PRACTICAL WORK BOOK. Basic Electrical & Electronics Engineering BE-104

Transcription

1 PRACTICAL WORK BOOK Basic Electrical & Electronics Engineering BE-104 Name: Enrollment No: Branch: Semester: Batch: Institute: Department of Electrical Engineering

2 I N D E X S.NO. EXPERIMENT NAME DATE 1 Measurement of active and reactive power in single phase A.C. Circuit. 2 Separation of Resistance and Inductance of Choke Coil 3 Measurement of Impedance of R-L, R-C,R-L-C & study of resonance phenomena 4 Measurement of power & power factor in a single phase AC circuit using three Ammeter Method 5 Perform load test on a single phase transformer 6 Study of transformer name plate Rating & Determination of ratio 7 Open circuit & Short circuit test on 1- Φ Transformer 8 Study of constructional feature s of a D.C. M/C 9 Measurement of various line and phase quantities for a three phase ac circuit. 10 Verification of KCL and KVL 11. Verification of Superposition Theorem SIGNATURE & REMARKS 2

3 EXPERIMENT NO. 1 Object:- Measurement of active and reactive power in single phase A.C. Circuit. Apparatus Required:- S.No. Name of equipments Quantity Specification range Type 1. Wattmeter A,125V Dynamo 2. Ammeter Amp. A.C. 3. Voltmeter V A.C. 4. Rheostat ohms - 5. Inductance(choke) Auto-transformer V A.C. 7. Connecting wires Theory:- Wattmeter has two coils, one is called current coil and other is pressure coil. The current coil carries the load current and pressure coil carries a current proportional to and in phase with supply voltage. The deflection of wattmeter depends upon the currents in the two coil and upon the P.F. of the circuit. In the wattmeter the current coil are arranged for different ratings i.e. 2.5,5 Amp.etc. and similarly voltage coils are rated for 125 V, 250V, 500Vetc. While doing the experiment the proper range should be selected according to the load voltage and current. The reading should be multiplied by a factor called the Multiplying Factor. Active power:- Multiplying factor of wattmeter = Current range * Voltage range Full scale reading of wattmeter It is the power, which is actually dissipated in the circuit Resistance. Reactive Power:- P = I 2 R = VI Cos ф watts It is the power developed in the inductive reactance of the circuit. Q = I 2 X L 3

4 Circuit Diagram:- 4

5 0-1A A + C + + W M V L R 230V A.C. Variable Supply V 0-230V L Active and Reactive Power Ckt.diagram From diagram:- It is clear Sinф = X L /Z Or, ZSinф =X L Q = I 2 X L = I*I*X L Q = I*I*Z Sinф Q = V*I* Sinф Q= V*I* Sinф Volt Ampere Reactive (VAR) 5

6 Power Triangle:- VI Cos φ VI φ VI Sin φ Impedance Triangle:- R φ X L Z Z = (R 2 ) + (X L 2 ) Procedure:- (1) Connect the circuit according to the circuit diagram. (2) For different value of supply voltage takes the various observations. (3) Take at least 3 sets or reading. Observation Table:- Multiplying factor of wattmeter =... S. NO. V I Wattmeter reading Actual power Wattmeter Reading*M.F. P.F.=Power/(V*I) Active Power P= V*I*Cosф Reactive Power(Q)= V*I*Sinф

7 Calculation: - Report: - Precautions: - Following precautions should be taken care of while performing this experiment. 1. All connections should be tight. 2. The zero setting of all the meters should be checked before connecting them in the circuit. 3. The current through ammeter should never be allowed to exceed the current rating of rheostat and load used. 4. Do not increase the current beyond the rated value of wattmeter. 5. The wattmeter should be connected properly in the circuit. Tutorials: Q.1 Define power factor in A.C. Ckt. & its importance? Q.2 What power is consumed in D.C. Ckt.? Q.3 What are the two components of power exist in A.C. Ckt.? Q.4 How many coils are there in wattmeter? Write down their names. Q.5 How dual current range is obtained in wattmeter? Q.6 Name such a Ckt. In which apparent& active power is equal? Q.7 Which coil in wattmeter, is moving & which type is stationary one? Q.8 Which type of wattmeter is generally used for measuring power in A.C.Ckt s? Q.9 What do you mean by purely inductive &purely capacitive load? Q.10 What power is consumed by purely inductive load? REFERENCES Books: 1. Fundamentals of Electrical engineering by Ashfaq Husain. 2. A Textbook of Electrical Technology by B.L Thereja. 3. Electrical Science by J. B. Gupta URLS:

8 EXPERIMENT NO. 2 Object :- Separation of resistance and inductance of choke coil. Apparatus Required :- S.No. Name of equipments Quantity Range Type 1. Voltmeter v Ac 2. Ammeter amp. Ac 3. Wattmeter amp,125v Dynamometer 4. Choke coil Auto transformer v Ac 6. Connecting leads Theory :- 1. Choke coil :- A coil of thick wire wound on a laminated iron core as negligible resistance is known as choke coil. 2. Resistance:- Resistance may be defined as that property of a substance oppose(or resists) the flow of an electric current (or electrons) through it. It is denoted by R. 3. Inductance:- The property of a coil due to which it oppose the change of current flowing through itself is called inductance of the coil. It is denoted by L. 4. Impedance:- The impedance of the circuit may be defined as the total opposition offered to the flow of alternating current. It is measured in ohms and denoted by Z. 5. Power Factor:- Cosine of angle between angle voltage and current is known as power factor. It is denoted by Cos Φ. 8

9 Procedure :- 1. Connect the circuit according to circuit diagram 2. For different values of supply voltages take the various observations. 3. Take at least three sets of readings. 4. After the observation switch off the supply and disconnect all the equipments. Circuit Diagram :- Observation table:- S. No. Voltage V Current I Power P Power factor CosΦ = P/ VI Circuit Impedance Z = V/I Circuit Resistance R =Z*CosΦ Circuit Reactance XL = Z²-R² Circuit Inductance L = XL / 2πf Calculation :- 9

