Reciprocating HP compressor efficiency

[p0lytr0pic]

Hello!

I try to calculate the energy efficiency (from electric plug to the output of the compressor) of this compressor when it produces air@300bars :
BAUER V 500-11-5 F

See specifications in attachment.

FAD : 500 l./min @ 200 bars
Power : 11kW
Filling time for 1 liter@300bars : 0.60 min.

Please, could someone validate/correct the following calculation ?

Efficiency = output energy / input energy

* Output energy :
Isothermal expansion from 1 bar to 300 bars:
W = 30 MP x 0.001 m^3 x ln 300 = 0.1711 MJ = 47.5 Wh

* Input energy:
11000 W x 0.6/60 = 110 Wh

Efficiency = 47.5 / 110 Wh = 43%

It seems quite low... Perhaps the motor doesn't use the max.specified power... but if I take let's say 10500W, I get efficiency = 45%.

What is today the best efficient HP compressor @300 bars or 420 bars we can find on the market ?
What is its efficiency in % ?


Thanks by advance,

Regards,

Steve.

 

[dcasto]

the polytropic efficency should run 75% to 80%.A little less.

 

[p0lytr0pic]

Hi Dcasti! Hi all !

This doesn't answer exactly my question... (

Perhaps I can simplify it:

Does a multi-stage high pressure compressor only filling a buffer tank usually work at constant torque, power and speed ?

...or does it work harder while pressure goes up in the buffer tank because, by example, it's harder to open the output valve of the last cylinder ?

In other words, can I evaluate its electricity consumption like this:

energy used = power of the motor x running time

In fact, I guess the motor is not used at its maximum power... but what % ? 90% ?
The best thing to do is of course to measure the used energy at the plug with a wattmeter by it's not easy before buying !

I'd like to evaluate the electricity cost of a compressor before buying...
Any suggestion ?

Thanks by advance !

Regards,

Steve.

 

[pmover]

i believe you have no formal engineering education or there is a language problem . . .

"* Output energy :
Isothermal expansion from 1 bar to 300 bars:"

this statement makes no sense at all. secondly, the process of expansion or compression may not be isothermal.

furthermore, since you want efficiency from electric plug to compressor outlet, you need to know and understand all process state changes and their values. i do not believe you have this information. there are other energy losses that you have not considered.

i suggest you direct your question to the manufacturer and obtain your objectives from them.

good luck!
-pmover

 

[p0lytr0pic]

Hi Pmover !

Absolutely yes, "expansion from 1 bar to 300 bars" is an enormous stupidity !
...that shouldn't have occured if I had reread my post before validating it. (
But I didn't see how to modify my 1st post.

Of course, as you can guess, I was talking about "Isothermal expansion from 300 bars to 1 bar".

The idea with isothermal expansion is to get maximum "potential" energy you have in the full tank.

But perhaps it's a stupidity too ? Not yet...


About the electric consumption, you're right, the best thing is to ask manufacturer.

But it's a pity it's not always written in leaflets. (

Sometimes it is, like there,

http://www.bauer-kompressoren.de/en...ichtung/hochdruck/i_baureihe_90_350/index.php

and sometimes not like in the doc in my 1st post.

Though it's very important to choice the most efficient device for economical and environmental reasons,isn't it ? ;o)

Regards,

Steve.

 

[25362]

It is apparent that pOlytrOpic estimated the compression work using a formula for a reversible isothermic path, desirable but not always attainable. A reasonable assumption should be somewhere between adiabatic and isothermic compression.

BTW, megapascals are MPa, not MP.

 

[seymours2571]

You are basing your electric consumption on the drive size. The drive size is the nominal size of the electric motor being used. The actual electric consumption will be a different value all together that will be sensitive to several motor specific variables, running speed, motor loading, etc.

As stated by some of the other responses the best thing to do would be to provide the manufacturer with your specific operational conditions and have them determine the efficiency for you. There would be no easy way to determine the efficiency values unless you had all of motor and system information.

Good luck.

Stephen Seymour, PE
Seymour Engineering & Consulting Group

http://www.seymourecg.com

 

[p0lytr0pic]

Hi all!

Thanks for your answers. )

To Seymour:

Is the torque usually constant during the compression work or greater at the end when the pressure reaches the maximum in the tank ?

To chemical, physics engineers:

The max. theorical "potential" energy in the tank can be calculated with isothermal expansion formula, can't it ?
In real world, the useful energy will depend on the efficiency of the mechanical device connected to the tank, won't it ?

Isothermal expansion of 1 liter of air from 300 bars to 1 bar (PV=nRT) gives:
W = 30 MPa x 0.001 m^3 x ln 300 = 0.1711 MJ = 47.5 Wh

Please, could an expert give the calculation for an isothermal expansion of air as real gas with the best model (Van Der Waals if I believe Miss Google ;o) ) ?
...with 3 variables: initial and final pressure, initial volume.
The calculation seems very tricky for an non-physics engineer like me! (

I guess the work will be lower but what % ? 5%? 10%? More?

Thanks by advance.

Regards,

Steve.

 

[seymours2571]

The torque will vary as the compression ratio varies. So in your instance as the downstream pressure (or receiving tank pressure) increases so will the torque requirements. This is assuming a basic functioning compressor. Some compressors have unloader valves that effectively reduce power requirements when there is no demand.

Most real process are somewhere between adiabatic and isothermal. For your calculations you could use the equation representing the polytropic process:

p1V1^n=constant=p2V2^n, where assuming a reversible process (i.e. no losses) and adiabatic conditions n would = k or the specific heat ratio for air, typically taken as 1.4 at room temperature.

Search around for example calculations. Also, these calcs should be in the first half of any thermodynamics book. I would also have someone else double check your calculations before comitting $$$ to anything.

Good luck.


Stephen Seymour, PE
Seymour Engineering & Consulting Group

http://www.seymourecg.com