CERTIFICATES OF COMPETENCY IN THE MERCHANT NAVY MARINE ENGINEER OFFICER EXAMINATIONS ADMINISTERED BY THE SCOTTISH QUALIFICATIONS AUTHORITY ON BEHALF OF THE MARITIME AND COASTGUARD AGENCY STCW 78 as amended MANAGEMENT ENGINEER REG. III/2 (UNLIMITED) 040-33 - ELECTROTECHNOLOGY THURSDAY, 20 JULY 2017 0915-1215 hrs Examination paper inserts: Notes for the guidance of candidates: 1. Non-programmable calculators may be used. 2. All formulae used must be stated and the method of working and ALL intermediate steps must be made clear in the answer. Materials to be supplied by examination centres: Candidate s examination workbook Graph paper [OVER
ELECTROTECHNOLOGY Attempt SIX questions only. All questions carry equal marks. Marks for each part question are shown in brackets. 1. Fig Q1 shows a ring main distributor fed at one point at 440 volts. The distances between the various loads are given in metres and the two cables has a go and return distance of 0.02 Ω per 100 metres. Determine EACH of the following: (a) the current in the cable between the 30 A and 70 A loads; (b) the lowest p.d across any of the loads; (c) the total power loss in distributor. (8) Fig Q1 [OVER
2. When connected to a 20 v d.c. supply a relay starts to operate 0.52 ms after switching on the supply, at which time the instantaneous current is 200 ma. The relay coil has a time constant of 5 ms. (a) Calculate EACH of the following: (i) (ii) the final steady state relay current; the resistance and inductance of the relay coil. (b) To increase the operating time a 40 Ω resistor is connected in series with the relay coil. Calculate the new time delay assuming the instantaneous current is still 200 ma. 3. In the two stage voltage amplifier shown in Fig Q3 both the npn and pnp transistors have high current gains. Transistor T 1 has a volt drop of 0.7 V between baseemitter and transistor T 2 has a volt drop of 0.3 V between base-emitter. Calculate EACH of the following: (a) the voltage between collector and emitter for each transistor; (b) the power dissipated in each transistor. (12) Fig Q3
4. A 40 kva 400 V /110 V single phase transformer has an iron loss of 0.9 kw. Maximum efficiency occurs at 75% full load and 0.8 p.f. lag. Calculate EACH of the following: (a) the copper loss at full load; (b) the efficiency at full load and 0.8 p.f. lag; (c) the efficiency at half full load and unity p.f. 5. Three identical coils are delta connected to a 3 ph, 440 V, 60 Hz supply and consume a total power of 9 kw at a power factor of 0.8 lag. (a) Calculate the resistance and inductance of EACH coil. (b) If the same three coils are now connected in star to the same supply, calculate the current in each line if: (i) one coil is short circuited; (ii) one coil is open circuited. 6. A three phase, six pole, delta connected induction motor is supplied at 380 V, 60 Hz. It draws a current of 45 A at a power factor of 0.85 lag. The stator losses are 4 kw and the windage and friction losses total 3 kw. It runs at 19 rev/sec. Calculate EACH of the following: (a) the rotor copper loss; (b) the shaft output power; (c) the shaft output torque. (8) [OVER
7. (a) Describe the FOUR conditions which have to be met before an alternator can be connected to live busbars. (b) Explain the process by which kw load can be delivered by a newly synchronised alternator. (c) Describe the effect of increasing the excitation of an alternator which is sharing a load without increasing the power input to the machine. 8. With reference to a 1 ph power transformer with air cooling: (a) sketch a labelled diagram of the basic construction; (b) explain the principle of operation; (c) state why it is rated in KVA rather than KW; (d) explain why it may overheat if operated at reduced frequency; (e) explain how operation at reduced frequency can be compensated for. (2) 9. (a) Describe, with the aid of a detailed sketch, the construction of a double cage rotor for a squirrel cage induction motor. (b) Explain how the rotor current is distributed between the two windings as the machine runs up from standstill to full speed. (c) Sketch a torque/slip curve for each cage during the run up period.
