FluidPower.Pro

How to select a hydraulic cooler

At this article, I show my vision on how to:

  • determine the value of the heat needs to be rejected from the system;
  • calculate and select the right cooler size.

…and will provide an example of cooler calculation and selection.

What is a cooler and what is a heat exchanger?

The hydraulic cooler is one of the heat exchangers type. But, what is a heat exchanger? The best definition of the heat exchanger is:

Heat exchanger is a device that transfers heat between two fluids.

The simple sentence but a very good description, because fluids can be either oil, air, water, etc., and because transferred heat can be for either cooling or heating target.

There are two most popular types of heat exchangers in hydraulic systems:

Plate heat exchangers

Plate heat exchanger

Plate heat exchanger

This type has the best value of efficiency/reliability and designed for both cooling and heating applications and a very good for low-viscosity fluids. Pairs of plates can be removed individually for maintenance, cleaning, or replacement. Another advantage of the plate is exchangers is their low initial cost, as well as easy and inexpensive operation.

Fan radiators (air-cooled)

Fan radiators (air-cooled)

Fan radiators (air-cooled)

This type is the only option where water is unavailable or expensive for a delivery. From benefits: low maintenance and operating costs and the only option for oil cooling in mobile applications.

Theory of cooler selection

Heat balance

Heat exchanger selection starts from the calculation of heat, generated by a hydraulic system, that needs to be rejected by the cooler. The right estimation of this value can keep balance of heat in a hydraulic system and prevent from both overcooling and overheating of a hydraulic system. It lets hydraulic components work in ideal temperature/viscosity conditions with maximum performance what makes hydraulic system reliable and efficient.

So, the question #1: where the heat comes from in the hydraulic system?

The short response is pressure losses:

  • Friction between fluid and conduit’s wall transfers part of energy into the heat and this is why making pressure drop on the conduit.
  • Cavitation in all hydraulic components (pumps, valves, fittings, etc.) due kinetic energy transformation generates major heat in the hydraulic system. Consider all components in the hydraulic systems as orifices, with pressure drop across equal to generated heat.
Elbow at the cylinder generates heat

Elbow at the cylinder generates heat

Here is a good example: Take a look at the infrared picture at the right – the simple elbow at the cylinder generates heat due cavitation caused rapidly changed flow direction.

The next question is how to evaluate the required heat needs to be rejected?

This is a really good question because there no specific formula to calculate the exact value. As you understand, heat is generated by absolutely all components in the system, and even if you try to calculate the major parts it can take a while… Therefore, we usually take the next values just from historical experience:

20-25% of input power for close loop systems.

25-30% of input power for open loop systems.

For example, we have a hydraulic system with both open loop and close loop circuits. The power, needs to be provided to open loop is 90 HP and power, needs to be provided to open loop is 80 HP. In this case, we can assume generated and required to be rejected heat around 45..47 HP:

## p_{heat}=P_{op}\times\frac{30}{100}+P_{cl}\times\frac{25}{100}=90\times\frac{30}{100}+80\times\frac{25}{100}=47\;HP ##

Another method to get generated by the system heat value is more precised and perfect for complex systems where open and close loops are mixed but requires more inputs. The idea is in a calculation of a difference between input and output power. All wasted power is transferred into heat, the value that we exactly looking for. Couple words about this method.

Any hydraulic system calculation starts from a technical design task where shown required functions performances: speed/force for cylinder or rpm/torque for motors. This data is a good start point to estimate the required system output power for the worst-case scenario: when max functions work at the same time.

The next steps are calculations of the input power needs to be provided from the electrical motor or diesel engine to the pumps using overall efficiency of all components in loops such as pumps, motors, and using pressure drop on valves or pressure losses at hoses, and etc. Detailed hydraulic system calculation I will provide in future articles.

For my projects, I usually do calculations by both these methods and take the highest value of the required rejection heat for cooler selection.

These are the simplest way to calculate rejected heat to keep heat balance in the system. In reality there many other factors like environment temperature and heat transfer via components surfaces (for example hydraulic tank) that affect heat balance. But for the regular hydraulic system, it is not required to dig so deep in calculations.

