controller

Report
Electrical ENgg. Department
Simulation of elevator using Simulink
Any drive system consists
of:
- Power Source
- Electronic Converter
- Motor
- Mechanical Load
- Controller
- Feedback loops
The scheme of elevator structure is shown in Fig. A drive belt is
looped over a drive pulley with the ends anchored to the top of
the elevator shaft. The electric drive is rigidly attached to the drive
pulley with the use of gearing. The car and counterweight, which
ride on pulleys, provide belt tension ensuring the toothed belt and
drive pulley do not slip. The elevator car and counter-weight are
of different weight .
GOAL: IS TO HAVE CONSTANT SPEED AND
RESPONSE TIME REGARDLESS OF THE LOAD
TORQUE
Mathematical Model of Elevator
Electric Drive .

equations that control operation of the elevator can be divided two
systems :

- Mechanical system .

- Electrical System .
Mathematical Model of Mechanical
System’s

The motion equation of the entire system from the motor’s perspective is:


 = 


+ .  + 



where:

 is the motor’s moment of inertia

 is the angular speed of the rotor

B is the friction coefficient of the motor

 is the load torque placed on the motor’s shaft
(1.1)

The load torque  that is placed on the drive pulley which is mounted on
the motor’s shaft is expressed in equ 1.2

 =   + 

 is the angular speed of the pulley

 is the pulley’s moment of inertia

 is the radius of the drive pulley
 is the force exerted on the drive pulley


(1.2)

If the elevator car is moving upwards the load force is defined by equ 1.3

 =   −  + 

where:

 is the gravitational constant

 is the mass of the car

 is the mass of the counter-weight

 is the linear speed of the car


(1.3)

The car speed expressed in terms of motor speed is expressed as:

 =  

In Equ. 1.3 the force  is affected by the gravitational force and the
inertia of the elevator car  , both of which are expressed as:

 =   − 

 = 

The car speed expressed in terms of motor speed is expressed as:

 =  


(1.4)
(1.5)
(1.6)
(1.4)

Equ. 1.2 The load torque is now expressed as:

 =    −  + 2 

Substituting the load torque equation for when the elevator car is moving
upwards (Eqn. 1.7) into the motor’s motion equation (Eqn. 1.1), the motion
equation of the entire system is obtained as:


+ 


(1.7)



 = 
(1.8)


+ .  +    −  + 2 


+ 



and the load torque due to gravity is:

 =    − 

where the equivalent moment of inertia is:

 =  + 2 


Using gear ratio a

 =  (1.11)

Mathematical Model of Electrical
System’s :

Why to choose series motor ?

1- Highest starting torque.



2 - Draws less current and power from the
source compared to a shunt or
compound motor.

In general case we can derive an expression for the motor which is

 =   + 

 is Terminal voltage of DC motor

 is armature resistance

 is armature inductance

And  = 

K is constant

Let  =   + 


+  +   + 

(2.1)

 is back e.m.f Voltage
 is Filed resistance
 is magnetic field


From 2.1,2.2 and 2.3 we can get

 =   +   +  + 

 Filed current
(2.2)
(2.3)


 Filed inductance
(2.4)
 is output motor speed [rad/s]

Closed loop control is when the firing angle of the power processing unit is
varied automatically by a controller to achieve a reference speed or
torque

This requires the use of sensors to feed back the actual motor speed and
torque to be compared with the reference values.

•Actual motor speed measured using the tachogenerator (Tach) is
filtered to produce feedback signal

•The reference speed is compared to motor obtain a speed error signal

•The speed (PI) controller processes the speed error and produces the
torque command Te

•Ia is compared to actual current Ia to obtain a current error signal

•The current (PI) controller processes the error to alter the control signal vc

• vc modifies the firing angle to be sent to the converter to obtained the
motor armature voltage for the desired motor operation speed
We Built The Drive System with:
Dc Series Motor ( The parameters of Motor were taken from the library )
Max passenger weight = 600 Kg
Cabinet weight = 100 Kg
Counter weight = 400 Kg
Pully Diameter = 30 cm
Vf consat = 300 V
Flux assumed to be = .2
Matlab Blocks:
UpWard
Case 1 : Max : 600 Kg passenger weight ( Total Of 700 Kg with Cabinet )
UpWard

Case 2 : 300 Kg Passenger weight ( total Of 400 Kg = Same as Counter
weight )
UpWard

Case 3 : 0 Kg passenger weigh (Total of 100 Kg less than Counter weight )
Downward

Case 4 : Max : 600 Kg passenger weight ( Total Of 700 Kg with Cabinet )
Downward

Case 5 : 300 Kg Passenger weight ( total Of 400 Kg = Same as Counter
weigh )
Downward

Case 6 : 0 Kg Passenger weight ( total Of 100 Kg = less than Counter
weight

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