SCOTTISH QUALIFICATIONS AUTHORITY MARKERS REPORT FORM PART I SUBJECT: Electrotechnology 040-33 DATE: 20 th July 2017 General Comments on Examination Paper This paper, differing only in two question from that set for 041-33, was of the correct standard and conformed to the modified syllabus for the new style of examination. General Comments of Specific Examination Questions Question 1.. A ring type distributor which posed no great problems, but a few candidates misunderstood the wording of Parts (b) and (c), I suspect, and calculated the minimum volt drop between load and the power delivered to the distributor rather than lost in the distributor. This a clearly a language problem Question 2. Q. This too produced many correct answers, but a few (too many! ) candidates made the mistake of assuming that the final current would still be 2 A. after extra resistance was added to the circuit. Question 3. Well answered, the only serious mistake was to subtract the 0.3V to the base voltage of the PNP transistor To obtain the emitter voltage, it should be added. Question 4. This few was answered well but a few students used KVA rather than KW to calculate the efficiency, an elementary mistake. Question 5. The only troubles which arose in this were in Part b(i) where some candidates added the two line currents arithmetically to find the third line current, these are phasor quantities and not in phase. Question 6.. By and large this held no terrors except for a few candidates who still insist on adding the various losses as they progress through the machine, producing a motor of more than 100% efficiency! A handful calculated the shaft torque using the shaft output in KW but forgot to multiply by 10 3, producing a sickly torque of a fraction of a Nm. Question 7. Part (a) of this question was answered well by a large number of students but the explanations for parts (b) and (c) were garbled and well nigh impossible to understand. This failure to give sensible and concise answers may well be a language difficulty rarher than a failure of understanding.
Question 8. Every candidate chose a core type construction (nothing wrong with that) but every one showed the primary on one limb and the secondary on the other. This is fine in The Boy s Book of Electricity but at this level they should have been told that in a real, practical, transformer, the windings are shared between the two limbs, the lower voltage winding innermost. No mark has been deducted for this, as they had clearly been ill informed but a mark has been deducted it the core is described as steel rather than iron and no mention is made of laminations. The most students made a fair go of the remaining parts, Question 9.. Only a handful tackled this question and the answers were poor. Some showed a sashbar construction and others showed the cages completely surrounded by iron with no slot to control the reluctance of each cage. Overall this earned few marks for most candidates.
CERTIFICATES OF COMPETENCY IN THE MERCHANT NAVY MARINE ENGINEER OFFICER EXAMINATIONS ADMINISTERED BY THE SCOTTISH QUALIFICATIONS AUTHORITY ON BEHALF OF THE MARITIME AND COASTGUARD AGENCY STCW 78 as amended MANAGEMENT ENGINEER REG. III/2 (UNLIMITED) 040-33 - ELECTROTECHNOLOGY THURSDAY, 30 MARCH 2017 0915-1215 hrs Examination paper inserts: Notes for the guidance of candidates: 1. Non-programmable calculators may be used. 2. All formulae used must be stated and the method of working and ALL intermediate steps must be made clear in the answer. Materials to be supplied by examination centres: Candidate s examination workbook Graph paper
ELECTROTECHNOLOGY Attempt SIX questions only. All questions carry equal marks. Marks for each part question are shown in brackets. 1. A 525 m, two core distributor cable is fed at one end with 240 V.d.c and at the other end with 250 V.d.c. The following roads are applied at distances measured from the 240 V end: Load 1 10 A at 100 m Load 2 100 A at 250 m Load 3 70 A at 450 m Load 4 75 A at 500 m The cable resistance (go and return) is 0.16 Ω per 100 m. Calculate EACH of the meeting: (a) the current supplied at each end of the distributor; (b) the voltage of each load point; (c) the power delivered at each end of the distributor. (8) (2) 2. A relay coil has a resistance of 200 Ω and the current required to operate the relay is 150 ma. When the coil is connected to 50 V d.c. it takes 40 ms for the relay to operate. (a) Calculate EACH of the following: (i) the steady state relay current; (ii) the time constant for the coil; (iii) the inductance of the coil. (2) (b) To increase the operating time for the relay, a 50 Ω resistor is connected in series with the coil. Calculate the new operating time for the relay. [OVER
3. The p.d. between base and emitter for the transistor shown in Fig Q3 is 0.3 V and the steady state output voltage V c is 6 V. Determine EACH of the following, assuming that the base current is small enough to be ignored: (a) the voltage at the base with respect to earth; (b) the p.d. between emitter and collector; (c) the value of the load resistor R L; (d) the power dissipated in the 200 Ω resistor; (e) the power dissipated in the transistor. 12 kω R L 16 V 4 kω 200 Ω 6 V Fig Q3 4. (a) A 400 V/110 V transformer has 3468 turns on each primary phase winding. If the volt drop in the windings are negligible, calculate the number of turns of each secondary phase winding for EACH of the following connections: (i) Delta/delta; (ii) Delta / star. (b) Explain why the two transformers described in Q4(i) and Q4(ii) cannot be operated with their primaries and secondaries connected in parallel.
