volumetric efficiency hydraulic pump quotation

Your final drive includes a hydraulic motor and that motor has a certain level of efficiency associated with it. Over time, that efficiency can drop -- so find out how efficiency is measured, what the source of losses are, and how to minimize them.

No system, no matter how well it’s designed, is going to be 100% efficient. High-quality, well-maintained radial piston motors are about 95% efficient while axial piston motors are about 90% efficient--which is likely why you see these two types of hydraulic motors used in the vast majority of final drive motors.

The definition of efficiency depends on what type of system you’re talking about, and even then there can be some variations. For a hydraulic motor, there are three ways efficiency can be measured or estimated: volumetric, mechanical/hydraulic, and overall efficiency.

Volumetric efficiency looks at the theoretical flow rate and the actual flow rate and provides information about leakage and wear. The theoretical flow rate is pretty easy to calculate: theoretical flow = (pump displacement per revolution) x (revolution speed).

This works much better in SI units, too. If the displacement is in cc/rev and the speed is in rpm, the results will be in liters/minute. Actual flow is then measured using a flow meter. The efficiency is then actual flow / theoretical flow x 100 to get efficiency as a percent.

Mechanical efficiency is based on actual work done and theoretical work done, both per revolution. This is based on theoretical torque and the actual torque, and in most hydraulic motors it’s about 0.9 (or 90%). Actual torque can be measured with a dynamometer, but is rarely done. The losses related to mechanical efficiency are directly tied to mechanical friction between mating parts.

Overall efficiency combines volumetric and mechanical efficiency. It"s simply the product of these two values: overall efficiency = mechanical efficiency x volumetric efficiency, and gives you an overall idea of how efficient your hydraulic motor is.

Some degree of internal leakage is normal and actually beneficial, but past a certain point it becomes a problem. Excess internal leakage most often results from wear. For example, the size of key clearances in a hydraulic motor can, over time, become larger because of abrasive wear and lead to internal leakage. That type of wear usually results from contaminated hydraulic fluid but can also result from normal wear and tear.

Friction is another major source of losses. Rough surfaces where they should be smooth cause friction issues with the hydraulic fluid, reducing the amount of power that can be transferred. There are other ways that friction can be introduced, however. For example, anti-friction bearings or plane bearings that are wearing out will be a source of friction.

One of the keys to preventing hydraulic motor losses relates to good maintenance practices, such as keeping the hydraulic fluid clean, replacing hydraulic filters, and not ignoring hydraulic leaks. It"s also important to look for symptoms of potential problems with the bearings, such as new noises, excessive vibration, and overheating.

Hydraulic losses relates to the construction of the pump or fan and is caused by the friction between the fluid and the walls, the acceleration and retardation of the fluid and the change of the fluid flow direction.

Mechanical components - like transmission gear and bearings - creates mechanical losses that reduces the power transferred from the motor shaft to the pump or fan impeller.

Due to leakage of fluid between the back surface of the impeller hub plate and the casing, or through other pump components - there is a volumetric loss reducing the pump efficiency.

The overall efficiency is the ratio of power actually gained by the fluid to power supplied to the shaft. The overall efficiency can be expressed as: η= ηh ηm ηv(4)

The losses in a pump or fan converts to heat that is transferred to the fluid and the surroundings. As a rule of thumb - the temperature increase in a fan transporting air is approximately 1oC.

An inline water pump works between pressure1 bar (1 105 N/m2)and 10 bar (10 105 N/m2).The density of water is 1000 kg/m3. The hydraulic efficiency is ηh= 0.91.

600 gpm of water is pumped a head of 110 ft. The efficiency ofthe pump i s 60% (0.6) and the specific gravity of water is 1. The pump shaft power can be calculated as

The shaft power - the power required transferred from the motor to the shaft of the pump - depends on the efficiency of the pump and can be calculated as Ps(kW) = Ph(kW)/ η (3)

Calculation of preliminary cooler capacity: Heat dissipation from hydraulic oil tanks, valves, pipes and hydraulic components is less than a few percent in standard mobile equipment and the cooler capacity must include some margins. Minimum cooler capacity, Ecooler = 0.25Ediesel

At least 25% of the input power must be dissipated by the cooler when peak power is utilized for long periods. In normal case however, the peak power is used for only short periods, thus the actual cooler capacity required might be considerably less. The oil volume in the hydraulic tank is also acting as a heat accumulator when peak power is used.

The system efficiency is very much dependent on the type of hydraulic work tool equipment, the hydraulic pumps and motors used and power input to the hydraulics may vary considerably. Each circuit must be evaluated and the load cycle estimated. New or modified systems must always be tested in practical work, covering all possible load cycles.

Hydraulic pumps convert mechanical energy into hydraulic energy. A high-performance piston pump can convert mechanical energy into hydraulic energy with an efficiency of 92 percent.

If the pump drives a piston motor, the motor is able to convert this hydraulic energy back into mechanical energy with an efficiency of 92 percent. The overall efficiency of this hydraulic drive, without considering flow losses, is 85 percent (0.92 x 0.92 x 100 = 85).

