EEE Machines Lab Manual PDF

Heaven’s Light is Our Guide Department of Electrical and Electronic Engineering LAB SHEET Electrical Machine-I Sessional EEE2106 (1.5 Credit...

Heaven’s Light is Our Guide Department of Electrical and Electronic Engineering LAB SHEET Electrical Machine-I Sessional EEE2106 (1.5 Credit) SL Name of Experiments 01 Polarity test and observation of voltage and current relation between primary and secondary of single phase transformer. 02 Open circuit test and short circuit test of a single phase transformer. 03 To determine the regulation of a transformer under different power factor. 04 Parallel operation of transformers. 05 Construction of three-phase transformer using three single-phase transformer and observation of line voltage & phase voltage and line current & phase current relation in primary and secondary winding with balance and unbalance loading. 06 Study the efficiency of Three-Phase Transformers. 07 Observation of speed torque characteristic curve of 3-phase induction motor. 08 Blocked rotor test & no load test of 3-phase induction motor and determination of circuit parameters. 09 Study of a 1-φ induction motor. 10 Measurement of circuit parameters of single phase induction motor and determination of efficiency. 11 Determination of value of capacitor for maximum starting torque of capacitor split- phase induction motor. 12 Starting an Induction Motor by applying star-delta starter.  LAB INFORMATION In the laboratory we use : 01. DC Generator. 02. DC Motor. 03. Induction Motor 04. Synchronous machine 05. Starter. 06. Techo Meter. 07. Resistors. 08. Voltmeter. 08. Ammeter. 10. Wattmeter 11. Variac. Study of the rating of different machines : DC Generator: The machine, which can transform mechanical energy to electrical energy, is called generator. If the output is DC voltage, then that machine is called dc generator. DC motor: The machine which can transform electrical energy to machine energy is called motor. If the input is DC, then this motor is called DC motor. Rating Table: DIRECT CURRENT MACHINE (DC GENERATOR / MOTOR) KW 2 kW Volts- 250/250 V AMP – 8.0 A RPM 1450 WOUND COMPOUND FLD-AMPS 0.467 A AS SH GEN FLD-OHMS 25 C 350 Ω DUTY CONT 60 C RISE ENCL DP SERVICE FACT 1.5 ARMATURE RESISTANCE - 1.68 Ω DIRECT CURRENT MACHINE (DC GENERATOR / MOTOR) KW 4 1/2 kW Volts- 250/250 V AMP – 18.0 A RPM 1450 WOUND COMPOUND FLD-AMPS 1.0 A AS SH GEN FLD-OHMS 25 C 152.8 Ω DUTY CONT 60 C RISE ENCL DP SERVICE FACT 1.15 SUIT AS 5 HP 1500/ 3000 RPM 240 V ARMATURE (dc) RESISTANCE - 1.72 Ω SYNCHRONOUS MACHINE ( CONSTANT SPEED MACHINE) TYPE SJ FRME 254Z 1500 RPM 50 CYCLE 50 C RISE GENERATOR 4KVA 0.8 P.F 220/240 V 3 PHASE 19.2 /9.6 A FL AMP MOTOR 6 HP 1.0 P.F 120 /240 V 3 HP 22 / 11 A FL AMP FILD-AMP MAX 2.1 FILD VOLTS 125 V Line to line stator terminal (dc)Resistance = 0.75 Ω Induction Motor: An AC motor, whose speed is varied with load. 3 – PHASE INDUCTION MOTOR HP 3 SERVICE FACTOR 1.15 FL RPM 1450 VOLTS 220 /440 PHASE 3 FL AMPS 9.1 /4.55 A CYCLE 50 TYPE M FRAME 254 U C RISE 40 TIME RATING CONT SEC VOLTS 145 SEC AMP 10 Line to line stator terminal (dc)Resistance = 0.75 Ω SINGLE PHASE INDUCTION MOTOR : TYPE DE 1028 RATED VOLTAGE 220 V RATED CURRENT 6.6 A RATED POWER 1 kW RATED SPEED 2820 RPM FREQUENCY 50 HZ INSULATION CLASS E P.F. COS φ = 0.90 Main Winding (dc) Resistance = 3.95 Ω Auxiliary Winding (dc) Resistance = 9.32 Ω Experiment No: 01 Date: Name of the Experiment: Polarity test and observation of voltage and current relation between primary and secondary of single phase transformer. Theory: Transformer is a device for transferring electrical energy from one circuit to another circuit without a change in frequency. The Transformer in which primary voltage is greater than secondary voltage is called Step-up Transformer. Transformer in which secondary voltage is greater than primary voltage is called Step-down Transformer. There are many transformers still in service whose terminal does not have the standard markings. In order to interconnect these transformers with others, either for single-phase parallel operation or for poly-phase operation, it is necessary to determine the relative polarities of the terminals. A simple test is enough to accomplish this. For the circuit connections shown in the circuit diagram, when the voltmeter reading is equal to the sum of the secondary coil voltage, the coils are in additive polarity. Similarly when the voltmeter reading is the difference between the voltages of the secondary coil the coils are connected in subtractive polarity. This is how polarity of the coils with respect to a coil can be determined. We know that the relation between primary and secondary voltage is, E1 N 1  E2 N 2 V 1 N1 or,  V 2 N2 I1 N 2 and the relation between primary and secondary current is, ,  =K I 2 N1 R2 and secondary side resistance referred to primary is, R1= 2. K Apparatus: 1. Transformer (150 V per coil, 1A). 