electrical &electronics lab manuel by Purna_Yalamanchili

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									                                                    ELECTRICAL AND ELECTRONICS ENGINEERING LAB



                                                Exp. No: 1


                                     SWINBURNE’S TEST
AIM: To pre-determine the efficiency of a D.C. shunt machine considering it as a generator or as a
motor by performing Swinburne’s test on it.

APPARATUS:

  S. No          Equipment                    Type/ Rating                     Quantity

    1           DC voltmeter                  (M.C) 0-300v                       1 NO.

    2           DC Ammeter                     (M.C) 0-20A                       1 NO.

    3           DC Ammeter                      (M.C) 0-2A                       1 NO.

    4           Tachometer                     0-2000RPM                         1 NO.

    5        Variable Rheostat                 300Ω/1.5A                         1 NO.




NAME PLATE DETAILS:

THEORY:

Testing of D.C .machines can be divided into three methods:

         Direct test
         Regenerative test
         Indirect. test
        Swinburne’s Test is an indirect method of testing a dc machine. In this method, the constant
losses of the DC machine are calculated at no-load. Hence, its efficiency either as motor or as a
generator can be pre-determined. In this method, the power requirement is very small. Hence, this
method can be used to pre-determine the efficiency of higher capacity dc machines as a motor and as a
generator.

Disadvantages:


Department of ECE, DJRCET                                                                        Page 1
                                                        ELECTRICAL AND ELECTRONICS ENGINEERING LAB



               (i)       Efficiency at actual load is not accurately known
               (ii)      Temperature rise on load is not known and
               (iii)     Sparking at commutater on load is not known.
Power input at No-load = Constant losses + Armature copper losses (which is negligible)

Power input at No-load = Constant losses

Power input              = Va Ia + Vf If

LOSSES IN A DC MACHINE:

      The losses in a D.C. machine can be divided as 1) Constant losses 2) Variable losses, which
changes with the load.

CONSTANT LOSSES:

Mechanical Losses: Friction and Wind age losses are called mechanical losses. They depend upon the
speed. A dc shunt machine is basically a constant speed machine both as a generator and as a motor.
Thus, the mechanical losses are constant.

Iron Losses: For a dc shunt machine, the field current hence the flux per pole is constant (Neglecting the
armature reaction which reduces the net flux in the air gap). Hence, hysterics and eddy current losses
(which are also called as iron losses) remain constant.




Field Copper Losses: Under normal operating conditions of a D.C. shunt machine, the field current
remains constant. Thus, power received by the field circuit (which is consumed as field copper losses) is
constant.

Constant losses in a dc shunt machine=Mechanical + losses Iron losses+ Field cu. Losses.

VARIABLE LOSSES:

The power lost in the armature circuit of a dc machine increases with the increase in load. Thus, the
armature copper losses are called as variable losses.

EFFICIENCY OF A DC MACHINE:


Department of ECE, DJRCET                                                                           Page 2
                                                               ELECTRICAL AND ELECTRONICS ENGINEERING LAB



                            Outputpower
          % Efficiency =                X 100
                             Inputpower

As a generator Input power Pin = Pout + Constant losses+ Armature copper losses at a

                                           Given load I2a Ra

                           Pout = VLIL

                   Where Ia = IL + If                       Self excited generator

                             Ia= IL                       Separately excited generator

 As a motor Input power Pin           = VL IL + Vf If

          Output power Pout = Pin – Constant losses – Armature Copper losses.




                             IL= Ia + If                 Self excited motor

                                                        (Vf If is not accounted for Pin)

                           IL = Ia                                 Separately excited motor>

Note: While calculating the armature copper losses on load condition, the hot resistance of the
armature= 1.2 Ra (Normal Temperature) is considered.