10 Result :- 1. Resistance of choke coil = 2. Inductance of choke coil = Precautions :- 1. Make the connections as shown in figure. 2. Connections should be tight. 3. Ensure that auto transformer should be at zero position before switching on the main supply. 4. Please get the circuit checked before switching on the main supply. 5. Do not touch any live wire. 6. Supply should be off after the completion of experiment LAB TUTORIALS 1. What do you understand by the term power factor in reference to a.c. circuits? 2. What is the importance of power factor? 3. What do you mean by a lagging power factor? 4. What do you mean by an Instantaneous and RMS value of a.c. current? 5. Draw the phasor diagram of RL circuit. REFERENCES Books: 1. Fundamentals of Electrical engineering by Ashfaq Husain. 2. A Textbook of Electrical Technology by B.L Thereja. 3. Electrical Science by J. B. Gupta URLS:

11 Object :- EXPERIMENT NO. 3 Measurement of impedance of R-L, R-C & R-L-C series circuit. Study of Resonance phenomenon. Apparatus Required :- S.no. Name of Equipment Quantity Range Type 1 Voltmeter V A.C. 2 Voltmeter V A.C. 3 Ammeter A A.C. 4 Resistance Inductance Capacitance Auto-Transformer V A.C. 8 Connecting Wires Theory :- (A) R-L Series Circuit :- It consists of a resistance of R -ohms & inductance of L- henry connected in series. V s = R.M.S. Value of supply voltage V r = R.M.S. Value of resistance voltage drop = I R V L = R.M.S. Value of Inductance voltage drop = I X L I = R.M.S. Value of current In R-L Series circuit V r = I R and V L = I X L V S = (V r 2 + VL 2 ) = (I R) 2 + (I X L ) 2 = I (R) 2 + (X L ) 2 I = V / R 2 + X L 2 = V / Z 11

12 Where, Z = R 2 + X L 2 is the impedance of the circuit X L = 2 π f L is the inductive reactance of the circuit L = X L / (2 π f) X L = V L / I (B) Series R C Circuit :- It consist of R Ohm and C Farads connected in series with source. V s = R.M.S. Value of supply voltage V r = R.M.S. Value of resistance voltage drop = I R V C = R.M.S. Value of capacitance voltage drop = I X C I = R.M.S. Value of current In R-C Series circuit V r = I R and V C = I X C V S = (V r 2 + VC 2 ) = (I R) 2 + (I X C ) 2 = I (R) 2 + (X C ) 2 I = V / R 2 + X C 2 = V / Z Where, Z = R 2 + X C 2 is the impedance of the circuit X C = 1 / 2 π f C is the capacitance reactance of the circuit C = 1 / 2 π f X C X C = V C / I 12

13 (C) R-L-C Series circuit :- It consists Resistance of R ohms, Inductance of L- henry and Capacitance of C - Farads connected in series. V s = R.M.S. Value of supply voltage V r = R.M.S. Value of resistance voltage drop = I R V C = R.M.S. Value of capacitance voltage drop = I X C V L = R.M.S. Value of Inductance voltage drop = I X L I = R.M.S. Value of current In R - L - C Series circuit V r = I R V C = I X C V L = I X L V S = V r 2 + (VL - V C ) 2 = (I R) 2 + (I X L - I X C ) 2 = I R 2 + (X L - X C ) 2 Where, Z = R 2 + (X L - X C ) 2 is the impedance of the circuit X L = 2 π f L is the inductive reactance of the circuit L = X L / (2 π f) X L = V L / I X C = 1 / 2 π f C is the capacitance reactance of the circuit C = 1 / 2 π f X C X C = V C / I V r = I R R = V r /I 13

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15 0-1A R L C A + V + + V V V 0-150V 0-300V 230V A.C. Variable Supply V 0-230V R-L-C Series Ckt.diagram 0-1A R L A + + V + V 0-75V 0-150V 230V A.C. Variable Supply V 0-230V R-L Series Ckt.diagram 15

16 R C A + + V + V 0-75V 0-300V 230V A.C. Variable Supply V 0-230V R-C Series Ckt.diagram Resonance: - Resonance in R-L-C series circuit: When capacitive reactance X C is equal to the inductive reactance X L then the circuit is said to be in resonance. The current will maximum, power factor is unity and lie in to phase with the supply voltage. V S = V r 2 + (VL - V C ) 2 = (I R) 2 + (I X L - I X C ) 2 = I R 2 + (X L - X C ) 2 At resonance, X L = X C Then I = V / R (Max) Now, X L = W C X C = 1 / W C W L = 1 / W C, 2 π f L= 1/2 π f C Fr = 1/2 π L C Fr = Resonance Frequency 16

17 Procedure :- (1) Connect the circuit diagram connecting R-L, R-C & R-L-C Series as shown in the circuit diagram. (2) The Auto-Transformer to zero position and switch on supply. (3) Adjust the Auto-Transformer till a suitable voltage is applied. (4) Take the reading from the voltage for V R,V L, V C & V S respectively. (5) Note the reading of ammeter. (6) Repeat step (3) Varying the supply voltage and record the reading in observation table. Observation Table :- (A) For R-L Series Circuit :- S. No S. No S. No Voltmeter Reading V r in V L in V S in Volt Volt Volt Voltmeter Reading V r in V C in V S in Volt Volt Volt Voltmeter Reading V r in V C in V L in Volt Volt Volt Ammeter Reading (A) in Amp I Circuit Impedance Z = V / I Circuit Resistance R = V r / I (B) For Series R C Circuit :- Ammeter Reading (A) in Amp I V S in Volt Circuit Impedance Z = V / I Circuit Resistance R = V r / I (C)For R-L-C Series circuit:- Ammeter Reading (A) in Amp I Circuit Impedance Z = V / I Circuit Resistanc e R = V r / I Circuit Reactance X L = V L /I Circuit Reactance X C = V C /I Circuit Reactanc e X C = V C /I Circuit Inductance L = X L /2πf Circuit Inductance C = 1/2πfX C Circuit Reactance X L = V L /I 17

