Thursday, December 17, 2015

homework and exercises - Is Volumetric flow rate constant?


I'm working on a project and I need to know something. My project is about a toy CD hover craft. It's a balloon, connected to a CD via a tube or a pipe, like this:


enter image description here


I want to know that is the volumetric flow rate in the pipe always equal or not?



Answer



That's an awesome project - I did something like that many, many years ago (using a carefully sanded wooden disk - CDs had not yet been invented).


The flow rate depends on the size of the pipe and the pressure in the balloon.


Now interestingly, a smaller balloon has a slightly higher pressure than a large balloon - see for example my answer to this earlier question.


The flow rate through a pipe is a function of the length and area. As the pressure from the balloon will change with its size, so will the pressure and the flow rate.



There is another point though - the effect of the spacing between the hovercraft (CD) and the surface it hovers over. Now as long as the pipe provides a more severe constriction, the spacing of the disk to the surface will be less important. This is the principle behind using a large series resistor to create a poor man's constant current source: if you have for example a battery in series with a 10k resistor, the current it produces will be almost the same whether you then connect a 1 ohm or a 10 ohm resistor across it.


If the balloon is bigger, the time that the toy can hover will increase - by a surprisingly large amount. Using the result from my other answer that pressure (and thus flow rate) scales with $1/r^2$, and volume scales with $r^3$, then time (which is the time it takes for the balloon to deflate) will scale with $r^5$. In other words, a bigger balloon will allow for much longer floating time, assuming that the flow rate is proportional with the pressure (and that the pressure of the fully inflated balloon is still large enough to keep the craft floating). That's an interesting result I was not expecting.


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