LAB2 FLUIDS

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VR RTEPPING MOTOR

ELECTROMECHANICAL AUTOMATION

MOHAMED HAFEZ 11154275

10/1/2011


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Electromechanical Automation Mohamed Hafez 11154275

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TABLE OF CONTENTS

Introduction ......................................................... ................................................................. .................................. 2

Objective ...................................................... ................................................................. .................................. 2

METHODOLOGY......................................................................................................................................................... 2

ACKNOWLEDGEMENTS................................................................................................................................................ 2

EQUIPMENTS.............................................................................................................................................................. 2

Figure1. Wound Stator Unwound Rotor ........................................................................................................ 3

Theory .......................................................... .............................................................. ............................................. 4

Figure2. Free body diagram for the weight ................................................................................................... 4

Figure3. Time Vs Acceleration ....................................................................................................................... 7

Figure4. Time Vs Position and Speed ................................................................ ............................................. 7

Figure5. Time Vs Acceleration ....................................................................................................................... 8

Figure6. Time Vs Position and Speed ................................................................ ............................................. 8

Measurements and Results..11

Moment of inertia ............................................................. ...............................Error! Bookmark not defined.

Single step response ........................................................................................Error! Bookmark not defined.

Figure 7. : Display the encoder counter position and velocity outputs ..........Error! Bookmark not defined.

Variable reluctance stepping

motor............................................................................................................. Error! Bookmark not defined.

Conclusion ................................................................................................................Error! Bookmark not defined.

Bibliography ..............................................................................................................Error! Bookmark not defined.

Appendix: Lab2 note...15

OBJECTIVEThe most important aims for this report are to be able to:

Calculate and measure the moment of inertia of a rotor. Calculate and observe the single step response of a singly excited electromagnetic system.


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Model and simulate a simple electromechanical system. Compare the calculations with measurement. Set up and observe the operation of a variable reluctance stepping motor with two types of Excitation: single step and 50 Hz.

METHODOLOGY:

It can be found in the lab 2 note in the appendix

ACKNOWLEDGEMENTS:

All work in this report has been done according to the requirements in the Lab Manual. The help of

the Lab tutor and the Lecture Notes Given by Dr Quang Ha.

EQUIPMENTS:

Lybotec bench with power supplies. 2-pole, salient pole stator with a rated coil current of 2A . 2-pole, salient pole rotor as used in Lab 1. 36-pole, 3phase, wound stator. 24-tooth (24 pole), unwound rotor. Shaft encoder. Shaft encoder counter. Commutator-mounted torque arm. Fluxmeter. Connecting leads.


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a

M

g

Figure 2: Free body diagram for the weight

THEORY

(I)..Showing that ( )

1ST

by using Newtons3rd

law and showing that

And since, ((m g) - = (m a)

and that s = (1/2)(at2) with ( r) = J (

)

Then by using Newtowns 2ndlaw : F = m a

Where + (m a) = (m g) gives (m g) - = (m a) hence

Since the torque = F r sin

= F r [Max Torque]

By assuming the string and the shaft face will be perpendicular to one another.

the Torque = F r

= (m a r) = (m r2 () ) = ( )

r = J ()where a = =( ) hence, r = m r (g-a)

= J a ( ) = m r ( g - ( ))

= [ J (

) ( )]

J = [ (m r2)( )( g - ( )) ] kg/m2Therefore resulting in;

Where is the string tension and a is the uniform acceleration


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Shaft

dr

D/2

r

L

L

W

d

x

y

x

w/2

r

-w/2

-/2 /2

Note: r2

= x2

+ y2

Rectangular Rotor

(II)..Verify that 8 w Since the rotor is a cylindrical shaft plus a rectangular prism,

we have;

J=

where dM p dv pLrdr

Since p = ( ) = [ (

)]

