Insane Hydraulics

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Magic Number Two or Thoughts About Specific Heat of Hydraulic Oil


After I finished writing this post it occurred to me that a short disclaimer is needed, since I was born and raised in the SI units land, and measure mass in kilograms and temperature in C, while our American friends find it more intuitive to use pounds and Fahrenheit. Well - the point of the article still stands - knowing the magic number is very helpful, so if you last through the post - I challenge you to come up with an imperial magic number of your own. Don't forget to let me know what it is!

Also - bear in mind that the numbers I'm using here are approximate values for normal operating conditions.

06 October, 2019

I overhauled an excavator pump the other day, and two days later the client called and complained about overheating. When questioned about the condition of the oil cooling system the man was adamant it was one hundred percent operational, so I went there to check things out.

When I arrived, I asked the owner about what made him so sure that the cooling system was OK. He said - "Well, we can do some digging and you can see for yourself! I've never seen the cooler work so well!" - " All right, let's see it!"

After about half an hour of digging, when the hydraulic oil got pretty hot, the man opened the radiator door and put his hand on the oil line returning from the radiator to tank: "Look!" - he said - "It's cold! Now check out the inlet line" - and he put his hand over the line that was feeding the radiator - "Burning hot!" - and then continued - "See? Inlet hot at the outlet cold - the radiator is cooling the heck out of the oil! You must've done some bad adjustment to the pump or something for her to run that hot!"

It took some time and effort to convince the man that the fact that the return line of the radiator is so cold meant quite the opposite to what he was stating and that the cooling system was, in fact, not working at all! A short investigation showed that the radiator check valve broke and was by-passing most of the return flow directly to the tank.

I find it interesting that intuitively people tend to think that there should be a big temperature gradient between a radiator's inlet and outlet for it to work properly. The main job of a cooler is to take the heat out of the system, i.e. to heat the environment via the heat transferring fluid, be it air or water - and you can't heat the environment with a cold radiator! In that case, the radiator indeed was super efficient in cooling the tiny partial return flow that was passing through it, but this didn't mean that it made it an efficient heat removing system.

Which brings me to the point of this article - specifically - the specific heat of oil (pun intended), and how it is important to know the "magic number 2" and use it in our hydraulic life to asses temperature gradients in heat exchangers and other things.

Number two?! Yes, number two! Very easy to remember, by the way, because people are born with two eyes, two arms, two legs, and two ears. And, transferring this number to hydraulics - it is also the approximate amount of joules (two joules) that it takes to increase the temperature of one gram of average mineral hydraulic oil by one degree K (or one degree C, or 1.8 degrees Fahrenheit). Also, inversely - this is the amount of energy the oil will give out if it's temperature is lowered by one degree.

You may ask - well, why should I care? And I agree - nobody working with hydraulic systems cares about measly grams and joules. What's a joule anyways? One joule per second is one watt - my grandmother's flashlight has more power than that! I spit on your number two!.. Wait, this one didn't come out right, did it?

Anyhow - if one joule per second is one watt, a thousand joules per second is one kilowatt. So, if we transform our grams in kilograms, we'll see that it takes two thousand joules to raise a temperature of a kilogram of average hydraulic oil by one degree C.

Let us go a little further now, and let us express the flow through our cooler in kilograms per second instead of liters per minute. If we have a flow of one kilogram per second (which for hydraulic oil is about 70 liters per minute) - and its temperature is reduced by a single degree C as it passes through the cooler - it means that the cooler is already dissipating the equivalent of 2 kW.

Now, let us consider a "decent" medium-sized mobile hydraulic system, with a 70kW (100 HP) prime mover, which "employs" a 140 l/minute pump running at 280-ish bar. I particularly like the flow of 140 liters per minute because it equals 2 kilograms per second, which makes calculations easier. So, if we imagine that our flow of 2 kg per second drops one degree C as it passes through a cooler - it's dissipating 4kW. And if our cooler can drop the temperature by 4-5 degrees - it's dissipating 16-20 kW of heat load, which is about 25% of the power installed and sounds about right for a reasonably efficient hydraulic system of this size.

If we take into consideration the fact that normal running temperature of a mobile hydraulic system is around 70 C, which is burning hot to touch, we'll see that "hand" temperature measurement can't help much, because if our cooler has its outlet pipe at 65C it is still burning hot to touch, so it may seem that the cooler is "doing nothing", even though it's providing an absolutely adequate cooling.

See why putting numbers on things is great? - It gives you concrete clues when you need them, plus - whenever you pull out a pen and a sheet of paper, oh wait - this was 20 years ago - I'll rephrase - whenever you fire up the calculator App - you instantly become more "professional" in the eyes of bystanders. It's even better with hydraulic systems, where 99 times out of 100 all you need to do is ballpark a value to reach a conclusion.

So - if you remember that changing a temperature of a kilogram of oil by one degree takes about two (two!) thousand joules and a thousand joules per second equals one kilowatt - you can perform all sorts of thermal calculation on the spot!

Another example - the temperature of a 100 liters of oil rose by 30 degrees C in 5 minutes - it means that it was injected with 2000 times 100 times 30 joules in 300 seconds, or, in other words, was heated with an equivalent of a (2000 * 100 * 30)/300 = 20 kW heater. Not a bad way to ballpark thermals, don't you think? And, once again, all you need to remember is the magic number two (two thousand joules per kilogram Kelvin) and have an idea that a thousand joules per second equal one kilowatt, and yes - don't forget to convert the flow into kilograms per second.

A couple of interesting facts - water has twice the specific heat of oil - about 4 j/gK, which means that with water-oil coolers you will see even less of a temperature differential on the water side. Steel has a specific heat of only 0.5 j/gK, which is four times less than that of oil. So, when you're ballparking a system's thermal load by evaluating its temperature increase in a given time - look at the main "heavy" components, "guesstimate" their weight, divide it by four and add it to the amount of oil that's in the system for a better precision (or just throw an extra 25 percent on top of the oil mass - should land about right...).

Remember - the magic number is two! Two!