Electrical Test part 3
ELECTRICAL TEST and MEASURING EQUIPMENT 3
(PRACTICAL MARINE ELCTRICAL KNOWLEDGE; Hall, Witherby)
QUESTION 21
What are the likely consequences of attempting to close the incomer's breaker when the generators are not in synchronism?
What are the likely consequences of attempting to close the incomer's breaker when the generators are not in synchronism?
ANSWER 21
At the instant of closing the breaker, the voltage phase difference causes a large circulating current between the machines which produces a large magnetic force to 'pull' the generators into synchronism. This means rapi6 acceleration of one rotor and deceleration of the other. The large forces may physically damage the generators and their prime movers and the large circulating current may trip each generator breaker. Result? Blackout, danger and embarrassment!
At the instant of closing the breaker, the voltage phase difference causes a large circulating current between the machines which produces a large magnetic force to 'pull' the generators into synchronism. This means rapi6 acceleration of one rotor and deceleration of the other. The large forces may physically damage the generators and their prime movers and the large circulating current may trip each generator breaker. Result? Blackout, danger and embarrassment!
QUESTION 22
What indication is available to show the optimum synchronised condition?
What indication is available to show the optimum synchronised condition?
ANSWER 22
The incoming generator ammeter will show very little 'kick' when correctly synchronised.
The incoming generator ammeter will show very little 'kick' when correctly synchronised.
QUESTION 23
How could you monitor the correct instant for synchronising without the aid of a synchroscope or synchronising lamps?
How could you monitor the correct instant for synchronising without the aid of a synchroscope or synchronising lamps?
ANSWER 23
Connect a pair of 500V voltmeter probes across one phase of the incoming machine circuit-breaker. Adjust the generator speed until the voltmeter slowly fluctuates from zero to maximum.
Connect a pair of 500V voltmeter probes across one phase of the incoming machine circuit-breaker. Adjust the generator speed until the voltmeter slowly fluctuates from zero to maximum.
Figure
Close the breaker when the voltmeter passes through zero.
QUESTION 23.1
Gen. 1 delivers 500kW at 0.8 power factdr lag, and Gen.2 delivers 400kW and 35OkVAr lag.
Estimate (a) kVAr loading of Gen.1.
(b) p.f. of Gen.2.
(c) Total bus-bar loading in kW, kVAr and p.f.
QUESTION 23.1
Gen. 1 delivers 500kW at 0.8 power factdr lag, and Gen.2 delivers 400kW and 35OkVAr lag.
Estimate (a) kVAr loading of Gen.1.
(b) p.f. of Gen.2.
(c) Total bus-bar loading in kW, kVAr and p.f.
ANSWER 23.1
Figure
(a) CosΦ = 0.8, so Φ 1 = 36.9°, and Q1 = P1 . tan Φ 1. 500. tan 36.9° = 375 kVAr
(b) TanΦ 2 = Q2/P2 = 350/400 = 0.875, so Φ 2 = 41.2°
Then pf2 = cos Φ 2 = cos4l.2° = 0.75 Lag
(c) Total P = 500 + 400 = 900kW, Total Q = 375 + 350 = 725kVAr
TanΦ = Q/P = 725/900 = 0 81
SoΦ = 38.9° and Total p.f. = cos 38.9° = 0.78 Lag.
(b) TanΦ 2 = Q2/P2 = 350/400 = 0.875, so Φ 2 = 41.2°
Then pf2 = cos Φ 2 = cos4l.2° = 0.75 Lag
(c) Total P = 500 + 400 = 900kW, Total Q = 375 + 350 = 725kVAr
TanΦ = Q/P = 725/900 = 0 81
SoΦ = 38.9° and Total p.f. = cos 38.9° = 0.78 Lag.
QUESTION 24
Two generators are load sharing equally in parallel when a total loss of excitation occurs on No.2 machine. What is the likely outcome?
Two generators are load sharing equally in parallel when a total loss of excitation occurs on No.2 machine. What is the likely outcome?
ANSWER 24
Generator No.2 will run as an induction generator drawing its excitation from No.1. Both generator currents will rise rapidly with No.1 becoming more lagging while No.2 runs with a leading p.f. (indicated on p.f. meter). A 'loss of excitation' trip (if fitted) or an over-current trip should trip No.2 generator possibly causing an overload on No.1. Alternatively, No.1 trips on over-current (current) which deprives No.2 of excitation and its breaker trips out on under-voltage. Result = total power failure!
