Electrical World

Hello Engineers. Today we are sharing alternator interview questions with answer. Q. 1. What are the two types of turbo-alternators ? Ans.  Vertical and horizontal. Q. 2. How do you compare the two ? Ans. Vertical type requires less floor space and while step bearing is necessary to carry the weight of the moving element, there is very little friction in the main bearings. The horizontal type requires no step bearing, but occupies more space. Q. 3. What is step bearing ? Ans. It consists of two cylindrical cast iron plates which bear upon each other and have a central recess between them. Suitable oil is pumped into this recess under considerable pressure. Q. 4. What is direct-connected alternator ? Ans. One in which the alternator and engine are directly connected. In other words, there is no intermediate gearing such as belt, chain etc. between the driving engine and alternator. Q. 5. What is the difference between direct-connected and direct-coupled units ? Ans...

Several theories have been developed to explain the nature of electricity. The modern electron theory of matter, propounded by scientists Sir Earnest Rutherford and Niel Bohr considers every matter as electrical in nature. According to this atomic theory, every element is made up of atoms which are neutral in nature. The atom contains particles of electricity called electrons and protons. The number of electrons in an atom is equal to the number of protons. The nucleus of an atom contains protons and neutrons. The neutrons carry no charge. The protons carry positive charge. The electrons revolve round the nucleus in elliptical orbits like the planets around the sun. The electrons carry negative charge. Since there are equal number of protons and electrons in an atom, an atom is basically neutral in nature. If from a body consisting of neutral atoms, some electrons are removed, there will be a deficit of electrons in the body, and the body will attain positive charge...

Why star delta starter is preferred with induction motor? Star delta starter is preferred with induction motor due to following reasons: • Starting current is reduced 3-4 times of the direct current due to which voltage drops and hence it causes less losses. • Star delta starter circuit comes in circuit first during starting of motor, which reduces voltage 3 times, that is why current also reduces up to 3 times and hence less motor burning is caused. • In addition, starting torque is increased and it prevents the damage of motor winding. State the difference between generator and alternator Generator and alternator are two devices, which converts mechanical energy into electrical energy. Both have the same principle of electromagnetic induction, the only difference is that their construction. Generator persists stationary magnetic field and rotating conductor ...

Let us consider the following two cases: Equal voltage ratios. Unequal voltage ratios. 1.39.1 Equal Voltage Ratios Assume no-load voltages E A and E B are identical and in phase. Under these conditions if the primary and secondary are connected in parallel, there will be no circulating current between them on no load. Figure 1.48 Equal Voltage Ratios Figure 1.48 shows two impedances in parallel. Let R A , X A and Z A be the total equivalent resistance, reactance and impedance of transformer A and R B , X B and Z B be the total equivalent resistance, reactance and impedance of transformer B . From Figure 1.48, we have E A = V 2 + I A Z A      (1.71) and           E B = V 2 + I B Z B      (1.72) ∴       I A Z A = I B Z B ∴     Equation (1.73) suggests that if two transformers with different kVA ratings...

Electrical Engineering Interview Questions Part 1 Electrical Engineering Interview Questions Part 2 Electrical Engineering Interview Questions Part 3 Electrical Engineering Interview Questions Part 4 Electrical Engineering Interview Questions Part 5 Electrical Engineering Interview Questions Part 6 Electrical Engineering Interview Questions Part 7 Electrical Engineering Interview Questions Part 8 Electrical Engineering Interview Questions Part 9 Electrical Engineering Interview Questions Part 10 Electrical Engineering Interview Questions Part 11 Electrical Engineering Interview Questions Part 12 Electrical Engineering Interview Questions Part 13 Electrical Engineering Interview Questions Part 14 Electrical Engineering Interview Questions Part 16

To determine the rise of maximum temperature of a transformer, its load test is of utmost importance. Using suitable load impedance, small transformers can be put on full load. The full-load test of large transformers is not possible because considerable wastage of energy occurs and it is difficult to get a suitable load for absorbing full-load power. Sumpner’s test is used to put large transformer on full load. This test can also be used to determine the efficiency of a transformer. Figure 1.45 shows the schematic diagram of Sumpner’s test. This test is also known as back to back test or load test . This test requires two identical transformers. The two primaries are connected in parallel and are energized at rated voltage and rated frequency. The wattmeter W 1 records the reading of core loss of both the transformers. Next the two secondaries are connected in series in such a way that their polarities are in phase opposition and the reading of the voltmeter V 2 becomes ...

The ratio of output in watts to input in watts is called commercial efficiency of a transformer. Distribution transformers are used for supplying lighting and general networks. Distribution transformers are energized throughout the day. Their secondaries are at no load most of the time in a day except during the hours of lighting period. Core loss occurs throughout the day. Copper loss occurs only when they are loaded and hence is less important. To judge their performance, all-day efficiency or operational efficiency is calculated. The all-day efficiency is defined by The all-day efficiency is less than the commercial efficiency of a transformer. Example 1.16 A 200 kVA single-phase transformer is in circuit throughout 24 hours. For 8 hours in a day, the load is 150 kW at 0.8 power factor lagging and for 7 hours, the load is 90 kW at 0.9 power factor. Remaining time or the rest period, it is at no-load condition. Full-load Cu loss is 4 kW and the iron loss is 1.8 kW. Calcu...