2. Voltmeter(0-450V) 3. Ammeter(0-2A) 4. AC supply(220V) 5. Variac (0- 250V) Circuit Diagram: FIG:STEP-UP TRANSFORMER FIG:STEP-DOWN TRANSFORMER  I1 A I2 a _ V1 V2 R2 n N Determination of Polarity: The primary side was given rated supply (150V). So, voltage of each coil was 150 volt. As in circuit-01 voltmeter reading was 0, the coils were in subtractive polarity. So, terminal X1 and X2 have opposite polarity and also terminal X2 and X4. In circuit-02 Voltmeter reading was 300. So, coils were connected is additive polarity. Thus, X 1 and X3 have same polarity. Similarly X2 and X4 have same polarity. So, if X1 and X3 are positive, X2 and X4 are negative and vice versa. Data Table: SL NO Primary side Secondary side Turns Ratio,N2/ VoltageV1 VoltageV2 N1 01 100 50 0.5 02 150 75 0.5 03 200 150 0.5 SL Primar Secondary Primary Secondary Turns Secondary Equivalent R1 NO y side side side side Ratio, side resistance =R2/ Voltag VoltageV2 Current CurrentI2 K=N2/ N1 Resistance referred to eV1 I1 = I1 / I2 R2= V2/ I2 Primary side R1= V1/ I1 =R2/ 01 100 50 0.25 0.50 0.50 100 400 400 02 100 50 0.30 0.60 0.50 83.33 333.33 333.32 03 100 48 0.37 0.70 0.53 68.57 270 244.1 04 100 45 0.42 0.80 0.526 56.25 238 203.3 Calculation: V1 R1= I1 V2 R2= I2 R R1= 22 K Precautions: 01. Care should be taken so that current in the coils does not exceed the rated value. 02. Winding should be given the rated or less than the rated supply. Discussion: Experiment No: 02 Date: Name of the Experiment: Open circuit test and short circuit test of a single phase transformer. Theory: We know that there the two types of losses of in transformer, such as copper loss and core loss. Copper loss is a variable loss and depends on current and core loss is a fixed loss and depends upon the voltage. Now the efficiency of a transformer is define as the ratio of output power to input power. Again the input power is equals to the sum of output power and losses. The losses can be fined out with open circuit test and short circuit test. The copper loss can be obtained by short circuit test and core loss can be obtained by the open circuit test. For Short Circuit Test first 0 voltage is applied to high voltage side and then the voltage is increased until for witch the rated current flows in low voltage side. Now the data or reading of the meters are taken. Again for Open Circuit Test 0 voltage is applied to low voltage side and then the voltage is increased until for which the rated voltage is obtained from high voltage side. Now the data or reading of the meters are taken. R1 X1 X 2´ R2 ´ X0 R0 Fig. Equivalent circuit of Transformer Apparatus: 01. Transformer (150 V per coil, 1A) 01. Voltmeter (0-450V). 02. Ammeter (0-2A). 03. Wattmeter (0-1200W). 04. Variac (0-240 V) Circuit Diagram: Data Table: Test Voltage Current Power Short circuit test 7.56 1.00 3.5 Open circuit test 100 0.21 16.5 Calculation: For short circuit test the applied voltage is very low and core loss is negligible. So it is assumed that current through RO & XO is zero ' 2 So we can write, RSCT = R1+ R 2 = P/ISCT = ZSCT = VSCT/ISCT = XSCT = X1+ X '2 = √ (ZSCT2 - RSCT2) = For Open circuit test Cu Loss is negligible i.e. loss in R1 is negligible and voltage drop at R1 & X1 is negligible. In this case Loss is assumed only in the core circuit 2 V OCT ROCT = RO = = P 1 GOCT = = R OCT V OCT ZOCT = ZO= = I OCT 1 YO = = ZO BO = √ (YO2 – GO2) = 1 XO = = BO % Efficiency =( Output/ input)×100% %Efficiency = (Input- Losses)×100/ Input %= (200-20)  100/200 =90 % Efficiency = Output/ (Output + losses)= Discussion: Experiment No: 03 Date: Name of the Experiment: To determine the regulation of a transformer under different power factor. Theory: Regulation is an indication of voltage changes due to change in load. Any equipment is said to have good regulation if this change of voltage is less. It is defined as For a transformer, for constant primary voltage as load increases, the voltage at the load decreases, as there is voltage drop due to internal resistance and reactance of the transformer. If we know the resistance and reactance of the transformer, its regulation can be determined under various load conditions. Apparatus: 1. Universal Power Supply Module 2. 1PH Transformer 3. AC Ammeter Module 0-1A 4. AC Voltmeter Module 0-250 V 5. 1PH Wattmeter Module 6. Resistive Load Module 7. Inductive Load Module 8. Capacitive Load Module 9. Connecting Cables Connection Diagram: Short Circuit Test: From the Short Circuit Test done in Experiment 2, note the value of R 01, X01 referred to the H.T. side. Procedure: 1. Make sure all the switches (I1, I2, I3, I4, I5, I6) on the Power Supply are turned OFF (downwards). 2. Make sure all the variable knobs (K3, K4) are at the min, counter clock wise (CCW) position. 3. Make connections according to the above diagram. 