Draw the force of the spring as a function of time

Question

CLOSING OF A CONTACTOR

1. Objective

The main objective of this work is to simulate the time evolution of the closure of a contactor (simplified model), i.e., the evolution from the rest position to the position 'closed'.

2. Fundamentals of the work

A bistable contactor bases its operation with a neodymium permanent magnet so that position ‘open' (x = x₁) and position ‘closed' (x = x₂) it is not required to consume energy in order to remain in the same position (without coil feed). In order to change the status of the contactor there is a spring (which force opposes the motion of the butt of the contactor) and also energy is supplied to the system through the coil which has N = 1200 turns and a resistance R of 5 Ω. The coil is fed with a voltage U of 24 V. This energy is also used to overcome friction (not shown in the figure) and can be considered negligible. There are also elements of non-magnetic material (rubber) to cushion the blows and have a thickness of x₂ = 0,5 mm. The core (fixed) and the butt of the contactor are made with the same material which permeability is μ_i. The system is such that the equilibrium position of the contactor in the ‘open' state (no power in the coil) is x₁ = 17,65 mm.

It can be considered that the section of the magnet, the iron section and the section of the air gap are equal and its value 1 cm². The length of the magnet (l_m) is 4 cm and the total average length of the iron (li) is 15 cm. The characteristics of the magnet at the working temperature are:

B_r = 1,24 T; H_c = 923,1 kA/m; μ_m = 1,069• μ₀; remember that μ₀ = 4Π10^-7 H/m The expression that governs the force of the spring is:

F_mo = -k. (x - x₀)

Where k = 4400 N/m and x₀ = 25 mm.

The butt of the contactor has a mass m = 0,1 kg. As a consequence of the finite value of magnetic permeability of the core will exist a certain leakage flux, which is considered negligible.

The equivalent circuit of the contactor is:

 

3. Development of the simulation

Implement differential equations that govern the system using Matlab/Simulink and simulate the system. Be especially careful to introduce the physical limits of the system. One of the elements (nonlinear) that can help is the saturation block, it is represented in the following figure and serves to limit a variable, such as the position x.

Another block you might be interesting is the "switch" block in order to switch from the speed that provides the integrator to the speed of 0 when the contactor is in the closed position (x = x₂).

In order to see the closure of the contactor transient response, the simulation can begin in the starting position for a short time (100 ms) and then feed the coil with the voltage U. One of the ways to achieve this it is represented in the following figure, based on a comparison block with a Boolean output (0 if t ≤ 0,1 s and 1 if t > 0,1 s).

To impose the initial conditions for the integration of differential equations it is needed to 'double-click' on the integrator of the variable in question and write the initial value. The figure below shows the parameters of the integrator of the speed (its out is the position) and note the value of the initial position (initial condition). It is very important to use the equilibrium position x1 = 17,65 mm as the initial condition when the contactor is open.

4. What have to be reported

The following questions (minimum) have to be reported. You (optionally) can add additional results that you think are interesting.

- Draw the force of the spring as a function of time indicating main values.
- Draw the magnetic force as a function of time indicating main values.
- Draw the speed of the butt of the contactor as a function of time indicating main values.
- Draw the position of the butt of the contactor as a function of time indicating main values.
- Draw the current consumed by the system as a function of time indicating main values.
- What is the value of the voltage U under which the contactor cannot be closed?