Insane Hydraulics
Bold and audacious blog about fluid power
You all know that this blog is of a more practical rather than theoretical nature, and that most of my advice is of the 'hands-on, do-it-at-your-own-risk' type, but I really wanted to talk about orifices - specifically, about the pressure drop that happens when hydraulic oil flows across them, and I feel that I must begin my orifice talk with theory - i.e. - with figuring out how such a pressure drop can be calculated, and only then move on to the "hands-on" part.
Now, as a hydraulic technician, I use and come across orifices all the time. But do I ever find myself in need of calculating a pressure drop across them? Not really. In fact, I can't even recall the last time I did it - I must have been estimating a flushing flow or something similar. But my point is this - the theory may not seem that exciting (although I secretly think that it totally is), but if you manage to wrap your head around a tiny bit of math, you will "register" important points that will explain a lot of flow-related phenomena you will encounter down the road. Also, in this day and age, when AIs do all the thinking for us, understanding things with our own heads is an increasingly rare skill, so a little bit of mental exercise should always be welcome.
But don't worry - there will be a practical session, too. Next week I will show something really cool, I promise! Now - a note - I am going to be including Math formulae here, and I did my best to display them in the correct format, if, however your browser does not render them correctly, please let me know.
Let us do a concrete example now, just to get the ball rolling. Say, you have an orifice with a diameter of 2 millimeters (0.079 inches) and 30 liters per minute (7.9 gal/min) of "normal" mineral oil flowing across it. How much pressure drop would it introduce? Danfoss has a neat calculator just for that right here - and if you punch in the numbers - you'll get 288 bar. Quite a pressure drop, isn't it? Let us try somebody else, shall we? For example - the best hydraulic calculator in the world from Hawe Hydraulik. It so happens that they also have a very neat calculator for the pressure loss through an orifice - and if you punch in the same numbers (and accept the suggested defaults) you'll get 181 bar. Very different from the 288 bar we just saw. You may also come across charts that would estimate a pressure drop of 300+ bar for the same flow rate and orifice diameter!
What is going on here? Quite obviously, even though these calculators (and the charts) use the same core formula, the big discrepancy in the results comes from the different default values of the so-called Discharge Coefficient (sometimes referred to as the orifice or resistance coefficient). The Danfoss calculator suggests using $C_d = 0.62$, while the Hawe calculator defaults to $C_d = 0.78$. So, let us see if we can figure out what this coefficient thing is by deriving the formula from the basic principles of fluid mechanics.
Let us begin with the good old
Bernoulli's equation states that for an incompressible, frictionless fluid flowing along a streamline, the total energy remains constant. Here's a shortened verision (without the elevation terms - $\rho g h$ - becasue, as "proper" industrial techs, we work with pressures high enough to completely disregard the flimsy pressure fluctuations introduced by the variations in height): $$P + \frac{1}{2}\rho v^2 = Constant$$ Where:
If you still remeber your high school physics lessons, you may recognize this as the sum of the static and dynamic pressures, but I really like to look at this through the lens of energy conservation. Energy conservation makes total sense, doesn't it? But how can pressure be related to energy? Well - it actually is very closely related to energy because pressure represents the potential energy the fluid can release per unit volume. This is how I explain it to myself: if you would want to quantify how much work can be performed by a volume of pressurized fluid, you would need to know its work per volume value, right? And if you look at this further - you'll see that $$Work = Joule (J) = N \cdot m$$ and when you divide work (energy) by volume: $$\frac{Work}{Volume} = \frac{N \cdot m}{m^3} = \frac{N}{m^2} = Pa$$ - you get exactly the units of pressure! So, pressure is energy per volume! (How about that?!)
And what about the second part - the dynamic pressure ($\frac{1}{2}\rho v^2$)? Once again, just like the static pressure, it represents the amount of energy per unit volume, only this time it is the kinetic energy contained in the moving mass of the fluid. The kinetic energy of a mass $m$ moving at speed $v$ is:$$KE = \frac{1}{2} m v^2$$ To convert this to work per volume, we divide by the volume $V$: $$\frac{KE}{V} = \frac{1}{2} \frac{m}{V} v^2$$ And since density $\rho$ is defined as mass per unit volume ($\rho = \frac{m}{V}$), we can substitute $\rho$ into the equation: $$\frac{KE}{V} = \frac{1}{2} \rho v^2$$ - which is exactly the formula for the dynamic pressure!
So both terms in Bernoulli's equation represent different forms of energy per unit volume - static pressure (potential energy) and dynamic pressure (kinetic energy). And since their sum is constant, you can see now why, when the speed of the fluid increases (when it passes through a constricted area) the static pressure goes down.