4. Verify the connection by your Lab Teacher 5. Turn ON Switch I1 (upwards). 6. Turn Key K2 Clock Wise Once, the Indicator Lamp L1 becomes Green. 7. Turn Knob K3 at min (CCW) 8. Turn ON switch I3 (upwards). 9. Keep all the Loads at OFF position 10. Apply voltage 110V on the LT side. 11. Note the readings on AC Voltmeter, Ammeter and Wattmeter Module 12. Now turn ON all the Loads 13. Note the readings on AC Voltmeter, Ammeter and Wattmeter Module Procedure: 1. Follow the procedure mentioned on Resistive Load for the above Diagram With R-C Load: Procedure: 1. Follow the procedure mentioned on Resistive Load for the above Diagram Report: Draw the vector diagrams under unity, lagging and leading pf and calculate analytically the regulation in each case. Compare the value of regulation found analytically with that of experimental value. Comment on the regulation under leading pf is it something different? Comment on this value. Short circuit test WSC = ISC = VSC Calculation: With Resistive Load Terminal Voltage = Load current = With R-L Load Terminal Voltage = Load current = Load power = With R-C Load Terminal Voltage = Load current = Load power = Experiment No: 04 Date: Name of the Experiment: Parallel operation of transformers. Theory: When in a power system the demand for load increases, a single transformer may not be able to supply the extra load. In that case, another transformer is connected in parallel with the existing transformer to share the load. In order to make two transformers parallel, some conditions have to be fulfilled. These are: 1) Terminal voltages on the primary and secondary side should be identical. 2) The relative polarities on the primary and secondary sides should be identical. 3) Preferably R/X ratio of both the transformers should be same. 4) Primary windings of the transformer should be suitable for the supply system voltage and frequency. Apparatus: 1. 1PH Transformer 2 pieces 2. AC Ammeter Module 0-1A 3. AC Voltmeter Module 0-250 V 4. DC ammeter (0 to 600 mA) 5. Load 100 watts - 2 pieces. 6. Clamp-on-meter -1 piece 7. DC Supply – 9 V 8. Connecting Cables Connection Diagram: Procedure: 1. Select two 1-φ transformers of identical manufacturer. 2. On the secondary side of the transformer, determine R and X with an R-L-C meter. 3. Determination of Relative Polarity: Connect a D.C. mA with polarity as shown on the secondary side, and a battery through a switch on the primary side. Push the switch (give a kick). If the mA deflects in the positive direction, then H1 and X1 have the same mode of winding. Repeat the step for TR2 and confirm that the modes of windings are identical. 4. Connection for Parallel Operation:  a) Connect a voltmeter and connect the transformers as shown. If the polarities are correct, the voltmeter should read zero. b) Disconnect the supply, remove the voltmeter and connect the two secondary terminals (X1, X1) of TR1 and TR2. With a clamp-on-meter, check if there is any circulating current. If there is any, note this circulating current. 5. Now, re-connect TR1 and TR2 as in step 4(a). On the secondary side, connect 2 load of 100W each. With clamp-on-meter, determine the currents supplied by TR1 and TR2 For TR1: R1 = X1 = For TR2: R2 = X2 = Voltmeter reading on step 4: clamp-on-meter reading on step 5: Currents supplied by TR1 = Currents supplied by TR2 = 1. Discuss on the value of circulating current found in step 4(c). 2. Why is parallel operation necessary? Why is relative polarity test necessary for parallel operation? 3. Why the voltmeter gives zero reading if the polarities are same at step 4(a)?  Experiment No: 05 Date: Name of Experiment: Construction of three-phase transformer using three single-phase transformer and observation of line voltage & phase voltage and line current & phase current relation in primary and secondary winding with balance and unbalance loading. Theory: A three phase transformer of both ∆ and Y configuration can easily be constructed by using single phase transformer. For connecting the transformers of single phase we need to know the polarity of each coil. After knowing the polarity of each coil 3- phase connection can be made properly. There are many transformers still in service whose terminal does not have the standard markings. In order to interconnect these transformers with others, either for single- phase parallel operation or for polyphase operation, it is necessary to determine the relative polarities of the terminals. Y- Y connection without primary neutral grounded works satisfactorily only if the load is balanced. With the unbalanced load to the neutral, the neutral point shifts thereby making the three line-to-neutral (i.e. phase) voltages unequal. The effect of unbalanced loads can be illustrated by placing a single load between phase (or coil) a and the neutral on the secondary side. The power to the