#Pump calculations pdf pdf#

Now we add our times ПЂ/4 times gW and get the c(D - c) x gW x ПЂ We can say that d = 2c + 2D - 3D = 2c - D Since W = c + D and we know from above that d = 2W - 3D, Now we add our times ПЂ/4 times gW and get the (2D - W) x (W - D) x gW x ПЂ If you know D and the center distance c: So we can calculate our displacement now using the hollow cylinder formula above, like so:

Since W = D + d + (D-d)/2, you can say that d = 2W - 3D, (ПЂDВІ/4 - ПЂdВІ/4) x gW, which simplifies to (DВІ - dВІ)/4 x gW x ПЂ If you know D and the bore width W: If you know D and d, the volume of our hollow cylinder is: Now let us do some algebra (it's very simple, I promise, but if you don't feel like looking at equations - jump directly to the ' why use one over the other' part below): To make things worse - let us use the formulae on an imaginary pump that has gears with outer diameter D of 4 cm, root diameter d of 2 cm, the center distance c of 3 cm, the body width W of 7 cm, and the gear width gW of 1 cm: Is measuring the body better than measuring the gear? And how about the center distance? We are all experienced and time hardened professionals here, but I can see newcomers puzzled by the different approaches to the way you should take your measurements. So, let us talk about these formulae, shall we? As you can see - there's already confusion. ((2D - W) x (W - D)/2) x gW x 6.03 (my guess is 6.03 was supposed to be 6.30, which is Pi times two)

#Pump calculations pdf pdf#

((2D - W) x (W - D)/2) x gW x 6 (here the Pi is essentially replaced with 6/2 for some reason)ДЄnd also another very respectable source states the same formula (in a designated pdf that is at least 12 years old!) as: (2D - W) x (W - D) x gW x ПЂ, which, can also be found as: Then there's a quite popular formula that uses the body's bore width (W): Then you find another formula, that uses the gear's root diameter (d): (DВІ - cВІ) x gW x 1.57, which is essentially the same (the ПЂ /2 is replaced with 1.57) The very first formula to show up uses the distance between shaft centers (c), and goes like:ДЄnother very respectable source in 'hydraulic word' states this same formula as: I would love to post screenshots, or maybe, direct links to the sources, but posting screenshots will cause copyright issues, and I am not sure that some of these links will stay alive for long, so you will have to take my word for it (or Google it yourself). All of them use the gear width and external diameter, coupled to a measurement made on a single gear, a body or a set of gears, and before listing the formulae, let us first designate the measurements with letters: On the day of the writing of this article (), a Google search for 'external gear pump displacement calculation' returns several formulae.

Google can answer any question, right? Unfortunately, no - the information is indeed out there, but it is scattered, incomplete, and, in some cases even incorrect, so my goal here is to put it all together in one place, and explain why the gear pump calculation is done the way it is done.ДЄllow me to explain what I mean by 'scattered, incomplete, and incorrect' now. Sounds very simple, especially in this day and age when Mr. I am referring to the gear pump displacement calculation - something I do all the time because we sell gear pumps every day, and very often the ones that are brought to us don't have any identification, so the only way to determine their size is to open them, take measurements, and do the math. There's one 'calculation', however, that I have been 'wanting to write home about' for a long time, even with it being as simple as it can possibly be. Nothing to write home about, so to speak. There's more, of course, but it all boils down to 'simple'. A simple power/torque calculation every once in a while (to explain to a client why installing a group three pump in his 5 HP log splitter to make it 'faster' is a bad idea), basic speed/flow calculations, occasional hose sizing (there's a perfect interactive nomogram for that, in case you didn't know), a heat load maybe (to properly size an oil cooler) - in all - pretty simple stuff. This doesn't require any but most basic math. I repair broken components and I sell new hydraulic stuff to clients that bring over their old hydraulic stuff. Do I use a lot of math in my day to day 'hydraulic challenges'? No, not really.