Insane Hydraulics
Bold and audacious blog about fluid power
Do I use a lot of math in my day to day "hydraulic challenges"? No, not really. I repair broken components and I sell new hydraulic stuff to clients that bring over their old hydraulic stuff. This doesn't require any but most basic math. 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. There's more, of course, but it all boils down to "simple". Nothing to write home about, so to speak.
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. 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.
Sounds very simple, especially in this day and age when Mr. 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.
Allow me to explain what I mean by "scattered, incomplete, and incorrect" now. On the day of the writing of this article (26/07/2020), a Google search for "external gear pump displacement calculation" returns several formulae. 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:
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).
The very first formula to show up uses the distance between shaft centers (c), and goes like:
(D² - c²) x gW x π /2
another very respectable source in "hydraulic word" states this same formula as:
(D² - c²) x gW x 1.57 , which is essentially the same (the π /2 is replaced with 1.57)
then you find another formula, that uses the gear's root diameter (d):
((D² - d²)/4) x gW x π
then there's a quite popular formula that uses the body's bore width (W):
(2D - W) x (W - D) x gW x π, which, can also be found as:
((2D - W) x (W - D)/2) x gW x 6 (here the Pi is essentially replaced with 6/2 for some reason)
and 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)/2) x gW x 6.03 (my guess is 6.03 was supposed to be 6.30, which is Pi times two)
So, let us talk about these formulae, shall we? As you can see - there's already confusion. 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. Is measuring the body better than measuring the gear? And how about the center distance?
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:
Formula one gives us 10.99 cm³
Formula two gives us 9.42 cm³
Formula three gives us 9,42 cm³
This adds to the confusion! 9.42 seems to be the "winner" here, but the formula that is most popular on Google and that comes from two very respectable sources gives a different result! Now - I know, I know! - we do not require "millimetric" precision, we need to size the pump to choose a new one with similar displacement. But still - 9,42 and 11 are almost 20% apart, which is a lot! It is clear that something here needs rectification, don't you agree?
Before all, let us take the "geometry" out of the way, and decide once and for all how exactly we can calculate the displacement of our gear unit.
If you look at the chambers that trap and transport the working fluid - you will see that the size of each chamber roughly equals the size of the gear tooth, and since there are two gears transporting the fluid, we can estimate that to calculate the volume that the pump transports per revolution, we need to calculate the volume of the hollow cylinder, formed by the outer and the root diameters (the D and d) of the gear, with the height of (gW):
Now, if we consider the geometry of the two meshed gears, and the fact that with gear pumps the backlash and the play is minimal, the dimensions, used in the above formulae, namely - the root diameter (d), the bore width (W) and the center distance (c) are related, so if you know D and any of the other three, you can calculate all of them:
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):
(πD²/4 - πd²/4) x gW, which simplifies to (D² - d²)/4 x gW x π
since W = D + d + (D-d)/2, you can say that d = 2W - 3D,
so we can calculate our displacement now using the hollow cylinder formula above, like so:
our (D² - d²) is now
D² - (2W - 3D)² =
D² - (4W² - 12WD + 9D²) =
4(3WD - W² - 2D²) =
4(-W² + 2WD - D² - D² + WD) =
4(-(W - D)² - D² + WD) =
4(D(W - D) - (W - D)²) =
4(W - D)(2D - W)
now we add our times π/4 times gW and get the (2D - W) x (W - D) x gW x π
since W = c + D and we know from above that d = 2W - 3D,
we can say that d = 2c + 2D - 3D = 2c - D
and our (D² - d²) is now
D² - (2c - D)² =
D² - (4c² - 4cD + D²) =
4cD - 4c² =
4c(D - c)
now we add our times π/4 times gW and get the c(D - c) x gW x π
This last formula is the correct one, and for our example will also give the result of 9.42 cm³. I am not sure where the so popular (but "not entirely correct") formula from the web came from. As it turns out - for real-life gears, the ratio of the gear's D and d most of the time produces an "accurate enough" result. Still - "geometrically speaking", the most popular formula from Google is wrong.
Now, let us add some common sense to the mix, and discuss why you would use one formula over the other.
The first formula is the best because it is very simple and intuitive, which means that it is easy to remember, and you only require a single gear to be able to size a gear pump. I use it all the time
Sometimes, when gears have an uneven number of teeth, or, maybe the teeth are very worn - it may not be that convenient to measure the D and d, and in this case measuring the body to determine the D and W can be easier. This is when you will use the formula number two. Also if the only thing that you have at your disposal is the pump's body, you will definitely need to use the second formula (the gW can sill be measured because the body will be marked by the gears). I honestly can't remember the last time I used it.
Now - what about the third formula? The one that uses the center distance between the shafts? Well - in my opinion it is useless. You will need two gears to measure the center distance, and measuring center distance between two shafts is an awkward procedure. Since you already have two gears in you hand, why don't you use one of them and apply the formula number one? It is good to know that it exists - and never use it.
So yes, it's all in one place now... Enjoy!
P.S.
As a follow-up to this article I created a small Gear Pump Displacement Calculator App that allows you to choose what you want to measure on a gear pump, and also exports the dimensions and the result in a pdf file. Do use.