Estimation of flow through the cooler

The next step is a selection of place in the system where the cooler has to be installed and estimation a flow at this place.

There are three most popular places where heat exchanger can be installed:

  • in the return line, between return manifold collector and a tank. This palce most popular for open loop system. To estimate flow trough the cooler, first, you need to figure-out flow for all functions that can work at the same time and, next, subtract drain flow, calculated from components via their volumetric efficiency.
  • in the drain line, between case drain manifold collector and tank. Most popular for close loop system.To estimate flow through cooler, you need to figure-out drain flow, calculated from all components via their volumetric efficiency.
  • as a separate loop with additional downstream filtration. Can be used for both open loop and close loop systems. To estimate flow trough cooler, first, you need to get the value of the rejected heat. Next, using vendors diagrams select flow from the cooler. And finally, select cooler loop pump size to provide required flow trough the cooler.

Adjusting the rejected heat to use vendor’s curves

Each application has specific conditions and cooler manufacturers can not provide performances for all of them. Therefore, in the cooler manufacture catalogs, you can find diagrams/curves for each cooler at specific conditions it was tested. For example, AKG coolers tested and published curves based on:

  • Oil viscosity: 50 SUS
  • Entering Temperature Difference (ETD): 100°F

Another example, Emmegi coolers tested and published curves based on:

  • Oil viscosity: 16 cSt SUS
  • Entering Temperature Difference (ETD): 50°F

Here the Entering Temperature Difference (ETD) is a difference between Entering oIl Temperature and Entering aIr Temperature to the cooler. Of course, your conditions will be different and to use vendor’s chart you need to calculate “Desired” ETD for your system. Your “Desired” ETD is the difference between Max ambient temperature (in the worst case) your system will be work and the highest desired entering oil temperature.

For example, for Texas I usually use 115°F, for Alaska – 70°F for Max ambient temperature (if the customer does not provide preffered value).

You need to estimate by yourself the expected highest desired entering oil temperature. For the drain line it can rise to 176°F, for return lines – around 160°F, but, again, it varies from the application.

As soon you know your “Desired” ETD you can make an adjustment for rejected power and use this value in manufacturer’s curves. You can find the formula in each vendor’s catalog for adjusted rejected power:

## p_{adj}=P_{heat}\times\frac{100^{\circ}F}{ETD_{des},^{\circ}F} ##

Preliminary select the cooler.

As soon you get two values: flow trough cooler and adjusted heat you can use vendor’s curves to cooler preliminary selection. Try to find the cooler model close to the middle of the curve.

Checking pressure drop across cooler

This is a very important step and this is why the previous step was called “preliminary”.

It is important to check pressure drop across the cooler because in some cases you can damage hydraulic components if pressure is too high. For example, if cooler in the case drain line and pressure drop across it more than 50 psi you can destroy bearings in some pumps or motors (you need to check max available case pressure for them in the catalogue). Another issue is a back pressure that cooler can create if placed in the return line. In both these cases, the bypass valve integrated into a cooler will not save the system from overpressure because it passes through only a small part of the flow. Do not forget, the cracking pressure is a value when the valve just starts to open with a very small cross-section for flow. I had a case in my practice when cooler with a bypass valve craking pressure 20 psi rises back pressure up to 90 psi at cold start and up to 45 psi at normal work temperature and it isn’t applicable.

I do not know why, but for some reasons, some manufacturers published cooler performance diagrams at very low viscosity, not realistic for real operation. This is why the pressure drop at the curve looks so nice.

AKG correction factor curve

AKG cooler correction factor curve

Also, by some strange reasons, at the latest AKG catalogs there no info about pressure drop across cooler depends on the flow. So, I use older catalogs for the same cooler model (what is not right, because cooler modified and newest may not have the same performances). Emmegi continues to provide this useful info in their catalogs.

So, to get an actual pressure drop across the cooler you need to use the correction factor, depends on oil viscosity in your application. (You can use online converter if viscosity or temperature is different from units in vendors catalog).

It was the last step and if you satisfied by pressure drop across the cooler you selected, you can proceed with it, if not – try to check different model or brand of the cooler.

