Jump to content

The Fountain Pen Network uses (functional) cookies. Read the FPN Privacy Policy for more info.  To remove this message, please click here to accept the use of cookies


Registration on the Fountain Pen Network

Dearest Visitor of the little Fountain Pen Nut house on the digital prairie,

Due to the enormous influx of spammers, it is no longer possible to handle valditions in the traditional way. For registrations we therefore kindly and respectfully request you to send an email with your request to our especially created email address. This email address is register at fountainpennetwork dot com. Please include your desired user name, and after validation we will send you a return email containing the validation key, normally wiithin a week.

Thank you very much in advance!
The FPN Admin Team






Photo

Cartridge: How To Determine The Maximum Size Of The Hole?

cartridge fluid dynamics formula

  • Please log in to reply
3 replies to this topic

#1 MrkLch

MrkLch

    NOS (New Old Stock)

  • Member - Silver

  • PipPip
  • 15 posts

Posted 30 July 2018 - 19:02

Does anyone know if there is a formula for calculating the maximum hole size of an ink cartridge without the fluid coming out?

In this site I proposed the problem but nobody could find a solution. At the moment.

https://physics.stac...er-stay-in-an-i

Sketch203213430.png

Sponsored Content

#2 inkstainedruth

inkstainedruth

    Ancient Artifact

  • Member - Gold

  • PipPipPipPipPipPipPipPipPipPipPipPip
  • 16,920 posts
  • Flag:

Posted 30 July 2018 - 21:53

I'm not sure what you're asking.  Cartridges generally have a seal that gets punctured when they're installed and the size of the hole depends on what brand(s) take that particular style/brand of cartridge.  There are lots of threads in which there are discussions about which pens take International Standard cartridges and converters, for instance (vs. proprietary brands such as Parker).  If you're trying to refill a cartridge and then store it for future use, some people use glue guns; I used to use 100% silicone grout (the stuff used to re-grout around your bathtub).  If you're trying to design a cartridge (say, to be made on a 3D printer, again, it's going to depend on the brand of pen and what size nipple is on the back of the feed.

If you're trying to calculate how much capillary action an inverted open cartridge is going to have, though, I'm betting "not much"....

Ruth Morrisson aka inkstainedruth

 

edited for typos


Edited by inkstainedruth, 30 July 2018 - 21:54.

"It's very nice, but frankly, when I signed that list for a P-51, what I had in mind was a fountain pen."

#3 JollyCynic

JollyCynic

    Extremely Rare

  • Member - Gold

  • PipPipPipPipPipPip
  • 263 posts

Posted 31 July 2018 - 01:33

This question is answered all over the Internet, but in a slightly different form.  Instead of looking for the stability of water's surface tension in ink cartridges, look for the stability of water's surface tension in straws by diameter.  Keep in mind that the answer will vary by the force of the ink (its mass acted on by gravity) and the temperature, which affects the ink's surface tension.



#4 Parker51

Parker51

    Collectors Item

  • Member - Gold

  • PipPipPipPipPipPipPipPip
  • 1,294 posts
  • Location:North America, U.S.A., Ohio, Delaware Cty

Posted 31 July 2018 - 02:45

As I recall, you will need to assume that the equation is in equilibrium in regard to temperature, and calculate the balance of forces. SH will vary based on more than one factor. One will be the mass of the ink, and of course, the force of gravity. Calculated, that will tell you how much force is pushing the ink down. Surface tension is a major force holding the ink in place. Another factor is the air pressure outside the cartridge, which times the area of SH tells you how much force is pushing up. And still another will be the air pressure within the cartridge times the size of the area of the top of the column gives you the additional force pulling the liquid up, as well as the surface tension inside the cartridge. This means is that the answer will vary based on more than one unknown. Now the question is do you have enough information to do the calculations? Let's see. You have the air pressure outside the cartridge and inside it. Surface tension can be looked up to calculate for water, but you don't know the area of either of the places where it is in effect.You know the force of gravity. You do not know the mass, or the area of the column. But, do you need to have all this information? No, because the limiting factor for the size of SH is based upon reality, not but based on a set of theoretical unknown relationships which need to be calculated. The theoretical largest SH is the same as the diameter of the column, if the internal pressure is low enough and the mass of the ink is low enough given typically experienced air pressures. Increase the mass of the ink and eventually you will get a column of ink which will have a mass great enough to overcome the forces holding it in place. The column would get quite large, and you put no maximum on your cartridge size. Reality is however that a cartridge can not be bigger than a regular size pen, so the ink will not flow out based upon mass. However, if the external pressure drops, say by flying higher and higher with eventual exposure to space, then the air pressure would get so week that the ink would flow out. But again if we assume external pressure is that at sea level and remains constant, well, you get the picture, you could assume some variables are actually constants and do your calculations. It is my experience that actually the largest size of SH is the diameter of the cartridge, for even a nearly full cartridge. The mass of ink is typically so low that the forces within the cartridge and outside the cartridge, unless surface tension is mechanically broken, is sufficient so as to hold the ink in place. It really isn't a physics question, it is an engineering question, with a touch of practical fluid mechanics, but if you like multi variable math calculations, start by taking some real measurements to get your theoretical maximums and minimums, calculate based on them and find out if there is a way to create an ink massive enough that it's mass can overcome the other forces and cause SH to be less than the diameter of the cartridge, but I doubt it, unless you find a way to put some very massive atoms into your ink. Who knows, maybe it can be done with depleted uranium, but then how do you keep the surface tension the same as water? Gosh I miss Engineering, it has been too long, 38 years too long.

Edited by Parker51, 31 July 2018 - 02:49.






Also tagged with one or more of these keywords: cartridge, fluid dynamics, formula



Sponsored Content




|