Compressor question

Time may be a little less the rate that your using is at 100 psi. The tank will not have any pressure on it when started so the out put from the compressor will be a little greater.

Is that all the compressor will be used for? Spray Painting uses a lot more and keeps a compressor running hard to keep up. My compressor seldom shuts down when I am painting.

A couple of things don't make sense here.

1. A gallon is 0.134 cu ft., so the volume of a 20 gallon tank is 2.67cu ft.

check out http://www.about-air-compressors.com/SCFM.html

2. Ignoring the details of temperature and water vapor (which are hardly negligible details), A Standard Cubic Foot has a rough pressure of 15psi. When compressed to 100psi and cooled back to the same temperature, it will have a volume of 15/100 of a cubic foot, so it will take 100/15, or 6.7 SCF to fill 1 cu ft of tank. Then it would take 6.7*2.67, or 17.8 SCF to fill the tank.In practice the air in the tank is nearly always going to be significantly hotter than the outside air, so the tank will reach 100psi sooner, and require less than 17.8 SCF. Now depending on the humidity of the incoming air, some fraction of the water will condense to liquid inside the tank (slightly reducing the effective size of the tank), requiring more air to replace the condensed water vapor; perhaps you used that to come up with your 20.86 cu ft...

If I have missed something here, please correct me!

Hello mrunal

Your 1^st part is slightly too long time. Atmosphere = 14.7psia, 100 psig = 114.7psia and you seem to have got 20.86 ft³ from 20gal x 114.7/14.7. But the tank starts at 14.7psia so only takes 20gal x 100/14.7 = 18.2 ft³ to get to 100psig.

5scfm compressor most likely to be positive displacement type so output varies only slightly with discharge pressure.

Duty cycle depends on air consumption. If it's 2.5scfm cycle is 50%, if it's 1.25scfm cycle is 25%.

You also need to consider compressor start frequency. Max start frequency occurs when consumption = 1/2 compressor output, and it's usually best to design for that as worst case. As output and tank volume fixed, it only depends on start/stop differential, ΔPpsi.

Max start frequency = compressor output*14.7/(4*tank volume*ΔP). You might get max acceptable start frequency from the supplier if you've just bought it new, but I'd suggest 20 per hour tops. ΔP 25psi gives 16.5 starts/hour.

Cheers.........Codey

I think you are spot on.

Taking your figures of 20.86 galls for the volume at face value (USA & UK gallons are different) but assuming it converts to 2.67 cu.ft (Dkwarner #3).

The pressure ratio is (100+14.7)/14.7 = 7.8. The tank therefore hold 2.67 x 7.8 = 20.83 cu.ft of free air when filled to 100 psig. This confirms your figures.

The compressor delivers 5 cfm (the free air rating) thus it takes 4.17 mins to fill - as you said.

As others have said, you need to take temperature into account to be precise, but assuming the air in the reciver cools to atmosphere, then things will equal out and you will be back to 4.17 mins.

You need to check the '5 cfm' because this might be displacement (the actual volume of the cylinders) and that means (probably) 4 cfm of actual air delivered. Thus your tank will take 5.2 mins to fill.

In fact, timing how long it takes to fill (as long as you are sure about the volume) will give you a pretty good idea of the compressor output.

Neglecting the temperature rise, I calculate the pressure of the tank at the end of:

At the end of First (1) minute as 42.22 psi

At the end of second (2) minute as 69.76 psi

At the end of third (3) minute as 97.28 psi

At the end of 3.25 minute as 104.1 psi

And the answer is:

1. 20 gal /7.48 = 2.6738 ft³ with one atmosphere (14.7psia)
2. Add 6.8 (100/14.7) atmospheres to 114.7 psia or 100 psig.
3. Then an additional (2.67 x 6.8 =) 18.16 scf will raise tank to 100 psig.
4. 18.6 scf / 5 scfm = 3.63 minutes at the initial flow rate.
5. The incoming flow will decrease as the pressure differential decreases. Then:

P tank = 100 (1-e ^t/tc) t_^c = 3.63 minutes

1. At 3.63 min tank presssure is 63.2 psig
2. At 7.26 min (two time constants) tank pressure is 100 x (1-e^-2) or 86.5 psig
3. At 10.89 min.(three time constants) pressure will be 100 x (1-e^-3) or 95 psig
4. At 14.52 min pressure is 98 psig
5. At 18.15 min (five time constants) pressure is 99.3 psig [close enough]

Air is compressible. You can't poke it through a hole into a tank at a constant rate unless you increase the supply pressure in relationship to the differential pressure.

This is pretty easy to try. If your pressure is approx 63 psig at 3.63 minutes I could be right. Tom

Thanks everyone for the detailed replies.. they are really helpful. esp when you guys include the explanation. I have better basic understanding now.

Tom,

Thanks for your input. I found a compressor 4.2 scfm @ 115 psig, 8 gallon tank. ( 8 gallon tank @115 psig = 9.44 cf) Calculated time to fill the tank - just basic - 9.44/4.2 = 2.25 mins

I drained the tank completely and then turned the compressor ON. It took about 1 min and 40 secs to reach 115 psig. As Ozzb as mentioned (first reply in this forum), it took less time to fill the tank.

mrunal Thank you for being kind. I really went upside down on my first reply and didn't have time to correct it after posting. In your example above there is one atmosphere in the tank so you only add 8.37 cf to reach 115 psig. 8.37/4.2 = 1.99 min. I have wondered why the same compressor is rated for more SCFM at lower pressure. Common sense, which escaped at a gallop in my first reply, would suggest that your volume would fill at a faster rate while the pressure was low. Sticking with the natural log curve and dividing the flow rather than pressure into time constants yeilded 1.74 min or 1 min 44.4 seconds. Still high from your actual results (possible heat expansion) but not as ridiculus as my other attempt. Thanks again. Tom

mrunal To apply the technique that seems to provide results close to your actual test to your original scenairo: (20 gallon /7.48) x 100/14.7* = 18.19 SCF *With one atmosphere in the tank at 0 psig add only 6.8 atmospheres to raise the pressure to 100 psig. If max. flow, 5 SCFM @ 100 psig and the flow during the fill interval follows the exponential curve it can be divided into 5 time contants (Tc) as follows: 5 = F(1-e^-Tc) For Tc 1, F(1) = 5 / .632 = 7.91 SCFM Tc 2 F(2) = 5 / .865 = 5.78 SCFM Tc 3 F(3) = 5 / .950 = 5.26 " Tc 4 F(4) = 5 / .982 = 5.09 " Tc 5 F(5) = 5 / .993 = 5.03 " F(1 thru 5) = 29.08 29.08 / 5 = 5.816 SCFM average flow rate from 0 to 100 psig Then 18.19 / 5.816 = 3.127 Minutes (3 min 7.6 sec.) That would be close enough for horse shoes, hand grenades and influenced minds. Tom

Thanks Tom.