Servo valves are a close relative of the proportional valve and are based on an electrical torque motor which produces a small deflection proportional to the electrical current through its coil. They commonly use feedback between the main and pilot spools to give precise control. A typical device is shown on Figure 4.40. This consists of a small pilot spool connected directly to the torque motor. The pilot spool moves within a sliding sleeve, mechanically linked to the main spool.
The fight-hand end of the main spool is permanently connected to the pilot pressure line, but because of the linkage rod its area is reduced to an annulus of area A. Pressure at the left-hand end of the spool is controlled by the pilot valve. There is no area restriction at this end, and the valve is designed such that the spool has an area of 2A.
If the same pressure P is applied to both ends, the spool experiences a left force of P x A and a right force of 2P x A causing a net force of P x A to the fight, resulting in a shift of the spool to the fight.
If a pressure of P is applied to the right-hand end and 0.5P is applied to the left-hand end, equal and opposite forces of P x A result and the valve spool is stationary.
With a pressure of P on the right-hand end and a pressure less than 0.5P on the left-hand end, net force is to the left and the valve spool moves in that direction.
The pilot valve can thus move the main spool in either direction, in a controlled manner, by varying pressure at left-hand end of the main spool from zero to full pilot pressure.
The mechanical linkage between main spool and pilot sleeve controls the flow of fluid between pilot valve and main valve, and hence controls pressure at the left-hand end of the main spool. Suppose the electrical control signal causes the pilot spool to shift left. This increases the pressure causing the main valve to shift right which in turn pushes the sleeve left. The main valve stops moving when the hole in the pilot sleeve exactly aligns with the land on the pilot spool. A change in electrical signal moving the pilot spool to the fight reduces pressure at the left-hand end of the main spool by bleeding fluid back to the tank. This causes the main spool to move left until, again, pilot sleeve and pilot lands are aligned. The main valve spool thus follows the pilot spool with equal, but opposite movements.
Figure 4.41 illustrates the construction of a different type of servo valve, called a jet pipe servo. Pilot pressure is applied to a jet pipe which, with a 50% control signal, directs an equal flow into two pilot lines. A change of control signal diverts the jet flow giving unequal flows and hence unequal pressures at ends of the main spool. The main spool is linked mechanically to the jet pipe, causing it to move to counteract the applied electrical signal. Spool movement ceases when the jet pipe is again centrally located over the two pilot pipes. This occurs when the main valve spool movement exactly balances the electrical control signal.
The servo valve in Figure 4.42 is called a flapper servo and is really the inverse of the jet pipe servo. Here, pilot pressure is applied to both ends of the main spool and linked by orifices to small jets playing to a flapper which can be moved by the electrical control signal. Pressure at each end of the main spool (and hence spool movement) is determined by the flow out of each jet which, in turn, is determined by flapper position and electrical control signal.
Servo valves are generally used as part of an external control loop in a feedback control system. The principle of a feedback control system is shown in Figure 4.43 where some plant variable (velocity or position, for example) is to be controlled. The plant variable is measured by a suitable transducer, and electronically compared with the desired value to give an error signal. This is amplified and used as the control signal for the servo valve.
It can be appreciated that, with small movements of the pilot spool (in Figure 4.40) and the fine jets (in Figures 4.41 and 4.42), servo valves are particularly vulnerable to dirt. Cleanliness is important in all aspects of pneumatics and hydraulics, but is overwhelmingly important with servo valves. A filtration level of 10um is normally recommended (compared with a normal filtration of 25 um for finite position valve systems).
Servo valves which are stationary for the majority of time can stick in position due to build-up of scum around the spool. This is known, aptly, as stiction. A side effect of stiction can be a deadband where a large change of control signal is needed before the valve responds at all.
Figure 4.44 shows a purely mechanical servo used as a mechanical booster to allow a large load to be moved with minimal effort. The pilot valve body is connected to the load, and directs fluid to the fixed main cylinder. The cylinder, and hence the load, moves until pilot spool and cylinder are again aligned. Variations on the system in Figure 4.44 are used for power steering in motor cars.