CIRCUIT DIAGRAM:




Department of ECE, DJRCET                                                                         Page 3
                                                            ELECTRICAL AND ELECTRONICS ENGINEERING LAB



PROCEDURE:

           1. Connections are made as per the circuit diagram.
           2. At the time of starting keep the field rheostat of the motor at minimum position.
           3. Motor is started under no load using the 3-point starter.
           4. Motor is brought to the rated speed by varying the field rheostat.
           5. At the rated speed No load readings of the ammeters and voltmeter are noted down.
OBSERVATIONS:

At no-load (separately excited dc motor):

                                                                               Constant losses =

                                                                                   (Pin - I2aRa)
   V              IL             IF         Ia        Pin= V IL       I2aRa
                                                                                     (watts)
 (volts)       (Amps)        (Amps)      (Amps)        (watts)       (watts)




EFFICIENCY AS A MOTOR:

Let us assume that the current drawn by the armature =

Input to the motor            = input to the armature + input to the field =

Total losses                = constant losses + armature copper losses =

Output                      = Input – total losses      =

                                 Output
Efficiency             ηm    =          X 100 =
                                  Input




Department of ECE, DJRCET                                                                          Page 4
                                                  ELECTRICAL AND ELECTRONICS ENGINEERING LAB



EFFICIENCY AS A GENERATOR:

Let us assume that the current delivered by the armature =

Output            =

Total losses      =constant losses + armature copper losses =

Input            = Output + Losses =

                        Output
Efficiency      ηg =           X 100     =
                         Input




AS A MOTOR:

                          Power        Copper     Total         Power
         S.      IL
                           Input        Loss      Loss           input    Efficiency
         No.   (Amps)
                          (watts)      (watts)   (watts)        (watts)

         1.

         2.

         3.

         4.

         5.




Department of ECE, DJRCET                                                              Page 5
                                                   ELECTRICAL AND ELECTRONICS ENGINEERING LAB



AS A GENERATOR:

                          Power       Copper       Total      Power
        S.        IL
                          Input        Loss        Loss       Output        Efficiency
        No.    (Amps)
                         (watts)      (watts)     (watts)    (watts)

        1.

        2.

        3.

        4.

        5.




CONCLUSION:

   1. The power required to conduct the test is very less as compared to the direct loading test.
   2. Constant losses are calculated from this method are used to compute the efficiency of a dc
       machine as a generator and as a motor without actually loading it.
   3. Hence, this is an economic method.



MODEL GRAPH:




Department of ECE, DJRCET                                                                       Page 6
                                                    ELECTRICAL AND ELECTRONICS ENGINEERING LAB



RESULT:




VIVA-VOCE:

1. Why do we conduct this test?

2. What are the different types of losses present in a DC Machine?

3. Why armature of a DC Machine is laminated?

4. What are the advantages of Swinburns test?

5. Swinburne’s test is applicable to what type of machines and why?




Department of ECE, DJRCET                                                              Page 7
                                                      ELECTRICAL AND ELECTRONICS ENGINEERING LAB



                                         Exp. No: 2

                        BRAKE TEST ON A DC SHUNT MOTOR

AIM: To obtain the performance characteristics of a DC Shunt motor by a load test.


APPARATUS:

   S. NO            EQUIPMENT                         TYPE/ RATING              QUANTITY

    1               DC voltmeter                      (M.C) 0-300v                 1 NO.

    2               DC Ammeter                         (M.C) 0-20A                 1 NO.

    3               DC Ammeter                         (M.C) 0-2A                  1 NO.

    4               Tachometer                         0-2000RPM                   1 NO.

    5              Spring balance                        0-10Kg                    2 NO.

    6                 Rheostat                           0-200Ω                    1 NO.




NAME PLATE DETAILS:




THEORY:

        This is a direct method of testing a dc machine. It is a simple method of measuring motor
output, speed and efficiency etc., at different load conditions A rope is would round the pulley and its
two ends are attached to two spring balances S1 and S2 . The tensions provided by the spring balances S1
and S2 are T1 and T2 the tensions of the rope can be adjusted with the help of swivels.

        The force acting tangentially on the pulley is equal to the difference between the readings of the
two spring balances in kg- force.

        The induced voltage Eb =V-Ia Ra and Eb= KΦN, Thus, KΦ=Eb /N

Department of ECE, DJRCET                                                                         Page 8
                                                    ELECTRICAL AND ELECTRONICS ENGINEERING LAB



V= applied voltage, Ia =armature current, Ra =Armature resistance.