The inefficiencies or losses in a hydraulic drive can be divided into two categories: hydraulic-mechanical, which comprise flow and mechanical friction losses, and volumetric, which comprise leakage and compressibility losses (Figure 1).

The advantages of a hydraulic drive, which include high-power density (high-power output per unit mass), variable speed control, simple overload protection and both rotary and linear motion, are possible from a single system.

As Table 1 shows, a key disadvantage of a hydraulic drive is that it is far less efficient than a mechanical drive. What’s worse, the wear process decreases a hydraulic drive’s volumetric efficiency (and therefore total efficiency) causing the drive to slow down and more energy to be given up to heat.

The hydraulic pump is usually the hardest working component of a hydraulic system. As the pump wears in service, internal leakage increases and therefore the percentage of theoretical flow available to do useful work (volumetric efficiency) decreases. If volumetric efficiency falls below a level considered acceptable for the application, the pump will need to be overhauled.

In a condition-based maintenance environment, the decision to change-out the pump is often based on remaining bearing life or deterioration in volumetric efficiency, whichever occurs first.

Volumetric efficiency is the percentage of theoretical pump flow available to do useful work. In other words, it is a measure of a hydraulic pump’s volumetric losses through internal leakage and fluid compression. It is calculated by dividing the pump’s actual output in liters or gallons per minute by its theoretical output, expressed as a percentage. Actual output is determined using a flow-tester to load the pump and measure its flow rate.

Because internal leakage increases as operating pressure increases and fluid viscosity decreases, these variables should be included when stating volumetric efficiency. For example, a hydraulic pump with a theoretical output of 100 GPM, and an actual output of 94 GPM at 5,000 PSI and 46 cSt is said to have a volumetric efficiency of 94 percent at 5,000 PSI and 46 cSt.

In practice, fluid viscosity is established by noting the fluid temperature at which actual pump output is measured and reading the viscosity off the temperature/viscosity graph for the grade of fluid in the hydraulic system.

When calculating the volumetric efficiency of a variable displacement pump, internal leakage must be expressed as a constant. Consider this example: I was recently asked to give a second opinion on the condition of a large, variable displacement pump. My client had been advised that its volumetric efficiency was down to 80 percent and based on this advice, he was considering having the pump overhauled.

The hydraulic pump in question had a theoretical output of 1,000 liters per minute at full displacement and maximum rpm. Its actual output was 920 liters per minute at 4,350 PSI and 46 cSt. When I advised my client that the pump’s volumetric efficiency was in fact 92 percent he was alarmed by the conflicting assessments. To explain the disparity, I asked to see the first technician’s test report.

The technician had limited the pump’s displacement to give an output of 400 liters per minute (presumably the maximum capacity of his flow-tester) at maximum rpm and no load. At 4,350 PSI the recorded output was 320 liters per minute. From these results, volumetric efficiency had been calculated to be 80 percent (320/400 x 100 = 80).

To help understand why this interpretation is incorrect, think of the various leakage paths within a hydraulic pump as fixed orifices. The rate of flow through an orifice is dependent 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.

Note that in the above example, the internal leakage in both tests was 80 liters per minute. If the same test was conducted with pump displacement set to 100 liters per minute at no load, pump output would be 20 liters per minute at 4,350 PSI - all other things equal.

This means that this pump has a volumetric efficiency of 20 percent at 10 percent displacement, 80 percent at 40 percent displacement and 92 percent at 100 percent displacement. As you can see, if actual pump output is measured at less-than-full displacement (or maximum rpm), an adjustment needs to be made when calculating volumetric efficiency.

In considering whether it is necessary to have this hydraulic pump overhauled, the important number is volumetric efficiency at 100 percent displacement, which is within acceptable limits. If my client had based his decision on volumetric efficiency at 40 percent displacement, his company would have paid thousands of dollars for unnecessary repairs.

Hydraulics are essential in many industrial applications; they use mechanical energy to force liquid into a system. Within the category of hydraulics, there are many different types of hydraulic pumps that accomplish various tasks within industrial fields. Let’s take a look at some of them.

An axial piston hydraulic pump is also a positive displacement pump. Axion pumps have cylinders that are assembled around a central axis (cylinder block). Within each cylinder, there are pistons which will attach to a swashplate or wobble plate. These swashplates then connect to the rotating shaft, which moves the pistons and pulls them in and out of the cylinders.

Axial piston pumps can also be made into variable displacement piston pumps, which provide more control over speed. In variable pumps, the swashplate is used to set the depth of the pistons, which creates a length variation to affect the discharge volume. This design helps maintain constant discharge rates in industrial applications.

Another hydraulic pump type is the radial piston pump. As the name suggests, the pistons are arranged along the radius of the cylindrical block, which includes the pintle and rotor. The rotor powers the pistons through the cylinders and forces hydraulic fluid in and out of the cylinder.

Both axial and radial piston pumps are used for high-operating pressures as they can withstand much higher pressures than hydraulic pump types. They are often used in ice and snow control applications, as well as on truck-mounted cranes.

