Equivalent energy stored in compressed gas

[bitman]

Hi All

I'm trying to work out a method to determine the efficiency of a compressed air motor.

Obviously I can measure the input pressure and flow rate of the compressed air, and I can measure the output power of the motor with a dynomometer.

What I want to know is how much energy is stored in the incoming gases in order to get the conversion efficiency (if this is the correct way to do it).

If anyone has done this or knows how to, your help would be much appreciated.

Many thanks

Bitman

 

[pixel01]

First you should know the energy stored in one volume unit of compressed aire:

E = RnT(ln(V)-ln(v))
where R = gas constant
T = temp (K)
v = 1 (m3) at the working pressure P
V= volume of v but at 1 bar (via ideal gas law
n= the number of moles of V (via ideal gas law)

with E you can identify the power via flow rate (m3/sec ...)

PS. the above equation is an approximate one because I consider the process is adiabatic. In reality, it is not.

 

[bitman]

Hi pixel01

Thanks for your reply. Could you just clear up a few points for me.

Is n the number of moles of gas in 1 m3 ?

Is E the answer in Watts ?

Presumably I need to work out the amount of energy for the incoming gas and subtract the amount of energy left in the exhaust gas taking into account the differing temperatures.

Thanks for your help.

Bitman

 

[pixel01]

Hi pixel01

Thanks for your reply. Could you just clear up a few points for me.

Is n the number of moles of gas in 1 m3 ?

Is E the answer in Watts ?

Presumably I need to work out the amount of energy for the incoming gas and subtract the amount of energy left in the exhaust gas taking into account the differing temperatures.

Thanks for your help.

Bitman

E = energy stored in 1 m3 of compressed air, so it is in Joules
n= the number of moles in 1 m3
The flow rate then is in in m3/sec.
For calculating the energy in exhaust gas, you should take into account the dynamic energy.
The temperature is assumed constant, say 298 K.

 

[bitman]

Hi pixel01

Many thanks for a clear and comprehensive answer.

Best Regards

Bitman