Hydraulic systems are used in many different machineries, equipment and facilities in processing units. The hydraulic sets and actuators are employed in many key machineries, equipment and devices such as processing equipment, material handling units, control valves, machinery packages, automation systems and production facilities. Hydraulic systems commence with the need for a set of actuators (such as hydraulic cylinders, hydraulic actuators, etc) to exert a certain force, or in the case of a rotary hydraulic motor for a specified torque, and for a certain operating speed to be achieved. Such systems contain hydraulic pumps, hydraulic actuators, piping, hydraulic valves, control systems, hydraulic fluid reservoirs and others. This article discusses hydraulic systems for processing facilities. The focus is on practical notes and useful guidelines to properly specify, purchase, operate and maintain hydraulic units.
A hydraulic system can be provided either by a high-pressure system using a small flow of fluid, or a low pressure one with a large flowrate. Either could be arranged to give the prescribed force or torque at the hydraulic actuators.
The choice of working pressure is a major decision for a hydraulic system. Some engineers encourage high pressures as this can reduce sizes and lead to overall compact sets and devices. Pressure loss is not directly dependent on the working pressure. Therefore, the higher the pressure, the smaller the loss will be in proportion, and the higher the efficiency. On the other hand, there have always been some limits on the hydraulic pressure. The working pressure cannot be limitlessly high. If the use of a very high pressure is involving stronger and presumably more costly equipment, it is not obvious whether there will be a net gain. Only a detailed optimization can indicate the optimum pressure.
Many engineers and experts believe there is no inherent advantage in the use of very high working pressures. To better explain, working pressure should be high and suitable for each hydraulic application, but not very high. Some applications and services need a high working pressure and compactness of high-pressure equipment is often an advantage. However, there are nearly always some limits. An optimization should be done to find the best and optimum working pressure for each application. The selection of working pressure tends to be governed by the maximum working pressure of available cost-effective and reliable equipment, components and parts. Nowadays working pressures above 100 Bar have been commonly used in hydraulic systems. Pressure as high as 250 Barg, 350 Barg, 400 Barg, or more are used in different compact hydraulic units. Once the working pressure has been determined, this usually fixes the rate of flow. It is then necessary to decide on the sizes for the piping, valves and others.
Sizing and selection
As a rough indication, the flow velocity between 3.5 m/s and 5.5 m/s have been used in hydraulic circuits. Although flow velocity outside of this range has also been successfully used. This rough rule can be employed to check the pipe diameter, valve sizes and sizes of other parts and components. Although useful for checking or rough estimates, relying on such simplistic procedure cannot be suitable in many cases. A key point is the pressure drop is not only the function of sizes. Different parts, inlines, valves, devices and components present different performance, functionality and pressure drop patterns, which cannot be determined using simplistic rules as above.
For the sizing of piping, valves and other hydraulic items, it is recommended to use an estimate of the allowable pressure loss. In other words, in the sizing and selection, firstly the overall allowable pressure loss in the hydraulic circuit should be decided based on operational and functional requirements. Then it is determined how the pressure loss should be distributed between the items and devices (piping, fittings, valves, etc) to achieve the best performance and lowest overall cost. As another indication, the ratio of the pressure loss at full working speed to the maximum pressure can be used for such estimates and sizing.
Hydraulic pumps are important components in any hydraulic system. Usually a positive displacement pump type is used for a hydraulic system. There are many types and models of hydraulic pumps such as reciprocating pumps, piston pumps and gear pumps. Many of these pumps can be arranged as horizontal pumps or vertical pumps. An axial piston pump is a positive displacement pump that has a number of pistons in a circular array within a cylinder block. This cylinder block is driven to rotate about its axis of symmetry by an integral shaft. This type of hydraulic pump is used in high pressure applications. Gear pumps have been used in so-called medium pressures.
There are different configurations and arrangements for hydraulic pumps. Some systems use two pumps in operating-standby configuration. Some others may use “n+1” (“n” operating and one standby) pumps. Often, it is easier to provide independent streams of flows rather than splitting the flow generated by a pump. Therefore, using multiple pumps or double pumps (two pump casings driven by a driver) are popular in some hydraulic systems.
The speed of the hydraulic pump is an important factor that should be selected with great care. Each type of pump has a specific speed range for the optimum operation. Driving the pump at a high speed decreases its required displacement, thereby reducing its size, weight, and torque requirement. On the other hand, if the speed is too high, higher than optimum ones, the life and reliability of pump is reduced and high maintenance can be expected.
Power rating of drivers
The driver of a hydraulic pump should provide sufficient power at the best efficiency. Power rating of the driver should be calculated with great care. Many drivers are undersized, which results in malfunction and operational problems. On the other hand, if the driver is too large and the power rating is far higher than needed, the pump unit will work most of the time at a point far from its best efficiency range. This is also problematic and inefficient. It results in waste of money and even reliability issues. The optimum selection of hydraulic pump and its driver is the key for efficient and reliable operation.
In some applications, the hydraulic pump should provide power for a set of actuators working at different loading over the time. In other words, power required would vary over the time. The required power rating should be calculated carefully considering all possible operating cycles and cases. This is particularly applicable for repetitive hydraulic actuators where there are different power ratings for each segment of the cycle. For instance, the power rating needed for a hydraulic actuator of repetitive work is 12.7 kW for 35 second, 8.6 kW for 15 second and 2.9 kW for 25 second (total cycle of 75 second). It might be assumed that the minimum power rating should be 12.7 kW and a 15-kW electric motor is needed. This is a conservative approach and overall the electric motor would work far from its best efficiency point. Using, for example, root-mean-square (RMS] method, the required power rating is estimated as 9.5 kW. In this way, 11 kW electric motor can be selected.