Overall efficiency accounts for the loss of mechanical energy due to friction, and the loss of hydraulic energy due to the leakage flow. The clearances between moving parts can be thought of as an orifice. Fluid on one side of the orifice has a high pressure (up to the maximum pressure that exists at the pump outlet), and fluid on the other side of the orifice has a pressure equal to the pressure in the pump housing (generally less than 50 psi). This ?P multiplied by the leakage flow gives an estimate of the hydraulic power loss.
Overall efficiency for the Hydreco Model 1919 gerotor pump is plotted in Fig. 4.22a (1200 rpm) and Fig. 4.22b (1800 rpm). Efficiency is near a maximum at 1000 psi, so the comparison will be made at this pressure. Efficiency is 87.5% at 1200 rpm and 88% at 1800 rpm. The trend is the same as that found for volumetric efficiency; however, the increase in overall efficiency at the higher rpm is smaller.
The energy loss (fluid energy converted to heat) due to fluid moving past a stationary part, or a part moving in a fluid, is proportional to velocity squared . We expect the energy loss due to friction to increase as operating speed increases. This friction loss, however, is only part of the total power loss. Leakage flow represents a loss of hydraulic power. This flow is raised to some pressure, but it is not part of the output flow, and thus it is not included in the output power from the pump. Realizing that those who have done an in-depth study of fluid machinery will question the simplification, total power loss will now be discussed as the sum of two losses,
Total power loss = Friction power loss + Leakage power loss
A simple experiment might help the reader better understand the partitioning of total loss in this manner. A schematic for this experiment is shown in Fig. 4.23. Suppose a variable displacement pump is set for zero displacement and operated at a given speed. Fluid at the outlet has an unimpeded pathway back to the inlet, so developed pressure (?P across the pump) will be very small. Outlet flow will be small, because the pump displacement is set at zero. Delivered hydraulic power (small pressure × small flow) is very small and can be taken to be zero. The measured input power is the friction power loss.
Input power from the electric motor is given by
Since no hydraulic power is delivered by the pump, this input power is the total power required to replace losses. (Speed is constant, so inertia is not a factor.). Friction power loss is defined by
It is instructive to calculate the overall efficiency for a gerotor pump based on leakage power loss only (neglecting the friction loss component) and compare this calculated efficiency with the overall efficiency reported by the manufacturer. This comparison reveals the magnitude of the two components for the gerotor design. Data from the 1200 rpm, 1500 psi test will be used.
As previously calculated, Ql = 3.53 GPM. Because of the gerotor pump design, it is difficult to measure pressure inside the housing at the point where the fluid is leaking from the outlet back to the inlet. We simply assume that the leakage ?P equals the pressure rise across the pump. Leakage power loss is
The overall efficiency based on the input horsepower curves given by the manufacturer is 85.4%. This overall efficiency includes both leakage and friction losses. When we calculated leakage loss, we assumed that leakage ?P equaled the ?P across the pump. Actual leakage ?P is smaller, thus our estimate of leakage loss was too high. It is true that the leakage power loss is by far the dominant term for the gerotor design.
Key comparisons between volumetric and overall efficiencies for the gerotor pump are given in Table 4.1. The fact that the overall and volumetric efficiencies are approximately equal for each speed and pressure reinforces our conclusion that leakage loss is by far the dominant term for the gerotor design.
Overall efficiency at 2500 psi increases from 77% at 1200 rpm to 81.5% at 1800 rpm for the gerotor pump. Because leakage losses are such a significant term for this type of gear pump design, it is generally good policy to use a smaller displacement pump and operate at higher rpm to get a given flow.
If a gerotor motor with similar efficiency, around 80%, is used to convert the hydraulic power back to mechanical power, the overall efficiency (pump and motor combination) is
Less than two-thirds of the input power is delivered to the load. Mechanical transmissions can have efficiencies of 95% or better, so it is readily apparent that a decision to transmit mechanical energy with fluids must carefully weigh the advantages and disadvantages.
