Transcript lethwala pavan

Dr. S &S.S GHANDHY GOVERNMENT
ENGINEERING COLLEGE
Electrical Department
SUBJECT: Electromechanical Energy
Conversion
SUBMITED BY :-
GUIDED BY: M.R.VASAVDA
NAME
ENROMENT NO.
LETHWALA PAVAN
130230109024
ASHISH MALKIYA
130230109025
MARATHE POOJA
130230109026
MEHTA KARNAV
130230109027
GANDHI MITESH
130230109028
Basics - electromechanical energy conversion
 The conversion of electrical energy into mechanical energy or
mechanical energy into electrical energy is called
Electromechanical energy conversion
 An electrical generator is a machine which converts mechanical
energy into electrical energy
 An electrical motor is a machine which converts electrical energy
into mechanical energy
Important Parts in electromechanical energy
conversion
1) An electrical system
2) An mechanical system
3) A coupling field
•
a) electrical
or
•
b) magnetic
Electromechanical Energy Conversion
• Consider the block diagram depicted below.
Coupling
Field
Electric
System
WE
Energy
supplied by
an electric
source
=
We
+
Energy transferred to the
coupling field by the
electric system
WeL
Mechanic
System
+
Energy losses of the
electric system.
Basically, I2R
WeS
Energy stored in the
electric o magnetic field
4
coupling medium
• There are two types of coupling medium in electromechanical
conversion.
1. Magnetic field as a coupling device
•
In Generator, On Application Of Mechanical Force On The
Conductor, The Conductor Moves In The Opposite Direction Of
The Magnetic Force Over Coming This Opposition The Energy .
• The Magnetic Field Therefore Plays The Part Coupling Link.
•
•
In Motor, The Current Carrying Conductor Is Placed In The
Magnetic Field. The Conductor Experience A Force That Tends To
Move It.
Now Conductor Is Free To Move In The Direction Of This Force.
Thus Magnetic Field, In This Case , Helps The Conversion Of
Electrical Energy , Acting As Coupling Medium.
CONTD….
2. Electric field as a coupling medium
• In electronic microphones and in electrostatic voltmeters the
electric field plays the part of coupling medium for energy
conversion.
• Taking a case of a capacitor having plates charged oppositely,
separated by a dielectric. A force of attraction exists between the
two plates. The plates tends to move towards each other. If one
plate is allowed to move in, the direction of force. Electrical
energy gets converted into mechanical energy.
CONTD….
• If we compare these two coupling media, very small
force is developed in electric field use but a large
force in case of magnetic field media
Fmech > Fele
Electro mechanical energy
conversion devices
• Devices Which Carry Out Electrical Operations By Using Moving Parts
Are Known As Electromechanical Devices.
•
Common Devices Are Generators And Motors
a) MOTOR
c) Car Starter
b) Generators
d) relay
Important points Worth noting in
Electromechanical Energy Conversion
Electromechanical Energy Conversion can be analyses by using principle of energy
conservation and electrical circuits and network.
Ignoring loss of energy in system is ideally a reversible process as show in fig.
In practical, during conversion some energy is lost forever due to copper loss, iron
loss, mechanical loss.
In electro-mechanical devices some air-gap is to be maintained between static and
moving parts.
Reluctances of air-gap is much more. Majority of the ampere turns of winding are
required to overcome air gap reluctance. Most of the energy is store in the air gap,
this is retuned to the electric source when the field is reduced.
Cont…
 Taking into account electrical loss (I2 R). Field loss in core due to change
in magnetic filed and loss due to rotation. i.e. friction and windage loss
(mech loss) the modified electro-mechanical energy conversion system is
represented as shown in fig.
V= Supply Voltage, I = Current, R = Resistance, loss input, output in
watts
Derivation of Magnetic Energy
Stored
V = Input applied Voltage;
R = Resistance of the coil;
N = Number of turns of the coil;
I = Current in the coil;
e = Self or back developed in coil;
Ø = flux; Ψ = mmf i.e. Ψ = NØ
Supply voltage V has to send the current in the coil overcoming its resistance(r)
and opposing emf ‘e’.
but……
… Ψ= NØ
---------(1)
This is voltage balance equation.
Cont…
Similarly to find power and energy balance equations. Multiplying equ. By I then
This is power balance equations. Now multiply by time dt .
This is energy balance equ.
