METHOD FOR TESTING SAFETY CHARACTERISTICS OF AN ELEVATOR

Abstract

The method for testing safety characteristics of an elevator comprises performing a balance check of the elevator according to a request.

Claims

1. A method for testing safety characteristics of an elevator, the method comprising performing a balance check of the elevator according to a request.

2. The method according to claim 1, wherein the request is generated by a remote entity, such as a cloud service or a remote server.

3. The method according to claim 1, wherein the request is based on a predetermined schedule.

4. The method according to claim 3, wherein the request is based on a time schedule.

5. The method according to claim 3, wherein if the balance check is not performed in accordance with the predetermined schedule, then normal elevator operation is prevented.

6. The method according to claim 1, wherein the result of the balance check is compared to predetermined criteria.

7. The method according to claim 6, wherein the predetermined criteria are based on logbook values collected during earlier tests.

8. The method according to claim 6, wherein if the result of the balance check fulfils the predetermined criteria, then permission for conducting a brake test is granted.

9. The method according to claim 6, wherein a balance error margin is used when determining whether the balancing satisfies the predetermined criteria.

10. The method according to claim 6, wherein if the result of the balance check does not fulfil the predetermined criteria, then normal elevator operation is prevented.

11. The method according to claim 6, wherein if the result of the balance check does not fulfil the predetermined criteria, then a degraded elevator operation is permitted.

12. The method according to claim 11, wherein in the degraded elevator operation, then the maximum load of the car is decreased.

13. The method according to claim 1, wherein the balance check comprises establishing a power model of the elevator, the power model comprising the motor power fed to the motor (PM) and power parameters of the motor and moved components in the shaft (PK, PP, PFr, PCu, PFe), performing a test run of the elevator, determining mid power values (PME,mid,up and PME,mid,down) for the up and down direction of the elevator, said mid power values defining the power fed to the motor only at the instant when the car is moving through the middle of the travelling path of the car in the up and down direction with constant velocity, calculating the power difference between the mid power values in the up and down direction, calculating the balancing weight difference (mB) from said power difference.

14. The method according to claim 1, wherein a brake test of the machinery brakes of the elevator is performed.

15. The method according to claim 14, wherein the brake test of the machinery brakes is a motor-assisted brake test.

16. A computer program product comprising program instructions, which, when run on a computer, causes the computer to perform a method as claimed in claim 1.

Description

DRAWINGS

[0032] The invention will in the following be described in greater detail by means of preferred embodiments with reference to the attached drawings, in which

[0033] FIG. 1 shows a side view of an elevator,

[0034] FIG. 2 shows a side view of an elevator machinery brake system,

[0035] FIG. 3 shows a flow diagram for testing safety characteristics of an elevator.

DETAILED DESCRIPTION

[0036] FIG. 1 shows a side view of an elevator.

[0037] The elevator may comprise a car 10, an elevator shaft 20, hoisting machinery 30, ropes 42, and a counterweight 41. A separate or an integrated car frame 11 may surround the car 10.

[0038] The hoisting machinery 30 may be positioned in the shaft 20. The hoisting machinery may comprise a drive 31, an electric motor 32, a traction sheave 33, and a machinery brake 100. The hoisting machinery 30 may move the car 10 in a vertical direction Z upwards and downwards in the vertically extending elevator shaft 20. The machinery brake 100 may stop the rotation of the traction sheave 33 and thereby the movement of the elevator car 10.

[0039] The car frame 11 may be connected by the ropes 42 via the traction sheave 33 to the counterweight 41. The car frame 11 may further be supported with gliding means 27 at guide rails 25 extending in the vertical direction in the shaft 20. The gliding means 27 may comprise rolls rolling on the guide rails 25 or gliding shoes gliding on the guide rails 25 when the car 10 is moving upwards and downwards in the elevator shaft 20. The guide rails 25 may be attached with fastening brackets 26 to the side wall structures 21 in the elevator shaft 20. The gliding means 27 keep the car 10 in position in the horizontal plane when the car 10 moves upwards and downwards in the elevator shaft 20. The counterweight 41 may be supported in a corresponding way on guide rails that are attached to the wall structure 21 of the shaft 20.

[0040] The car 10 may transport people and/or goods between the landings in the building. The elevator shaft 20 may be formed so that the wall structure 21 is formed of solid walls or so that the wall structure 21 is formed of an open steel structure.

[0041] FIG. 2 shows a side view of an elevator machinery brake system.