load has to be supplied by primary phase (or coil) A. This primary coil A cannot supply the required power because it is in series with primaries B and C whose secondaries are open. Under these conditions, the primary coils B and C act as very high impedances so that primary coil can obtain but very little current through them from the line. Hence, secondary coil a cannot supply any appreciable power. In fact, a very low resistance approaching a short-circuit may be connected between point a and the neutral and only a very small amount of current will flow. This, as said above, is due to the reduction of voltage Ean because of neutral shift. In other words, under short-circuit conditions, the neutral is pulled too much towards coil a. This reduces Ean but increases Ebn and Ecn (however line voltage EAB EBC and ECAare unaffected). On the primary side, EAN will be practically reduced to zero whereas EBN and ECN will rise to nearly full primary line voltage. This difficulty of shifting (or floating) neutral can be obviated by connecting the primary neutral (shown dotted in the figure) back to the generator so that primary coil A can take its required power from between its line and the neutral. Apparatus: 01. Transformer -3 (1-φ, 150V, 1A each coil) 02. 3- phase Variac. (0-50V) 03. Voltmeter. (0-450V) 04. Ammeter. (0-2A) 05. Variable Resistor(107 Ω, 2A) Circuit Diagram: A a A a + + A a + + + + _ _ _ _ _ _ B b B b B b + + + + + + V _ _ _ _ _ _ C c C c C c + + + + + + N n _ _ _ _ _ _ Fig1: Y-Y Connection, Fig2 polarity test for ∆-∆ connection Fig2 : ∆-∆ connection Fig. Construction of three phase transformer using single phase transformer:  Fig1: Unbalanced load in Y-Y connection without primary neutral grounded. Precautions: 01. Care should be taken so that current in the coils does not exceed the rated value. 02. Winding should be given the rated or less than the rated supply. 03. Polarity of each coil should always be tested before construction of 3-φ transformer. Data Table: For Y-Y connection Sl. VAN VBN VCN VAB VBC VCA Van Vbn Vcn Vab Vbc Vca NO. 01 158 170 164 290 290 290 84 88 86 144 144 144 02 134 132 134 229 230 232 67 66 67 114 114 115 For ∆-∆ connection Sl. VAB VBC VCA Vab Vbc Vca Ia Ib Ic Ian Ibn Icn NO. 01 210 215 214 106 108 108 0.97 1.0 0.94 0.55 0.58 0.57 For balance /unbalance loading, in Y-Y connection with / without primary neutral grounded Conditions VAN VBN VCN VAB VBC VCA Van Vbn Vcn Vab Vbc Vca Icn Balanced/pr imary 158 170 164 290 290 290 84 88 86 144 144 144 neutral grounded Unbalanced & primary 250 198 86 290 290 290 128 100 38 144 144 144 neutral ungrounded Balanced/pr imary 134 132 134 229 230 232 67 66 67 114 114 115 neutral grounded Unbalanced & primary 207 178 52 230 231 233 103 88 25 114 114 115 neutral ungrounded Discussion: Experiment No: 06 Date: Name of Experiment: Study of the efficiency of Three-Phase Transformers. Theory: In an ideal transformer, the power in the secondary windings is exactly equal to the power in the primary windings. This is true for transformers with a coefficient of coupling of 1.0 (complete coupling) and no internal losses. In real transformers, however, losses lead to secondary power being less than the primary power. The degree to which a real transformer approaches the ideal conditions is called the efficiency of the transformer: where Pout and Pin are the real output and the input powers. Apparent and reactive powers are not used in efficiency calculations. Transformers may be connected in parallel to supply currents greater than rated for each transformer. Two requirements must be satisfied: 1) The windings to be connected in parallel must have identical output ratings; 2) The windings to be connected in parallel must have identical polarities. Severe damage may be made to circuitry if these requirements are not satisfied. Single-phase transformers can be connected to form 3-phase transformer banks for 3-phase power systems. Four common methods of connecting three transformers for 3-phase circuits are Δ-Δ, Y Y, Y-Δ, and Δ-Y connections. Equipments: 1. 1PH Transformer 2 pieces 2. AC Ammeter Module 0-1A 3. AC Voltmeter Module 0-250 V 4. DC ammeter (0 to 600 mA) 5. Load 100 watts - 2 pieces. 6. Clamp-on-meter -1 piece 7. DC Supply – 9 V 8. Connecting Cables Procedure: 1. Select three 1-φ transformers of identical manufacturer. 2. Make sure all the switches (I1, I2, I3, I4, I5, I6) on the Power Supply are turned OFF (downwards). 3. Make sure all the variable knobs (K3, K4) are at the min, counter clock wise (CCW) position. 4. Make connections according to the above diagram. 5. Verify the connection by your Lab Teacher 6. Now verify the advantages for each type of combination. 7. Keep all the Loads at OFF position 8. Apply voltage 110V on the LT side. 9. Note the readings on AC Voltmeter, Ammeter and Wattmeter Module. 10. Now turn ON all the Loads. 