Total power input to the motor Pin =Field circuit power + Armature power

                              = VfIf + Va Ia

CIRCUIT DIAGRAM:




PROCEDURE;-

    1. Connections are made as shown in the circuit diagram
    2. Before giving supply, the field rheostat is kept in minimum position.
    3. During starting load is not applied and by varying the field rheostat reading make the motor to
        run at its rated speed and note down the no load values.
    4. The Motor is slowly loaded with the help of spring balances, tightening the rope.
    5. At every equal increment of load, ammeter, voltmeter, and spring balances are noted.
    6. The efficiency is calculated for different load conditions and maximum efficiency point is
        indicated.




Department of ECE, DJRCET                                                                     Page 9
                                               ELECTRICAL AND ELECTRONICS ENGINEERING LAB



OBSERVATION TABLE:

S.No Voltage Speed Load         Field     W1   W2   Torque Input   Output           =
                        Current Current
     (V)        (rpm)                     (kg) (kg) (N-M)   (Watt) (watt)   (output/input)*100
                        IL(A)   If(A)




MODEL GRAPHS:




Department of ECE, DJRCET                                                          Page 10
                                                   ELECTRICAL AND ELECTRONICS ENGINEERING LAB



RESULT:




VIVA VOCE

   1. What are the disadvantages of this test?
   2.     Why do interpoles in DC machines have tapering shape?
   3. If the field of DC shunt motor gets opened while the motor is running then what happens?
   4. What will happen if the filled of a DC shunt motor is open?
   5. What happens when a Dc motor is connected across AC supply?




Department of ECE, DJRCET                                                                  Page 11
                                                     ELECTRICAL AND ELECTRONICS ENGINEERING LAB



                                                 Exp. No: 3

                    OPEN CIRCUIT AND SHORT CIRCUIT TESTS

                           ON A SINGLE PHASE TRANSFORMER
AIM:

       1. To predetermine the efficiency and regulation of the given single phase transformer at
            different power factors.
       2. To draw equivalent circuit referred to the primary.
APPARATUS REQUIRED:

    S.No:        Apparatus             Type               Rating           Quantity

       1.      Ammeter         Moving iron                0-20A                1

       2.      Voltmeter       Moving iron               0-300V                1

       3.      Wattmeter       Dynamometer             5A/150V,LPF             1

       4.      Wattmeter       Dynamometer            20A/150V,UPF             1




NAME PLATE DETAILS:




THEORY:

TRANSFORMER:
        Transformer is a static device which transfers electrical power from one circuit to another circuit
without any electrical connection but through magnetic medium. It works on the principle of Faraday
law of electro magnetic induction. It transfers energy without change in frequency but usually change in
voltage and current.

Department of ECE, DJRCET                                                                         Page 12
                                                      ELECTRICAL AND ELECTRONICS ENGINEERING LAB



        The performance of transformer can be pre determined with reasonable accuracy the
parameters of its equivalent circuit. The parameters and losses in a transformer can be determined by
conducting two simple tests to estimate the efficiency and voltage regulation without actually loading
the transformers.

They are

               Open circuit or no load test.
               Short circuit test.
        Open circuit test is conducted by opening the secondary (H.V) of the transformer.

By using open circuit test the core loss of the transformer can be determined.

From open circuit test data

                                                   RO= Vo / Ic

                                                   Xm=Vo / Im
        Short circuit test is conducted by short circuiting the secondary (L.V) of the transformer. By
using short circuit test the copper loss of the transformer can be determined.

From short circuit test data:

                                                    Zh=Vsc/Isc

                                                   RH=Wsc/Isc2

                                                  XH=(ZH2-RH2 )1/2




Department of ECE, DJRCET                                                                         Page 13
                                                  ELECTRICAL AND ELECTRONICS ENGINEERING LAB



CIRCUIT DIAGRAM:
Open Circuit Diagram:




Short Circuit Diagram:




PROCEDURE:

Open circuit test
    1. Connections are made as per the circuit diagram
    2. HV side is kept open and rated voltage is applied to the low voltage winding by adjusting the
        autotransformer
    3. The meter readings are noted down and tabulated




Department of ECE, DJRCET                                                                  Page 14
                                                       ELECTRICAL AND ELECTRONICS ENGINEERING LAB




Short circuit test
    1. Connections are made as per the circuit diagram
    2. LV side is short circuited and adjusting the autotransformer, rated current is send on the HV side
    3. The meter readings are noted down and tabulated



OBSERVATIONS:

Open circuit test:


                      Io                                                        Im=Io*Sino
              Vo             Wo       Coso=Wo/(Vo*Io)           Ic=Io* Coso
             (V)      (A)    (W)                                        (A)         (A)




Short circuit test:

                      Vsc                Isc                   Wsc

                      (V)                (A)                  (Watts)




Efficiency curve:

Efficiency at any load (X times full load) and at a given power factor can be calculated as
follows.

