# Compressed Air Energy Storage, Entropy and Efficiency

The basic operating principle behind Compressed Air Energy Storage (CAES) is extremely simple. Energy is supplied to compress air, and when energy is required this compressed air is allowed to expand through some expansion turbines. But, as and when we approach this simple theory, it starts becoming more complex because of the thermodynamics involved.

Air gets heated up when it is compressed. This could be easily seen had you ever used a bicycle pump. Depending upon how air is compressed, it could be broadly classfied according to two thermodynamic processes, Adiabatic and Isothermal.

Adiabatic Compression: In this process, the heat of compression is retained, that means, there is no heat exchange resulting in zero entropy change. So the compressed air becomes very hot.

Isothermal Compression: The temperature of the gas is kept constant by allowing the heat of compression to get transferred to the environment. The entropy of the gas decreases as it gives out heat, but the entropy of the surroundings get increased by the same amount as it is accepting heat. Since both are equal, the net entropy change is zero.

Pure adiabatic and isothermal processes are very difficult to achieve. Practical compressors are somewhat in between these two. Let me put it in simple words. Take a bicycle pump, insulate the cylinder using a rubber sheet and compress it very fast in one second, that would be more of an adiabatic compression. Touch the cylinder of the pump, you could feel it. Where as, take the same pump, put it in water so that it remains cool. Compress it slowly say by 10% of the cylinder length, allow it to cool, continue compression and cooling a few times. Let the whole process take 1 minute instead of 1 second, that would be more of an isothermal compression.

The same holds good while expansion also, if the gas is not allowed to take heat from outside, then it would be adiabatic expansion resulting in a drop of temperature. But, in isothermal expansion, the gas is allowed to expand by taking heat from the surroundings and keeping the temperature constant.

In practice, isothermal compression is achieved very similar to the second bicycle example given above. Compress the air with a small compression ratio, allow it to cool without changing the volume, repeat this cycle until the required compression is achieved.

We could see that a reversible Isothermal compression is,
$Isothermal = \lim_{R \to 0, N \to \infty} \sum_{1}^N Adiabatic_N + Isochoric_N$

In effect, repeat an infinitesimal Adibatic compression followed by an Isochoric (Constant Volume) cooling, N times so that the temperature does not change. Applying Limit, when N tends to infinity the process becomes an ideal Isothermal compression. Here R is the compression ratio of the Adiabatic compression. It could be easily seen that, multiplying each compression ratio R of every cycle would give the total compression ratio. But, in normal systems a definite number of compressor stages are used with intercoolers as heat exchangers between stages to provide isochoric cooling and drop in pressure.

Expansion process is a bit different. The air goes through adiabatic expansion of multiple stages with heat exchangers in between stages. These heat exchangers are used to perform opposite function of the compressor intercoolers, that is to reheat the air by taking heat from the surroundings and there by an increase in pressure. It is assumed that all the stages use adiabatic expansion using a constant volume ratio, with the exception of the last stage. In the last stage, adiabatic expansion at constant pressure ratio is used so that the output is of ambient pressure. If constant volume ratio is used, the output of the last stage expander would have much lower pressure than that of the surroundings. The discharged air from the last stage subsequently gets expanded and heated up by taking surrounding heat using an Isobaric (Constant Pressure) process. This could be compared to the expansion and heat rejection processes of a Brayton cycle gas turbine.

### Efficiency of Processes.

Theoretically both pure adiabatic and isothermal processes are reversible. That means whatever energy supplied during compression could be retrieved back during expansion, that implies 100% efficiency. Entropy change justifies both. In adiabatic there is no entropy change at all. Whereas in Isothermal, the entropy change of the system and surroundings are opposite with the same value because heat is exchanged at constant temperature, so that there is no net entropy change. But in practical compressed air scenario it is far from correct because of many reasons.

• Pure adiabatic or isothermal processes are not possible.
• If adiabatic storage is used, air temperature and pressure could be very high for higher compression ratio. The container should handle this large pressure and temperature.
• If isothermal storage using mixed adiabatic and isochoric stages are used, it would lead to reduction in efficiency.
• Efficiency could be improved by increasing the number of stages, but that would increase cost and complexity.
• Air is not an ideal diatomic gas
• All intercoolers and heat exchanges do a mixed Isochoric and Isobaric heating or cooling.
• Mechanical parts are subject to friction and other inefficiencies.

### A few examples

In all these examples ambient air at 25C and 1atm (100kPa) with an initial volume of 1.0m3 is used. All compression and expansion are assumed to be adiabatic. And heat transfer are through Isochoric process except the last expansion stage which uses Isobaric process.

Attachment: In order to do calculations, I wrote a python script. It could be downloaded from here. (Again wordpress is not allowing me to upload a python text file. So I uploaded it as an ODT file. Download and save it as thermo.py with execute permissions). It could be invoked with parameters like number of stages, compression ratio etc.

Example 1:
Compression: A single stage compression using volume ratio 100, followed by isochoric cooling.
Expansion: A single stage expansion using a pressure ratio 100 followed by isobaric heating.

Reference Isothermal Process: Pure isothermal compression requires 460.5kJ of work, reaching a pressure of 100atm and volume of 0.01m3. The heat rejected is the same as work done and the entropy change of the system and surroundings are equal at 1545J/K. An ideal isothermal expander could get back the same energy during expansion process.

But, if adiabatic compression is employed, keeping the the same volume ratio, the compressor has to do 1327kJ of work. This work reflects in increasing both pressure and temperature to 631atm and 1607C. Since it is an adiabatic process, there is no change in entropy, so an ideal adiabatic expander would get back the same energy as work.