18 Calculation :- Calculation the various quantities as follows : (1) R = V r / I (2) X L = V L /I & L = X L / (2 π f) Result:- (3) X C = V C /I & C = 1/ 2 π f X C (1)Resistance (R) = Ώ (2)Capacitance Reactance = Ώ (3)Inductive Reactance (4)Inductor (L) (5)Capacitor (C) = Ώ = Henry = μf Precaution :- (1) Do not touch any live wire. (2) Remove parallax error while taking reading from various instruments (3) All connections should be tight. (4) The zero setting of all the meters should be checked before connecting them in the circuit. (5) The current through ammeter should never be allowed to exceed the current rating of rheostat and load used. 18

19 LAB TUTORIALS 1. What do you understand by the term power factor in reference to a.c. circuits? 2. What is the importance of power factor? 3. What do you mean by a lagging power factor? 4. What do you mean by a leading power factor? 5. Define phase difference. REFERENCES Books: (1) Fundamentals of Electrical engineering by Ashfaq Husain. (2) A Textbook of Electrical Technology by B.L Thereja. (3) Electrical Science by J. B. Gupta URLS:

20 OBJECT:- EXERCISE NO. 4 To measure power factor in a single phase A.C. circuit using Three Ammeters. APPARATUS REQUIRED:- Voltmeter 0-300V, MI; ammeters 10A, 5A, 5A, MI; single phase inductive variable load, rheostat 100Ω,5A; variac 230V, 10A. THEORY:- The circuit to be used for measurement of power in an A.C. circuit using three ammeter is shown in circuit diagram. We know in a D.C. circuit the power is given by the product of voltage and current, whereas,in A.C. circuit it is given by the product of voltage, current and power factor. For this reason, it is not possible to find power in an A.C. circuit simply from the readings of a voltmeter and ammeter. In A.C. circuits power is normally measured by wattmeter. However, this method demonstrates that the power in a single phase A.C. circuits can be measured by using three ammeters. From the circuits shown in fig 1.we can write. Power consumed by load =P =VI 3 Cos Φ (1.1) 20

21 Where I 3 is current through load and V is the voltage across load. The phasor diagram of this circuit can be drawn by taking the supply voltage V as the reference phasor diagram is shown in fig. 01 θ I 2 v Φ I 1 phasor diagram we can write I 3 From the I 1 2 =I 2 2 +I I 2 I 3 CosΦ.. (1.2) Power factor, CosΦ = I 2 1 -I 2 2 -I 2 3 /2I 2 I 3. (1.3) I 2 = V/R [R is a known resistance]. (1.4) Now from eq. 1.1 I 3 CosΦ = P/V Put this value in eq. 1.2 I 2 1 = I 2 2 +I I 2 I 3 CosΦ I 2 1 = I 2 2 +I I 2 P/V 2I 2 P/V= I 2 1 -I I 3 or P= (I 1 2 -I 2 2 -I 3 2 )V/2I 2 From eq. (1.4) we put the value of I 2 21

22 So P = (I12-I22-I32) (V.R)/(2V) P = (I12-I22-I32) (R/2). Eqs. 1.3 and 1.5 show that we can find the power and power factor in an a.c. circuits by using 3- single phase ammeters, instead of a wattmeter. PROCEDURE:- The stepwise procedure for conducting this experiment is given below: 1. Make the connections as per the fig. 2. Keep the rheostat at its maximum value. 3. keep the variac at its mini9mum position 4. Switch on the supply. 5. Increase the voltage applied using variac slowly, so that the reading of voltmeter and ammeter, A1 are appreciable. 6. Decrease the resistance R (rheostat) so that ammeter A2 gives suitable reading. 7. Take down the readings of voltmeter and three ammeters. 8. Change the position of rheostat and repeat step 7 a number of times. OBSERVATION:- Record your observation as shown in table Table: Observation and calculations for 3-Ammeters method. S. No. Observations Calculation V Volts I 1 Amp I 2 Amp I 3 Amp P Cos Φ CALCULATION:- For each set of observation calculate the power consumed (Eq 1.5) and the power factor (Eq 1.3). Next take the average of all the set of calculation for Cos Φ i.e., power factor; and P i.e., power consumed in the load. 22

23 RESULT:- The power factor of the circuit and the power consumed in circuit should be recorded here. PRECAUTIONS:- Following precautions should be taken care of while performing this experiment. 1. All connections should be tight. 2. The zero setting of all the meters should be checked before connecting them in the circuit. 3. The current through ammeter should never be allowed to exceed the current rating of rheostat and load used. LAB TUTORIALS 1. What is the impact of low power factor in power system? 2. Define phase and phase difference. 3. Define term: Alteration, Frequency, Time period and cycle. REFERENCES Books: (1) Fundamentals of Electrical engineering by Ashfaq Husain. (2) A Textbook of Electrical Technology by B.L Thereja. (3) Electrical Science by J. B. Gupta URLS:

24 EXERCISE NO. 5 Object: - To perform load test on a single-phase transformer & to determine the following: (a)- Efficiency at different loads & to plot graph between efficiency Vs load currents. (b)- Regulation of the transformer & to plot graph between regulation Vs load currents. Apparatus Required:- S.No. Name of equipments Range Quantity Type 1. voltmeter 0-150V 1 A.C. 2. Ammeter 0-10 A 1 A.C. 3. Wattmeter 250V,2.5A 1 A.C. 4. Lamp bank load 250V,1Kw ф Transformer 230\115V ф Variac 0-260V 1 - Theory: - Wattmeter has two coils, one is called current coil & other is called as pressure coil. The current coil carries the load current & the pressure coil carries a current proportional to and in phase with supply voltage. The deflection of wattmeter depends upon the currents in the two coils and upon the P.F. of circuit. In the wattmeter the current coils are arranged for different ratings i.e. 2.5,5 Amp etc. and similarly voltage coils are rated for 125V, 250V, and 500V etc. While doing the experiment the proper range should be selected according to the load voltage & current. The reading should be multiplied by a factor called the Multiplying factor. Multiplying factor of wattmeter = Current range * Voltage range Full scale reading of wattmeter Performance of the transformer can be determined as follows from the observation of load test- Efficiency of the transformer can be determined as ratio of the power output to the power input. Let power input to the transformer = W 1 Power output to the transformer = VI Cos φ Thus the efficiency of particular load ή = (VI Cos φ / W 1 )*100 % 24

