Measuring Bag Volume 1

[AJDC]

I have a bag of irregular shape connected to an air compressor. I put an initial pressure in the compressor. I release the air into the bag, where the initial pressure drops/stabilizes to an equilibrium pressure. How would I measure the volume of the bag using the Ideal Gas Law, i.e. pV = constant? I think this is harder than you think.

 

[sailoday28]

The pressure in the bag should increase.
Can we assume the bag expands and that this is over a short period of time?
Does the air compressor discharge pressure and temperature stay constant?
Can you measure the equilibrium pressure and temperature within the bag?

With the above questions answered, a model may be set up.

Regards

 

[AJDC]

Yes, bag expands over short period, i.e. I bring compressor to 200 kPa and release valve to let air into bag. It takes about one second fro bag to fill.
Temp. of gas should be constant. The bag is definitely cold after bag is filled. I assume that means the air from the hose that I'm using to fill the compressor with is cold. I don't think the air could cool that fast in a second.
I can get the equilibrium pressure from the compressor's gauge, which I know is accurate. I can't exactly get the temp. of the gas in the bag.

 

[blackwed]

The rub is that PV for the tank isn't constant in your case. There's a different mass of gas in the tank after filling the bag than before. PV=nRT? Maybe you could figure out the number of moles of gas released from the tank based on its new pressure (assuming all other parameters remain constant and that all the gas went in the bag)then plug that into the gas law for the bag at atmospheric pressure. At standard conditions (0 deg C and 1.0 A, a mole of any gas occupies 22.4 liters.

DB

 

[itsmoked]

If you let the tank reach equilibrium temperature (several hours)and read its pressure. Then open the valve. Let the new system stabilize (several hours) and read the pressure again, you should be able to calculate the volume difference in the compressor tank. That difference is in the bag.

Someone check me here.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[AJDC]

Actually, I'm starting to think something is going on with the temp. before and after. After filling the tank, the temp. inside should be room temp., right? But upon releasing the valve quickly the, gas is cold. Maybe temp. does play a role in this.

 

[sailoday28]

If the compressor outlet (or tank inside ) temp and pressure remain constant AND the bag inner surface does not transfer much heat (which is probably true for the short time you are stating---

U = internal inergy Kj
m = mass kg
H = enthalpy kj
p= pressure N/sq meter
V = volume cu. meters

energy balance of the air bag yields

dU = ho dm - pdV

dH - pdV -Vdp = ho dm -p dV ho is stagnation enthalpy of source (tank, compressor) and next will be assumed constant.

integration yields H - Hi - integ Vdp = ho (m-mi)

where subscript i relates to initail conditions in bag.

mh - mi* hi - integ Vdp = ho (m-mi)

mi, hi, pi, should be known. Problem is to integrate Vdp

Let us assuume mi in bag is negligible

mh - integ Vdp = ho* m
or m(h-ho) = integ Vdp

m and integ Vdp will be positive numbers.
Therefore h - ho = positive number and final specific enthalpy will increase within bag.
For a perfect gas, the temp within bag will increase.


Regards

 

[itsmoked]

You bet something is happening to the temperature!! The air shoved into the tank is hot! It feels cold as you let it out due to local cooling from expansion. (but it was hot!!) That is why I say let it sit for long enough to reach room temperature.

Furthermore on opening the tank to the bag there will be cooling of the tank air! So again you must let it sit. This all requires a leak free system BTW.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[quark]

If your compressor is supplemented by a receiver of known water volume, you can do it easily by creating isothermal conditions of compressed air transfer. (I appreciate itsmoked for understanding and putting this thing in simple words)

Using absolute pressure units, if P1 is the pressure of receiver initially, P2 is final pressure and V is water volume, the total volume of air transferred to bag from the receiver (expanded to atmospheric pressure)V_trans is (P1-P2)*V/P_atm.

Now the bag reached a pressure of P2 and contains air V_trans at atmospheric pressure. So water volume of the bag is P_atmxV_trans/P2. (considering unflattened and inelastic bag)

As for the cooing of air, search for Joule Thomson Effect.