Work machine

Abstract

To provide a work machine that can appropriately prevent engine lug-down and stall.SOLUTION: A work machine is equipped with a discharge line in which an opening area of a flow path that directs hydraulic oil from a hydraulic pump to a tank is changed by an actuator control valve and a flow control valve, and a controller that controls the actuator control valve and the flow control valve based on the amount of operation of an operating device. The controller calculates a required flow rate of the hydraulic pump based on an operation amount of the operating device, calculates an opening area of the discharge line based on the operation amount of the operating device, predicts the maximum pressure of a hydraulic pump discharge pressure that can rise when a hydraulic actuator suddenly stops based on the required flow rate and the opening area, and controls an engine rotation speed and the hydraulic pump volume based on the predicted maximum pressure so that a torque of the hydraulic pump does not exceed the allowable torque of the engine when it is assumed that the hydraulic pump discharge pressure rises to the maximum pressure.SELECTED DRAWING: Figure 7

Description

[Technical Field]

[0001] The present invention relates to a work machine such as a hydraulic excavator. [Background technology]

[0002] A technique has been proposed for preventing engine lug-down or stall when the load on a hydraulic excavator suddenly increases during operation (see Patent Document 1). Patent Document 1 discloses a work machine that includes a target rotational speed setting means for setting a target rotational speed for the engine, a rotational speed detection means for detecting the actual rotational speed of the engine, and a main controller that calculates the rotational deviation between the target rotational speed and the actual rotational speed, and, if the rotational speed deviation becomes large, performs speed sensing control by reducing the discharge flow rate of the hydraulic pump in accordance with the amount of deviation. [Prior art documents] [Patent documents]

[0003] [Patent Document 1] Japanese Patent Application Laid-Open No. 2012-202219 Summary of the Invention [Problem to be solved by the invention]

[0004] In the control method described in Patent Document 1, the command value (target value) for the displacement of the hydraulic pump is reduced after the actual rotational speed of the engine begins to decrease. Also, the actual displacement of the hydraulic pump decreases with a delay after the command value decreases. For this reason, the control method described in Patent Document 1 may not be able to prevent engine lug-down or stall, depending on the specifications of the hydraulic pump and engine, operating conditions, etc.

[0005] When engine lug-down occurs, the hydraulic pump cannot supply the required flow rate of hydraulic oil to the hydraulic actuator, resulting in a decrease in work efficiency. When engine stall occurs, the hydraulic pump stops, resulting in a decrease in work efficiency. Therefore, in order to prevent a decrease in work efficiency, it is necessary to appropriately prevent engine lug-down and stall.

[0006] An object of the present invention is to provide a work machine that can appropriately prevent the occurrence of engine lug-down and stall. [Means for solving the problem]

[0007] A work machine according to one aspect of the present invention comprises an engine, a hydraulic pump driven by the engine and discharging hydraulic oil, a hydraulic actuator driven by hydraulic oil supplied from the hydraulic pump, an actuator control valve controlling the flow of hydraulic oil supplied from the hydraulic pump to the hydraulic actuator, a flow control valve controlling the flow rate of hydraulic oil discharged from the hydraulic pump to a tank through the actuator control valve, a discharge line in which the opening area of ​​a flow path that guides hydraulic oil from the hydraulic pump to the tank through the actuator control valve and the flow control valve is changed by the actuator control valve and the flow control valve, an operating device that operates the hydraulic actuator, and a controller that controls the actuator control valve and the flow control valve based on the amount of operation of the operating device. The controller calculates the required flow rate of the hydraulic pump based on the amount of operation of the operating device, calculates the opening area of ​​the discharge line based on the amount of operation of the operating device, predicts the maximum pressure of the discharge pressure of the hydraulic pump that can increase if the hydraulic actuator suddenly stops based on the required flow rate and the opening area, and controls the rotational speed of the engine and the displacement of the hydraulic pump based on the predicted maximum pressure so that the torque of the hydraulic pump does not exceed the allowable torque of the engine when it is assumed that the discharge pressure of the hydraulic pump has increased to the maximum pressure. [Effects of the Invention]

[0008] According to the present invention, it is possible to provide a work machine that can appropriately prevent the occurrence of engine lug-down and stall. [Brief explanation of the drawings]

[0009] [Figure 1] FIG. 1 is a side view of a hydraulic excavator according to a first embodiment. [Figure 2] FIG. 2 is a schematic diagram of a hydraulic system provided in a hydraulic excavator. [Figure 3] FIG. 3 is a flowchart illustrating an example of the flow of processing executed by the controller. [Figure 4] FIG. 4 is a diagram showing an example of the first opening table Ta. [Figure 5] FIG. 5 is a diagram showing an example of the second opening table Tb. [Figure 6] FIG. 6 is a diagram showing an example of the torque table Tt. [Figure 7] FIG. 7 is a calculation block diagram showing the flow of command value calculation processing executed by the controller. [Figure 8] FIG. 8 is a time-series change diagram of the operational data of a hydraulic excavator according to a comparative example of this embodiment. [Figure 9] FIG. 9 is a time series diagram showing changes in operational data of the hydraulic excavator according to this embodiment. [Figure 10] FIG. 10 is a calculation block diagram showing the flow of command value calculation processing executed by a controller according to the second embodiment. DETAILED DESCRIPTION OF THE INVENTION

[0010] A construction machine according to an embodiment of the present invention will be described with reference to the drawings. In this embodiment, an example will be described in which the construction machine is a crawler-type hydraulic excavator. The construction machine performs work such as civil engineering work, construction work, demolition work, and dredging work at a work site.

[0011] First Embodiment - Hydraulic excavator configuration - FIG. 1 is a side view of a hydraulic excavator 100 according to the first embodiment. As shown in FIG. 1, the hydraulic excavator 100 includes a vehicle body (machine body) 105 and a working device 104 attached to the vehicle body 105. The vehicle body 105 includes a crawler-type running body 102 and a revolving body 103 that is rotatably provided on the running body 102. The running body 102 travels by driving a pair of left and right crawlers with traveling motors 102A. The revolving body 103 is connected to the running body 102 via a revolving device that has a revolving motor 103A, and is driven by the revolving motor 103A to revolve relative to the running body 102.

[0012] The rotating body 103 includes an operator's cab 118 in which an operator sits, and a machine room 119. The machine room 119 houses, for example, an engine 2 (see FIG. 2) serving as a prime mover, and hydraulic equipment such as a hydraulic pump driven by the engine 2.

[0013] An electric operating device 13 (see FIG. 2) is provided in the operator's cab 118 to operate the hydraulic actuators (111A, 112A, 113A, 103A, 102A) of the working device 104, the revolving body 103, and the traveling body 102. Also provided in the operator's cab 118 is a controller 120, which is a control device that controls the operation of each part of the hydraulic excavator 100.

[0014] The working device 104 is an articulated working device attached to the revolving unit 103, and includes multiple hydraulic actuators and multiple drive target members that are driven by the multiple hydraulic actuators. The working device 104 has a configuration in which three drive target members (a boom 111, an arm 112, and a bucket 113) are connected in series. The base end of the boom 111 is rotatably connected to the front of the revolving unit 103 via a boom pin. The base end of the arm 112 is rotatably connected to the tip of the boom 111 via an arm pin. The bucket 113 is rotatably connected to the tip of the arm 112 via a bucket pin.

[0015] The boom 111 is driven to rotate by the extension and retraction of a boom cylinder 111A, which is a hydraulic actuator. The arm 112 is driven to rotate by the extension and retraction of an arm cylinder 112A, which is a hydraulic actuator. The bucket 113 is driven to rotate by the extension and retraction of a bucket cylinder 113A, which is a hydraulic actuator. By operating the work device 104, the hydraulic excavator 100 can perform work such as excavating earth and sand, leveling work, and compacting the ground.

[0016] Since the boom cylinder 111A, the arm cylinder 112A, and the bucket cylinder 113A have the same configuration, they will be hereinafter also collectively referred to as hydraulic cylinder 11S.

[0017] -Hydraulic system configuration- Fig. 2 is a schematic diagram of the hydraulic system 106 provided in the hydraulic excavator 100. Of the multiple hydraulic actuators provided in the hydraulic excavator 100, Fig. 2 shows only the configuration for driving the boom cylinder 111A as a representative, and does not show the configurations for driving the other hydraulic actuators. In Fig. 2, relief valves, check valves, etc. are not shown.

[0018] 2, the hydraulic system 106 includes a hydraulic pump 1, a tank 12, a cylinder control valve (actuator control valve) 3, a center bypass control valve (flow control valve) 5, and a controller 120. The controller 120 controls the engine 2, the hydraulic pump 1, the cylinder control valve 3, and the center bypass control valve 5 based on the operation amount L of the operation device 13. The controller 120 controls a pressure reducing valve (not shown), which is an electromagnetic valve, to thereby control the cylinder control valve 3. The tank 12 stores hydraulic oil.

[0019] The boom cylinder 111A is provided with a piston that divides the interior of the cylinder tube into a head-side oil chamber (bottom-side oil chamber) 110a and a rod-side oil chamber 110b. The piston is provided with a piston rod. One end of the cylinder tube of the boom cylinder 111A is attached to the rotating body 103, and the piston rod protruding from the other end of the cylinder tube is attached to the boom 111. The boom cylinder 111A is driven by hydraulic oil supplied from the hydraulic pump 1.