25 Note: For resistive load Cos φ is Unity. Efficiency of transformer will be maximum if Iron losses = Copper losses. Regulation of transformer determined as The change in secondary terminal voltage from no load to full load with respect to no load voltage is called voltage regulation of the transformer. Let E 2 = Secondary terminal voltage at no load (Bulb off) V 2 = Secondary terminal voltage at full load Then Voltage regulation = ((E 2 -V 2 ) / E 2 )* 100 CIRCUIT DIAGRAM M L 0-10A 230V AC Wattmeter C 115 LOAD 220 V 2 25

26 Circuit Diagram:- 26

27 Circuit Diagram:- M C + 0-5A L + W PM1 A + V AM1 SW1 200W R1 - + TR1 230 V A.C. VG1 230 V Variable H.V. S upply N1 N2 115 V L.V. + V VM1 SW2 200W R2 SW3 200W R3 SW4 200W R4 SW5 200W R5 R e s i s t i v e L o a d Load test on a Single PhaseTransformer Procedure:- 1. Connect the diagram as shown in fig. 2. Ensure that there is no load on the secondary winding of the transformer. 3. Switch on the A.C. supply & record no load voltage across the secondary winding. 4. Adjust approximately 10% of full load current in the secondary by switching on certain lamp bank load. Record the reading of the entire meter. 5. Reduce the load on the transformer by switching off the bulbs in the lamp bank load. 6. Switch off the A.C. supply. 27

28 Observation table:- Multiplying factor: - S.No Input V 1 V 2 E 2 I 2 V 2 I 2 % ή %Regulation Load Calculation:- Calculate the efficiency using the formula ή = (Output power / Input power) *100 = (V 2 I 2 /W 1 )*100 % The Voltage Regulation %Voltage Regulation = ((E 2 -V 2 )/E 2 )*100 % Report:- Precaution:- The efficiency of the transformer on full load = The regulation of the transformer = 1)-Connection should be tight. 2)-Load on the transformer should nit increase beyond its capacity. 28

29 DISCUSSION:- By calculating the voltage regulation the figure of merit which determines the voltage characteristics of a transformer can be determined. Also the transformer efficiency can t be determined with high precision since the losses are of order of only 1 to 4%. The best and accurate method of determining the efficiency of a transformer would be to compute losses from open circuit and short circuit test and then determine the efficiency. LAB TUTORIALS 1. What does the reading of wattmeter indicate in case of short- circuit test on transformer? 2. How do the copper losses vary with load on the transformer? 3. What is the magnitude of no load current as compared to full load current? 4. What is the power factor of a transformer under no load test situation? REFERENCES Books: 1. Fundamentals of Electrical engineering by Ashfaq Husain. 2. A Textbook of Electrical Technology by B.L Thereja. 3. Electrical Science by J. B. Gupta. 4. Electrical Engineering Fundamentals by Del Toro. URLS:

30 Lab-Quiz: 1. The transformer are rated in (a) KVA (b) KW (c) KV (d) none of these 2. Which part of transformer is subjected to maximum heating (a) Frame (b) Core (c) Winding (d) Oil 3. In a transformer the losses at full load are i) Copper losses =800 W ii) Iron losses =600 W. Copper losses and iron losses respectively at ¼ full load will be (a) 200W, 600W (b) 200W,150W (c) 50 W, 150W (d) 50 W,600W 4. The maximum efficiency in a transformer occurs when (a) Copper losses = Iron losses (b) Eddy current loss = Hyteresis loss (c) Copper losses in the primary =Iron losses (d) Copper losses in the secondary =Iron losses 5. A transformer transforms (a) Current (b) Voltage (c) Power (d) Voltage and current 6. For an ideal transformer the winding should have (a) Max. resistance on primary side and least resistance in secondary side (b) Least resistance on primary side and max. resistance on secondary side (c) Equal resistance on primary and secondary side (d) No ohmic resistance 7. Full load copper losses in a transformer are 400W. Copper losses at half load (a) 400W (b) 200W (c) 50W (d) 100W 1. Transformer core is laminated to (a) Improve cooling (c) Reduce eddy current loss (b) Reduce weight of steel (d) Reduce hysteresis loss 9. Hysteresis losses is proportional to- (a) f (b) f½ (c) f² (d) f³ 10 While carrying out OC test for a 10 kva, 110 / 220 V, 50 Hz, single phase transformer from LV side at rated voltage, the watt meter reading is found to be 100 W. If the same test is carried out from the HV side at rated voltage, the watt meter reading will be (a) 100 W (b) 50 W (c) 200 W (d) 25 W 30

31 EXERCISE NO. 6 OBJECT: 1. Study & construction of single phase transformer. 2. Name plate rating of single phase transformer 3. Determination of transformation ration. APPARATUS REQUIRED: S.NO. Name of equipment Quantity Range Type 1. 1-ΦTransformer 1 230/115V,1KVA Shell type 2. Auto Transformer ,10A Variac type 3. Voltmeter V Moving iron 4. Voltmeter V Moving iron THEORY: Study & construction of single phase transformer: The main elements of a transformer are two copper coils & laminated silicon steel core. A transformer is a static device or a machine that transforms electrical energy from one circuit to another electrical circuit through the medium magnetic flux. And without a change in frequency. The electrical circuit which receive energy from the supply mains is called primary winding and the other circuit which,which delivers electrical energy to the load,is called secondary winding.theoretically it may seem that transformers may be built to handle any voltage or current. But in reality there are limits to both the voltage & current. The name plate rating of a power transformer : The name plate rating of a power transformer usually contains Volt ampere rating of transformer in KVA. Voltage ratio or turn ratio in V 1 /V 2... Frequency of 1-Φ or 3-Φ. Equivalent impedance of a transformers in %. A typical name plate of a 1-Φ transformer is as follows: 230 Volts/115Volts, 50 Hz, 1KVA, Shell type, 10 Amp. Here 1 KVA is the rated output at output terminals.230/115means when 230V. is the applied to the primary,the secondary voltage on full load at specified power factor is 115volts.The ratio of V 1 & V 2 is not exactly equal to N 1 /N 2, because of voltage drop in primary & secondary. Rated primary & secondary current can be calculated from the rated KVA and corresponding rated voltage thus Rated (Full load) primary current = KVA /V 1 = 1000/230 = 4.35 Amps Rated (Full load) secondary current = KVA /V 2 = 1000/115 = 4.35 Amps 31