Actually, AKG TS provides with cooler selection online tool, that you can find by the link:

http://78.94.222.53:18000/cl1/

Example of cooler selection

The example below is the real calculation for the real unit (Blender Trailer) that is working now somewhere in the north of Texas.

Technical task

  • Mobile unit is designed to work in a place where the max average ambient temperature is: 45°C = 113°F
  • The oil going to be used: ISO VG46
  • System input power: 120 HP
  • The hydraulic system contains close loops only
  • Cooler is in a drain line where estimated flow is approx. 42 GPM
  • The highest expected entering oil temperature in drain line: 80°C = 176°F
  • The cooler has to be electrically driven, 24VDC

Step 1. Calculation of the heat needs to be rejected

First, I assume in the worst case the heat can be generated is 25% of input power (as for close loop systems):

## p_{heat}=120\times\frac{25}{100}=30\;HP ##

Step 2. Determine Desired ETD

## ETD_{des}=T_{oil}-T_{amb}=176-113=63\;^{\circ}F ##

Step 3. Calculation of the adjusted rejected heat

Adjusted HP for cooler selection at 100°F ETD (for AKG coolers):

## p_{adj}=P_{heat}\times\frac{100^{\circ}F}{ETD_{des}, ^{\circ}F}=30\times\frac{100^{\circ}F}{63^{\circ}F}=50\;HP ##

## 50\;HP=127000\;BTU/h ##

Adjusted HP for cooler selection at 50°F ETD (for Emmegi coolers):

## p_{adj}=P_{heat}\times\frac{50^{\circ}F}{ETD_{des}, ^{\circ}F}=30\times\frac{50^{\circ}F}{63^{\circ}F}=25\;HP ##

## 25\;HP=63600\;BTU/h ##

Step 4. Preliminary cooler selection

Let’s select DCS series cooler from AKG. As you can see at the performance curves below, the best choice is model DCS-60:

AKG DCS Cooler curves

AKG DCS Cooler curves

Checking coolerweight (111 lb), max work pressure (250 psi), electrical motor current (2x10A=20A), dimensions, cost, etc.

Let’s select DC series cooler from Emmegi. As you can see at the performance curves below, the good choices are models HPV-36 and SBV-6:

Emmegi DC Cooler curves

Emmegi DC Cooler curves

For this calculation, I selected model SBV-6 because both coolers AKG DCS-60 and Emmegi SBV-6 look similar, have the same dimensions and almost the same price. So it is a good choice for future comparison.

Checking cooler weight (82 lb), max work pressure (280 psi), electrical motor current (2×10.4A=20.8A), dimensions, cost, etc.

Step 5. Checking pressure drop across both coolers

From the DCS-60 series cooler curve above, for 42 GPM flow, I get approx. 22 psi pressure drop. But this is a pressure drop is test results for oil with viscosity 50 SUS. For our ISO VG46 oil at work temperature 40°C estimated viscosity is 212 SUS. Lets figure-out correction factor for our conditions from the curve in AKG catalog:

AKG correction factor curve

AKG cooler correction factor curve

So, the correction factor is equal to 3, therefore the real pressure drop across the cooler to be 22 PSI × 3 = 66 PSI.

Let’s check Emmegi SBV-6. From the curve above I get pressure drop around 7 psi for oil with viscosity 16 cSt at 42 GPM flow. Now, we need to get a correction factor, which is not published in the Emmegi catalog. But you are lucky, I was made a request to Emmegi and their technical specialist provided me with an approx curve that we can use for calculation:

Emmegi correction factor curve

Emmegi cooler correction factor curve

So, the correction factor is equal to 2.3, therefore the real pressure drop across the cooler to be 7 PSI × 2.3 = 16 PSI

As a result, my final selection is Emmegi SBV-6, because:

  • Emmegi cooler provides much less pressure drop at the estimated flow (what is good for a drain line)
  • Emmegi cooler weight is 25% lighter than AKG analog (what is critical for the mobile application)
  • Other criteria (el. motor current, dimensions, cost, etc) are similar.

Easy? Oh yes, but some nuances exist…

1 Comment

  1. YANG

    This article has explained it very clearly, I feel I have gained a lot
    ! Thank you very much to the author of this article! I am so lucky!

    Reply

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