The Hydreco Model 1919 gerotor pump has a displacement of 4.53 in3/rev. The manufacturer gives performance data up to a maximum speed of 3000 rpm. The maximum pressure curve is 2500 psi.
When driven at 1200 rpm, the volumetric efficiency drops from 92% at 500 psi to 78% at 2500 psi (Fig. 4.21a). At 1800 rpm, the volumetric efficiency drops from 93% at 500 psi to 84% at 2500 psi (Fig. 4.21b). Why is the volumetric efficiency higher at the higher speed?
This question can be answered by comparing performance at a specific pressure. In this case, the comparison will be made at 1500 psi. The leakage flow is
Leakage flow is primarily a function of pressure, with rotational speed being a much less significant factor. Suppose that leakage flow is the same for the performance test run at 1800 rpm, 1500 psi.
Measured volumetric efficiency was evp = 88.7%, slightly less than the efficiency calculated by assuming constant leakage flow. There is some increase in leakage at higher speeds due to the increased turbulence of the fluid. This explains why the volumetric efficiency is less than that calculated by assuming constant leakage flow as speed increases. It is generally good design practice to select a smaller pump and operate it at higher speed to achieve a higher volumetric efficiency.
A volumetric efficiency in the 85 to 90% range is achievable with a gerotor pump if pressures are below 1500 psi. Remember, however, that volumetric efficiency decreases as operating temperature increases. Viscosity of the fluid decreases, and more leakage occurs through the clearances in the pump.
Load sensing was developed to improve the efficiency of a circuit. It requires a variable displacement pump; thus, it is appropriate to discuss it here.
An example will illustrate the advantage provided by load sensing. An application requires a maximum flow of 20 GPM at a maximum pressure of 2500 psi. To meet this requirement, a pump must be selected that can supply 20 GPM at 2500 psi. Most of the time, however, the application requires less than 20 GPM at less than 2500 psi. The percentage of total operating time the system operates under reduced load, and the way the system responds to this condition, is a key issue. The specific reduced load situation we will consider is the activation of a cylinder. A metered flow rate of 9 GPM at 1300 psi is required. To understand the advantage of load sensing, it is helpful to review
the operation of an open-center and closed-center circuit that meets the functional objective and then consider a closed-center circuit with load sensing.
The principle of operation of the radial piston pump is shown in Fig. 4.16. In this case, the cylinders are positioned radially around the axis of rotation. As the shaft rotates, the connecting rods push the pistons back and forth in the cylinders to develop the pumping action. The design is used to pump both liquids and gases but, in the fluid power industry, it is more commonly used for pneumatic systems.
The axial piston pump has a series of cylinders (typically 7 or 9) mounted parallel to the axis of rotation. (The arrangement is similar to shell chambers in a revolver.) Pistons are installed in the cylinders. Each piston has a spherical end that mounts in a shoe (Fig. 4.12).
The shoe is held against a swashplate by a spring in the cylinder block (not shown in Fig. 4.12). The swashplate remains stationary as the cylinder block rotates with the input shaft. When the swash plate is at an angle to the shaft (as shown in Fig. 4.12), it moves the pistons back and forth in the cylinders as the cylinder block rotates. This movement provides the “pumping action.”
It is helpful to follow the motion of one piston as the cylinder block makes one revolution. As shown in Fig. 4.13, the piston moves to the left as the cylinder block rotates 180° to place the piston at the bottom of the cylinder. It moves to the right as the cylinder block returns to the original position. The reader can readily visualize that, if proper porting is provided, fluid will flow into the cylinder during the first 180° of rotation, and this fluid will be forced out during the second 180° of rotation. Note the bar graph in the figure that shows when the inlet port is open and when the outlet port is open.
Implementation of the design in an actual pump is shown in Fig. 4.14. Three key components, not previously discussed, are identified in this figure.