32 Rated frequency is the frequency for which the transformer is designed to operate. TRANSFORMATION RATIO: The turn ratio of the single phase transformer can be found by measuring the primary & secondary voltage. Let V 1 &V 2 is the primary and secondary voltage at on load. 1/K = V 1 / V 2 = N 1 /N 2 = I 2 /I 1 = Turn Ratio Induced E.M.F. in primary winding, E 1 = 4.44f N 1 Φ Volts Induced E.M.F. in secondary winding, E 2 = 4.44f N 2 Φ Volts For ideal transformer E 1 = V 1 and E 2 = V 2 Hence, Transformation Ratio K = V 2 / V 1 = N 2 / N 1 = I 1 / I 2 PROCEDURE: 1. Connect the circuit as per figure & set up auto transformer to zero position. 2. Switch on A.C. supply and adjust the Auto transformer till a suitable voltage. 3. Record voltage, V 1 across the primary and V 2 across the secondary winding. 4. Vary the Auto transformer and repeat above step, take at least 3 readings. 5. Switch off the supply. OBSERVATION: S.NO. Primary Voltage V 1 Secondary Voltage V 2 K = V 2 / V

33 CIRCUIT DIAGRAM: 33

34 V A.C. Variable S upply + V 230 V Primary H.V. 115 V + N1 N2 Se condary L.V. V Ckt Diagram of Turn Ratio on Single PhaseTransformer RESULT: The transformation ratio of given transformer is.. PRECAUTION: 1. Connection should be tight. 2. Do not touch on live wire. 3. Load on the transfer should not increase beyond its capacity. Tutorials: Q.1 What materials are used for the construction of different parts of transformer? Q.2 What are the types of transformer on the basis of output voltage? Q.3 Draw & explain core type & shell type transformer? Q.4 What will happen if a D.C. supply is supplied to a transformer Q.5 What information is obtained by determining the turn ratio of transformer? Q.6 Define transformation ratio? Q.7 Define all day efficiency of transformer. Q.8 Why transformer s rating is given in KVA? 34

35 OBJECT: EXPERIMENT NO To calculate the complete parameter of the equipment of 1- Φ transformer. 2. To determine iron & copper losses. 3. To calculate efficiency & voltage regulation at 1/4, 1/3, 1/2, 3/4 full load and 1.25times full load at 0.8 P.F. lagging. 4. To plot the efficiency curve v s load. APPARATUS REQUIRED: S.NO. Name of equipment Quantity Range Type 1. 1-ΦTransformer 1 230/115V,1KVA Shell type 2. Auto Transformer ,10A Variac type 3. Voltmeter V Moving iron 4. Ammeter A Moving iron 5. Ammeter A Moving iron 6. Wattmeter 1 2.5/0.5A,125/250/500V Dynamometer 7. Connecting leads THEORY: These two test on transformer help to determine- 4. The parameters of equipments circuit of 1- Φ transformer. 5. The voltage regulation of 1- Φ transformer. 6. The efficiency of 1- Φ transformer. OPEN CIRCUIT TEST OR NO LOAD TEST: In this test voltmeter, Ammeter & Wattmeter are connected on low voltage side of transformer. The high voltage is left open circuited. The rated voltage applied to the primary. The ammeter reads no load current, or the exciting I 0.Since I 0 is quite small (2 to 6% of rated current) the primary leakage impedance drop is almost negligible and for all practical purpose the applied voltage V 1,is equal to induced E.M.F V 1.The input power (iron loss) is given by wattmeter reading, consist of core loss and ohmic loss.since the exciting current is very small, the ohmic losses during open circuit test is negligible as compared to normal core loss. 35

36 CIRCUIT DIAGRAM: - M C + + W L A + V V A.C. Variable S upply + V 115 V L.V. N1 N2 230 V H.V. O P E N C K T. Open Ckt test on a Single PhaseTransformer M C + + W L 0-5A A + V V A.C. Variable S upply + V 230 V H.V. N1 N2 115 V L.V. S H O R T C K T. Short Ckt test on a Single PhaseTransformer 36

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38 CALCULATION: Applied rated voltage on low voltage side = V 1 Exciting Current or no load current = I 0 Wattmeter reading, W o / Iron loss, P C = V O I O CosΦ o No load power factor, CosΦ o = P C / V O I O Working component, I W Magnetizing component, IΦ = I O CosΦ o = I O SinΦ o Core loss Resistance, R C = P c I 2 W = V 1 /I W =V 1 /I 0 CosΦ 0 Magnetising reactance, XΦ = V 1 /I Φ = V 1 /I o Sin Φ o Thus open circuit test gives the following information: 1. Core loss at rated voltage & frequency. 2. The shunt branch parameter of equivalent circuit i.e., XΦ & R C. SHORT CIRCUIT TEST: The low voltage side of the transformer of the transformer is short circuited & instrument are placed on H.V. side. Apply the low voltage on H.V. side & with the help of autotransformer go on increasing the applied voltage till the rated current starts flowing in the short circuited winding(l.v. side).the primary voltage 10% to 12% of its rated value is sufficient to circulate the rated current in short circuited winding. Since the core flux induces the voltage, which is 1% to 6% of its rated value hence core loss can be neglected. The wattmeter records only the ohmic loss is both, the primary & secondary winding. 38

39 CALCULATION: V sc, I sc & P sc are the voltmeter ammeter & wattmeter reading Z SC = V SC /I SC R SC = P SC /I 2 SC X SC = Z 2 SC R 2 SC Thus the short circuit test gives the following information 1. Ohmic loss at rated current and frequency. 2. Equivalent resistance and leakage reactance and leakage impedance. Load x P.F. The efficiency at any load, η = X 100 % Load x P.F.+ W o + I o 2 R o PROCEDURE FOE OPEN CIRCUIT TEST: 1. Connect the circuit diagram as shown in figure and set up the autotransformer at zero position. 2. Adjust the supply voltage with the help of autotransformer to 230 volts with secondary winding terminal open. 3. Record the ammeter, voltmeter, wattmeter reading. 4. Vary the supply voltage with the help of the auto transformer and enter the reading in observation table. OBSERVATION TABLE FOR OPEN CIRCUIT TEST S.NO. Primary Voltage Voltmeter Reading 1. Input Current Ammeter Reading Input power in watts Wattmeter reading 39