11. Note the readings on AC Voltmeter, Ammeter and Wattmeter Module. 12. With constant resistive load determine the efficiency for each combination. With Δ-Δ connection Terminal Voltage = Load current = Load power = With Y- Y connection Terminal Voltage = Load current = Load power = With Y-Δ connection Terminal Voltage = Load current = Load power = With Δ-Y connection Terminal Voltage = Load current = Load power = Experiment No: 07 Date: Name of Experiment: Observation of speed torque characteristic curve of 3-phase induction motor. Theory: From the general torque equation, the running torque may be written in an equation: It is clear that When s = 0,T = 0. At normal speed, close to synchronous speed the term sX2 is small and hence negligible with respect to R2 So, Or T s if R2 is constant. Hence, for low values of slip, the torque/slip curve is approximately a straight line. As slip increases (for increasing load on the motor), the torque also increases and becomes maximum when. This torque is known as 'pull-out or 'breakdown' torque Tb or stalling torque. As the slip further increases (i.e. motor speed falls) with further increase in motor load, then R 2 becomes negligible as compared to sX2 Therefore, for large values of slip Hence, the torque/slip curve is a rectangular hyperbola. So we see beyond the point of maximum torque, any further increase in motor load results in decrease of torque developed by the motor. The result is that the motor slows down and eventually stops. In fact, the stable operation of the motor lies between the values s= 0 and that corresponding to maximum torque. Apparatus: 1) Three phase induction motor (440V, 4.55 A, 1420 rpm). 2) Ammeter (0-5 A). 3) voltmeter (0-450v). 4) D.C generator.(250v, 8A, If=0.467A, 1450 rpm) 5) Tachometer. 6) Variable Resistor -4 (107 Ω, 2A). Circuit Diagram: A A FR1 L1 T1 I FR2 V L T2 G F1 RL 2 M G L3 T3 F2 Data Table: S. VDC IL IF Ia PDC= PLF= PL= N PW+F PL0SS= PM T= NO VDC×IL VDC×IF Ia2  Ra = VDC×If. K×N + Ra+ N-m PW+F 01 207 0.93 0.37 1.30 192.51 76.59 1.45 1460 146 225.43 417.94 2.73 02 206 0.99 0.37 1.36 203.94 76.22 1.65 1458 145.8 225.13 429.07 2.81 03 205 1.08 0.37 1.45 221.4 75.85 1.96 1456 145.6 224.98 446.38 2.93 04 203 1.15 0.37 1.52 233.45 75.11 2.22 1455 145.5 224.49 457.94 3.01 05 202 2.25 0.36 1.61 252.5 72.72 2.63 1452 145.2 222.27 474.74 3.12 06 200 1.38 0.36 1.74 276 72 3.20 1451 145.1 222.19 498.19 3.28 07 198 1.56 0.36 1.91 308.88 69.3 4.09 1447 144.7 220.13 529.01 3.40 Maximum torque or Breakdown Torque Point Maximum torque or Breakdown Torque Rated T (Torque) Operating Rated Torque Point Starting Torque Rated Speed S (Sleep) N (Speed) Calculation: PDC=VDC×IL IL=Ia+ IF PLOSS includes friction & windage loss in DC Generator and induction motor (PW+F) which is proportional to speed, Field losses (VDC×If) & armature Cu Losses ( Ra) in DC generator. PL0SS= VDC×IF + Ra+ PW+F = PW+F= K×N= (Assume, K=0.10) For the case when DC generator Load current is zero PM = PDC + PLOSS= Again, PM = T×ω = T× T= = Discussion:  Experiment No: 08 Date: Name of Experiment: Blocked rotor test & no-load test of 3-phase induction motor and determination of circuit parameters. Theory: Blocked Rotor Test: It is also known as locked-rotor or short-circuit test. This test is used to find 1. short-circuit with normal voltage applied to stator 2. power factor on short-circuit 3. total leakage reactance X01, of the motor as referred to primary (i.e. stator) 4. total resistance of the motor R01 as referred to primary. In this test, the rotor is locked (or allowed very slow rotation) and the rotor windings are short-circuited at slip-rings, if the motor has a wound rotor. Just as in the case of a short-circuit test on a transformer, a reduced voltage (up to 15 or 20 per cent of normal value) is applied to the stator terminals and is so adjusted that full-load current flows in the stator. As in this case s=1, the equivalent circuit of the motor is actly like a transformer, having a short-circuited secondary. The values of current, voltage and power input on short circuit are measured by the ammeter, voltmeter and wattmeter connected in the circuits. W = total power input on short-circuit Vs= line voltage on short-circuit IS = Line current on short-circuit. Now, the motor input on short-circuit consists of mainly stator and rotor Cu losses. Core-loss, which is small due to the fact that applied voltage is only a small percentage of the normal voltage. No load test: In this test rated voltage is applied to the induction motor without any mechanical load and the speed is assumed synchronous speed. In practice, it is not feasible to run the induction motor synchronously for getting R0 and X0. Instead the motor is run without any external mechanical load on it the speed of the rotor would not be synchronous, but very much near to it; so that for all