[0020] The hydraulic pump 1 is driven by the engine 2, sucks hydraulic oil from a tank 12, and discharges it into a discharge oil passage. The hydraulic pump 1 is a variable displacement hydraulic pump whose discharge capacity (displacement volume per rotation) can be changed. The hydraulic pump 1 is, for example, a swash plate type or bent axis type piston pump.

[0021] The volume (discharge capacity) q of the hydraulic pump 1 is controlled by a regulator 1a attached to the hydraulic pump 1. The regulator 1a has a volume control valve which is an electromagnetic valve, a command pressure chamber to which a pump control command pressure generated by the volume control valve is input, and a volume variable mechanism which changes the volume q according to the pump control command pressure input to the command pressure chamber. For example, the controller 120 may input a required volume q to the volume control valve. req When a pump control command corresponding to the pressure is output, the volume control valve generates a pump control command pressure. When the pump control command pressure is input to the command pressure chamber, the volume q of the hydraulic pump 1 is changed by the volume variable mechanism to the required volume q req The control is performed so that

[0022] The cylinder control valve 3 is a 6-port, 3-position directional flow control valve. The cylinder control valve 3 controls the flow of hydraulic oil supplied from the hydraulic pump 1 to the boom cylinder 111A. The cylinder control valve 3 has a pump port 3a connected to the hydraulic pump 1, a tank port 3b connected to the tank 12, a center bypass inlet port 3c connected to the hydraulic pump 1, a head side port 3d connected to the head side oil chamber 110a of the boom cylinder 111A, a rod side port 3e connected to the rod side oil chamber 110b of the boom cylinder 111A, and a center bypass outlet port 3f connected to the tank 12 via the center bypass control valve 5.

[0023] Pressure receiving chambers (external input sections) 3g, 3h to which operating pressure (external input) is input, and a centering spring are provided at each end of the cylinder control valve 3. When no operating pressure is input to the pressure receiving chambers 3g, 3h at both ends, that is, when the pressure in the pressure receiving chambers 3g, 3h at both ends is tank pressure, the centering spring holds the cylinder control valve 3 in a neutral position. In the neutral position, the center bypass opening PT of the cylinder control valve 3 is fully open, and the meter-in opening PC and meter-out opening CT are fully closed.

[0024] When operating pressure is input to one of the pressure receiving chambers 3g, 3h at both ends, the cylinder control valve 3 moves to one side (to the left in the figure), and the boom cylinder 111A extends. This causes the boom 111 to move in the raising direction. When operating pressure is input to the other of the pressure receiving chambers 3g, 3h at both ends, the cylinder control valve 3 moves to the other side (to the right in the figure), and the boom cylinder 111A retracts. This causes the boom 111 to move in the lowering direction.

[0025] A pressure reducing valve (not shown) reduces the primary pressure of a pilot hydraulic source (not shown) in response to a control signal (control current) from controller 120, and outputs the generated secondary pressure as an operating pressure (command pressure). The pilot hydraulic source is driven by engine 2, and draws hydraulic oil from tank 12 and discharges it to the pilot line. The pilot hydraulic source is, for example, a fixed-displacement hydraulic pump (pilot pump) with a constant discharge capacity (volume). Cylinder control valve 3 is driven by the operating pressure (external input) output from the pressure reducing valve, and controls the flow rate and direction of hydraulic oil supplied from hydraulic pump 1 to boom cylinder 111A.

[0026] The controller 120 controls the pressure in the head-side oil chamber 110a and the rod-side oil chamber 110b of the boom cylinder 111A by adjusting the magnitude of the operating pressure supplied to the cylinder control valve 3 using a pressure reducing valve in accordance with the amount of operation of the operating device 13. This controls the operating speed (extension speed and retraction speed) of the boom cylinder 111A.

[0027] As the operating pressure acting on one of the pair of pressure receiving chambers 3g, 3h increases, the opening areas of the meter-in opening PC and the meter-out opening CT increase, and the opening area A of the center bypass opening PT a When the operating pressure exceeds a predetermined value, the opening area A of the center bypass opening PT decreases. a becomes 0, and all the hydraulic oil discharged from the hydraulic pump 1 is supplied to the hydraulic cylinder 11S. a corresponds to the cross-sectional area of ​​the flow path of the hydraulic oil flowing from the hydraulic pump 1 to the tank 12 in the cylinder control valve 3.

[0028] 2 shows only one cylinder control valve 3, but in reality, the hydraulic system 106 is equipped with a plurality of cylinder control valves 3 that control the hydraulic cylinder 11S. The plurality of cylinder control valves 3 are connected in tandem by a center bypass line Lc that connects the hydraulic pump 1 and the tank 12, and are also connected in parallel by parallel lines (not shown) branching from the center bypass line Lc.

[0029] A center bypass control valve 5 is provided at the most downstream of the center bypass line Lc. That is, the center bypass control valve 5 is provided between the cylinder control valve 3 and the tank 12. The center bypass control valve 5 is a 2-port 2-position flow control valve. The center bypass control valve 5 controls the flow rate of hydraulic oil discharged from the hydraulic pump 1 through the cylinder control valve 3 to the tank 12.

[0030] The center bypass control valve 5 has a pump port 5a connected to the hydraulic pump 1 via the cylinder control valve 3, and a tank port 5b connected to the tank 12. The center bypass control valve 5 operates in response to a control signal (control current I) output from the controller 120 to adjust the opening area A of the internal passage. b It is an electromagnetic proportional valve that can continuously change the opening area A b corresponds to the cross-sectional area of ​​the flow path of the hydraulic oil flowing from the hydraulic pump 1 to the tank 12 in the center bypass control valve 5.

[0031] When the operating device 13 is not operated, the controller 120 does not output a control current I to drive the center bypass control valve 5. When the control current I is not input to the center bypass control valve 5, the opening of the internal passage that connects the pump port 5a and the tank port 5b is fully opened by a spring. The controller 120 increases the control current I as the operation amount L of the operating device 13 increases. The opening area A of the internal passage of the center bypass control valve 5 b The larger the control current I, the smaller the opening area A of the internal passage of the center bypass control valve 5. b becomes 0 when it exceeds a predetermined value. The center bypass control valve 5, together with the cylinder control valve 3, adjusts the flow rate of hydraulic oil flowing from the hydraulic pump 1 to the tank 12 through the center bypass line Lc, that is, the so-called bleed-off flow rate.

[0032] Opening area A of center bypass control valve 5 bBy reducing this, the discharge pressure of the hydraulic pump 1 can be increased. By providing the center bypass control valve 5, it is possible to reduce bleed-off loss and improve operability. Note that as the flow rate and pressure of the hydraulic oil passing through the center bypass control valve 5 increase, the thrust required to drive the valve element increases. In this case, it is preferable to use a pilot-driven center bypass control valve 5.

[0033] As described above, the center bypass line Lc is provided with the cylinder control valve 3 and the center bypass control valve 5. Therefore, the opening area A of the flow path of the center bypass line Lc that guides the hydraulic oil from the hydraulic pump 1 to the tank 12 is c is changed by the cylinder control valve 3 and the center bypass control valve 5.

[0034] The hydraulic system 106 is equipped with a plurality of pressure sensors. The plurality of pressure sensors detects the pressure (hereinafter also referred to as discharge pressure) P P and a discharge pressure sensor 9 that detects the pressure (operating pressure) P of the hydraulic oil in the pressure receiving chambers 3g and 3h of the cylinder control valve 3. a ,P b The pressure sensors (9, 6a, 6b) are connected to a controller 120 and output signals representing the detected pressures (discharge pressure, operating pressure) to the controller 120. The operating pressure P acting on the cylinder control valve 3 is a ,P b There is a certain relationship between the operating pressure and the operation amount L of the operating device 13. Therefore, the operating pressure sensors 6a and 6b can also be said to be operation amount sensors that detect the operation amount L of the operating device 13.

[0035] An operating device 13 that operates the boom cylinder 111A is connected to the controller 120. The operating device 13 has an operating lever 13a that is operated by an operator, and an operating amount sensor 13b that detects the operating amount L of the operating lever 13a (lever operating amount). The operating amount sensor 13b outputs a signal that indicates the detected operating amount L to the controller 120.

[0036] The controller 120 is composed of a computer equipped with a processing device 121 such as a CPU (Central Processing Unit), MPU (Micro Processing Unit), or DSP (Digital Signal Processor), non-volatile memory 122 such as ROM (Read Only Memory), flash memory, or hard disk drive, volatile memory 123 known as RAM (Random Access Memory), an input interface, an output interface, and other peripheral circuits. These pieces of hardware work together to run software and realize multiple functions. The controller 120 may be composed of one computer or multiple computers.

[0037] The nonvolatile memory 122 stores programs capable of executing various calculations. In other words, the nonvolatile memory 122 is a storage medium (storage device) from which programs that realize the functions of this embodiment can be read. The volatile memory 123 is a storage medium (storage device) that temporarily stores the results of calculations performed by the processing device 121 and data input via the input interface. The processing device 121 is a device that loads the programs stored in the nonvolatile memory 122 into the volatile memory 123 and executes the calculations, and performs predetermined calculations on data taken in from the input interface, the nonvolatile memory 122, and the volatile memory 123 in accordance with the programs.