40 PROCEDURE FOE SHORT CIRCUIT TEST: 1. Connect the circuit diagram as shown in figure and set up the autotransformer at zero position. 2. Adjust the supply voltage with the help of autotransformer (keep in mind that 10-12% of rated voltage is sufficiency) with secondary winding terminal short circuited and circulate full rated current in short circuited winding. 3. Record the ammeter, voltmeter, wattmeter reading. 4. Vary the supply voltage with the help of the auto transformer and enter the reading in observation table. 5. Three readings adjust at 50%, 86.6% & 100% rated full load current. OBSERVATION TABLE FOR SHORT CIRCUIT TEST S.NO. Primary Voltage Voltmeter Reading Input Current Ammeter Reading Input power in watts Wattmeter reading RESULT: PRECAUTION: 1. In open circuit test, the H.V. side should be open circuited (left side). 2. In open circuit test, low voltage should be apply to the H.V. side & it should be increased gradually to circulate the rated current in H.V. side. 3. Connection should be tight. 4. Do not touch on livewire. 40

41 DISCUSSION: The open circuit and Short circuit test have been performed and various parameters like R 01, X 0, Z 01 and X 01 are calculated. LAB TUTORIALS 1. Why indirect testing of large size transformers is necessary? 2. When a transformer is energized what types of losses occur in the magnetic frame of transformer? 3. Why do you understand by all day efficiency of a transformer? 4. What quantity is ignored in the S.C test on a transformer? REFERENCES Books: 4. Fundamentals of Electrical engineering by Ashfaq Husain. 5. A Textbook of Electrical Technology by B.L Thereja. 6. Electrical Science by J. B. Gupta URLS:

42 Lab-Quiz: 1. Full load copper losses in a transformer are 400W. Copper losses at half load (a) 400W (b) 200W (c) 50W (d) 100W 2. The frequency of the secondary voltage of a transformer wills be. (a) Less than the frequency of the primary voltage. (b) Equal to the primary voltage. (c) Greater than the frequency of the primary voltage. (d) Very much greater than the frequency of the primary voltage. 3. For an ideal transformer the winding should have (a) Maximum Resistance on primary side and least resistance in secondary side (b) Least resistance on primary side and maximum resistance on secondary side (c) Equal resistance on primary and secondary side (d) No Ohmic resistance on primary and secondary side 4. At Full load Iron losses in a transformer are 100W. Iron losses at half load (a) 400W (b) 200W (c) 50W (d) 100W 5. The maximum efficiency in a transformer occurs when (a) Copper losses = Iron losses (b) Eddy current loss = Hysteresis loss (c) Copper losses in the primary = Iron losses (d) Copper losses in the secondary = Iron losses 6. A transformer transforms (a) Current (b) Voltage (c) Power (d) Voltage and current 7. An ideal transformer has 300 turns in primary and 2000 turns in secondary, the transformer is connected to 220V, 50HZ supply. Determine the secondary voltage.. (a) 2.50 KV (b) 1.46 KV (c) 4.6 KV (d) 1000V 42

43 Object:- Study of constructional features of D.C. machines. Apparatus Required:- D.C. Machines model. EXPERIMENT NO. 8 Theory:- A d.c. machine is an Electro-Mechanical energy conversion device. It can convert mechanical power into d.c. electrical power and is known as a d.c. generator. On the other hand, when it converts d.c. electrical power into mechanical power it is known as d.c. motor. Contructional Details:- There are two main parts of a d.c. machine:- Field System: - a. Electromagnetic Poles b. Yoke c. Field Winding Armature: - a. Armature Core b. Armature Winding c. Commutator Magnetic Frame or Yoke :- The outer cylindrical frame to which main poles and inter poles are fixed and by means of which the machine is fixed to the foundation is called the Yoke. It serves two purposes: a) It provides mechanical protection to the inner parts of the machine. b) It provides a low reluctance path for the magnetic flux. The yoke is made of cast iron for smaller machines and larger machines; it is made up of cast steel. Pole core and Pole shoes:- The pole core and pole shoes are fixed to the magnetic frame or yoke by bolts. They serve the following purpose: a) They support the field or exciting coils. b) They spread out the magnetic flux over the armature periphery more uniformly. c) Since pole shoes have large X-section, the reluctance of magnetic path is reduced. Usually, the pole core and pole shoes are made of thin cast steel. 43

44 Field or Exciting coils:- Anamelled copper wire is used for the construction of field or exciting coils. The coils are wound on the former and then placed around the pole core. When direct current is passed through the field winding, it magnetizes the poles which produce the require flux. The field coils of all the poles are connected in series in such a way that when current flows through them, the adjacent poles attain opposite polarity. Armature core:- It is cylindrical in shape and keyed to the rotating shaft. At the outer periphery slots are cut, which accommodate the armature winding. The armature core serves the following purpose: (i) It houses the conductors in the slots. (ii) It provides an easy path for magnetic flux. Since armature is a rotating part of the machine, reversal of flux takes place in the core, hence hysterisis losses are produced. To minimize these losses silicon steel material is used for its construction. The rotating armature cuts across the magnetic field which induces an e.m.f. in it. The e.m.f circulates eddy currents which results in eddy current losses in it. To reduce these losses armature core is laminated, in other word we can say that about 0.3 to 0.5 mm thick stampings are used for its construction. Each lamination or stamping is insulated from the outer by varnish layer. Armature Winding:- The insulated conductors housed in the armature slots are suitably connected. This is known as armature winding. The armature winding is the heart of d.c. machine. It is a place where conversion of power takes place i.e. in case of generator, mechanical power is converted into electrical power and in case of motor, electrical power is converted into mechanical power. On the basis of connections, there are two types of armature winding names as:- (a) Lap Winding (b) Wave Winding. Commutator:- It is the most important part of d.c. machine and serves the following purposes:- (i) It connects the rotating armature conductors to the stationary external circuit through brushes. (ii) It convert the alternating current induced in the armature conductor into unidirectional current in the external load circuit in generator action whereas, it converts the alternating torque into unidirectional torque produced in the armature motor action. The commutator is of cylindrical shape and is made up of wedge-shaped hard drawn copper segments. The segments are insulated from each other by a thin sheet of mice. The segments are 44

