# The relation between Internal Pressure and flow rate in a balloon [duplicate]

This question already has an answer here:

I'm working on a project. We have two balloons, connected by a hose with a valve.We know when we open the valve, by according to internal pressure of each balloon, air goes form the higher pressure balloon to lower pressure balloon. I want to know the relation between internal pressure of the balloons and the flow rate between these two balloons. Pay attention, I don't want the air transfer direction, I want the FLOW RATE.

Thanks.

## marked as duplicate by John Rennie, ACuriousMind♦, Kyle Kanos, Brandon Enright, JamalSJan 30 '15 at 6:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

If you know electric stuff, this will help: Pressure is like voltage, air flow is like current. A pressure difference in the two balloons is like a voltage difference. The tube and valve connecting the balloons has a "resistance" to air flow. The air flow between the baloons will be $F = (P2-P1)/R$ where $F$ is the air flow rate, $P2$ and $P1$ are the pressures in the two balloons, and $R$ is the resistance to air flow of the tube and valve between the two balloons.

What is $R$? Exercise left to the reader. Given what I'm telling you, you can think of a few ways of measuring it, and even determine what its units are. But I will point out the following:

1. $R$ is higher for a thin tube, lower for a wide tube.
2. $R$ is higher for a long tube, lower for a short tube.
3. If the valve has large uninterrupted room for airflow through it, it won't increase $R$ much, if it is more restrictive than the tubes, it will dominate $R$.