5 Swinburnes

Electrical & Electronics LabExperiment VSWINBURNE'S TEST ON DC SHUNT MACHINEAim:To conduct Swinburnes test on a DC shunt machine and predetermination of efficiencies when the machine is running as motor and generator.

Name Plate Details:Type Shunt

Power5 H.P

Armature /Field Voltage230 V

Armature current20 A

Field Current1A

Apparatus Required:

S.NoName of the apparatusTypeRangeQuantity

1.AmmeterMC(0-2)A1

2.AmmeterMC(0-30)A1

3.VoltmeterMC(0-30)V1


5.RheostatWire wound370/1.7A1

6.TachometerDigital0-10000R.P.M1

7.Connecting wires-0-20ARequired

Theory:

Circuit diagram:

Observations:

Tabulation to find out the constant losses:Voltage No load currentShunt Field CurrentNo load armature currentNo load Cu. LossesConstant losses

VI0IshIa0(Ia0)2RaWi=VI0- (Ia0)2Ra

Tabulation to find out the efficiency running as motor:Fractionof loadLoad currentField CurrentArm.currentArm Cu. lossTotal lossesMotor InputMotor OutputEfficiency

XIL=XIratedIshIa=IL-IshWcu=Ia2RaWT=Wc+WcuPin=VILPout=Pin - WT%=Pout/Pin*100

1

Tabulation to find out the efficiency running as generator:

Fractionof loadLoad currentField CurrentArm.currentArm Cu. lossTotal lossesGeneratorOutputGenerator InputEfficiency

XIL=X*IIshIa=IL+IshWcu=Ia2RaWT=Wc+WcuPin=VILPout=Pin+ WT%=Pout/Pin*100

1

Expected graph:

Calculations:Case1. To fine Constant Losses of DC machine:Here VI0 = iron losses + mechanical losses + shunt field copper losses + armature copper lossesIn above losses iron losses, mechanical losses, shunt field copper losses are constant of all loads. Let the constant losses occurring in the machine be represented by Wi.VI0= Wi + armature copper loss at no load.Thus constant losses occurring in the dc machine is Wi = VI0 Ia02Ra (Where Ia0 = I0 Ish)Where Ra is the resistance of armature in ohmsCase 2. Efficiency of a DC machine as motor:

Let the full load current drawn by motor be IL amps, the shunt field current will remain constant as Ish.Armature current :Ia = IL IshArmature copper losses :Wcu = Ia2RaInput to the motor :Pin= VILTotal losses :WT = Wi + Ia2RaOutput = Input losses :Pout= V*IL (Wi + Ia2Ra)% Efficiency as a motor :%m= (Pout /Pin)*100

Case 3. Efficiency of DC machine as generator:

The full load current of generator be IL which can be known from the rating of generator.Full load output generator :Pout = VILArmature current :Ia = IL + IshArmature copper losses :Wcu= Ia2Ra Total losses :WT = Wi + Ia2RaInput = Output + losses :Pin= (VIL )+ (Wi + Ia2Ra)% Efficiency as a generator: %g= (Pout /Pin)*100

Procedure:

1. Connect the circuit as per the circuit diagram2. Switch ON the supply and close the DPST switch 3. Slowly start the motor with the help of 3-point starter 4. Make the motor to run at rated speed by adjusting the field rheostat.5. Take the meter readings at no load condition 6. Now bring back the Rheostat to initial position 7. Open the DPST switch and Switch-OFF the supply to the motor.8. Measure the armature resistance with voltmeter and ammeter method.Precautions:

1. Motor field rheostat should be kept at minimum resistance position.2. The motor should be at no load condition throughout the experiment.

Result: The efficiency of a given DC shunt machine by conducting swinburns test is obtained in different loads motor as well as generator.

Experiment VBRAKE TEST ON DC SHUNT MOTOR.Aim:To perform brake test on the given dc shunt motor and obtain the characteristics of the motor from the test observations.

Name plate details:

TypeShunt

Power3 H.P

Armature /Field Voltage220 V

Armature current12 A

Field Current1A

Apparatus:

S.NoName of the apparatusTypeRangeQuantity

1.AmmeterMC(0-30)A1


3.RheostatWire wound300/1.7A1

4.TachometerDigital0-1000 rpm1


Theory:In this method the motor is subjected to direct mechanical loading the attaching a brake drum and water cooled pulley to the motor shaft. A rope or belt is wound the pulley at its two ends. The two ends are connected to spring balances S1 & S2. The tension of the belt can be adjusted by tightening it on the pulley. The tangential force on the pulley is equal to the difference of the two spring balance readings.Circuit Diagram:

Observations:VoltageCurrentSpeedSpring balance readingsTorque(T)Angular VelocityOutputPower(watts)InputPower(Watts)Efficiency

VINS1S2S=S1~S2T=9.81Sr=2N/60Pout=TPin=VI%

Expected Graph:

Calculations: Torque on the pulley :T = 9.81 (S1~S2) r N-mPower Output:Pout = 2NT/60 wattsPower input:Pin = VI watts% Efficiency:% = (Pout/Pin) 100

Where r is the radius of the pulley in metersS1 & S2 are spring balance readings in Kg N is the speed in RPMV is the voltage across the motor I is the current drawn from the motor

Procedure: 1. Connect the circuit as per the circuit diagram2. Check the belt on the pulley is free so that there is no load on the pulley 3. Closed DPST switch and start the motor slowly using the 3-point starter4. Adjust the field current with field rheostat so that the motor runs at its rated speed5. Apply load on the pulley gradually in steps, tightening the belt around it.6. Take the readings of the ammeter and voltmeter connected to the motor and two spring balance readings and the speed at every step.7. Continue the experiment till full load of the motor is reached. 8. Remove the load and switch of the supply9. Tabulate the observation

Precautions: 1. The motor field rheostat should be kept at minimum resistance position.2. At the time of starting, the motor should be in no load condition.3. Cool the pulley by using water while the experiment is performed.4. While measuring the radius of the pulley effective radius must be considered.