45 held together by means of 2 V-shaped rings that fit into the V-grooves cut into the segments. Each armature coil is connected to the commutator segment through riser. Brushes:- The brushes are pressed upon the commutator and from the connecting link between the armature winding and the external circuit. They are usually made of high grade carbon because carbon is conducting material and the same time in powdered form provides lubricating effect on the commutator surface. The brushes are held in particular position around the commutator by brush holders. End housings:- End housings are attached to the ends of the main frame and support bearings. The front housing supports the bearing and the brush assemblies whereas the rear housing usually supports the bearing only. Bearings:- The ball or roller bearings are fitted in the end housings. The function of the bearings is to reduce friction between the rotating and stationary parts of the machine. Mostly high carbon steel is used for the construction of bearings as it is very hard material. Shaft :- The shaft is made of mild steel with a maximum breaking strength. The shaft is used to transfer mechanical power from or to the machine. The rotating parts like armature core, commutator, cooling fan etc. are keyed to the shaft. 45

46 46

47 Lab Tutorials: Q.1 What function do the yoke perform in D.C. Motor.? Q.2 What do you mean by commutation in D.C. Motor.? Write down its advantages. Q.3 Write down the material used for the construction of different parts of D.C. Machine.? Q.4 Draw & explain different characteristics of D.C. Generator. Q.5 What are the two main function performed by brushes? Q.6 What advantages D.C. Machine have over A.C. Machine.? Q.7 What are the different type of D.C.Generator? REFERENCES Books: 1. Fundamentals of Electrical engineering by Ashfaq Husain. 2. A Textbook of Electrical Technology by B.L Thereja. 3. Electrical Science by J. B. Gupta URLS:

48 EXPERIMENT NO. 9 Object :- Measurement of various line and phase quantities for a three phase ac circuit. Apparatus Required :- S.No. Name of equipments Quantity Range Type 1. Voltmeter v Ac 2. Ammeter amp. Ac 3. 3-phase Auto transformer v Ac 4. Connecting leads Resistive load Theory:- Three phase supply system:-an A.C. system having three equal voltages of same frequency arranged to have equal phase difference between adjacent e.m.f. is called a three phase supply system. Three phase supply system is divided into two categories. (1) 3-phase balanced system: - A supply system is said to be balanced when the tree voltages of the same frequency have equal magnitude and one displaced from another by equal time angle i.e. 120 electrical degree with respect to each other is called three phase balanced system. (2) 3-phase unbalanced system:- A three phase supply system will be unbalanced when either of the three phase voltages are unequal in magnitude or the phase angle between them is not equal to 120 electrical degree with respect to each other is called three phase unbalanced system. Procedure:- 1) Connect the voltmeter and ammeter to the load through three phase auto transformer as shown in figure and set the autotransformer at zero position. 2) Switch on the three phase supply and adjust the auto transformer till a suitable voltage. Note down the readings of voltmeter and ammeter. 3) Vary the voltage by the autotransformer and note down the various readings. 4) Now after the observation switch off and disconnect all the equipments. 48

49 Circuit Diagram :- 49

50 Observation table:- (1) FOR STAR CONNECTED LOAD:- S. No. Voltmeter Reading Ammeter Reading Line Voltage Phase Voltage Line Current Phase Current (2) FOR DELTACONNECTED LOAD:- S. No. Voltmeter Reading Ammeter Reading Line Voltage Phase Voltage Line Current Phase Current Calculation:- Result:- (1) In star connected load :- Line voltages are 3 times of phase voltages Line currents are equal to phase currents (2) In delta connected load :- Line currents are 3 times of phase currents Line voltages are equal to phase voltages 50

51 Precautions:- (1) Make the connections as shown in figure. (2) Connections should be tight. (3) Ensure that auto transformer should be at zero position before switching on the main supply. (4) Please get the circuit checked before switching on the main supply. (5) Do not touch any live wire. (6) Supply should be off after the completion of experiment REFERENCES Books: 1. Fundamentals of Electrical engineering by Ashfaq Husain. 2. A Textbook of Electrical Technology by B.L Thereja. 3. Electrical Science by J. B. Gupta URLS:

52 EXPERIMENT NO. 10 Object: VERIFICATION OF KCL AND KVL Kirchhoff s Current Law (KCL) and Kirchhoff s Voltage Law (KVL) Appararus Required: Apparatus has been designed to calculate the value of current flowing through different branches of the circuit. The instrument comprises of the following built in parts : 1. DC Regulated power supply of 0-5V. 2. Four type of wire wound resistances, each of 5watt. (5W, 10W, 22W & 33W ) are mounted on front panel. 3. Circuit for Kirchhoff laws is engraved on front panel. 4. Two meters are provided on the front panel to measure corresponding Voltage & Current with connections brought out on sockets. THEORY In simple circuits,the resistance and potential difference are calculated with the help of ohms law. But in actual practice, we come across complicated circuits which contain a large number of resistances alongwith several sources of e.m.f. In such cases, the effective resistance and the e.m.f. can not be calculated easily from ohm s law. In order to solve such networks, Kirchhoff gave two laws which are known as Kirchhoff s laws. First Law Kirchhoff s Current Law: According to Kirchhoff s first law. The algebriac sum of the various currents meeting at a junction in a closed electrical circuit is Zero. i 1 - i 2 - i 3 + i 4 + i 5 = 0 or i 1 + i 4 + i 5 = i 2 + i 3 Sign Convention : The currents flowing towards a junction are taken as positive while the currents flowing away from the junction are taken as negative. Let us consider a junction 0. Where a number of conductors meet. i1, i2 are the currents flowing through them in the directions shown in Fig. (1). Applying Kirchhoff first law, we get. 52