Re- arranging the terms we get,
(V-ir) i.dt = idΨ
as V-ir =e
In equ e.idt is electrical energy output or electrical energy is converted into
magnetic energy. This is called as “co-energy”
”
Co-energy = e.idt
and I dΨ = Magnetic energy or stored energy
Cont…
• We can find total energy stored (wfield) by
integration
• Total energy stored (wfield) =
=
=
Total energy stored
=
Graphical representaion of Co-energy
and field energy (in open state)
W
F1
i1
0
0
co=co-energy in area OAC=
W
field =field energy in area OAB=
 dF   di
1
1
0
0
 Fd   id
Total energy  co - energy  field energy
Graphical representaion of Co-energy and
field energy (in closed steady state)
• In this case , there is no
air-gap and flux path is
through only iron core
which has magnetic
saturation limit.
Graphical representaion of Co-energy and
field energy (in closed steady state) cont…
Comparision of open and close
energy
• As current increases the
flux-linkage decreases. so
magnetization is no more
a linear but a curve.
• In (1) is open steady state
energy and (2) is close
steady state energy
• In store energy
close steady state energy >
open steady state energy
Classification of Excitation system
1. Singly excited: in this a single coil is used which is supported
on fixed part and necessary excitation is produced which is
magnetizes fixed as well as movable part.
• Ex. Magnetic relays , induction machine. These are used for
motion through limited distances or through small angle.
.
induction machine
Magnetic relays
Classification of Excitation
system(contd..)
2. Doubly excited: have two exciting coils ,dc motors ,
commutator motors and synchronous motors are doubly
excited type machine.
D.C motor
synchronous motor
Classification of Excitation
system(contd..)
3. Multiple Singly excited: stator part excited by 3 winding with
3-phase supply and rotor receives power by transformer
action i.e. 3-phase induction motor.
3-phase induction motor.
Classification of Excitation
system(contd..)
• 4. Multiple doubly excited: 3-phase synchronous motor’s 3phase windings are excited by 3-phase AC supply at the
same time rotor single windings is excited by a separate d.c.
voltages
3-phase synchronous motor
Singly Excited Magnetic Field
System
Induction machine
• Single coil is used in singly
excited magnetic system.
• It is supported on fixed
part.
• Necessary excitation is
produced which
magnetizes fixed and
moveable part. Ex
Magnetic relays,
Induction machine.
Voltage Equation of Singly Excited Magnetic
System
•Take the basic construction of a static core and a
rotable part rotor.
•Stator is wound with an exciting coil whereas
rotor has no coil.
•It is simply a magnetic material part supported
on shaft and rotate.
•Initial position is 1.
•Putting on the voltage V the coil carries the
current I and a magnetic flux ø is produced. Stator
is magnetized.
•The same flux passes through rotor body and it
gets with opposite polarities by induction.
• Being opposite stator and rotor poles, rotor is attracted by a force
and takes position 2.
• This is because 2 position has less reluctance than 1 position of
the rotor.
• Once it comes to position 2 rotor stops rotating
V= Supply Voltage, I = Current, R = Resistance , e = Self induce emf
of coil
Therefore,
V = I*R + e
e = N* dø/dt
Where,
N = number of turns of coil
Ø = magnetic flux
Hence,
V = I*R + d (N. Ø)/dt
V = I*R + d ψ/dt
• Multiplying above equation with I on both sides,
V*I = I*I*R + I* d ψ/dt
• Multiplying above equation with time dt,
V*I dt = I*I*R dt + I* dψ/dt
Integrating above equation,
We = Wloss + Wf
Total electrical input = Electrical energy loss + Electrical energy
converted into
mechanical energy
Now,
Wf = We(energy stored
+ Wme(energy spent in doing
in magnetic field)
mechanical work)
From above relation we can write static equation when rotor is
Steady and Wme = 0
Energy stored is Wf = integration of I dψ = (L*I*I)/2
Where ,
ψ = L*I
Doubly Excited Magnetic Field System
• The system is excited by two
different coils.
• One coil is wound on the
static part ‘stator’ and the
rotor also has coil wound on
it as shown in figure.
• For example DC motors ,
Commutator motor and
synchronous motor.
Commutator motor
Energy Equation For Doubly Excited
Magnetic System
• Stator coil has resistance R1
and rotor coil has resistance
R2.
• Respective exciting voltages
are V1 and V2 and currents
in coils are I1 and I2.
• Ø1 and Ø2 are respective
fluxes by coil 1 and coil 2.
• Ψ1 and Ψ2 are total flux
linkages of coil 1 and coil 2.
