# How much time to pressure balance two air chambers

In summary, a smooth flexible tube with a diameter of 1mm and a length of 1 foot is connected to a tank via an open valve. The tube is peristaltically compressed, forcing all the air into the tank and pressurizing it to twice the normal pressure. Once the valve is closed, the tube is allowed to naturally reflate to atmospheric pressure. When the valve is opened, there is a fast release of air from the tank into the tube and out the open end. The time it takes for this process to occur is dependent on factors such as the speed of sound in air, the resistance of the tube, and the increase in pressure for a given volume of added air.
A smooth flexible tube 1mm in diameter and 1 foot in length holds air at normal / atmospheric pressure (14.7psi).

the tube is open at one end. the other end is connected to a tank by an open valve.

the tube is peristaltically compressed so all the air in the tube is forced into the tank.

the tank is dimensioned so that the air in the tank is pressurised to twice normal pressure (29.4psi)

the valve to the tank is now closed trapping the air in the tank .

the tube is allowed to naturally reflates to atmospheric pressure.

the valve to the tank is now opened:

Clearly the pressure differential between the tank and the tube results in air within the tank venting into the tube and out the open end - until the pressure in the tube and the tank stabilise at normal pressure.

MY QUESTION

Clearly that process is fast. But how fast?

I would like to know how much time that takes.

Can anyone tell me?

Thanks

Hi Adam and welcome to PF.
Could you clear up some questions about the experimental details? Is the tank pressurised just by the volume of air that was originally in the tube? that would mean its volume is similar to the tube volume. Does the temperature of the tank return to room temperature after it's been pressurised? For the sake of the experiment, is the tube 'rigid' or are you allowing it to be 'rubber' (implied by the peristaltic method for emptying the tube into the tank)?
The problem is not uncommon and it has various different facets. The time taken for an initial disturbance to make itself felt at the other end of the tube will be related to the 'speed of sound' in the air (as in post #2) but there are other factors. The pressurised tank will be discharging its excess pressure through the tube and the drag of the air will provide a back pressure. The situation would be analogous to a Capacitor discharging through a resistor with an exponential decay where.
Vat time t = Vstart exp(-t/RC)i
The basic maths is the same and t's a simpler situation.
The time constant (time for V to fall to 1/e) of the start V will be RC. You now want an equivalent quantity for R and C. R, the resistance of the tube, will depend on the radius and the length and the equivalent to C will be the the pressure increase for a given volume of added air.
I will leave that with you to see if it makes sense so far.

sophiecentaur said:
Hi Adam and welcome to PF.
Could you clear up some questions about the experimental details? Is the tank pressurised just by the volume of air that was originally in the tube? that would mean its volume is similar to the tube volume. Does the temperature of the tank return to room temperature after it's been pressurised? For the sake of the experiment, is the tube 'rigid' or are you allowing it to be 'rubber' (implied by the peristaltic method for emptying the tube into the tank)?
The problem is not uncommon and it has various different facets. The time taken for an initial disturbance to make itself felt at the other end of the tube will be related to the 'speed of sound' in the air (as in post #2) but there are other factors. The pressurised tank will be discharging its excess pressure through the tube and the drag of the air will provide a back pressure. The situation would be analogous to a Capacitor discharging through a resistor with an exponential decay where.
Vat time t = Vstart exp(-t/RC)i
The basic maths is the same and t's a simpler situation.
The time constant (time for V to fall to 1/e) of the start V will be RC. You now want an equivalent quantity for R and C. R, the resistance of the tube, will depend on the radius and the length and the equivalent to C will be the the pressure increase for a given volume of added air.
I will leave that with you to see if it makes sense so far.

Thank you for your reply. I am very poor with equations and so perhaps it would help if I reframed the question in simpler terms:NOTE: This is a question where all the elements are ideal; there is no need to consider real world issues e.g. temperature / flow resistance etc.

A smooth tube 1mm in diameter and 1 metre in length holds air at normal / atmospheric pressure (14.7psi).
the tube is open at one end to the atmosphere. The other end of the tube is connected to a rigid tank via a valve which is closed.
The valve is 1mm in diameter.
The tank volume is half that of the tube.
The air pressure in the tank is twice that of the tube (29.4psi)
the valve to the tank is now opened:
(assume that this opening is instantaneous)
Because the tank holds air at twice atmospheric pressure one half of this air will vent into the tube.
The air already in the tube will then be displaced into the atmosphere.
This air movement will last for as long as it takes for the air pressure in the tank and the tube to settle at 14.7psi.
My assumption is this process would take only fractions of a second.
My question is: can this time be calculated.
(in the abstract – ie ignoring tube friction etc.)

And if so - what is the answer?

## 1. How do you determine the time needed to pressure balance two air chambers?

The time needed to pressure balance two air chambers depends on several factors such as the volume and pressure of the air chambers, the size and flow rate of the connecting tube, and the type of valves used. A mathematical equation, known as the Bernoulli's equation, can be used to calculate the time needed for pressure equalization.

## 2. How does the volume of the air chambers affect the time needed for pressure balance?

The larger the volume of the air chambers, the longer it will take to pressure balance them. This is because there is more air that needs to be transferred between the two chambers to reach equal pressure. Therefore, larger air chambers will require more time for pressure equalization compared to smaller ones.

## 3. Can the flow rate of the connecting tube affect the time needed for pressure balance?

Yes, the flow rate of the connecting tube can greatly affect the time needed for pressure balance. A larger flow rate means that more air can be transferred between the two chambers per unit of time, resulting in a faster pressure equalization. On the other hand, a smaller flow rate will result in a longer time for pressure balance.

## 4. What type of valves are best suited for pressure balancing two air chambers?

The type of valves used can also impact the time needed for pressure balance. Ball valves or gate valves are commonly used for pressure balancing air chambers as they allow for a more precise control of air flow. Other types of valves such as butterfly valves or globe valves may also be used, but they may result in longer pressure equalization times.

## 5. Is there a way to speed up the time needed for pressure balance?

Yes, there are a few ways to speed up the time needed for pressure balance. One way is to increase the pressure difference between the two chambers, as this will result in a faster flow of air between them. Another way is to increase the flow rate of the connecting tube, either by using a larger diameter tube or by increasing the pressure on one side of the tube. However, it is important to note that these methods may also result in higher energy consumption and should be carefully considered.

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