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SPEED CONTROL OF DC SHUNT MOTOR AIM To Control the speed of the DC Shunt motor by armature control (varying armature voltage with field current kept constant) and field control (varying field current with armature voltage kept constant) methods THEORY The voltage equation of a DC shunt motor is given by Va = Eb + IaRa Volts Eb = ∅ZN P Volts 60 A From the above relation speed of DC Shunt Motor is given by relation N = (Va - IaRa)/KØ Where N = speed (rpm) Va = Armature voltage (volts) Ia = armature current (amps) Ra = resistance of the armature (ohms) K = motor design constant φ = strength of the field flux (weber) Z = Total number of armature conductors P = Number of field Poles A = Number of parallel paths If = Field Current (amps) From the relation of speed above, the speed of DC shunt motor can be varied by varying 1. Armature resistance Ra 2. The Field Flux Ø PROCEDURE Field Control Method Make the Connection as per circuit diagram shown below 1. The armature rheostat should be kept at the maximum position so that there should be maximum resistance in series with armature. 2. The field rheostat should be at minimum position. 3. Switch on the supply by closing DC On/Off Isolator and DC MCB 4. Adjust the armature voltage to some fixed value. 5. Change the speed by varying the field rheostat in steps and note down the corresponding values of speed and field current. 6. Repeat the steps 5 and 6 for different armature voltages 7. Plot field current V/s Speed graph. Tabulation S.No. Armature Voltage Va =____ (V) Field Current Speed If(A) N (rpm) 1. 2. 3. 4. 5. Model Graph Armature Voltage Va =____ (V) Field Current If(A) Speed N (rpm) Armature Control Method 1. The armature rheostat should be kept at the maximum position so that there should be maximum resistance in series with armature 2. The field rheostat should be at minimum position. 3. Switch on the supply by closing DC On/Off Isolator and DC MCB 4. Adjust the field current to some fixed value and change the speed in steps by varying armature rheostat. 5. Note down the corresponding values of armature voltage, armature current and speed. 6. Repeat step 5 for different values of field current. 7. Plot Back Emf v/s Speed. Tabulation Field Current If =____ (A) S.No. Armature Armature Voltage Current Va (V) Ia (A) 1. 2. 3. 4. 5. Model Graph Back Emf Eb (V) Speed N (rpm) Field Current If =____ (A) Armature Voltage Va (V) Armature Back Current Emf Ia (A) Eb (V) Speed N (rpm) SWINBURNES TEST AIM To pre-determine the percentage efficiency of the given DC Machine whenworking as a generator and as a motor. THEORY In Swinburne’s method the d.c. machine (generator or motor) is run asa motor at no-load and losses of the machine are determined. Once the losses of the machine are known, its efficiency at any desired load can be determined in advance. It may be noted that this method is applicable to those machines in which flux is practically constant at all loads. E.g., Shunt and Compound machines. As a motor, Motor input power Pi(M) = VLoILo (watts) No load armature currentIao= (ILo- Ish) (amps) Armature Copper Loss Wcu = I2ao Ra(watts) Constant losses Wc =VLoILo – I2ao Ra (watts) Gross power developed by armaturePo(M) = VLoIao-I2ao Ra = (VLo -Iao Ra)Iao Po(M) = EbIao Percentage efficiency, η(M) = (Po(M) / Pi(M)) *100 Where, VLo = Motor input voltage (volts) ILo = No-Load current (Amps) Iao = No load armature current (Amps) Ish = Motor Shunt field current (Amps) Ra = Armature resistance (Ohms) As a generator, Generator output Po(G) = VIL (Watts) Armature copper lossWcu = I2a Ra(Watts) Armature current Ia = IL + Ish (Amps) Total losses Wt = Wc+ I2a Ra (Watts) Generator input Pi(G) = Po(G)+ Wt(Watts) Percentage efficiency, η(G) = (Po(G) / Pi(G)) *100 = (VIL) / (VIL+ Wc+ (IL + Ish)2 Ra) * 100 Procedure 1. The armature rheostat should be kept at the maximum position so that there should be maximum resistance in series with armature. 2. The field rheostat should be at minimum position. 3. Switch on the supply by closing DC On/Off Isolator and DC MCB. 4. Bring the motor to its rated speed by adjusting motor field rheostat. 5. Note down the corresponding motor input voltage (V), No-Load current (ILO) and field current (Ish). Tabulation S.No. Input Voltage V (Volts) No load Current ILO (Amps) Field Current Ish (Amps) No Load Armature Current Iao (Amps) Constant Loss Wc (Watts) Calculation As a motor S.No. Assumed Load Current IL(Amps) Armature Current Iao (Amps) Armature Copper Loss Wcu (watts) Motor input power Pi(M) (watts) Motor output power Po(M) (watts) Efficiency η(M) (%) Armature Current Ia (Amps) Armature Copper Loss Wcu (watts) Generator output Po(G) (watts) Generator input Pi(G) (watts) Efficiency η(G) (%) 1. 2. 3. 4. As a Generator S.No. Assumed Load Current IL(Amps) 1. 2. 3. 