Generator No.2 will run as an induction generator drawing its excitation from No.1. Both generator currents will rise rapidly with No.1 becoming more lagging while No.2 runs with a leading p.f. (indicated on p.f. meter). A 'loss of excitation' trip (if fitted) or an over-current trip should trip No.2 generator possibly causing an overload on No.1. Alternatively, No.1 trips on over-current (current) which deprives No.2 of excitation and its breaker trips out on under-voltage. Result = total power failure!
QUESTION 25
What is the synchronous speed of a 6-pole motor supplied at 60Hz.
What is the synchronous speed of a 6-pole motor supplied at 60Hz.
ANSWER 25
20 rev./sec. or 1200 rev./min.
20 rev./sec. or 1200 rev./min.
QUESTION 26
How is the rotor direction reversed?
How is the rotor direction reversed?
ANSWER 26
Simply by swopping over any two supply line connections at the stator terminal box. This reverses the direction of the rotating magnetic field.
Simply by swopping over any two supply line connections at the stator terminal box. This reverses the direction of the rotating magnetic field.
QUESTION 27
If a 6-pole motor is supplied at 60Hz and runs with a slip of 5%, what is the actual rotor speed?
If a 6-pole motor is supplied at 60Hz and runs with a slip of 5%, what is the actual rotor speed?
ANSWER 27
The synchronous speed is 1200 rpm, and the rotor slips by 5% of 1200, i.e. by 60 rpm so the rotor runs at 1140 rpm.
The synchronous speed is 1200 rpm, and the rotor slips by 5% of 1200, i.e. by 60 rpm so the rotor runs at 1140 rpm.
QUESTION 28
Why is it essential that contactors S and D on page 4/8 are not both closed at the same time?
Why is it essential that contactors S and D on page 4/8 are not both closed at the same time?
ANSWER 28
If you check the starter diagram, you will see that a full short-circuit is applied across the supply in this condition.
If you check the starter diagram, you will see that a full short-circuit is applied across the supply in this condition.
QUESTION 29
Estimate and compare the likely starting current surges for a motor that takes 200A on full load when started (a) DOL, (b) Star-Delta, (c) Autotransformer with a 50% tapping.
Estimate and compare the likely starting current surges for a motor that takes 200A on full load when started (a) DOL, (b) Star-Delta, (c) Autotransformer with a 50% tapping.
ANSWER 29
(a) When starting DOL the initial surge current is about 5 x FL, i.e.1000A.
(b) A star-delta starter reduces the initial starting surge to one-third of the equivalent DOL value, i.e. to about 330A in this case.
(c) The autotransformer method reduces the initial starting surge to (x)2 . IDOL where x =tapping point. In this example x = 0.5. so the surge current level is 0.52 x 1000A = 250A.
The DOL starter is simple and cheap but causes a large starting surge. Star-delta starting reduces the surge but is somewhat more complex, requiring three contactors and a timer. The autotransformer method can be arranged to match the motor surge current and run-up period to meet the supply limitations by a suitable choice of voltage tapping. This starter is considerably more expensive than the other two starter type.
(a) When starting DOL the initial surge current is about 5 x FL, i.e.1000A.
(b) A star-delta starter reduces the initial starting surge to one-third of the equivalent DOL value, i.e. to about 330A in this case.
(c) The autotransformer method reduces the initial starting surge to (x)2 . IDOL where x =tapping point. In this example x = 0.5. so the surge current level is 0.52 x 1000A = 250A.
The DOL starter is simple and cheap but causes a large starting surge. Star-delta starting reduces the surge but is somewhat more complex, requiring three contactors and a timer. The autotransformer method can be arranged to match the motor surge current and run-up period to meet the supply limitations by a suitable choice of voltage tapping. This starter is considerably more expensive than the other two starter type.
QUESTION 30
What causes the large current surge in open transition starters when going from the start to the run condition?
What causes the large current surge in open transition starters when going from the start to the run condition?
ANSWER 30
All motors generate a back emf against the supply voltage when they are running. When the supply is removed from a running induction motor the magnetic field does not immediately collapse. The motor begins to slow down but still generates an emf. When the supply is reconnected, the supply voltage and motor emf are not necessarily in phase (the condition is similar to synchronising a generator onto the bus-bars). So each time the starter is operated a different surge current will be produced - sometimes very large, sometimes quite small. Large surges could cause appreciable voltage dip on the supply and so affect other equipment. Closed transition starters overcome this because the motor is never disconnected from the supply during the starting cycle.
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