0
$\begingroup$

Compressed air energy storage is looked down upon because of the inefficiencies it brings along. Every article I've found till today, makes use of the stored energy by converting it into mechanical torque of a shaft, and more often than not, converting the mechanical to electrical energy.

Is there a reason why, on a domestic scale, we cannot use the cooling of the air as it expands for air conditioning, along with whatever energy it provides of the mechanical kind? Please provide some sample calculations too if you can.

My own thinking would be to adiabatically pressurize a cylinder buried in a bunch of sand, which would absorb whatever low grade heat is given off, and then to decompress through a turbine to turn a shaft, while the exhaust cool air can be used for air conditioning?

What aspect will be uneconomical or unrealistic? Space required, efficiency, safety standards, what?

EDIT: Leakages from valves when decompressing are always counted as an inefficiency. If we simply pass the leakages along with exhaust cool air from the components to come (turbine etc.) through a heat exchanger, that can be used for air conditioning! Why is it a loss?

Similarly, heat loss when compressing can be captured by the aforementioned sand (or water etc) which will be preheated for heating applications elsewhere. I understand the compressed air will never give us back that energy when decompressing, but the capturing media will!

$\endgroup$
4
  • 1
    $\begingroup$ Will you be accepting an answer if you get any? Just so we can be sure our effort will be justified. $\endgroup$
    – Solar Mike
    Nov 5, 2022 at 8:21
  • $\begingroup$ @SolarMike Of course, why not? I've awarded many since I joined! <3 $\endgroup$
    – El Flea
    Nov 5, 2022 at 8:56
  • $\begingroup$ Must be one you missed : engineering.stackexchange.com/q/43488/10902 $\endgroup$
    – Solar Mike
    Nov 5, 2022 at 8:57
  • $\begingroup$ @SolarMike Oh yes! Haven't been around here since some time. When I last checked, there weren't any satisfying ones. I'll check right away. EDIT: accepted the most reasonable one! $\endgroup$
    – El Flea
    Nov 5, 2022 at 8:59

2 Answers 2

2
$\begingroup$

You could certainly make a compressed air storage system as you describe, using the waste air for cooling. One big source of inefficiency however would be that you need some kind of mechanical work to compress the air in the first place. If the main goal is to generate mechanical work, why go through all the extra steps of compressing and storing the air? Sometimes it makes sense to do this, like to run air tools such as an impact wrench or air grinder, but that's only because an air turbine is preferable to an electric motor for those applications.

As for refrigeration, you could technically use air as a working fluid by just compressing and expanding it to move heat around, but it wouldn't be very efficient without undergoing a phase change between liquid and gas. Refrigerants are chosen because they change phase at conditions that are close to those of the working environment. Air (gaseous nitrogen) will liquify only at relatively cold temperatures, at least below -150­°C. So, it wouldn't work for typical domestic applications.

As an example, say we have a tank of compressed air at 300 Kelvin being used to drive a turbine. The inlet pressure to the turbine is regulated to be a constant 500kPa, and the air expands to 200 kPa in the process. Assuming no heat transfer between the surroundings and the expanding air (probably not a very good assumption), the exiting air temperature can be calculated using the isentropic equation of state:

$$ T_{exit} = T_{inlet}\left( \frac{P_{exit}}{P_{inlet}} \right) ^ \frac{\gamma-1}{\gamma} = 300K\cdot\left( \frac{200 kPa}{500 kPa} \right) ^ \frac{1.4-1}{1.4} = 231 K $$

The work done per unit mass of air would be equal to the change in enthalpy of the air, times the turbine efficiency (assume ~50%):

$$ w=-\eta \Delta h=-\eta c_p \Delta T=-\eta c_p \left( T_{exit} - T_{inlet}\right) = -0.5\cdot 1000 \frac{J}{kg\cdot K} \cdot \left( 231K - 300K\right) = 35.5\frac{kJ}{kg} $$

Obviously I've made some simplifying assumptions here, but this should be enough to get you started with the basic concepts.

$\endgroup$
1
$\begingroup$

In few word: because of COP.

COP = kW of "cold" / kW of electricity energy.

Most air conditioners (with freons) have a COP of 2.3 to 3.5. [https://en.m.wikipedia.org/wiki/Coefficient_of_performance].

Air thermodynamic cycles have limit of COP = 0.8 or worse.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.