Calculating Electric Power Output of Compressed Air System

In summary, a cu.m/min of compressed air at 100 psi gives B watts of power. You can use electricity to power devices using air driven motors.
  • #1
sharp81
32
0
Hello,
I am interested in finding out the electric power output from a compressed air system. i see that there is a formula for calc. it with
the amount of compressed air required to generate about 1 W with a pressure of 1 bar is given by

Compressed air(cu.mm/s) = (1000 N-mm/s) / (0.1 N/sq.mm)

But since air is compressible fluid , I think we need to consider temp and pressure ratios. But what I am also wanting to know is if there is a general rule like A cu.m/min of compressed air at 100 psi gives B watts of power
 
Engineering news on Phys.org
  • #2
I can't see how you can directly calculate electric power output from the expansion work done by a compressed gas. The usual calculation done in this area is the amount of electrical power required to compress the air to given pressure at a specific flow rate.

What are you trying to accomplish here? What is the basis for this calculation you are trying to perform?
 
  • #3
try and consult a thermodynamics book for this... the formula needs to be sorted out there. here it is to discuss concepts...
 
  • #4
what i am trying to do is to generate electric power from compressed air. So I would like to find out the amount of compressed air required for producing about 1 W of power.
 
  • #5
Well, you can do mass flow times delta-P to get the energy in the air. But then you need the efficiency of your turbine. It could be anywhere from 30-70%. Then your generator is somewhere around 90%.
 
  • #6
sharp81 said:
what i am trying to do is to generate electric power from compressed air. So I would like to find out the amount of compressed air required for producing about 1 W of power.
Generate power using what device? Electricity does not magically appear from compressed air.
 
  • #7
I am thinking of a motor or a generator to be driven by the compressed air which in turn produced electricity
 
  • #8
How far can I drive on a gallon of gas?




What, you don't have an answer for me?

Until you provide detailed information about your generator and how you propose to drive it no one can give you an answer.
 
  • #9
Hi Sharp,
Air driven motors (also known as expanders) use the energy of a compressed gas to do useful work. The power output is simply the inlet enthalpy minus the discharge enthalpy times your mass flow rate.

The inlet enthalpy is a known state which you can calculate knowing the inlet pressure (100 psig) and temperature (70 F). The discharge enthalpy isn't as easy to determine. The way this is done is to consider a perfect expander or air motor which is operating at 100% isentropic efficiency (ie: no change in entropy of the gas as it expands) and then compare that to what it actually gets. That value is provided by the manufacturer as Russ eludes to. It will typically be well below 100%, and I'd go along with the rough values Russ has provided (30% - 70%) but the manufacturer will give that as an efficiency.

For a perfectly isentropic expansion (100% efficiency), I created a function of the difference in enthalpy between some inlet pressure and a discharge pressure of 14.7 psia (atmospheric pressure) using a properties database. That function is:
dh(theoretical) = 16.8 Ln (p) - 21.8
where dh(theoretical) = change in enthalpy (Btu/lbm)
p = inlet pressure (psig)
Assumptions: Inlet temperature is 70 F. Discharge pressure is 14.7 psia. Inlet pressure between 10 psig and 100 psig.

The power output is a function of your air motor efficiency.

P = dh(actual) * mdot
Where P = power output (Btu/second)
dh(actual) = actual enthalpy difference (Btu/lbm)
mdot = flow rate (lbm/second)

I'll leave it to you to change Btu/s to Watts.

To determine dh(actual) you need efficiency.

dh(actual) / dh(theoretical) = Eff(%)

You can generally obtain the isentropic efficiency from the motor manufacturer.

Once you know the output power from the air motor, you can determine how much of that power will be converted into electricity from the efficiency of the generator, and including any losses you have in transmitting that power from the motor to the generator.

Try and Google "air motor" to give you an idea of efficiency. They may also provide the power output on a chart instead of having to go through the above calculation.

What you'll find is that trying to use compressed air as a source of power is highly inefficient. This has to do with the loss of energy during compression more than anything. In general, using compressed air as a source of energy is only worthwhile if you aren't worried about wasting huge amounts of energy.
 
  • #10
can you direct me to a good literature on the same. I think my knowledge on enthalpy and entropy needs to be improved. And is there a direct conversion to find the flow rate of the pressed air(in cfm) that is converted to flowing electric power of about 1 W when the efficiency of the whole system is 100%
 
Last edited:
  • #11
Just take the dh(theoretical) and multiply by mass flow rate, then convert your units.
 
  • #12
I have a question about dh(theoritical) ...what is with the values

16.8 ln(p) - 21.8

I don't understand how you got those values ...