[0038] The input interface converts signals input from various devices (such as the discharge pressure sensor 9, the operating pressure sensors 6a and 6b, and the operating amount sensor 13b) into data that can be calculated by the processing device 121. The output interface also generates an output signal according to the calculation result in the processing device 121, and outputs the signal to various devices (such as the pressure reducing valve for driving the cylinder control valve 3 and the volume control valve of the regulator 1a).

[0039] -Hydraulic system operation- The operation of the hydraulic system 106 will be described with reference to Figure 2. Rotational power generated by the rotation of the engine 2 is transmitted to the hydraulic pump 1, which rotates the hydraulic pump 1. As the hydraulic pump 1 rotates, hydraulic oil is sucked into the hydraulic pump 1 from the tank 12 and discharged downstream of the hydraulic pump 1.

[0040] The discharge flow rate Q of the hydraulic pump 1 is determined by the volume q of the hydraulic pump 1 and the rotation speed Np of the hydraulic pump 1. The rotation speed Np of the hydraulic pump 1 and the rotation speed N of the engine 2 are in a proportional relationship. For the sake of convenience, it is assumed below that the rotation speed Np of the hydraulic pump 1 and the rotation speed N of the engine 2 are equal. Discharge pressure P P is determined by the load on the hydraulic cylinder 11S and the pressure loss in the piping downstream of the hydraulic pump 1. The torque (also referred to as the shaft input torque) T for driving the hydraulic pump 1 is calculated by multiplying the discharge pressure P P , volume q, and pump efficiency η. The engine 2 supplies rotational power (rotational power=rotational speed N×shaft input torque T) to the hydraulic pump 1.

[0041] The cylinder control valve 3 is operated by a pressure reducing valve (not shown) in response to an operation amount L of the operating device 13. a ,P b and controls the direction and flow rate of the pressure oil discharged from the hydraulic pump 1.

[0042] When the boom raising operation is performed by the operation device 13, the controller 120 outputs a control signal (control current) to the pressure reducing valve (not shown) to extend the boom cylinder 111A. ais input to the pressure receiving chamber 3g, the spool of the cylinder control valve 3 moves to the left in the figure. This connects the hydraulic pump 1 to the head side oil chamber 110a of the boom cylinder 111A, and connects the rod side oil chamber 110b of the boom cylinder 111A to the tank 12. Pressurized oil discharged from the hydraulic pump 1 flows into the head side oil chamber 110a of the boom cylinder 111A, and hydraulic oil in the rod side oil chamber 110b of the boom cylinder 111A flows out to the tank 12. As a result, the boom cylinder 111A extends, and the boom (controlled object) 111 is raised by the boom cylinder (hydraulic cylinder) 111A.

[0043] On the other hand, when a boom lowering operation is performed by the operating device 13, the controller 120 outputs a control signal (control current) to the pressure reducing valve (not shown) to contract the boom cylinder 111A. b is input to the pressure receiving chamber 3h, the spool of the cylinder control valve 3 moves to the right in the figure. As a result, the hydraulic pump 1 and the rod-side oil chamber 110b of the boom cylinder 111A communicate with each other, and the head-side oil chamber 110a of the boom cylinder 111A communicates with the tank 12. Pressurized oil discharged from the hydraulic pump 1 flows into the rod-side oil chamber 110b of the boom cylinder 111A, and hydraulic oil in the head-side oil chamber 110a of the boom cylinder 111A flows out to the tank 12. As a result, the boom cylinder 111A contracts, and the boom (controlled object) 111 is lowered by the boom cylinder (hydraulic cylinder) 111A.

[0044] - Overview of control for improving energy efficiency (fuel efficiency) - In order to reduce the fuel consumption of the hydraulic excavator 100, it is important to increase the volume q of the hydraulic pump 1 and reduce the rotation speed N of the engine 2 as much as possible. However, when the volume q of the hydraulic pump 1 is increased, the torque (shaft input torque) T of the hydraulic pump 1 increases. Here, as the rotation speed N of the engine 2 decreases, the allowable torque T max In this case, the lower the rotation speed N is, the smaller the allowable torque T maxTherefore, when using such an engine 2, the torque T of the hydraulic pump 1 becomes smaller than the allowable torque T max Therefore, the volume q must be adjusted taking this into consideration.

[0045] The controller 120 according to this embodiment calculates the required flow rate q of hydraulic oil to the hydraulic cylinder 11S (the required flow rate of hydraulic oil discharged from the hydraulic pump 1) determined from the operation amount L of the operation device 13. req , the rotation speed-torque characteristics of the engine 2, and the discharge pressure P of the hydraulic pump 1 detected by the discharge pressure sensor 9. P The rotation speed N of the engine 2 and the volume q of the hydraulic pump 1 are controlled using this control. However, with this control alone, when the engine 2 is operating at a low rotation speed, if the tip of the working implement 104 hits a rock, the load in the hydraulic circuit suddenly increases, and the discharge pressure P P If the torque T of the hydraulic pump 1 rises suddenly, the allowable torque T of the engine 2 rises suddenly. max In other words, there is a risk of engine 2 lagging down. In particular, as the rotation speed N decreases, the allowable torque T max If the engine 2 has a characteristic that reduces the allowable torque T max Therefore, even if the volume q of the hydraulic pump 1 is reduced after detecting that lag-down has occurred, the rotation speed N of the engine 2 cannot be increased (returned to the command value) immediately due to the response delay of the hydraulic pump 1, and there is a possibility that the engine 2 will stall.

[0046] - Overview of control to prevent engine lag-down - Therefore, the controller 120 according to this embodiment determines the opening area A of the center bypass line (discharge line) Lc that guides the hydraulic oil from the hydraulic pump 1 to the tank 12 through the cylinder control valve 3 and the center bypass control valve 5 based on the operation amount L of the operating device 13. c The controller 120 calculates the required flow rate Q reqand opening area A c Based on this, the maximum pressure P of the discharge pressure of the hydraulic pump 1 that can rise when the working implement 104 comes into contact with a rock or the like and the hydraulic cylinder 11S suddenly stops is calculated. max Furthermore, the controller 120 predicts the predicted maximum pressure P max Based on this, the discharge pressure P of the hydraulic pump 1 P is the maximum pressure P max The torque T of hydraulic pump 1 when it is assumed that the torque T of engine 2 is increased to the allowable torque T max The rotation speed N of the engine 2 and the volume q of the hydraulic pump 1 are controlled so that the rotation speed N does not exceed the value.

[0047] -Functions and processing flow of the controller according to the first embodiment- The processing performed by the controller 120 will be described in detail with reference to Figs. 3 to 7. Fig. 3 is a flowchart showing an example of the flow of processing executed by the controller 120. The flowchart shown in Fig. 3 is started, for example, when the ignition switch of the hydraulic excavator 100 is turned on, and is repeatedly executed at a predetermined control period. For convenience of explanation, in the equations described below, the flow coefficient, efficiency, and coefficients for unit conversion are omitted, and the tank pressure is assumed to be 0. Furthermore, although the hydraulic pump 1 may be connected to the engine 2 via gears or the like, in this embodiment, the description will be given assuming that the rotation speed Np of the hydraulic pump 1 is equal to the rotation speed N of the engine 2.

[0048] 3, in step S110, the controller 120 executes a process of acquiring the sensor values. In the process of acquiring the sensor values, the controller 120 acquires the operation amount L detected by the operation amount sensor 13b, the discharge pressure P detected by the discharge pressure sensor 9, P , and the operating pressure P detected by the operating pressure sensors 6a and 6b a ,P b The larger of these is the operating pressure P c Obtain as.

[0049] In the next step S115, the controller 120 refers to a predetermined flow rate table and determines a required flow rate Q, which is a command value (target value) for the flow rate of the hydraulic oil discharged from the hydraulic pump 1, based on the manipulated variable L acquired in step S110. req The flow rate table is calculated by dividing the operation amount L and the required flow rate Q req The flow rate table is a data table that defines the relationship between the required flow rate Q and the operation amount L, and is stored in the nonvolatile memory 122. req In addition, the operating amount L and the required flow rate Q are calculated using a formula instead of a flow rate table. req The relationship may be specified.

[0050] In the next step S120, the controller 120 calculates the required flow rate Q calculated in step S115. req The maximum rotation speed N of hydraulic pump 1 max The required volume at maximum rotation speed, q, is calculated by dividing req (N max That is, the controller 120 calculates the required volume q at the maximum rotation speed using the following equation (1): req (N max ) is calculated. Required volume q at maximum rotation speed req (N max ) means that the hydraulic pump 1 is rotated at the maximum speed N max When rotating at , the required flow rate Q req is the volume (discharge capacity) of the hydraulic pump 1 required to discharge the hydraulic oil.

[0051]

number

[0052] In the next step S125, the controller 120 calculates the required volume q req (N max ) and the discharge pressure P obtained in step S110 P By multiplying this, the required torque T of hydraulic pump 1 is req (N maxThat is, the controller 120 calculates the required torque T req (N max ) is calculated. req (N max ) is the maximum rotation speed of hydraulic pump 1 N max When rotating at , the required flow rate Q req is the torque of the hydraulic pump 1 required to discharge the hydraulic oil.