practical purposes, the speed may be assumed synchronous. The no load test is carried out with different values of applied voltage below and above the value of normal voltage. The power input is measured by two watt meters, As motor is running on light load, the p.f. would be low i.e. less than 0.5, hence total power input will be the difference of the two wattmeter readings W1 and W2.The readings of the total power input WO and total current I0 and voltage V0. The no-load input W0 to the motor consists of small stator Cu loss 3Io 2 R1 , stator core loss and friction and windage. The stator core loss and friction and windage losses are collectively known as fixed losses, because they are independent of load. Apparatus: 1) Three phase induction motor (440V, 4.55 A, 1420 rpm). 2) Ammeter (0-10 A). 3) voltmeter (0-450v). 4) Wattmeter.(450 V, 10A) 5) Ohm meter. Circuit Diagram: L1 T1 L1 T1 A A W W V V A V L2 L2 T2 R T2 IM IM I A C W L3 W L3 T3 T3  R1 X1 X 2´ R 2´ X0 1  R0 RL´ = R 2  '  1 s  Fig. Electrical equivalent circuit of 3- φ induction motor. Data Table: For Blocked rotor Test S.NO. VS IS W1 W2 WT R01= Z01= V0/ X01=√ P/3IS2 3 I0 (Z012 – R012) 01 99 4.55 255 140 395 6.36 12.56 10.83 No load test 2 S.NO. VO Io W1 W2 WT PCORE R0= Z0= V0/ B0=√ (Y0 1 2 2 XO = V /PO O 3 I0 – G0 ) BO 01 440 2.15 200 820 620 458.47 422.27 118.16 8.125  103 123.07 02 400 1.95 110 690 580 03 354 1.80 35 580 545 04 270 1.6 70 420 490 DC resistance test of Stator winding : RT-T= 4.18 Ω Calculation: For Block rotor test the applied voltage is lower Than the rated voltage and core loss is negligible. So it is assumed that current through RO & XO is zero ' 2 So we can write, RS = R1+ R 2 = P/3IS = RT For Y connected induction motor R1dc = = 2.09 2 R1=1.4  R1dc=2.926 So, R '2 = RS - R1= ZS = VS/ 3 IS = XS = X1+ X'2 = √ (ZS2 - RS2) = It is assumed that X1= X '2 = XS 2 For No Load test there are mainly core loss, very little Cu Loss and friction & windage loss.It is assumed that the voltage drop at R1 & X1 is negligible. In this case Core Loss is PF+W= K×N= 148 ( N= 1480, Assuming, K=0.10) Stator Cu loss = IO  R1 2 PCORE=WT – IO2  R 1 -PF +W= 620 - 2.152  2.926 -148 =458.47  2 2 V RO = 3V OCT = OCT = 3P CORE P CORE 1 GO = = RO VO ZO= = 3IO 1 YO = = ZO BO = √ (YO2 – GO2) = 1 XO = = 123.07 BO The rated voltage and current of the tested induction motor are 440V and 4.55 A respectively. So If it is assumed that at rated condition p.f is 0.80 Then the input power is = 3 V L IL cos  = 2774.05 W And Losses = PCU +PCORE + PF+W =1001.5 W % Efficiency =( Output/ input)×100% %Efficiency = (Input- Losses)×100/ Input %= 63.9 % Efficiency = Output/ (Output + losses)= 63.9% Discussion: Experiment No: 09 Date: Name of Experiment: Study of a 1-φ induction motor. Theory: The single-phase induction motor when fed from 1-φ supply, its stator winding produces a flux which is alternating i.e. which alternates along one space axis only. An alternating or pulsating flux acting on a stationary squirrel-cage rotor cannot produce rotation. That is why single-phase induction motor is not self-starting. But if the rotor of such machine is given initial start by hand or somewhat other means in either direction, then immediately torque arises and accelerates to its final speed. To make the motor self starting another winding called a starting/auxiliary winding is placed 90 electrical degrees apart from the main or running winding and connected in parallel across the 1-φ supply. AA/ = Main winding. BB/ = Auxiliary winding. Two theories have been put forward to explain the operation of 1-phase Induction motor. The theories are: 1) Double Revolving Field theory. 2) Cross-field theory. Equipments: 1. LC meter 2. 1-φ Variac 3. Wattmeter 4. AC Voltmeter (0-300 V) 5. AC Ammeter (0-2.5 A) 6. Tachometer 7. Wire for connection Connection Diagram: Procedure: A) To determine the terminals of running and starting winding: Normally 3 or 4 terminals are available on the motor terminal box. If one terminal of both starting and running windings are internally connected, then 3-terminals are available on the terminal box. The auxiliary winding is designed to have higher resistance/reactance ratio than the main winding so that the current in the auxiliary winding leads the current in the main winding. 1) Take any two-coil terminals and determine its resistance by a sensitive ohm-meter. R1= ……… ohms. 2) Similarly determine the inductance of the coil by an L-C meter. L1=…………henry, XL1= 2πfL1=…………… ohms. 3) Now take the remaining two coil terminals and measure its resistance and inductance. R2=…………ohms, L2=…………….henry, XL2=……………ohms. Determine R/XL1 and R/XL2. From this ratio determine the running winding and starting winding terminals. B) Running of the motor by the main winding: 1) Connect the terminals of the main winding as per following figure. C) Running of the motor with the auxiliary winding: 1) Connect the instruments as per following figure. Report: 1. In step-B why did the motor rotate in both directions? 