experiments ems ii

EMS-II LABS

OSAMA TAHIR09-EE-88

INTRODUCTION TO AC MACHINES

EXPERIMENT NO. 01

TYPES OF AC MACHINES1) SYNCHRONOUS MOTORS External source is required to initiate the

machine. Doubly excited machine Requires both AC and DC once started runs with constant speed.

2) ASYNCHRONOUS MOTORS Self excited machine Follow law of electromagnetic induction Singly excited Speed decreases with increase in load 95% motors are induction motors

CIRCUIT DIAGRAM

What is induction Motor?

SLIP“The difference between synchronous speed

and actual speed of rotor is called slip”S= Ns-NWhere

Ns=(120f)/Pf=frequency of supply

P=no. of polesN=actual speed of rotorNs=synchronous speed

AT STAND STILL:S=Ns-NS=Ns-0S=Ns

S=1(Maximum)AT RUNNING CONDITION:

s= Ns-NIf Ns=N

S=0

Percentage Slip:% slip= (Ns-N) *100

NsEffects: (Factors effected by slip)a)ROTOR VOLTAGE:

Es = SE2If s=1

Es=E2 (voltage maximum)If s=0

Es=0 (voltage is zero)b) Supply Frequency

f=sfrIf s=1 then fr=f

OBSERVATIONS AND CALCULATIONS

Ns=(120f)/P, Here f=50Hz,P=4So Ns= 1500 r.p.m

S. No.

Rotor Speed

N(volts)

Rotor Speed

N(r.p.m)

Ns(r.p.m)

SLIPS= Ns-N

%s=Ns-N *100

Ns

1.

2.

3.

4.

5.

6.

RESULT

TO DRAW SLIP-TORQUE CHARACTERISTICS OF 3-Ø SLIP RING

INDUCTION MOTOR

EXPERIMENT NO. 02

CIRCUIT DIAGRAM

TORQUE IN INDUCTION MOTOR

There are two types of torque in induction motor

Starting torqueRunning torque

STARTING TORQUE“The torque developed by motor at the

instant of starting is called starting torque”Tst ∝ E2I2cosØ --------- (1)

E2 = rotor emf/phase at stanstill

R2 = Rotor resistance/phase

X2 = rotor reactance/phase

Z2 =√ (R22+X2

2) = rotor impedance/phase

X2Z2

R2

Ø

A B

C

I2 = E2/Z2 = E2/ √ (R22+X2

2)

cosØ = R2/Z2 = R2/ √ (R22+X2

2)

From equation (1)Tst = K1 E2I2cosØ

= K1E2 [E2/ √ (R22+X2

2) ][R2/ √ (R22+X2

2) ]

Tst = [K1 E22 R2/ (R2

2+X22) ] --------- (2)

Tst = [K2R2/ (R22+X2

2) ] ---------- (3)

Let K2= K1 E22

Differentiating equation(2) w.r.t R2

d Tst/dR2 = d/dR2 [K2R2/ (R22+X2

2) ]

= K2 (R22+X2

2) (1) - R2 (2 R2+0)

(R22+X2

2)

Put d Tst/dR2 =0 for maximum torque

0 = K2 (R22+X2

2-2R22)

-R22+X2

2 = 0

R22 = X2

2

R2= X2 --------------- (4)

This is the condition for achieving maximum starting torque.

RUNNING TORQUE

Tr = [K1S2 E22 R2/ (R2

2+X22) ]

R2= SX2 ----------------- (5)

RELATIONSHIP BETWEEN SLIP AND TORQUE

AT NO LOAD AT NORMAL LOAD AT OVERLOAD

N is high N is low N is low

S is low S is high S is high

fr = sf is low fr = sf increases fr = sf increases

Xr = sX2 is low Xr = sX2 increses but less than R2

Xr = sX2 increases and greater than R2

R2 is constant R2 is constant R2 is constant

Torque decreases Torque increases Torque decreases

T ∝ s T∝ s T∝(1/ s)


Synchronous speed = Ns = ________ rpmS. No.

Rotor speed

N(rpm)

SlipS=(Ns-N)/Ns

ForceF

(newton)

TORQUET=F*rWhere r=0.26(N-m)

RESULT

TO DRAW CIRCLE DIAGRAM OF AC INDUCTION MOTOR

EXPERIMENT NO. 03

To draw circle diagram of induction motor there are three types of tests to be performed.

1.No load test2.The DC test for stator resistance3.The locked or blocked rotor test

NO LOAD TEST:

The no load test of an induction motor measures the rotational losses of the motor provides information about its magnetization current. The only load on the motor are the frictional and the windage losses. So all input power in this test is consumed by mechanical losses and the slip of the motor in this case is very small.

CIRCUIT DIAGRAM

OBSERVATIONS AND CALCULATIONSI1= __________

I2= __________

I3= __________

I0=( I1+I2+I3)/3= ________

V1= _________

V2= _________

V3= _________

V0= (V1+V2+V3)/3= ________

W0=no load losses

= wattmeter reading*multiplying constant= ______________

AsW0 = √3V0I0cosØ0

cosØ0 = W0/(√3V0I0)

Ø0 = cos-1 [W0/(√3V0I0)]

THE DC TEST FOR STATOR RESISTANCE

In the DC test, a DC voltage is applied to the stator winding of an induction motor. As current has DC value there is no induced voltage in the rotor circuit and no resulting rotor current flow. Also the reactance of the motor is zero at Dc current, therefore the only quantity limiting current flow in the motor is the stator resistance that can be determined.

CIRCUIT DIAGRAM


Req = V/IReq = Ra(Ra+Ra)

3RaReq = (2Ra2)/3Ra

= (2/3)RaRa = (3/2) Req = ___________

THE LOCKED OR BLOCKED ROTOR TEST

This test corresponds to the short circuit test on a transformer. In this test we utilize the variac transformer to supply the motor. Rotor is locked or blocked so that it cant move. Resulting values of voltage current and power are measured.

CIRCUIT DIAGRAM

OBSERVATIONS AND CALCULATIONSI1= __________

I2= __________

I3= __________

ISC=( I1+I2+I3)/3= ________

V1= _________

V2= _________

V3= _________

VSC= (V1+V2+V3)/3= ________

WSC= Total losses= ___________

Wsc= √3VscIsccosØsc

cosØsc = Wsc/(√3VscIsc)

Øsc = cos-1 [Wsc/(√3VscIsc)]

= _____________Now we convert all values to the normal voltage value

asISN = Isc*(VN/Vsc)= __________

Where ISN= short circuit current at normal voltage(VN=110volts)

WSN = Wsc* (VN/Vsc) = __________

Stator Cu losses = IN2 * Ra = ___________

Rotor Cu losses = WSN – stator Cu losses = _________

PURPOSE OF CIRCLE DIAGRAM:

Circle diagram is used to calculateMotor outputStator Cu lossesCore lossesMotor inputSlipEfficiency

IMPORTANT FORMULEA:

1. Motor Input = Output + losses

2. Slip = (rotor Cu losses)/Rotor Input

3. Efficiency = (Output/Input)*100

HOW TO DRAW CIRCLE DIAGRAM OF INDUCTION MOTOR?

INTRODUCTION TO SYNCHRONOUS MOTOR

EXPERIMENT NO. 04

Synchronous motors are like induction motors in that they both have a stator winding which produces a rotating magnetic field. Unlike an induction motor the synchronous motor is excited by an external DC source and therefore requires slip ring and brushes to provide current to the rotor.

CONSTRUCTIONA synchronous motor is composed of the following

parts:1) STATOR

The stator is the outer shell of the motor, which carries the armature winding. This winding is spatially distributed for poly-phase AC current. This armature creates a rotating magnetic field inside the motor.

2) ROTORThe rotor is the rotating portion of the motor.

It carries field winding, which may be supplied by a DC source. On excitation, this field winding behaves as a permanent magnet. Some machines use permanent magnets in the rotor.

3) SLIP RINGSThe slip rings on the rotor, to supply

the DC to the field winding. 4) STATOR FRAME

The stator frame contains and supports the other parts and may include bearing housings.

Large machines may include additional parts for cooling the machine, supporting the rotor, lubricating and cooling the bearings, and various protection and measurement devices.

CIRCUIT DIAGRAM

PRINCIPLE OF OPERATION The operation of a synchronous motor is simple to

imagine. The armature winding, when excited by a poly-phase (usually 3-phase) supply, creates a rotating magnetic field inside the motor. The field winding, which acts as a permanent magnet, simply locks in with the rotating magnetic field and rotates along with it. During operation, as the field locks in with the rotating magnetic field, the motor is said to be in synchronization.

Once the motor is in operation, the speed of the motor is dependent only on the supply frequency. When the motor load is increased beyond the break down load, the motor falls out of synchronization i.e., the applied load is large enough to pull out the field winding from following the rotating magnetic field. The motor immediately stalls after it falls out of synchronization.

STARTING METHODSSynchronous motors are not self-starting motors.

This property is due to the inertia of the rotor. When the power supply is switched on, the armature winding and field windings are excited. Instantaneously, the armature winding creates a rotating magnetic field, which revolves at the designated motor speed. The rotor, due to inertia, will not follow the revolving magnetic field. In practice, the rotor should be rotated by some other means near to the motor's synchronous speed to overcome the inertia. Once the rotor nears the synchronous speed, the field winding is excited, and the motor pulls into synchronization.

The following techniques are employed to start a synchronous motor:

A separate motor (called pony motor) is used to drive the rotor before it locks in into synchronization.

The field winding is shunted or induction motor like arrangements are made so that the synchronous motor starts as an induction motor and locks in to synchronization once it reaches speeds near its synchronous speed.

Reducing the input electrical frequency to get the motor starting slowly.

CHARACTERISTICS

It is a doubly excited machine.It is used for power factor improvement.It will be started as induction motor but

shifted to synchronous speed later on.Stator is same as induction motor but rotor

becomes permanent magnet after the application of DC supply.

USES

Synchronous motors find applications in all industrial applications where constant speed is necessary.

Improving the power factor as Synchronous condensers

Low power applications include positioning machines, where high precision is required, and robot actuators.

Mains synchronous motors are used for electric clocks.

ADVANTAGESSynchronous motors have the following

advantages over non-synchronous motors:Speed is independent of the load, provided an

adequate field current is applied. Accurate control in speed and position using

open loop controls, eg. stepper motorsThey will hold their position when a DC current is

applied to both the stator and the rotor windings. Their power factor can be adjusted to unity by

using a proper field current relative to the load. Also, a "capacitive" power factor, (current phase leads voltage phase), can be obtained

by increasing this current slightly, which can help achieve a better power factor correction for the whole installation.

Their construction allows for increased electrical efficiency when a low speed is required (as in ball mills and similar apparatus).

They run either at the synchronous speed or they do not run at all.

EXAMPLES

brushless DC electric motorstepper motorThree-phase AC synchronous motorsSwitched reluctance motorSynchronous brushless wound-rotor doubly-

fed electric machine

V-CURVE CONSTRUCTION OF A SYNCHRONOUS MOTOR

EXPERIMENT NO. 05

APPARATUS

AmmetersWattmetersVoltmetersSynchronous motor set

CIRCUIT DIAGRAM

THEORY

“The V-curves of a synchronous motor shows how armature current varies with its field

current when motor input is kept constant”These are obtained by plotting AC armature

current against dc field current while motor input is kept constant and are so called because of their shape. There is a family of such curves each corresponding to a definite power intake.


S.No Ia

(Amp)

If

(Amp)

W=P

(watts)

VL

(volts)

*cosØ

P= √3VLILcosØ

cosØ= (P/√3VLIL)

GRAPH

If

Ia

Ia

COMMENTS AND CONCLUSIONSFor each curve, the minimum armature current

occurs at unity power factor, when only real power is being supplied to the motor.

At any other point on the curve, some reactive power is being supplied to or by the motor as well.

For field current less than the value giving minimum Ia the armature current is lagging consuming Q

For field currents greater than the value giving the minimum Ia the armature current is leading supplying Q to the power system as a capacitor would.

Therefore by controlling the field current of a synchronous motor the reactive power supplied to or consumed by the power system can be controlled.

TO FIND EFFICIENCY OF AN INDUCTION MOTOR BY IEEE METHOD

EXPERIMENT NO. 06

APPARATUS3 phase slip ring induction motor DC shunt generatorWattmeter setAmmeterVoltmeterLoads (Tungsten bulbs)Connecting leads

THEORYThe efficiency is given by

η= (output/input)*100now losses can be constant or variable. Core

losses are example of constant losses while copper losses are example of variable losses.

To find the efficiency of induction motor following tests are performed

1.No load test2.Load test3.DC test for stator resistance measurement

1. THE DC TEST FOR STATOR RESISTANCE:

In the DC test, a DC voltage is applied to the

stator winding of an induction motor. As current has DC value there is no induced voltage in the rotor circuit and no resulting rotor current flow. Also the reactance of the motor is zero at Dc current, therefore the only quantity limiting current flow in the motor is the stator resistance that can be determined.

CIRCUIT DIAGRAM


Req = V/IReq = Ra(Ra+Ra) 3RaReq = (2Ra2)/3Ra= (2/3)RaRa = (3/2) Req = ___________

NO LOAD TEST

The no load test for an induction motor measures the rotational losses of the motor and provides information about its magnetization current.

CIRCUIT DIAGRAM:

A

A

A

V

C.C

P.C

P.C

C.C

VARIAC

T/F

Ф1

Ф2

Ф3


Vo = _____________ voltsIL = ___________ AmpInput power = _____________ wattsConstant losses = input – stator Cu losses

________ (A)WhereTotal stator Cu losses = 3(IL/√3)2Ra

= 3IL2Ra = _________

LOAD TESTIn this test we apply load on induction motor

and by changing the value of loads we calculate Pin, losses and Pout for each load .After this efficiency of motor can be find out by using the formula

%η = (output/input)*100Where

Output = Input-losses

CIRCUIT DIAGRAM


IL

(Amp)

V

(volts)

Pin

(watts)

Stator

Cu

losses

Ns

(rpm)

N

(volts)

N

(rpm)

slip F

(N)

T=

F*r

r=0.26

Const.

losses

Rotor input

Pin-Pst.Cu

Rotor Cu

loss

S*Prot

Total

loss

Pout %η

TABLE A:

TO FIND EFFICIENCY OF AN INDUCTION MOTOR BY

DYNAMOMETER METHOD

EXPERIMENT NO. 07

In this method we can find out the efficiency of induction motor directly. For this purpose we will use the observations of TABLE A given in Experiment No. 06.

EXPLANATION

1. Find out input power at each load2. Calculate output for each load by using

the formulaOutput = (2Πnt/60)

3. Calculate Efficiency as%η = (output/input)*100

MEASUREMENT OF SLIP OF AN INDUCTION MOTOR BY COMPARING

ROTOR FREQUENCY AND SLIP FREQUENCY

EXPERIMENT NO. 08

APPARATUS

3-phase slip ring induction motorGalvanometerStop watchConnecting wires

CIRCUIT DIAGRAM

% S= (slip speed/Ns)*100%= [(Ns-N)/Ns]*100%

fr = sfes= (fr/fe)


S.No.

fr(cycle/sec)

f (Hz)

S = (fr/f)

Ns (rpm)

N (rpm)

S= (Ns-N)/N

RESULT

PARALLEL OPERATION OF 3-Ø ALTERNATORS

EXPERIMENT NO. 09

APPARATUS

Two generatorssynchronizing lamp device prime mover

CIRCUIT DIAGRAM

THEORY

Now a days an isolated synchronous generator supplying its own load independently of other generators is very rare. For all usual generator applications there is more than one generator operating in parallel to supply the power demanded by the loads.

ADVANTAGES OF PARALLEL OPERATIONSeveral generators can supply a bigger load

than one machine by itself.It will increase the reliability of the power

system, since the failure of any one of them does not cause a total power loss to the load.

Having many generators operating in parallel allows one or more of them to be removed for shutdown and prevent maintenance.

The process of load sharing of several machines increase the efficiency and reliability of the system.

CONDITIONS REQUIRED FOR PARALLELING

There are three conditions that must be considered when the alternators are connected in parallel.

1. PHASE SEQUENCEIt means sequence in which the phase voltages peak in the two generators must be same.

An easy method to check the phase sequence is through “Three Light bulb method”. The circuit diagram for this is given below.

SYNCHRONIZING LAMP METHODIn this method when the bulbs are

alternately ON and OFF the phase is correct, otherwise not. Another method to check phase sequence is through synchroscope. The switch used in this lamp is TPST (triple pole single throw).