Result: The efficiency of a given DC shunt motor by conducting brake test is obtained.Experiment VIOC & SC TESTS ON SINGLE-PHASE TRANSFORMERAim: To conduct O.C. & S.C. test on a given transformer and predeterminations of1) Efficiency,2) Regulation,3) Parameters of equivalent circuit

Name plate Details:

Type of Machine1- Transformer

Power2 KVA

Low Voltage230 V

High Voltage415 V

Frequency50Hz

Apparatus Required:S.No.EquipmentTypeRangeQuantity

1.AmmeterMI(0-2)A1

2.AmmeterMI(0-10)A1

3.VoltmeterMI(0-300)V1

4.VoltmeterMI(0-75)V1

5.WattmeterDynamo300V,5A,LPF1

6.WattmeterDynamo75V,10A,UPF1

7.1- auto transformer-230/(0-270)V,3KVA1


General Theory: These two tests on a transformer helps to find determine 1) The parameters of equivalent circuit 2) The voltage regulation3) EfficiencyComplete analysis of the transformer can be carried out once its equivalent circuit parameters are known. The power required during these two tests is equal to the appropriate power loss occurring in the transformerO.C.Test: This test is conducted by opening the secondary of a transformer. The core loss of the transformer can be determined from the test. It also gives the no-load current I0, which is used to calculate the parameters R0, Xm of the magnetizing circuit. The transformer is connect as indicated in the circuit diagram. One of the windings usually the low voltage winding is connected to the supply voltage source while the high voltage winding is kept open..This ensure magnification of the no-load current I0 , The rated voltage applied to the transformer using auto- transformer, the ammeter gives the total power loss and the ratio of voltmeter readings V2/V1 gives the ratio of the turns.S.C. Test: This test gives the full load copper loss. In this test, secondary side low voltage winding is short circuited. A small voltage applied to the primary and increased carefully till the current (Isc) in the primary winding reaches the rated full-load value. Under these conditions, the copper loss in the in the winding is same as that on full load.

Circuit Diagrams:O.C. Test

S.C. Test

Procedure:O.C.Test:

1. Make the connections as per circuit diagram.2. Give the supply by using DPST switch.3. Gradually increase the voltage by using auto transformer till voltmeter reads the rated Voltage on LV side.4. Note down the readings of voltmeter, Ammeter & Wattmeter.5. Slowly bring the auto transformer to its initial position & switch OFF the supply.

S.C.Test:1. Make the connections as per the circuit diagram.2. Give the supply by closing the DPST switch.3. Gradually increase the current by using auto transformer till ammeter reads the rated current4. Note down the readings of voltmeter, Ammeter &Wattmeter.5. Slowly bring the auto transformer to its initial position & switch OFF the supply

Observations:

O.C. Test.S.No.V1(Volts)Io(Amps)Wo(Watts)

S.C.TestS.No.Isc(Amps)Vsc(Volts)Wsc(Watts)

Tabulation to find the efficiency: at cos=0.8

Fractinal load (X) O/P(watts)

Iron losses(Wi)Cu losses(Wcu)Total losses%

1

Tabulation to find out the regulation: at full load ,X=1

P.F(Cos)%Reg(lag)%Reg(lead)

1

0.8

0.6

0.4

0.1

Calculations:O.C. Test:Core Loss:WoNo load power factor:Cos0 = W0/( I0 V1)Magnetizing component of current:Iw = I0 Cos 0Core Loss component of current:Iu = I0 Sin 0No load resistance:R0 =V0/ Iw in ohmsNo load reactance:X0 = V0/ Iu in ohms

Where W0=open circuit power in wattsI0=Open circuit current in AmpsEquivalent impedance referred to HV side Z02 = Vsc/Isc in ohms.Equivalent resistance referred to HV side R02 = Wsc/I2sc in ohms.Equivalent reactance referred to HV side X02 =(Z202- R202) in ohms.V1=Open circuit voltage in Volts

S.C. Test:Equivalent Circuit of 1- transformer referred to LV side:Cos0 = W0/( I0 * V1Iw = I0 Cos 0Iu = I0 Sin 0Z02 = Vsc/Isc, R02 = Wsc/I2sc, X02 =(Z202- R202)Transformation ratio (K) = V2/V1Equivalent resistance referred to LV side (R01) = R02 / K2Equivalent reactance referred to LV side (X01) = X02/K2 Efficiency & Regulation of 1- transformer:Output power = (X*KVA*Cos )Where X = fraction of load. KVA = power rating of transformer, Cos = power factor Iron losses (Wi) = W0Copper losses (Wcu)= X2*WscTotal losses = Cu losses + Iron losses. Output powerEfficiency = ------------------------ * 100 ( output power+ losses) X.Isc[ R02Cos X02Sin ] Regulation = ------------------------------------- * 100 V2 Where + for lagging. - for leading.

Expected graphs:

Result:

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5 Swinburnes