And since,

J= r p L r d r = p L rdr where R=D/2 pL )r4]

pL ) ()4

] - [ () (0)4

] pL ]4 pL ]= [ (4 m1 d2L L ]= )m1D2

Since the density p = (m2 / v) = [ (m) / (wdL) ] and

Where dm2 = p dxdydzWe have;

p x y dx dy dz-

-

* +

*

+

* +

= p(

3w) + (w3 d

= [ (

3w) + (w3 d= ) m2 2+w2)..................................... 2)

Then by adding the two equation of (1) & (2)We will get;


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J = [ )m1D2+ ) m2 2+ w2) ]pis the mass density of steel at 7860kg/m3. and since we are given; 4 0-3m L 50 0-3m 48 0-3m d 5 0-3mw =

78 0-3m

Subbing these values into the expression of the Rotor inertia equation we have;

J = [ (1/8)m1D2+ (1/12) m2 (2+ w2) ]= [ (1/8)(m14 0-3)2 + (1/12)(m2)( (48 0-3)2+ w2) ].

(III)Write a differential equation or a state space model to obtain the motor step response given

the rotor moment of inertia, the electromagnetic torque, and the assumed load torque.

From the Hints in the question we can find that

The State Equation: T J ) +B() + Tc Sign()the Eectroechanica torque is T - |T|sign,and sign(x) = +1 if x > 0 or -1 if x < 0

We have Torque T F r sin TLFor ax torque wi resut in: T F r TLSo the load Torque is TL= Tc Sign (w) + Bw

TL Tc Sign () + B ()Since the electro-echanica Torque > F r J ) J

)(

)

J r ) J ) J

resulting in : T J ) + Tc Sign () + B ()


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(IV)Calculate the rotor angular acceleration , speed , and position against time for

approximately 5 oscillations. Use the chart function of the spreadsheet to plot the results.

For B = 0.005

0 0.5 1 1.5 2 2.5

-200

-150

-100

-50

0

50

100

Time, sFigure 1 Time Vs Acceleration

0 0.5 1 1.5 2 2.5-1

0

1

2

Time, s

0 0.5 1 1.5 2 2.5-20

-10

0

10

Time, s

Acceleration, rad/

Position, rad

Speed, rad/s

Figure 2: Time Vs Position and Speed


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(v)Experiment with the coefficients in your spreadsheet. Find out what the simulated

angle and velocity settle to and what happens when B=0.005 and when B=0.02 Nms/rad.

Comment.

For B = 0.02

0 0.5 1 1.5 2 2.5

-200

-150

-100

-50

0

50

100

Time, s

0 0.5 1 1.5 2 2.5

-1

0

1

2

Time, s

0 0.5 1 1.5 2 2.5-20

-10

0

10

Time, s

Acceleration, rad/

Position, rad

Speed, rad/s

Figure 3 Time Vs Acceleration

Fi ure 4 Time Vs Position and S eed


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10

15

(vi)

The step angle is calculated by:

60 60 4 5

(vii)

In one electrical cycle, there are 2m steps at a frequency supply frequency. Since in 1s there are 4

cycles or at 2mf steps/sec.

5 0 0 0steps/seconds

As the number of steps in one revolution is s=mNr then

We can use ( (2mf)/(s) ) = [ (2mf)/(m Nr) ] = (2f)/(Nr) rev/sec

Hence,

So in 1 minute, the number of rev/min (rotational speed) where

6 0

0

050

450

(viii)

The Time duration for ts, is determined by;

60 60 5 0 4

60

8000 0.00 sec

. 0s .


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3

3

3

A

B

C

3

3

3

22.2mH

22.2mH

22.2mH

B

A

C

When supplied

by AC supply ie)

Inductors

exists

When supplied

by DC supply ie)

Inductors

short

(ix)

(x)

0.00

0.050.00 6.8

Req = Rwwiinnddiinngg++RRextra, and also, since;

.5.

= 6.813 = 3.81 ohms

Since input supply is DC supplied to each phase,

we result in the inductors acting as a shortcircuit through a DC supply.