Also is the mass flow rate the rate of pressed air from the compressor ...
 
  • #13
Mass flow is the volume times the density of the air at any point in your system (it will be constant throughout).
 
  • #14
Hi sharp. The equation I provided is just a curve fit for the change in enthalpy from some pressure between 10 and 100 psig air at 70 F expanding isentropically to atmospheric pressure. It's very easy to put a pressure and temperature into a database, have it give you enthalpy and entropy, then use that entropy at a lower pressure to determine the new temperature and enthalpy. Computerized databases like that are tremendously useful in all sorts of thermodynamic analysis. I don't use first principals to calculate properties, it would take way too long.

Regarding the flow rate, as Russ points out, the flow you want is the flow through the air motor.
 
  • #15
I found in one of the books which says , "At a delivery pressure of 6 bar, the power required is 2.08 kW per 1 l/s of compressed air"
I have a doubt if we can actually consider this as a reversible process. ie is it same as 1 l/s of compressed air is required for producing an electrical power of 2,08 kW.

Is it also that the mass flow rate = power/dh(theoritical) since you mentioned that the power is mass flow rate times the dh(theoritical).

And what is this mass flow rate´s unit??
 
Last edited:
  • #16
The difference in enthalpy between an isentropic expansion from some pressure (ex: 100 psig) at ambient temperature to ambient pressure is not the same as the isentropic compression from ambient pressure and temperature to some higher pressure (100 psig). The isentropic compression will result in a very hot gas, whereas the isentropic expansion results in a very cold gas. The compression requires much more energy input than the expansion will produce.

Mass flow rate is in units of mass per unit time. For the equation I provided above, it is in lbm/s.
 
  • #17
Can you tell me if any univ or industry is working on some similar product
 
  • #18
Hi sharp. If you're asking if there are any universities or industries researching the use of compressed air as an energy storage medium, I think the answer is no because it's so inefficient. If you're asking if there are any industries working with and using compressed air to transmit energy, then yes, there are lots of those. The list of uses for air driven machinery would be huge. You'll find them in car repair shops, dental and medical offices, factories, etc... In each case, you'll find the benefits of using compressed air, such as compact devices or non-explosive motors, outweigh the drawbacks - especially the low energy efficiency.
 
  • #19
What other alternatives do you suggest in an environment where I would use compressed air from my electropneumatic actuator to actually drive it. I was looking at ABB´s magnetic coupling method but I think that's for very low power. Probabaly we would need a energy of about 500 mW.
Can you give some information with respect to the expenses
 
  • #20
Not sure what you're asking. What's this about an electropneumatic actuator? What's the ABB coupling you're referring to?
 
  • #21

Related to Calculating Electric Power Output of Compressed Air System

1. What is the equation for calculating electric power output of a compressed air system?

The equation for calculating electric power output is P = Q x ΔP / 33000, where P is power in horsepower, Q is flow rate in cubic feet per minute, and ΔP is pressure difference in pounds per square inch.

2. How do I determine the flow rate of a compressed air system?

The flow rate of a compressed air system can be determined by measuring the volume of air that is delivered in a given period of time, typically in cubic feet per minute (CFM). This can be done using a flow meter or by measuring the volume of a designated container filled with air in a specific amount of time.

3. What is the significance of pressure difference in calculating electric power output?

Pressure difference, also known as pressure drop, is an important factor in calculating electric power output of a compressed air system because it represents the amount of energy that is lost due to friction and resistance within the system. The greater the pressure difference, the more power is required to maintain the desired flow rate.

4. Can the electric power output of a compressed air system be increased?

Yes, the electric power output of a compressed air system can be increased by either increasing the flow rate or decreasing the pressure difference. This can be achieved by optimizing the system design, using more efficient components, or reducing leaks and obstructions within the system.

5. How can I improve the efficiency of a compressed air system?

To improve the efficiency of a compressed air system, it is important to regularly maintain and monitor the system for leaks and obstructions. Additionally, using energy-efficient components, optimizing the system design, and implementing proper air storage and distribution can also help improve efficiency and reduce energy consumption.

Similar threads

  • Mechanical Engineering
Replies
6
Views
1K
  • Mechanical Engineering
Replies
13
Views
470
  • Mechanical Engineering
Replies
2
Views
1K
  • Mechanical Engineering
Replies
5
Views
2K
  • Mechanical Engineering
Replies
3
Views
2K
  • Mechanical Engineering
Replies
4
Views
2K
Replies
6
Views
2K
Replies
6
Views
1K
  • Mechanical Engineering
Replies
20
Views
1K
  • Mechanical Engineering
Replies
3
Views
1K
Back
Top