Sizing a air accumulator 1

[milena02]

I am trying to size an air receiver tank for the pneumatic operation of 16 valves. There are:
ten 12" valves that require 0.58 gallons for a full stroke (at atmospheric pressure)
6 24" valves that require 2.29 gallons for a full stroke (at atmospheric pressure)
The operating pressure of the valves is 80 psi.
The air receiver tank will be 120 psi.
I believe the calc is as follows:
Volume required for valves at pressure (Vact):
(0.58gal*10 + 2.29gal*6)*((80psi+14.7psi)/14.7psi)= 125.88 gal
Pressure available in tank(Ptank)=Pmax-Pmin=120psi - 80psi=40psi
Minimum Vol of Air Tank = Pact*Vact/Ptank=125.88gal*80psi/40psi.
The solenoid valve is located at the actuators, so the entire pipeline will be pressurized, and piping could count as part of the tank.
Please let me know if this is correct. I would greatly appreciate any input.

 

[milena02]

I forgot to say that the tank will supply air to handle the valves in a air supply cut off.

All valves need to operate once.

 

[ARenko]

So this is a one shot deal when air supply is lost? If so...

Your 125.88 gal calculation looks fine (assuming standard temperature), but your min volume is incorrect. Replace your Pact in your min volume calculation with Patm = 14.7 and it should be right.

Vact = 125.88 gal
Vtank = 125.88 x 14.7/40 = 46.3 gal

Look at it this way...

a 46.3 gal tank at 120 psi gives 46.3*134.7/14.7 = 424 gal free air

at 80 psi you get 46.3*94.7/14.7 = 298 gal free air

Subtracting gives 424-298 = 126 gal free air

 

[milena02]

Hi DJV , thanks .

 

[quark]

This is incorrect procedure or you don't get any advantage out of it, atleast. The main ingredient you forgot is time. The air in the receiver at 40psi is useless to you as you can't actuate the valves, so calculating that value is not necessary.

If you are sizing the receiver equal to one stroke of all the valves together then you have to refill the receiver every time. This is like connecting your compressor directly to the valve actuators.

Go through this link on Air Receiver to get an understanding.

 

[katmar]

quark, the 40 psi is not an actual pressure - it is a differential pressure.

When the air pressure in the accumulator is decreased from 120 psi to 80 psi it must have released 125.88 gallons of free air (but at 80 psi).

Let the accumulator volume be V_acc. Before the valves are actuated the air in the accumulator in terms of free air (i.e. at atm press) is V_acc x 134.7/14.7

After the accumulator pressure is let down to 80 psi the remaining air, in terms of free air, is V_acc x 94.7/14.7

The change in air contained between these 2 conditions is 125.88 gallons i.e.

(V_acc x 134.7/14.7) - (V_acc x 94.7/14.7) = 125.88

V_acc / 14.7 x (134.7 - 94.7) = 125.88

V_acc / 14.7 x (40) = 125.88 (This is where the 40 psi comes in)

V_acc = 46.3



Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[quark]

katmar,

I don't see any problem with the calculation but what I am talking about is usefulness of it.

Say, you decided to go with a 46.3gal receiver and rised the pressure upto 120 psi. So, the free air in the receiver will be 424 gal. Start drawing 19.54 gal of free air for one time actuation and now you will have approximately 404 gal of free air in the receiver. If your valves are to be actuated then your receiver should always contain 46.3*94.7/14.7 = 298 gal of free air, otherwise the receiver pressure will drop below 80 psi.

Now the difference of 424 and 298 is sufficient for approximately 6 valve operations and then you have to replenish air into the receiver. Suppose your valves cycle for 6 times an hour then your receiver capacity is sufficient for one hour. This is the time I was talking about. Further, you have to take the compressor capacity into picture so as to check whether you can accumulate air into the receiver or not.

Secondly, the OP seems to have been misinterpreting the data. When the manufacturer says volumetric flowrate requirement is 0.58 (free air) gal/valve and the actuation pressure is 80 psi then actual volume rate should be 0.58*14.7/94.7 = 0.09 gal.

 

[quark]

Harvey,

Amends to my post. You are right and I should have said air below 80psi is of no use. I thought you were replying to the OPs question as well.

 

[katmar]

quark, yes I was also very confused by the way the volumes were expressed. The values given are just too small to be free air volumes for those size valves. The 0.58 and 2.29 must be the "swept volumes" rather than the volume of free air. Also, the OP has multiplied the 19.54 gallons by ((80+14.7)/14.7) so he must be converting actual volumes at 80 psi to free air volumes.

"A well stated question is 50% of the answer"


Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ARenko]

I took the gallon numbers to be swept volumes as well. I thought about time initially and compressor capacity, but I understood that all the valves are actuated at one time by the charge in the receiver when the machine goes into shutdown due to loss of air supply from the compressor. In this case we don't have to worry about refilling the receiver between actuations. We just have to refill it before starting up once the loss of air supply problem is corrected.