Let us see the associated isochoric cooling. During the cooling process, the entire 1327kJ of heat is rejected to the surroundings as expected. The entropy of air is reduced by 1545J/K, the same as that of ideal isothermal compression. But, on the other side, the entropy of surroundings got increased by 4449J/K resulting in net entropy increase of 2904J/K. Looking carefully, it took 1327kJ instead of 460.5kJ of an ideal isothermal process, giving a mere 34.6% efficiency. Since the entropy change of the air is same in both cases, the maximum work that could be extracted from this air is also the same, that is 460.5kJ.

Coming back to the expansion side. During the adiabatic stage, the air is brought back to 1atm and the volume is increased to 0.268m3 but at a temperature of -193C, also 183kJ of work could be extracted from the process. After that, the air undergoes isobaric heating and expansion taking 256kJ from the surroundings to come back to ambient condition. A part of this heat which is the same as 183kJ is used for increasing the internal energy of the air and the remaining 73kJ is used as work done. During the isobaric process, the entropy of the air got increased by the same 1545J/K and the entropy of surroundings got dropped by 858J/K giving a net increase of 685J/K. So, looking at the whole cycle, the net efficiency is 183kJ/1327kJ = 13.8%, with a net 3589J/K entropy production. This is a near impossible scenario because of the high and low temperature involved. For comparison the melting point of Steel and Iron is 1535C on the hot side, the boiling point of air is -195C on the cold side

Example 2:
Compression: Two stages of compression using volume ratio 10, two stages of isochoric cooling.
Expansion: expansion using volume ratio 10, then isochoric heating followed by another expansion using pressure ratio 10 and isobaric heating.

Here both ideal isothermal stages should have taken 230.3kJ, so the total work done is the same as that of the above example at 460.5kJ. But as adiabatic process is employed, each stage uses 378kJ, but better than the first example. Each isochoric cooling stage rejects the same 378kJ of heat, so the compression efficiency is 230.3kJ/378kJ at 61%.

After the entire compression and expansion process, the round trip efficiency is improved to 35.8%. The total entropy of the air goes through 772J/K at each stage coming back to zero. But the net entropy change of the system and surroundings together got increased by 1464J/K.

Example 3:
Like Example 2, but with 4 stages

Here, it could be seen that the total efficiency is up to 59.3% with a net 704J/K entropy production.

### What could we see from this

As the compression ratio is reduced by increasing the number of stages the difference in work done between adiabatic and isothermal processes decreases. Looking at the entropy front, the net entropy change of the system that is the air under consideration remains the same after the whole cycles, but the entropy of the surrounding increases amounting to an overall entropy increase. As the compression ratio decreases the net entropy production also decreases, correspondingly efficiency increases. That is the beauty of the greatest law of nature, the second law of thermodynamics. The entropy law governs everything.

Pure adiabatic and isothermal process do not add any net entropy, so they have no loss. Actual entropy production takes place during isochoric and iobaric heating or cooling. In these examples, when the number of stages increases, net entropy production decreases improving efficiency.

If there is some way by which the heat is retained instead of dissipating to the surroundings, the overall efficiency could be improved.

### References

1. Compressed Air Energy Storage – How viable is it?
http://canada.theoildrum.com/node/3473

2. Ideal Gases under Constant Volume, Constant Pressure, Constant Temperature, & Adiabatic Conditions
http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/ideal_gases_under_constant.htm

3. Wikipedia for general information on different thermodynamic processes

### Equations

As equations are generally disliked, I moved them to the bottom.

Universal Gas Law

$PV = nRT$

The Heat Capacity Ratio

$\gamma = C_p/C_v$
$C_p - C_v = R$

Adiabatic Compression

For Adiabatic Process $\delta Q = 0$ so $\delta W = \delta U$
$PV^\gamma = K_a$
$\delta T = K_a (V_f^{1-\gamma} - V_i^{1 - \gamma} / nC_v(1 - \gamma)$
$T_f$ could also be computed using Universal Gas Law
$Work\ Done = K_a (V_f^{1-\gamma} - V_i^{1 - \gamma} /(1 - \gamma) = n C_v \delta T$
$\delta S_{system} = 0$
$\delta S_{surroundings} = 0$

Isothermal Compression

For Isothermal Process $\delta U = 0$, so $\delta W = \delta Q$
$Work\ Done = P_f V_f ln\frac{P_i}{P_f}$
$\delta S_{system} = -\frac{|Work \ Done|} {T_{ambient}}$
$\delta S_{surroundings} = \frac{|Work \ Done|} {T_{ambient}}$

Isochoric Cooling

For Isochoric Process $\delta W = 0$, so $\delta U = \delta Q$
$\delta Q = n C_v \delta T$
$\delta S_{system} = -|n C_p ln\frac{T_f}{T_i} -R ln\frac{P_f}{P_i}| = -|n C_v ln\frac{T_f}{T_i}|$
$\delta S_{surroundings} = \frac{|\delta Q|} {T_{ambient}}$

Isobaric Heating

$\delta U = n C_v \delta T$
$\delta W = n R \delta T = P \delta V$
$\delta Q = n C_v \delta T + n R \delta T = n C_p \delta T$
$\delta S_{system} = n C_p ln\frac{T_f}{T_i}$
$\delta S_{surroundings} = -\frac{|\delta Q|} {T_{ambient}}$

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## Published by Anand Sivaram (आनन्दः )

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## 3 thoughts on “Compressed Air Energy Storage, Entropy and Efficiency”

1. Des Spence says:

Why don’t we use this as compressed air – air-con – Simple solar electric pump and steel tanks – simple and cheap?

2. great post thanks