With that out of the way, let us look at two points: one before the orifice (Point 1) and one right after the orifice (Point 2). According to Bernoulli's equation: $$P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$$ $$P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)$$
or $$\Delta P = \frac{1}{2}\rho (v_2^2 - v_1^2)$$
With our tiny orifices the $v_2$ is orders of magnitude higher than the $v_1$, so we can assume that $(v_2^2 - v_1^2) \approx v_2^2$, and write this formula as: $$\Delta P \approx \frac{1}{2}\rho v_2^2$$
Now, if we are given a flow rate $Q$ (m³/s), we can express the velocity $v_2$ in terms of $Q$ and the orifice area $A_0$ (m²): $$v_2 = \frac{Q}{A_0}$$
And thus, the pressure drop across the orifice becomes: $$\Delta P = \frac{1}{2}\rho \left( \frac{Q}{A_0} \right)^2$$
Great! Now, let us punch in the numbers that we used at the beginning of this article - the orifice with the diameter of 2 mm, the flow rate of 30 l/min, and since we, like all normal people, use mineral oil, let us use the density of 870 kg/m³:
$$\Delta P = 0.5 \cdot 870 \cdot \left( \frac{0.0005}{3.1416 \times 10^{-6}} \right)^2$$$$\Delta P = 435 \cdot (159.15)^2$$$$\Delta P = 435 \cdot 25328$$$$\Delta P \approx 11,018,679 \text{ Pa} \approx 110.2 \text{ bar}$$
Hmm... If you worked with similar orifices and flow rates - you will immediately see that this number is a bit "off" - pretty low for what we usually see in the real world, but don't worry - we'll be correcting this in a minute, but for now the most important thing is to grasp the theory:
In an ideal lossless system, some of the potential energy would be transformed into kinetic energy inside the orifice, and then back into potential energy downstream it. So, the 30 l/min of mineral oil flowing through a 2-mm orifice, would result in a pressure drop of 110 bar in the narrowest point (the term for this point is Vena Contracta - Latin for "contracted vein"), and the pressure would fully recover back to the upstream static pressure once the flow left the constriction point.
Pretty cool if you think about it - with an ideally good orifice we would register no pressure drop across it at all because the pressure would fully recover downstream! But... doesn't this mean that if we had an ideally bad orifice - i.e. - an orifice that would convert all of the dynamic pressure energy into heat, we would be registering the 110-bar pressure drop across it?
Well... with an ideally bad orifice, this would be the case, but, as it turns out, real-life orifices are actually much worse than ideally bad orifices, because the Vena Contracta is much smaller than the orifice. As the fluid rushes toward the hole, it comes from the sides, and creates momentum that prevents it from turning a sharp 90° corner instantly - and the jet continues to shrink after it passes through the hole. It is that, and, of course, the friction, too. This is why we need a coefficient, that will bring our theoretical area closer to the real-life area:
$$\Delta P = \frac{\rho}{2} \left( \frac{Q}{C_d A_0} \right)^2$$ Where
In fact, the discharge coefficient corrects for more than just the reduced Vena Contracta, it also corrects for the friction and other dynamic losses, but in my head that is the most physically intuitive way to think about it.
So, where does the recommended 0.62 value (for a sharp-edged orifice) come from? It's purely empirical, it varies with the orifice size and shape, and as you just saw at the beginning of the article, even the tiniest correction of this factor produces vastly different results, which means that if you, for some reason, need to create a pressure drop of a certain magnitude with an orifice - you will use the formula to estimate the approximate diameter of the hole to drill, and then you will need to test it exhaustively in real life to see if it is any close to what you need.
Now, to finish our "session" off, let us convert the formula so that it contains the diameter of the orifice rather than the area:
$$A_0 = \frac{\pi d^2}{4}$$ $$\Delta P = \frac{1}{2}\rho \left( \frac{Q}{C_d \left( \frac{\pi d^2}{4} \right)} \right)^2$$ $$\Delta P = \frac{1}{2}\rho \left( \frac{4Q}{C_d \pi d^2} \right)^2$$ $$\Delta P = \frac{1}{2}\rho \left( \frac{16Q^2}{C_d^2 \pi^2 d^4} \right)$$ $$\Delta P = \rho \left( \frac{8Q^2}{C_d^2 \pi^2 d^4} \right)$$ $$\Delta P = \frac{8\rho Q^2}{C_d^2 \pi^2 d^4}$$
Notice that I squared every term inside the parentheses, to clearly show two very important things:
And there you have it - the full derivation of the orifice pressure drop formula! I hope you found this useful and interesting. Thank you for reading!
The sources used for the derivation and explanation of the orifice flow formula: