Air consumption and compressor flow rate at low atm pressure 2

[rulmismo]

Hi,

I have a project in Mexico city, >2000m above sea level, 80% of standard atm pressure.

My pneumatics supplier says that there is no matter for consumption calculation or compressor nominal rate (X normal liters/min) relating the altitude, altought we are really very near the limit. Thay have done the calculations with 1atm pressure.

The pneumatic actuators (cilinders etc.) are calculated with two steps
a)the "dead volume" that must be increased from atm press. to working pressure and
b) the working volume that must be filled from void to working pressure when the cylinder makes its stroke.

I have some dobuts due to my initial reasoning was:
1- the "dead volume" is initially at atm pressure(P0)
2- So to get final pressure (P1)) I need more air that would be needed if atm pressure were P0'>P0
3- The working volume must be filled from void when the cylinder makes its stroke to P1. In this case I need same air quantity no matter P0 to get P1.

About 2-3 I start to get lost, cause the main objective is get the same cylinder effort no matter P0.
But this is: Cylinder effort= P1xA - P0xA - Return spring

If P0 is less than in standard atm maybe I don´t need really get P1, maybe need a little less P1'<P1 so really point 2-3- above are not quite.

Regarding the compressor if it gives C l/min at normal atm. in Mexico will give Cx0.8 (Boyle).
But again if the result is that the pressures that I will really need are a little lower than in standard atm maybe everything cancels and my supplier is right.¿?

So by this point you can see that I am winding myself, plese help me to unwind ;-)

Regards

 

[zekeman]

At first glance I would think it is almost a wash, but not linear.

If you know P1-P0, then you get P1 and the new mass flow is proportional to the new P1 and < P1 initial
For adiabatic compression , the work done per unit of mass flow is given by approximately

CpT[(P1/P0)^{(k-1)/k}-1]

Since mass flow rate is proportional to P1, the power is proportional to

CpT[(P1/P0)^{(k-1)/k}-1]*P1

Depending on the values of P1,P0 you may need more or less power for this.

 

[rulmismo]

"If the rated discharge is 10.3 barg then instead of 11.36 bara you'll be getting 11.16 bara (98%). So the capacity of the compressor is only decreased by 2% (q2=q1*P1/P2). If the discharge pressure is lower then the impact is greater (at 5.2 barg, it is 3.5%)."

Sorry but I don´t see it clear,
Imagine I have a volume V at 1bara that I compress to 10bara.
The new volume will be V'=V(1/10)=0.1V

If now I go to Mexico city, I have V at 0.8bara but I compress it to 10bara cause have a nice absolute pressure transducer to control my compressor. In this case the new volume will be V'=V(0.8/10)=0.08V

These "discrete" compressions, but n times per minute is what my compressor does, so the flow rate discharge changes 20%

I assume that something is wrong in my argumentation but can´t see it :-(

Regards

 

[Compositepro]

In a piston air compressor, if you reduce inlet pressure to 80% of sea level you will reduce flow capacity by 20%(this assumes constant motor speed). The affect of the dead volume of the cylinder will be to reduce flow capacity even further due to the lower starting pressure.

 

[rulmismo]

So I assume that I am not saying nonsenses by supposing that have a reduction of 20% flow rate.

Now my second derivative is:

Ok I have 20% less flow rate BUT for getting a constant effort (Freq) in my cylinder I really get

Freq=5N A=1^-5m2 Return_spring=-4N (1bar = 10^5Pa = k)

At sea level my cylinder work in nominal conditions:
P1 = 10bara P0 = 1bara so F = 10.k.A - 1.k.A - 4N = 5N

At Mexico city
P1' = ? P0 = 0,8bara so F = 5N = P1'.k.A - 0,8.k.A - 4N -> P1' = 9,8bara
I don´t need so much pressure...
(0,2/10=0,02 2% less is not a real difference though!)

And third derivative:
I have some pneumatic springs that support load L at height h = h1 = constant, nominal working pressure P1.
At atmospheric pressure (P0) they don´t support anything (h gets=0).

With nominal load L

At sea level
P1 = 5bara P0 = 1bara h = h1. Nominal supplier conditions.

At Mexico
P1 = ? P0 = 0,8bara h = h1, which pressure P1 should I have?

My supposition is that P0 cancels due to it affects both load and airspring in every direction, so what I really need to support the load is
L = P1 x Spring area at h=h1.

So in this case P1 seems independent of atmospheric pressure (not like in cylinders case!).

If you guys see some error please let me note it, I tend to cheat myself!

 

[Compositepro]

I'm sorry but I don't understand your questions And don't have the patience right now. I did finally figure out that Freq is probably "force required" and not frequency.

 

[rulmismo]

First thanks for your replies. I really appreciate

I words what I mean is that lower atm pressure means that you need less cylinder "high side" pressure to get the same effort, but this "less" represents only in my case 0.2bar from about 10bar (working pr.), so is really not representative.

On the other hand the 20% reduction that CompositePro mentioned for the compressor flow rate is VERY representative and must be took in account.

About the air spring problem mentioned, thinking again about it I see that it is not really different from cylinder case.