A definition of heat –
If we look to our friend Noah Webster, for a definition of heat, he would tell us that” Heat is a form of energy associated with the motion of atoms or molecules in solids and capable of being transmitted through solid and fluid media by conduction, through fluid media by convection and through empty space by radiation.” And he might mention something about animals and certain times during the year when they are friendlier to each other than others, but we will leave that definition to the biologists.

Anyway, for our use in hydraulic applications, we need to translate the definition above into a more workable statement that will help us better understand the physics behind this phenomenon called heat. So here it goes “Anytime fluid flows from a high pressure to a lower pressure without producing mechanical work output heat is generated.”

So, now that we have a definition for heat lets look at some of the things that occur in a hydraulic system that would cause heat to be generated.



  • Flow restriction or throttling – The use of flow controls, proportional, reducing, relief, reducing/relieving, counterbalance and servo valves all create a pressure drop in order to do their job.    
  • Excessive flow Velocities – Incorrect sizing of fluid conductors can cause the generation of heat. For example, with ½ inch OD pipe, a flow rate of 10 GPM generates heat at the rate of about 25 BTU/FT-HR. Doubling the flow rate to 20 GPM increases heat generation 8 times to about 200 BTU/FT-HR. Here are some rules of thumb when sizing hydraulic conductors velocities.

               PUMP SUCTION LINES should be sized for 2 – 4 FT/ S
               RETURN LINES should be sized for 10 – 15 FT/S
               MEDIUM PRESSURE LINES (500 – 2000 PSI) should be sized for 10 – 15 FT/S
               HIGH PRESSURE LINES (3000-5000 PSI) Should be sized for 20-30 FT/S

  • Slippage in pumps – As pumps wear, the internal leakage or “slippage” increases.  On fixed displacement pumps this leakage flows from the high-pressure outlet back through the pump to the low-pressure inlet.  In a pressure compensated pump this flow is forced out through the case drain.  As this occurs fluid is taken from a high pressure to a low pressure without doing any mechanical work thereby creating heat.    
  • Internal leakage in valves – As valves wear they develop leakage paths that allow high-pressure oil to leak to a low-pressure port creating heat.
  • Gas-filled accumulators – Pulsating accumulators may develop high pressures on the gas side.  This heat can transmit back into the oil raising the temperature and creating a hot spot in your hydraulic system.
  • Non-regenerative release of potential energy – When a load is lifted hydraulically, potential energy is stored in the load.  Release of the load usually involves non-regenerative throttling, which generates heat.


Effects of heat on the systems
Heat has many detrimental effects on the hydraulic system components.  But the most detrimental effect of heat is the breakdown of the oil.  Oil temperatures should be maintained at 120F for optimum performance, and should never be allowed to exceed 150F.  At high temperatures, oxidation of the oil is accelerated.  This oxidation shortens the fluid’s useful life by producing acids and sludge, which corrode metal parts.  These acids and sludge clog valve orifices and cause rapid deteriation of moving components.

The chemical properties of many hydraulic fluids can change dramatically by repeated heating/cooling cycles to extreme temperatures.  This change or breakdown of the hydraulic media can be extremely detrimental to hydraulic components, especially pumping equipment.

Another effect of heat is the lowering of the oil’s viscosity and its ability to lubricate the moving parts of the pump and related hydraulic equipment effectively.

Useful heat calculations
HP = GPM X PSI / 1714    

HP = horsepower            GPM = gallons per minute            PSI = pounds per square inch    

1HP = 2545 X BTU/hr            HP X 746 = KW                    

KW X 3413 = BTU/hr            KW X 1341 = HP            

Dissipation of heat from steel reservoir

HP (heat) = 0.001 X ≅T X A

A = the surface area of the reservoir in sq. ft.  The surface area of the bottom of the reservoir can only be used in the calculations if the tank sits 6.0 inches off of the ground.

≅T = the difference in degree Fahrenheit between surrounding air and oil temperature inside the tank.

Recommendations to reduce heat generation
Unload the pump during intervals when pressure is not required

~    This can be accomplished by adding a solenoid vented relief valve on fixed displacement pumps and a solenoid vented control on pressure compensated pumps.  This will remove the high-pressure component of the definition above.

Use the largest reservoir that is practical for your application.  

~    In order to gain the most surface area or cooling capacity from the reservoir take into consideration the calculations listed above.

Set the main system relief to the lowest value that will still do the work.

~    This setting is usually 200 –250 PSI above the maximum pressure needed in the system to do the work.

Place the tank in a location that will give it assess to the greatest amount of airflow.  

~    By enclosing the tank you greatly reduce the tanks capacity to radiate heat and in some applications can cause the system to prematurely overheat.

Install or designs heat exchangers into the system will help remove excess heat.  

~    Heat exchanges can be used to remove the excess heat in a hydraulic system.  The implementation of heat exchangers has many variables that need to be taken into account.  Rules of thumb when sizing a heat exchanger is as follows:

  • Simple circuit with minimal valves: 25%
  • Simple circuit with cylinders: 28%
  • Simple circuit with fluid motors: 31%
  • Hydrostatic transmissions: 35 – 40%
  • Servo based systems: 60 – 75%
  • Low-pressure fluid transfer systems: 15%

Multiply the input horsepower (motor hp) by the percentage listed above that best describes the system parameters.  For example, if your system is a simple circuit with fluid motors and has an electrical motor input horsepower of 30hp:

30hp X 0.31 = 9.3hp  

The tank needs to dissipate at least 9.3 horsepower or the system will overheat.  Another rule to keep in mind is if your system pressure is above 1000 PSI and your tank is sized for 3 times or less pump output you WILL need a heat exchanger.


There are many more aspects of thermal characteristics within a hydraulic system than this paper was meant to cover.  With this information; however, you should be able to make educated decisions when working with an existing system or new design in order to combat heat generation.  With this information you should also feel comfortable calling a specialist to discuss ways to minimize the heat you may experience in your system.  

When in doubt, consult your local fluid power professional.

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