[0053]

number

[0054] In the next step S130, the controller 120 refers to a predetermined first opening table Ta (see FIG. 4) and calculates the operating pressure P c Based on this, the opening area A of the center bypass opening PT of the cylinder control valve 3 a Calculate the following.

[0055] 4 is a diagram showing an example of the first opening table Ta. The first opening table Ta shown in FIG. 4 is created in advance based on experimental data and the like, and is stored in the nonvolatile memory 122 of the controller 120. The controller 120 according to this embodiment uses the operating pressure P detected by the operating pressure sensor 6a. a A positive value of the operating pressure P c and the operating pressure P detected by the operating pressure sensor 6b b is a negative value of the operating pressure P c As shown in FIG. 4, the first opening table Ta is obtained by the operation pressure P c When is 0, the opening area A a is the maximum opening area A ax and the operating pressure P c The larger the absolute value (magnitude) of a In addition, the operating pressure P c and opening area A a The relationship between the opening area A and the a may be calculated.

[0056] As shown in FIG. 3, in step S130, the opening area A a When the calculation process of step S135 is completed, the process proceeds to step S135. In step S135, the controller 120 refers to a predetermined current table and calculates the control current (external input value) I of the center bypass control valve 5 based on the operation amount L acquired in step S110. The current table is a data table that defines the relationship between the operation amount L and the control current I of the center bypass control valve 5, and is stored in the non-volatile memory 122. The current table defines the characteristic that the control current I increases as the operation amount L increases. Note that the relationship between the operation amount L and the control current I may be defined by a formula instead of the current table, and the control current I may be calculated using the formula.

[0057] In the next step S140, the controller 120 refers to a predetermined second opening table Tb (see FIG. 5) and determines the opening area A of the center bypass control valve 5 based on the control current I calculated in step S135. b Calculate the following.

[0058] 5 is a diagram showing an example of the second opening table Tb. The second opening table Tb shown in FIG. 5 is created in advance based on experimental data and the like, and is stored in the nonvolatile memory 122 of the controller 120. As shown in FIG. 5, the second opening table Tb stores the opening area A b is the maximum opening area A bx As the control current I increases, the opening area A b In place of the second aperture table Tb, the control current I and aperture area A are calculated by a formula. b The relationship between the opening area A and the b may be calculated.

[0059] As shown in FIG. 3, in step S140, the opening area A b When the calculation process is completed, the process proceeds to step S145. In step S145, the controller 120 calculates the opening area Aa and the opening area A calculated in step S140. b Based on this, the opening area A of the center bypass line Lc c Calculate the opening area A of the center bypass line Lc c For example, the opening area A a Opening and opening area A b The synthetic aperture area when the apertures are arranged in series is calculated using formula (3).

[0060]

number

[0061] In addition, the opening area A of the center bypass line Lc c The method of calculating the opening area A is not limited to the case where formula (3) is used. a and opening area A b The smaller of these is the opening area A of the center bypass line Lc. c It may be considered as such.

[0062] In the next step S150, the controller 120 refers to a predetermined torque table Tt (see FIG. 6) and calculates the required rotation speed N , which is the command value of the rotation speed of the engine 2 calculated in the immediately previous control cycle. req Based on this, the allowable torque T of engine 2 max The required rotation speed of engine 2, N req The initial value of is the maximum rotation speed N max is.

[0063] Fig. 6 is a diagram showing an example of the torque table Tt. The torque table Tt shown in Fig. 6 is determined based on the specifications of the engine 2 and is stored in the non-volatile memory 122 of the controller 120. As shown in Fig. 6, the torque table Tt is a relationship between the rotation speed of the engine 2 and the allowable torque T maxAs shown in the figure, in this embodiment, the rotation speed-torque characteristic of the engine 2 is a linear characteristic. The torque table Tt defines a relationship in which the allowable torque T max becomes larger, and in the range of the predetermined speed N0 or more, the allowable torque T max The torque table Tt is replaced by a mathematical formula to determine the rotation speed of the engine 2 and the allowable torque T max The relationship between the torque and the allowable torque T max may be calculated.

[0064] In this embodiment, the predetermined speed N0 is set to the maximum rotation speed N max The range of the rotation speed of the engine 2 according to this embodiment is set as follows: min Maximum rotation speed N max The range is as follows (N min >0,N max =N0,N min <N max Therefore, the allowable torque T max is the minimum rotation speed of engine 2, N min to maximum rotation speed N max In the entire range up to , the lower the rotation speed of the engine 2, the smaller the value.

[0065] As shown in FIG. 3, in the next step S160, the controller 120 calculates the required rotation speed N req and the required volume q, which is the command value (target value) output to hydraulic pump 1. req In the command value calculation process (S160), the controller 120 calculates the discharge pressure P P , the required flow rate Q calculated in step S115 req , the required volume q calculated in step S120 req (N max ), the required torque T calculated in step S125req (N max ), the opening area A of the center bypass line Lc calculated in step S145 c , and the allowable torque T calculated in step S150 max Based on this, the required rotation speed of engine 2 is N req , and the required volume q of hydraulic pump 1 req Calculate the following.

[0066] The command value calculation process (S160) will be described in detail with reference to Fig. 7. Fig. 7 is a calculation block diagram showing the flow of the command value calculation process executed by the controller 120. As shown in Fig. 7, the controller 120 functions as an excess torque calculation unit 131, an increased volume calculation unit 132, a pressure prediction unit 133, an upper limit volume calculation unit 134, a maximum volume calculation unit 136, a minimum value selection unit 137, and a command value calculation unit 138.

[0067] As shown in FIG. 7, the surplus torque calculation unit 131 calculates the allowable torque T max The required torque T at the maximum rotation speed calculated in step S125 req (N max ) to obtain the surplus torque T mrg Calculate (T mrg =T max -T req (N max )) Reserve torque T mrg is the allowable torque T max The required torque T req This is the torque that can be increased from the maximum torque limit, i.e., the margin value.

[0068] The increased volume calculation unit 132 calculates the surplus torque T mrg The discharge pressure P of the hydraulic pump 1 acquired in step S110 P The increase in volume q can be calculated by dividing inc That is, the increased volume calculation unit 132 calculates the increaseable volume q using the following equation (4): inc Calculate the volume q incis the torque T of hydraulic pump 1 when engine 2 is rotated at its maximum rotation speed, and is the maximum torque T of engine 2. max This is the volume (discharge capacity) of the hydraulic pump 1 that can be increased within a range not exceeding .

[0069]

number

[0070] The maximum volume calculation unit 136 calculates the incrementable volume q calculated by the increment volume calculation unit 132. inc The required volume q at the maximum rotation speed calculated in step S120 req (N max ) to obtain the maximum volume q inc(max) Calculate the maximum volume q inc(max) is the current discharge pressure P P As it is, the torque T of hydraulic pump 1 is equal to the allowable torque T of engine 2. max This is the maximum volume that does not exceed.

[0071] The pressure prediction unit 133 refers to a predetermined pressure prediction map Mp and calculates the opening area A of the center bypass line Lc calculated in step S145. c and the required flow rate Q calculated in step S115 req Based on this, the maximum pressure (predicted value) P max Calculate the maximum pressure P max is the discharge pressure P that can be reached when the load suddenly increases, for example, when the bucket 113 comes into contact with a rock or the like during work. P The pressure prediction map Mp is created in advance based on experimental data or the like, and is stored in the nonvolatile memory 122 of the controller 120.

[0072] The pressure prediction map Mp is c The pressure prediction map Mp includes a plurality of pressure prediction tables Tp1 to Tp5 corresponding to the magnitude of the required flow rate Q req The larger the maximum pressure P max becomes larger, and the opening area A c The smaller the maximum pressure P maxIn the figure, the opening area A c The pressure prediction table Tp1 with the largest opening area A is shown by a dotted line. c The pressure prediction table Tp5 with a small opening area A c As the flow rate increases, the required flow rate Q req Maximum pressure P max The slope (rate of change) of becomes smaller.

[0073] The pressure prediction unit 133 calculates the opening area A of the center bypass line Lc. c Select the pressure prediction table Tp corresponding to the magnitude of the required flow rate Q req Based on the maximum pressure P max For example, the pressure prediction unit 133 calculates the opening area A c When the opening area is the maximum, the pressure prediction table Tp1 with the dotted line is selected, and the required flow rate Q is calculated by referring to the selected pressure prediction table Tp1. req Based on the maximum pressure P max Determine.

[0074] In addition, the required flow rate Q is calculated by a formula instead of the pressure prediction map Mp. req , opening area A c , and maximum pressure P max The relationship between the maximum pressure P max For example, the pressure prediction unit 133 may calculate the maximum pressure P max may be calculated.

