Dear JimLee,

I think that it depends on 2 factors:

1)the max air comsumption

2)your air compressor size

Best Regards

■

Dear JimLee,

air comsumption is Function(time),air compressor is constant,so the tank size=MAX{¡Æf(t)-Ct}

Best Regards

■

Air receivers store compressed air for peak demand in excess of compressor flow rate. They increase the cooling of the compressed air and collect possible residual condensate and oil droplets.

The pressure variations in the air net will be equalised and short cycle loading and unloading of the compressor minimised.

So you have to work out the maximum system demand & its duration to know what mass of air you must store.

The equations are the normal Gas Laws. ■

Dear JimLee

Proceed as follows:

-Calculate the required air volume of the installation at a standard conditions.

Q-install. in standard m3/min

-Determine the installation pressure

p-install.(absolute)

-Then, decide the pressure fluctuation in your air tank as

p-max(abs) and p-min(abs)

-Then, decide the usage factor of your compressor (alpha)

-Then decide the ON/Off cycle time of your compressor ( T in minutes)

Then calculate :

Tank volume =

(273 + t-tank) * ( 1 – alpha ) * T * (Q-install.)

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

(273 + t-install.) * (p-max – p-min) * p-install

Your compressor volume Q-compr. = Q-install / alpha * (273 + t-amb) / (273 + t-install).

Example :

Q-installation = 50 nm3/min

Alpha = 0.7

T = 10 minutes

p-installation = 8 bar(a) = 7 bar(o)

p-max = 10 bar(a)

p-min = 8 bar(a)

t = 10 degr. Celcius

Calculated is :

Tank volume = 9.38 m3

Q-compressor = 71.43 nm3/min

If you are interested, I can send you a simple spreadsheet calculation.

Best regards ■

it helps me a lot,TEUS,would you like to give me a copy o f your calculation?

my email:zpewljg@yahoo.com.cn ■

Dear Teus,

I don't understand your equation for the tank volume calculation,could you clarify it?

Regards

■

Dear Jack,

Just substitute the figures into the equation :

Tank volume =

(273+10) * (1 - 0.7 ) * 10 * 50

-------------------------------------------- = 9.38 m^3

(273 + 100 * (10 - 8) * 8

Q-compressor =

50 *(273 + 10)

------------------------ = 71.43 m^3/min

0.7 * (273 + 10)

That is all

all for now ■

dear JimLee,

let me know your e-mail address, so that I can

send you the file :

"compressed-air-vessel.xls"

hear from you ■

Dear Teus,

you misunderstood what i said.what i mean is that how you deduct to this equation.Or is the equation from a referance document?

I don't understand WHY (p-max - p-min) in the equotion.

COULD YOU EXPLAIN IT?

thanks

■

Dear Jack,

Indeed, I misunderstood your question.

The compressor switched off at 10 bar(a) and switches on, after the tank pressure has dropped to 8 bar(a)

The larger this proportional band, the bigger the tank should be.

In my example the ON-time of the compressor is 7 minutes and the OFF time is 3 minutes

The cycle time of the compressor is then 7+3=10 minutes

The usage factor, then is 7minutes/10 minutes = 0.7

BR ■

Dear Teus,

two questions about your equotion:

1)when you design a new system ,how did you get the value of usage factor ? Depend on your experience?

2)why the (Pmax-Pmin) is inverse proportional to the surge tank volume?

BEST REGARDS

Jack

■

Dear Jack,

The usage factor should be discussed with the compressor manufacturer.

F.i. piston compressor usage factor = 0.6 – 0.8

Oil filled screw compressor usage factor = 0.8 – 0.9

But, also the fluctuations in air demand of the installation are important.

Peak demands can be delivered at a higher usage factor for a limited period of time.

The difference between p(max) and p(min) determines the amount of available air without the compressor running.

If you take half the pressure drop p(max) - p(min), you need twice the volume, to have the same amount of compressed air at your disposal for a certain period of time

If you let me know your e-mail address, I will send you the simple spreadsheet calculation.

Have a nice day ■

Dear JimLee

Can you specify your question about “the right volume of the blow vessel”.

? blow vessel

? system

Best regards ■

Dear JimLee

May be you posed a general question regarding the influence of the blow vessel size on the overall capacity off a pneumatic conveying system.

A blow tank has to be :

1)filled

2)pressurized

3)emptied

4)pipeline purged

5)valves switched

The items 1), 4) and 5) are constant and not related to the blow tank size (tconstant)

The items 2) and 3) are proportional to the blow tank size.

Tpress = C2 * Tanksize

Tempty = C3 * Tanksize

The vessel content = C1 * Tanksize

The overall capacity becomes then :

C1 * Tanksize

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

(tconstant) + (C2 + C3) * Tanksize

From the equation you can learn:

-The bigger the blow vessel size, the higher the capacity (and the investment)

-When tconstant << (C2 + C3) * Tanksize, then the overall capacity reaches its maximum.

-Tank size is a trade off between capacity and investment.

Have a nice day ■

Dear Teus,

My email:yuwenjie2002@msn.com

thanks

■