2. FREQUENCYIf flickering rate is high the frequency

difference is high. If it is low difference will be small.

3. VOLTAGETo check that voltages are same it is to be

noted that if the upper bulb is dark and the lower bulbs are bright than the voltage is same. This method is known as “one dark two bright lamp method” and the other method that can be applied is “All dark all bright method”.

SEPERATION OF CORE LOSSES AND FRICTIONAL LOSSES OF INDUCTION

MOTOR

EXPERIMENT NO. 10

APPARATUS

Variac transformerAmmeterVoltmeterInduction motorConnecting wires

CIRCUIT DIAGRAM

THEORY:There are two types of constant losses/no load

losses in induction motorFriction and windage lossesCore losses

Wo = Wo(no load) – I2R(stator)

Constant losses = Core + Windage lossesFrictional and windage losses depend on the

speed of motor. At no load frictional and windage losses are nearly constant as the speed of motor does not change.

Core losses depends on voltage, as the input voltage varies core losses vary with the changing voltage.


S.No. Voltage (volts) Current (Amp) WNL

Power

No load

(Watts)

Cu losses of

stator=

IL2R

Const.

Losses

=WNL-IL2R

V1 V2 V3 VL I1 I2 I3 IL

RESULT

EXPERIMENT NO. 11

TO DRAW O.C.C AND S.C.C OF A SYNCHRONOUS GENERATOR

CIRCUIT DIAGRAM

THEORY

The behavior of a real synchronous generator can be determined by three quantities

Relationship b/w If and Ø(and therefore b/w Ia and Ea)

Synchronous reactanceArmature resistance

“OPEN CIRCUIT CHARACTERISTICS” Or “OPEN CIRCUIT TEST”

To perform this test the generator is turned at rated speed, the terminals are disconnected from all loads and If is set to zero. Then field current is gradually increased in steps and terminal voltage is measured at each step.

OBSERVATIONSS. No. If

(Amp)

E

(Volts)

N

(rpm)

“SHORT CIRCUIT CHARACTERISTICS” Or “S.C.C TEST”

Adjust the If to zero again and short circuit the terminals of generator. Then the armature current or line current is measured as If is increased.

OBSERVATIONSS. No. If

(Amp)

Ia

(Amp)

N

(rpm)

SYNCHRONOUS IMPEDANCE“The whole of the voltage E is being used to

circulate the armature short circuit current Ia against the synchronous impedance Zs”

Where Zs = E(at O.C.C) Ia(at S.C.C)

But value found by this method is very much greater than real value.

SHORT CIRCUIT RATIO“the ratio of the field current required for the

rated voltage at open circuit to the field current required for the rated Ia at short circuit”

For turbo generators it is small 0.5 to 0.6For salient pole 1 to 1.5If some generators have high SCR it mea

lower Zs and vice versa

ADVANTAGES OF LOWER ZsStability Ps = (3VphEphSinθ)/(Xs or Zs)Voltage regulation

DISADVANTAGEHigh SCR

EXPERIMENT NO. 12

VOLTAGE REGULATION OF AN ALTERNATOR BY SYNCHRONOUS IMPEDANCE METHOD

As Zs = E/IaXs = √(Zs2 – Ra2)Knowing Xs and Ra (Ra is already known, it is

always have smaller value and in bigger machines it can be neglected) we draw vector diagram for any load and any power factor

NowE = √ [(VCosØ+IaRa)2 + (VSinØ±IaXs)2

_ Sign for capacitice case+ sign is for inductive caseNowRa = 0.7ΩIa = 7.9 AmpV = 220/√3 = 127.02 voltsZs = E/Ia = ______Xs = √ (Zs2 – Ra2)

CASES

CosØ = 0.8 lagging _____ SinØ have +ve signCosØ = 0.8 lagging ______SinØ have _ve signCosØ = unity______SinØ have +ve signNow

% Reg = (E-V)/V *100

GRAPH B/W %Reg and Ia

We can also draw graph b/w these two parameters. As

Ia 25% 50% 75% 100%2A 4A 6A 8A

And by finding corresponding values of %Reg for current values mentioned.

GRAPH

% Reg

Ia

RESULT

experiments ems ii

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