[0075]

number

[0076] In equation (5), ρ is the density of the hydraulic oil, and C is the flow coefficient. As can be seen from equation (5), it is assumed that the bucket 113 comes into contact with a rock or the like during work, causing the hydraulic cylinder 11S to suddenly stop, and the entire flow rate of the hydraulic oil discharged from the hydraulic pump 1 is discharged to the tank 12 through the center bypass line Lc. As a result, for example, if the tip of the working implement 104 collides with a rock, the pressure inside the hydraulic cylinder 11S will suddenly rise and the discharge pressure P P It is possible to predict the maximum pressure in the hydraulic circuit (the maximum pressure in the oil passage connecting the hydraulic pump 1 and the cylinder control valve 3) when the pressure suddenly rises.

[0077] The operation amount L is large, and the opening area A of the center bypass line Lc c When the working implement 104 collides with a rock, the discharge pressure P P If the pressure in the hydraulic circuit increases suddenly, the pressure in the hydraulic circuit will increase to the set pressure of the relief valve (not shown) provided in the hydraulic circuit. On the other hand, the opening area A of the center bypass line Lc c is greater than 0, the hydraulic oil discharged from the hydraulic pump 1 can be discharged to the tank 12 through the center bypass line Lc. P If the pressure in the hydraulic circuit rises suddenly, it is assumed that the pressure in the hydraulic circuit will not rise to the set pressure of the relief valve. P The maximum pressure P max Predict.

[0078] The upper limit volume calculation unit 134 calculates the allowable torque T max The maximum pressure P calculated by the pressure prediction unit 133 max By dividing by q, the upper volume max That is, the upper limit volume calculation unit 134 calculates the predicted maximum pressure P max Based on this, the upper volume q max Calculate the upper limit volume q max is the discharge pressure P of hydraulic pump 1 Pis the maximum pressure P max Assuming that the torque T of the hydraulic pump 1 is increased to the allowable torque T of the engine 2, max This is the upper limit of the volume (discharge capacity) that cannot be exceeded.

[0079]

number

[0080] The minimum value selection unit 137 selects the maximum volume q calculated by the maximum volume calculation unit 136. inc(max) , and the upper limit volume q calculated by the upper limit volume calculation unit 134 max Select the smaller of the two.

[0081] The command value calculation unit 138 calculates the required flow rate Q req The volume q selected by the minimum value selection unit 137 s (=Min(q inc(max) ,q max )) to obtain the required rotation speed N req That is, the command value calculation unit 138 calculates the required rotation speed N req Calculate.

[0082]

number

[0083] Here, for example, when the operating device 13 is being finely operated, the required rotation speed N req is the minimum rotation speed N min The command value calculation unit 138 calculates the required rotation speed N req is the minimum rotation speed N min In the following cases, the minimum rotation speed N min Request rotation speed N req Determine as (N req =N min In this case, the command value calculation unit 138 calculates the required flow rate Q req Request rotation speed N req (=Nmin ) to obtain the required volume q req That is, the command value calculation unit 138 calculates the required volume q using the following equation (8): req Calculate the following.

[0084] Also, volume q is selected volume q s In this case, even if the rotation speed of the engine 2 is increased to the maximum rotation speed, the discharge flow rate may be insufficient. In this case, the calculation result is req is the maximum rotation speed N max The command value calculation unit 138 calculates the required rotation speed N req is the maximum rotation speed N max In the above cases, the maximum rotation speed N max Request rotation speed N req Determine as (N req =N max In this case, the command value calculation unit 138 calculates the required flow rate Q req Request rotation speed N req (=N max ) to obtain the required volume q req That is, the command value calculation unit 138 calculates the required volume q using the following equation (8): req Calculate the following.

[0085]

number

[0086] The command value calculation unit 138 calculates the required rotation speed N req is the minimum rotation speed N min Larger, maximum rotation speed N max If it is smaller, the volume q selected by the minimum value selection unit 137 s The requested volume q req That is, the command value calculation unit 138 determines the required volume q using the following equation (9): req Calculate the following.

[0087]

number

[0088] The command value calculation unit 138 calculates the required rotation speed N req to the fuel injection device of the engine 2. In other words, the command value calculation unit 138 calculates the actual rotation speed N of the engine 2 based on the requested rotation speed N req The controller 120 may control the fuel injection device via an engine controller. The command value calculation unit 138 calculates the required volume q req to the regulator 1a of the hydraulic pump 1. In other words, the command value calculation unit 138 calculates the actual volume q of the hydraulic pump 1 in accordance with the required volume q. req The controller 120 controls the regulator 1a so that the following holds: The controller 120 may control the regulator 1a via a pump controller.

[0089] The controller 120 may control the engine 2 and the hydraulic pump 1 by feedforward control, or may control the engine 2 and the hydraulic pump 1 by feedback control based on the rotational speed N detected by a rotational speed sensor of the engine 2 and the volume q detected by a volume sensor of the hydraulic pump 1.

[0090] As described above, the command value calculation unit 138 calculates the volume q s The volume q of the hydraulic pump 1 is controlled based on the above. max If selected (q s =q max ), the command value calculation unit 138 calculates the upper limit volume q max and the required flow rate Q of hydraulic pump 1 req Based on this, the required rotation speed N of engine 2 is req Then, the command value calculation unit 138 calculates the upper limit volume q max Based on this, the volume q of the hydraulic pump 1 is controlled, and the required rotation speed N req On the other hand, the minimum value selection unit 137 selects the maximum volume q inc(max) If selected (qs =q inc(max) ), the command value calculation unit 138 calculates the maximum volume q inc(max) and the required flow rate Q of hydraulic pump 1 req Based on this, the required rotation speed N of engine 2 is req Then, the command value calculation unit 138 calculates the maximum volume q inc(max) Based on this, the volume q of the hydraulic pump 1 is controlled, and the required rotation speed N req The rotation speed N of the engine 2 is controlled based on the above.

[0091] The effects of this embodiment will be described in comparison with a comparative example with reference to Figs. 8 and 9. Fig. 8 is a time series change diagram of operational data of a hydraulic excavator according to a comparative example of this embodiment. Fig. 9 is a time series change diagram of operational data of the hydraulic excavator 100 according to this embodiment. The horizontal axis of each graph represents time. The vertical axis of graph (a) represents the operation amount L of the operating device 13. The vertical axis of graph (b) represents the rotation speed N of the engine 2. The vertical axis of graph (c) represents torque. In graph (c), the solid line represents the torque T of the hydraulic pump 1, and the dashed dotted line represents the allowable torque T of the engine 2. max The vertical axis of graph (d) represents the flow rate (discharge flow rate) of hydraulic oil discharged from the hydraulic pump 1. In graph (d), the solid line represents the actual flow rate Q of hydraulic oil discharged from the hydraulic pump 1, and the dotted line represents the required flow rate Q, which is the command value (target value) of the flow rate of hydraulic oil discharged from the hydraulic pump 1. req The vertical axis of graph (e) represents the volume of the hydraulic pump 1. In graph (e), the solid line represents the actual volume q of the hydraulic pump 1, and the dotted line represents the required volume q, which is the command value (target value) of the volume of the hydraulic pump 1. req represents.

[0092] The controller according to the comparative example of this embodiment, like the conventional technology, controls the rotation speed N of the engine 2 to a constant value. Also, like the conventional technology, the controller of the comparative example controls to reduce the displacement of the hydraulic pump 1 when the rotation speed N of the engine 2 falls below a predetermined threshold, that is, when the difference between the actual measurement value and the target value of the rotation speed of the engine 2 exceeds a predetermined threshold.

[0093] As shown in Figures 8 and 9, the operation amount L increases from time t0 and reaches a predetermined operation amount that is smaller than the maximum operation amount at time t1. At the predetermined operation amount, the opening area A of the center bypass line Lc c is a predetermined value greater than 0. Time t2 is the time when the working implement 104 comes into contact with a rock or the like and the hydraulic cylinder 11S suddenly stops. In other words, time t2 corresponds to the time when the load begins to increase suddenly.

[0094] As shown in Fig. 8, in the comparative example, the torque T of the hydraulic pump 1 increases and the rotation speed N of the engine 2 decreases from time t2. As the rotation speed N of the engine 2 decreases, the allowable torque T max At time t30, when the rotation speed N of the engine 2 falls below a predetermined threshold, the controller adjusts the required volume q req In this way, in the comparative example, the required volume q req Furthermore, due to the response delay of the hydraulic pump 1, the required volume q req It takes a predetermined time for the volume q to actually decrease after the volume q is decreased. Therefore, the actual volume q of the hydraulic pump 1 starts to decrease at time t40, when a predetermined time has elapsed since time t30.

[0095] The flow rate Q of the hydraulic oil discharged from the hydraulic pump 1 starts to decrease from time t2 as the rotation speed N of the engine 2 decreases, and further decreases from time t40 as the actual volume q of the hydraulic pump 1 starts to decrease. As the actual volume q of the hydraulic pump 1 decreases from time t40, the torque T of the hydraulic pump 1 decreases. As the torque T of the hydraulic pump 1 decreases, the rotation speed N of the engine 2 begins to increase. As the rotation speed N of the engine 2 increases, the allowable torque T max As the rotation speed N of engine 2 increases, the required volume q req The torque T of the hydraulic pump 1 decreases and the allowable torque T max increases, so at time t50 the torque T of hydraulic pump 1 becomes the allowable torque Tmax After that, the flow rate Q of the hydraulic oil discharged from the hydraulic pump 1 increases with the increase in the rotation speed N of the engine 2 and the increase in the actual volume q, and the required flow rate Q req reaches.