To determine the rise time tr of the transient

process of the current when we switch to DC

supply, we have;

j

Therefore resulting in a rise time of;


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MEASUREMENTS AND RESULTS

MOMENT OF INERTIA

m = 0.5 kg

s = 0.85 m

r = 0.025 m

g = 9.8 m/s2

t = 1.8 s

0.5 0.05 9.8.80.85 0.0055 .

Which is quite close to The theoretical value of J is 0.00565 kg.m2

SINGLE STEP RESPONSE

Following, the display of the CRO for the position and velocity versus time are shown, as well as the

calculated results graph.

Conversion factors

Speed: 100 RPM = 3.33 rad/s = 1 V Position: 18 = 0.1 = 1 V

The minimum peak of the velocity waveform is about -1.2 V with respect to the

origin.

1.2 * 3.33 = - 12.5 rad/s

The reading from the graph obtained from the theory part is about - 11 rad/s

The minimum peak of the position waveform is about -2.3 V with respect to the

origin.

2.3 * 0.1 = - 0.72 rad

The reading from the graph obtained from the theory part is about - 0.9 rad

-3.00

-2.50

-2.00

-1.50

-1.00

-0.50

0.00

0.50

1.00

1.50

2.00

2.50

0.0 0.5 1.0 1.5 2.0 2.5CH1 Volt

CH2 Volt

Figure 7: Display the encoder counter position and velocity outputs


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VARIABLE RELUCTAN CE S TEPPING MOTO R

f..Voltage across each phase = 4.62 V

Current through motor = 1.5 ATherefore resistance per phase = V/I = 4.62/1.5 = 3.08 ohms

g..In the lab note in the appendix

h..Step angle = 5 degrees

i..Step angle = 5 degrees

j..In the lab note in the appendix

k..Phase current = 0.97 A

Speed = 250 RPM

l..Phase current = 0.55 A

m..In the lab note in the appendix


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CONCLUSION AND DISCUSSION

Throughout my laboratory work I was able to come to some conclusions of what I found and

determined through my analysis.

I was able to determine that a stepper motor is also abrushless,synchronouselectric motor

that can be divided into a full rotation into a large number of steps. The motor's position can

be controlled precisely, without any feedback mechanism. Stepper motors are similar to

switched reluctance motors.

I was able to observe that a stepper motor basically operates differently from normal DC

motors, which basically rotate when a voltage is applied to their terminals. Stepper motors

in the other hand, have effectively multiple "toothed" electromagnets arranged around a

central gear-shaped piece of iron like the one we used in the lab. The electromagnets are

energized by an external control circuit, such as a micro controller.

I was able to understand that to make the motor shaft turn, one of the electromagnet is

given power, which makes the gear's teeth magnetically attracted to the electromagnet's

teeth. When the gear's teeth are aligned to the first electromagnet, they are slightly offset

from the next electromagnet.

So when the next electromagnet is turned on and the first is turned off, the gear rotates

slightly to align with the next one, and from there the process is repeated respectively. Each

of those slight rotations is called a "step," hence, number of steps making a full rotation. In

that way, the motor can be turned by a precise angle.

Comparing pre work to calculated work determined in the lab, I could successfully say there

were slight differences in calculations, possibly due to experimental errors.

But overall, I was very pleased with the results produced and the analysis obtained from the

laboratory experiment and by what I learned in the lab.
http://en.wikipedia.org/wiki/Brushless_DC_electric_motorhttp://en.wikipedia.org/wiki/Electric_motorhttp://en.wikipedia.org/wiki/Reluctance_motorhttp://en.wikipedia.org/wiki/Reluctance_motorhttp://en.wikipedia.org/wiki/Electric_motorhttp://en.wikipedia.org/wiki/Brushless_DC_electric_motor


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BIBLIOGRAPHY Lecture notes on Electromechanical Energy Conversion. Lab notes on Electromechanical Energy Conversion. Slemon, F 2005, Electrical machines and drives. Centikunt, S., Mechatronics, John Wiley and Sons, 2007, Chapter 8.

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LAB2 FLUIDS