2. What is the purpose of capacitor in step-C ? Explain. 3. What have you noticed when you switched-off the SPST in step-C ? 4. Why does the motor continue to run in step-C, when the SPST was switched-off? Experiment No: 10 Date: Name of Experiment: Measurement of circuit parameters of single phase induction motor and determination of efficiency. Theory: As in a polyphase induction motor, the equivalent-circuit parameters of a single-phase induction motor can be measured from No-load and Blocked-rotor tests and from stator winding resistance test. Blocked rotor test: The single-phase voltage, applied to the stator main winding, is increased gradually from zero so that rated current flows in the main winding. Under these conditions voltmeter, ammeter and wattmeter readings respectively denoted by VSC , ISC , WSC, are recorded. Then DC resistance of main stator winding is measured. This dc resistance should be multiplied by a factor 1.4 so as to obtained effective AC resistance r 1 at line frequency with rotor stationary, the slip S=1, and the voltage required to circulate full load current is very low. Therefore, the flux is small and the magnetizing current flowing through Xm is also very low. In view of this magnetizing reactance can be neglected. From the test data, the equivalent impedance referred to stator is ZSC= The equivalent resistance RSC is given by RSC= r1+ = Since the resistance of main stator winding R1 is already measured effective rotor resistance r2 at line frequency is R2 =RSC – r1 = - r1 From equivalent circuit for blocked rotor test , XSC =x1+ =x1 +x2 Since the leakage reactance x1 and x2 can’t be separated out, it is a common practice to assume So, x1= x2 x1= x2= or, x1= x2= = In blocked rotor test , note that only main winding is excited, auxiliary winding is left open. No-Load Test: The single phase induction motor, under no load condition is run at rated voltage and frequency. With the motor running at no load, the slip is very close to zero. It may therefore assumed that S. Under this condition, become infinity and becomes several time smaller than In view of this approximation and (across ) may be neglected. Under no load test let the voltmeter, ammeter, and wattmeter readings be V1 , Inl and Wnl respectively. Then no load power factor is Cosθnl = No load equivalent impedance is Xnl = Znl = Znl sin Again, Znl = Rnl +j Xnl = ] r1 =Xnl Since x1 and x2 are already known from blocked rotor test , magnetizing reactance X m can be calculated from the above equation.  r1 Fig. for blocked rotor test r1 Fig. Electrical equivalent circuit of single phase induction motor : Fig. for No load test Apparatus: 1) Single phase induction motor(220V, 6.6 A , 1 kW, 2820 rpm, p.f. 0.90). 2) Ammeter (0-5 A). 3) Wattmeter (0-10A, 0-450v). 4) Voltmeter (0-450v). 5) Variac(0 -250V) Circuit Diagram: Main Auxiliary Winding Winding Xa W A V Auxiliary Winding A R C V Vs Xm Vs I Main Main Winding A winding 220 V C Fig: Circuit diagram of single phase induction motor. Data Table: Test V I P Pin Pout η= Pout / Pin (v) (A) (W) (W) (W) No Load Test 220 3.3 180 Blocked 96 6.5 400 Rotor Test Calculation: From blocked rotor test: VSC ZSC= = 14.77 Ω ISC P RSC= r1+r2 = 2SC = 9.47 Ω ISC XSC=x1+ x2 = ZSC 2  R SC 2 = 11.33 Ω 1 x1 = x2= XSC = 5.67 Ω 2 r1dc = 3.95  r1 = 3.95  1.4 = 5.53 Ω r2=RSC-r1= 3.94 Ω From no load test: Pnl = 0.248 Cosθnl = V nl Inl sin θnl= 1  cos 2nl = 0.97 Znl = Vnl = 66.67 Ω Inl Xnl = Znl sin θnl = 64.67 1 Now, Xnl =x1 + (x2 +Xm) 2  Xm = 112.33 Ω r2 Now the rated core and friction & windage losses = Wnl - I2nl (r1 + ) 4 = 109.05 Watt.  Rated Cu loss and core, friction & wnidage losses = 400+109.05 Watt = 509.05 Watt. Rated p.f. =0.90 Rated input = 220  6.5  0.9 = 1287 Watt η= Pout  100 % = Pin  Plosses  100 %= 60.45 % Pin Pin Precautions: 1. the connection of C.C and P.C should be correct. 2. The readings of the various meters should be taken carefully. Discussion: Experiment No: 11 Date: Name of Experiment: Determination of value of capacitor for maximum starting torque of capacitor split-phase induction motor. Theory: A single phase induction motor with main stator winding has no inherent starting torque, since main winding produces only stationary, pulsating air-gap flux wave. For the development of starting torque, rotating air-gap field must be produced. Several methods which have been developed for starting of single phase induction motor, ma be classified as follows; 1) split-phase starting. 2) Shaded-pole starting. 