• L1 ,L2 and M are self and
mutual inductances of coil 1
and coil 2.
• Flux linkages with coil 1 and coil2,
Ψ1 = L1*I1 + M*I2
Ψ2 = L2*I2 + M*I1
• Voltage equations,
V1 = I1*R1 + d Ψ1/dt
V2 = I2*R2 + d Ψ2/dt
• Putting the value of Ψ1 in voltage equation,
V1 = I1*R1 + d (L1I1 + MI2)/dt
V1 = I1*R1 + L1dI1 /dt + I1 dL1 /dt + MdI2 /dt + I2 dM /dt
• Multiplying with I1 on both sides,
V1*I1 = I1* I1*R1 + L1* I1 dI1 /dt + I1 * I1 dL1 /dt + M *
I1 dI2 /dt + I1 * I2 dM /dt
• Similar equation for rotor side,
V2*I2 = I2* I2*R2 + L2* I2 dI2 /dt + I2 * I2 dL2 /dt + M * I2 dI1 /dt +
I1 * I2 dM /dt
• Adding both equations and multiplying with dt,
(V1*I1 + V2*I2)dt = (I1* I1*R1 + I2* I2*R2 )dt + (L1* I1 dI1 +
L2* I2 dI2 + M * I1 dI2+ M * I2 dI1 +
2* I1 * I2 dM + I1 * I1 dL1 + I2 * I2 dL2 )dt
(V1*I1 + V2*I2- I1* I1*R1 - I2* I2*R2 )dt = (L1* I1 dI1 + L2* I2 dI2 +
I1 * I1 dL1 + I2 * I2 dL2 )dt
+ (M * I1 dI2+ M * I2 dI1 +
2* I1 * I2 dM)dt
Useful electrical energy input = Field energy stored in the system
+ Electrical to Mechanical energy
transfer
• From above derived relations we can find the stored magnetic
energy in doubly excited system considering steady state where
mechanical output is treated as zero the entire electrical energy
is stored in the system in magnetic energy form.
• Steady terms in equation dL1 , dL2 , dM is zero and after
integrating the equation,
Wfe = ½*L1*I1*I1 + ½*L2*I2*I2 +M*I1*I2
Electromagnetic Torque Equation
For Doubly Excited System
• Stored energy is used to mechanical work.
• Mechanical work done = Rate of change of stored energy
= dWfe/dt
dWfe/dt = d (½*L1*I1*I1 + ½*L2*I2*I2 +M*I1*I2 )/dt
= ½*L1*d(I1*I1)/dt +½*I1*I1*dL1/dt + ½*L2*d(I2*I2)/dt
+½*I2*I2*dL2/dt +M*I2*dI1/dt+M*I1*dI2/dt+I1*I2dM/dt
= L1*I1d(I1)/dt + ½*I1*I1*dL1/dt + L2*I2(dI2/dt)
+ ½*I2*I2*dL2/dt+M*I2*dI1/dt+M*I1*dI2/dt
+I1*I2dM/dt
Integrating above equation w.r.t dt,
 Wfe = (L1*I1dI1 + L2*I2dI2 +M*I1*dI1+M*I1*dI2)+( ½*I1*I1*dL1
+ ½*I2*I2*dL2+I1*I2*dM)
 Wfe = Wstored + Wme
 Wme = ½*I1*I1*dL1 + ½*I2*I2*dL2+I1*I2*dM
In rotation while mechanical energy is produced the terms dL1 ,
dL2, dM are not constant but vary with θm.
Differentiating above equation w.r.t θm ,
Torque equation = dWme/dθm =½*I1*I1*dL1/dθm+½*I2*I2*dL2/dθm
+I1*I2*dM/dθm
Putting I2 = 0, we get torque equation for singly excited system,
Torque equation = ½*I1*I1*dL1/dθm
Comparison between Singly and
Doubly Excited Systems
Singly Excited System
Doubly Excited System
• One coil takes active part in
energy conversion process.
• It has winding on stationary
part only.
• It works on induction or
asynchronous principle.
• Used in constant speed
application.
• Produce useful startup
torque.
• E.g. Magnetic relays, D.C.
generator,
Induction
machine
• Two coils take active part in
energy conversion process.
• It has winding on stationary
and rotating part.
• It works on synchronous
principle.
• Used in variable speed
application.
• Do not produce useful
startup torque.
• E.g. DC motor , Commutator
and Synchronous motor
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