That depends on the insulation of the cylinder or, in other words, on the temperature of the gas after the compression. If the cylinder is fully insulated, the gas heats up reversibly and no heat escapes and you loose nothing, except for mechanical losses. If the latter losses are nearly zero, the process is fully reversible without losses (adiabatic and reversible). If the cylinder takes up some heat because it is not insulated or has a lower temperature than the gas after compression, this heat is lost for the gas. That means: the thermodynamical losses are largest if the cylinder wall keeps the temperature of the gas before compression, because then the internal energy or enthalpy of the gas keeps despite you compress it. There may be other (flow/pressure) losses due to compression via a tube with sudden expansion inside the tube or cylinder.

To complete: The energy loss is 100% if the gas has the same temperature  as before the compression. But one thing has changed: The gas has now a higher pressure (and density) than the ambient gas, meaning that the easily usable energy, called exergy, is higher now, at reduced non-easily usable energy, called anergy. If you open the cylinder, the gas expands and gets cooler. The energy that you can gain is again in the temperature (difference). The energy of a gas depends only on the temperature but consists of anergy and exergy, and their shares depends on the pressure.

Markus J. Kloker,

Mark, thanks for responding. You state that the energy loss is 100% if the gas has the same temperature as before the compression. But is that so? because now the gas itself is compressed. Doesn't the compressed gas have Energy Of Compression (EOC)? And as long as the gas remains compressed the energy of compression (EOC) will remain unchanged even when the system reaches ambient temperature. Wasn't this energy (EOC) supplied by the overall energy that was applied initially to the whole system. So how can it be correct that " the energy loss is 100% if the gas has the same temperature as before the compression". The (EOC) was not lost even when the system reached ambient temperature. Markus, if the above is correct what is the percentage of (EOC) compared to the total energy applied to compressing the gas? 

Don

Don,

I am not sure whether you want to challenge me or you want a real answer for understanding. I assume the latter: In classic thermodynamics the energy e  of a calorically perfect gas (ideal gas with p=rho*RT) at rest depends only on T, the temperature: e=c_v*T, c_v the specific heat at constant volume. The enthalpy is e+p/rho=c_p*T, taking into account the pressure additionally. Now, if you compress the gas in a cylinder that is insulated you have typically a so-called isentropic process (adiabatic, reversible) where p/(rho** gamma) is constant; gamma is the ratio of the specific heats.T goes up accordingly, reflecting the energy/enthalpy gain. Since the process is fully reversible you get the same work out that you put in by decreasing p to the uncompressed value.

If the compression heat is conducted away p decreases somewhat and you have eventually T_initial; then all the compression energy is lost formally, with the enthalpy also having the initial value. But you still have a higher pressure because of the reduced gas volume. If you let the gas no expand to p_init, you gain energy from the reduction of T that gets no below T_init. This can be defined as EOC but is less on principle then the energy put into the compression because you lost heat in between (and increased the entropy).

This is the description of the physical process. There are formulas giving the numbers for the loss or the efficiency factor, that can be looked up. They depend on the respective definition.

Don,

ok here a number for the process: First there is isentropic compression, then isochoric expansion (heat loss at constant volume until initial T with some loss of pressure), and finally you expand the gas to the initial pressure again isentropically.

The ratio of output to input energy is:

eta= - (1-eps**a)/(1-eps**-a

Don,

ok here a number for the process: First there is isentropic compression, then isochoric expansion (heat loss at constant volume until initial T with some loss of pressure), and finally you expand the gas to the initial pressure again isentropically. 

The ratio of output to input energy is:

eta= - (1-eps**a)/(1-eps**(-a))

eps= compression ratio = V_0/V_1

a= (1-gamma)/gamma,  gamma: ratio of specific heats.

example: air -> gamma=1.4,  eps=10   ->  eta=0.518.

Markus,

Yes, you assumed correctly it is the latter. I think at this point I need to clarify a few issues concerning the experimental setup I initially had in mind. An empty, non insulated, cylinder at room temperature (RT) is filled with a gas to a high pressure of 10,000 PSI. The cylinder, of constant volume, will become hotter and hotter as the cylinder pressure reaches 10,000 PSI. Then if the cylinder is allowed to cool down it will reach (RT) if given enough time. The cylinder, if now at (RT), is still under high pressure even though at a somewhat lower pressure due to the cylinder having cooled down to (RT). If we now limit the the discussion to just the above experimental setup the questions I'm asking are 1) what is the total energy supplied to compressing the gas into the cylinder. 2) What is the percentage, from the total energy involved, that is absorbed by the gas molecules and 3) what is the percentage of energy lost as heat due to the cylinder cooling to (RT). Now Markus, correct me if I'm wrong, if we were to add the two percentages together the sum should be approximately equal to the total energy that was supplied to initially compressing the gas into the cylinder.

Thanks for all your input!  

Don

Don,

I am sorry but this is not correct.

To answer exactly your questions: If we forget about the heating of the cylinder material all the compression energy - for instance applied by a non-slow piston movement compressing the gas in the cylinder from room pressure - goes into the (translational) motion of the molecules, reflected by the temperature (reversible adiabatic, viz. isentropic, compression). If the system cools down afterwards at constant volume (piston position held fixed) until it reaches again room temperature, all the energy used for compression escapes in the form of heat. Hence the loss is 100%,. As I already said the system however has still a higher pressure than before compression, and is able to provide energy easily. If you let the piston do work now, again not slowly, until it stops because the gas reaches the room pressure again, the gas has cooled down distinctly below room temperature. This temperature decrease stands for the work/energy you get out which is less then you put in by compression.  The ratio of these two energies (output to input) is given in my previous answer by eta. Note that the energy of a (n ideal) gas does not depend on pressure, only on temperature; the pressure just indicates whether energy can be taken out more easily.

The compression energy is : E=m*c_v*(T_1-T_0), m - mass of gas [kg], T - temperature  0 - room, 1 - after compression, c_v -specific heat at constant volume; 

(T_1-T_0)=T_0*(eps**(-a) -1),  for a and eps see my previous answer.

Best luck, Markus

Markus,

You said, If the system cools down afterwards at constant volume (piston position held fixed) until it reaches again room temperature, all the energy used for compression escapes in the form of heat. Hence the loss is 100% Then how is it that the cylinder still remains at a high pressure?. Isn't the cylinder still at high pressure because higher pressure forced more gas molecules into the cylinder and in this way increased the potential energy in the system. So the total energy used can't be equal to just the heat energy lost because the system has not lost it's increased potential energy.

Don

Since the piston is fixed, there in no work done, and you loose the work done previously in the adiabtic compression that had been transformed into heat. After the cool down the pressure is higher and the density also, but the ratio of the two is the same as before the compression (equation of state: p=rho *R*T or p*V=mRT).

You may define something like a potential energy that is reflected by the pressure. But  thermodynamically seen the pressure does stand for the energy, only for the easiness of using the energy of the system. If you compress a spring, there is potential energy because there is nothing like a temperature, and the "pressure" of the spring reflects the energy stored.  With a gas it is different. Example: The gas has a certain temperature, and a high pressure, so you can expand easily and the energy comes from the decrease of temperature. But what if the temperature can no more decrease further? (For example because the gas transforms to the solid phase at low T, or you are near the absolute zero T .) Then the over-pressure is useless.

Ok, I will stop here, thank you for the discussion.

Markus,

You said, " If you compress a spring, there is potential energy because there is nothing like a temperature, and the "pressure" of the spring reflects the Energy Stored.  With a gas it is different."

I respectfully disagree that with a gas it is different !. 

Thanks for the discussion,

Don

It should be the wall-tangent friction force times the traveled distance of the piston.

Note that the piston's contact area is not showing explicitly here.

Don,

you seem somewhat reluctant to accept the internationally founded definitions and physics of thermodynamical systems.

As I said, you may define something like a potential energy using the pressure, but it has to include that the real parameter for the energy is T.

I provided a formula for the output-to-input energy that you wanted to know, but you did not use it, maybe because the compression ratio eps is involved and not the pressure ratio. You can calculate eps by  eps=(p_1/p_0)**(1/(1-a)); ** means to the power of.

Example for the values you provided:

Compression from 14.5 psi (1 bar) to 10000 psi (689.7 bar).

1. Gas is air (gamma=1.4), then eps=161.4 and output-to-input energy ratio eta=0.234, id est 76.6% loss.

2.  Gas is a noble gas (gamma=5/3), then eps=106.6 and eta=0.154, id est  84% loss.

The formula for eta provided already earlier is

eta= - (1-eps**a)/(1-eps**(-a))

eps= compression ratio = V_0/V_1=; V_0 - volume before compression, V_1 volume after compression

a= (1-gamma)/gamma,  gamma: ratio of specific heats of the gas.

In classical thermodynamics the loss is 100% after the adiabatic compression and the isochoric escape of the heat generated by the compression. If you then let the piston do work you get something out, but the heat for it comes from the surrounding energy.