A Great Lie About Hydraulic Tanks

how to heat hydraulic oil safelyDuring a recent conversation I had with a member who builds hydraulic power units, this observation was made: “Oil coolers are generally resisted by customers because of initial cost and maintenance, so fitting a bigger reservoir, when possible, is often the preferred solution.”

I recall coming across the very same resistance when I was building hydraulic power units for end-user customers many moons ago. In fact, with some customers their desire was no oil cooler and the smallest tank possible! Yes an oil cooler does increase the initial capital cost of the system. But so too does increasing tank size. At say $3 per liter, an extra 100 liters adds $300 dollars to the initial fill cost. And that’s without considering the fabrication cost of a larger tank, or the additional cost each time the oil is changed.

The economics aren’t as simple as they appear. And on a practical level, the idea that increasing tank oil volume can eliminate the need for an oil cooler is flawed-on all but the smallest of systems, in terms of input power. This is of course a customer (hydraulics end-user) education issue. And the overarching message should be: if your hydraulic machine runs too hot, its reliability WILL suffer. In other words, pay now or pay later.

How much heat can the hydraulic tank get rid of? Short answer: not much. The formula for calculating heat convection from the tank, in SI units, is:

P = deltaT x A x h / 1000

P = heat rejected in kW
deltaT = temperature difference of the oil and air in °C
A = area of the tank, excluding base in m2
h = convective heat transfer coefficient for air in W/m2°C
Use: 12 for a normally ventilated space;
         24 for forced ventilation;
          6 for poor air circulation.

Let’s consider a tank with an oil volume of 200 liters and area (excluding its base) of 1.7 m2, an ambient air temperature of 35°C and an operating oil temperature of 85°C. In a ‘normally ventilated’ space, the theoretical heat rejection of the tank is:

(85-35) x 1.7 x 12 /1000 = 1 kW

For the purposes of illustration, let’s say this calculation is too conservative, so we double the above number. In other words, we expect the 200 liter tank will dissipate 2 kW of heat. Working backwards from this number, if we want the hydraulic system to have installed cooling capacity of 25% of input power-and the tank is the only installed cooling capacity, then the maximum, continuous input power allowable is just 8 kW! (2 ÷ 0.25 = 8).

As you can see, the concept of a big (or bigger) tank in lieu of an oil cooler is not practical in the vast majority of applications. Believing otherwise can be a costly mistake. And to discover 6 other costly mistakes you want to be sure to avoid with your hydraulic equipment, get “Six Costly Mistakes Most Hydraulics Users Make… And How You Can Avoid Them!” available for FREE download here.

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