Output at X times full load =X*rated KVA*PF

Iron loss = Wo

Copper loss at x times full load =X2 *full load copper loss

Department of ECE, DJRCET                                                                       Page 15
                                                            ELECTRICAL AND ELECTRONICS ENGINEERING LAB



%Efficiency = output*100/(output + losses)

Efficiency at different assumed loads for a given power factor are calculated and tabulated as follows

 S.No.      Load(X)        Output    Iron          Copper      Input=output + losses   %Efficiency
                                     loss          loss
                           (watts)                                 (watts)
                                     (watts) (watts)




Regulation curve:


Percentage regulation= r Cos ± X sin
r, percentage resistance=I*R1*100/V

x, percentage reactance=I*X1*100/V

I, rated LV side current

V, rated LV side voltage

Positive sign for lagging power factor and negative sign for leading power factor percentage regulation
for full load for different power factors are calculated and results are tabulated as shown bellow.




  S.No                Cos                  sin          %Regulation




Department of ECE, DJRCET                                                                            Page 16
                                                  ELECTRICAL AND ELECTRONICS ENGINEERING LAB



PRECAUTIONS:
   1. Wattmeter connections must be done as per the rating of the transformer
   2. LPF wattmeter to be used for open circuit test
RESULT:




VIVA VOCE:

      1. What is the operating principle of transformer?
      2. Why OC and SC tests are convenient and very economical?
      3. What is the main purpose of OC and SC test?
      4. Why iron losses are negligible during SC test of a transformer?
      5. What is the primary reason to conduct OC test only on the Low voltage winding of the
             transformer?




Department of ECE, DJRCET                                                                 Page 17
                            ELECTRICAL AND ELECTRONICS ENGINEERING LAB




Department of ECE, DJRCET
      6.                                                      Page 18
                                                     ELECTRICAL AND ELECTRONICS ENGINEERING LAB



                                               Exp. No: 4

                            SPEED CONTROL OF DC SHUNT MOTOR

AIM: To control the speed of DC shunt motor by

1. Armature control method

2. Field control method



APPARATUS REQUIRRED:

               Sl.no   Apparatus                   Type                 Range               Quantity

                1.     Voltmeter                    MC                 (0-300)V                 1

                2.`    Ammeter                      MC                  (0-5)A                  1

                3.`    Ammeter                      MC                  (0-2)A                  1

                4.     Rheostat                                     250 ohm/1.5A                2

                5      Tachometer                 Digital                                       1




Fuse Rating:

40% of rated current

40*17/100=6.8≈10 A

THEORY:

 Flux Control method

 The speed of the DC motor is inversely propositional to the flux per pole, when the armature voltage is
kept constant. By decreasing the flux the speed can be increased and vice –versa. Hence the main flux of
field control method the flux of a DC motor can be changed by changing field current with help of a shunt
field rheostat. Since shunt field current is respectively small shunt field rheostat has to carry only a small
amount of current which means I2R losses is small so that rheostat is small in size .This method is very
efficient.



Department of ECE, DJRCET                                                                       Page 19
                                                         ELECTRICAL AND ELECTRONICS ENGINEERING LAB



Armature Control method

 This method is used when speed below the no load speed are required.As the supply voltage is normally
constsnt the voltage across the armature is varied by inserting a variable rheostat in series with the
armature circuit. As conductor resistance is increased potential difference across the armature is decreased,
herby decreasing the armature speed. F or a load of constant torque speed is approximately propositional to
the potential difference across the armature.

PRECAUTIONS:

1. Armature rheostat must be kept at maximum resistance position.

2. Field rheostat should be kept at minimum resistance minimum position.

PROCEDURE:

ARMATURE CONTROL METHOD:



1. Connect as per the circuit diagram

2.Close the DPST switch

3. Start the motor using three point starter

4.By keeping the field current(If) as constant value, adjust the armature rheostat and note down the
corresponding armature voltage and motor speed.