4. Model Graph LOAD TEST ON SINGLE PHASE TRANSFORMER AIM To find out the efficiency of the given single phase transformer by conducting direct load test in it. THEORY The transformer is a device which transfers energy from one electrical circuit to another electrical circuit through magnetic field as coupling medium. In this process it does not change the frequency of voltage or current. It works on the basic principle of electromagnetic induction (mutually induced EMF). Being a static device it has a very high efficiency as compared to rotating machine of same rating as the losses are less. When primary winding of transformer is energized with source of voltage V pan EMF Esis induced across the secondary winding and it is also equal to secondary terminal voltage Vs till there is no load across secondary winding. As soon as load is applied across the secondary winding the terminal voltage is decreased from Es to Vs. Percentage Efficiency Ƞ = [Output Power (Ws) / Input Power (Wp)]*100 Vp - Primary Voltage (Volts) Vs - Secondary Voltage (Volts) Ip -Primary Current (Ampere) Is - Secondary Current (Ampere) Wp -Power consumed by primary winding (Watts) Ws -Power consumed by secondary winding (Watts) Ƞ - Percentage Efficiency (%) PROCEDURE 1. Make the connections as per the circuit diagram. 2. Keep the switch on secondary side open so that load is zero to measure no load voltage. Also keep knob of auto transformer at zero output voltage position. 3. Now increase the voltage through auto transformer until voltage in voltmeter V2 reads rated value of secondary winding & read no load voltage E2. 4. Switch onresistive load such that secondary winding current be approximately 10% of the rated current of secondary side. 5. Take the corresponding readings from WattmetersWs&Wp,Voltmeters Vs& Vp and AmmetersIs& Ip 6. Increase the load current in steps of 10% of the rated value by switching on few more loads and take the readings of the Wattmeter, Ammeter & Voltmeter till it reaches 120-125% of rated value. 7. Reduce the load to zero by switching of the loads one-by-one. 8. Switch off the Supply. TABULATION S.No. Primary Primary Primary Secondary Secondary Secondary Efficiency Voltage Current Power Voltage Current Power (%) Vp (V) Ip (A) Wp (W) Vs (V) Is (A) Ws (w) 1. 2. 3. 4. 5. MODEL GRAPH +++ OPEN CIRCUIT AND SHORT CIRCUIT TEST ON SINGLE PHASE TRANSFORMER AIM To obtain the equivalent circuit parameters of a single phase transformer by conducting open circuit and short circuit test in it. THEORY OPEN CIRCUIT TEST The shunt branch parameters can be determined by performing this test. Since, the core loss and themagnetizing current depend on applied voltage, this test is performed by applying the rated voltageto one of the windings keeping the other winding open (generally HV winding is kept open and ratedvoltage is applied to LV winding). Since,the secondary terminals are open (no load is connected across the secondary), current drawn from thesource is called as no load current. Under no-load condition the power input to thetransformer is equal to the sum of losses in the primary winding resistance Rand coreloss. Since, no load current is very small, the loss in winding resistance is neglected. Hence, on no loadthe power drawn from the source is dissipated as heat in the core. SHORT CIRCUIT TEST Suppose the input voltage is reduced to a small fraction of ratedvalue and secondary terminals are short-circuited. A current will circulate in the secondary winding.Since a small fraction of rated voltage is applied to the primary winding, the flux isgenerated in the core and hencethe core loss is very small. Hence, the power input on short circuit is dissipated as heat in the winding.In this test, the LV terminals of thetransformer are short circuited. The primary voltage is gradually applied till the rated current flows inthe winding. Since, the applied voltage is very small (may be of the order of 5-8%). FORMULA OPEN CIRCUIT TEST (LV Side) 1. Iron or core loss, Wi=Woc 2. No load power input power, Woc = Voc* Ioc* cos Φ oc 3. No load power factor,Cos Φ oc = Woc/(Voc*Ioc) 4. Working component of no load current,Iw = IoccosΦoc 5. Magnetizing component of no load current,Iµ = Ioc sin Φoc 6. No load Resistance R0=Voc/Iw 7. No load Reactance X0=Voc/ Iµ Voc = Open circuit voltage (Volts) Ioc = Open circuit current (Amps) Woc = Open circuit power (Watts) SHORT CIRCUIT TEST (HV Side) 1. Equivalent impedance of winding referred to HV sideZ02=Vsc/Isc 2. Equivalent resistance of winding referred to HV side, R02 =Wsc/ Isc 3. Equivalent Reactance of winding referred to HV side, X02= √( Z202– R202 ) 4. Transformation Ratio, K= V1 / V2 5. Equivalent resistance of winding referred to LV side, R01= R02 / K2 6. Equivalent reactance of winding referred to LV side, X01= X02 / K2 V1 = LV side or Primary side voltage (Volts) V2 = HV side or secondary side voltage(Volts) TABULATION OPEN CIRCUIT TEST S.No. No load Voltage (Voc) No load Current (Ioc) No load Power (Woc) SHORT CIRCUIT TEST S.No. Short Circuit Voltage (Vsc) Short Circuit Current (Isc) Short Circuit Power (Wsc) PROCEDURE OPEN CIRCUIT TEST 1. Make the connections as per the circuit diagram. 2. Switch on the AC breaker and isolator. 3. Adjust single phase auto transformer till the voltage connected across the primary winding reads rated primary voltage. 4. Note down the corresponding open circuit voltage (Voc), no load current (Ioc) and no-load power (Woc) indicated by the respective voltmeter, ammeter and wattmeter. 5. Reduce the auto transformer voltage to zero. 6. Switch off the AC breaker and isolator. SHORT CIRCUIT TEST 1. Make the connections as per the circuit diagram. 2. Switch on the AC breaker and isolator. 3. Adjust single phase auto transformer till the ammeter reads rated load current. 4. Note down the corresponding short circuit voltage (Vsc), short circuit current (Isc), and short circuit power (Wsc) indicated by the respective voltmeter, ammeter and wattmeter. 5. Reduce the auto transformer voltage to zero. 6. Switch off the AC breaker and isolator. EQUIVALENT CIRCUIT (referred to primary)