5. A balanced star connected three phase load has a coil of inductance 0.2 H and resistance 50 Ω in each phase. It is supplied at 415 V, 50 Hz. Calculate EACH of the following: (a) the line current; (b) the power factor; (c) the value of EACH of three identical delta connected capacitors to be connected across the same supply to raise the power factor to 0.9 lag; (d) the new value of the line current. 6. A 6 pole 3 phase squirrel cage induction motor runs on 380 V 60 Hz supply. It draws a line current of 80 A at a power factor of 0.8 lag. The shaft speed is 19 rev/sec. If the iron losses are 2 kw, the stator copper loss is 1 kw and the windage and friction loss is 1.5 kw, calculate EACH of the following: (a) the slip as a per unit value; (b) the rotor copper loss; (c) the shaft output power; (d) the efficiency. 7. (a) Sketch the reverse voltage/current characteristic for a low power Zener diode with a breakdown voltage of 10 V. (b) Sketch a simple voltage regulator circuit using a Zener diode. (c) State which factors determine the value of the series resistor used in the circuit described in Q7(b). (d) State which factors determine the power rating of the Zener diode in the circuit described in Q7(b). [OVER
8. (a) Explain the term power factor correction. (b) State TWO advantages of power factor correction. (c) Explain, with the aid of a circuit diagram, how power factor correction can be effected in a three-phase circuit using capacitors. (d) State ONE method, other than the use of capacitors, by which power factor correction can be effected in a 3 ph circuit. 9. (a) Explain how torque is produced in a 3 phase squirrel cage induction motor. (b) State why the starting current is several times higher than the full load current. (c) State why the power factor is very low on starting. (d) Describe ONE method of construction by means of which the starting power factor may be raised, the starting current lowered and the starting torque improved.
SCOTTISH QUALIFICATIONS AUTHORITY MARKERS REPORT FORM SUBJECT: 040-33 Electrotechnolgy DATE: 30 th March 2017 General Comments on Examination Paper I have no reservations about the standard of the paper, a number of scores in the nineties indicate that the standard was correct and the spread of questions appropriate. Comments of Specific Examination Questions Q.1. A straightforward distribution network which attracted a large number of correct solutions. Most of those who failed to score full marks made mathematical errors or confused the total power supplied to the network with the power wasted in the distribution cables. Q2. This question was also well answered. Most of those who failed to score full marks did so because they did not realise that the time constant had changed in the second part of the question because extra resistance had been added. Q.3. No problems here. One or two candidates added the base emitter volt drop rather than subtracting it and got an incorrect value for the emitter voltage. Q.4. Almost all candidates got the 10 easy marks by correctly calculating the secondary turns having due regard to the appropriate phase voltage in the secondary. Q.5. This caused a few problems for those who did not correctly resolve the line current into in phase and quadrature components before finding the reduced quadrature component and hence the phase current for each capacitor. Q6. A stock induction motor question, but a few candidates are still convinced that the power increases as you move through the motor from input to final shaft power! Q.7. There were some very droopy reverse characteristics! The sudden increase in reverse current at the Zener breakdown point is almost right angled.. The question asked for a simple voltage regulator circuit, a few candidates offered a full excitation circuit for an alternator. Very few students were able to pinpoint the factors determining the value of the series resistor or the appropriate rating of the Zener diode.
Q.8. At least half the candidates explained the term power factor not power factor correction. The question requires the candidate to mention lagging KVA and the need to add leading KVA to raise the power factor. Full marks will only be obtained for the last part of the question if the candidate states an over excited synchronous motor. Q.9. There were some very hazy explanations of the production of torque in the induction motor! A fair number of candidates had a pulsating magnetic field rather than a rotating one and a fair number also thought that the reluctance of the air gap was responsible for the high starting current and poor starting power factor. A method of construction to improve the starting p.f. and reduce the starting current does not include the star-delta starter or the auto transformer starter but does include the wound rotor as well as the Boucherot, trislot and sashbar types of rotor.