# Calc L/min air compression to different pressures & temps in my compressor

• marcophys
In summary, the conversation discusses the testing of a new compressor rated at 540 L/min @ 7 bar 150L tank. The speaker tested the compressor at 23 degC, timing compression at 2 bar intervals and found that it maintained 34 seconds at each interval up to 8 bar. A spreadsheet was set up to understand different compression rates at different temperatures, but there were concerns about the accuracy of the calculations. The conversation also delves into the effects of rising back pressure and increasing power consumption of the electric motor as the tank pressure rises. The conversation concludes with a discussion on the impact of air temperature on the compression process and the variance between tank fill temperature and water temperature during a dive.
marcophys
I have acquired a new compressor rated at 540 L/min @ 7 bar 150L tank
I decided to test this at 23 degC, timing compression at 2 bar intervals (0-2, 2-4 etc.)

I was surprised to find that it maintained 34 seconds at each interval up to 8 bar (but this could be poor experimental controls)...(300L/34)*60 = 529.4 L/min

I thought... that's pretty good, we don't know the exact size of the tank... so I guess it is a true 540 L/min.
However, the @ 7 bar seems redundant, because there was no apparent change.

I set up a spreadsheet to understand different compression rates at different temperatures, but I may have failed.

---------------------

V1 = normal liters per minute = V1 = (P2/P1*T1/T2)*V2
V2 = liters per minute = V1/(P2/P1*T1/T2)*V2
T1 = temperature at normal conditions = 273.15K
T2 = actual temperature of air = T1 + air temp in Celcius
P1 = pressure at normal conditions = 101.325 kPa
P2 = actual pressure = say 7 bar = 709.275 or 810.6 Kpa (not sure - I chose 810.6kPa)
if V2 = 540 ...&... T2 = 273.15 (0C)
V1 = 4320 NL/min

I thought... knowing V1 @ 273.15K I could then find V2 @ 23 degC
But V2 = 585.47 L/min
V2 should less than 540 L/min

I then ran
if V2 = 540 ...&... T2 = 296.15 (23C)
V1 = 3984 NL/min

(V1 @ 0C) - (V1 @ 23C) = 336 NL/min
There is 336 NL/min less @ 23 deg C.

Using V1 = 336 NL/min
@ 23 C = 46 L/min

Therefore the compressor pumps 46L/min less at 23 degrees C as compared to pumping at 0 degrees C

-------------------

I don't know if a manometer read @ 7 bar should be calculated at 7 or 8 bar.
But more than that...
I just don't know whether this maths is correct or not.

Can anybody confirm or deny this?

If this is a reciprocating piston compressor it is compressing exactly the same amount of air at atmospheric pressure on every stroke regardless of the tank back pressure upon discharge to the tank. Th element that makes the 8 bar pressure rating relevant is that (apart from its physical design to withstand the forces and pressure physically imposed on the unit at 8 bar) is that it has sufficient horsepower to deliver the air into the tank at 8 bar. The critical element that you are not measuring is the increasing power consumption of the electric motor or engine driving the compressor as the tank pressure rises.

marcophys
JBA said:
The critical element that you are not measuring is the increasing power consumption of the electric motor or engine driving the compressor as the tank pressure rises.

Point taken re the rising back pressure.
RE your comment on the rising power consumption of the electric motor... strangely enough, I'd never considered that.
Thanks for raising this issue, as it is an important consideration.

But I'm still left with the question of the changing rate of compression with changing temperature., and whether my calculations are correct?

From a quick look your basic mathematics appears correct but you should understand the the fill volume of positive displacement compressor is based upon the intake air and ambient temeprature, not the compressor discharge air temperature. The amount of effect the discharge temperature has upon the compression process is dependent upon the rate of cooling throughout the piping and the storage vessel during the filling cycle. What this results in is a temporary premature compressor pressure switch trip at the combined increased pressure/temperature condition that results in compressor restarts and cycling as the storage system cools to the surrounding ambient temperature, which at that point the total delivery is that expected at the temperature of the intake air.

A short time back I was asked to develop a program that would accurately predict the number of scuba tanks that could be filled from a particular size air storage vessel. In the end the program was developed, but with all other factors held equal, one of the remaining variables was "how fast will you try to fill the set of tanks" because we were dealing with the opposite situation you are addressing, in that, as each scuba tank was filled, the air temperature and therefore pressure in the supply tank would drop so although there remained sufficient air in the supply vessel to fill another scuba tank at ambient temperature there was not enough pressure at the lower temperature to do so. In a parallel, the amount of air that would be delivered to a customer was also dependent upon the scuba tank air temperature when the target tank pressure was achieved. What I learned during this project that the variance between tank fill temperature and water temperature during a dive similarly affects the tank's available dive time in the same manner.

marcophys
JBA said:
From a quick look your basic mathematics appears correct but you should understand the the fill volume of positive displacement compressor is based upon the intake air and ambient temperature, not the compressor discharge air temperature. The amount of effect the discharge temperature has upon the compression process is dependent upon the rate of cooling throughout the piping and the storage vessel during the filling cycle. What this results in is a temporary premature compressor pressure switch trip at the combined increased pressure/temperature condition that results in compressor restarts and cycling as the storage system cools to the surrounding ambient temperature, which at that point the total delivery is that expected at the temperature of the intake air.

Ah ha!
The fading of the attained pressure, at motor off.

Thanks for raising this mission critical information / reasoning.
I was thinking that this cannot be due to the pressure release (for the compression side), as the drop was too great.

Of course, as soon as the explanation is made, it is 'kick yourself time'.

The effect is highlighted by the greater pressure drop when compressing from 1 bar to 8 bar, as compared to, from 6 bar to 8 bar.
The motors are that much hotter.

I noticed around 0.25 bar drop @ 8 bar... for a 100L tank (800L) that's 25L of air that isn't there.
Not a huge percentage, but really we are mainly looking at the top two bar = 200L

Therefore 25L (per 100L tank) is quite substantial.
It is yet to be seen what the drop will be, when the compressor runs continually.

Thanks for sharing your experience :)

## What is the formula for calculating L/min air compression to different pressures and temperatures in a compressor?

The formula for calculating L/min air compression is: L/min = (P1 x V1) / (P2 x V2) x (T2 / T1), where P1 is the initial pressure, V1 is the initial volume, P2 is the final pressure, V2 is the final volume, and T1 and T2 are the initial and final temperatures, respectively.

## How does pressure affect the L/min air compression in a compressor?

The higher the pressure, the lower the volume of air that can be compressed in a given amount of time. This is because as the pressure increases, the molecules of air are packed closer together, making it more difficult for the compressor to compress them further.

## What is the role of temperature in L/min air compression in a compressor?

Temperature plays a crucial role in L/min air compression as it affects the density of air. Warmer air is less dense, meaning there are fewer air molecules in a given volume. This makes it easier for the compressor to compress the air and results in a higher L/min air compression rate.

## How can I convert L/min air compression to different units of measurement?

To convert L/min to other units of measurement, you can use conversion factors. For example, 1 L/min is equal to 0.0353 cubic feet per minute (CFM), 60 liters per minute (LPM), and 0.0017 cubic meters per minute (m3/min).

## What are the potential limitations to consider when using the L/min air compression calculation in a compressor?

There are a few limitations to consider when using the L/min air compression calculation. These include the efficiency of the compressor, the type of gas being compressed, and environmental factors such as humidity and altitude. It is also important to regularly maintain and monitor the compressor to ensure accurate and safe L/min air compression rates.

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