[0096] In this way, in the comparative example, the time from time t2 to time t40 when the actual volume q of the hydraulic pump 1 starts to decrease is long, so between time t30 and time t40, the torque T of the hydraulic pump 1 decreases to the allowable torque T max If the rotation speed N of engine 2 further decreases due to lag down, the allowable torque T max This also reduces the engine speed, which may result in stalling.

[0097] As shown in FIG. 9, in this embodiment, as in the comparative example, the torque T of the hydraulic pump 1 is small from time t0 to time t2. In other words, a light load operation is being performed. In the comparative example, the rotation speed N of the engine 2 is constant regardless of the load (see FIG. 8). In contrast, in this embodiment, the rotation speed N of the engine 2 is reduced during a light load operation, resulting in good energy efficiency (fuel economy). However, as the rotation speed N of the engine 2 is reduced, the allowable torque T max The size has also become smaller (see Figure 6).

[0098] Light-load operations include, for example, the operation of moving an empty bucket 113 to the excavation start point when repeatedly excavating an excavation target, the operation of moving the empty bucket 113 toward the loading target after loading when scooping up the loading target with the bucket 113 and transporting it to a dump truck for loading, and the operation of dredging the ground surface during ground leveling work.

[0099] If the bucket 113 comes into contact with a rock or the like at time t2 during light load operation, the load will rise suddenly. However, in this embodiment, the volume q of the hydraulic pump 1 is controlled in advance with a margin. Specifically, from time t1 onwards, the volume q of the hydraulic pump 1 is controlled to the upper limit volume qmax. For this reason, although the torque T of the hydraulic pump 1 increases from time t2, the rate of increase (slope) of the torque T is smaller than in the comparative example. For this reason, the required volume q req When the torque T of the hydraulic pump 1 is reduced, even if a predetermined time has passed until the time t31 when the volume q actually starts to decrease, the torque T of the hydraulic pump 1 is max It will never exceed.

[0100] The flow rate Q of the hydraulic oil discharged from the hydraulic pump 1 starts to decrease from time t2, and further decreases from time t31 as the actual volume q of the hydraulic pump 1 starts to decrease. The actual volume q of the hydraulic pump 1 decreases to the required volume q at time t41. req Since this coincides with the flow rate Q of hydraulic oil discharged from the hydraulic pump 1, the flow rate Q of hydraulic oil discharged from the hydraulic pump 1 increases as the rotation speed N of the engine 2 increases from time t41, and the required flow rate Q req reaches.

[0101] -Effects of the first embodiment- According to the first embodiment, the following effects are achieved.

[0102] (1) A hydraulic excavator (work machine) 100 includes an engine 2, a hydraulic pump 1 driven by the engine 2 and discharging hydraulic oil, a hydraulic cylinder (hydraulic actuator) 11S driven by hydraulic oil supplied from the hydraulic pump 1, a cylinder control valve (actuator control valve) 3 that controls the flow of hydraulic oil supplied from the hydraulic pump 1 to the hydraulic cylinder 11S, a center bypass control valve (flow control valve) 5 that controls the flow rate of hydraulic oil discharged from the hydraulic pump 1 through the cylinder control valve 3 to a tank 12, a center bypass line (discharge line) Lc in which the opening area of ​​the flow path that guides hydraulic oil from the hydraulic pump 1 to the tank 12 through the cylinder control valve 3 and the center bypass control valve 5 is changed by the cylinder control valve 3 and the center bypass control valve 5, an operating device 13 that operates the hydraulic cylinder 11S, and a controller 120 that controls the cylinder control valve 3 and the center bypass control valve 5 based on the amount of operation of the operating device 13. The controller 120 calculates the required flow rate Q of the hydraulic pump 1 based on the operation amount of the operation device 13 (in this embodiment, the operation amount L detected by the operation amount sensor 13b). req The controller 120 calculates the amount of operation of the operating device 13 (in this embodiment, the operating pressure P c and the operation amount L detected by the operation amount sensor 13b, the opening area A of the center bypass line Lc is calculated. c The controller 120 calculates the required flow rate Q req and opening area A c Based on this, the maximum pressure P of the discharge pressure of the hydraulic pump 1 that can rise when the hydraulic cylinder 11S suddenly stops is max The controller 120 predicts the predicted maximum pressure P max Based on this, the discharge pressure of the hydraulic pump 1 is the maximum pressure P max The torque T of hydraulic pump 1 when it is assumed that the torque T of engine 2 is increased to the allowable torque T max The rotation speed N of the engine 2 and the volume q of the hydraulic pump 1 are controlled so that the rotation speed N does not exceed the value.

[0103] In this configuration, the predicted maximum pressure P maxThe rotation speed N of the engine 2 and the volume q of the hydraulic pump 1 are controlled using the above formula, so that the allowable torque T max Therefore, even if the load actually increases suddenly and the hydraulic cylinder 11S suddenly stops, the torque T of the hydraulic pump 1 is greater than the allowable torque T of the engine 2. max Therefore, according to this embodiment, it is possible to appropriately prevent lug-down and stalling of the engine 2. As a result, it is possible to supply hydraulic oil at a required flow rate from the hydraulic pump 1 to the hydraulic cylinder 11S, and therefore it is possible to prevent a decrease in work efficiency caused by lug-down and stalling of the engine 2.

[0104] (2) As shown in FIG. 7, the controller 120 (upper limit volume calculation unit 134) calculates the predicted maximum pressure P max Based on this, the discharge pressure P of the hydraulic pump 1 P is the maximum pressure P max Assuming that the torque T of the hydraulic pump 1 is increased to the allowable torque T of the engine 2, max The upper limit of the volume that does not exceed the upper limit volume q max The controller 120 (command value calculation unit 138) calculates the upper limit volume q max If is selected, the upper limit volume q max and the required flow rate Q of hydraulic pump 1 req Based on this, the required rotation speed N of engine 2 is req The controller 120 calculates the upper limit volume q max The controller 120 controls the displacement q of the hydraulic pump 1 based on the required rotation speed N req The rotation speed N of the engine 2 is controlled based on the above.

[0105] In this configuration, the volume q of the hydraulic pump 1 can be made as large as possible, and the rotation speed N of the engine 2 can be made as low as possible. This reduces losses that depend on the rotation speed N, and allows the hydraulic pump 1 to be used in its most efficient range, thereby improving energy efficiency (fuel efficiency).

[0106] (3) The hydraulic excavator 100 is configured to operate at a discharge pressure P P As shown in FIG. 7, the controller 120 (an excess torque calculation unit 131, an increased volume calculation unit 132, and a maximum volume calculation unit 136) calculates the discharge pressure P detected by the discharge pressure sensor 9. P Based on this, the torque T of the hydraulic pump 1 is equal to the allowable torque T of the engine 2. max The maximum volume q is the maximum volume that does not exceed inc(max) The controller 120 (the minimum value selection unit 137 and the command value calculation unit 138) calculates the maximum volume q inc(max) and upper limit volume q max The volume q of the hydraulic pump 1 is controlled based on the smaller one of the above.

[0107] In this configuration, the operation amount L is small, and the opening area A of the center bypass line Lc c When is sufficiently large, there is little pressure and there is no risk of lag down occurring, so the volume q is set to the maximum volume q inc(max) On the other hand, if the operation amount L is relatively large and the opening area A of the center bypass line Lc is c When is small, it is predicted that the pressure will rise suddenly, so the volume q is set in advance to the upper limit volume q max By adjusting it to this, you can prevent lag down.

[0108] (4) The hydraulic excavator 100 can improve energy efficiency by reducing the rotation speed N of the engine 2. As shown in FIG. 6, the operating range of the engine 2 according to this embodiment is such that the allowable torque T of the engine 2 decreases as the rotation speed N of the engine 2 decreases. max is set to the rotation speed range where N min ≦N≦N max Therefore, if the load suddenly increases while the engine 2 is being operated in a low rotation speed range in order to improve energy efficiency, the engine 2 is likely to lag down. However, as described above, the controller 120 according to this embodiment can calculate the maximum pressure P maxBy predicting this, the volume q can be adjusted in advance to prevent lug-down of the engine 2. Therefore, even if the engine 2 has characteristics that make it prone to lug-down (see FIG. 6), lug-down can be effectively prevented.

[0109] (5) The controller 120 controls the operating pressure P c Based on this, the opening area A, which is the cross-sectional area of ​​the flow path of the hydraulic oil flowing from the hydraulic pump 1 to the tank 12 in the cylinder control valve 3, is a The controller 120 calculates the opening area A, which is the cross-sectional area of ​​the flow path of the hydraulic oil flowing from the hydraulic pump 1 to the tank 12 in the center bypass control valve 5, based on the control current I, which represents the operation amount of the operating device 13. b The controller 120 calculates the opening area A of the cylinder control valve 3. a and the opening area A of the center bypass control valve 5 b Based on this, the opening area A of the center bypass line Lc c In this configuration, the controller 120 calculates the opening area of ​​the cylinder control valve 3 and the center bypass control valve 5, respectively, to calculate the opening area A of the center bypass line Lc. c can be appropriately determined.

