ECE 4363 Electromech. Energy Conv.

1) Calculate the value of the

Fig 1. energy conversion cycle OABO

1R for Fig 1:

42

142

24Energy ElectricalInput Total Energy Mechanical Converted

1,3,,,1,1,,

,,,

1

+

=

+=

=

BAOAO

OBAO

SSS

R

2R for Fig 2:

214

224

0

Energy ElectricalInput Total Energy Mechanical Converted

3,,,1,1,,

,,,,

2

=

++=

=

BAOAO

OCBAO

SSS

R

2)

O

4

1

3

Electromech. Energy Conv. Solution to

Calculate the value of the 1R and 2R for both electromagnetic relays.

Energy ElectricalInput Total Energy Mechanical Converted

=R

energy conversion cycle OABO Fig 2. energy conversion cycle OABCO

4.024

2()(

)()()(

)(EnergyEnergy

=

+

=

+

=

ef

m

ee

m

BAWAWBAW

BAWAWBAW

7.0241

2232

()()()()(

EnergyEnergy

=

+

+

+

=

+

=

ef

m

ee

m

AWAWW

CBAWAWCBAW

A

B

i 4

O

1

3

2

C

Solution to Homework 02

electromagnetic relays.

Fig 2. energy conversion cycle OABCO

)B

)())(

0

+

e

m

CBWBABAW

43421

A

B

i 4

3)

4)

5)

6) Calculate the value of the magnetic force in position of the

Based on the given data on the problem, the value of the current is constant. So the best way is choosing the co-energy formulamathematical equation of the cobe explained as a function of currentdata of the problem because the flux linkage

and winding current (i), see

][)1(4 3

1

iig

diW f ++

==

Calculate the value of the magnetic force in position of the )(1 mg =

][)1(4 3

1

iig

++

=

Based on the given data on the problem, the value of the current is constant. So the best way is formula. The another reason for choosing the co

of the co-energy = diW f . In this equation be explained as a function of current. You can see that these conditions is verified by the given

the flux linkage ( ) is an explicit function of the displacement][)1(

4 31

iig

++

= .

]43

32[)1(

4 34

23

iig

di ++

=

) and )(8 Ai = .

Based on the given data on the problem, the value of the current is constant. So the best way is The another reason for choosing the co-energy is, the

di , flux linkage must nditions is verified by the given function of the displacement (g)

]8438

32[)1(

4]8438

32[)1(

4]43

32[)1(

4 34

23

234

23

)(8

34

23

.

++

=

+

+

=

+

+

=

=

==

gggii

gggW

FAiConsti

f

)(08.27]8438

32[)11(

4)1( 34

23

2 NgF =++

==

7) In order to find the direction of the magnetic force we need to plot flux streamlines. In the below figure the streamlines are shown by blue color. Now, we can find the magnetic poles N and S at each edge. At each edge both magnetic poles is shown. So, we the plunger is under two magnetic forces. The magnetic poles in axis-x create the magnetic force Fx (north) and the magnetic poles in axis-y create Fy (west). Hence, the average value of the magnetic force would be,

22yx FFF +=

Now, Find the Fx and Fy. In order to find the F, please attention to these notes:

The winding is excited by AC voltage source.

x

a a

a

a

a

x y

y

+ N

S

N S

Fx

Fy

Fx

Fy

F

The total reluctance of the magnetic circuit or the total inductance of the winding is a function of both x and y, because by moving the plunger both displacement x and y will be changed in both direction. In order to find the independent displacement variable of the reluctance or inductance, move the plunger or armature in direction of the force and then see the value and direction of changes in displacement (x and/or y). In this problem we found the direction of average force (red color). If you change the plunger you can see that the air gap length will change in both directions (x and y). Therefore we have:

),(circuit magnetic theof reluctance Total yxR= or ),( winding theof inductance Total yxL= . Based on the lecture, we have two main formulas for F. Using the AC current (RMS or

Maximum values) and total winding inductance, which is,

x

yxLIx

yxLIF rmsavx

=

=

),(41),(

21 2

max2

)( (because rmsII 2max = ) and

yyxLI

yyxLIF rmsavy

=

=

),(41),(

21 2

max2

)( .

In the above equations, ),(),(2

yxRNyxL = .

Another way is calculation F is based on the AC flux (RMS or Maximum values) and total reluctance of magnetic circuit,

x

yxRx

yxRF rmsavx

=

=

),(41),(

21 2

max2

)( (because rms 2max = ) and

yyxR

yyxRF rmsavy

=

=

),(41),(

21 2

max2

)( .

To calculation of ),( yxR , attention to the both air gap which is shown by blue color, because we have two series air gap. As you can see the air gap ( l ) and cross section area ( A ) are the functions of variables x and y,

ayax

axa

yyxRyxRyxR )()(),(),(),( 0021 +

=+=

.

Now the question is which way fit on the given data. According to the problem, the winding is excited by the AC voltage, tVtv e cos2)( = . Finding the value of the current form the voltage in this problem is possible but takes a long time. If you choose the second way, we need find the maximum or RMS value of the flux by using the AC voltage tVtv e cos2)( = . Remember in the class, I strongly recommend that keep this equation in your mind: The relationship among RMS value of the magnitude excitation voltage, maximum flux in magnetic

core and frequency of the voltage source explain by:

== maxmax 44.42 NNVe

=

NVe2

max

+

=

+

=

=

)(1

)(1

21

)()(2

41),(

41

20

22

2

00

22max)(

yaxay

aNV

ayax

axa

yxN

Vx

yxRF

e

eavx

2220

2

022

2

2022

2

2)(

24121

)2

(1

)2

(21

21

aNV

aaNV

aa

aa

a

aNVF eeeayxavx

=

=

+

===

+

=

+

=

=

20

22

2

0022

22max)(

)()(11

21

)()(21),(

41

yax

xaaNV

ayax

axa

yyN

Vy

yxRF

e

eavy

2220

2

2022

2

2)(

2

)2

(2

)2

(11

21

aNV

aa

a

aa

aNVF eeayxavy

=

+

===

So, 2220

2

2

)(2

)(2

aNVFF e

ayxavyayxavx

==

==

==

Then, 2220

2

2)(

2

2)(

2)(

222aN

VFFFF eayxavxayxavyavx ==+=

====