1. Cylinder block spring. This spring holds the block in position so that the piston shoes are always held in contact with the swashplate. This spring rotates with the cylinder block.
2. Yoke spring assembly. This spring holds the swashplate against the actuator piston.
3. Activator piston. This activator functions like a small hydraulic cylinder. When fluid flows into the cylinder, the piston extends and reduces the angle of the swashplate. The yoke spring assembly is compressed when the swashplate angle is reduced. Like the swashplate, both the actuator piston and the yoke spring assembly are stationary. [Smaller pumps (<15 hp) are actuated directly and do not have the activator piston, unless they are pressure compensated.]
Like the vane pump, an axial piston pump can be configured as a pressurecompensated pump (Fig. 4.15). The outlet pressure (high pressure), Ps, is incident on the end of the compensator valve spool. This pressure multiplied by the area of the spool gives a hydraulic force, Fh, which is opposed by the spring force, Fs, produced by the compensator valve spring. When Ps increases to the point where Fh equals Fs, the spool shifts downward, and fluid flows to the actuator piston.
The pressure at the actuator piston is Pc = Ps ? ?P, where ?P is the pressure drop across the orifice formed when the compensator valve cracks open. As Ps increases, the compensator valve opens more, ?P decreases, and Pc approaches Ps. The increase in Pc increases the hydraulic force produced by the actuating piston, and eventually it rotates the yoke until it is perpendicular to the shaft and pump displacement is zero. The pump will hold this pressure and deliver no flow until something is done to lower the pump outlet pressure.
The pressure-compensated axial piston pump, like the pressure-compensated vane pump, can be used in a circuit without a relief valve. It is good design practice to include the relief valve. The first reason a relief valve is needed is to clip pressure spikes due to load dynamics. The second reason is readily apparent with a more careful examination of Fig. 4.15. Suppose the spool in the compensator valve sticks. [Spool-type valves will “silt-up” if they are actuated infrequently. The silting phenomenon is caused by tiny particles in the fluid (contaminants) being forced into the clearances in the valve. Eventually, the valve spool sticks and can be shifted only with the application of a sizeable force.] If this happens, the system pressure can continue to increase above the deadhead pressure set by the compensator valve spring. A relief valve protects the circuit if this happens.
A vane pump (Fig. 4.7) has a series of vanes that slide back and forth in slots. There are springs in these slots that push the vanes out until the tip contacts the cam ring. (Some designs port pressurized fluid into the slots to force the vanes out.) A chamber is formed between adjacent vanes and the cam ring. As the rotor turns, the chamber decreases in size. Fluid flows into this chamber when it is a maximum size and exits during some ?? of rotation when it is a minimum size. This change in chamber size provides the pumping action.
The principle of operation of a vane-type variable displacement pump is shown in Figs. 4.7 and 4.8. These illustrations are not to scale and are incomplete. Certain features are not shown. The cam ring is held in position in Fig. 4.7 with a threaded rod turned with a hand wheel. This ring will slide to the left when the hand wheel is turned. In the position shown in Fig. 4.6, the cam ring is centered on the axis of rotation. The chambers are equal size at the inlet and outlet, so no fluid is pumped (displacement is zero.) The rotor turns at the same speed, the vane tips are in contact with the cam ring, but no fluid is pumped.
The vane pump can be converted to a pressure-compensated pump by replacing the hand-wheel adjustment with a spring as shown in Fig. 4.9. A small cylinder, identified as the compensator, is placed on the opposite side. Outlet pressure acting on the compensator piston creates a hydraulic force that opposes the spring force. When the outlet pressure rises to a certain point, the hydraulic force becomes greater than the spring force, and the cam ring shifts to the left. As pressure continues to rise, the ring shifts more to the left until it eventually is centered on the axis of rotation (as shown in Fig. 4.8). At this pressure, known as the deadhead pressure, the pump displacement is almost zero. Some flow is produced to replace leakage.
It is now clear how the circuit in Fig. 4.6 can operate without a relief valve. The maximum pressure the pump can develop is limited by the compensator spring in the pressure-compensated pump.