[0042] The car 10 may hang on a first side of the traction sheave 33 and the counterweight 41 may hang on an opposite second side of the traction sheave 33. The hoisting ropes 42 may pass from the car 10 over the traction sheave 33 and to the counterweight 41. The traction sheave 33 may be driven by the electric motor 32 which may be formed of a permanently magnetized synchronous electric motor. The machinery brake 100 may comprise two electromagnetic brakes 110, 120 acting on the traction sheave 33. The electromagnetic brakes 110, 120 may be controlled by a machinery brake controller 200. The electric motor 32 may be driven by a drive 31 e.g. a converter. The elevator, the drive 31 and the brake controller 200 may be controlled by a main controller 300. The traction sheave 33 may be provided with a motion measuring device 130, which may be formed of e.g. a tachometer.

[0043] The machinery brakes 110, 120 may as a first alternative be tested by a brake test according to method in EP 1 701 904.

[0044] The method in said EP patent comprises the following steps.

[0045] A predefined test load is set to apply to the drive machine of the elevator. This may be done by loading the elevator car with a test load whose weight is sufficiently reliably known. The weight of the test load depends on the amount of overload that the machinery brakes are required to tolerate. When the machinery brakes are required to withstand a 25% overload, i.e. a 125% load, then the test load must be 75% of the nominal load of the elevator. If the machinery brakes are required to withstand an overload of P %, then the test load must be 50%+P %. The test load may alternatively be provided by the torque of the drive motor without additional weights in the car.

[0046] The torque of the motor is then increased until it is established that the car starts moving. The torque value of the motor that starts the movement is measured and stored in memory. This torque value is denoted as a first torque value.

[0047] At least one of the machinery brakes is then closed.

[0048] The empty elevator car is then driven in the upward direction with the first torque value.

[0049] Movement of the car is then checked.

[0050] If movement of the car is detected, then the at least one machinery brake is regarded as defective.

[0051] The machinery brakes 110, 120 may as a second alternative be tested by a brake test according to method in EP 1 915 311.

[0052] The method in said EP patent comprises the following steps.

[0053] Only one machinery brake is engaged at an end of an elevator run and the other machinery brakes are engaged with a delay.

[0054] The state of the motion of the elevator and any slipping of the machinery brake with one machinery brake engaged is monitored.

[0055] The elevator is prevented form starting if the machinery brake is detected as slipping in the previous step.

[0056] EP 2 774 885 B1 discloses a method for performing a balance check of an elevator. The balance check determines the balancing weight difference of the elevator. The balancing weight difference is the difference between the weight of the empty elevator car and the counterweight of the elevator.

[0057] The method in said EP patent comprises the following steps.

[0058] Establishing a power model of the elevator, the power model comprising the power fed to the motor (PM) and power parameters of the motor and the moved components in the shaft (PK, PP, PFr, PCu, PFe).

[0059] Performing a test run of the elevator.

[0060] Determining mid power values for the up (PME,mid,up) and down (PME,mid,down) direction of the car, said mid power values comprising the power fed to the motor only at the instant when the car is moving through the middle of the travelling path of the elevator car in an up and down direction with constant velocity.

[0061] Calculating the power difference between the mid power values in the up and down direction.

[0062] Calculating the balancing power difference (mB) from said power difference between the mid power values in the up and the down direction.

[0063] The balance check for the elevator may be performed by using a simplified power model of the elevator which comprises the motor power fed to the motor (PM) and the power parameters of the motor and the moved components in the shaft (PK, PP, PFr, PCu, PFe). With such a model the behaviour of the elevator system can be simplified as to retrieve the balancing weight difference (=weight difference between car and counterweight) in an easy manner.

[0064] The power model may be defined in the following way:

PM=PK+PP+PFr+PCu+PFe

[0065] Where PM=power fed to the elevator, PK=kinetic power of the moved elevator components, PP=potential power of the moved elevator components, PFr=frictional losses of the elevator components, PCu=internal losses in the winding resistance of the motor, PFe=internal iron losses of the motor.

[0066] The power model simplifies an elevator system by modelling the power flow in said system. The necessary information for the balance check may be retrieved by a test run in which the car is driven in at least one closed loop to the upper end and to the lower end of the travelling path of the car.

[0067] The power difference in the up and down direction of the car may be determined at a point in which the elevator is driving with constant speed. This makes it possible to disregard the kinetic power PK of the system. The kinetic power PK=mi*v*a, where mi is the mass of the moved components of the elevator system, v is the velocity of the car, a is the acceleration of the car. The acceleration of the car is zero when the elevator runs at constant speed.

[0068] The power difference in the up and down direction only in the middle of the travelling path is considered. All moved elevator components except the car and the counterweight are balanced in the middle of the travelling path where the car is aside of the counterweight. The weight portion of these components can thus be disregarded in the middle of the travelling path. These components are e.g. suspension ropes, hoisting ropes or compensation ropes. The relevant components for the balance check that remain are thus the car and the counterweight, which are the essential weight components for the balance check.

[0069] The power model may thus be simplified considerably. All components which are based on acceleration can be disregarded. All components which are independent of the travelling direction, e.g. friction losses and iron losses will cancel out as the difference of said values in the up and in the down direction is used in the calculations.

[0070] The kinetic power PK is zero in the middle of the travel path of the elevator car.

[0071] The power parameter of the copper losses can easily be calculated from the motor current IM and the motor winding resistance RS (PCu=IM.sup.2*RS). These copper losses can be subtracted from the motor input power PME=PM−Pcu. PME denotes the amended motor power reduced by the copper losses of the windings.

[0072] The power model may thus be simplified to the following model:

PME=PP+PFr+PFe

[0073] The difference between the power values for the motor in the upwards and in the downwards direction may then be determined. The friction losses PFr and the iron losses PFe are further assumed to be independent of the travelling direction of the elevator. The difference between the power values is then:

PME(up)−PME(down)=PP(up)−PP(down)

[0074] The power difference in the up and the down direction is thus only dependent on the potential power parameter. The potential power parameter contains all elevator components which are moved vertically in the elevator shaft, such as e.g. the car, the counterweight, the hoisting ropes, the suspension ropes, and the compensation ropes.

[0075] The power difference is only considered for the middle of the travelling path of the car where the elevator car is positioned aside of the counterweight i.e. on the same level. The weight of the other moved elevator components except the car and the counterweight is at the middle of the travelling path balanced. The other moved components are formed e.g. by the hoisting ropes, the suspension ropes, and the compensation ropes. Only the weight of the car and the counterweight are thus relevant in the middle of the travelling path.

[0076] The difference between the power values in the middle of the travelling path can then be calculated in the following way:

PME(mid,up)−PME(mid,down)=mB*g*Vnom−mB*g*(−Vnom)

[0077] Where mB is the balancing weight difference or balance of the elevator system in kilogram, Vnom is the nominal speed of the elevator, and g is the gravitational acceleration g=9.81 m/s.sup.2.

[0078] The balancing weight difference mB may thus be calculated from the following equation:

mB=(PME,mid,up−PME,mid,down)/(2*g*Vnom)

[0079] The drive unit is thus able to calculate the elevator system balance at the middle point of the shaft by calculating, during constant speed run, the motor current from which the copper losses of the windings are removed in the up and down directions and dividing the difference with the nominator=2 times g times the nominal velocity Vnom of the elevator.

[0080] Instead of using one power value in the middle of the elevator shaft, the mean value of several test runs can be used. The arithmetical mean value of the test runs may then be used.

[0081] FIG. 3 shows a flow diagram for testing safety characteristics of an elevator.

[0082] Step 501 comprises performing a balance check according to a predetermined schedule.

[0083] Step 502 comprises comparing the result of the balance check with predetermined criteria.

[0084] Step 503 comprises, if balancing fulfills the predetermined criteria, then step 505, else step 504.

[0085] Step 504 comprises, preventing normal elevator operation.

[0086] Step 505 comprises, granting permission to perform brake test.

[0087] The use of the invention is not limited to the elevator disclosed in the figures. The invention can be used in any type of elevator e.g. an elevator comprising a machine room or lacking a machine room, an elevator comprising a counterweight or lacking a counterweight. The counterweight could be positioned on either side wall or on both side walls or on the back wall of the elevator shaft. The drive, the motor, the traction sheave, and the machine brake could be positioned in a machine room or somewhere in the elevator shaft. The car guide rails could be positioned on opposite side walls of the shaft or on a back wall of the shaft in a so called ruck-sack elevator.

[0088] It will be obvious to a person skilled in the art that, as the technology advances, the inventive concept can be implemented in various ways. The invention and its embodiments are not limited to the examples described above but may vary within the scope of the claims.