing Air: Pressure and Temperature

In summary: Actually what I was trying to do was 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 don't think I will come up with an accurate number. Andrew..if the work done on the gas was immediately transferred to the surroundings as the gas is compressed (by cooling the chamber) how would you re-write your euqations??
  • #1
tonyzzztony
3
0
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
 
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  • #2
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
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
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
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

Thank you,
Tony
 
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  • #6
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
Would it not be feasible 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
Peter.E said:
Would it not be feasible 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 :)
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:

[tex]P_1V_1^\gamma = P_2V_2^\gamma[/tex] where [itex]\gamma = C_p/C_v[/itex]

Since P = nRT/V:

[tex]T_1V_1^{\gamma-1} = T_2V_2^{\gamma-1}[/tex]

AM
 
  • #9
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 don't think I will come up with an accurate number. Andrew..if the work done on the gas was immediately transferred to the surroundings as the gas is compressed (by cooling the chamber) how would you re-write your euqations??
Thank you
 
  • #10
tonyzzztony said:
Andrew..if the work done on the gas was immediately transferred to the surroundings as the gas is compressed (by cooling the chamber) how would you re-write your euqations??
Thank you
In that case, temperature would be constant (ie remains at T1). It is just:

[tex]P_1V_1 = nRT_1 = P_2V_2[/tex]

AM
 

Related to Exploring Effects of Compressing Air: Pressure and Temperature

1. What is the relationship between pressure and temperature in compressed air?

The relationship between pressure and temperature in compressed air is known as the ideal gas law, which states that pressure and temperature are directly proportional to each other when volume and number of moles of gas are held constant. This means that as pressure increases, temperature also increases, and vice versa.

2. How does compressing air affect its properties?

Compressing air causes its molecules to be forced closer together, resulting in an increase in pressure and temperature. This also leads to a decrease in volume, making the air more dense and increasing its energy potential.

3. What factors can affect the pressure and temperature of compressed air?

The pressure and temperature of compressed air can be affected by factors such as the volume of the container, the amount of air being compressed, the type of gas being compressed, and the rate at which the air is being compressed.

4. How can we measure the pressure and temperature of compressed air?

Pressure can be measured using a pressure gauge or manometer, while temperature can be measured using a thermometer or a thermocouple. These measurements can be taken before and after compressing the air to observe any changes in pressure and temperature.

5. What are some practical applications of studying the effects of compressing air?

The study of compressing air has various practical applications, such as in the design and operation of engines, turbines, and compressors. It is also used in the production of compressed air for industrial and household use, as well as in the manufacturing of products such as aerosol cans and airbags.

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