9 replies to this topic

[sheiko]

Hi,

First, let me wish you a happy new year! Be it full of articles, tips and advice from our mentors here in Cheresources Forums!

Now my first question of the year:

I have noticed in several texts on air receivers, a formula to determine their capacity, that is:

V = t*C*Pa/(P1-P2)
With,
V: receiver capacity
t: time required to allow safe shutdown of critical equipment
C: total air requirement (at standard conditions)
Pa: atmospheric pressure
P1: receiver operating pressure
P2: receiver final pressure

I was wondering how is it derived? since i don't succeed in finding the same doing a material balance (the formula i get namely requires the temperature)...

Edited by sheiko, 07 January 2011 - 03:12 PM.

[daryon]

I've seen a similar equation before and used it as a rough sizing for air recievers. I have checked the result against a more accurate method using the ideal gas law as you suggest, and the result are fairly close together. I think this equation is fine for sizing air recievers, although I think the equatuion should acutally be:

V= t.Q / 0.57(P1 - P2)
where:
Q = flow rate in Sm³/h
P1= Norm Op pressure in kPag
P2= Min Air pressure in kPag
t = Storage time in min
V = Volume Required (m³)

I'm not sure how it's derived and I have never sat down and tried (I prefer drinking beer to deriving equations), but the reference I have for this equation is a Stat Oil process design handbook from 1984 - see attached extract for reference.

Attached Files

•  Pages from PHB5.pdf   48.32KB   398 downloads

[Zauberberg]

Sheiko,

Best wishes for the New Year.

The formula used is derived from the Ideal Gas Law (or some form of it), as we are normally interested in calculating the surge volume of air that will be available from the Air Receiver during transition from P1 (initial pressure) to P2 (final pressure). This surge volume must be equal to the C*t, or (Demand)*(Time) as you wrote in your post.

The key thing is to convert total demand (during transition from P1 to P2) into kmole of Air. Then, we need to calculate what is the receiver volume that will contain a certain quantity (kmole) of Air at P1, and certain quantity (kmole) of Air at P2, so that the difference between these two quantities equals our (Demand)*(Time) figure, converted into moles of Air. See attached spreadsheet.

Temperature and Gas constant are excluded from the equation as we normally assume isothermal conditions.

Attached Files

•  Air Volume.xls   20KB   425 downloads

[sheiko]

Dear Daryon and Zauber, thanks.

Zauber, i do understand the material balance. What i don't understand is the origin of the formula i gave in my first post.

When doing my material balance i get:

accumulation = in - out + source
As no IN and no SOURCE, acc = - out
=> dn/dt = -F (n is number of mole and F is molar flowrate)

n = PV/(RT) => dn/dt = (V/RT)*dP/dt (as V and T are constants)
So, (V/RT)*dP/dt = -F

As F = C/Vmol (Vmol is molar volume at a given standard condition)
=> V = t*R*T*C/((P1-P2)*Vmol)

As, Vmol = R*T0/Pa (with T0 is a standard temperature), i finally find:
V = t*C*Pa*T/((P1-P2)*T0)

That is, i find the same formula than in my first post, but with a temperature correction factor equal to T/T0!

Edited by sheiko, 28 January 2011 - 02:20 PM.

[Zauberberg]

Exactly. And, as we have concluded in the previous post, the term (T/To) can be eliminated as the conditions are very close to isothermal. There is certain amount of cooling due to expansion (plus heat transfer with ambient) but this is normally not taken into account, and it does not affect receiver size in significant proportions.

[sheiko]

I believe the isothermal assumption is already taken into account in the derivation by assuming a constant T. So to me, assuming T = T0 would be another assumption (for example, T0 may be equal to 0°C or 15C depending on the Normal or Standard conditions...). Do you see what i mean and do you guys agree with my analysis?

Edited by sheiko, 07 January 2011 - 12:23 PM.