A pressure-compensated pump can maintain deadhead pressure with very little energy input. Hydraulic power output is proportional to pressure × flow. If flow is zero, then hydraulic power output is zero. Some input energy is required to maintain deadhead pressure because of friction and leakage. The advantage of a demand-flow circuit (Fig. 4.6) as compared to a constant-flow circuit (Fig. 4.3) is that pressure is available at the instant the DCV is shifted; it does not have to build from zero.
Typical flow vs. pressure characteristics for a pressure -compensated variable displacement vane pump is shown in Fig. 4.10. When pressure reaches 2900 psi, identified as the cutoff pressure, the cam ring begins to shift, and the pump flow decreases. The rate of decrease (slope of the curve) is set by the spring constant of the compensator spring.
Figure 4.11 shows a pressure compensated variable displacement pump in an exploded view. The cam ring (pressure ring), vanes, and compensator are readily visible. Modern designs, like the one shown, do not use a compensator spring; rather, a specific pilot pressure on a bias piston that holds the pressure ring in place. As the pressure developed at the pump outlet rises, the force developed by the compensator piston eventually becomes large enough to equal the bias piston force and center the pressure ring.
Variable displacement pump circuits are called demand flow circuits. A simple example is shown in Fig. 4.6. (The astute reader will quickly observe that this circuit is the first circuit we have studied with no relief valve. Although no relief valve is shown, it is recommended that a relief valve always be used to ensure that pressure can never reach an unsafe level. Rapid deceleration of a large load can produce a dangerous pressure spike.) Before understanding how a demand flow circuit operates, it is necessary to learn about the operation of a particular variable displacement pump—the vane pump.
A simple circuit with fixed displacement pump is shown in Fig. 4.3. Circuits with fixed displacement pumps are called constant-flow circuits. The key concept is stated below.
Each revolution of a fixed displacement pump delivers a certain volume of fluid to the circuit. This fluid ultimately returns to the reservoir, either as a return flow, or a leakage flow.
In the circuit shown in Fig. 4.3, the fluid passes through the open center directional control valve (DCV) back to the reservoir. The only pressure developed is that required to overcome pressure drops in the lines and through the DCV. When an operator shifts the DCV, flow is diverted to the cylinder. At this instant, the pressure in the circuit is the pressure to overcome the pressure drops. We now see a key disadvantage of a constant-flow circuit. The pressure must build from a very low level to the level required to accelerate the load. In a manufacturing application, cycle time is important; thus, acceleration of the load is a key functional requirement.
Most readers will have observed an operator controlling the speed of cylinder extension by controlling the opening of the DCV. An example would be a backhoe operator who swings the boom around quickly until it nears the target and then smoothly decelerates to stop the boom at the correct location. If the boom circuit is the circuit shown in Fig. 4.1, what happens to the “constant” flow of fluid being delivered by the pump?
To answer the above question, it is necessary to consider the operation of a spool-type directional control valve (DCV). When the spool is shifted to connect pump flow to port A and connect the reservoir to port B, orifices are created at ports A and B (Fig. 4.4). The pressure drops across these orifices are designated ?PDCVA and ?PDCVB , respectively. These orifices raise the total pressure at the relief valve until it cracks open and diverts part of the flow back to the reservoir, thus slowing the actuator speed. It is helpful to consider a specific example.
The relief valve in Fig. 4.3 has the characteristics shown in Fig. 4.5. The cracking pressure is 900 psi, and full flow pressure is 1000 psi, meaning that, when the system pressure reaches 1000 psi, all flow is diverted over the relief valve back to the tank.
The bore of the cylinder is d = 4 in., and the cylinder area ratio is 2:1. Thus, the cap end area is Ac= 12.56 in2 , and the rod end area is Ar= 12.56/2 =6.28 in2. Total flow from the pump is Q = 50 in3/s. When the DCV is fully open to extend the cylinder, the pressure drops in the circuit are as follows:
A no-load test was conducted to determine the friction force. (The seals around the piston and rod produce a friction force that opposes movement of the piston. This force is referred to as the friction force .) During no-load extension, the following pressures were measured at the cylinder ports.