53 Thus the total amount of current flowing into a junction must be equal to the total current flowing out of the junction. Clearly, according to this law, electric current cannot accumulate at any point. Unlike charge, current cannot be stored. It must flow on. Second law -Kirchhoff s Voltage Law (KVL) : According to second law, in a closed loop (closed circuit or mesh) the algebriac sum of the emfs is equal to the algebric sum of the products of the resistances and the respective currents flowing through them. The distribution of currents in the circuit is shown in the diagram given below according to Kirchhoff s first law. NOTE : Resistance tolrance is in +10%. PROCEDURE 1. Connect the circuit as show in Fig. (1a) i.e. connect voltmeter across the positive to positive and negative to negative Current meter (ma) is connected to positive to positive and negative to negative. 2. Set output voltage 2.5 volts and connect the input through patch cord. 3. Short the A and B or C and D point through patch cord. E and F or G and H point will be open. 4. Switch on the instrument and note down the current in (ma). Applying Kirchhoff s second law to the closed mesh A B C D, We get R1 i 1 +R2i 1 = voltage (R 1 +R 2 ) i 1 = voltage i 1 = V /(R 1 +R 2 ) Amp. i 1 =?Amp (convert to miliamp.) (i+ x 1000 = ma) V 1 = R 1 i 1 V 2 = R 2 i 1 V=V 2 +V 1 Compare this calculated value to observed value at current meter. 53

54 Compare this calculated value to observed value at current meter. Calculation of i 2 current: 1. Connect the circuit as show in fig 1b i.e. connect voltmeter across the positive to positive and negative to negative Current meter (ma) 2. Set output voltage 2.5 volts and connect the input through patch cord. 3. Short the E and F or G and H point through patch cord. A and B or C and D point will be open. 4. Switch ON the instrument and note down the current in (ma). Applying Kirchhoff s second law to the closed mesh E F G H, We get R 3 i 2 +R 4 i 2 = voltage (R 3 +R 4 ) i 2 = voltage i 2 = V /(R 3 +R 4 ) Amp. i 2 =?Amp (convert to miliamp.) (i 1 x 1000 = ma). V 3 = R 3 i 2 V 4 = R 4 i 2 V=V 3 +V 4 Calculation of total current i : Connect the circuit as shown in Fig. (1c) i.e. connect point A and B or C and D or E and F or G and H. Also connect current meter (ma) and voltmeter positive to positive and negative to negative. R Total Current I =i1 + i2 convert the miliampere (i x1000= ma) 54

55 Observation Table. Record the observation as per the table given below: Verification of Kirchhoff s Current Law S.No. I 1 amp I 2 amp I amp (I 1 +I 2 ) = I amp Verification of Kirchhoff s Voltage Law S. No. CURRENT I 1 AMP VOLTAGE V Volt VOLTAGE V 1 Volt VOLTAGE V 2 Volt V 1 + V 2 = V? VOLTAGE V 3 Volt VOLTAGE V 4 Volt V 4 + V 3 = V? Calculations. 1. Kirchhoff s Current Law (KCL): Add the readings in column 2 and 3 of table and record it in column 5 of this table.check that the entry A3 in column 4 and (A1+A2 ) in column 5 of this table should be equal. 2. Kirchhoff s Voltage Law (KVL): Add the voltage v 1 and v 2 recorded in column 3 and 4 of table and record the same in the last column of this table. Check that the voltage v 3 in column 5 agrees with the voltage (v 1 + v 2 ) recorded in the last column of this table. 55

56 Results. 1. Kirchhoff s Current Law is verified. 2. As The Voltage V 1 + V 2 = V And V 4 + V 3 = V Are varified As Shown In Table, The Kirchhoff s Voltage Law Is Verified. Precautions: Following precautions should be observed while performing this experiment: (1) All the connection should be tight. (2) Before connecting the instruments, check their zero reading. (3) The terminals of the rheostats should be connected properly. (4) The direction of currents should be identified correctly. 56

57 Lab-Quiz: (1) Is the Kirchhoff s current law applicable to a.c. circuits too? (2) how many laws were developed by Kirchhoff related to electrical circuit? (3) State Kirchhoff s current law applicable to electrical circuit. (4) what do you mean by the term node in reference to electric circuits? (5) In what respect, Kirchhoff s laws are better than Ohm s law. (6) State kirchhoff s voltage law as applicable to electrical circuits. (7) How kirchhoff s voltage law can be applied to A.C. circuits? (8) Define the term loop. (9) What do you mean by algebraic sum? (10) What is the difference between a mesh and a loop? REFERENCES Books: 1. Fundamentals of Electrical engineering by Ashfaq Husain. 2. A Textbook of Electrical Technology by B.L Thereja. 3. Electrical Science by J. B. Gupta URLS:

58 AIM: To verify Superposition Theorem. EXPERIMENT NO. 11 OBJECTIVE: To apply the principle of Superposition Theorem for electrical network containing independent Dc sources. APPARATUS: Digital Multimeter, Power Supply, Resistance (wire wound), Connecting Wires THEORY OF EXPERIMENT: Superposition theorem states that in a linear network containing several independent sources, the overall response at any point in the network equals the sum of responses due to each independent source considered separately with all other independently sources made inoperative. To make a source inoperative, it is short circuited leaving behind its internal resistance or impedance, if it is a voltage source and it is open circuited leaving behind its internal resistance or impedance, if it is a current source. In most electrical circuit analysis problems, a circuit is energized by a single independent energy source. In such cases, it is quite easy to find the response (i.e. current, voltage, power) in a particular branch of the circuit using simple network reduction techniques(i.e. series parallel combination, star delta transformation etc.). However, in the presence of more than one independent sources in the circuit, the response cannot be determined by direct application of network reduction techniques. In such a situation, the principle of superposition may be applied to a linear network, to find the resultant response due to all the sources acting simultaneously. The superposition theorem is based on the principle of superposition. The principle of Superposition states that the response (a desired current or the voltage) at any point in the linear network having more than one independent source can be obtained as the sum of responses caused by the separate independent sources acting alone. The validity of principle of superposition means that the presence of one excitation does not affect the response due to other excitations. 58