EAT216 Computer Aided Engineering

Question:

This diagram shows how to build a DC motor drive system that is separately excited.

Where are the system variables?

Va and Ia represent the DC Armature voltage and current (V) respectively.

Ea is the Armature’s back electromagnetic field (V).

Vf, If are the DC Field voltages (V) or current (A).

Te and TL represent the electrical drive and load torques, respectively (Nm).

o is the drive speed in revolutions/sec

And The Parameter Constants

Resistance to armature

Section A1

Section A3

Rf

Field resistance

Lf

Field inductance

50 H

J

Drive systems are in a moment of inertia

B

Utilizing elemental analysis, prove that the differential equation describing motor’s armature can be used to determine its validity

Ra is calculated using the last 3 digits of University of Sunderland Student ID numbers.

The relationship (Keis, a constant) defines the back emf of a DCmotor.

You can use this relationship to build a Simulink simulation of the motor’s electrical current. This is because the armature voltage of the motor is kept at 100V, and the field current as well as the drive speed are constants.

The A2 above test was conducted on an actual system. It was found out that the current of the armature took just 5ms to reach its maximum value of 63.2%.

Your simulation will allow you to determine La, the value of the inductance for the armature.

From the equation A1, explain how theoretically the maximum armature current can be found and the value of lacould.

The differential equation describes the motor’s drive systems.

Te is the electrical force supplied by the DC motor. This can be defined as:

You can expand the simulation to include the drive system.

A plot is created when theDrive’s speed changes with time under constant load torques of 0 to 100Nm, keeping the armature voltage set at 100V. This can be done in 10Nm steps.

Comment upon the response variation and the maximum load torque that can be applied before stalling occurs.

Section C – Speed Control

The Integral control system can be used to control the speed of the field by driving its voltage through a voltage/voltage convertor.

The following describes the system:

The error e, sp is the error and the set point is the constant speed of the system.

Construct the system.

This simulation calculates a controller gain value KI. This provides 10% overshoot when (Va=100V, TL=0 ) experiences an increase in setpoint from 0 to 10.

The controller C2 is used to change the set point from 20 to 30 rads/sec. (Va= 100V; TL = 0) This ensures that the system is stable both before and during the test.

Compare the performance with C2 and comment.

Answer:

Below is the equivalent circuit diagram for a DC motor that has been separately excited.

Field resistance Rf = 75

Field inductance LF = 50 H

J = Moment of Inertia Of Drive = 0.2 Kgm2

B = friction coefficient of drive system = 0.067 kgm2/s

Ia = armature current

Ea = back EMF generated at the Armature

Vf = DC Field Voltage and If = Current (A)

Te = electric drive torque and Tl = loads torques (Nm).

o = drive speed expressed in rads/sec

Ra = armature resistance; La = armature ductance. Rf = Field resistance. Lf = Field inductance.

KVL analysis can be used to determine the cause.

Va = V(Ra), V(La), and Ea

Va = Ia*Ra – Ia*(dIa/dt), Ia*(dIa/dt), & Ea.

Now, Ra’s value is (2*xyz/999) + 1.

Here, ‘xyz’ is 304.

Ra = 2*304/999 + 1, = 1.6086?

Now, the DC motor’s Back emf is provided by.

Ea = Ka*If*o

As shown below, you can now build this in Simulink.

The drive speed w and field current If are set at 0 respectively. A controlled voltage source maintains the field voltage at 100 volt.

Plot Of Field Current

The DC machine block contains the parameters for the dcmotor.

Plot of Armature Current

As a result, the steady state value for armature current as measured from the simulation scope is 62.16 Amps.

In one simulation, the armature current reached 63.22% of its final value within 5 msec (or 0.005 sec).

This simulation requires the La armature inductance to be found.

Here is the question. This is a trial and error method where armature Inductance La is varied. It is used to calculate the approximate simulation. Next, the value La which gives close 63.2% of the final armature curent in 0.005 sec is recorded.

Now, 63.2% final armature current = 62.2 *0.632 = 39.2788.

Simulating the waveform at La=0.0015 H yields an estimate of the specified value for the armature present.

Therefore, the value for the armature-inductance at which the current reaches 63.2% @ 0.005 sec is 0.0.0015 H.

The equation A1 can be found by

When the motor’s back EMF is zero, and inductance is zero, then the armature current reaches its maximum.

This happens at start of motor.

This means that at the start of the motor, the armature current will be given by Ia = Va/Ra= 100 / 1.6086 = 62.16659 Amps.

The following procedure can be used to calculate La armature inductance.

The inductance in an armature coil’s armature coil can be given by

L = N*/I

N is the number turn count in the coil.

= total magnetic flux in Weber, Wb

A = Cross sectional Area of Coil.

Section B

The motor’s differential equation is found in,

Te = J*(do/dt), Bo + T

Here, Te = electric torque = Ke*If*Ia; and TL = loads torque

Now, put all the values for the inertia consta J, B, Te.

0.5*If*Ia = 2.067*o + T

0.2*(do/dt), = 0.5*If*Ia, 0.067*o, TL

The block diagram can be viewed below.

In the Simulink diagram, the armature voltage is at 100 Volts. The step field current is 2A. The motor speed is plotted using the load torque of 0 to 100 Nm in 10 steps.

Speed and load torque:

It is evident from the torque and speed curves that motor speed decreases or motor stalls when the torque exceeds 60 Nm.

60 Nm is the motor’s stalling torque.

This torque is beyond which the motor speeds will decrease, and at some torque levels it will drop to zero.

This happens when the torque generated by the motor’s load torque exceeds the maximum torque it can generate.

Section C: Speed control

The integrative control system of the field’s voltage through voltage-to-voltage converter is given below.

sp stands for the desired system’s constant speed.

Below is the Simulink design for Integral Speed Control.

Now, the simulation shows that the Integral constant (KI) is calculated using trial and error. This means that the speed exceeds 10% of the final steady-state value.

The set speed is a step input, which increases from 0 to 10rad/sec after 1 second.

The armature voltage is kept at 100 volts. The load torque TL = 0.

It is discovered that KI = 0.05 causes the speed to exceed 10% of its steady-state value.

Speed Curve

It can be seen that motor speed curves at the given value of KI are 11 rad/sec for speeds between 18 and 30 secs. This is 10% more than the steady-state speed of 10 rad/secs.

The set point value is now increased from 20 to 30 R/sec.

This is the final steady state speed and the overshoot value.

In the Simulink block, as shown below, you can specify the set point step input.

The speed curve above shows that the final steady-state speed is now 30 rad/sec. At t = 18 seconds, it becomes 33 rad/sec. This 10% increase is not noticeable.

Leave a Reply

Your email address will not be published.

Get 20% off your first purchase

X
Open chat
1
Hi
We are here at your service.
Order today and save 30% Discount code 12HOURACE20