[0110] Second Embodiment A hydraulic excavator 100 according to a second embodiment of the present invention will be described with reference to Figures 3 and 10. Note that components that are the same as or equivalent to those described in the first embodiment are given the same reference symbols, and differences will be mainly described. Figure 10 is similar to Figure 7 and is an operational block diagram showing the flow of command value calculation processing (S160 in Figure 3) executed by a controller 220 according to the second embodiment.

[0111] Depending on the specifications of engine 2, when adjusting the discharge flow rate Q of hydraulic pump 1, the required flow rate Q of hydraulic pump 1 may be req In this case, when the rotation speed N is reduced, the required flow rate Q reqAs a result, more hydraulic oil than necessary is discharged to the tank 12 through the center bypass line Lc, resulting in energy loss. On the other hand, when the rotation speed N is increased, the required flow rate Q req As a result, the speed of the working device 104 decreases, and the working efficiency decreases.

[0112] Therefore, in this embodiment, in order to suppress the delay in the change in the rotation speed N of the engine 2, the area change rate ΔA c The required rotation speed N is calculated using the gain G according to req This is explained in detail below.

[0113] -Functions and processing flow of the controller according to the second embodiment- 10, the controller 220 according to the second embodiment has the same functions as the controller 120 according to the first embodiment. Furthermore, the controller 220 according to the second embodiment has the functions of a change rate calculation unit 241 and a gain calculation unit 242.

[0114] The change rate calculation unit 241 calculates the opening area A of the center bypass line Lc calculated in step S145 (see FIG. 3) of this control cycle. c The opening area A of the center bypass line Lc calculated in step S145 of the control cycle immediately preceding (t) c The area change rate ΔA is calculated by dividing the value obtained by subtracting (t-1) by the time width Δt of the control cycle. c That is, the controller 220 calculates the area change rate ΔA using the following equation (10): c Calculate the opening area A c (t),A c The t in (t-1) is the number of control steps representing the control period, where t represents the current control period and t-1 represents the immediately preceding control period.

[0115]

number

[0116] The gain calculation unit 242 refers to a predetermined gain table Tg and calculates the area change rate ΔA calculated by the change rate calculation unit 241. c The gain G is calculated based on the above.

[0117] The gain table Tg is created in advance based on experimental data, etc., and is stored in the nonvolatile memory 122 of the controller 220. c When increases, the area change rate ΔA c is a positive value, and the opening area A c When decreases, the area change rate ΔA c The gain table Tg is a negative value. c When is 0, the gain G is 1, and the area change rate A c The larger the absolute value (magnitude) of , the larger the gain G becomes. Note that the area change rate A c The relationship between the gain G and the frequency may be defined, and the gain G may be calculated using a mathematical formula.

[0118] The command value calculation unit 238 calculates the required flow rate Q calculated in step S115 (see FIG. 3). req The volume q selected by the minimum value selection unit 137 s The required rotation speed N is obtained by multiplying the value divided by the gain G. req That is, the command value calculation unit 238 calculates the required rotation speed N req Calculate.

[0119]

number

[0120] The command value calculation unit 238 calculates the required rotation speed N req is the minimum rotation speed N min In the following cases, the minimum rotation speed N min Request rotation speed N req Determine as (Nreq =N min In this case, the command value calculation unit 238 calculates the required flow rate Q req Request rotation speed N req (=N min ) to obtain the required volume q req (see equation (8)). The command value calculation unit 238 also calculates the required rotation speed N req is the maximum rotation speed N max In the above cases, the maximum rotation speed N max Request rotation speed N req Determine as (N req =N max In this case, the command value calculation unit 238 calculates the required flow rate Q req Request rotation speed N req (=N max ) to obtain the required volume q req is calculated (see equation (8)).

[0121] The command value calculation unit 238 calculates the required rotation speed N req is the minimum rotation speed N min Larger, maximum rotation speed N max If it is smaller, the volume q selected by the minimum value selection unit 137 s The requested volume q req (See equation (9)).

[0122] Note that equation (11) corresponds to the right side of equation (7) described in the first embodiment multiplied by a gain G. Therefore, the command value calculation unit 238 according to the second embodiment multiplies the required rotation speed N req By multiplying by the gain G, the required rotation speed N req The command value calculation unit 238 corrects the corrected required rotation speed N req The rotation speed N of the engine 2 is controlled based on the above.

[0123] -Effects of the second embodiment- According to the second embodiment, in addition to the same effects as the first embodiment, the following operational effects are achieved.

[0124] (6) As shown in FIG. 10, the controller 220 (change rate calculation unit 241) calculates the area change rate ΔA c The controller 220 (gain calculation unit 242) calculates the area change rate ΔA c Based on this, the area change rate ΔA c The controller 220 (command value calculation unit 238) calculates the gain G based on the required rotation speed N calculated using equation (7). req By multiplying by N, the required rotation speed is req The controller 220 (command value calculation unit 238) corrects the corrected required rotation speed N req The rotation speed N of the engine 2 is controlled based on the required rotation speed calculated using the equation (11).

[0125] In this configuration, the time change in the rotation speed N of the engine 2 can be accelerated using the gain G. That is, according to this embodiment, the responsiveness of the engine 2 can be improved using the gain G, thereby improving energy efficiency and work efficiency.

[0126] <Modification of the second embodiment> In the second embodiment, when adjusting the discharge flow rate of the hydraulic pump 1, the required flow rate Q req In contrast to this, in this modified example, the required flow rate Q of the hydraulic pump 1 is req A control method suitable for the case where there is a delay in the time change of the volume q of the hydraulic pump 1 relative to the time change of the volume q will be described.

[0127] Depending on the specifications of hydraulic pump 1, when adjusting the discharge flow rate Q of hydraulic pump 1, the required flow rate Q of hydraulic pump 1 may be req In this case, when the volume q is reduced, the required flow rate Q reqAs a result, more hydraulic oil than necessary is discharged to the tank 12 through the center bypass line Lc, resulting in energy loss. On the other hand, when increasing the volume q, the required flow rate Q req As a result, the speed of the working device 104 decreases, and the working efficiency decreases.

[0128] Therefore, in this modified example, in order to suppress the delay in the change in the volume q of the hydraulic pump 1, the area change rate ΔA c The required volume q is calculated using the gain G according to req This is explained in detail below.

[0129] -Functions and processing flow of the controller according to the modified example of the second embodiment- The command value calculation unit 238 calculates the required rotation speed N req is the minimum rotation speed N min In the following cases, the minimum rotation speed N min Request rotation speed N req Determine as (N req =N min In this case, the command value calculation unit 238 calculates the required flow rate Q req Request rotation speed N req (=N min ) and multiplying it by the gain G, the required volume q req That is, the command value calculation unit 238 calculates the required volume q using the following equation (12): req The command value calculation unit 238 calculates the required rotation speed N req is the maximum rotation speed N max In the above cases, the maximum rotation speed N max Request rotation speed N req Determine as (N req =N max In this case, the command value calculation unit 238 calculates the required flow rate Q req Request rotation speed N req (=N max ) and multiplying it by the gain G, the required volume qreq That is, the command value calculation unit 238 calculates the required volume q using the following equation (12): req Calculate the following.

[0130]

number

[0131] The command value calculation unit 238 calculates the required rotation speed N req is the minimum rotation speed N min Larger, maximum rotation speed N max If it is smaller, the volume q selected by the minimum value selection unit 137 s By multiplying by the gain G, the required volume q req That is, the command value calculation unit 238 calculates the required volume q using the following equation (13): req Calculate the following.

[0132]

number

[0133] Note that equation (12) corresponds to the right side of equation (8) described in the first embodiment multiplied by the gain G. Therefore, the command value calculation unit 238 according to the modified example of the second embodiment multiplies the required volume q calculated by equation (8) in the first embodiment by req By multiplying by the gain G, the required volume q req Similarly, equation (13) corresponds to the right side of equation (9) described in the first embodiment multiplied by the gain G. Therefore, the command value calculation unit 238 according to the modified example of the second embodiment corrects the required volume q calculated by equation (9) in the first embodiment. req By multiplying by the gain G, the required volume q req The command value calculation unit 238 corrects the corrected required volume q req The volume q of the hydraulic pump 1 is controlled based on the above.

[0134] In this modification, the response of the hydraulic pump 1 can be improved by using the gain G, but in order to prevent the command value from being exceeded, it is preferable to make the gain G smaller as the actual measurement value approaches the command value. For example, the controller 220 detects the current volume of the hydraulic pump 1 using a volume sensor, and compares the detected actual volume (actual measurement value) q with the selected volume q s (For example, the upper limit volume q max ) and calculate the difference Δq (Δq=q s The controller 220 reduces the gain G as the absolute value (magnitude) of the difference Δq decreases.

[0135] -Operation and effect of the modified example of the second embodiment- According to this modification of the second embodiment, in addition to the same effects as those of the first embodiment, the following operational effects are achieved.