[Art Montemayor]

Sheiko:

You remind me of me – when I was in your shoes. I never trusted any formula or equation that wasn’t backed up by a detailed derivation or acknowledge and recognized source. I still don’t.

In April of 1964, when I was on my way to a foreign assignment in Peru, I needed resources and engineering backup since I was going to be on my own, so I purchased a little handbook in Chicago, Illinois that was titled “Compressed Air Data” by the Compressed Air Magazine Company, New York City. It cost all of $3.00 and was a very popular and recommended book. I still have that little book and on page 147 it reads:

“Time Required to Pump up an Air Receiver
It is often desirable to know the approximate time required for a compressor to pump up a receiver to a certain pressure. The formula below can be used for this purpose.

T = (V) (P2-P1) / [(14.7) (PD) (VE)]

In which,

T = time required to pump up receiver from pressure P1 to pressure P2, in minutes;
V = Volume of receiver, in cubic feet
P2 = Final or desired pressure in receiver, psia
P1 = Initial pressure in receiver, psia
PD = Piston displacement of compressor, ft3/min
VE = Average volumetric efficiency of compressor, (as a decimal).

Note that (PD)(VE) is the actual delivery of the compressor, in cfm. (In those days, every air compressor was a reciprocating type.) If this value is known it may be substituted in the formula for the two terms PD and VE.

These formulae will give only approximate results, since factors such as temperature changes are not considered in the formula and cannot be accurately measured without elaborate apparatus. In most cases these factors would tend to slightly decrease the pump up time from that shown by the formula.”

After analyzing the above equation and explanation, it is obvious that this is the same equation you have – term for term.

The original derivation was done to find the time for pumping up an air receiver. Your use of it is to do the opposite: to find the air receiver capacity in a certain amount of time. Either way, the equation is applicable – within the restraints noted above.

I have used this equation – in Peru, and many other countries since. But what I usually do is to optimize the application, taking into consideration the type, size, and capacity of compressor being used and the storage pressure being applied in the receiver. There are a lot of variables involved. To illustrate what I state, read the attached copy I made of another excellent book that is no longer in print (it seems that most of all the engineering books I have in my library at home are out of print): “Compressed Air and Gas Data” by Charles W. Gibbs of the Ingersoll-Rand Company (1969). Note how Gibbs goes about resolving the problem of selecting an air receiver. The answer is not to be found in just one, simple, and convenient equation. It involves other parameters.

I hope this information is of value and use for you.

Have a Happy and Prosperous New Year.

Attached Files

•  The Engineered Air Receiver.doc   42.5KB   371 downloads

[sheiko]

Note that (PD)(VE) is the actual delivery of the compressor, in cfm.

Ok, i got it.
Many thanks!

Edited by sheiko, 07 January 2011 - 12:28 PM.

[Zauberberg]

I believe the isothermal assumption is already taken into account in the derivation by assuming a constant T. So to me, assuming T = T0 would be another assumption (for example, T0 may be equal to 0°C or 15C depending on the Normal or Standard conditions...). Do you see what i mean and do you guys agree with my analysis?

You have assumed an isothermal process i.e. T = constant, and not a function of time/pressure. Being a constant, it does not affect the final equation.

The term T/To is something different, and - as you derived correctly - it takes into account the difference in volume if air receiver operates at different temperature (different in general, and not during the transition period), where the number of moles of air is different that if receiver operates at To (for any given volume). It means that, if receiver is hotter than To, you'll need an additional air/receiver volume because the required quantity of moles (or Nm3) of air, is less when receiver is hotter than To.

The empirical equation, in my opinion, simply neglects any temperature effects as it is quite common to have air receiver operating at temperatures that are very close to the referenced temperature(s).

[sheiko]

Makes sense. Moreover, ISO 1217 (for air compressors) indicates a reference temperature T0 equal to 20C, which seems close enough to the operating temperature of any air receivers (i guess).
http://www.engineeri...sors-d_848.html

Edited by sheiko, 08 January 2011 - 09:04 AM.