Polarity testing of transformers is vital before connecting them in parallel. Otherwise, with incorrect polarity, it is not possible to connect them in parallel. The rated voltage is applied to the primary and its two terminals are marked as A 1 and A 2 , respectively, as shown in Figures 1.44(a) and 1.45(b), respectively. The secondary winding terminals are also marked as a 1 and a 2 , shown in Figures 1.44(a) and 1.45(b), respectively. Now a voltmeter is connected across A 2 and a 2 . if it measures the difference of E 1 and E 2 , A 2 and a 2 are of the same polarity. If it measures the addition of E 1 and E 2 , A 2 and a 2 are of opposite polarity. Figure 1.44 Polarity Test of a single-phase Two Winding Transformer

Due to the losses in a transformer, its output power is less than the input power. ∴ Power output = Power input – Total losses ∴ Power input = Power output + Total losses = Power output + P i + P Cu The ratio of power output to power input of any device is called its efficiency ( η ).        Output power of a transformer at full-load = V 2 I 2ft cos θ , where cos θ is the power factor of the load, I 2 ft is the secondary current at full load and V 2 is the rated secondary voltage of the transformer. Full-load copper loss of the transformer = I 2 ft R 02 . ∴     Efficiency of the transformer at full load is given by Now V 2 I 2 ft = VA rating of the transformer. ∴     i.e.,     Efficiency of the transformer at any load m is given by where m = and P Cuft is the Cu loss of the transformer at full load. 1.34 CONDITION FOR MAXIMUM EFFICIENCY During working of a transformer at constant vol...

Two types of losses occur in a transformer: Core loss or iron loss occurs in a transformer because it is subjected to an alternating flux. The windings carry current due to loading and hence copper losses occur. 1.32.1 Core or Iron Loss The separation of core losses has already been introduced. The alternating flux gets set up in the core and it undergoes a cycle of magnetization and demagnetization. Therefore, loss of energy occurs in this process due to hysteresis. This loss is called hysteresis los s ( P h ), which is expressed by P h = K h B m 1.6 f V W       (1.59) where K h is the hysteresis constant depending on the material, B m is the maximum flux density, f is the frequency and V is the volume of the core. The induced emf in the core sets up eddy current in the core, and hence eddy current loss ( P e ) occurs, which is given by P e = K e B m 2 f 2 t 2 W per model W       (1.60) where K e is the ed...

Kapp had designed a diagram shown in Figure 1.42 to determine the regulation at any power factor. The description of the construction of the diagram is shown below. Figure 1.42 Kapp’s Diagram Load current ( I 2 ) is taken as a reference phasor. OA representing V 2 is drawn at angle θ 2 with I 2 . AB represents I 2 R 02 drawn parallel to I 2 , whereas BC represents I 2 X 02 drawn perpendicular to AB, i.e., I 2 . Here OC represents secondary emf ( 0 V 2 = E 2 ) at no-load. The circle 1 known as often circuit EMF circle is drawn with O as centre and OC as radius. The line OO ′ is drawn parallel to AC representing I 2 Z 02 . With O ′ as centre and OA as radius, the circle 2 known as terminal voltage circle is drawn, which intersects with circle 1 at the points D and E. The region above and below the reference line represents the lagging and leading power factors region, respectively. The point D is the point corresponding to zero regulation. The intercept FG gives the maximum ...

The voltage regulation up is expressed mathematically by Positive sign is for lagging power factor and negative sign is for leading power factor. 1.31.1 Zero Voltage Regulation For lagging power factor and unity power factor, 0 V 2 > V 2 . Therefore, we get positive voltage regulation. For leading power factor, V 2 starts increasing. At a certain leading power factor, 0 V 2 = V 2 and hence regulation becomes zero. If the load power factor is further increased, 0 V 2 becomes less than V 2 and hence regulation becomes negative. For zero voltage regulation, we have 0 V 2 – V 2 = 0 i.e.,      I 2 ( R 02 cos θ – X 02 sin θ )=0             Equation (1.56) shows the leading power factor at which voltage regulation becomes zero. 1.31.2 Condition for Maximum Voltage Regulation Maximum voltage regulation can be obtained for lagging power factor. For maximum voltage regula-tion, we have i.e.,    ...

Full-load voltage of a transformer can be expressed as a fraction of the full-load terminal voltage. Let I 1 fl be the full-load primary current, I 2 fl be the full-load secondary current, V 1 be the rated pri-mary voltage and V 2 be the rated secondary voltage. Per unit resistance drop of a transformer ∴       Per unit reactance drop of a transformer Per unit reactance drop of a transformer is called per unit reactance and it is given by Per unit impedance drop of a transformer is called per unit impedance and it is given by