Pi = pressure at inlet = 25 psi
Po = pressure at outlet = 15 psi
The friction force is
For this example, we assume that the friction force is constant. A more detailed analysis would require that we deal with the fact that it is not consistent from initiation of movement to full extension.
The load is 10,000 lbf. We are now ready to determine the total pressure required to extend the cylinder. Pressure at the cylinder outlet port is
Assuming that the operator wishes to reduce this speed to 2 in/s, what must occur? Flow to the cylinder must be reduced from 50 in3/s to 25 in3/s to reduce the cylinder speed to 2 in/s. Referencing Fig. 4.5, the pressure at the relief valve must increase to 950 psi for 25 in3/s (half the pump flow) to be diverted to the tank.
The operator partially closes the DCV to reduce the orifice size at ports A and B. The new pressure drops are
?PDCVA = 80 psi
?PDCVB = 80 psi
The force balance on the cylinder now gives the following pressure at the inlet:
With a slight adjustment to the DCV spool position, the operator will be able to increase the ?PDCV pressure drops and get Prv = 950 psi. The speed of the cylinder will then be 2 in/s. Experienced operators develop such a “feel” for the system that they can operate a cylinder almost like they move their own arms.
In cases where the load is such that operating pressure is close to the cracking pressure of the relief valve, it is possible to have the pump flow divided into three flows.
1. Flow through relief valve back to reservoir
2. Flow to load
3. Leakage flow through DCV back to reservoir (smallest of three flows)
When this happens, the operator, functioning as a feedback loop, has to readjust the DCV position.
Flow across an
orifice converts hydraulic energy to heat energy and thus reduces the efficiency of the circuit. To achieve speed control with the circuit shown in Fig. 4.1, this loss is unavoidable. Since the fluid is being heated by the speed control method, heat rejection from the circuit must be planned accordingly.
Vane and piston pumps are available as fixed or variable displacement pumps. If the cam ring is fixed in position (vane pump), or the swashplate is fixed in position (piston pump), the displacement is fixed. Gear pumps, however, are available only with a fixed displacement. These pumps are less expensive, so they are widely used.
As will be seen in the next section, the leakage in some gear pumps is high, and this leakage increases as the pump wears over time. Ownership cost of fixed-clearance gear pumps is lower, but operating cost is higher over the life of the pump. Total cost (ownership + operating) for the entire design life may or may not be lower for a pump with lower initial cost.
Pressure-balanced gear pumps have volumetric efficiencies that exceed, or at least, rival those of piston pumps. Some of these gear pumps are rated for pressures in excess of 4000 psi.
An external gear pump is shown in Fig. 4.1. One gear is driven with the input shaft, and it drives the second gear. Fluid is captured by the teeth as they pass the inlet, and this oil travels around the housing and exits at the outlet. The design is simple, and it is inexpensive. It is also apparent that there are opportunities for leakage all along the housing as the fluid travels from the inlet to the outlet.
A gear pump design is available with a moveable side plate. When this side plate is out, the fluid is not confined, and no pressure is built. When it is pressed in against the rotating gear, the fluid is confined, and pressure builds. These pumps are used on over-the-road trucks. They are driven by an auxiliary shaft from the engine referred to as the power-takeoff (PTO) shaft. There is no clutch, so the PTO) shaft turns whenever the engine turns. The pump is engaged (pumps fluid) when the side plate is pressed in against the rotating gear.
A gerotor (gear) pump has an inner drive gear and an outer driven gear (Fig. 4.2). As the inner gear rotates, it drives the outer gear. The inner gear has one less tooth than the outer gear, and this feature creates chambers of decreasing volume, and thus the “pumping action.” A port plate ensures that fluid enters the chamber when it is largest and exits when it is smallest.