Air Compressor approximate flow

Dear all,
I have air compressor having flow rate 3 cubic meter per minute as mentioned in the name plate.
I have started this compressor at 9:00AM and stop at 5:00PM.
At 9:00AM it shows
Load hours: 100
Unload hours: 200

At 5:00PM
Load hours: 104
Unload hours: 208

It looks like that air compressor is loaded for only four hours out of eight hours.
Can we say that total air produced can be calculated as
3x240=720 cubic meter
And
Average air flow is 1.5 cubic meter per minute.
Please suggest

Your math is wrong, your concept is OK, depending on what you want the flow rate for. Checking compressor performance - not good enough
Energy estimate - probably OK

For any problem of this sort you need to write the units in the calculation and make sure they correctly cancel algebraically. See attached example.

Or, in this case, by inspection you could observe that the compressor is loaded 1/3 of the time so the rate is 1 m3/min. The first method ALWAYS works though.

Dear Tom,
Compressor loaded for only 4 hours.
104-100= 4 hours
So in the first calculation, we will take the difference of load hours.
Ideally if a compressor produced 3 m3/min than it can produce 3x60= 180m3/hr
And during the 8 hour duty, if it runs continuously than 180x8=1440 m3 can produced.
But during 8 hours duty it only run for 4 hours, so 4x180= 720m3
Sorry for not mentioning the units as I am using android phone right now.

ssana168 said:

Dear Tom,
Compressor loaded for only 4 hours.
104-100= 4 hours
So in the first calculation, we will take the difference of load hours.
Ideally if a compressor produced 3 m3/min than it can produce 3x60= 180m3/hr
And during the 8 hour duty, if it runs continuously than 180x8=1440 m3 can produced.
But during 8 hours duty it only run for 4 hours, so 4x180= 720m3
Sorry for not mentioning the units as I am using android phone right now.

Click to expand...

I misunderstood your values, but it doesn't matter. Using the method of cancelling units shown if you plug in the correct values you will get the correct answer.

Tom, he is doing a delta not a summation. Since he is in effect performing a ratio calculation (time/time) all other units remain the same. From what I can tell he has the right numbers in the right way...kind of.

The thing that caught my eye is that in an 8 hour period his compressor is reporting 4 load hours (104-100) and 8 unload hours (208-200). That is a total of 12 hours in an 8 hour period. It leads me to believe he can't trust the basic numbers being reported.

Keith

Thanks Tom,
I need to know the approximate flow rate of my air compressor. Any member of this forum having knowledge of compressed air are requested to solve this matter.
My air compressor display shows only running hours and loaded hours. And I need to find out the approximate value of compressed air generation.
Recently I have examined the air compressor by filling 1000 litre tank from it. It has taken 185 seconds to reach the air tank pressure from 0 to 9 bar.
The flow rate of air compressor is 3m3/min as written on its name plate.

Sorry kamenges,
It was written as RUNNING HOUR not UNLOAD HOURS.
So running hours at 9am is 200 and load hours is 100.
And running hours at 5pm is 208 and load hours is 104.
I think this is clear now, that compressor operates for 8 hours but loaded only for 4 hours. Means 4 hours standby

The 3m^3/min shown on the compressor is the volumetric flow rate. It is based solely on the dimensions of the pump and does not account for pressure. That is why it take 3 minutes for a 3m^3/min pump to fill a volume of 1m^3. The volume of air being drawn into the pump compresses into a smaller volume .

I assume you are looking for the "normal cubic meters per minute" which converts the air volume at pressure to the air volume at "normal" conditions. "Normal" is not real well defined but in this particular case it is the reference pressure that is the biggest contributor to the calculation. Your reference pressure is very likely 1 bar.

If you are looking for "normal" flowrate your listed compressor volumetric flowrate should get you pretty close. Just keep in mind that, as you saw with the tank filling exercise, you can't simply use the volumetric flowrate to determine a volume fill time without accounting for pressure.

Keith

Dear Kamenges,
I have done two more exercises, as stated below.

1. I have started air compressor and noted time. The compressor has taken 80 seconds to fill the 1m3 tank from 0 bar to 4bar.

2. In the second exercise, i have used the same tank and same compressor and allow it to fill till 8 bar, than it has taken approximately 165 seconds to reach the pressure from 0 to 8 bar.

It looks like that at 1 bar, the tank can store 1m3 of air.
1 m3 at 2 bar = 2m3 at 1 bar
1 m3 at 3 bar = 3m3 at 1 bar

Is the relation right????
It looks like directly relation keeping temperature constant.

Refer to Boyle's Law.

The proverb goes:

"give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime"

We are trying to teach you to fish.

Tom is right; check out Boyle's Law.

Keith