What is the most efficient hydraulic pump?

What is the most efficient hydraulic pump?

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What is the most efficient hydraulic pump?

What is the most efficient hydraulic pump?
What is the most efficient hydraulic pump?
Hydraulic Pumps and Motors: Considering Efficiency
Brendan Casey

In a condition-based maintenance environment, the decision to change out a hydraulic pump or motor is usually based on remaining bearing life or deteriorating efficiency, whichever occurs first.

Despite recent advances in predictive maintenance technologies, the maintenance professional’s ability to determine the remaining bearing life of a pump or motor, with a high degree of accuracy, remains elusive.

Deteriorating efficiency on the other hand is easy to detect, because it typically shows itself through increased cycle times. In other words, the machine slows down. When this occurs, quantification of the efficiency loss isn’t always necessary. If the machine slows to the point where its cycle time is unacceptably slow, the pump or motor is replaced. End of story.

In certain situations, however, it can be helpful, even necessary, to quantify the pump or motor’s actual efficiency and compare it to the component’s native efficiency. For this, an understanding of hydraulic pump and motor efficiency ratings is essential.

There are three categories of efficiency used to describe hydraulic pumps (and motors): volumetric efficiency, mechanical/hydraulic efficiency and overall efficiency.

Volumetric efficiency is determined by dividing the actual flow delivered by a pump at a given pressure by its theoretical flow. Theoretical flow is calculated by multiplying the pump’s displacement per revolution by its driven speed. So if the pump has a displacement of 100 cc/rev and is being driven at 1000 RPM, its theoretical flow is 100 liters/minute.

What is the most efficient hydraulic pump?

Actual flow has to be measured using a flow meter. If when tested, the above pump had an actual flow of 90 liters/minute at 207 bar (3000 PSI), we can say the pump has a volumetric efficiency of 90% at 207 bar (90 / 100 x 100 = 90%).

Its volumetric efficiency used most in the field to determine the condition of a hydraulic pump – based on its increase in internal leakage through wear or damage. But without reference to theoretical flow, the actual flow measured by the flow meter would be meaningless.

A pump’s mechanical/hydraulic efficiency is determined by dividing the theoretical torque required to drive it by the actual torque required to drive it. A mechanical/hydraulic efficiency of 100 percent would mean if the pump was delivering flow at zero pressure, no force or torque would be required to drive it. Intuitively, we know this is not possible, due to mechanical and fluid friction.

Table 1. The typical overall efficiencies of hydraulic pumps, as shown above, are simply the product of volumetric and mechanical/hydraulic efficiency. Source: Bosch Rexroth

Like theoretical flow, theoretical drive torque can be calculated. For the above pump, in SI units: 100 cc/rev x 207 bar / 20 x p = 329 Newton meters. But like actual flow, actual drive torque must be measured and this requires the use of a dynamometer. Not something we can – or need – to do in the field. For the purposes of this example though, assume the actual drive torque was 360 Nm. Mechanical efficiency would be 91% (329 / 360 x 100 = 91%).

Overall efficiency is simply the product of volumetric and mechanical/hydraulic efficiency. Continuing with the above example, the overall efficiency of the pump is 0.9 x 0.91 x 100 = 82%. Typical overall efficiencies for different types of hydraulic pumps are shown in the Table 1.

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System designers use the pump manufacturers’ volumetric efficiency value to calculate the actual flow a pump of a given displacement, operating at a particular pressure, will deliver.

As already mentioned, volumetric efficiency is used in the field to assess the condition of a pump, based on the increase in internal leakage due to wear or damage.

When calculating volumetric efficiency based on actual flow testing, it’s important to be aware that the various leakage paths within the pump are usually constant. This means if pump flow is tested at less than full displacement (or maximum RPM) this will skew the calculated efficiency – unless leakage is treated as a constant and a necessary adjustment made.

For example, consider a variable displacement pump with a maximum flow rate of 100 liters/minute. If it was flow tested at full displacement and the measured flow rate was 90 liters/minute, the calculated volumetric efficiency would be 90 percent (90/100 x 100). But if the same pump was flow tested at the same pressure and oil temperature but at half displacement (50 L/min), the leakage losses would still be 10 liters/minute, and so the calculated volumetric efficiency would be 80 percent (40/50 x 100).

The second calculation is not actually wrong, but it requires qualification: this pump is 80 percent efficient at half displacement. Because the leakage losses of 10 liters/minute are nearly constant, the same pump tested under the same conditions will be 90 percent efficient at 100 percent displacement (100 L/min) – and 0 percent efficient at 10 percent displacement (10 L/min).

To help understand why pump leakage at a given pressure and temperature is virtually constant, think of the various leakage paths as fixed orifices. The rate of flow through an orifice is dependant on the diameter (and shape) of the orifice, the pressure drop across it and fluid viscosity. This means that if these variables remain constant, the rate of internal leakage remains constant, independent of the pump’s displacement or shaft speed.

Overall efficiency is used to calculate the drive power required by a pump at a given flow and pressure. For example, using the overall efficiencies from the table above, let us calculate the required drive power for an external gear pump and a bent axis piston pump at a flow of 90 liters/minute at 207 bar:

External gear pump: 90 x 207 / 600 x 0.85 = 36.5 kW

Bent axis piston pump: 90 x 207 / 600 x 0.92 = 33.75 kW

As you’d expect, the more efficient pump requires less drive power for the same output flow and pressure. With a little more math, we can quickly calculate the heat load of each pump:

Drive power for a (non-existent) 100% efficient pump would be: 90 x 207 / 600 x 1 = 31.05 kW

So at this flow and pressure, the heat load or power lost to heat of each pump is:

External gear pump: 36.5 – 31.05 = 5.5 kW

Bent axis piston pump: 33.75 – 31.05 = 2.7 kW

No surprise that a system with gear pumps and motors requires a bigger heat exchanger than an equivalent (all other things equal) system comprising piston pumps and motors.


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