# Calculate the thrust Fdr developed by the linear motor necessary to overcome the aerodynamic force at above-mentioned constant speed if A = 3(3 m2 .

### electric engineering

1. A Magnetically Levitated (MAGLEV) train “Transrapid 07” is driven by a synchronous linear motor with the speed of 500 km/h . The aerodynamic drag force the train experiences can be estimated as 2

0.046

Lw

fCAu

=××

in N , where Cw = 0.4 is the drag coefficient, A is the train cross-sectional area in m2 and u is the train speed in km/h .a) Calculate the thrust Fdr developed by the linear motor necessary to overcome the aerodynamic force at above-mentioned constant speed if A = 3(3 m2 . Find the mechanical power developed by the motor.

b) Write the motion equation for the train, draw the block diagram and do the simulation. In simulation the train starts from 0 speed being driven by the force equal to 2xFdr (Fdr is found in question a). Then, after reaching constant speed the linear motor goes into regenerating mode developing the negative force equal to (–100 u2). Assume that the mass of the train is 100,000 kg and no other friction force exists. Record the driving force, load force, mechanical power, and train speed. What will be the speed of the train after 10 minutes .

Note: 1) Use the switch block which allow you to switch the driving force from positive to negative when the train reaches constant speed.

2) Draw all four characteristics in one graph using scale for some of quantities to make the characteristics readable. Use the simulation time, long enough for the train to reach the steady state.

2. The elevator shown schematically in Fig.1 is driven by the electric motor.

The data of the mechanical system are as follows:

· Mass of the counterweight M1 = 200 kg,

· mass of the cabin with a load M2 = 350 kg,

· the combine moment of inertia of the rotor and pulley is (Jp + JR) = 1.5 kg(m2,

· pulley radius R = 0.2 m,

· friction coefficient of the rotating part B = 0.2 N(m/(rad/sec).

a. Find the electromagnetic torque Tem to lift up the elevator with the steady linear speed u = 1 m/s.

b. Write the motion equation of the whole system when the elevator is moving up. Draw the appropriate block diagram and simulate the speed response to the step torque Tdr = 1.3Tem driving the elevator during the first 2 seconds and then Tdr drops to Tem, where Tem is the torque found in question a. Record the linear speed of the elevator and 0.01Tdr in the same graph for the period of time when speed reaches its constant value. What is the maximum acceleration amax of the car?

c. When the elevator is moving freely down the motor develops the braking torque described by the function Tem = -50(. Write the motion equation, draw the appropriate block diagram and simulate the dynamic process when the elevator starts to go down with the initial speed equal to 0. Record the linear speed u and electromagnetic torque Tem in the same graph.

Find the speed of the elevator u and the motor ( in steady–state condition.

Hint: The force attached to the pulley circumference is steady and equal to Td = g((M2 – M1). The elevator is not driven by the motor (the motor becomes braking generator), so the inertia of the masses (M2 and M1) is not taken into account.

Hint: For part (b) equivalent moment of inertia: 222

2

()1.50.235015.3

eqMP

JJJRMkgm

=++=+×=

For part (b) equivalent moment of inertia: 2

1.5

eqMP

JJJkgm

=+= w

T

T

L

em

u

R

Rotor (motor)

J

r

J

p

M

1

M

2

Pulley

Fig.1