A rotary vane pump is also a type of positive displacement pump. It uses rigid vanes rather than the rotor hubs. These vanes are placed around an eccentric rotor device, which moves around the inside wall of the housing container. This movement forces the hydraulic fluid through the discharge port, and, in some applications, can be adjusted to align with the rotational axis of the motor.

Vane pumps are often used in utility vehicles but have lost popularity over the years in favor of gear pump hydraulic systems. However, they used to be very common in aerial buckets and ladders along with other mobile, truck-mounted hydraulic applications.

Gear pumps have become the most common hydraulic pump type that’s used in industrial applications today. The gear pump has fewer moving parts than piston or vane pumps, which makes it easy to service and relatively inexpensive compared to other hydraulic pumps. They are also less likely to be contaminated during use.

An external gear pump uses two gears on the outside of individual shafts to aid in movement and push both thin and thick fluids through the gears. These external pumps are commonly used in fixed-displacement applications and high-pressure environments.

Internal gear pumps place gears on the insides of the shafts rather than on the outside as found in external gear pumps. That makes them self-priming and non-pulsing, and can even be run without liquid for short periods of time—although they can’t be left dry for too long.

Additionally, internal gear pumps are bi-rotational, meaning that one can be utilized to both unload and load devices. And, with only two moving parts, they are considered to be one of the most reliable types of hydraulic pumps.

Volumetric efficiency is the amount of output the pump actually produces as a percentage of its theoretical production.The higher the percentage the more efficient the pump.

Among the factors affecting volumetric efficiency are leakage and fluid compressibility (ability for volume to be reduced under pressure). These issues can cause the pump to lose efficiency and waste energy as that energy converts to heat.

Still have questions about hydraulic pumps or their parts and repairs? Contact Panagon Systems today. We’ve been a leading hydraulic pump manufacturer in the U.S. for over two decades, and can help you find the best solution for your application. You can view our full line of pumps here. To request a consultation or quote, please fill out our online form.

Knowing how to right-size an electric motor for your hydraulic pump can help reduce energy consumption and increase operational efficiency. The key is to ensure the pump motor is operating at peak continuous load. But how can you know how much power is needed?

Before you can choose the correct electric motor, you must know how much horsepower (Hp) is required to drive the pump shaft. Generally, this is calculated by multiplying the flow capacity in gallons per minute (GPM) by the pressure in pounds per square inch (PSI). You then divide the resulting number by 1714 times the efficiency of the pump, for a formula that looks like this:

If you’re not sure how efficient your hydraulic pump is, it is advisable to use a common efficiency of about 85% (Multiplying 1714 x 0.85 = 1460 or 1500 if you round up). This work-around simplifies the formula to:

The above formula works in most applications with one notable exception: If the operating pressure of a pump is very low, the overall efficiency will be much lower than 85%. That’s because overall efficiency is equal to mechanical efficiency (internal mechanical friction) plus volumetric efficiency.

Internal friction is generally a fixed value, but volumetric efficiency changes depending on the pressure used. Low-pressure pumps have high volumetric efficiency because they are less susceptible to internal leakage. However, as the pressure goes up and internal fluids pass over work surfaces such as pistons, port plates, and lubrication points, the volumetric efficiency goes down and the amount of torque required to turn the pump for developing pressure goes up.

This variance makes it very important to know the efficiency of your pump if you’re using it at low pressure! Calculations that do not take low pressure into account will lead to a failed design.

If you calculate 20 GPM @ 300 PSI with an assumed overall efficiency of 89%, you would probably select a 5 Hp electric motor. However, if you calculate the same 20 GPM @ 300 PSI with the actual overall efficiency of 50%, you would know that you should be using a 7.5 Hp motor. In this example, making an assumption about the efficiency of your pump could result in installing a motor that is too large, driving up your overall operating cost.

There are many contributors to the overall efficiency of a hydraulic pump, and it pays to be as accurate as possible when choosing a motor. A best practice for proper sizing is to use published data from the pump vendor that shows actual input torque vs. pressure or overall efficiency vs pressure. Note that efficiency is also affected by RPM.

Identifying a right-sized motor for your hydraulic pump does not always ensure you are using the most efficient motor. Be sure to read Part 2 of this post to learn how RMS loading and Hp limiting can help you scale down the size of your electric motor to save money while maximizing efficiency.

Gear pumps have very few moving parts. They consist of two intermeshing gears. These pumps have a constant flow rate. They operate at pressures generally between 50 and 210 bar. Gear pumps operate at the highest speeds of any pumps at up to 3000-6000 rpm.

In an external-gear pump, only one of the gear wheels, the drive gear, is connected to the drive. The other gear wheel, the driven gear, rotates in the opposite direction, so that the teeth of the rotating gear wheels interlock.

There are also double external-gear pumps, which combine two gear pumps driven by the same coupling shaft. A double external-gear pump has the advantage of supplying two independent hydraulic circuits, and also provides more flow to one circuit.