[0136] (7) As shown in FIG. 10, the controller 220 (change rate calculation unit 241) calculates the area change rate ΔA c The controller 220 (gain calculation unit 242) calculates the area change rate ΔA c Based on this, the area change rate ΔA c The larger the volume q selected by the minimum value selection unit 137, the larger the gain G is calculated. s is the upper limit volume q max In this case, the controller 220 (command value calculation unit 238) calculates the gain G using the required volume q req (i.e., upper limit volume q max ) to obtain the upper limit volume q max The controller 220 (command value calculation unit 238) corrects the corrected upper limit volume q max (Required volume q calculated by equation (13) req ) based on which the volume q of the hydraulic pump 1 is controlled.

[0137] The following modified examples are also within the scope of the present invention, and it is possible to combine the configuration shown in the modified example with the configuration described in the above embodiment, to combine the configurations described in the different embodiments above, or to combine the configurations described in the different modified examples below.

[0138] <Variation 1> In the above embodiment, the controller 120 detects the operation pressure (operation amount) P c Based on this, the opening area A of the center bypass opening PT of the cylinder control valve 3 a However, the controller 120 calculates the lever operation amount (operation amount) L and the opening area A a and calculates the opening area A based on the operation amount L detected by the operation amount sensor 13b by referring to a data table that defines the relationship between the operation amount L and the opening area A a However, as described above, the operating pressure P detected by the operating pressure sensors 6a and 6b may be calculated. c Based on the opening area A a is more preferable because the calculation results are more accurate.

[0139] <Variation 2> In the above embodiment, the controller 120 controls the opening area A of the center bypass control valve 5 based on the control current I, which is an external input value input to the center bypass control valve 5. b However, the controller 120 calculates the lever operation amount (operation amount) L and the opening area A b and calculates the opening area A based on the operation amount L detected by the operation amount sensor 13b by referring to a data table that defines the relationship between the operation amount L and the opening area A b may be calculated.

[0140] <Variation 3> In the above embodiment, an example has been described in which the rotation speed-torque characteristic of the engine 2 is a linear characteristic (see FIG. 6). However, the rotation speed-torque characteristic of the engine 2 may be a curved characteristic.

[0141] <Variation 4> In the above embodiment, the controller 120 refers to the torque table Tt in FIG. 6 and calculates the required rotation speed N req Based on this, the allowable torque T of engine 2 max However, the controller 120 refers to the torque table Tt in FIG. 6 and calculates the allowable torque T of the engine 2 based on the actual rotation speed of the engine 2 detected by the engine speed sensor. max may be calculated.

[0142] <Variation 5> In the above embodiment, an example in which there is one hydraulic pump 1 has been described, but there may be provided a plurality of hydraulic pumps 1 for driving the hydraulic actuator. However, when a plurality of hydraulic pumps 1 are used, it is necessary to consider the torque distribution of each hydraulic pump 1, etc.

[0143] Although the embodiments of the present invention have been described above, the above embodiments merely illustrate some of the application examples of the present invention, and it is not intended that the technical scope of the present invention be limited to the specific configurations of the above embodiments. [Explanation of symbols]

[0144] 1...hydraulic pump, 2...engine, 3...cylinder control valve (actuator control valve), 5...center bypass control valve (flow control valve), 6a, 6b...operation pressure sensor (operation amount sensor), 9...discharge pressure sensor, 11S...hydraulic cylinder (hydraulic actuator), 12...tank, 13...operation device, 13a...operation lever, 13b...operation amount sensor, 100...hydraulic excavator (work machine), 104...work device, 105...body (machine body), 106...hydraulic system, 111...boom, 111A...boom cylinder (hydraulic actuator), 11 2A...Arm cylinder (hydraulic actuator), 113A...Bucket cylinder (hydraulic actuator), 120...Controller, 121...Processing device, 122...Non-volatile memory (storage device), 123...Volatile memory (storage device), 131...Excess torque calculation unit, 132...Increased volume calculation unit, 133...Pressure prediction unit, 134...Upper limit volume calculation unit, 136...Maximum volume calculation unit, 137...Minimum value selection unit, 138...Command value calculation unit, 220...Controller, 238...Command value calculation unit, 241...Change rate calculation unit, 242...Gain calculation unit, A a ...opening area (cross-sectional area of ​​the flow path of hydraulic oil from the hydraulic pump to the tank in the actuator control valve), A b …Opening area (cross-sectional area of ​​the flow path of hydraulic oil flowing from the hydraulic pump to the tank in the flow control valve), ΔA c …area change rate (time change rate of the opening area of ​​the discharge line), G…gain, I…control current, L…lever operation amount (operation amount), Lc…center bypass line (discharge line), Mp…pressure prediction map, N…engine rotation speed, N max …Maximum engine speed, N min …minimum engine speed, N req …requested rotation speed (command value (target value) of engine rotation speed), P a ,P b ,P c …Operating pressure (operating amount), P max …Maximum pressure, P P …Discharge pressure, PT…Center bypass opening, q Volume (actual volume) q…Volume, Q…Discharge flow rate, Q req ...Required flow rate, q inc(max) ...maximum volume, q inc …increasable volume, q max …Upper limit volume, q req…required volume, q s ...Volume (the smaller of the upper limit volume and the maximum volume), T...Torque (axial input torque of the hydraulic pump, pump torque), Ta...First opening table, Tb...Second opening table, Tg...Gain table, T max …allowable torque, T mrg ... margin torque, Tp... pressure prediction table, T req …required torque, Tt…torque table

Claims

1. The engine and a hydraulic pump driven by the engine and discharging hydraulic oil; a hydraulic actuator driven by hydraulic oil supplied from the hydraulic pump; an actuator control valve for controlling the flow of hydraulic oil supplied from the hydraulic pump to the hydraulic actuator; a flow control valve for controlling the flow rate of hydraulic oil discharged from the hydraulic pump through the actuator control valve to a tank; a discharge line in which an opening area of ​​a flow path that guides hydraulic oil from the hydraulic pump through the actuator control valve and the flow control valve to the tank is changed by the actuator control valve and the flow control valve; an operating device for operating the hydraulic actuator; a controller that controls the actuator control valve and the flow control valve based on an operation amount of the operating device, The controller Calculating a required flow rate of the hydraulic pump based on an operation amount of the operating device; calculating an opening area of ​​the discharge line based on an operation amount of the operation device; predicting a maximum pressure of the discharge pressure of the hydraulic pump that can increase when the hydraulic actuator suddenly stops, based on the required flow rate and the opening area; based on the predicted maximum pressure, controlling the rotation speed of the engine and the displacement of the hydraulic pump so that the torque of the hydraulic pump does not exceed the allowable torque of the engine when it is assumed that the discharge pressure of the hydraulic pump has increased to the maximum pressure; A work machine characterized by:

2. 2. The work machine according to claim 1, The controller calculating an upper limit value of a volume at which the torque of the hydraulic pump does not exceed an allowable torque of the engine when it is assumed that the discharge pressure of the hydraulic pump has increased to the maximum pressure based on the predicted maximum pressure; calculating a required rotation speed of the engine based on the upper limit value of the volume and the required flow rate of the hydraulic pump; controlling the displacement of the hydraulic pump based on the upper limit value; controlling the rotation speed of the engine based on the required rotation speed; A work machine characterized by:

3. 3. The work machine according to claim 2, The controller Calculating a time rate of change of the opening area of ​​the discharge line; calculating a gain that increases as the time rate of change of the opening area of ​​the discharge line increases based on the time rate of change of the opening area of ​​the discharge line; correcting the required rotation speed by multiplying the gain by the required rotation speed; controlling the rotation speed of the engine based on the corrected required rotation speed; A work machine characterized by:

4. 3. The work machine according to claim 2, The controller Calculating a time rate of change of the opening area of ​​the discharge line; calculating a gain that increases as the time rate of change of the opening area of ​​the discharge line increases based on the time rate of change of the opening area of ​​the discharge line; correcting the upper limit value of the volume by multiplying the upper limit value of the volume by the gain; controlling the displacement of the hydraulic pump based on the corrected upper limit value of the displacement; A work machine characterized by:

5. 2. The work machine according to claim 1, a discharge pressure sensor for detecting the discharge pressure of the hydraulic pump; The controller calculating a maximum value of a volume at which the torque of the hydraulic pump does not exceed an allowable torque of the engine based on the discharge pressure detected by the discharge pressure sensor; calculating an upper limit value of a volume at which the torque of the hydraulic pump does not exceed an allowable torque of the engine when it is assumed that the discharge pressure of the hydraulic pump has increased to the maximum pressure based on the predicted maximum pressure; The displacement of the hydraulic pump is controlled based on the smaller of the maximum value and the upper limit value. A work machine characterized by:

6. 2. The work machine according to claim 1, The operating range of the engine is set to a rotation speed range in which the allowable torque of the engine decreases as the rotation speed of the engine decreases. A work machine characterized by:

7. 2. The work machine according to claim 1, The controller calculating a cross-sectional area of ​​a flow path through the actuator control valve for hydraulic oil flowing from the hydraulic pump to the tank and a cross-sectional area of ​​a flow path through the flow control valve for hydraulic oil flowing from the hydraulic pump to the tank based on an operation amount of the operating device; calculating an opening area of ​​the discharge line based on a cross-sectional area of ​​a flow path of the actuator control valve and a cross-sectional area of ​​a flow path of the flow control valve; A work machine characterized by:

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