Maximum hydraulic pressure and maximum torque
In many applications, there is a maximum load (force, torque, etc) needed at hydraulic actuators to start the operation, break-in or passing through a working barrier. An important consideration is the maximum hydraulic pressure needed for the application. This is linked to the maximum torque required to drive the hydraulic pump. In other words, in addition to the confirmation of power rating, the maximum torque required to drive the hydraulic pump plus a reasonable margin can be provided by the driver.
In general, the torque requirement is an important factor in the sizing and selection of the driver. Speed is less critical, because if a pump runs slowly, it will still pump hydraulic fluid. However, if the driver does not develop enough torque to drive the hydraulic pump, the pump will not produce any output flow at maximum pressure. For electric motors, this issue is not often a major problem as in each electric motor power rating (in each electric motor family), there are several available models with different torque capabilities. However, this can be a major issue in other types of drivers such as engines or turbines.
Electric motor drivers
The curve of the electric motor should be checked and its suitability should be verified. When an electric motor starts and begins to accelerate, the torque in general decreases until it reaches a low point at a certain speed (pullup torque) before the torque increases until it reaches the highest torque at a higher speed point (breakdown torque). In other words, upon receiving power, the electric motor develops an initial, locked-rotor torque, and the rotor turns. As the rotor accelerates, torque decreases slightly, then begins to increase as the rotor accelerates further. This dip in the torque curve generally is referred to as the pull-up torque. Torque eventually reaches a maximum value. This is the motor's break-down torque. As rotor speed increases beyond this point, torque applied to the rotor decreases sharply. This is known as the running torque, which becomes the full-load torque when the motor is running at its rated full-load speed.
The pull-up torque is critical for many hydraulic applications that need load to go through some working barriers achieving the operation. Typically, the electric motor's pull-up torque is checked to be greater than the maximum torque required with a working margin. An ample margin is usually needed due to many reasons, one is the voltage variation. For instance, an electric motor running 8% low on voltage will produce only 85% of rated pull-up torque.
Common practice is to ensure that torque required from the electric motor will always be less than the breakdown torque. Applying torque equal to or greater than breakdown torque will cause the motor's speed to drop suddenly and severely, which will tend to stall the electric motor and most likely burn it out. If the electric motor is already running, it is possible to momentarily load an electric motor to near its breakdown torque. If the load increases, motor speed will decrease and torque will increase to a value higher than full-load torque; although this is less than breakdown torque.
Many hydraulic systems use gas, diesel or gasoline engines as drivers of the hydraulic pumps. The sizing and selection of the engine driver is more of a challenge and the rated power of the engine is usually more than an electric motor driver for the same hydraulic system application. In other words, specifying a gasoline, gas or diesel engine to drive a hydraulic power unit follows a different procedure than that for specifying an electric motor. As a rough indication, the power rating of an internal-combustion engine can be two times or more than an equivalent electric motor. This is because internal combustion engines have different torque-speed relationships than electric motors. This means an engine exhibits a much less variable torque output throughout its speed range. An engine usually exhibits a similar torque curve throughout their operating speed range. Depending on their characteristics, diesel engines with the same power ratings may generate slightly higher or lower torque at lower speeds than gasoline engines.
For engine drivers, usually, the maximum required hydraulic torque depending on the maximum hydraulic pressure dictates the power rating of the engine. Maximum torque produced by an engine usually occurs at speeds near the rated speed, but is still well below the maximum hydraulic torque required by the pump if the power rating was calculated as per common rules employed for electric motor drivers.
Even if the engine produced enough torque at the speed near the rated speed, the driver is inadequate because high torque is needed at the lower speed. Moreover, manufacturers recommend that engines only operate continuously at about 85% of their maximum rated values to prevent seriously shortening of their service life. This is another factor for the maximum torque and power ratings.
Depending on the details and characters, an engine’s optimum fuel efficiency often occurs at a speed other than where it produces maximum torque. For instance, a 30-kW engine achieves the best fuel efficiency at about 2,200 rpm. At its maximum speed 3,000 rpm, the engine would be considerably less fuel efficient.
For the speed selection, an operating speed where the engine produces maximum torque generally takes priority. This is because if the engine cannot generate enough torque at its optimum fuel efficiency speed, a larger engine would be required. However, a larger engine consumes more fuel, which would defeat the purpose of trying to conserve fuel by operating at a specific speed. In addition, hydraulic pumps generally have a speed range at which they are most efficient. Therefore, even if an engine operates 10% to 20% above or below its optimum fuel efficiency speed, torque produced and pump dynamics generally have a more pronounced effect on the overall efficiency of the pump-driver unit. Usually an overall optimum speed is between the optimum fuel efficiency speed and maximum torque speed.
It is important to choose a pump speed that offers the best combination of pump and engine performance. Proper selection of the hydraulic pump and correct speed match are vital. Operating a hydraulic pump unit at higher speed more closely matches the engine performance to the application by increasing torque produced by the engine and reducing the torque required by the pump. More specifically, selecting a high-speed pump and operating the pump at relatively higher speed can reduce the maximum torque required by the hydraulic system and increase the produced torque by the engine. This, overall, eases the sizing and selection.
An alternative would be to provide a gear unit or other type of speed changing unit between the engine and pump. Although this would add components to the power unit, it would increase torque and reduce speed while allowing both the engine and the pump to operate at their optimum speeds. The additional cost of the speed reducer may be offset by the lower cost of a smaller, lighter, and less-expensive engine.