5.Repeat the step four till the motor reaches the rated speed.



OBSERVATIONS



                     Field Current If = 0.6 A              Field Current If = 0.5 A

                     Armature                              Armature
         SL.NO.                                Speed N                            Speed N
                     Voltage Va                            Voltage Va
                                                RPM                                   RPM
                              V                                    V

               1.




Department of ECE, DJRCET                                                                      Page 20
                                                   ELECTRICAL AND ELECTRONICS ENGINEERING LAB



               2.



               3



               4



               5



               6




FLUX CONTROL METHOD:

1. Connect as per the circuit diagram

2.Close the DPST switch

3. Start the motor using three point starter

4.By keeping the armature voltage as constant value, adjust the field rheostat and note down the
corresponding field current and motor speed.

OBSERVATIONS



                      Armature Voltage Va= 180 V          Armature Voltage Va = 170 V

         SL.NO.       Field Current            Speed N    Field Current            Speed N

                            If                  RPM             If                   RPM

                            A                                  A

               1.




Department of ECE, DJRCET                                                                    Page 21
                                                    ELECTRICAL AND ELECTRONICS ENGINEERING LAB



               2.



               3



               4



               5



               6




GRAPHS:

1.Field current Vs speed

2. Armature voltage Vs speed

RESULT:



VIVA QUESTIONS:



1.How does the speed of a DC shunt motor vary with armature voltage and field current?



2. Compare the resistance of the armature and field winding.



3.What is the importance of speed control of DC motor in industrial application?



4.Which is of the two method of speed control is better and why?

5.Why is the speed of DC shunt motor practically constant under normal load condition?


Department of ECE, DJRCET                                                                Page 22
                            ELECTRICAL AND ELECTRONICS ENGINEERING LAB




Department of ECE, DJRCET                                     Page 23
                                                     ELECTRICAL AND ELECTRONICS ENGINEERING LAB



                                         EXP.NO. 5

       LOAD TEST ON 3-PHASE SQUIRREL CAGE INDUCTION MOTOR
AIM: To draw the performance characteristics of 3-phase squirrel cage induction motor by conducting
load test.

APPARATUS REQUIRED:

S.No                Name of             Range               Type                 Qty.
                    apparatus

1.                  Ammeter             (0-5)A              MI                   1

2.                  Voltmeter           (0-600)V            MI                   1

3.                  Wattmeter           (600V,5A)           UPF                  2

4.                  Tachometer                              Digital              1

5.                  3-Ф                                                          1
                    autotransformer



FUSE RATING;

125% of 4.8A=6A=10A



THEORY:

A 3-phase induction motor consists of stator and rotor with the other associated parts. In the stator, a 3-
phase winding is provided. The windings of the three phase are displaced in space by 120º.A 3-phase
current is fed to the 3-phase winding. These windings produce a resultant magnetic flux and it rotates in
space like a solid magnetic poles being rotated magnetically.




Department of ECE, DJRCET                                                                        Page 24
                            ELECTRICAL AND ELECTRONICS ENGINEERING LAB




Department of ECE, DJRCET                                     Page 25
                                                               PRECA
                                                      ELECTRICAL AND ELECTRONICS ENGINEERING LAB



UTIONS-

1.TPST switch is kept open initially.

2.Autotransformer is kept at min. voltage position.

3.There must be no load when starting the load.

PROCEDURE-

1.Connections are given as per circuit diagram.

2.3-Ф induction motor is started with DOL starter.

3. If the pointer of one of the wattmeter readings reverses, interchange the current coil terminals and
take the reading as negative.

3.The no load readings are taken.

4. The motor is loaded step by step till we get the rated current and the readings of the voltmeter,
ammeter, wattmeters, spring balance are noted.

FORMULAE USED-

      1)      % slip= (Ns-N/Ns)*100
      2)      Input Power = (W1+W2)watts
      3)      Output Power = 2∏NT/60 watts
      4)      Torque = 9.81*(S1-S2)*R N-m
      5)      % efficiency = (o/p power/i/p power)* 100
GRAPHS-

         1)   Output Power vs Efficiency
         2)   Output Power vs Torque
         3)   Output Power vs Speed
         4)   Output Power vs %s
RESULT




Department of ECE, DJRCET                                                                        Page 26

								
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