3) Repulsion motor starting and 4) Reluctance starting. All the split-phase motors have two stator winding, a main (or running) winding and an auxiliary (or starting ) winding. Both this windings are connected in parallel but there magnetic axes are spaced displaced by 90 electrical. There are four types of split-phase motor a) Resistor split-phase motor b) Capacitor split-phase motor c) Capacitor start and run motor. d) Capacitor run motor. In capacitor split phase motor the main winding is designed by taking into consideration the required performance of this motor. The time phase displacement between auxiliary winding current Ia and main winding current Im is obtained by putting a suitable capacitor in series with the auxiliary winding. A centrifugal switch CS is placed in the auxiliary winding which disconnect the auxiliary winding when the motor reaches at about 70 to 80% of synchronous speed. By use of suitable starting capacitor, the highest possible starting torque can be achieved. But size and hence cost of the starting capacitor; then, would be quite high. The auxiliary winding and starting capacitor are in circuit for a short time only, these can therefore, be designed for a minimum cost. The maximum starting torque can be obtained if the value of capacitor (C) is such that 1 Xc=  Xa  r a r m 2fC Zm  Xm Where , Xc= Capacitive reactance for maximum starting torque. Xa=Inductive reactance of the auxiliary winding Xm= reactance of the main winding. Zm=Impedance of the main winding. rm=resistance of main winding. ra=resistance of main winding. In this experiment a reduced voltage is applied at the auxiliary winding and main winding is applied separately. Then current and voltage reading are taken. Apparatus: 1) Single phase induction motor(220V, 6.6 A , 1 kW, 2820 rpm, p.f.-0.90). 2) Ammeter (0-5 A). 3) Wattmeter (0-10A, 0-450v). 4) voltmeter (0-450v). 5) Variac (0 -250V) Circuit Diagram: W A V A Main winding / Vs R V Auxiliary Winding 220 v I A C Fig: Circuit diagram of single phase induction motor Data Table: Xc= C= ra rm Xa  1 Winding V I P r=P/I2 Z=V/I X=√Z2-R2 Zm  X m 2f X C F Main 100 7.25 430 8.18 13.79 11.11 27.54 115.58 Aux 100 3.7 200 14.61 27.03 22.74 Calculation: P ra= 2 =14.61 Ia P rm= 2 = 8.18 Im V Za= a =27.03 Ia V Zm= m = 13.79 Im Xa= Za2  X a2 = 22.74 Xm= Z2m  X 2m = 11.11 Xc= X a  r a r m = 27.54 Ω Zm  X m 1 C= = 115.58 F ( f=50 Hz) 2f X C Discussion: Apparatus: 6) Single phase induction motor(220V, 6.6 A , 1 kW, 2820 rpm, p.f.-0.90). 7) Ammeter (0-5 A). 8) Wattmeter (0-10A, 0-450v). 9) voltmeter (0-450v). 10) Variac (0 -250V) Circuit Diagram: W A V A Main winding / Vs R V Auxiliary Winding 220 v I A C Fig: Circuit diagram of single phase induction motor Data Table: Xc= C= ra rm Xa  1 Winding V I P r=P/I2 Z=V/I X=√Z2-R2 Zm  X m 2f X C F Main Aux Calculation: P ra= 2 = Ia P rm= 2 = Im V Za= a = Ia V Zm= m = Im Xa= Za2  X a2 = Xm= Z2m  X 2m = Xc= X a  ra rm = Zm  X m 1 C= = ( f=50 Hz) 2f X C Discussion: r2 Now the rated core and friction & windage losses = Wnl - I2nl (r1 + ) 4 =  Rated Cu loss and core, friction & wnidage losses =. Rated p.f. =0.90 Rated input = 220  6.5  0.9 = η= Pout  100 % = Pin  Plosses  100 % Pin Pin Precautions: 3. the connection of C.C and P.C should be correct. 4. The readings of the various meters should be taken carefully. Discussion: Experiment No: 12 Date: Name of Experiment: Starting an Induction Motor by applying star-delta starter. Theory: The induction motor must be built to run normally with a mesh-connected stator winding. At starting, the winding is connected temporarily in star. The phase voltage is thus reduced to 1/√3 = 0.58 of normal, and the motor behaves as if the auto-transformer were employed with a ratio x = 0.58. The starting current per phase is Is = 0.58Ix , the line current is (0.58)2Ix =0.33Ix , the starting torque is one-third of short-circuit value. The method is cheap and effective, so long as the starting torque is not required to exceed about 50 percent of full load torque. It can therefore be used for machine-tools, pumps, motor generators etc. The method is unsuitable for motors at voltages exceeding 3000 V. because of the excessive number of stator turns needed for delta running. Where induction motors are required to run for considerable periods on small loads, a star-delta switch permits the machine to be star- connected during these periods, with reduction of magnetizing current and increase in efficiency. This is a starting method that reduces the starting current and starting torque. At starting the Induction motor is star connected after it reached the approximate operational speed it is switch to delta. The motor must be delta connected during a normal run, in order to be able to use this starting method. This starting method only works when the application is light loaded during the start. If the motor is too heavily loaded, there will not be enough torque to accelerate the motor up to speed before switching over to the delta position. Apparatus: 1) 3-φ Induction Motor 2) Star-Delta Starter 3) Connecting Wires 4) Table: (Readings ) Conclusion: