1 Structural principle of electromagnetic clutch

 

The compressor uses two electromagnetic clutches: 4 poles and 6 poles. They are similar in structure except for the number of poles. Both are composed of coils, pulleys, and drive plates. The coil is press-mounted on the shaft diameter of the front cover, the pulley is seated on the front cover through a double row angular contact ball bearing, and the drive disc is press-mounted on the main shaft. After the coil is energized, electromagnetic force is generated to attract the drive disc and the pulley to drive the main shaft to rotate, and the compressor works. After the coil is powered off, the drive disc and the pulley are disconnected, the pulley is idling, and the compressor does not work.

 

1.1 Magnetic pole

 

The electromagnet has south (S) and north (N) poles. The same pole repels the other pole and attracts each other. The two magnetic poles that attract each other produce an attractive force, which is called a pair of magnetic poles. The compressor uses two types of clutches-4-pole and 6-pole clutches, 4 poles have 4 pairs of magnetic poles, and 6 poles have 6 pairs of magnetic poles. According to the electromagnetic theory, the magnetic flux is generated after the coil is energized. We use the magnetic line of force to express it. The magnetic line of force forms a closed circuit along the smallest reluctance path in the magnetic circuit material. The end face of the pulley and the end face of the drive plate have annular magnetic isolation rings, so the magnetic lines of the 6-pole pulley enter the drive plate 6 times and 6 times from the drive plate into the pulley, and the 4-pole pulley magnetic line enters the drive plate 4 times and 4 times from the drive plate into the pulley .

 

1.2 Air gap

 

The clutch has 3 types of air gaps: suction air gap Δ0-the air gap between the pulley and the end face of the drive plate, Δ0 = 0.3-0.7, its function is to prevent friction between the pulley and the drive plate, the value is large, and it overcomes the spring when pulling in If the force is larger, the drive torque is reduced, and the value is smaller, the machining accuracy must be improved. Outer annular gap Δ1-the radial (unilateral) gap between the outer diameter of the coil and the inner wall of the pulley. Inner annular gap Δ2——The gap between the pulley hole seat and the coil shell. Theoretical analysis and calculation results prove that 80% of the magnetic pressure drop is consumed in the inner and outer annulus, so the inner and outer annulus should be reduced, usually Δ1, Δ2≤0.5.

 

2 The key factors that affect the clutch's carrying capacity are the flatness of the end surface of the pulley and the drive plate

 

The electromagnetic clutch transmits torque by friction, so whether the end face of the pulley and the end face of the drive plate fit well has a significant impact on the clutch's carrying capacity. The author has done a comparison test of 4 groups of different combinations of planes. Although the test conditions are simple, the influence of the plane conditions is obvious. A set of pulleys has a convex end face of 0.04 and a driving disc end face of 0.045, which has a high load capacity. Because the end face of the pulley and the end face of the drive plate form a pair of good fitting surfaces, the second group is just the opposite, it is convex to middle convex, forming a small area contact, the load capacity is small, the third and fourth groups are convex 0.01 to middle convex concave The load capacity of 0.02 is average. In design, the flatness of the working end surface of the pulley and the drive plate is required to be less than 0.05, and the belt pulley is generally concave 0.01～0.03. The problem is the drive plate, which has poor rigidity, riveting stress, and deformation after riveting. Low, convex at the midpoint of the connection between the two rivets, and the entire plane forms a complex curved surface with three protrusions, which is very unfavorable for torque transmission.

 

To improve flatness, the following measures are taken:

 

1) Increasing the thickness of the drive disc: If the thickness is increased to 4.9±0.1, the rigidity of the drive disc can be increased by 40% for level 6 and 17% for level 4.

 

2) Control the hole distance and hole precision of the yellow plate, drive plate and drive plate to reduce riveting deformation.

 

3) Reduce the deformation of the drive plate caused by external forces during assembly, handling, installation, and maintenance. Deformation has a significant impact on the load capacity. There are two common deformations: (1) The drive plate is twisted, and the end faces of the drive plate and the pulley have only a few points or Small contact area cannot transmit torque.

 

(2) Deflection or retraction, when the drive disc is in contact with the end face of the pulley when idling, the drive disc is ground off, and the drive disc is in poor contact with the pulley after electrification and cannot work.

 

3 ampere turns

 

The product of the number of coil turns n and the current A is called the number of ampere turns. Under the same conditions, the clutch load capacity is proportional to the square of the number of ampere turns. Usually the voltage is given. To increase the number of ampere turns, you can only increase the diameter of the enameled wire. If the diameter of the enameled wire does not increase, only increasing the number of turns cannot increase the number of ampere-turns, because the number of turns increases, the coil resistance increases in the same proportion, and the number of ampere-turns remains unchanged.

 

4The influence of the drive disk size

 

A large drive plate has a large magnetic pole area and a large moment arm, so it can transmit a large torque.

 

5 Influence of air gap size

 

The previous qualitative analysis of the influence of the size of the air gap on the carrying capacity, how to calculate this influence, the following examples in electrical engineering. Known magnetic flux φ=3×10-3Wb, core material D21, armature material cast iron, find the number of ampere turns and suction. This is the known magnetic flux to find the number of ampere turns and the attraction, which is a straight solution, the solution is. Attraction: F=0.5B2S/μ0 N where: B is the magnetic induction intensity (Wb/m2); S is the magnetic pole area (m2); μ0 is the air permeability μ0=4π×10-7.

 

Substituting the test data into 2F=0.8B2S×106=0.8×12×3×10-3×106=2 400 N In summary, the clutch load capacity formula can be written; M=0.4B2SnfR×106 Nm where: n Is the number of clutch pole pairs, 4 poles n=4, 6 poles n=6; f is the friction coefficient, f=0.3, see the "Machinery Design Handbook" Volume 1**, Chemical Industry Press; R is the average radius of the drive disc (m) . The magnetic pressure drop in the air gap is very large, the magnetic induction intensity decreases, so the load carrying capacity drops rapidly. The calculation results prove this: when the internal and external air gaps are both 0.45mm, the torque is 44N.m, and the internal and external air gaps are both 0.55mm. When the torque is 32N.m, a decrease of 27%, the internal and external air gaps are both 0.65mm, and the torque is 25N.m, a decrease of 43%.

 

6 Influence of low voltage and high temperature

 

Under normal conditions, the design capacity of the clutch can meet the requirements, but the following clear conditions may be exceptions.

 

1) The voltage is insufficient. Assuming that the rated coil voltage is 12V and the actual voltage is 10V, according to Ohm’s law, the coil current drops to 10/12=0.83, the magnetic induction decreases to 0.832=0.69, and the rated torque M0=40N.m, voltage 10V When M=0.69M 0 =27.6N.

 

m; 2) High temperature, if the room temperature resistance is 3.06Ω and 4.24Ω at 1150C, assuming the actual working temperature of the coil is 1000 C, the resistance is 3.06+(4.24?3.06)×100/115=4.09Ω. As the resistance increases, the current decreases To 3.06/4.09=75%, it can be approximately considered that the torque is reduced to 0.752M0=22.4Nm